[seqfan] Re: Looking for a simpler pattern for that matrix

[Gottfried Helms]

Seqfans, dear Alexander,

In answering Alexander's request for more detailed explanation
I'll restate the computation/decomposition of A in more detail here.

We'll denote the given (infinite) matrix of coefficients A

 A =                                                              (1)
     2        .          .          .           .          .
    -5        3          .          .           .          .
    10      -22          4          .           .          .
   -17      108        -57          5           .          .
    26     -432        504       -116           6          .   ...
   -37     1520      -3510       1600        -205          7   ...
   ...
and their elements as a{r,c} for row- and column-index, the indices
beginning at 1.

The Pascalmatrix P

 P=                                                               (2a)
  1  .   .   .   .  .  .   ...
  1  1   .   .   .  .  .   ...
  1  2   1   .   .  .  .   ...
  1  3   3   1   .  .  .   ...
  1  4   6   4   1  .  .   ...
  1  5  10  10   5  1  .   ...
  1  6  15  20  15  6  1   ...
  ...

containing the binomial-coefficients p{r,c} = b(r-1 , c-1).
The n'th power of the Pascalmatrix, P^n, has the elements

   p^(n){r,c} = n^(r-c) * b(r-1,c-1)                              (2b)



Then I gave the basic scheme, with some "magic" coefficients in
the column-vectors on the rhs

    A_{,1} = P^1 * [2,3, 2, 0, ...                    ]            (3)
    A_{,2} = P^2 * [0,3,10,12,  0, ...                ]
    A_{,3} = P^3 * [0,0, 4,21, 36, 0, ,...            ]
    A_{,4} = P^4 * [0,0, 0, 5, 36, 80, 0, ,...        ]
    A_{,5} = P^5 * [0,0, 0, 0,  6, 55,150, 0, ,...    ]


We rewrite the above columnvectors of coefficients as matrix M

 M =                                                               (4a)
  2   .   .   .    .   .  . ...
  3   3   .   .    .   .  . ...
  2  10   4   .    .   .  . ...
  0  12  21   5    .   .  . ...
  0   0  36  36    6   .  . ...
  0   0   0  80   55   7  . ...
  0   0   0   0  150  78  8 ...
... ... ... ... ... ... ... ...
where
  m{c  ,c} =  ( c+1) * c^0                                          (4b)
  m{c+1,c} =  (2c+1) * c^1
  m{c+2,c} =  ( c+1) * c^2

and all other entries are zero                                      (4c)
--------------------------------------------------------------------------------
Then
                  c+2
 abs(  a{r,c} ) = Sum  p^(c){r,k} * m{k,c}                          (5a)
                  k=c


                  c+2
 abs(  a{r,c} ) = Sum  c^(r-k) * b(r-1,k-1)    * m{k,c}             (5b)
                  k=c

The sum notation and elements m expanded :

 abs(  a{r,c} ) =   c^(r-c  ) * b(r-1,c-1)    * ( c+1)*1            (5c)
                 +  c^(r-c-1) * b(r-1,c  )    * (2c+1)*c
                 +  c^(r-c-2) * b(r-1,c+1)    * ( c+1)*c^2

and one more simplification gives
====================================================================
   the complete evaluation for the signless a{r,c}:

      abs(  a{r,c} ) =   c^(r-c) * (   (c+1)*( b(r-1,c-1)            (6a)
                                              +b(r-1,c  )
                                              +b(r-1,c+1))
                                    +      c* b(r-1,c  )
                                     )
   and finally introducing the signs

                    s = (-1)^(r-c)                                   (6b)

       a{r,c} = s * abs(a{r,c})                                      (7)

=====================================================================
=====================================================================
Example:

      abs(   a{4,3} ) =    3^1   * (     4 *( b(3,2) + b(3,3)+ b(3,4)  )+ 3*b(3,3))
                      =    3     * (     4 *( 3 + 1 + 0 )     + 3*1 )
                      =    3*19
                      = 57
      The sign is
              (-1)^(4-3) = -1
      so
             a{4,3} =  -57

 which matches the entry in the matrix A.


Gottfried

Am 07.05.2014 02:28 schrieb Gottfried Helms:
> Dear Alexander / dear Frank -
> 
>  thank you for your help! Meanwhile I tried it myself hard and found
>  a solution which can be nicely expressed in terms of row- and column-
>  indexes.
>