Exact and Unique solution for base e^(1/e)
#13
Hi,

these coefficients occur exactly in the powers of the matrix of Stirling numbers 2'nd kind, as I described it in my posting about iteration for e^x-1 and which must be used in a factorial scaled version to obtain the results.

Your coefficients in rows for a certain iteration are the coefficients of the second column of the related power of St2 (second column only will be used for the scalar result in my notation).

Code:
VE(St2,8,6):  \\ extraction of 8 rows, 6 columns
  1   .    .     .     .     .   ...
  0   1    .     .     .     .
  0   1    1     .     .     .
  0   1    3     1     .     .
  0   1    7     6     1     .
  0   1   15    25    10     1
  0   1   31    90    65    15
  0   1   63   301   350   140   ...
  ...                            ...

second iteration = second power
Code:
VE(St2^2,8,6):
  1     .      .      .      .     .   ...
  0     1      .      .      .     .
  0     2      1      .      .     .
  0     5      6      1      .     .
  0    15     32     12      1     .
  0    52    175    110     20     1
  0   203   1012    945    280    30
  0   877   6230   8092   3465   595   ...
  ...                            ...

third iteration = third power
Code:
VE(St2^3,8,6):
  1       .       .       .       .      .
  0       1       .       .       .      .
  0       3       1       .       .      .
  0      12       9       1       .      .
  0      60      75      18       1      .
  0     358     660     255      30      1
  0    2471    6288    3465     645     45
  0   19302   65051   47838   12495   1365
  ...                            ...


P.s. If you need a free numb-theoretical package with arbitrary precision you might consider Pari/GP I've also written a convenient user-interface for this usable in win-xp ("Pari-TTY")

Gottfried
Gottfried Helms, Kassel


Messages In This Thread
RE: Exact and Unique solution for base e^(1/e) - by Gottfried - 08/13/2007, 03:45 PM

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