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



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6 Formulae BBP in base 2 : s  (-  N  ,v = pq  , x = 12n-  in Y

6.1 The considered integrals

In the worry of decomposing our integrals  integral  1lnk(y)R (y)
   ----------dy
 0    Q (y)  in simpler to calculate integrals, we are interested to the following "prime" integrals :

pict
pict
pict
pict
pict
                  V~ -                      V~ -
       1         --3            1        --3
o`u s1 = 2 (1+ i)+ 2 (1- i), s2 = 2 (1+ i)- 2 (1- i)
(83)


    1          V~ 3           1          V~ 3
s3 = 2 (1 - i)+ 2--(1 + i), s4 = 2 (1- i)--2-(1+ i)
(84)

pict

In [3], Broadhurst establish some relation between the different sums of polylogarithm. However he does not seem to consider the link with integrals. We will try to do so.  Which is why we introduce the following integrals :

pict

The link between the J(k)
 n  and the is the following :

pict
pict


6.2 The method

The calculation of the integrals I(nk)  will give linear combination of constants of order k + 1  like pk+1  or z(k + 1)  thanks to their expression under polylogarithm form of order k+ 1  . But furthermore, we can obtain BBP formula with the    I(kn)  by using what Gery Huvent calls the denomination tables and which are just the expressions I(k)
 n  in the form of integrals        integral 1 P(y)lnk(y)
       0   yn- a  whom we have seen the direct expression under BBP serie form with the formula (69). We just need to obtain BBP series for the precise constant that we are interested in, which boils down to use a certain linear combination of  (k)
In  . For this, we intoduce normally the linear form                 11
L  (a  ,...,a  ) =  sum  a I(k)
 k   1    11   n=1 n n  . Then we impose some relation between the a
 n  so to cancel out the coefficients in fronts of the unwanted constants. We then get a BBP serie for the remainder, p  or log(2)  or a lot more other things! Here is the denominator table. The method is then detailled with an example.

pict
pict
pict
pict

where P  is a polynomial with integers coefficient which depends both of n  and of the chosen denominator.

6.3 Formulae for p  , ln(2)  ln(3)  and ln(5)

In the case where k = 0  , we get, by a calculation of integrals

pict

6.3.1 BBP formulae applications for p

So to get some formula for p,  we impose the following relations :
(-a2 - a3- a4 - 2a7- 2 a8- 4a10 -a1) = (a2 + a7) = (a3 +a10)
= (2a6 - 2a11) = 0  .
So we get

pict

So to get some BBP formula with few terms, we can in a first of all fix a  = a   = a = a  = a = 0
 11   10    9   8    7  , which gives

a5-p = - a I(0)+ a I(0)+ a I(0)
 2       4 1    4 4     5 5
(113)

We then consult the denominator table. To simplify a sum that calls I1,I4  and I5,  we can write each integral under the form  integral 1Pn8(y)4dy,
0 y -2  n = 1,4,5.  Hence

a5-       (0)     (0)     (0)   integral  1-P (y)    -1-     ( 4       )
2 p = -a4I1  + a4I4 + a5I5  =  0 y8- 24dy = 24 BBP0  2 ,8,P (y)
(114)

where P  is a polynomial whose coefficients depends on a4  and a5  . If we fix a5 = 1  and a4 = - 4r- 1  we get the formula by Adamchik-Wagon (cf [6] )

     sum  oo    (                                              )
p =    -1-  8r+4-  -8r--- -4r-+  -8r-2+ -2r-1 + -2r-1 + -r--
    i=016i  8i+1   8i+2   8i+3   8i+4    8i+5    8i+6    8i+7
(Adamchik-Wagon)


The choice of a4 = -a5 = 1  give the formula by Plouffe (61). We can now look for a denominator in y24- 212  . We take then the formula (111) and we simply fix a11 = a10 = 0  .

pict

We choose the other coefficients so to cancel out the most coefficients of P  (by imposing a5 /= 0  ).
The best choice seems to be
[ai] = (a1,...,a11) = (0,- 1,0,1,1,0,1,-1,-1,0,0)  which gives

pict

We can also look for a denominator in y40- 220  by fixing a  = a  = a = 0,
  9   8    7  the best choice seems to be
[a ] = (a ,...,a  ) = (0,0,0,0,-2,1,0,0,0,0,1)
  i     1    11  which gives

pict

In fact it turns out that the last formula -P40(y)20
y  -2   can be simplified to -Q20(y)10
y +2   , which gives the alternated formula :

 p = -1BBP0  (- 210,20,2y17- 5y14- 8y13- 8y9- 128y5- 160y4 + 512y)
     26
  1-  oo  sum  (-1)i( -512--  -160-  -128-  --8---  --8---  --5---  --2--)
= 64    210i  20i+2 - 20i+5 - 20i+6 - 20i+10 - 20i+14 - 20i+15 + 20i+18
    i=0
(Bellard)

