This is worth 5 points extra credit if done properly and turned in by the last class.
Newton-Raphson Division is a means to divide by multiplying. Why would you want to do that? Well, even for computers division is harder (read slower) than multiplying.The trick is that $p/q$ is simply $p \cdot (1/q)$, so finding a means to compute a reciprocal by multiplying will reduce division to multiplication. (This trick is used by yeppp, a high performance library for computational mathematics.)
Well suppose we have $q$, we could try to use Newton's method to find $1/q$, as it is a solution to $f(x) = x - 1/q$. The Newton update step $x - f(x) / f'(x)$ simplifies to:
$$~ x - (x - 1/q)/ 1 = 1/q. ~$$That doesn't really help, as Newton's method is just $x_{i+1} = 1/q$ – that is it just jumps to the answer, the one we want to compute by some other means. That isn't so helpful.
Let's try again, simplify the update step for a related function: $f(x) = 1/x - q$ and then write the julia code for one step of Newton's method (your answer should look something like $x = 1/2 \cdot (x + q/x)$, though that solves $\sqrt{q}$):
Now for $q$ in the interval $[1/2, 1]$ we want to get a good initial guess. Here is a claim. We can use $x_0=48/17 - 32/17 q$. Let's check:
Make a graph of both $y=1/x$ and $y=48/17 - 32/17 \cdot x$ on this interval. If you look closely you should see that the largest difference between the two graphs is at the endpoints and the midpoint. Can you see that from your graph?
From your graph estimate the largest difference:
Here is something that people have shown to be true: for any $q$ in $[1/2, 1]$ and with the initial guess $x_0 = 48/17 - 32/17q$, Newton's method will converge to 16 digits of accuracy in no more than this many steps:
$$~ \log_2(\frac{53 + 1}{\log_2(17)}). ~$$What is the smallest integer as large as this value?
For $q = 0.80$, find $1/q$ using the above. Write your commands showing all the steps you needed.
Finally, any thoughts on what to do to find $p/q$ this way if $q$ is not in the interval $[1/2, 1]$?