pi

with this formula

π = n sqrt {(sin 360 over n)^2 + (1 - cos 360 over n)^2} over 2

where n is the number of sides of a shape inscribed in a circle. it is correct to say that a n value of infinity will give a perfect mesurement of pi to infinte decimal places?

Well obviously you can't just substitute infinity into the equation, but it does converge to pi in the limit going to infinity.

http://en.wikipedia.org/wiki/Computing_π

Why even bother with [sin(360) / n]^2 = [sin(0) / n]^2 = [0 / n]^2 = 0^2 = 0?

You sure your formula is correct?

>>4
He meant [sin(360/n)]²
Yes, in the limit, it's still 0.

by 360 you mean 2pi

sage op here, nvr mind, i lost intrest