Stirling polynomial
The sequence Sk(x-1) is of binomial type, since Sk(x+y-1)=∑ki=0(ki)Si(x-1)Sk-i(y-1). Moreover, this basic recursion holds: Sk(x)=(x-k)Sk(x-1)x+kSk-1(x+1).
These are the first polynomials:
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1.
S0(x)=1;
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2.
S1(x)=12(x+1);
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3.
S2(x)=112(3x2+5x+2);
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4.
S3(x)=18(x3+2x2+x);
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5.
S4(x)=1240(15x4+30x3+5x2-18x-8).
In addition we have these special values:
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1.
Sk(-m)=(-1)k(k+m-1k)Sk+m-1,m-1, where Sm,n denotes Stirling numbers of the second kind. Conversely, Sn,m=(-1)n-m(nm)Sn-m(-m-1);
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2.
Sk(-1)=δk,0;
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3.
Sk(0)=(-1)kBk, where Bk are Bernoulli’s numbers;
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4.
Sk(1)=(-1)k+1((k-1)Bk+kBk-1);
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5.
Sk(2)=(-1)k2((k-1)(k-2)Bk+3k(k-2)Bk-1+2k(k-1)Bk-2);
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6.
Sk(k)=k!;
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7.
Sk(m)=(-1)k(mk)sm+1,m+1-k, where sm,n are Stirling numbers of the first kind. They may be recovered by sn,m=(-1)n-m(n-1n-m)Sn-m(n-1).
Explicit representations involving Stirling numbers can be deduced with Lagrange’s interpolation formula:
Sk(x)=k∑n=0(-1)k-nSk+n,n(x+nn)(x+k+1k-n)(k+nn)=k∑n=0(-1)nsk+n+1,n+1(x-kn)(x-k-n-1k-n)(k+nk). |
These following formulae hold as well:
(k+mk)Sk(x-m)=k∑i=0(-1)k-i(k+mi)Sk-i+m,mSi(x), |
(k-mk)Sk(x+m)=k∑i=0(k-mi)sm,m-k+iSi(x). |
Title | Stirling polynomial![]() |
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Canonical name | StirlingPolynomial |
Date of creation | 2013-03-22 15:38:36 |
Last modified on | 2013-03-22 15:38:36 |
Owner | kronos (12218) |
Last modified by | kronos (12218) |
Numerical id | 9 |
Author | kronos (12218) |
Entry type | Definition |
Classification | msc 05A15 |