Q 80 - Calculus

Question

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 $\left|x_{k+1}-x_k\right|<0.001.$

$\sqrt{2}$

Step-by-Step Solution

Answer

The required approximate value is :-

$\sqrt{2}=1.414214$.

Step 1. Given Information

We have given the following square root :-

$\sqrt{2}$.

We have to find the value of $\sqrt{2}$ by using the recursion formula with initial approximation $x_0=1$. Also we have to stop our process of find iteration while $\left|x_{k+1}-x_k\right|<0.001$.

Step 2. To find the value of 2

In the previous example we find the following recursion formula :-

$x_{k+1}=\frac{1}{2}\left(x_k+\frac{a}{x_k}\right)$, to find the approximate value of $\sqrt{a}$.

Here $a=2$ and $x_0=1$, then the first iteration is :-

$$\begin{gathered} x_1=\frac{1}{2}\left(1+\frac{2}{1}\right) \\ \Rightarrow x_1=\frac{1}{2}\times 3 \\ \Rightarrow x_1=1.5 \end{gathered}$$

then second iteration is :-

$$\begin{gathered} x_2=\frac{1}{2}\left(1.5+\frac{2}{1.5}\right) \\ \Rightarrow x_2=\frac{1}{2}\left(2.8334\right) \\ \Rightarrow x_2=1.41667 \end{gathered}$$

Then third iteration is :-

$$\begin{gathered} x_3=\frac{1}{2}\left(1.41667+\frac{2}{1.41667}\right) \\ \Rightarrow x_3=1.414216 \end{gathered}$$

Then fourth iteration is :-

$$\begin{gathered} x_4=\frac{1}{2}\left(1.414216+\frac{2}{1.414216}\right) \\ \Rightarrow x_4=1.414214 \end{gathered}$$

Here $\left|x_4-x_3\right|=\left|1.414214-1.414216\right|=0.000002<0.001$

So $x=1.414214$ is the required approximate value of $\sqrt{2}$.