Devise a Newton iteration formula for computing $\sqrt[3]{R}$ where $R>0$. Perform a graphical analysis of your function $f(x)$ to determine Devise a Newton iteration formula for computing $\sqrt[3]{R}$ where $R>0$. Perform a graphical analysis of your function $f(x)$ to determine the starting values for which the iteration will converge.
I know that $\sqrt[3]{R}$ is the root of $f(x)=x^3-R$, with which, the iteration formula we need is $x_{n+1}=x_n-(\frac{x_n^3-R}{3x_n^2})$, here I have the graphs for when $R=1,2,3$. But I'm not sure where to start the method so that convergence is assured, it will be in the $[0,1]$ interval since according to the graph the roots go through there? Thank you very much.


You can prove by induction that the Newton iteration sequence is decreasing and always greater than $\sqrt[3]{R}$ if you start at $x_1 > \sqrt[3]{R}$:


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*If $x_n > \sqrt[3]{R}$ then $x_{n+1}-x_n=-(\frac{x_n^3-R}{3x_n^2}) < 0$, hence $x_{n+1} < x_n$.

*As $f$ is convex for $x > \sqrt[3]{R}$, you get that $x_{n+1} > \sqrt[3]{R}$.


As a decreasing lower bounded sequence converges, if you start with $x_1 > \sqrt[3]{R}$, the Newton iteration sequence converges. As $f$ is continuous, it can in that case only converge to $l= \sqrt[3]{R}$. 
