Approximating Pi using GeometryDate: 08/12/98 at 00:33:38 From: Rimi Ghosh Subject: Value of pi I need to know a simple method to find the approximate value of pi (3 < pi < 4) using elementary geometry. Date: 08/12/98 at 13:48:02 From: Doctor Rick Subject: Re: Value of pi Hi, Rimi. I had fun with this problem a while back in connection with a Geometry Problem of the Week on the Math Forum. I'll share a partial answer from that work. Of course, you realize that "simple" and "fast" are not the same thing - the methods that are really used to approximate pi converge much faster than what I will show you, but they require some higher knowledge than geometry, like Taylor series. The geometrical method works like this. We will try to approximate the area of a circle of radius 1 -- the area is pi. We do this by noting that the area of an inscribed polygon must be less than the area of the circle, and the area of a circumscribed polygon must be greater than the area of the circle. Here is what they look like when the polygons are squares: +----------------******----------------+ | * /\ * | | * / \ * | | * / \ * | | * / \ * | | * / \ * | | * / \ * | |* / \ *| * / \ * */ \* *\ /* * \ / * |* \ / *| | * \ / * | | * \ / * | | * \ / * | | * \ / * | | * \ / * | | * \/ * | +----------------******----------------+ First take the circle and inscribe a square in it. The inscribed square has side sqrt(2) [that's the square root of 2], because its diagonal is a diameter of the circle. Next inscribe a regular octagon. Its area turns out to be 2*sqrt(2) = 2.82843. It isn't very close to pi yet, but we can keep going. For this purpose I derived a formula for the area of a regular (2^n)-gon (a polygon with (2^n) sides) inscribed in a circle of radius 1. It turns out that this area is: A(n) = 2^(n-2) * s(n-1) where s(n-1) is the edge length of the inscribed (2^(n-1))-gon. For example, the area of the octagon (n = 3) is 2^1 * s(2) where s(2) is the edge length of the square: s(2) = sqrt(2) so A(3) = 2*sqrt(2), as I just said. I won't try to prove the formula here, just leave it as an exercise for the reader. I still need a formula for the edge length s(n). I derived a formula for a(n) in terms of s(n-1), so we can compute, for example, the edge length of the inscribed octagon from the edge length of the square. The formula is: s(n) = sqrt(2 - sqrt(4 - s(n-1)^2)) (I have discovered a truly remarkable proof which this e-mail is too small to contain.) For example, knowing that the edge length of the inscribed square is sqrt(2), we find that the edge length of the inscribed octagon is: s(3) = sqrt(2 - sqrt(4 - s(2)^2)) = sqrt(2 - sqrt(4 - sqrt(2)^2)) = sqrt(2 - sqrt(2)) = 0.76536 Using these two formulas, here are the areas of some regular polygons inscribed in a circle of radius 1: Square 2 Octagon 2*sqrt(2) = 2.82843 Hexadecagon 4*sqrt(2 - sqrt(2)) = 3.06147 32-gon 8*sqrt(2 - sqrt(2 + sqrt(2))) = 3.12144 64-gon 16*sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2)))) = 3.13655 I'm on my way to computing a very good approximation to pi. I leave it to you to derive formulas for the area of circumscribed polygons in a similar manner. The difference between the areas of the inscribed and circumscribed polygons will indicate how good the approximation is, and it will keep getting smaller. This may be more than you wanted, Rimi, but it makes an interesting project. It's nice to know that we don't have to take pi on faith - it can be computed using geometry alone. - Doctor Rick, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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