# Archimedes and numerical roots

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?

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 $\sqrt3$ using the method given in the
problem. I know that $\sqrt3$ is between 1 and 2 because 1
^{2} < (?3) ^{2} < 2 ^{2} or 1 < 3
< 4.

I know that the approximation of ?3 correct to five decimal places is: $$\sqrt{3} \approx {1.73205}$$Now I show each of the approximation steps:

First approximation: $$\sqrt{3} \approx {2}$$Second approximation: $$\sqrt{3}\approx {{{3\over{2}} + 2} \over {2}} ={1.75}$$Third approximation: $$\sqrt{3} \approx {{{3\over{1.75}} + 1.75} \over {2}} = {1.732142857}$$ Fourth approximation: $$\sqrt{3} \approx {{{3\over{1.732142857}} + 1.732142857} \over {2}} = {1.73205081}$$ So, four approximations are sufficient to approximate $\sqrt{3}$ correct to 5 decimal places.You could think of the above as $$ \sqrt{a^2}\approx
{{{a^2\over{n}} + n} \over {2}} ={m}$$ 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 the next approximation $$\quad = {{{a^2\over{a+k}} + a+k} \over {2}}$$The next approximation = $${{{a^2\over{a+k}} + a+k} \over {2}}$$But $${{{a^2\over{a+k}} + a+k} \over {2}} = {{2a^2 + 2ak + k^2} \over{2(a+k)}}$$ and $${{2a^2 + 2ak + k^2} \over{2(a+k)}} = {{2a(a+k)+ k^2} \over{2(a+k)}}= {{2a(a+k)} \over{2(a+k)}} + {{k^2} \over{2(a+k)}} = a + {{k^2} \over{2(a+k)}}$$While a is positive, $${{k^2} \over{2(a+k)}}$$must be positive as k is numerically less than a.

So $$a< {{{a^2\over{a+k}} + a+k} \over {2}}$$ But the same equation could be written as: $${{2a^2 + 2ak + k^2} \over{2(a+k)}} = {{a^2+ (a+k)^2} \over{2(a+k)}}= {{(a+k)^2} \over{2(a+k)}} + {{a^2} \over{2(a+k)}} = {{a+k} \over{2}} + {{a^2}\over {2(a+k)}}$$The following number is equal to a+k: $${{a+k} \over{2}} + {{a^2}\over {2(a+k)}} + {{2ak+k^2}\over{2(a+k)}} = {{(a+k)^2 + a^2 + 2ak + k^2}\over{2(a+k)}} = {{a^2 + k^2 + 2ak + a^2 + 2ak + k^2}\over{2(a+k)}} = {{2(a^2 + 2ak + k^2)}\over{2(a+k)}}={{(a+k)^2}\over{(a+k)}} = (a+k)$$ This means that $${{{a^2\over{a+k}} + a+k} \over {2}}< {a+k}.$$From the two inequalities I obtain that: $$a< {{{a^2\over{a+k}} + a+k} \over {2}}< {a+k}.$$ This means that the solution obtained goes closer and closer at each step to the real value of whether k is positive or negative.