
Re: Prime numbers and pi
Posted:
Apr 13, 2009 8:26 PM


On Mon, 13 Apr 2009 06:47:09 EDT Franz Gnaedinger <discussions@MATHFORUM.ORG> wrote:
The infinite pi multiplication of Wallis mentioned by James A. Landau can be given as a stairway approximation:
2 x 3/2
2 x 4/3 x 7/6
2 x 4/3 x 16/15 x 11/10
2 x 4/3 x 16/15 x 36/35 x 15/14
2 x 4/3 x 16/15 x 36/35 x 64/63 x 19/18
2 x 4/3 x 16/15 x 36/35 x 64/63 x 100/99 x 23/22
and so on. The numbers of the last terms of each line increase by 4, and are then replaced by new terms whose numbers are bigger by 1x1, 3x3, 5x5, 7x7, 9x9, and so on. I am trying to find a regular pattern for the infinite pi product of Euler involving the primes, but I think I won't succeed.
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ I hereby supply you with the following totally useless algorithm for computing pi:
For n = 2, 3, ...
Define P(n) as the product of all prime numbers from n+1 through 2n, e.g.
P(10) = 11 x 13 x 17 x 19
Define T(n) recursively as follows:
T(2) = 2 For n > 2 If n is prime and 2n1 is prime, T(n) = 2 T(n1) If n is prime and 2n1 is composite, T(n) = (4n2)T(n1) If n is composite and 2n1 is prime, T(n) = 2 T(n1) / n If n is composite and 2n1 is composite, T(n) = (4n2)T(n1)/n
T(3) = 2T(2) = 4 T(4) = 2T(3)/4 = 2 T(5) = 18T(4) = 36 T(6) = 2T(5)/6 = 12 T(7) = 2T(6) = 24 T(8) = 30T(7)/8 = 90 T(9) = 2T(8) /9 = 20 T(10) = 2T(9)/10 = 4
(Note: the "^" symbol is being used below to signify exponentiation).
As n increases, 2^(2n) / (n [P(n) T(n)]^2) converges to pi.
Warning: convergence is glacial. For n = 25, the approximation is 3.26
If you would like to figure out for yourself how this algorithm works, stop here. Else read on and I will reveal the trick.
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Answer: P(n) is a red herring. As far as I am aware, there is no significant connection between pi and prime numbers.
The product P(n) T(n) is the binomial coefficient C(2n, n) which is also "2n choose n" or "2nCn" or the number of combinations of 2n things taken n at a time.
C(2n, n) is also the middle term of the nth row of Pascal's triangle (which I am informed is also known as Tartaglia's Triangle or Yang Hui's Triangle).
1 1 1 1 2 1 middle term is 2 1 3 3 1 1 4 6 4 1 middle term is 6 1 5 10 10 5 1 1 6 15 20 15 6 1 middle term is 20
The terms of the nth row of Pascal's Triangle, when divided by the rowsum which is 2^n, are the binomial probabilities for the binomial case (n, p = q = 1/2).
In The Elements of Chance (1733), Abraham De Moivre showed that as n increases, the binomial distribution approaches what we now call the normal distribution, with standard deviation sqrt(n/2).
Hence, as n increases, the middle term of the nth row divided by 2^n approaches the mean of the normal density function, which is
1/(sigma sqrt(2 pi)) e^0 = 1 / (sqrt(n pi)
I simply rearranged things so that I would be computing pi rather than 1/(sqrt(2 pi)
It is possible to compute e with the same approach. Given 1/sqrt(n pi) find the onesigma points which will have the value
1/(sqrt (n pi) e^(1)
The results are not worth the effort.
I do not claim anything original in this algorithm, as all I have done is to take De Moivre's result and rearrange it to form a parlor trick.
James A. Landau Test Engineer NorthropGrumman Information Technology 8025 Black Horse Pike, Suite 300 West Atlantic City NJ 08232 USA
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