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



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5 Formulae BBP : The technique

In 1995, Bailey, Borwein and Plouffe discovered the famous formula

  oo     (                              )
 sum   -1-  --4---- --2---- --1---- --1---
k=1 16k   8k+ 1   8k+ 4   8k+ 5   8k+ 6
(B.B.P.)

which , as we now know, goes further than it's simple form by giving us a method to find the n-th digit of Pi without knowing the previous one. Even better, it revolutionised the view of Pi by putting it amoung the constants that we can generate nearly like a chaotic dynamic system, which Bailey and Crandall turn into their famous conjecture in 2000 [11] linking the concept of normality to the passage of a digit to the next.

And with the function Y  , quickly, we will notice that the BBP formulae are a particular case of relations of function class that we call polylogarithms. A nice introduction to those extraordinarly simple but promising function is the work of e Gery Huvent that I adapt here :

5.1 s = 1,  v = 0   : The polylogarithmes

5.1.1 Definition

For z  (-  C  ,|z|< 1  and n  (-  N  , we define the polylogarithm of order n  by

        + sum  oo  zk
Ln (z) =   kn-
        k=1
(32)

Hence

pict

For n > 2,  note that the convergence is normal of the closed unity disc.
We have the following integral representation

 integral 
  1 lnn-(y)-        n+1
 0  y-  1z dy = (-1)  n!Ln+1(z)
(35)


which allow us to extend the polylogarithm, for n > 2,  on C\]1,+ oo [  to a holomorphe function.
The polylogarithm can be also defined by succesive integration, in fact

dLn-(z)-  Ln-1-(z)-
  dz   =    z
(36)

as well

         integral 
          zLn--1(t)
Ln (z) = 0    t   dt
(37)

and for example

          integral               integral   integral 
           zln(1--t)      z  t1 --1--
L2(z) = - 0     t   dt = 0  0  t1- ududt
(38)

5.1.2 Remarquable values

The first remarquable values are, for n > 2,

pict

where z  is Riemann's zéta function.
In particular

         2               2
L2 (1) = p--et L2(- 1) = - p
        6               12
(Euler)

For n > 1,

pict

where       +o o      k
b(n) =  sum --(-1)-n
      k=0(2k +1)  . In particular

pict

Note 1 We know the exact value of z (2n)  and of b (2n + 1).

5.1.3 Duplication formula

On we have C\]1,+o o [  ,  integral   n        integral   n       integral        (          )      integral       (      )           integral   n
 01ln-(y1)dy +  10 ln-(y1)dy = 01lnn (y)--11+ --11  dy =  10 lnn(y) -22y1-  dy = 221n  10 ln-(u1)du
    y- z        y+z               y- z  y+ z                y- z2    u=y      u- z2  . Which prove the duplication formula, that when the three terms have a direction :

                   1    (  )
Ln (z)+ Ln(- z) =-n-1Ln  z2
                 2
(45)

5.1.4 Euler and Landen's formulae for the dilogarithm

Euler's Formulae On ]- 1,1[,  we have

L'n(x) = Ln--1(x)
           x
(46)

This equality allow us to prove through differentiation Euler's identity for the dilogarithm

 A x  (-  ]0,1[, L2(x)+ L2(1 -x) = z(2)- ln (1 - x)ln (x)
(47)

In particular with x = 1 ,
   2  we get the following result, thanks to Euler

   (1 )   + sum  oo   1    p2   1
L2   -  =    -n-2 = ---- -ln2(2)
     2    k=12 n    12   2
(Euler)

Landen's Formulae We also have the following relation said to be Landen's identity (that we prove using differentiation)

    [     ]           (      )
 A x  (-  - 1, 1 , L2 (x)+ L2-x-- =  -1 ln2(1 -x)
         2              x- 1      2
(48)

This identity is true on       [    ]
C\RU   -1, 12

Particular Values With     1-
x = f2  where    1    V~ 5-
f = 2 + 2  is the golden number (which satisfies  2
f + f = 1  ), since -x-
x-1 = x- 1  in this case, we obtain