(119)
Finnaly if we considered the formula (111) in all generality, it can be written 1 (a  + a )p =  integral 1-P(y)-dy
2   11    5     0y120- 260  , we just need to choose the right (ai)i  . The choice that seems to lead to a formula with the least terms possible is
[ai] = (a1,...,a11) = (0,0,0,0,0,1,0,0,- 2,0,1)  and gives
     1       (60         )
p = 256BBP0   2 ,120,P (y) o`u P (y) a 42 coefficients non nuls
(120)

We can explain P  by writting that
                            integral 1-P(y)--
L0 (0,0,0,0,0,1,0,0,-2,0,1) = 0 y120- 260dy  .
To finish off, the choice of [a] = (0,0,0,0,0,1,0,0,-2,0,1)  leads to

pict

And [a] = (0,0,0,0,2,0,0,0,-2,0,0)  gives

pict

6.3.2 BBP formulae for ln (2), ln (3)  and ln(5)

We can apply the same method to obtain some BBP formula for ln(2)  , ln (3)  and ln(5)  .

6.4 Case of polylogarithms of order 2   : Formulaes of order 2

We can notice that for order 1, that is for the BBP formulae giving p  or for example, we had to deal with integrals with rational fractions. If we now want to get BBP series giving p2  or p *ln(2)  or ln(2)2  , we need to introduce a logarithm to the numerator of the integral. More precisely, for a BBP formula of order n  , we need to consider integrals of type  integral  ln(x)n-1P(x)dx
     Q(x)  . This is due to the fact that we then obtain polylogarithm combination of order n  .

This is what was done once more by Gery Huvent [12], noting as well that the first series were found by Plouffe and that Broadhurst [3] provided a few as well. The interest in order 2 is to be able to find BBP series for a famous constant which is Catalan's constant defined by G =  sum o o -(-1)n-
      n=0(2n+1)2   . This shows, if you are not convinced, that Catalan's constant is "homogenuous" to an order 2, that is that it is without doubt of the same nature as   2
p  or p *ln(2)  concerning the spread of its digit in base 2 or 16.

6.4.1 The classical expression :  (1)  (1)  (1)  (1)
I1 ,I2 ,I3 ,I4  and  (1)
I5

A classical result by Euler is

       2    2
I(11) = p--  ln-(2)-
      24     4
(123)

Which allows us to write that

                        (  )
 (1)    p2-  ln2(2)  1    1
I2  = -24 +   4   + 4L2  4
(124)

Kummer's equation for the polylogarithm of order 2  is written (cf [4])

pict

The inverse formula is

  (  )
    1             p2-  ln2(-z)
L2  z  = - L2(z)-  6 -    2
(Formule d’inversion)

and finally the duplication formula in the general case :

L (z)+ L (- z) =--1-L  (z2)
 k      k       2k-1  k
(Formule de duplication)


By applying Kummer's equation for x = 1-2-i, y = 12  then x = 1+2i, y = 12  , we get two equality which added gives

pict

By using the inverse formula for z = - 1+ i  and z = - 1- i  and the duplication formula , we get

pict

By duplication, we also have

  (     )     (     )     (      )      (      )       (   )
    1+-i        1--i        -1--i-       --1+-i    1      1
L2   2    + L2   2    + L2    2    + L2    2     = 4L2  - 4
(128)

We therefore deduce

pict

Similarly, Kummer's equation for x = 1-2i, y = 12  and x = 1+2i, y = 12  , gives two equality which when taken away fives a new equality. By then using the inverse formula for z = - 1+ i  and z = - 1- i  and the duplication formula for           1+2i  and 1-2i,  we obtain

   (     )     (     )
L   1-+-i - L   1---i - (L  (i)- L (- i))+ ip-ln(2) = 0
  2   2       2   2       2      2          4
(131)

But          p2-
L2(i) = - 48 + iG  where G  is Catalan's constant. Hence

 (1)   (   (1 - i)     (1 + i))   p ln (2)
I5 = i  L2  -2--  - L2  -2--   = ---4-- - 2G
(132)

6.4.2 Calculation of I(71),I(18)  and I(91)

Proposition 2 We have

       2      (  )
I(17)=  p-+ 1 L2  1
      72  4     4
(133)

Proof. Kummer's equation with         V~ 
x = 1-i2-3, y = 12   gives the equality

pict

which gives straight away the wanted result because
L  (1)= p2 - ln2(2)
  2 2    12     2   , L (1)+ L  (- 1)= 1L  (1),
 2 2     2   2   2 2  4
   (1+i V~ 3)    (1-i V~ 3)    (-1+i V~ 3)    (--1- i V~ 3)
L2    2   + L2    2   + L2     2    +L2     2
    (  (     V~ -)     (     V~ -))
= 1  L2  -1+i-3 + L2  -1-i-3
  2        2            2 by the duplication formula and Kummer formula for           V~ -
x = y = 1-i-3
         2   gives    (     V~ -)    (     V~ -)     2
L2  -1+i2-3  +L2  -1-i2-3  = - p-.
                             9   _

