Finding the roots through Newtons method

Telemachus
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Homework Statement


In each of the following items approximate the zeros of f using Newton's method. Continue iterating until making two successive approximations differ at most in 0.001.

The Newton iteration: x_{n+1}=x_n-\displaystyle\frac{f(x_n)}{f'(x_n)}

Well, I have a doubt about this. I'm not sure if it's asking me to iterate till |x_{n+1}-x_n|\leq{0.001}|, or if I should apply some of this:

k_1&gt;0, |f&#039;(x)|\geq{k_1} and |f&#039;&#039;(x)|\leq{}k_2 for all x\in{<b>}</b>, then:

|x_{n+1}-r|&lt;\displaystyle\frac{k_2}{2k_1}|x_n-r|^2

If r\in{}[r-\delta,r+\delta]\subset{[a,b]}, and \delta&lt;2(\displaystyle\frac{k_1}{k_2})

|x_{n+1}-r|&lt;\displaystyle\frac{2k_1}{k_2}(\displaystyle\frac{\delta}{\displaystyle\frac{2k_1}{k_2}})^2n
 
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Stop when |x_{n+1} - x_n| &lt; 0.001.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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