# Newtons method help

#### house2012

##### New member
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 $$1-\frac{R}{x^n}$$ and to $$x^n-R$$ for determining $$(R)^{\frac{1}{n}}$$ results in 2 similar, but different iterative formulas, with $$n \ge 2$$ and $$R >0$$

#### CaptainBlack

##### Well-known member
First the wording of this question makes it very difficult to understand what is required. I will assume that you are asked to use Newton's method with:

$f(x)=1-\frac{R}{x^n}$

to find $$R^{1/n}$$.

First applying Newton's method to $$f(x)$$ if it works finds a solution to $$f(x)=0$$.

Putting $$f(x)=1-\frac{R}{x^n}=0$$ rearranges to $$\frac{R}{x^n}=1$$ or $$x^n=R$$ so $$f(x)$$ is of the correct form for finding $$R^{1/n}$$.

Now Newton's iteration to find a root of $$f(x)=0$$ is:

$x_{k+1}=x_k-\frac{f(x_k)}{f'(x_k)}$
which in this case reduces to:

$x_{k+1}=x_k-\frac{(x_k^n-R)x_k}{Rn}$

Now the question as asked does not indicate where to go from here.

CB

Last edited:

#### Amer

##### Active member
the itirating
$x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)}$

for first one

$x_1 = x_0 - \frac{((x_0)^n - R)}{n(x_0 ^{n-1})}$

second one

$x_1 = x_0 - \frac{1 - \frac{R}{x_0 ^{n}} }{ \frac{-nR}{x_0 ^{n+1}}}= x_0 - \frac{x_0(x_0 ^n-R)}{-nR}$

these are different