Proposition 3 We have

       2    2
I(81)= 5p--  ln-(2)-
     36      2
(134)

and

I(91) = -2G-
        3
(135)

Proof. Kummer's equation for     1+i     1-i V~ 3
x = -2-, y =--2--   gives, considering
                (         )
L2 (1 - i) = p126- i G + p-ln4(2) :

                         (          )      (    )     (      )
L  (s )+ L  (s ) = 17-p2- i G + p-ln-(2)  - L   e-ip6  - L   e-5ip6-
 2  1    2  2    144              4       2           2
(136)

similarly, Kummer's equation for                V~ 
x = 1-2i, y = 1-i2-3   give

                 17      (     pln(2))     ( ip)     (  5ip)
L2 (s3)+ L2 (s4) = ---p2 + i G+  ------ - L2  e 6  - L2  e 6
                 144             4
(137)

With the help of the inversion formula

pict

Which allows us to conclude that

pict

and

pict

So we just now need to calculate the last sum of polylogarithm. But we have gain in simplicity because those logarithm uses roots of unity.
We therefore use the multiplication formula

         q sum -1  (  2ikp-)
L2(zq) = q  L2  e q z
         k=0
(Formule de multiplication)

which gives with z = -i  and q = 3

1          ( ip)     (  5ip)
-L2 (i) = L2 e 6  + L2  e 6  + L2 (- i)
3
(144)

then with z = i  and q = 3

1           (   ip-)     (   5ip)
-L2(- i) = L2  e- 6  + L2  e- 6   + L2(i)
3
(145)

and allows us to easily conclude.  _

6.4.3 Calculating  (1)
I10 ,  relation between  (1)
I6  and  (1)
I11

Kummer's equation for x = -1  and y = 1+ i  and for x = -1  and y = 1 -i  gives two equality which when added gives

pict

Which is

               (  1)   1        p2
J(51)=  2J(13)+ L2  - -  + -ln2(2)- ---
                  4    4        16
(147)

and allows us to confirm that

       2     2         (   )
I1(10)= 2p- - ln-(2)+ 1L2  - 1
      15     2     4      4
(148)


If, instead of adding them, we substract them, we get

pict

which gives us

 (1)    ( (1)    (1))  p-ln-(2)
J6 = 2  J2 - J4   +    4
(150)

i.e.

I(1)+ I(1)= p-ln-2- 12G
 6    11     4     5
(151)

6.4.4 Application to the determination of BBP formulae

We now consider the linear form L (a ,a ,...,a  ) = a I(1)+ ...+ a I(1)
 1  1  2    11     11         1111  . Taking into account the equalities (123), (124), (129), (130), (132), (133), (134), (135) and (148), we have

pict

Formulae for p2   So to obtain BBP formulae for p2,  we fix the following equality :
(- 1a  + 1a  + 1a - 1 a - 1 a - 1a  )= (- 2a  - 2a -  12-a  )
   4 1   4 2   4 3  4  4  2  8  2 10        5   3 9   5  11
= (a + a  ) = (a  + a ) = (a + a ) = (a -a  ) = 0
    5    11     2    7     3   10     6    11  so to obtain the equality

pict

We now just need to use particular variables so to obtain simple formulae.

A few simple alread known formulae for p2

We obtain those formulae by choosing the integrals who give an denominator of a degree less than in the correspndance table.
Let us one last time show details an example :
So to obtain a denominator of the form yb- a  of smaller degree (in that case y24- 212  ), we choose a9 = a10 = 0  such that we cancel down I(11)0  and I(111)  . OSo we then get

pict

This equality allows to give the general formula with 3  parameters

(                  )
 -1     -1     -1     2   -1-     ( 12       )
 16a4 + 72 a7 + 18a8 p  = 211BBP1  2  ,24,P (y)
(160)

pict

So not to put too much on this page only the formulae with parameters that correspond to the denominator in y24 - 212  will be given. There exist some who are assoiciated to other denominators (for example y120- 260  ).
Let us look again at the equality (159), if we choose to fix a  = 0
 7  and a = 16,
 4  this equality becomes

    (1)     (1)        integral  1 y ln(y)       integral  1(2y- 2)ln(y)    2
-16I1  + 16I4  = - 16   2-y2--2dy+ 16    -y2--2y-+2--dy = p
                     0                0
(161)

We can look at this equuality under the form of the sum of BBP formula.

