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

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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.**