  ( 1 )     (  1 )    1    (1 )    1
L2  -2  + L2  ---  = -- ln2  --  = -- ln2 (f)
    f          f      2     f      2
(49)

But Euler's formula give us

  (  1)     (1 )
L2  -2  + L2  -- = z (2) -2 ln2(f)
    f         f
(50)

and the duplication formula

   (1 )     (  1 )   1  ( 1 )
L2  --  + L2 - -- =  -L2  -2
    f          f     2    f
(51)

By combining the three results, we get

pict

Complex values Landen's indentity can be extend through analysis to C\]1,+o o [.  With x = i,  we get

  (1 - i)   5p2   1  2    (1           )
L2  -2--  = -96 - 8 ln 2 +i 8p ln (2)- G
(54)

5.1.5 Landen's formula for trilogarithm

     ]   1]
 A x  (-  -1, 2

  (  x  )                                          1                1
L3  x--1- + L3 (x) + L3(1- x) = z (3) +z (2)ln(1- x) - 2 ln(z)2ln (1 - x)+ 2 ln3(1- x)
(55)

This formula can be deduced by Euler's and Landen's identity for the dilogarithm after differentiation.

5.1.6 Particular values

We deduce form Landen's identity for the trilogarithm the following remarquable values :

   (  )           2         3
L3  1   = 7z(3)- p--ln-(2) + ln-(2)- pour x = 1
    2     8         12       6            2
(56)

  (   )            2          3
L3  1-  = 4z(3)- p--ln-(f) + 2ln-(f)  pour x =-1
    f2    5         15        3             f2
(57)

5.1.7 A few classical integrals

The integral representation (35) immediatly give

 integral  1             integral  1              2
   ln-(u)du  =      ln-(1---t)dt = -p--
 0 1- u   u=1-t 0     t         6
     integral  12 ln(u)      p2  1  2
        1--udu = - 12- 2 ln  (2)
     0
(58)

(59)
The first integral is classical, the second not so!

Similarly

pict

5.2 The links between integrals, BBP formulae and polylogarithms

5.2.1 Notations

Let us start by simplifying notations! It would be great if we could talk about a BBP formula without having to copy down the formula every time...

Gery Huvent has offered to denote the sum        (              )
 sum o o  1i  sum b -2--al-k+1-
  i=0 a    l=0 (bi+l+1) , for
                         b-2
P (y) = a0 + a1y +...+ ab-2y  .

Plouffe's formula is now written

pict

Which simplifies a lot !

5.3 Integrals and BBP formulae

In this section we offer to get integrals equivalent to BBP series. It is often the case easier to go through integral calculation to find the result of a serie!

5.3.1 Integrals equivalent to BBP series

So, from the equality  A a  (-  Z  , a /= 1,  and a /= 0  ,  A y  (-  [0,1], A k > 1,

 integral  1ayk-1     integral  1 oo  sum  ybi+k-1     oo  sum  1---1--
 0 a -ybdy =  0      ai  dy =    aibi+ k
                i=0            i=0
(62)

we deduce that if                                      b- 2
P  (-  Qb-2 [X] , P = a0 + a1X + ...+ ab-2X  then

  integral  1                 b-2      integral  1           sum  oo   (b sum -1         )
a   a0 +-a1y-+...+-ab-2y--dy =    aP-(y)dy = -   -1     ---ak----
  0         yb- a              0 yb- a      i=0ai  k=0bi+ k+ 1
(63)


Similarly,

pict
pict

5.3.2 More general integrals and BBP formulae

Now, we have suspicion that it is not this kind of integrals that we would be able to calculate, and so we will not be able to deduce BBP formula, but we should deal with an integral of type  integral  1 k
   ln-(y)R(y)dy
 0    Q (y)  . But the whole trick is there !

