Thread: 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 $\displaystyle R >0$

Thanks for your help guys!

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
http://upload.wikimedia.org/wikipedi...ration_Ani.gif

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

A derivation can be found here Newton's method - Wikipedia, the free encyclopedia

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 $\displaystyle (R)^{1/n}$