Starting with x=3, find the third approximation to the root of the equation x^3 - 4x^9 = 0.
SOLUTION:
We apply Newton's method with f(x) = x^3 - 4x^9.
So the iterative equation becomes x_n+1 = x_n - (f(x_n) / f'(x_n)), where f'(x) = 3x^2 - 4.
With n=1, we have:
x_1 = x_0 - (f(x_0) / f'(x_0))
x_1 = 3 - ((3^3 - 4(3^9)) / (3(3^2) - 4))
x_1 = 3 - ((27 - 4(19683)) / (27 - 4))
x_1 = 3 - ((27 - 78732) / 23)
Then with n=2, we obtain:
x_2 = x_1 - (f(x_1) / f'(x_1))
x_2 = 2.743 - ((2.743^3 - 4(2.743^9)) / (3(2.743^2) - 4))
x_2 = 2.743 - ((20.052 - 4(19683)) / (22.743 - 4))
x_2 = 2.743 - ((20.052 - 78732) / 18.743)
To see how accurate the approximation is, look at which decimal place does not change. The first of these gives the number of decimal places to which the approximation is accurate.