pict

We can bring vack this integral to a denominator of the form yb - a  with the help of the corresponding table so to obtain

    integral      (                               )
32  1 ln(y)-y6--2y5--2-y4--8y3--4-y2--8y+-8-dy = p2
    0                y8 -16
(163)

which gives the following equality

32 BBP1 (16,8,y6- 2 y5- 2y4- 8 y3- 4y2- 8y + 8)= p2
(164)

or

      oo  sum    (                                                      )
p2 =    1-- --16-2-- --16-2-- --8-2-  -16-2-  --4-2- ---4-2 +---2-2
     i=0 24i  (8i+1)   (8i+2)   (8i+3)    (8i+4)    (8i+5)   (8i+6)   (8i+7)
(165)

This equality was already mentioned by Plouffe in [1].
The choice of a4 = 0, a7 = 72  gives

pict

which gives the following formula thanks to Plouffe :

pict

A few simple and new formulae

An other solution consist of keeping the integrals which gives denominator of the form  b
y - a  of high degree but adjust the parameters so to gave many nul coefficients in the BBP formulae.
A few good choice seems to be the following  :
[ai] = (a1,...,a11) = (1,1,0,0,0,0,- 1,0,0,0,0)  which gives the formula to 10  terms

pict

The polynomial P (y)  having only odd powers, this formula can be simplified to give

pict

[ai] = (2,1,0,1,0,0,- 1,- 1,0,0,0)  which gives the formula to 11  terms

pict

[a] = (- 2,-1,-1,- 2,0,0,1,0,0,1,0)
  i  which gives the formula to 43  terms (notice the 260  ) :

pict

If we are interested in formula with the least term, let us state that other formulae with 50  terms ([ai] = (- 1,- 1,0,0,0,0,1,0,0,0,0)  ) and with 55  terms ([ai] = (- 2,-1,0,-1,0,0,1,1,0,0,0)  ) exists.
We can even look for alternating series, for example
[ai] = (0,0,0,-2,0,0,0,,0,0,0)  gives

       9      (                                                )
p2 = - --BBP1  - 26,12,y10- 8y8- 12y7- 4y6 + 8y4 + 48y3 + 64y2- 32
       20
(Huvent)

Notice

The integrals equality allows to write in different ways  2
p  as the sum of BBP formula.
For example, the result by (Huvent) ([ai] = (-1,- 1,0,0,0,0,1,0,0,0,0)  ) gives the integral equality

pict

Which can be written

pict

Formulae for the constants G,p ln(2)  and  2
ln (2)  For p ln (2)

The same method leads to the general formula for pln(2)  (when we impose a denominator in  24   12
y  - 2  )

pln(2) =-1-BBP   (212,24,P (y)) o`u
        211    1
(Huvent)

pict

By adjusting the coefficients a4  and a8,  we have formulae with 17  termes of the form

pln(2) = BBP1 (212,24,P (y))

One of the simplest seems to be [ai] = (2,- 6,0,4,1,0,6,- 6,-3,0,0)

pict

An other formula with 18  termes is obtained for [a] = (0,0,0,0,1,0,0,0,-3,0,0)
  i

pict

To finish [ai] = (0,0,0,0,4,1,0,0,0,0,1)  gives

pict

For Catalan's constant G

Similarly, by imposing a denominator in y24- 212,  we obtain

pict

The most interesting case is obtain when all the ai  are zero except a9  which we let be equal to 1  . We then obtain

pict

The interest in this formula resides in the coefficients of P  which are all powers of 2  .
The choice of [ai] = (1,- 3,0,2,0,0,3,-3,1,0,0)  allows to write
G = BBP1  (212,24,P (y)) where P (y)  has 16  non-zero coefficients.
To finish [a ] = (0,0,0,0,- 1,1,0,0,-1,0,1)
  i  gives G = BBP   (230,60,P (y))
         1 where P  has 28  non-nul coefficients.

Warning 4 It seems that I was the first to have experimentaly discovery a real formula  BBP  for  G  (Mai 2000) without being able to find proof. The discussion between me and  David Broadhurst with no doubt, but that's not very important...

For ln2(2)

We obtain

(9a  + 2a + 8a )ln2(2) = - 1-BBP (212,24,P (y))
   4    7    8           29    1
(173)

pict

The simplest case is given by [ai] = (- 4,- 3,0,0,0,0,3,0,0,0,0)  which leads to

pict

The choice of [ai] = (- 42,-21,-20,- 40,0,0,21,0,0,20,0)  leads to   2          ( 60         )
ln (2) = BBP1 2  ,120,P (y) where P  has 52  non zero coefficients.

