Archimedes and numerical roots
Problem
This problem builds on Approximating Pi. This brilliant man Archimedes managed to establish that $3 1/10 < \pi < 3 1/7$.
The problem is how did he calculate the lengths of the sides of the polygons, which he needed to be able to calculate square roots? He didn't have a calculator but needed to work to an appropriate degree of accuracy. To do this he used what we now call numerical roots.
How might he have calculated $\sqrt{3}$?
This must be somewhere between 1 and 2. How do I know this?
Now calculate the average of $3/2$ and $2$ (which is 1.75) - this is a second approximation to $\sqrt{3}$. i.e. we are saying that a better approximation to $\sqrt{3}$ is $(3/n + n)/2$ where $n$ is an approximation to $\sqrt{3}$.
We then repeat the process to find the new (third) approximation to $\sqrt{3}$.
How many approximations do I have to make before I can find $\sqrt{3}$ correct to five decimal places?
Why do you think it works?
Will it always work no matter what I take as my first approximation and does the same apply to finding other roots?
Getting Started
Student Solutions
There was a correct solution from Andrei Lazanu (School 205, Bucharest). The first part is very clear but I have tried to simplify his solution to the second part for inclusion here. Perhaps someone could improve on this for us. Thank you for your hard work Andrei. First, I approximated $\sqrt{3}$ using the method given in the problem. I know that $\sqrt{3}$ is between 1 and 2 because $1^2 < (\sqrt{3})^2 < 2^2$ or $1 < 3 < 4$.
I know that the approximation of $\sqrt{3}$ correct to five
decimal places is:
Now I show each of the approximation steps:
First approximation:
You could think of the above as
Where n is the approximation to the root of a 2 (that is "a") and m the next approximation.
The first approximation (n) differs from a by k. I can therefore write n as a + k where k is numerically less than a (k could be negative).
So I have
This means that
From the two inequalities I obtain that:
This means that the solution obtained goes closer and closer at each step to the real value of whether k is positive or negative.