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I am asked to find $\sqrt{2012}$ using Newton-Raphson's Method with the following recursive method

$$x_{n+1} = \frac{1}{2} (x_n + \frac{a}{x_n}) $$

I notied that give same answers as using

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)} $$

This is easy, but the next part asks to find a similar recursive method to find $\sqrt[3]{2012}$. How do I find such a method?

UPDATE

I did

$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 12.7551$$

$$x_2 = 12.6257$$

$$x_3 = 12.6244$$

$$x_4 = 12.6244$$

$$x_4 ^ 3 = 2012.02$$

Which seems correct. But I didn't use a similar recursive method like the question asked?

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Hint: if you want to find $a$, then write it as a root of some function $f(x)$. In your case, write $\sqrt[3]{2012}$ as a root of a <fill_here> degree polynomial. Now, you can read Robert Israel's answer. –  user2468 Mar 14 '12 at 5:44
    
If you use the idea described by Robert Israel, and fool around with the resulting formula a bit, you will end up with something that has a shape very similar to the one for square root. –  André Nicolas Mar 14 '12 at 5:58
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1 Answer 1

up vote 4 down vote accepted

Hint: What does Newton's method say for $f(x) = x^3 - a$?

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