Question

Find the positive real root of x³ - x - 3 = 0 using Newton-Raphson method. If the starting guess (x₀) is 2, the numerical value of the root after two iterations (x₂) is __________ (round off to two decimal places).

Answer 1

Concept:

According to the Newton-Raphson method:

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

where x_n is starting guess.

Calculation:

Given:

f(x) = x3 - x - 3 

\(f'\left( x \right) = 3{x^2} -1\)

Starting guess (x0 = 2)

Now first iterations

\(x_1=x_0-\frac{(x_0^3\;-\;x_0\;-\;3)}{3x_0^2\;-\;1}\)

\({x_1} = {x_0} - \frac{{f\left( {{x_0}} \right)}}{{f'\left( {{x_0}} \right)}} = 2 - \frac{{{{\left( { 2} \right)}^3} -2-3 }}{{3{{\left( { 2} \right)}^2} - 1}} = 1.7273\)

Second iterations

\(x_2=x_1-\frac{(x_1^3\;-\;x_1\;-\;3)}{3x_1^2\;-\;1}\)

x₁ = 1.7273

\({x_2} = {x_1} - \frac{{f\left( {{x_1}} \right)}}{{f'\left( {{x_1}} \right)}} = 1.7273 - \frac{{{{\left( { 1.7273} \right)}^3} -1.7273-3 }}{{3{{\left( { 1.7273} \right)}^2} - 1}} = 1.67369\)

x₂ = 1.67