Archimedes pi

caraa · Mar 1, 2008, 03:49 PM

For my maths gsce coursework I have to write about how archimedes estimated the value of pi. I have completed all of it up to an a grade. To get an a star I have to give this in terms of n. I'm really stuck can anyone help?

galactus · Mar 1, 2008, 04:11 PM

There is lots and lots on the web about this. Why not just Google it?

It is rather extensive to show all the details,

Looking at fig. 3, we have $(1-x)^{2}+(s/2)^{2}=1........[1]$

and $x^{2}+(s/2)^{2}=t^{2}...........[2]$

Squaring [1] gives us $(1-2x+x^{2}+\frac{s^{2}}{4}=1$

or $-2x+(x^{2}+\frac{s^{2}}{4})=0$

Now use [2] to make a substitution:

$-2x+t^{2}=0$ or $x=\frac{t^{2}}{2}$

On the other hand, [1] yields us $(1-x)^{2}=1-\frac{s^{2}}{4}=\frac{1}{4}(4-s^{2})$

Square root of both sides:

$1-x=\frac{1}{2}\sqrt{4-s^{2}}$

$1-\frac{1}{2}\sqrt{4-s^{2}}=x=\frac{t^{2}}{2}$

or $\frac{t^{2}}{2}=1-\frac{1}{2}\sqrt{4-s^{2}}$

Therefore, we multiply by 2 and get:

$t^{2}=2-\sqrt{4-s^{2}}$ or $t=\sqrt{2\sqrt{4-s^{2}}}$

Finally, with $t=s_{n+1}, \;\ and \;\ s=s_{n}$ ,

we get a formula for the side of the (n+1)st polygon in terms of that of the nth polygon:

$s_{n+1}=\sqrt{2-\sqrt{4-s_{n}^{2}}}$

$s_{1}$ is the side of the inscribed square , so we get:

$s_{1}=\sqrt{2}$

Double the sides and get 8 sides:

$s_{2}=\sqrt{2-\sqrt{2}}$

Double it and get 16 sides:

$s_{3}=\sqrt{2-\sqrt{2+\sqrt{2}}}$ .

In general,

$s_{n}=\sqrt{2-\sqrt{2+\sqrt{2+.....\sqrt{2+\sqrt{2}}}}}$

Since the perimeter of the nth polygon is $p_{n}=2^{n+1}$ , an

approximation to $\pi$ is $\frac{1}{2}p_{n}=2^{n}s_{n}$

or

${\pi}_{n}=2^{n}\sqrt{2-\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{2}}}}}$ , where we have n radicals.

This can be written in a better way as ${\pi}_{n}=2^{n}\sqrt{2-2_{n-1}}$

For instance, if we use n=8 in the above formula, we get 3.14157294025

I believe Archimedes used up to a 96-sided polygon.

I hope this helps. Remember, Archimedes did not use trig as we know it.