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



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11 Centrals binomial series

An interesting particular case and yet not to har on factorial series involve the introduction to the central binomials coefficients in the series. After a few notes on those series in the first section we will plunge into serious stuff with several formulae and their proofs!

11.1 Combinations inversion

A funny thing is that we can obtain some series giving the same result but in one case with the central binomial coefficient on the numerator, while in the other on the denominator! This also confirrm why we obtain the same kind of results with a serie with a combination and one without:

Let us take the example of the arctan  function: first of all, we have the classical formula

            oo  sum  (-1)kx2k+1   x   (   2  1)
arctan(x) =    --2k+-1---= 2-Y  -x ,1,2
           k=0
(372)

We know from Euler (see Euler's page) that the arctan function can also be expressed as

                ( sum  oo      k    (   2  )k)
arctan(t) = --t--     ----4-----  -t---
           1+ t2 k=0 (2k + 1)Ck2k  1+ t2
(373)

Finaly, the coupled relation  arcsin(x) = 2 sum o o  -Ck2k--(x)2k+1
             k=0 (2k+1) 2  with                (      )
arctan(x) = arcsin  V~ -x-2
                  1+x   give

             2x    oo  sum      Ck2k     (  x2  )k
arctan(x) =  V~ 1-+-x2  22k+1(2k-+-1) 1-+-x2
                  k=0
(374)

A relation between the different series and the central binomial coefficients is hence the arctan :

pict

Suprising, no ? ! !

By deriving the serie from the center along x  , we can for example find the expression under the combination form of 11+x2   .

And so starting from this serie, we can find the combination expression of serie of type Machin under the form of combination of binomial formulae! We will see in the following section the 11.3factorials.

We can anyway signal for the arctan function

pict

The proof is obvious by decomposing the serie in even and odds...

Furthermore, we also have, still playing of the developpement of power series

(    oo                )2      oo 
  2 sum  --Ck2k-- (x)2k+1   = 1  sum  --1-- (2x)2k
   k=0(2k+ 1)  2          2 k=0k2Ck2k
(378)

Finally, we have according to the expression 13 and 11

( sum  oo    (x)k)2    sum  oo 
    Ck2k  --    =    xk
 k=0     4       k=0
(379)

which is ammusing as well !

11.2 Useful development

We remember the following development of power series which will be useful for the calculation of certain series and integrals. :

   1     + sum  oo  Ck  k
 V~ 1---x-=   -42kkx
         k=0
(380)

which, by composing by -x2  or x2  and by integrating, give respectively

pict

furthermore, Euler's developpement given under it's arcsin  form arcsin

pict

that we can hence proof by the method of differential equations.

11.3 First direct formulae

The four last developpement of the previous section give then classical series, with      1   V~ 2  V~ 3
u = 1,2,-2 , 2

pict

By using

     (    )        (    )
        V~ 5            V~ 10    p
arcsin  -5-  + arcsin   10-- =  4-
(389)

we can obtain, for example

  V~ -     sum  oo     n     (V ~      )
p 10 = 4    -n-C2n-----  2 + 1n
        n=0 20  (2n + 1)       2
(390)

With the developpement of     2
arcsinu(u)  , we obtain

       oo 
p2 = 2  sum -4n---
     n=1 n2Cn2n
(391)

 2     oo  sum  -2n---
p = 8    n2Cn
     n=1    2n
(392)

     9  oo  sum    3n
p2 = -    -2--n-
     2 n=1n C2n
(393)

         oo 
p2 = 18 sum  --1---
       n=1n2Cn2n
(Euler 1748)


The values sin(p8) , sin(p12),  sin(p5) gives the formulae

           (    V~ -)
  2     sum  oo  2-  2 n
p  = 32    -n2Cn----
       n=1      2n
(394)

        sum  oo  (2-  V~ 3)n
p2 = 72    ---2-n---
       n=1  n C 2n
(Grandall 1994)

         oo  (    V~ -)n
p2 = 25 sum  --5---5---
       n=12n+1n2Cn2n
(395)


We can also write that arcsin2( V~ u)-=  sum o o -222n-n1un
               n=1n C2n  which when differentiating gives

arcsin( V~ u)   oo  sum  22n-1  n- 1
 V~ uV ~ 1---u =   nCn u
            n=1   2n
(396)

and obtain pther series such as

pict

By differentiating again u d-arcsin2( V~ u)
 du  , we have

         sum  oo  2n+1
p = -2 +    Cn--
        n=1   2n
(400)

etc...

