Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 12

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.