Question

(a) Apply Newton’s method to the equation \({x^2} - a = 0\) to

derive the following square-root algorithm (used by the

ancient Babylonians to compute \(\sqrt a \) ):

\({x_{n + 1}} = \frac{1}{2}\left( {{x_n} + \frac{a}{{{x_n}}}} \right)\)

(b) Use part (a) to compute \(\sqrt {1000} \) correct to six decimal places.

Short answer

1. The algorithm \({x_{n + 1}} = \frac{1}{2}\left( {{x_n} + \frac{a}{{{x_n}}}} \right)\) is derived.
2. The value of \(\sqrt {1000} \) correct to six decimal places is \(31.622777\).

Step-by-step solution

Newton’s formula for nth approximation of root

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

(a)Step 2: Driving the Square root algorithm

Let,\(f\left( x \right) = {x^2} - a\)

For finding derivative differentiating the function with respect to x:

\(\begin{aligned}{l}f'\left( x \right) &= \frac{d}{{dx}}\left( {{x^2} - a} \right)\\f'\left( x \right) &= 2x\end{aligned}\)

Using Newtons Method:

\({x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\)

Substituting the values:

\(\begin{aligned}{l}{x_{n + 1}} &= {x_n} - \frac{{{x_n}^2 - a}}{{2{x_n}}}\\{x_{n + 1}} &= \frac{{2{x_n}^2 - {x_n}^2 + a}}{{2{x_n}}}\\{x_{n + 1}} &= \frac{{{x_n}^2 + a}}{{2{x_n}}}\\{x_{n + 1}} &= \frac{1}{2}\left( {{x_n} + \frac{a}{{{x_n}}}} \right)\end{aligned}\)

Hence the square root algorithm is derived.

(b)Step 3: Finding \(\sqrt {1000} \)

The algorithm derived in part (a) is:

\({x_{n + 1}} = \frac{1}{2}\left( {{x_n} + \frac{a}{{{x_n}}}} \right)\)

As we know that the algorithm is used to calculate the square root and we need to calculate\(\sqrt {1000} \), so\(a = 1000\).

We know\({32^2} = 1024\), which is very close to 1000, therefore taking the initial approximation\({x_1} = 32\).

\(\begin{aligned}{l}{x_2} &= \frac{1}{2}\left( {{x_1} + \frac{a}{{{x_1}}}} \right)\\{x_2} &= \frac{1}{2}\left( {32 + \frac{{1000}}{{32}}} \right)\\{x_2} &= 31.625\end{aligned}\)

\(\begin{aligned}{l}{x_3} &= \frac{1}{2}\left( {31.625 + \frac{{1000}}{{31.625}}} \right)\\{x_3} &= 31.6227766798\end{aligned}\)

\(\begin{aligned}{l}{x_4} &= \frac{1}{2}\left( {31.6227766798 + \frac{{1000}}{{31.6227766798}}} \right)\\{x_4} &= 31.6227766017\end{aligned}\)

Since the first 6 decimal places are not changing therefore rounding correctly to 6 digits \(\sqrt {1000} = 31.622777\).