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



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7 s  fixed integers,v = pq  , x = 31n   : BBP in base 3

So many results already ! But it's not over yet... in fact, even if we can not BBP formula for all bases, an important class of BBP formula was obtained in base 3, which correspond to taking x = 13n  . Here again, Gery Huvent explained the best the relations in [12].

7.1 Formulae for p V~ 3

We take the same kind of integrals than that of base 2, but this time it's the 3 that will be omnipresent in the polynomials.

An elementary calculation shows that

pict

By cancelling the coefficients of ln(2)  and ln(3),  we obtain a formula for   V~ 
p 3   :

pict

The following particular case is the most elegant (               )
 b = 0,d = -e = 13  :

  V~ -    2-  oo  sum  -1-(--34--   --34--   --33--  ---3--   --3---   ---1---)
p  3 = - 33   36i 12i+ 2 + 12i+ 3 + 12i+ 4- 12i+ 8 - 12i+ 9-  12i+ 10
           i=0
(Huvent)

7.2 Formulae of order : integrales with ln(y)

There exist reals (p,q,r)  such that

pict

( the two members are linear form in (a,b,c)  decompose on two different base of the dual of R3  )
Plouffe showed that 

pict

which under integral form, give

    integral              integral              integral 
     1 ln(y)y-      1 ln(y)y-      1ln(y)(2y--3)-    p2-
-2  0 2y2- 3dy +  0 2 y2 + 3dy + 0 y2 - 3y+ 3 dy = 18
(271)

and correspond to the equality                 p2-
I1(-2,1,1) = r = 18  . We can deduce the value of r  .

But we have

pict

where         sum  oo  k
L2(z) =   z2
       k=1k
We deduce from it

            (  )
3p- 2q = 1 L2 1
        4     9
(273)


But Ramanujan proved that

  (  )       (  )    2    2
L2  1  - 1L2  1   = p--  ln-(3)-
    3    6    9     18     6
(274)

By using

pict

We deduce from it

  (8    4  )     (1 )   1  (1 )
I1  -,- -,0  = L2  -  - -L2  -
    6   6          3    6    9
(276)

and

       2     2
-2q = p--- ln--(3)-
 3    18     6
(277)

From  (273) and (277) we get the values of p  and q  and the equality

                                                     ( )
                    p2-             ln2(3)-         L2-19--
I1(a,b,c) = (a- b+ 5c)36 + (- a+ b- 3c) 12  + (a+ 2b) 12
(278)

Note 7   (  4 2  )       1   1  (1 )    1     1  2
I1 - 3,3,0 = L2(- 3)- 3L2  9 = - 18p2 + 6 ln 3  is another relation proved by Ramanujan .

Note 8

pict
o`u P (y) = 4y11 - 9y10 + 81y9 + 117y7 + 81y6 + 972y5 + 243y4 + 1053y3 + 6561y- 2187

We deduce

pict

Note 9                  (         )
I (a,b,0) = (a- b) p2-  ln2(3) + (a+ 2b)L2(19)= a integral  12 ln(y)ydy + b integral 12 ln(y)ydy
 1                36    12            12      0  y2-3       0  y2+3  , with a = -2  and b = 1  and changing variables      2
u = y   , we get

 integral  1 ln-(u)(u-+-9)   1  2  1  2
 0    u2 - 9  du = 6 p - 2 ln (3)
(281)

and the equality

 sum  oo  1    oo  sum      1       2 2     2
    9ii2+    -i(---1)2 = 3p - 2 ln (3)
i=1      i=0 9  i+ 2
(282)

Note 10

  oo        oo 
 sum   1-+  sum  --(-1--)=2  ln 3
 i=1 9ii  i=09i i+ 12
(Huvent)

7.3 Formulae of order 3 : Integrales with ln2
 (y)

Here again, the order 3 introduce some formulae giving   3
p  and other     3
ln(2)  but also z(3)  .

