Copyright © University of Cambridge. All rights reserved.

'Archimedes and Numerical Roots' printed from http://nrich.maths.org/

Show menu


This problem builds on the one in May on calculating Pi. This brilliant man Archimedes managed to establish that $3\frac{10}{71} < \pi < 3\frac{1}{7}$. 

He needed to be able to calculate square roots first so that he could calculate the lengths of the sides of the polygons which he used to get his approximation for $\pi$. How did he 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 $\frac{3}{2}$ 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  $$\frac{(\frac{3}{n} + n)}{2}$$ where n is an approximation to $\sqrt 3$ .

We then repeat the process to find the new (third) approximation to $\sqrt{3}$ $$\sqrt{3} \approx {(3 / 1.75 + 1.75) \over {2}} = 1.73214... $$

to find a forth approximation repeat this process using 1.73214 and so on...

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?