91Ó°ÊÓ

The given term is $\sqrt{4}$and $x_0=1$.

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

Let us consider the function$f\left(x\right)=x^2-4$

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

Therefore,

$\begin{array}{rl}& f\left(x_k\right)=x_k^2−4\\ & f^{\mathrm{′}}\left(x_k\right)=2x_k\end{array}$

Substituting the values in equation (1)

$x_{k+1}=x_k−\frac{x_k^2−4}{2x_k}$

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

$\begin{array}{rl}& x_{0+1}=x_0−\frac{{x_0}^2−4}{2x_0}\\ & x_1=x_0−\frac{{x_0}^2−4}{2x_0}\end{array}$

Substitute $x_0=1$

$\begin{array}{rl}{x}_1& =\left(1\right)−\frac{(1{)}^2−4}{2\left(1\right)}\\ & =1−\frac{1−4}{2}\\ & =1−\frac{−3}{2}\\ & =\frac{1+3}{2}\\ & =\frac{4}{2}\\ & =2\end{array}$

Therefore,$x_1=2$

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

$x_2=x_1−\frac{x_1^2−4}{2x_1}$

Substitute$x_1=2$

$\begin{array}{rl}{x}_2& =\left(2\right)−\frac{(2{)}^2−4}{2\left(2\right)}\\ & =2−\frac{4−4}{4}\\ & =2−\frac{0}{4}\\ & =2\end{array}$

Therefore$x_2=2$

Now to find the value of $x_3$, substitute localid="1649345412906" $k=2$in equation (2)

$x_3=x_2−\frac{x_2^2−4}{2x_2}$

Substitute$x_2=2$

$\begin{array}{rl}{x}_3& =\left(2\right)−\frac{(2{)}^2−4}{2\left(2\right)}\\ & =2−\frac{4−4}{4}\\ & =2−\frac{0}{4}\\ & =2\end{array}$

Therefore,$x_3=2$

Now to find the value of $x_4$, substitute localid="1649345423154" $k=3$in equation (2)

$x_4=x_3−\frac{x_3^2−4}{2x_3}$

Substitute$x_3=2$

$\begin{array}{rl}{x}_4& =\left(2\right)−\frac{(2{)}^2−4}{2\left(2\right)}\\ & =2−\frac{4−4}{4}\\ & =2−\frac{0}{4}\\ & =2\end{array}$

Therefore,$x_4=2$

Here,

$\begin{array}{rl}|x_4−x_3|& =|2−2|\\ & =\left|0\right|\end{array}$

Since, role="math" localid="1649335839464" $\begin{array}{rl}|x_4−x_3|& <0.001\end{array}$let us stop the iteration.

Therefore, the approximate value of the root of $\sqrt{4}$is $2$