On the same model, Gery Huvent offered what seems an open conjecture, having not read stuff on this subject...

Conjecture 16

        sum  oo  nk2n            bk
 A k > 0,    Cn  = akp + bk o`u ak --> p
       n=1  2n
(401)

Example 17

 sum  oo  n42n
    Cn2n  = 113p + 355
n=1
(402)

Conjecture 18

        sum  oo  nk              b       V~ -
 A k > 0,   -n--= akp+ bk o`u -k---> 2p  3
       n=1 C2n              ak
(403)

Example 19

13130  V~ -        sum  oo  n6    196      V~ -
-----p  3+ 196 =    -n--et13130-- 2p  3 = - .000541 81
 729            n=1 C2n    729
(404)

Conjecture 20

                                    V~ -
        sum  oo  3nnk            -vk-   p--3
 A k > 0,   Cn2n = ukp + vk o`uuk -->   2
       n=1
(405)

Example 21

                                      V~ -
32524-  V~            sum  oo  3nn4  29496-  p--3            -7
 3   p 3 + 29496 =     Cn  et 32524 -   2  = -4.87× 10
                  n=1  2n      3
(406)

11.4 Formulae of greater order

The study of series of the kind  sum +oo  -xn-
  n=1 njCn2n  or  sum +o o  Cn2nxn
  n=1  nj  , for j > 1  complicated things a lit, because those sums comes from succevise integrations of functions where j = 1  . In fact, we will notice that

            integral                   integral   integral  u sum +o o  yn-1
+ sum  oo --xn--    x + sum  oo -un--1--      x-0---n=1nj-2Cn2n-dy
   njCn2n = 0     nj-1Cn2ndu =  0         u        du
n=1            n=1
(407)

and so on to go back for example up to j = 1  .

but of course, those functions don't let themself be easily integrated! And we often need to have a lot of patient and a few trick up our sleeves to see the end of it.

However, the results are there. We know Euler's formula

+ oo         V ~ 
 sum  --1--= p--3
n=1nCn2n     9
(408)

who I give a proof to the paragraph dedicated to 10.2calculus and the method of convergence acceleration by the finite difference, invented by Euler himself!

The greater order, it's a formula such as

+ sum  oo    1     p2
   n2Cn--=  18-
n=1    2n
(409)

but also

 sum  oo  (- 1)n    2
   --3-n-= - -z (3)
n=1n C 2n    5
(410)

and the formula by Comtet dating of 1974 whose proof was tedious...

+ oo             4
 sum   --1--=  17p-
n=1 n4Cn2n   3240
(411)

Obviously, this intrigued mathematicians to know wether a similar formula existed for order 5! Unfortunatly, we now know that this serie does not give the wanted result since if we recall       V~ -
r := ( 5 -1)/2  , we have according to [8]

pict

But this doesn't stop the higher order from being interesting.

11.4.1 A first formula with proof

I noticed in august 2001 that

216 sum  oo     (2i)!         3
-7-   16i(i!)2(2i+-1)3 = p
   i=0
(413)

which is transformed into hypergeometric form into

 3  216  oo  sum  -----(2i)!-----  216    (  12, 12, 12 1)
p =  7     16i(i!)2(2i+ 1)3 = 7  4F3    3, 3 ,4
        i=0                            2 2

It must be one of the simplest factorial formula and the presence of the 2i+ 1  make it seems like a BBP formula. The proof is  complicated and linked to the folowing integral calculation

 integral  12 ln(x)2      integral  p6        2
     V~ ----2dt =    ln (sin(t)) dt
 0   1 - x      0
(414)

because the development into power serie of  integral  12 l V~ n(x)2
 0  1-x2dt  give

 sum                        sum                      sum 
 o o i=02.16i((i2i!))2(!2i+1) ln2(2)+ o o i=0 16i(i(!)22i)(2!i+1)2 ln(2)+ o o i=0 16i(i!(2)2i)(!2i+1)3   which we know how to calculate the first two part.

We hence let more generaly the following integrals :

        integral  x (     ( ))n
Ln(x) =   ln  2sin  t     dt
        0          2
(415)

Note that we can clearly brings this integral back to 9.4.1in the sense where             |   (  )|
Cl1(x) = - ln|2sin x2 | and hence

              integral 
 n         n   x       n
L (x) = (- 1) 0 (Cl1(x)) dt

for x > 0  . There no coincidence !