6.4.5 A few composite formulae

In the determination of BBP formulae, we have systematicly cancel down the coefficients of    (1)    ( 1)
L2  4 , L2 -4 and  (1)
I6  . If we then decide to keep those term, we can obtain among those possible formulae, the following result :
[a] = (1,1,0,0,0,0,2,0,0,0,0)  give

1      (                      )   p2   3  (1 )
16-BP P1 26,12,3y11 + 29- 165 +32 = 36-+ 4L2  4
(174)

which can be simplify to

pict

This formula is equivalent to the formula (167). If we apply this idea to Catalan's constant, we obtain with [a] = (0,0,0,0,0,-1,0,0,1,0,0)  the equality

pict

Who under serie form gives the two following equality :

pict

The same idea leads, with [a] = (0,0,0,0,-4,8,0,0,- 12,0,0)  to

pict

and with [a] = (4,8,0,0,- 4,4,12,0,0,0,0)  to

pict

This kind of formula has not been systematically researched.

6.5 Cases of polylogarithms of order 3

The order 3 introduce of course formulae giving p3  and other ln(2)3  but mostly the famous constant z(3)  proved irrational by Apéry in 1978 [14] by developping a continued fraction of a factorial formula which will be mentioned later on.

This constant stays never the less mysterious, and the BBP formula intuitively shows that this constan is very likely not to be very different from a point of vue of the orderning of it's decimals and hence it's complexity.

Kummer's equation for the triologarithm is

pict

and the inverse formula

   ( )
L   1  = L  (z) + p2ln(-z) + 1ln3(-z)
 3  z      3     6          6
(181)

A classical formula allows to state that

 (2)   7      p2ln(2)   ln2(2)-
I1 =  8z(3)-    12  +    6
(182)

and by definition

       1   (  1)
I2(2)= - -L3  - -
       2      2
(183)

6.5.1 Calculation of  (2)
I4

Landen's equation for the trilogarithm is (cf [4]

pict

Applied to    1+i
z = 2 ,  we get  (with        --3z(3)+-ip3
L3(i) =     32  )

         (   (1 +i)      (1 -i))     35       5p2ln(2)  ln3(2)
I(42)= -2  L3  ----  + L3  ----    = --- z(3) + --------- ------
                2           2        16          48       12
(185)

result implicitely contained in [3].
We deduce by the duplication formula that

                2         3         (    )
I(32) = 35z(3)- 5p-ln(2)+  ln-(2)-  1L3  -1
      16         48       12     8     4
(186)

6.5.2 Calculation of I(82)

As for the calculation of I8(1),  we use Kummer's equation with                V~ 
x = 1+2i, y = 1-i2-3  then with                 V~ 
x = 1-2i, y = 1-i2-3.  We add then those two equation we obtain. We simplify those equations with the help of the valuer of L3(i)  and the equality   ( 1+ i V~ 3)   z(3)  5ip3
L3  -------  = ----+ ----
       2        3     162  . This allows us to state that

                  2         3
I(28)= - 119z(3)+ 5p--ln-(2) - ln--(2)-
       48          36        6
(187)

6.5.3 Relation between I(52)  and I(92),  values of I(210)

We take the first two equation for the calculation of I(81)  that we substract this time. We then use the inverse formula with z = 1+ i  and z = 1 - i  so that we make appear the term (L2 (1+2i ) - L2(1-2i)) . Finaly one last application of the inverse formula with z =  V~ 3-i
     2  and with z =  V~ 3+i
     2  leads to

pict

Which prove that

            3p3  p ln2 (2)
3I9(2)- I(25)=  ---- -------
            32      8
(189)


If we use Kummer's equation with x = -1  and y = 1+ i,  then with x = - 1  and y = 1 -i,  we obtain two equality that we substract. We simplify the obtained result with the inversion formula applied to - 1- i,- 1+ i,1+ i  and 1- i  so to obtain

pict

This equality is equivalent with the help of integrals to

  1 (2)   (2)    (2)   (2)    (2)  13 3  -7    2
- 2J6  +I6  = 2J4  - I5  - 2J2  + 64p - 16 pln (2)
(191)

and gives the relation

   (2)   (2)    (2)   23p3  p-ln2-(2)
-6I5  + 5I6  + 5I11 =  160 -    8
(192)


If, instead of substracting the equality obtained previously, we add them, we get :

pict

which gives

  1  (2)   (2)   15 2       5  3              (2)   (2)     (2)
- 2J5  - I3 =  32 p ln(2)- 8 ln (2)- 7z(3) -2J3 - I4 - 2J1
(194)

and furnish

                                      (    )
 (2)    959      2p2-ln-(2)   ln3-(2)-  1      1
I10 = - 400z(3)+    15    -   6   - 8L3  - 4
(195)

