Question

# Part A. Consider the nonlinear equation x5-x=15 Attempt to find a root of this equation with...

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

Part A:

Given: f(x) = x5 - x- 15.

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

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

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

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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