The result that interest us is

     108  (p )
p3 = ---L2  --
      7     3

which is equivalent to the serie through an immediate change of variable u = t2  in 414.

Raymon Manzoni, then Gery Huvent , contributed to give a proof for this formula, which I will present to you now, and also including a proof of the formula by Comtet done by Gery Huvent .

Proof. Consider, for integers p  and q  q > 1,  the function

pict

this function is holomorph on
For p = 0, C\]1,+ oo [
For p > 1  C  without ]1,+o o [  and de ]-  oo , 0]
Consider, for x  (-  [-p,p]  the path g(t) = eit, t  (-  [0,x]  and the integral  integral 
 gfp,q (z)dz  (with x /= ±p  if p /= 0  ).
Since   (      )    (     )
ln 1 -eit = ln 2sin t2 - i2 (p - t),  we have

 integral               integral  x (   (     )          )q
   f  (z)dz = ip  tp  ln  2sin t  - i (p - t)  dt
 g  p,q          0            2    2
(417)

If Fp,q  disign a primitive of fp,q  on the domain containing the path g,  we have

   integral  x (   (     )           )q        (  )
ip    tp  ln  2sin t - -i(p- t)  dt = Fp,q eix -Fp,q(1)
   0            2    2
(418)

The problem hence goes back to the calculation of a primitive !
The null primitive of fp,1  in z = 1  is

                         p
Fp,1(z) = (- 1)p p!z(p+ 2)+  sum  (- 1)k+1--p!---lnp-k(z)Lk+2 (z)
                        k=0        (p- k)!
(419)

Hence

 p+1  integral  x p( (     t)   i      )         p            sum p    k+1---p!--    ( ix)   p-k
i    0 t  ln  2sin 2  - 2 (p - t) dt = (-1) p!z(p+ 2)+   (-1)   (p- k)!Lk+2 e   (ix)
                                                    k=0

                                            (                                            )
 integral  x   (     t)     i(pxp+1    xp+2)             p            sum p    k+1 (ix)p-kp!    (   )
   tpln  2 sin2  dt = 2  p-+-1-- p+-2- + i-p-1 (- 1) p!z(p+ 2)+    (- 1)   -(p---k)!-Lk+2 eix
 0                                                           k=0
(420)

We therefore obtain a second expression of  integral xtpln(sint)dt,
 0  by comparing with it what we already obtained, for p = 0,1,2...  , we find again simply the value of Ln(eih)+  (- 1)nLn (e-ih) . This is nothing new, because we know the developpement in Fourier siries of the nth polynomial of bernoullli. This developpement gives the equality

                                   (   )
   (ih)      n   ( -ih)    (2ip)n    -h-
Ln e   + (-1) Ln  e    = -  n!  Bn  2p
(421)

The primitive of f0,q  which is nul in z = 1  is

 sum p    k+1   p!
   (-1)   (p--k)! lnp- k(1- z)Lk+1(1 -z)
k=0
(422)

hence

 integral  x(  (      )          )p        sum p
     ln  2sin t - -i(p- t)  dt = - i  (-1)k+1---p!--lnp-k(1- eix)Lk+1(1- eix)
 0          2    2                k=0       (p- k)!
(423)


In particular,

pict

We can refind that   (   (ix))  1 2   1     1 2
 R  L2 e    = 4x  - 2px+  6p  and that

 integral  x (     t)        (   (  ))
    ln  2sin -  dt = -  I  L2 eix
 0         2
(425)

And

   integral  x ( (     t)   i      )       (   )      (  )
-    t ln 2 sin -  - -(p -t)  dt = L3 eix - ixL2 eix  -z (3)
   0           2    2
(426)

from which

 integral  x  (     t)         (  )      (   )         (1     1    )
   tln  2sin -  dt = -L3 eix + ixL2 eix + z(3)- i  -x3- -px2
 0          2                                    6    4
(427)


Then we have

 integral  x(  (     )           )                             2
     ln  2sin t + -i(p- t) dt = ln(1- t)lnt+ L2 (1 - t)- p--
 0          2    2                                    6
(428)

which allows us to find Euler's formula for the dilogarithm

pict

hence

pict

Note 22 We can simplify this equality with Landen's formulae and find the classical result

 integral  p  (     t)     p3
    ln2  2sin 2  dt = 12
  0
(431)

result that we obtain with the developpement into Fourier's series of   ||    (x)||
ln 2sin 2 and Parseval's formula.

