The Jacobi-Stirling Numbers Eric S. Egge (joint work with G. Andrews, L. Littlejohn, and W. Gawronski) Carleton College
March 18, 2012
Eric S. Egge (Carleton College)
The Jacobi-Stirling Numbers
March 18, 2012
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The Differential Operator xD
y = y (x)
Eric S. Egge (Carleton College)
D=
The Jacobi-Stirling Numbers
d dx
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The Differential Operator xD
y = y (x)
D=
d dx
xD[y ] = 1xy 0 (xD)2 [y ] = 1xy 0 + 1x 2 y (2)
Eric S. Egge (Carleton College)
The Jacobi-Stirling Numbers
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The Differential Operator xD
y = y (x)
D=
d dx
xD[y ] = 1xy 0 (xD)2 [y ] = 1xy 0 + 1x 2 y (2) (xD)3 [y ] = 1xy 0 + 3x 2 y (2) + 1x 3 y (3)
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The Jacobi-Stirling Numbers
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The Differential Operator xD
y = y (x)
D=
d dx
xD[y ] = 1xy 0 (xD)2 [y ] = 1xy 0 + 1x 2 y (2) (xD)3 [y ] = 1xy 0 + 3x 2 y (2) + 1x 3 y (3) (xD)4 [y ] = 1xy 0 + 7x 2 y (2) + 6x 3 y (3) + 1x 4 y (4)
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The Jacobi-Stirling Numbers
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The Differential Operator xD
y = y (x)
D=
d dx
xD[y ] = 1xy 0 (xD)2 [y ] = 1xy 0 + 1x 2 y (2) (xD)3 [y ] = 1xy 0 + 3x 2 y (2) + 1x 3 y (3) (xD)4 [y ] = 1xy 0 + 7x 2 y (2) + 6x 3 y (3) + 1x 4 y (4)
(xD)n [y ] =
n X j=1
n x j y (j) n+1−j
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n n−1 n−1 = +j j j −1 j
The Jacobi-Stirling Numbers
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The Jacobi-Stirling Numbers
`α,β [y ] = −(1 − x 2 )y 00 + (α − β + (α + β + 2)x)y 0
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The Jacobi-Stirling Numbers
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The Jacobi-Stirling Numbers
`α,β [y ] = −(1 − x 2 )y 00 + (α − β + (α + β + 2)x)y 0
Theorem (Everitt, Kwon, Littlejohn, Wellman, Yoon) `nα,β [y ]
=
1 wα,β (x)
Eric S. Egge (Carleton College)
n X n j=1
j
(j) (−1)j (1 − x)α+j (1 + x)β+j y (j)
α,β
The Jacobi-Stirling Numbers
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The Jacobi-Stirling Numbers
`α,β [y ] = −(1 − x 2 )y 00 + (α − β + (α + β + 2)x)y 0
Theorem (Everitt, Kwon, Littlejohn, Wellman, Yoon) `nα,β [y ]
=
1 wα,β (x)
n X n j=1
j
(j) (−1)j (1 − x)α+j (1 + x)β+j y (j)
α,β
n n−1 n−1 = + j(j + α + β + 1) j α,β j − 1 α,β j α,β
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The Jacobi-Stirling Numbers
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What Does
n j
α,β
2γ − 1 = α + β + 1
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Count?
[n]2 := {11 , 12 , 21 , 22 , . . . , n1 , n2 }
The Jacobi-Stirling Numbers
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What Does
n j
α,β
2γ − 1 = α + β + 1
Count?
[n]2 := {11 , 12 , 21 , 22 , . . . , n1 , n2 }
Theorem (AEGL) For any positive integer γ, the Jacobi-Stirling number
n j γ
counts set
partitions of [n]2 into j + γ blocks such that 1
There are γ distinguishable zero blocks, any of which may be empty.
2
There are j indistinguishable nonzero blocks, all nonempty.
3
The union of the zero blocks does not contain both copies of any number. Each nonzero block
4
contains both copies of its smallest element does not contain both copies of any other number.
These are Jacobi-Stirling set partitions. Eric S. Egge (Carleton College)
The Jacobi-Stirling Numbers
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A Generating Function Interpretation z =α+β
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The Jacobi-Stirling Numbers
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A Generating Function Interpretation z =α+β S(n, j) := Jacobi-Stirling set partitions of [n]2 into 1 zero block and j nonzero blocks.
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The Jacobi-Stirling Numbers
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A Generating Function Interpretation z =α+β S(n, j) := Jacobi-Stirling set partitions of [n]2 into 1 zero block and j nonzero blocks.
Theorem (Gelineau, Zeng) The Jacobi-Stirling number nj is the generating function in z for S(n, j) z with respect to the number of numbers with subscript 1 in the zero block.
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The Jacobi-Stirling Numbers
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A Generating Function Interpretation z =α+β S(n, j) := Jacobi-Stirling set partitions of [n]2 into 1 zero block and j nonzero blocks.
Theorem (Gelineau, Zeng) The Jacobi-Stirling number nj is the generating function in z for S(n, j) z with respect to the number of numbers with subscript 1 in the zero block.
