Question

(a) Apply Newton’s method to the equation \(\frac{1}{x} - a = 0\) to

derive the following reciprocal algorithm:

\({x_{n + 1}} = 2{x_n} - a{x_n}^2\)

(This algorithm enables a computer to find reciprocals without actually dividing.)

(b) Use part (a) to compute \({1 \mathord{\left/

{\vphantom {1 {1.6984}}} \right.

\kern-\nulldelimiterspace} {1.6984}}\) correct to six decimal places.

Short answer

1. The algorithm \({x_{n + 1}} = 2{x_n} - a{x_n}^2\) is derived.
2. The value of\({1 \mathord{\left/
3. {\vphantom {1 {1.6984}}} \right.
4. \kern-\nulldelimiterspace} {1.6984}}\)correct to six decimal places is 0.588789.

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,\(f\left( x \right) = \frac{1}{x} - a\)

For finding derivative differentiating the function with respect to x.

\(\begin{aligned}{l}f'\left( x \right) &= \frac{d}{{dx}}\left( {\frac{1}{x} - a} \right)\\f'\left( x \right) &= - {x^{ - 2}}\end{aligned}\)

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{{{1 \mathord{\left/

{\vphantom {1 {{x_n}}}} \right.

\kern-\nulldelimiterspace} {{x_n}}} - a}}{{ - {x^{ - 2}}_n}}\\{x_{n + 1}} &= {x_n} + {x_n}^2\left( {{1 \mathord{\left/

{\vphantom {1 {{x_n}}}} \right.

\kern-\nulldelimiterspace} {{x_n}}} - a} \right)\\{x_{n + 1}} &= {x_n} + {x_n} - a{x_n}^2\\{x_{n + 1}} &= 2{x_n} - a{x_n}^2\end{aligned}\)

Hence the algorithm is derived.

(b) Step 3: Finding \({1 \mathord{\left/ {\vphantom {1 {1.6984}}} \right. \kern-\nulldelimiterspace} {1.6984}}\)

The algorithm derived in part (a) is:

\({x_{n + 1}} = 2{x_n} - a{x_n}^2\)

As we know that the algorithm is used to calculate the reciprocal and we need to calculate,\({1 \mathord{\left/

{\vphantom {1 {1.6984}}} \right.

\kern-\nulldelimiterspace} {1.6984}}\), so\(a = {1 \mathord{\left/

{\vphantom {1 {1.6984}}} \right.

\kern-\nulldelimiterspace} {1.6984}}\)

We know1.6984 is very close to 1.7 so, therefore taking the initial approximation\({x_1} = \frac{1}{{1.7}} = 0.5\)

\(\begin{aligned}{l}{x_2} &= 2{x_1} - a{x_1}^2\\{x_2} &= 2\left( {0.5} \right) - \left( {\frac{1}{{1.6984}}} \right){\left( {0.5} \right)^2}\\{x_2} &= 0.5754\end{aligned}\)

\(\begin{aligned}{l}{x_3} &= 2\left( {0.5754} \right) - \left( {\frac{1}{{1.6984}}} \right){\left( {0.5754} \right)^2}\\{x_3} &= 0.58848496425\end{aligned}\)

\(\begin{aligned}{l}{x_4} &= 2\left( {0.58848496425} \right) - \left( {\frac{1}{{1.6984}}} \right){\left( {0.58848496425} \right)^2}\\{x_4} &= 0.58878929143\end{aligned}\)

\(\begin{aligned}{l}{x_5} &= 2\left( {0.58878929143} \right) - \left( {\frac{1}{{1.6984}}} \right){\left( {0.58878929143} \right)^2}\\{x_5} &= 0.58878944889\end{aligned}\)

Since the first 6 decimal places are not changing therefore round up to 6 digits\({1 \mathord{\left/

{\vphantom {1 {1.6984}}} \right.

\kern-\nulldelimiterspace} {1.6984}} = 0.588789\).