### Archimedes and Numerical Roots

The problem is how did Archimedes calculate the lengths of the sides of the polygons which needed him to be able to calculate square roots?

Last year, on the television programme Antiques Roadshow... work out the approximate profit.

# Archimedes and Numerical Roots

##### Age 14 to 16Challenge Level

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: $$\sqrt{3} \approx {1.73205}$$

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 $$\mbox{The next approximation} = {{{a^2\over{a+k}} + a+k} \over {2}}$$ 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.