1. ## Newtons method help

Hey guys, need some help with this question. I am stuck and don't know what to do.

Q: Show that using newton's method to $\displaystyle 1-\frac{R}{x^n}$ and to $\displaystyle x^n-R$ for determining $\displaystyle (R)^{\frac{1}{n}}$ results in 2 similar, but different iterative formulas, with $\displaystyle n \ge 2$ and $\displaystyle R >0$

2. ## Re: Newtons method help

Ok so Newton's method is a way of determining a root of the equation
$\displaystyle f(x)=0$

in both of the cases above for $\displaystyle f(x)$ we obtain $\displaystyle f(x) =0 \Rightarrow x = (R)^{1/n}$ so we just need to apply Newton's method.

This works by calculating the tangent line at a point $\displaystyle x_n$ finding where it intersects the x-axis and assuming that this gives a better approximation to the root than $\displaystyle x_n$

Graphically you can see it at work here
$\displaystyle x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$
So work out the derivatives of the two functions you have above then plug it into the formula and you should have two iterative formulas which will converge on $\displaystyle (R)^{1/n}$