Curious binomial-identity /A002720

[Gottfried Helms]

Am 25.11.2006 22:53 schrieb Joseph Biberstine:
> The example you gave has an off-by-one row index as follows:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n+1)
> 
> Note that the row index is incremented.  Mathematica simplifies this
> fraction to LaguerreL[-(n+1), 1]/LaguerreL[n+1, -1].  This is not equal
> to Exp[1], but interestingly seems to approach Exp[1] monotonically from
> below as n approaches infinity.
> 
> I'm sure you meant:
> 
> Exp[1] = (exp. weighted sum of column n)/(exp. weighted sum of row n)
> 
> Mathematica simplifies this to LaguerreL[-(n+1), 1]/LaguerreL[n, -1].
> Although I can't get it to explicitly resolve that this is exactly
> Exp[1], the values do seem to bear this truth out.


I think I got it. As usual I like to write that problem in
matrix-mode.
So with

     V(x)~ * Pf = B

defined as the (infinite) matrix product as set of formal powerseries
in x:

                     1/0!   .     .     .
                     1/1!  1/0!
                     1/2!  2/1!  1/0!
                     1/3!  3/2!  3/1!  1/0!
                 .....
                   -------------------------------
[1 x x^2 x^3...]  = [ b0   b1    b2     b3 .... ]

we have, using the apostroph ' for derivative wrt x:

   b0  =            exp(x)      /0!
   b1  =  x*  [ x  *exp(x) ]'   /1!
   b2 =   x^2*[ x^2*exp(x) ]''  /2!
   b3 =   x^3*[ x^3*exp(x) ]''' /3!
 ...
   bk =   x^k*[ x^k*exp(x) ]'(k)' /k!

Expanding the derivatives gives

(1):

   b0 *0! / exp(x) =   1
   b1 *1! / exp(x) =   1 + x

   b2 *2! / exp(x) =   (2 + 4x + x^2)
                   =   2!/0! + 2*2!/1!  + 1*2!/2! x^2

   b3 *3! / exp(x) =   (6 + 18x + 9x^2 + x^3)
                        3! + 3*3!/1! x + 3*3!/2! x^2 + 1 *3!/3! x^3

   b4 *4! / exp(x) =   (24   + 96x  + 72x^2   + 16x^3 + 1x^4)
                       (1*4! + 4*4!/1! x + 6*4!/2!x^2 + 4*4!/3!x^3 + 1*4!/4!*x^4)



and the rhs written as matrix, where also the k! of the k'th row is cancelled , gives

(2):
   b0 / exp(x)      =   1                              *  [1,  x,  x^2,  x^3, ....]~
   b1 / exp(x)      =   1/0!  1/1!
   b2 / exp(x)      =   1/0!  2/1!  1/2!
   b3 / exp(x)      =   1/0!  3/1!  3/2!   1/3!
     ...            =     ....

If we now consider the result when letting x->1 we can further use
the horizontal symmetry of the binomial-coefficients to reorder
the rows and just get the matrix Pf again as coefficients

(3):

 [b0 b1 b2 b3...]~ /exp(1) =    1/0!                            * [1, 1, 1, 1, 1, ...]~
                                1/1!   1/0!
                                1/2!   2/1!   1/0!
                                1/3!   3/2!   3/1!   1/0!


 so

   V(1)~ * Pf = B~
 or
           Pf~ * V(1) = B
 and
           Pf  * V(1) = B / exp(x)

The systematic of the derivation of matrix Pf as resultant
of the derivative-proceeding in (1) and (2) needs now a better
systematic proof, but I think, that could be done...


--------------------------------------------------

One could proceed one further step and state

 for a current row and column m of Pf we have

(4):
    V(x)~ * Pf[0..inf,m] = exp(x) * Pf[m,0..m] * x^(2m) * V(1/x)

or (Pari:)

   suminf( k=0, x^k*Pf[k+1,m+1]) = exp(x) * x^(2*m) *  sum(k=0,m,Pf[m+1,k+1]/x^k)

where for a row r and col c the entry Pf[r+1,c+1] may be replaced by

      Pf[r+1,c+1] = ch(r,c) / (r-c)!          // for r>=c
                  = 0                         // for r<c

where ch(a,b) is the binomial-coefficient.

----------------

The common notation for the above observation may be given as:

    oo            x^k               m    m              x^(m-k)
   sum   ch(k,m)*------   = exp(x) x  * sum   ch(m,k)* ---------
    k=m          (k-m)!                 k=0              (m-k)!

                where ch(a,b) is the binomial-coefficient.


Gottfried Helms