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pict

Benoit Cloitre

Calculation of the decimals of Pi: Some new with old stuff!


January 4, 2009

To Sum up

By exploiting a simple geometric idea, Benoit, still as prolific, improves the efficients of Archimedes' method for the calculation of the decimals of p  . This approach mixes algorithms and series and allows in theory to obtain a convergence speed of as fast as we want it. The convergence stays for the moment linear.

1 The Formula

A modern version of Achimedes algorithms consiste of defining the algorithm      1
a1 = 2  and

                V~ ---------------
     1           1(     V~ -----)
a0 = 2  an+1 =   2 1 -   1- a2n
(1)

which allows to calculate p  given that

       n
nli-->m oo  6.2 an = p
(2)

The efficience of this method is not bad since the convergence is in O(4 -n)  . It is not extraordinary either and numerous other methods (algorithms, series...) also converges linearly, equlas it or are an improvement (without taking into account the algorithms by Salamin-Brent or the other Borwein's brother of course). We offer still to start from this old idea and to calculate the decimals of p  with a speed of how great we choose it to be (according to a pre calculation). Hence, for all values of n  we have the following formula which stay valid:

                (   )
   n sum  oo --a2kn+1--  2k
6.2    4k(2k +1)  k   = p
    k=0
(3)

2 Proof

We ask you to exactly calculate the strippy area below, which is a sector of a unit square.

pict

It is easy to see that this surface is worth

 integral  +AB/2            V~ ------2-
        f(x)dx - AB-  1-  AB--.
 -AB/2           2        4
(4)

By choosing AB2-= an  , we have for all n  the cercle divided into equal sectors and :

    (   integral  an V~ -----     V ~ -----)
6.2n  2      1- x2dx -an  1 - a2n  = p
        0
(5)

because we find again the area of the unit circle. On the other hand, we have the well known series :

 V~ ------      oo  sum     x2k   (2k)
  1- x2 = 1 -   4k(2k--1)  k
             k=1
(6)

i.e.  integral an V~ ---2-         sum o o --a2nk+1--(2k)
 0   1- x dx = an-   k=14k(4k2-1) k and    V~ ----2-       sum o o  -a2kn+1-(2k)
an  1- an = an-   k=1 4k(2k-1)  k and hence    V~ ------              2k+1 (  )
an  1- a2n = an-  sum o o k=1 4ka(n2k--1)-2kk . The equation 5 hence becomes

                        (  )
          n  oo  sum  --a2nk+1-- 2k
 A n > 0, 6.2    4k(2k+ 1)  k  = p.
            k=0
(7)

This gives a family of series converging as fast as we want towards p  and hence improves Archimedes' algorithm .

3 Trials

For n = 100  , if we calculate a100 = 4.310466040910874761824374062.10-31  , then the serie

  100  oo  sum  --a21k0+01--(2k)
6.2      4k(2k+ 1)  k
     k=0
(8)

gives however 60  good decimals of p  at each term.

Other ideas by Benoît Cloitre

e  and p  in a mirror 

p  and Log(2)  in a mirror 

A formula for Christmas for p   


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