6.5.4 Calculation fo  (2)
I7

Similarly to the calculation of  (1)
I7  ,  Kummer's equation for the polylogarithm of order 3  with        (1-i V~ 3- 1)
(x,y) =  --2--,2 , then the inverse formula with         V~ -
z = 1- i 3  leads immediatly to

pict

6.5.5 Application to the determination of BBP formulae

Let us now consider the linear form L2 (a1,a2,...,a11) = a1I(21)+ ...+ a11I(211)  . Then the previous result allows us to state that

pict

Formulae for p3   If we look for formulae for p3  we cancel down the coefficients of the constants
z (3),p2ln(2)  ... to obtain

pict

We can not choose a  = 0,
  9  which mean we have to use a denominator in y120- 260  for the BBP formulae giving     p3  . In all generality, we hence obtain a BBP formula for p3  with two parameters (a
 4  and a
 10  ), formula that the reader can establish. The simplest formula is hence obtain for [a ] = (0,0,0,0,5,-5,0,0,3,0,- 5)
  i  and with 90  termes. So to wrie it, we introduce the polynomials TTk  defined by  (0)    integral 1 TTk(y)
Ik  =  0y120-260dy  , for example         2y(y120-260)
TT1 (y) = ---y2- 2---  .
We then have

pict

This rquality can be written differently.
In fact L2 (a1,a2,...,a11) = L2 (a1,a2,a3 -a10,a4,a5,a6 -a11,a7,a8,a9,0,0)+ L2(0,0,a10,0,0,-a11,0,0,0,a10,a11)
But                                                integral 1
L2(a1,a2,a3- a10,a4,a5,a6- a11,a7,a8,a9,0,0) =  0 lny(y2)4-P1(21y2)   and
L2 (0,0,a10,0,0,- a11,0,0,0,a10,a11) =  integral 1ln(y4)0P2(2y0)
                                    0 y - 2   where P2  has 8  non-zero coefficients in general, and only 6  if a10 = 0  or a11 = 0  or a10 + a11 = 0  .
If we apply this here, we have
L2 (0,0,0,0,5,- 5,0,0,3,0,-5) = L2(0,0,0,0,0,5,0,0,0,0,-5)+ L2 (0,0,0,0,5,-10,0,0,3,0,0)  which gives

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This formula can also be written as

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Formula for    2
pln (2)  To obtain a simple BBP formula for    2
pln (2)  , we apply the same idea as that for   3
p  (the problem is the same, we need to use I11  which gives a denominator in  120   60
y   - 2  and gives an expression with two parameters)
We then obtain with [ai] = (0,0,0,0,67,-75,0,0,69,0,- 75)

pict

This formula can also be written

pict

Compare with those obtain for p3  .

Formulae for z(3)  For z(3),  we obtain, if we look for a denominator in  24   12
y  -2  ,

pict

Which give the general formula

(            )
  91a8   21a4        -1-     ( 12        )
   144  +  32   z(3) = 210BBP2 2  ,24,P (y)
(Huvent)

pict

The simplest BBP formula is obtain for
[a] = (-3,- 1,0,- 2,0,0,1,2,0,0,0)

pict

We can also use the following :
L2 (a1,a2,a3,a4,a5,a6,a7,0,0,0,0,) = L2 (0,a7,0,0,0,0,a7,0,0,0,0,)+ L2(a1,a2- a7,a3,a4,a5,a6,0,0,0,0,0,)
But                                 integral 1 ln2(u)          (   )
L2(0,a7,0,0,0,0,a7,0,0,0,0,) = a367 0 u+8-du = - a187L3 - 18 and
L2 (a1 + b1,a2 + b2,a3,a4,a5,a6,0,0,0,0,0,) =  integral 1P2(y)8ln2(4y)dy
                                           0  y -2
If we then impose a  = a
 8    10  in (203), we have L (7a , 9a ,0,a ,0,0,- 9a ,0,0,0,0) = -21a z(3)
 2 2 4 2 4    4      2 4           32  4
= L (7a , 18a ,0,a ,0,0,-9a ,0,0,0,0) + L (0,--9a ,0,0,0,0,-9a  ,0,0,0,0)
    22  4 2 4    4      2 4            2   2  4         2  4  which gives

pict

Finally the simplest formula for a denominator in  120   60
y   - 2  is obtained with
[ai] = (8,4,5,9,0,0,- 4,1,0,- 5,0)  and contains 67  termes.

Note 5 Similarly here, I thought I was the first to give a formula for z (3)  whose proof, a lot less simple but on the same model, was finished in june 200 with the help of Raymond Manzoni [5].

Formulae for p2ln(2)  We obtain the following general formula

(            )
 -3 a4 +-13a8  p2ln(2) =---1-- BP P2(212,24,P (y))
 80     360             5 × 210
(Huvent)

pict

With [ai] = (-8,- 3,0,- 5,0,0,3,5,0,0,0)  , we have the formula with 15  termes

pict

This relation is noteworthy, in fact the equality (204) is obtain for [ai] = (- 3,-1,0,-2,0,0,1,2,0,0,0)
This leads to fix

pict
pict
pict

Then

pict

Finaly the simplest formula with a denominator in  120   60
y   - 2  is obtain with
[ai] = (42,21,25,46,0,0,- 21,4,0,-25,0)  and has 67  termes.