in particular

 integral  p3 2(     t)     -7- 3
 0 ln   2sin 2  dt = 108p
(Gourevitch)

because

   ( ip-)   1       5
L3  e3  =  -z(3)+ ---ip3
           3      162
(432)

This  equality then gives with       (t)
u = sin 2 and a DSE

+ sum  oo       n
   ----C2n---- = -7-p3
n=116n(2n + 1)3   216
(Gourevitch)


We can also establish that

 integral  p   (    (  ))        (   (     ))
  2 ln2  2sin  t   dt = 2 I  L  1 - i   + -23 p3 + p-ln22-- G ln(2)
 0            2            3  2   2     192       16
(433)


We can also calculate the primitive of ln(z)ln2z(1--z)-= f1,2(z)  , we find

pict

This equality allows several applications :

The equality

 integral  12ln(z)ln2(1 - z)   1        p4
   --------------dz =- ln4 2- ---
 0       z           4       360
(435)

Then with

 integral  12 2               integral  1 2
   ln-(1--t)ln(t)dt-    ln-(1--t)ln(t)dt = 1 ln4(2)
 0       t           12      t          2
(436)

obtained by integration by parts and changing variables(u = 1 - t),  because  integral 1 2                   integral  1(          )
 12 ln(1-tt)ln(t)= - 12 ln42- 12 - ln(11--tt)ln2t dt  .
We have

 integral  1      2             4
   ln(z)ln--(1---z)dz = - p-
 0       z             180
(437)

which gives

               4
lim F1,2 (t) = - p-
t-->1           180
(438)


Finnaly, starting from

 integral              integral  x (  (     )          )2
   f1,2(z)dz = i  t  ln  2sin t - i (p- t)  dt
 g             0           2    2
(439)

that we developpe like in the previous example, we find

With x = p,
    3  and taking into account the value of   (    )
L3 e- ip3-

             2   (     V~ -)      (      V~  )
F1,2 (eip3)=  2p-L2  1-+-i-3  - 2L4  1+-i-3-  + -77-p4 + 8ipz(3)
            9        2               2       4860       9
(440)

which gives

                                       (      V~ -)      (      V~  )
  integral  p3 (  (    -t)   i      )2     2p2    1-+i--3         1+-i-3-    p4-  8ipz(3)
i 0  t ln  2 sin2  -  2 (p- t) dt =  9 L2     2     - 2L4     2    -  486 +   9
(441)

By developping the square under the integral and by taking into account the previous calculated values of  integral  p3 (     t)
0  tln 2 sin 2 dt,  it turns out (Oh miracle!) that we get

 integral  p3   (     t)     17p4
   tln2  2sin 2  dt = 6480
 0
(Comtet)


This last equality, gives with       (t)
u = sin 2 and the DSE of arcsin(u)   d-     2
 V~ 1-u2-= du arcsin (u),  the equality

+ oo             4
 sum   --1--=  17p-
n=1 n4Cn2n   3240
(442)

With x = p2,  the result is less simple, however we have

pict

By combining with 433, we have

 integral  p2(    )   (      )                                                   (  )
    p-- t ln2 2sin-t dt = 25-p4 + 5-ln2(2)p2 - 5-ln4(2) - 35 ln(2)z(3)-  5L4  1
 0  2             2       576     96          96        32           4    2
(444)

 _

Now the trick is to use this approach to developpe a particular method and as often we need to go through equivalent integrals, who we just saw a first example. All of this on an idea by Gery Huvent  :

11.4.2 Calculation of a primitive of       (  )
tpcoth at2

For a  (-  C  , we introduce the function

               integral  x     (at )
 A p > 1, Fp(x) = tpcoth  --  dt
               0         2
(445)

Proposition 23

               xp+1    2p!  sum p (ax)p-k    (    )   2p!
 A p > 1, Fp(x) = p-+-1- ap+1 (p---k)!Lk+1  e- ax + ap+1z (p + 1)
                          k=0
(446)

The previous equality is valid on R  if a / (-  iR
If a = ia  or a  (-  R  the previous equality is valid on ] p p [
- a,a.

