You may also like

Rationals Between...

What fractions can you find between the square roots of 65 and 67?

Root to Poly

Find the polynomial p(x) with integer coefficients such that one solution of the equation p(x)=0 is $1+\sqrt 2+\sqrt 3$.

Consecutive Squares

The squares of any 8 consecutive numbers can be arranged into two sets of four numbers with the same sum. True of false?

Archimedes and Numerical Roots

Age 14 to 16
Challenge Level

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?