Formulae for ln3 (2)  We have the general formula

(27a + 26a )ln3(2) =-3BP P  (212,24,P (y))
    4     8         28    2
(Huvent)

pict

and the simplest formula is ontained for
[ai] = (- 22,-9,0,-12,0,0,9,12,0,0,0)

pict

And the one with a denominator in  120   60
y  - 2  is given by
[ai] = (- 202,-99,-100,- 200,0,0,99,0,0,100,0)

6.6 Cases of polylogarithm of order 4

6.6.1 The relations

For the polylogarithm of order 4  and 5,  we only have one tool left, this is Kummer's equation. It is written, with e = 1- x,  n = 1 -y,

pict

and the inverse formula

           (  )    2            4
L4(z) = - L4 1  - p--ln2(-z)-  7p--- -1 ln4(-z)
             z    12          360   24
(214)


We then so apply the following method :
- Using Kummer's formula with a particualr couple (x,y)
- Eliminating the polylogarith with argument modulus greater than 1  with the inverse formula
- Eventual simplification with the help of the duplication formula or some value of L4  in 1,-1,i  and -i  .

  • With (x,y) = (- 1,1+ i),  we get
       (   (    )      (    ))     (   (    )      (    ))
        1--i        1-+i            1---i       1-+-i
22  L4    2   + L4    2     -3  L4    4   + L4    4
                                                    (1 )   81   (  1)   137          115       1697
                                               - 7L4  -  + --L4  - -  + ---ln2(2)p2 - ---ln4(2)- ---- p4 = 0
                                                      2    64      4    384          192       9216
    Which with the integrals and  (3)
Ik  and  (3)
Jk  becomes
    pict
  • With (x,y) = (- 1,i),  we get
    pict

    We obtain a second equality with conjugating (or with (x,y) = (-1,- i)  ). We add, or substract the two equality obtained and by replacing L4 (i)  by    4
-7p--+ ib(4)
11520  or         sum       k
b (4) =    -(-1)-4-
       k>0 (2k + 1)  , this gives

    pict

    and

    pict
  • With        (1-i V~ 3 1)
(x,y) =  --2--,2 ,  we obtain
               (   (      V~ -)     (      V~ -))
          9  L4  1--i--3  + L4  1+-i-3-
                    4              4
         57(   ( -1 + i V~ 3-)     ( - 1- i V~ 3))
       + --  L4  --------  + L4  --------
    (  ) 4     (  ) 2                2
      1    9    1     1- 2  2    217p4   -5  4
-5L4  2  - 8L4  4  +  12p ln (2)+  1620 - 24 ln (2) = 0
    (221)
    We then use the multiplication formula
               (   (          )     (          )        )
   (3)           -1 -i V~ 3         -1 + i V~ 3
L4  z  = 27 L4   ---2----z  + L4  ---2----z  + L4(z)
    (222)

    which with z = 1  allow us to confirm that

       (       V~ -)     (       V~ -)        4
L4   -1-+i--3  + L4  -1---i-3  = - 13p-
        2               2          1215
    (223)

    and that

               (  )      (  )    4
12I(3)- 5L4  1  -  9L4  1  - p--- 5-ln4(2)+  1-p2ln2(2) = 0
   7        2     8    4    54   24        12
    (224)

  • Finaly, with        (    V~     )
(x,y) = 1-i2-3, 1+2i , we obtain
         (  (   )     (   ))      (     )      (     )     (  )
0 = 9 L4  -1  + L4  1-   - 5L4  1+-i  - 2L4  1--i  - L4  1
          s3        s4            2            2         2
             - -349-p4 + -7-p2ln2(2)- -5-ln4(2)
               5(5296     768          384     )
                 9-- 3       1-   3     10
            +i   256 p ln(2)-  64 pln (2)-  3 b(4)
    (225)
    By considering the real part and the imaginary part of this expression, we have respectively
    pict

6.6.2 Application of the determination of the BBP formula

We now consider the linear form                      (3)          (3)
L3(a1,a2,...,a11) = a1I1 + ...+ a11I11  . With the previous formula giving

pict
pict

6.6.3 Formula for p4, p2ln2 (2)  and ln4(2)

For p4.  The simplest formula (and the only associated with a denominator in y24- 212  ) is obtained for [ai] = (37,9,0,26,0,0,-9,- 27,0,0,0)  and gives

pict

There exist a formula with two parameters with a denominator in  120   60
y  - 2  , the simplest is given by [ai] = (34,18,25,41,0,0,- 18,18,0,-25,0)

For  2  2
p ln  (2)  The simplest formula is obtained with
[ai] = (- 10381,-3303,0,- 6836,0,0,3303,6957,0,0,0)  and gives

pict

For  4
ln (2)  Wiht [ai] = (-18932,- 6849,0,- 11176,0,0,6849,11322,0,0,0),  we get

pict

The case of         3
b(4),pln (2)  and   3
p  ln (2)  It is not possible to determine formulae for those constant, we can only find two independant relation which are :

pict

This shows that we only need to find a BP  P  formula for one of those three constant and then we can deduce one for the two others.

