The world of Pi - Cloitre

www.pi314.net

History

Mathematicians

All formulas

Num. approx.

Softwares

Misc. math.

Digits

Poetry

Papers/videos

Delirium !

Pi-Day

Images

Music

Links

Bibliography

Boris Gourévitch
The world of Pi - V2.57
modif. 13/04/2013

Home

Version history

Guestbook

Who I am

Some pictures (fr)

Acknowledgements

Last modifications

Contact

Benoit Cloitre

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

January 4, 2009

To sum up
1 The formula
2 Proof
3 Trials
Other ideas by Benoit Cloitre

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.

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