Use the result of Exercise 79 to approximate the square roots in Exercises 80–83. In each case, start with $x_0=1$and stop when $\left|x_{k+1}-x_k\right|<0.001$.

81.$\sqrt{3}$

The given term is $\sqrt{3}$and $x_0=1$

Here, we have to find the root of the functions.

Let us consider the function$f\left(x\right)=x^2-3$

We have the equation $x_{k+1}=x_k-\frac{f\left(x_k\right)}{f\text{'}\left(x_k\right)}$.......Equation (1)

Therefore,

$\begin{array}{r}f\left(x_k\right)=x_k^2-3\\ f^{\mathrm{\prime }}\left(x_k\right)=2x_k\end{array}$

Substituting the values in equation (1)

$x_{k+1}=x_k-\frac{x_k^2-3}{2x_k}$

Now to find the value of $x_1$, substitute $k=0$in equation (2)

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

Substitute$x_0=1$

role="math" localid="1649317289580" $\begin{array}{rl}{x}_1& =\left(1\right)-\frac{(1{)}^2-3}{2\left(1\right)}\\ & =1-\frac{1-3}{2}\\ & =1-\frac{-2}{2}\\ & =\frac{2+2}{2}\\ & =\frac{4}{2}\\ & =2\end{array}$

Therefore,$x_1=2$

Now to find the value of $x_2$, substitute$k=1$ in equation (2)

Substitute$x_1=2$

$\begin{array}{rl}{x}_2& =\left(2\right)-\frac{(2{)}^2-3}{2\left(2\right)}\\ & =2-\frac{4-3}{4}\\ & =2-\frac{1}{4}\\ & =\frac{7}{4}\end{array}$

Therefore,$x_2=\frac{7}{4}$

Now to find the value of $x_3$, substitute localid="1649345349442" $k=2$in equation (2)

localid="1649317673477" $x_3=x_2-\frac{x_2^2-3}{2x_2}$

Substitute$x_2=\frac{7}{4}$

localid="1649317953513" $$\begin{gathered} x_3=\left(\frac{7}{4}\right)-\frac{{\left(\frac{7}{4}\right)}^2-3}{2\left(\frac{7}{4}\right)} \\ =\frac{7}{4}-\frac{\frac{49}{16}-3}{\frac{7}{2}} \\ =\frac{7}{4}-\frac{49-48}{16} \\ =\frac{4\left(7\right)\left(7\right)-2}{16\left(7\right)} \\ =\frac{196-2}{112} \\ =\frac{194}{112} \end{gathered}$$

Hencelocalid="1649318122927" $x_3=\frac{194}{112}$

Now to find the value of $x_4$, substitute localid="1649345362057" $k=3$in equation (2)

$x_4=x_3-\frac{{x_3}^2-3}{2x_3}$

Substitute,$x_3=\frac{194}{112}$

localid="1649318499813" $\begin{array}{rl}{x}_4& =\left(\frac{194}{112}\right)-\frac{{\left(\frac{194}{112}\right)}^2-3}{2\left(\frac{194}{112}\right)}\\ & =\frac{194}{112}-\frac{\frac{37636}{12544}-3}{\frac{388}{112}}\\ & =\frac{194}{112}-\frac{\frac{37636-37632}{12544}}{\frac{388}{112}}\\ & =\frac{194}{112}-\frac{\frac{4}{125444}}{\frac{388}{112}}\\ & =\frac{194}{112}-\frac{4}{125444}\times \frac{{\displaystyle 112}}{{\displaystyle 388}}\\ & =\frac{194}{112}-\frac{1}{28\left(388\right)}\\ & \begin{array}{rl}& =\frac{194\left(97\right)-1}{28\left(388\right)}\\ & =\frac{18817}{10864}\end{array}\end{array}$

Therefore,$x_4=\frac{18817}{10864}$

Here,

$\begin{array}{rl}|x_4-x_3|& =|\frac{18817}{10864}-\frac{194}{112}|\\ & =\left|\frac{18817-18818}{10864}\right|\\ & =\left|\frac{-1}{10864}\right|\\ & =0.00009\end{array}$

Since, $|x_4-x_3|<0.001$let us stop the iteration.

Therefore, the approximate value of$\sqrt{3}$is$\frac{18817}{10864}\approx 1.7320$