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Stirling transform

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In combinatorial mathematics, the Stirling transform of a sequence { an : n = 1, 2, 3, ... } of numbers is the sequence { bn : n = 1, 2, 3, ... } given by

b n = k = 1 n { n k } a k {\displaystyle b_{n}=\sum _{k=1}^{n}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}a_{k}} ,

where { n k } {\displaystyle \left\{{\begin{matrix}n\\k\end{matrix}}\right\}} is the Stirling number of the second kind, which is the number of partitions of a set of size n {\displaystyle n} into k {\displaystyle k} parts. This is a linear sequence transformation.

The inverse transform is

a n = k = 1 n ( 1 ) n k [ n k ] b k {\displaystyle a_{n}=\sum _{k=1}^{n}(-1)^{n-k}\leftb_{k}} ,

where ( 1 ) n k [ n k ] {\textstyle (-1)^{n-k}\left} is a signed Stirling number of the first kind, where the unsigned [ n k ] {\displaystyle \left} can be defined as the number of permutations on n {\displaystyle n} elements with k {\displaystyle k} cycles.

Berstein and Sloane (cited below) state "If an is the number of objects in some class with points labeled 1, 2, ..., n (with all labels distinct, i.e. ordinary labeled structures), then bn is the number of objects with points labeled 1, 2, ..., n (with repetitions allowed)."

If

f ( x ) = n = 1 a n n ! x n {\displaystyle f(x)=\sum _{n=1}^{\infty }{a_{n} \over n!}x^{n}}

is a formal power series, and

g ( x ) = n = 1 b n n ! x n {\displaystyle g(x)=\sum _{n=1}^{\infty }{b_{n} \over n!}x^{n}}

with an and bn as above, then

g ( x ) = f ( e x 1 ) {\displaystyle g(x)=f(e^{x}-1)} .

Likewise, the inverse transform leads to the generating function identity

f ( x ) = g ( log ( 1 + x ) ) {\displaystyle f(x)=g(\log(1+x))} .

See also

References

  • Bernstein, M.; Sloane, N. J. A. (1995). "Some canonical sequences of integers". Linear Algebra and Its Applications. 226/228: 57–72. arXiv:math/0205301. doi:10.1016/0024-3795(94)00245-9. S2CID 14672360..
  • Khristo N. Boyadzhiev, Notes on the Binomial Transform, Theory and Table, with Appendix on the Stirling Transform (2018), World Scientific.
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