Question

Part A.

Consider the nonlinear equation

x^{5}-x=15

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

Use a starting value of x_{0}=4 and apply Newton's
method once to find x_{1}

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

**Part B.**

Using the value for x_{1} obtained in Part A, apply
Newton's method again to find x_{2}

Note you should not round x_{1} when computing
x_{2}

Answer #1

Part A:

Given: f(x) = x^{5} - 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, x_{1} = x_{0} -
f(x_{0})/f'(x_{0}).

Further, f'(x) = 5x^{4} - 1.

=> x_{1} = 4 - f(4)/f'(4) = 4 - (4^{5} - 4 -
15)/(5*4^{4} - 1) = 4 - (1024 - 4 - 15)/(1280 - 1) = 4 -
(1005/1279) = **3.2142 (rounded)**

Part B:

x_{2} = x_{1} -
f(x_{1})/f'(x_{1}) = 3.21422298 -
f(3.21422298)/f'(3.21422298) = 3.21422298 - (3.21422298^{5}
- 3.21422298 - 15)/(5*3.21422298^{4} - 1)

= 3.21422298 - 0.6098 = **2.60437 = 2.6044**

(If you liked the answer, feel free to give it a thumbs up. Thank you so much!)

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