There exist reals (p2,q2,r2)  such that

pict

The calculation of r2  is elementary because

pict

Proposition 11 We have

13          (1 )   p2ln(3)     (  1 )  ln3(3)
-- z(3)- 6L3  -  -  -------+ 3L3  --  + ----- = 0
 2           3       2            3      2
(285)


This relation is equivalent to

pict

Proof. Kummer's equation for the polylogarithm of order 3  is written

pict

With x = 2
    3   and y = 1,
    3  we obtain

pict

Landen's equation is

                    (     )
                      -z---         p2-          1       2        1  3
L3(z)+ L3(1 - z) + L3  z- 1  = z(3)+  6 ln (1- z)- 2 ln(z)ln (1- z)+ 6 ln  (1 - z)
(290)

Applied to z = 3
   4   and to z = 2
    3   , it allows to express L3(3)
   4 and L3 (2)
    3 in functions of L3 (1), L3(-2), L3(- 1)
    3                3 and     L3(1)
       4 . The relation

  (1-)                 p2- ln3(-z)
L3  z  = L3(z)+ ln(-z) 6 +    6
(291)

applied to 3, 2,- 2
4  3  and - 3  allows to simplify the left member of (289).
Then all there is left to do is use the two following equalities

pict

to conclude.  _

With  integral 1 ln2(y)y      1  (1)
 02 y2-3 dy = - 2L3 3 and  integral  1 ln2(y)y     1   ( 1)
 02  y2+3 dy = -2L3  -3 ,  (285) can be written

pict

If we replace p2  by it's value compared to q2,  the right hand side of I2(a,b,c)  becomes a polynomial of degree   1  in         q2  whose coefficient of q2  is equal to  3
ln-(63)c  . Hence I2(a,b,0)  does not depend on q2  . After calculation, we get

   integral  1 2           integral  1  2                   (  )           (                 2    )
a    2ln2(y)ydy+ b    2ln2-(y)ydy = - (a+-2b)L3  1  + (--a+-b)  13z(3)+ ln3(3) - p-ln(3)-
   0  y - 3        0  y + 3          12       9       36                      36
(294)

In particular, we have with a = 1,b = - 1

    integral  1      2              3                (1)    2
12    ----yln-(y)----dy = - ln-(3) - 13z(3)+ L3-9-+  p-ln(3)
    0 (y2- 3)(y2 + 3)       18    18        24       18
(295)

which gives the equality

 oo         oo              (  2       )
 sum  -1-+  sum  ----1---- = 4-ln-(3)--p2-ln(3)+  52z(3)
i=1 9ii3  i=09i(i+ 1)3           3           3
                 2
(Huvent)

Note 12 We can also take a = -2,  b = 1  and to a change of variables u = y2   which leads to

 integral  1 2
   ln-(u)2(u+-9)du = -1 ln3 3+ 1p2 ln 3- 13z (3)
 0     u - 9         3       3         3
(296)

and to the same equality as for the series.

We just need to calculate the exact value of p2  and q2  .

Proposition 13 We have the equality

 integral 
  1ln2(y)(2y--3)-      26      5p2-ln-(3)   ln3-(3)-
 0  y2- 3 y+ 3 dy = - 9 z(3)+    36    -  12
(297)

Proof. Landen's equation applied to z =-1
   z1   where          V~ -
z1 = 3 + i-3
    2    2   is a root of y2 - 3y+ 3  (the other root is     z2   such that -1 = 1- -1
z2      z1   ) give

  ( 1 )     (     1)     (  2p)         5p2ln(3)  ln3(3)  2ip3
L3  z-  + L3  1- z-  + L3 ei 3  = z(3)- ---72---+ --24--+ -81-
     1            1
(298)


But    (   )
L3  ei2p3  = - 4z(3)+ 2ip3
             9       81   and                    (   (  )     (  ))
 integral  1ln2(y2)(2y-3)dy = - 2 L3-1 + L3  -1
 0  y -3y+3             z1       z2 . We deduce that

pict
 _

By using the equations (293) and (299), we deduce from it that

pict

And hence

                                                         (  )
I (a,b,c) = --a+-b--3-cln3(3)+ a---b+-5cp2ln(3) + --a--2bL   1  + 13-(--a-+-b--8c)z(3)
 2             36                36              24     3 9           36
(302)

Note 14 With a = - 2,b = c = 1  we get

          (                      )
   integral  1 ln2(y)-y4--8y3 +-18-y2--24y+-9   p2ln(3)  65
3  0   (y2- 3)(y2 + 3)(y2 - 3y+ 3)  dy =   18   - 36 z(3)
(303)

and the following BBP formula

pict


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