The world of Pi - Central binomial series

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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 :

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

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

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

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 $1 V~ 2 V~ 3 u = 1,2,-2 , 2$

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

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]

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

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,

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

hence

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

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

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,

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