Chapter 12.1, Problem 95AYU

To determine

To find: The square roots and compare the result with calculator.

Explanation of Solution

Given: The recursive formula is,

  $a_n=\frac{1}{2}\left(a_{n-1}+\frac{p}{a_{n-1}}\right)$

  $a_0=4$

Calculate $a_1$ .

  $\begin{array}{c}{a}_1=\frac{1}{2}\left(a_0+\frac{21}{a_0}\right)\\ =\frac{1}{2}\left(4+\frac{21}{4}\right)\\ =4.625\end{array}$

Calculate $a_2$ .

  $\begin{array}{c}{a}_2=\frac{1}{2}\left(a_1+\frac{21}{a_1}\right)\\ =4.58277\end{array}$

Calculate $a_3$ .

  $\begin{array}{c}{a}_3=\frac{1}{2}\left(a_2+\frac{21}{a_2}\right)\\ =4.58257\end{array}$ \

Calculate $a_4$ .

  $\begin{array}{c}{a}_4=\frac{1}{2}\left(a_3+\frac{21}{a_3}\right)\\ =4.582575\end{array}$

Calculate $a_5$ .

  $\begin{array}{c}{a}_5=\frac{1}{2}\left(a_4+\frac{21}{a_4}\right)\\ =4.5825757\end{array}$

The value of $\sqrt{21}$ from calculator is $4.58257$ .

Hence, both the result are equivalent.