Question

Part A.

Consider the nonlinear equation

x⁵-x=15

Attempt to find a root of this equation with Newton's method (also known as Newton iteration).

Use a starting value of x₀=4 and apply Newton's method once to find x₁

Enter your answer in the box below correct to four decimal places.

Part B.

Using the value for x₁ obtained in Part A, apply Newton's method again to find x₂

Note you should not round x₁ when computing x₂

Answer 1

Part A:

Given: f(x) = x⁵ - x- 15.

While applying Newton's method: for the (n+1)^th iteration of the process,

x_n+1 = x_n - f(x_n)/f'(x_n) where f(x_n) is the value of the function f(x) at x = x_n.

f'(x_n) is the first derivative of f(x) at x = x_n.

Thus, for the first iteration, x₁ = x₀ - f(x₀)/f'(x₀).

Further, f'(x) = 5x⁴ - 1.

=> x₁ = 4 - f(4)/f'(4) = 4 - (4⁵ - 4 - 15)/(5*4⁴ - 1) = 4 - (1024 - 4 - 15)/(1280 - 1) = 4 - (1005/1279) = 3.2142 (rounded)

Part B:

x₂ = x₁ - f(x₁)/f'(x₁) = 3.21422298 - f(3.21422298)/f'(3.21422298) = 3.21422298 - (3.21422298⁵ - 3.21422298 - 15)/(5*3.21422298⁴ - 1)

= 3.21422298 - 0.6098 = 2.60437 = 2.6044

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