Solve R Root with Newton's Method: n ≥2, R>0

[house2012]

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 $$R >0$$

Thanks for your help guys!

 

[CaptainBlack]

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

 

[Amer]

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