Chapter 8, Problem 1NT

To determine

(a)

To find:

The square root of 13 by Newton method up to 3 decimal.

Answer to Problem 1NT

Solution:

The solution is 3.606.

Explanation of Solution

Given:

Use Newton method for estimation.

Approach:

The guess taken is always positive number.

Calculation:

Let Guess 1 be $g_1=4$.

$n=13$

Step 1: By dividing $\frac{13}{4}$

Step 2: By adding $\frac{13}{4}+4$

Step 3: dividing $\frac{1}{2}\left(\frac{13}{4}+4\right)$

So, obtain guess 2, $g_2\approx 3.625$

Step 4: repeating procedure for guess 2.

$\begin{array}{rll}{g}_3& =& \frac{1}{2}\left(\frac{13}{3.625}+3.625\right)\\ & \approx & 3.605\\ g_4& =& \frac{1}{2}\left(\frac{13}{3.625}+3.625\right)\\ & \approx & 3.606\end{array}$

$\begin{array}{rll}{g}_5& =& \frac{1}{2}\left(\frac{13}{3.606}+3.606\right)\\ & \approx & 3.606\end{array}$

Hence square root of 13 up to 3 decimal place is 3.606.

To determine

(b)

To write:

The step of algorithm in a recursive formula.

Answer to Problem 1NT

Solution:

The recursive formula is

$g_{n+1}=\frac{1}{2}\left(\frac{a}{g_n}+g_n\right)$

Explanation of Solution

Given:

Use Newton method for estimation.

Approach:

The guess taken is always positive number.

Calculation:

The recursive formula formed to find $\sqrt{a}$ is

$g_{n+1}=\frac{1}{2}\left(\frac{a}{g_n}+g_n\right)$

$g_n$ is guess for $n$ term.

Step 1: Divide $a$ by $g_1$ (guess 1).

Step 2: Now add Guess 1 to quotient obtained in step 1.

Step 3: Divide the sum in step 2 by 2. The quotient becomes $g_2$.

Step 4: Repeat step 1-3 using Guess 2 to obtain successive guess.

Step 5: The process may be repeated until describe accuracy is achieved.