Question

Use Newton’s method to derive the recursion formula

$x_{k+1}=\frac{1}{2}\left(x_k+\frac{a}{x_k}\right)$

for approximating$\sqrt{a}$.

Short answer

We can derive the recursion formula as following :-

$$\begin{gathered} x_{k+1}=x_k-\frac{{x_k}^2-a}{2x_k} \\ \Rightarrow x_{k+1}=\frac{2{x_k}^2-{x_k}^2+a}{2x_k} \\ \Rightarrow x_{k+1}=\frac{{x_k}^2+a}{2x_k} \\ \Rightarrow x_{k+1}=\frac{1}{2}\left(\frac{{x_k}^2+a}{x_k}\right) \\ \Rightarrow x_{k+1}=\frac{1}{2}\left(x_k+\frac{a}{x_k}\right) \end{gathered}$$

Step-by-step solution

Step 1. Given Information

We have to derive a recursion formula by using Newton's method to approximate the value of$\sqrt{a}$.

Step 2. Derivation of recursion formula

We have to approximate the value of $\sqrt{a}$.

Let:-

$$\begin{gathered} x=\sqrt{a} \\ \Rightarrow x^2=a \\ \Rightarrow x^2-a=f\left(x\right)Say \end{gathered}$$

localid="1654268190999" $\Rightarrow f\left(x\right)=x^2-a$

Then:-

$f\text{'}\left(x\right)=2x$

We have the following newton's formula to approximate a value of a function:-

$x_{k+1}=x_k-\frac{f\left(x_k\right)}{f\text{'}\left(x_k\right)}$

Put the above values, then we have:-

localid="1654498543121" $\begin{array}{rcl}{x}_{k+1}& =& x_k-\frac{{x_k}^2-a}{2x_k}\\ & =& \frac{2{x_k}^2-{x_k}^2+a}{2x_k}\\ & =& \frac{{x_k}^2+a}{2x_k}\\ & =& \frac{1}{2}\left(\frac{{x_k}^2+a}{x_k}\right)\\ & =& \frac{1}{2}\left(x_k+\frac{a}{x_k}\right)\end{array}$

This is the required formula.