# Stirling polynomial

Stirling’s polynomials $S_{k}(x)$ are defined by the generating function  $\left({t\over{1-e^{-t}}}\right)^{x+1}=\sum_{k=0}{S_{k}(x)\over k!}t^{k}.$

The sequence $S_{k}(x-1)$ is of binomial type, since $S_{k}(x+y-1)=\sum_{i=0}^{k}{k\choose i}S_{i}(x-1)S_{k-i}(y-1)$. Moreover, this basic recursion holds: $S_{k}(x)=(x-k){S_{k}(x-1)\over x}+kS_{k-1}(x+1)$.

These are the first polynomials:

1. 1.

$S_{0}(x)=1$;

2. 2.

$S_{1}(x)={1\over 2}(x+1)$;

3. 3.

$S_{2}(x)={1\over 12}(3x^{2}+5x+2)$;

4. 4.

$S_{3}(x)={1\over 8}(x^{3}+2x^{2}+x)$;

5. 5.

$S_{4}(x)={1\over 240}(15x^{4}+30x^{3}+5x^{2}-18x-8)$.

In addition we have these special values:

1. 1.

$S_{k}(-m)={(-1)^{k}\over{k+m-1\choose k}}S_{k+m-1,m-1}$, where $S_{m,n}$ denotes Stirling numbers of the second kind. Conversely, $S_{n,m}=(-1)^{n-m}{n\choose m}S_{n-m}(-m-1)$;

2. 2.

$S_{k}(-1)=\delta_{k,0}$;

3. 3.

$S_{k}(0)=(-1)^{k}B_{k}$, where $B_{k}$ are Bernoulli’s numbers;

4. 4.

$S_{k}(1)=(-1)^{k+1}((k-1)B_{k}+kB_{k-1})$;

5. 5.

$S_{k}(2)={(-1)^{k}\over 2}((k-1)(k-2)B_{k}+3k(k-2)B_{k-1}+2k(k-1)B_{k-2})$;

6. 6.

$S_{k}(k)=k!$;

7. 7.

$S_{k}(m)={(-1)^{k}\over{m\choose k}}s_{m+1,m+1-k}$, where $s_{m,n}$ are Stirling numbers of the first kind. They may be recovered by $s_{n,m}=(-1)^{n-m}{n-1\choose n-m}S_{n-m}(n-1)$.

Explicit representations involving Stirling numbers can be deduced with Lagrange’s interpolation formula:

 $S_{k}(x)=\sum_{n=0}^{k}(-1)^{k-n}S_{k+n,n}{{x+n\choose n}{x+k+1\choose k-n}% \over{k+n\choose n}}=\sum_{n=0}^{k}(-1)^{n}s_{k+n+1,n+1}{{x-k\choose n}{x-k-n-% 1\choose k-n}\over{k+n\choose k}}.$

These following formulae hold as well:

 ${k+m\choose k}S_{k}(x-m)=\sum_{i=0}^{k}(-1)^{k-i}{k+m\choose i}S_{k-i+m,m}S_{i% }(x),$
 ${k-m\choose k}S_{k}(x+m)=\sum_{i=0}^{k}{k-m\choose i}s_{m,m-k+i}S_{i}(x).$
Title Stirling polynomial  StirlingPolynomial 2013-03-22 15:38:36 2013-03-22 15:38:36 kronos (12218) kronos (12218) 9 kronos (12218) Definition msc 05A15