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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.

Answer: Let $R>0$ and set $x=\sqrt[3]{R}$. Then $x$ is a root of the equation $x^3-R=0$. If we use Newton's method on the function $f(x)=x^3-R$, the iteration formula can be written as $$x_{n+1}=x_n-\dfrac{f(x_n)}{f'(x_n)}=x_n-\dfrac{(x_n^3-R)}{3x_n^2}=\dfrac{1}{3}\left( 2x_n+\dfrac{R}{x_n^2}\right).$$

Next, we will perform a graphical analysis of $f(x)=x^3-R$ to determine the starting values for which the iteration will converge.

Can someone explain what a graphical analysis is? I don't just want the answer. I want to under the process for this analysis.

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You can plot your iteration as a function $y=\frac 13\left(2x+\frac R{x^2}\right)$. Given $x_n$, you can find it on the $x$ axis and go up to find $x_{n+1}$. If you plot the line $y=x$ on the same plot you can go horizontally from your point $(x_n,x_{n+1})$ to intersect $y=x$ at the point $(x_{n+1},x_{n+1})$, then go vertically to intersect the other curve at $(x_{n+1},x_{n+2})$. If you do this for several steps you can see if you are getting closer to or further from the root. The plot below shows starting at $x_0=5$ as point $A$, then goes up to $B$, left to $C$, down to $D$, and left to $E$. It appears to be converging to $2$. You can try different starting points to see what happens.

enter image description here

Ross Millikan
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