Chapter 4.7, Problem 15E

To determine

To calculate: The roots of the equation,

  ${(x-2)}^2=\ln x$ .

Answer to Problem 15E

The roots are

  $x_2\approx 1.333333$

  $x_3\approx 0.816537$

  $x_4\approx 1.262929$

Explanation of Solution

Given information:

The equation is given as:

  ${(x-2)}^2=\ln x$ .

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,

  ${(x-2)}^2=\ln x$

Now,

  $\begin{array}{l}{(}^x-\ln x=0\\ f(x)={(}^x-\ln x\\ f\text{'}(x)=2(x-2)-\frac{1}{x}\end{array}$

Let the initial approximation $x_1=1$

Now, at $n=1$

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

  $x_2=x_1-\frac{{(x_1-2)}^2-\ln x_1}{2(x_1-2)-\frac{1}{x_1}}$

  $\begin{array}{l}{x}_2=(1)-\left(\frac{{(1-2)}^2-\ln (1)}{2(1-2)-\frac{1}{1}}\right)\\ x_2=(1)-\left(\frac{{(-1)}^2-0}{2(-1)-1}\right)\\ x_2=(1)-\left(\frac{1}{-2-1}\right)\end{array}$

  $\begin{array}{l}{x}_2=1-\left(\frac{1}{-3}\right)\\ x_2=1-(-0.333333333)\\ x_2=1.333333333\end{array}$

The second approximation is $x_2=1.333333$ .

Let $n=2$ ,

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

  $x_3=x_2-\frac{{(x_2-2)}^2-\ln x_2}{2(x_2-2)-\frac{1}{x_2}}$

  $\begin{array}{l}{x}_3=(1.333333)-\left(\frac{{(1.333333-2)}^2-\ln (1.333333)}{2(1.333333-2)-\frac{1}{1.333333}}\right)\\ x_3=(1.333333)-\left(\frac{-0.444444888-0.287681822}{-0.666666-0.750000187}\right)\\ x_3=1.333333-\left(\frac{-0.73212671}{-1.416666187}\right)\end{array}$

  $\begin{array}{l}{x}_3=1.333333-0.5167955\\ x_3=0.816537499\end{array}$

The third approximation with six decimal places is $x_3=0.816537$ .

Let $n=3$ ,

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

  $x_4=x_3-\frac{{(x_3-2)}^2-\ln (x_3)}{2(x_3-2)-\frac{1}{x_3}}$

  $\begin{array}{l}{x}_4=(0.816537)-\left(\frac{{(0.816537-2)}^2-\ln (0.816537)}{2(0.816537-2)-\frac{1}{0.816537}}\right)\\ x_4=(0.816537)-\left(\frac{1.400584672-(-0.202683052)}{-2.366926-1.224684246}\right)\\ x_4=0.816537-\left(\frac{1.603267724}{-3.591610246}\right)\end{array}$

  $\begin{array}{l}{x}_4=0.816537+0.446392457\\ x_4=1.262929457\end{array}$

The fourth approximation with six decimal places is $x_4=1.262929$ .

Therefore, the roots with six decimal places are :-

  $x_2\approx 1.333333$

  $x_3\approx 0.816537$

  $x_4\approx 1.262929$