Proof. We differentiate the right hand side,

pict

Then in x = 0  the right hand side is equal to 0  .  _

Application :Intégrales of kind Dirichlet Consider the equality (446) with a = 2i,  we obtain the expression of the integral  integral xtpcotan(t)dt
 0  with the help of polygarithms. An integration by part (where we integrate the cotangent) gives

pict

We then obtain the following equalities

pict

With a = 2,  we get

pict

which gives

pict

For the folks mathematics We then easily show that

 integral  12(  2        )
    42t - 18t+ 1 ln (sinpt) = 0
 0
(459)

The minimum of the trinomial          2
P (t) = 42t - 18t +1  is obtain in     3-
t = 14  , funny !

11.4.3 Uses of F1

Uses of the primitive for tcoth (t)  We have for p = 1  and a = 2

                integral  x                                            2
 A x > 0, F1 (x) =  tcoth (t)dt = 1x2 + x ln(1 - e- 2x)+ 1L2(e-2x)+  p--
                0             2                   2           12
(460)

Hence for        ( )
x = 12 ln 1u or u  (-  ]0,1[,  we obtain

 integral   1
  - 2ln(u)            1  2     1                1       p2-
 0       tcoth (t)dt = 8 ln (u)- 2 ln(u)ln(1- u)- 2L2(u)+ 12
(461)

Starting from

            integral  x            sum  oo     n  n
 A x  (-  [0,1], arcsinhy-dy =  -(--1)-C2n-x2n+1
           0    y        n=022n(2n+ 1)2
(462)

The change of variable t = arcsinh(y)  then gives

pict

As an application, we obtain for u = 1-
    f2

pict

As another application, u = 1
    2  gives

pict

Uses of the primitive of tcotan (t)  We have for p = 1  and a = 2i

     ]     [    integral  x              2     (        )      (    )    2
 A x  (-  - p, p ,  tcotan(t)dt = ix-+ x ln 1 - e- 2ix + i L2 e-2ix -  ip--
        2 2    0               2                  2            12
(466)

Starting from

            integral  x            oo  sum       n
 A x  (-  [0,1],  arcsin-ydy =    ---C-2n----x2n+1
            0   y        n=0 22n (2n + 1)2
(467)

the change of variabl t = arcsin (y)  gives

           oo  sum  ---Cn2n---- 2n+1   iarcsin2x-          (    - 2iarcsinx)   i  ( -2iarcsinx)  ip2
 A x  (-  [0,1],  22n (2n + 1)2x    =     2    + arcsinx ln 1 - e         + 2L2 e         -  12
          n=0
(468)

As an application, we have with x = 1

pict

With      V~ 
x = 22-

pict

11.4.4 Uses of F2

Uses for the primitive of t2 coth (t)  We have for p = 1  and a = 2

                integral  x 2          x3   2  (    -2x)      ( -2x)  1   (- 2x)   z(3)
 A x > 0, F1 (x) = 0 t coth (t)dt = 3 + x ln 1 -e    -xL2  e    - 2L3  e    +  2
(471)

Starting from

 integral  x     2        sum  oo     n+1 2n
   arcsinh-(y)=  1    (--1)---2--x2n
 0     y        4n=1   n3Cn2n
(472)

by change of variables, we obtain  A x  (-  [0,1],

   oo 
1 sum   (--1)n+122n-2n   arcsinhx3-   2  (    -2arcsinhx)      ( -2arcsinhx)  1   ( -2arcsinhx)  z-(3)
4      n3Cn2n   x   =     3    + x ln 1- e          -xL2  e         - 2 L3 e         +   2
 n=1
(473)

With x = 1
    2  , we get

pict

It's with this last equality that Apery proved the irrationality of z(3)  .
With x = -1 V~ -
    2 2  we have

pict

Uses of the primitive of t2 cotan (t)  By the same kind of method we obtain

pict

which with x = 1  gives

pict

then with      1
x =  V~ 2

pict

By combining those last two equalities with the one found by Apéry (474), we get

pict

11.4.5 Other formulae

Starting from the result  integral                 V~ -  ( )
 01 V~ -ln(x)2-dx = --28pG 14 2
     x(1-x )  stated in the Gradshteyn [9] (4.241), we get the formula

 oo  sum        n         V~ -- (  )2
   ----C-2n---- = -2p-G  1   =  V~ 1-pK(k1)
n=04n-1(2n+ 1)2    8     4       2
(484)

Nice formula, putting in evidence the arrival of a constant linked to elliptical integral of first sort, not bad...


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