6.7 Case of polylogarithm of order 5

6.7.1 The relations

Broadhurst in [3] shows relations between the integrals I
 k  and J
 k  with the help of Kummer's equation (relations (65)  to (67) . Then discover two equality through numerical mean (relations (68)  and (69)  ). He then prove the relation (68)  by using the hypergeometric series and Euler's sums. He deduce from it four equalities for z(5)  (relation (70)  )
We can establish those four equalities directly with the help of Kummer's formula which is written with n = 1- x, e = 1 - y, a = - x, b = - y
                        n       e

pict

We apply then the same method as that for polylogarithm of order 4.

  • With        (1 1-i)
(x,y) =  2, 2  ,  we obtain, taking into account
                         5
L5(i) = - 15-z(5)+ 5ip-,
         512      1236
    pict

    wich gives us

    pict

    Which can be simplified to, by using  (4)   (4)    3-  ( 1)
I3 + I4  = - 32L5  -4

    pict
  • With (x,y) = (- 1,1+ i)  we get
    pict

    which is equal to

    pict
  • With        (       V~ -)
(x,y) =  1+2i, 1-i2-3 , we get
    pict

    We simplify this equality with the help of    (    V~ -)
L5  1+i-3 =  25z (5) + 17-ip5
      2      54       5832  and the multiplication formula   (  )       (     V~  )       (     V~  )
L5 z3  = 81L5  --1+i2-3z  + 81L5 --1-2i3z  + 81L5(z)  , which with z = i  allows us to have    ( V~   )     (   V~   )
L5  -3-2i  + L5 ---32-i = 82654z(5))- 62205208ip5  .
    Hence

    pict
  • With        (        )
(x,y) =  -1, 1-i V~ 3 ,
             2  we obtain
    pict

    which gives us

    pict

Note 6 The relation proved by Broadhurst ([3]) correspond to the calculation of  (4)
I4 .  The calculation given here seems a lot more simple.

6.7.2 Application of the determination of the BBP formulae

We now consider the linear form                      (4)          (4)
L4(a1,a2,...,a11) = a1I1 + ...+ a11I11  . The previous results gives

pict

L3(a1,a2,...,a11) =
                                         (4)     (4)     (4)     (4)
                                    + a5I5  + a6I6 + a9I9  + a11I11 +
                         (  56575     6417    6417     3931389     403   )
                          -  2304 a8 - 256 a4 + 256 a3- 160000 a10- 864 a7 z(5)+
                         ( 2237       1      343      861       343  )
                           -----a8 + ---a7- ----a3 + -----a10 +----a4  p4ln(2)+
                           2(5920     324     3840     10000     3840)
                            - 5-a4 - -1-a7 + -5 a3- -7- a8- -19 a10 p2ln3(2)+
                              96     216     96     144     400
                            (-11                      -87-      -1-   )  5
                             240 a8 + 1/20a4 - 1/20a3 + 2000 a10 + 144 a7 ln (2)
                                 3           (   1)   3             (  1)
                               - - (a2 + a7)L5  - -  - --(a10 + a3)L5 - -
                                 2               2    32(               4                        )   (  )
                                                     +  15 a3- 15 a4- 411 a10- 17 a8- 3 a1- 7 a7  L5  1
                                                         2      2      50       2     2     3         2
(266)

6.7.3 Formulae for z(5),p4 ln(2),p2ln3(2)  and ln5(2)

We deduce from this some formulae for the constants z(5),p4ln(2),p2ln3(2)  and ln5(2)  . We remember that the polynomials are defined by Ik(0)=  integral 1-TT12k0(y)60dy
      0 y  -2
We hence obtain
with [ai] = (311416,168912,239375,382643,0,0,- 168912,96489,0,-239375,0)

pict

with [ai] = (22046536,11642616,15741875,26174255,0,0,-11642616,5323725,

0,-15741875,0)

pict

with [ai] = (32556,17892,25625,40413,0,0,-17892,10899,0,-25625,0)

pict

with [ai] = (12347068,6435126,8249375,14111819,0,0,- 6435126,2392497,

0,-8249375,0)

pict

6.7.4 Simplification of those formulae

Those formula can be simplified by making the term I3(4)+ I(14)0  appear. For example p4ln (2) = 311416I(14)+ 168912I2(4)+ 2× 239375I(34)+ 382643I(44)

        (4)       (4)        ( (4)   (4))
- 168912I7  + 96489I8  - 239375 I3 + I10 which gives

pict

Similarly

pict
pict
pict


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