91影视

The given term is $\sqrt{101}$and $x_0=1$

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

Let us consider the function$f\left(x\right)=x^2-101$

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

Therefore,

$\begin{array}{rl}& f\left(x_k\right)=x_k^2鈭�101\\ & f^{\mathrm{鈥�}}\left(x_k\right)=2x_k\end{array}$

Substituting the values in equation (1)

$x_{k+1}=x_k鈭�\frac{x_k^2鈭�101}{2x_k}$

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

$\begin{array}{rl}& x_{0+1}=x_0鈭�\frac{{x_0}^2鈭�101}{2x_0}\\ & x_1=x_0鈭�\frac{{x_0}^2鈭�101}{2x_0}\end{array}$

Substitute $x_0=1$

$\begin{array}{rl}{x}_1& =\left(1\right)鈭�\frac{(1{)}^2鈭�101}{2\left(1\right)}\\ & =1鈭�\frac{1鈭�101}{2}\\ & =1鈭�\frac{鈭�100}{2}\\ & =\frac{2+100}{2}\\ & =\frac{102}{2}\\ & =51\end{array}$

Therefore,$x_1=51$

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

localid="1649334770192" $x_2=x_1鈭�\frac{x_1^2鈭�101}{2x_1}$

Substitute$x_1=51$

$$\begin{gathered} x_2=\left(51\right)鈭�\frac{(51{)}^2鈭�101}{2\left(51\right)} \\ =26.4902 \end{gathered}$$

Therefore,$x_2=26.4902$

Now to find the value of $x_3$, substitute localid="1649345459279" $k=2$in equation (2)

$x_3=x_2鈭�\frac{x_2^2鈭�101}{2x_2}$

Substitute$x_2=26.4902$

$$\begin{gathered} x_3=\left(26.4902\right)鈭�\frac{(26.4902{)}^2鈭�101}{2\left(26.4902\right)} \\ =15.1515 \end{gathered}$$

Therefore,$x_3=15.1515$

Now to find the value of $x_4$, substitute localid="1649345472922" $k=3$in equation (2)

localid="1649335440301" $x_4=x_3鈭�\frac{x_3^2-101}{2x_3}$

Substitute$x_3=15.1515$

$$\begin{gathered} x_4=\left(15.1515\right)鈭�\frac{(15.1515{)}^2鈭�101}{2\left(15.1515\right)} \\ =10.9087 \end{gathered}$$

Therefore,$x_4=10.9087$

Now to find the value of $x_5$, substitute localid="1649334968584" $k=4$in equation (2)

$x_5=x_4鈭�\frac{{x_4}^2鈭�101}{2x_4}$

Substitute$x_4=10.9087$

$$\begin{gathered} x_5=\left(10.9087\right)鈭�\frac{(10.9087{)}^2鈭�101}{2\left(10.9087\right)} \\ =10.0837 \end{gathered}$$

Therefore,$x_5=10.0837$

Now to find the value of $x_6$, substitute localid="1649345491217" $k=5$in equation (2)

$x_6=x_5鈭�\frac{x_5^2鈭�101}{2x_5}$

Substitute$x_5=10.0837$

$$\begin{gathered} x_6=\left(10.0837\right)鈭�\frac{(10.0837{)}^2鈭�101}{2\left(10.0837\right)} \\ =10.0499 \end{gathered}$$

Therefore,$x_6=10.0499$

Now to find the value of $x_7$, substitute localid="1649345501090" $k=6$in equation (2)

$x_7=x_6鈭�\frac{{x_6}^2鈭�101}{2x_6}$

Substitute$x_6=10.0499$

$$\begin{gathered} x_7=\left(10.0499\right)鈭�\frac{(10.0499{)}^2鈭�101}{2\left(10.0499\right)} \\ =10.0499 \end{gathered}$$

Therefore,$x_7=10.0499$

Here,

$$\begin{gathered} \left|x_7鈭�x_6\right|=|10.0499鈭�10.0499| \\ =\left|0\right| \end{gathered}$$

Since,$\left|x_7鈭�x_6\right|<0.001$let us stop the iteration.

Therefore, the approximate value of $\sqrt{101}$is$10.0499$