In fact, if Q(y)  divides yb- a,   integral  1lnk (y) R(y)     integral  1 lnk(y)P (y)
   ----------dy =    ---b------dy
 0    Q(y)         0   y - a  where             (yb- a)
P (y) = R (y)------.
             Q (y)  This allows, when we know how to calculate the integral  integral  1 k
   ln-(y)R-(y)dy
 0    Q (y)  , to deduce a BBP formula. The same kind of remarque is applicable if Q (y)  divides yb + a  .

5.3.3 Calculations of intergrals

Now, lets talk about polylogarithms. The above remarque is true in the case where Q (y)  divides yb -a  to extract BBP series. But to calculate the integral? Well we can use a decomposition into simple elements of R-(y)
Q (y)  R-(y)
Q (y)  so to go back to known integrals. And that's where polylogarithms come in handy.

We remember that the integral expression of a polylogarithm of order n  for a non zero complex z  with modulus less than 1  .

 integral  1 n
    ln--(y)dy = (-1)n+1n!Ln+1(z)
 0  y-  1z
(68)

This allows, after decomposition into simple elements of R(y)-
Q(y)  , to express an integral of type  integral    k
  1ln-(y)R-(y)dy
  0   Q (y)  as sums of polylogarithms.

5.4 Function Y  and polylogarithm

As well as the obvious relation Y(x,s,0) = Ls(x)  , the previous paragraph puts light to the decomposition of a rational fraction in the integrals allows to bring BBP formula to polylogarithm. But furthermore, the formula (64), which we have seen the direct link to the BBP formulae if we have a polynomial Q (y)  dividing  b
y  -a  , can also be written

pict

that is in the form of combination of the functions Y  . In fact, the BBP formulae are nothing other than the combination of functions Y  where the parameter x  does not move and is the inverse power of an integer.

Hence, the Plouffe's formula

pict

Starting from here, and with order greater than 1 (k > 0)  , we have all the bits to link the polylogarithm to the BBP formulae and now the functions Y  .

At this stage, we need to note that the polylogarithms are of the style Ls(x) =   oo  sum  1-i
   isx
i=0  while the BBP formulae use more series of type  sum  oo           (     )
   -i+1psxi = Y x,s, pq
i=0(  q) . Intuitively, instead of having the parameter n = 0  as the equivalence Y(x,s,0) = Ls(x)  let us see, the polylogarithms are able to be "developed" by giving a linar relation between the functions Y  where     p
v = q  , basicly the BBP formulae. In this form with the parameters of the function Y  , we can see that this kind of generalisation of the arctan since we don't have v = 12  anymore but more generaly v = pq  .

5.5 The interest of the BBP formulae

This formulation has showed us that we can in base 2k  and 3k  calculate any digits of constants resulting from this kind of series, without knowing the previous. Why ? I will explain it a bit later on the page consacred of Plouffe but in an intuitive way, we can notice that the development  in base 10 of a number a0.a1a2a3...  is  sum o o -ann
  n=010  with an  an integer between 0 and 9. So by considering the series of the kind  sum o o -1n-= ln(2)
  n=12 i  we are not far from the development in base 2 ! All we need is to calculate the decimals in a certain place of 1
i  .

And this is what happened not only with ln(2)  but also with loads of other constants around p  , the Riemann function etc...

Futhermore, we can notice that they are rational series and so they are perfect target for simple calculation of the decimals of Pi. We know more immediatly that thanks to the term -1n
j  of the series how many decimals we can obtain in practise with n terms of the series. In fact, with n  decimales, in that case we obtain nlog(j)  decimales (and even a bit more if we take into account the term in --1-
an+b ).

Alternating intensive phases of exploration with more calm phases, the search in that direction was very fertile! A few elements in the following paragraphs

In a synthetic way, Gery Huvent put forward a few integrals useful to the decomposition that the previous paragraph talk about. We will present it here, with their application to the research into BBP formulae.This text takes with a few adaptations the article by Gery Huvent [12] which can not be more fundamental!


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