Corollary The leading coefficient in
Eric S. Egge (Carleton College)
n j z
is the Stirling number
The Jacobi-Stirling Numbers
n j . March 18, 2012
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Jacobi-Stirling Numbers of the First Kind
n
x =
n X n j=0
Eric S. Egge (Carleton College)
j
j−1 Y
(x − k(k + α + β + 1))
α,β k=0
The Jacobi-Stirling Numbers
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Jacobi-Stirling Numbers of the First Kind
n
x =
n X n j=0
n−1 Y
j
j−1 Y
(x − k(k + α + β + 1))
α,β k=0
n X n+j n (x − k(k + α + β + 1)) = (−1) xj j α,β
k=0
Eric S. Egge (Carleton College)
j=0
The Jacobi-Stirling Numbers
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Jacobi-Stirling Numbers of the First Kind
n
x =
n X n j=0
n−1 Y
j
j−1 Y
(x − k(k + α + β + 1))
α,β k=0
n X n+j n (x − k(k + α + β + 1)) = (−1) xj j α,β
k=0
j=0
n n−1 n−1 = + (n − 1)(n + α + β) j α,β j − 1 α,β j α,β
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The Jacobi-Stirling Numbers
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Jacobi-Stirling Numbers of the First Kind
n
x =
j
j=0 n−1 Y
j−1 Y
n X n
(x − k(k + α + β + 1))
α,β k=0
n X n+j n (x − k(k + α + β + 1)) = (−1) xj j α,β j=0
k=0
n n−1 n−1 = + (n − 1)(n + α + β) j α,β j − 1 α,β j α,β
Theorem (AEGL)
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−j −n
n+j
= (−1) γ
n j 1−γ
The Jacobi-Stirling Numbers
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Balanced Jacobi-Stirling Permutations
Theorem (AEGL) For any positive integer γ, the Jacobi-Stirling number of the first kind
n j γ
is the number of ordered pairs (π1 , π2 ) of permutations with π1 ∈ Sn+γ and π2 ∈ Sn+γ−1 such that 1
π1 has γ + j cycles and π2 has γ + j − 1 cycles.
2
The cycle maxima of π1 which are less than n + γ are exactly the cycle maxima of π2 .
3
For each non cycle maximum k, at least one of π1 (k) and π2 (k) is less than or equal to n.
Such ordered pairs are balanced Jacobi-Stirling permutations.
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The Jacobi-Stirling Numbers
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Unbalanced Jacobi-Stirling Permutations
Theorem (AEGL) Ifn2γ is a positive integer, then the Jacobi-Stirling number of the first kind j γ is the number of ordered pairs (π1 , π2 ) of permutations with π1 ∈ Sn+γ and π2 ∈ Sn such that 1
π1 has j + 2γ − 1 cycles and π2 has j cycles.
2
The cycle maxima of π1 which are less than n + 1 are exactly the cycle maxima of π2 .
3
For each non cycle maximum k, at least one of π1 (k) and π2 (k) is less than or equal to n.
Such ordered pairs are unbalanced Jacobi-Stirling permutations.
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The Jacobi-Stirling Numbers
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More Generating Functions
Σ(n, j) := all (σ, τ ) such that σ is a permutation of {0, 1, . . . , n}, τ is a permutation of {1, 2, . . . , n}, and both have j cycles. 1 and 0 are in the same cycle in σ. Among their nonzero entries, σ and τ have the same cycle minima.
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The Jacobi-Stirling Numbers
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More Generating Functions
Σ(n, j) := all (σ, τ ) such that σ is a permutation of {0, 1, . . . , n}, τ is a permutation of {1, 2, . . . , n}, and both have j cycles. 1 and 0 are in the same cycle in σ. Among their nonzero entries, σ and τ have the same cycle minima.
Theorem (Gelineau, Zeng) n
is the generating function in z for Σ(n, j) with respect to the number of nonzero left-to-right minima in the cycle containing 0 in σ, written as a word beginning with σ(0). j z
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The Jacobi-Stirling Numbers
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Where are the Symmetric Functions?
hn−j (x1 , . . . , xj ) = hn−j (x1 , . . . , xj−1 ) + xj hn−j−1 (x1 , . . . , xj )
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The Jacobi-Stirling Numbers
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Where are the Symmetric Functions?
hn−j (x1 , . . . , xj ) = hn−j (x1 , . . . , xj−1 ) + xj hn−j−1 (x1 , . . . , xj ) n = hn−j (1(1 + z), 2(2 + z), . . . , j(j + z)) j z
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The Jacobi-Stirling Numbers
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Where are the Symmetric Functions?
hn−j (x1 , . . . , xj ) = hn−j (x1 , . . . , xj−1 ) + xj hn−j−1 (x1 , . . . , xj ) n = hn−j (1(1 + z), 2(2 + z), . . . , j(j + z)) j z
en−j (x1 , . . . , xn−1 ) = en−j (x1 , . . . , xn−2 ) + xn−1 en−j−1 (x1 , . . . , xn−2 )
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The Jacobi-Stirling Numbers
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Where are the Symmetric Functions?
hn−j (x1 , . . . , xj ) = hn−j (x1 , . . . , xj−1 ) + xj hn−j−1 (x1 , . . . , xj ) n = hn−j (1(1 + z), 2(2 + z), . . . , j(j + z)) j z
en−j (x1 , . . . , xn−1 ) = en−j (x1 , . . . , xn−2 ) + xn−1 en−j−1 (x1 , . . . , xn−2 )
n = en−j (1(1 + z), 2(2 + z), . . . , (n − 1)(n − 1 + z)) j z
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The Jacobi-Stirling Numbers
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An Open q-uestion
Is there a q-analogue of the Jacobi-Stirling numbers associated with the q-Jacobi polynomials?
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The Jacobi-Stirling Numbers
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The End
Thank You!
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The Jacobi-Stirling Numbers
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