Friday, 30 July 2021

Pythagoras's and Newton's formulas for 𝝅

This fun video by Veritasium gives the story of Pythagoras and Newton and ##\pi## but it doesn't go into details. I wanted to know more.

Pythagoras calculated ##\pi## by placing regular n-agons just inside and outside a unit circle and calculating their perimeters. Some examples are shown below.
Apparently he stopped at a 96-agon. I reckon the result he got was$$48\sqrt{2-2\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{4}\sqrt{2+\sqrt3}}}}<\pi<\frac{96\sqrt{2-2\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{4}\sqrt{2+\sqrt3}}}}}{\sqrt{2+2\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{4}\sqrt{2+\sqrt3}}}}}$$It was accurate to four significant digits and not surprisingly he stopped! He didn't even have decimal numbers.

The second last person to use this method in anger was Ludolph van Ceulen in about 1600 who did a ##2^{62}##-agon and still only got 35 decimal places. Apparently it took him 25 years. Nutter.

Then along came Newton (about 1700) and he used calculus (which he had just invented) and the binomial theorem to get$$\pi\approx-\frac{3\sqrt3}{2}+6-\left(\frac{1}{4}+\frac{3}{320}+\frac{45}{53,760}+\frac{1,575}{22,855,680}\right)$$That's accurate to five significant digits and a good deal easier to calculate. The further you go the better it gets:
Here's how they did it Newtons formula for pi.pdf (10 pages).

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