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Ludolph van Ceulen and Pi


Date: 11/02/98 at 02:19:09
From: Ryan Hammond
Subject: Ludolph van Ceulen and pi

Dear Dr. Math,

For a class I was asked to find out about a certain mathematican. I'm 
looking for information on Ludolph van Ceulen. The Internet site I 
went to said he approximated pi to the first 35 digits by inscribing 
polygons in a circle.

1) How do you figure out pi this way?

The site also said he used polygons with 2 to the 62nd power sides.  
My friend said that making a polygon with this many sides is impossible 
by human hand. And then he gave a reasonable argument.  

2) So my second question is how did van Ceulen make such a large 
   polygon?


Date: 11/02/98 at 09:37:49
From: Doctor Rob
Subject: Re: Ludolph van Ceulen and pi

These are very good questions! Thanks for asking them.

2) He didn't have to actually construct the polygon, he just had to
calculate its perimeter. See below. You could say he used a virtual
polygon.

1) Since the value of Pi does not depend on the radius, let's use a 
circle with radius 1 and diameter 2. Inscribe and circumscribe squares. 
Their sides will be sqrt(2) and 2, respectively. Their perimeters will 
be 4*sqrt(2) and 8, respectively. Their perimeters bound the 
circumference of the circle, 2*Pi, so 5.656 = 4*sqrt(2) < 2* Pi < 8. 
We have just proved that 2.828 < Pi < 4.

A general formula for the perimeter of the inscribed regular n-gon is:

   p(n) = 2*n*sin(Pi/n)

and for the circumscribed regular n-gon, it is:

   P(n) = 2*n*tan(Pi/n)

Also p(n) < 2*Pi < P(n). Note that:

   p(4) = 2*4*sin(Pi/4) = 8*sqrt(2)/2 = 4*sqrt(2)
   P(4) = 2*4*tan(Pi/4) = 8*1 = 8

which agrees with what we found above. The problem with these formulas 
is that the computation of the sine and tangent are difficult.  
Fortunately there are recursive formulas relating p(2*n) and P(2*n) to 
p(n) and P(n). The formulas are:

   P(2*n) = 2*p(n)*P(n)/[p(n)+P(n)]
   p(2*n) = sqrt[p(n)*P(2*n)]

If you rewrite them in terms of 1/P(n) and 1/p(n) they become even
simpler:

   1/P(2*n) = [1/p(n)+1/P(n)]/2
   1/p(2*n) = sqrt([1/p(n)]*[1/P(2*n])

The first formula is an arithmetic mean, or average, and the second is
a geometric mean. The amount of calculation that has to be done with
numbers of high precision at each step is an add, a division by 2, a
multiply, and a square root. At the end you have to take a reciprocal
and divide by 2 again.

This allows one to double the number of sides with a feasible amount 
of calculation. Ludolph van Ceulen doubled the number of sides from
2^2 to 2^3, to 2^4, and repeated this step 60 times. That gave him 
2^62 sides, and the values of p(2^62) and P(2^62), and Pi was trapped 
between half these values. It turns out that:

   p(2^62)/2 = 3.14159265358979323846264338327950288395418471219027...
   P(2^62)/2 = 3.14159265358979323846264338327950288468313877374477...

Thus, to 35 decimal places,

   Pi = 3.14159265358979323846264338327950288...

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Geometry
High School History/Biography
High School Triangles and Other Polygons
Middle School Geometry
Middle School History/Biography
Middle School Pi
Middle School Triangles and Other Polygons

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