www.pi314.net


History
Mathematicians
All formulas
Num. approx.
Softwares
Misc. math.
Digits
Poetry
Papers/videos
Delirium !
 Pi-Day
Images
Music
Links
Bibliography



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



Pi-Day in
Pi Day Countdown
Google
Home Version history Guestbook Who I am Some pictures (fr) Acknowledgements Last modifications Contact

Cette page en français This page in English


Benoit Cloitre

e and in a mirror


Benoit Cloitre is a brilliant maths amateur when he wants.

1 Two mirrored sequences

Consider the sequences (un)n and (vn)n to be defined as

u1= 0, u2 = 1, un+2 = un+1 + un
-n-
v1= 0, v2 = 1, vn+2 = vn+1
 n + vn
Then
     n             2n
lnim--> oo  Un-= e et lnim--> oo  V-2= p
                    n

In general, for

u1= x, u2 = y, un+2 = un+1 + un
-n-

Then

lim n--> oo Un-
 n = x + y-2x
  e

By considering those two sequences, e and p are mirrored! However, the speed of convergence are very different... Un converges very quickly (faster and faster) while V n converges in a logarthmic way...

2 Prove

Benoit Cloitre first offered a direct prove, but one can prefer Gery Huvent's prove, from Lille, which allow us to find a general method to prove those sequences.

2.1 For (vn)n

Calculating the first few terms 0,1,1,32,32,158-,158- suggest that

 A p > 1, v2p = v2p+1

This property can be prove by induction on p, since it's true for p = 1. Suppose that v2p = v2p+1 then

v2p+2 = v2p +v2p = 2p+-1vp
        2p         2p

and

v2p+3 = v2p+2-
2p+ 1 + v2p+1 = --1---
2p+ 1 * 2p+-1-
 2pvp + v2p

= 2p+-1-
 2pvp = v2p+2
hence the result. We also notice that the sequence (v2p)p satisfy
v    =  2p+-1v
 2p+2    2p   p

which by induction give us

                     p-1
v2p = v2p+1 = (2p- 1) C2p-2
                    22p- 2

Taking into account the classical equivalence (obtain with Wallis)

Cn2n-  -1--
22n ~   V~ pn

we have

      V~ --
v  ~   2n-
 n     p

Note We can find a limited development with Stirling's formula, to get

      V~  2n  1 V~ -2-   1 V~ -2--  21 V~ --2-   (  1 )
vn =   ---- -  ---+ --  ---3- --   --5 + o -5-
       p    4  pn   32  pn    48   pn      n2
2.2 For (u )
  nn

Consider the function that generates the sequence (u  )
  nn

      + sum  oo 
f (x) =  unxn
      n=1

the given recurrent sequence can be written as

+ sum  oo          + sum  oo         + sum o o 
   un+2xn =    un+1xn +    unxn
k=1         k=1        k=1 n

or

         2                    integral  x
f (x)--u2x---u1-- f-(x)--u1x--   f-(t)dt = 0
      x2             x        0   t

by differentiating, we get

-x (x - 1) df(x)- (x2- x + 2)f(x)+ u x
----------dx------3--------------1--
                 x = 0

hence f is the solutiong to the differential equation

x (x - 1)y'+ (x2- x+ 2)y = u1x

which has the general solution

  xu1     Cx2e- x
------2 + ------2
(x - 1)   (x- 1)

taking into acount than u1 = 0 and u2 = 1 = f(2)(0)
  2 we find

      -x2e-x-
f (x) = (x - 1)2

Now we need to determine the development of the power serie of f so to find each term of the sequence. For this note that

        integral 
e-x--    xf-(t)
1- x =  0   t dt

( f(x)
--x = v2x + v3x2 + ....is defined in 0 )
Then, we remember that if

      + oo 
h(x) =  sum  a xn
      n=0 n

then

h(x)-  + sum  oo    n         n sum 
1- x =    Snx  où Sn =    ak
       n=0            k=0

using Cauchy's product of two series. Hence

          (         )
e-x    + sum  oo   sum n (- 1)k
1--x-=         -k!--  xn
       n=0 k=0

and

       + sum  oo  (  sum n    k)
f-(x)-=      n   (--1)-  xn-1
  x    n=0   k=0  k!

       + oo  (  n     k)
f (x) =  sum  n  sum  (-1)-  xn
       n=0   k=0  k!

and

              n
              sum  (-1)k
 A n > 1, un = n   k!
             k=0

more particularly

(   )
 un-   converge (très rapidement) vers 1
  n  n                            e

Note If we did the same for the sequence (vn)n, the generating function is a solution to

               '
x (x - 1)(x+ 1)y + (x+ 2)y = x(x+ 1)v1

       + sum  oo         x2   V~  1-+-x
g (x) =    vnxn = -2---- -----
       n=1       x - 1  1 - x

which satisfy

 integral  xg (x)      V~  1-+-x      1 + x
   --2- = 1-   ----- = 1-  V~ ----2-
 0  x          1 - x        1- x

with

   1     + sum  oo  Ck
 V~ -----=    -24kk xk
  1- x   k=0


back to home page