# root(100)(100) Use Newton's method to approximate the given number correct to eight decimal places.

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### Textbook Question

Chapter 4, 4.8 - Problem 12 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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lfryerda | High School Teacher | (Level 2) Educator

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Newton's method for the equation f(x)=0  is given by x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}  where x_n  is the nth iteration.  In this case, we need to consider the number (100)^{1/100} as equivalent to solving the equation x^{100}-100=0 which means that f(x)=x^{100}-100 and f'(x)=100x^{99}.

It is now necessary to set up an iteration, and use an initial guess for the answer.

x_{n+1}=x_n-\frac{x^{100}-100}{100x^99}      using the Newton's method formula

We can take any reasonable guess for the starting point, so we select x_1=1.1.  This gives the iterations:

x_2=1.089079822

x_3=1.078403357

x_4=1.068187729

x_5=1.058964293

x_6=1.051816165

x_7=1.048027084

x_8=1.047165635

x_9=1.047128613

x_{10}=1.047128548

This last value (1.047128548) is correct to 8 decimal places.

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