Hankel determinants, continued fractions, orthgonal polynomials, - - PowerPoint PPT Presentation

Hankel determinants, continued fractions,

• rthgonal polynomials,

and hypergeometric series

Ira M. Gessel with Jiang Zeng and Guoce Xin LaBRI June 8, 2007

Continued fractions and Hankel determinants

There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials.

Continued fractions and Hankel determinants

There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials. Why are we interested in these things?

Continued fractions and Hankel determinants

There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials. Why are we interested in these things? Continued fractions count certain “weighted Motzkin paths” which encode objects of enumerative interest such as partitions and permutations. (Flajolet)

Continued fractions and Hankel determinants

There is a close relationship between continued fractions, Hankel determinants, and orthogonal polynomials. Why are we interested in these things? Continued fractions count certain “weighted Motzkin paths” which encode objects of enumerative interest such as partitions and permutations. (Flajolet) Hankel determinants arise in some enumeration problems, for example, counting certain kinds of tilings or alternating sign matrices.

Orthogonal polynomials

A sequence of polynomials

• pn(x)
• n≥0, where pn(x) has degree

n, is orthogonal if there is a linear functional L on polynomials such that L

• pm(x)pn(x)
• = 0 for m = n, but L
• pm(x)2

= 0. We will assume that L(1) = 1.

Orthogonal polynomials

A sequence of polynomials

• pn(x)
• n≥0, where pn(x) has degree

n, is orthogonal if there is a linear functional L on polynomials such that L

• pm(x)pn(x)
• = 0 for m = n, but L
• pm(x)2

= 0. We will assume that L(1) = 1. The moments of the sequence

• pn(x)
• are µn = L(xn).

Orthogonal polynomials

A sequence of polynomials

• pn(x)
• n≥0, where pn(x) has degree

n, is orthogonal if there is a linear functional L on polynomials such that L

• pm(x)pn(x)
• = 0 for m = n, but L
• pm(x)2

= 0. We will assume that L(1) = 1. The moments of the sequence

• pn(x)
• are µn = L(xn). So

µ0 = 1.

Orthogonal polynomials

A sequence of polynomials

• pn(x)
• n≥0, where pn(x) has degree

n, is orthogonal if there is a linear functional L on polynomials such that L

• pm(x)pn(x)
• = 0 for m = n, but L
• pm(x)2

= 0. We will assume that L(1) = 1. The moments of the sequence

• pn(x)
• are µn = L(xn). So

µ0 = 1.

• Theorem. The sequence
• pn(x)
• f monic polynomials is
• rthogonal if and only if there exist numbers
• an
• n≥0 and
• bn
• n≥1, with bn = 0 for all n ≥ 1 such that

xpn(x) = pn+1(x) + anpn(x) + bnpn−1(x), n ≥ 1 with p0(x) = 1 and p1(x) = x − a0.

The connection

We then have the continued fraction

∞

• k=0

µkxk = 1 1 − a0x − b1x2 1 − a1x − b2x2 1 − a2x − · · · and the Hankel determinant det(µi+j)0≤i,j<n is equal to bn−1

1

bn−2

2

· · · b2

n−2b1.

If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials?

If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . .

If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments.

If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments. What are the moments of the classical orthogonal polynomials?

If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments. What are the moments of the classical orthogonal polynomials? What are their generating functions?

If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments. What are the moments of the classical orthogonal polynomials? What are their generating functions? Why do some, but not all of them, have nice exponential generating functions?

If we’re interested in Hankel determinants or continued fractions, what good are the orthogonal polynomials? There is a systematic treatment of all the “classical” orthogonal polynomials, that have explicit expressions as hypergeometric series, called the Askey scheme. In a comprehensive paper (or web site) by Koekoek and Swarttouw, you can look up all of these orthogonal polynomials and find out everything you might want to know about them. . . except the moments. What are the moments of the classical orthogonal polynomials? What are their generating functions? Why do some, but not all of them, have nice exponential generating functions? What are the orthogonal polynomials whose moments are the Genocchi numbers?

How do we find the moments of a sequence of orthogonal polynomials? The key is to find numbers α0, α1, α2, . . . such that we can evaluate L

• (x + α0)(x + α1) · · · (x + αk)
• for each
• k. (Usually this isn’t too hard to do.)

How do we find the moments of a sequence of orthogonal polynomials? The key is to find numbers α0, α1, α2, . . . such that we can evaluate L

• (x + α0)(x + α1) · · · (x + αk)
• for each
• k. (Usually this isn’t too hard to do.)

We then apply the following Lemma:

∞

• n=0

(x+α0)(x+α1) · · · (x+αn−1) tn (1 + α0t) · · · (1 + αnt) = 1 1 − xt .

• Proof. We have the indefinite sum

m

• n=0

(x + α0) · · · (x + αn−1) tn (1 + α0t) · · · (1 + αnt) = 1 1 − xt

• 1 − (x + α0) · · · (x + αn)

tm+1 (1 + α0t) · · · (1 + αnt)

• ,

which is easily proved by induction. The lemma follows by taking m → ∞.

Now suppose that L

• (x + α0)(x + α1) · · · (x + αn−1)
• = Mn.

We apply L to

∞

• n=0

(x+α0)(x+α1) · · · (x+αn−1) tn (1 + α0t) · · · (1 + αnt) = 1 1 − xt . to get

∞

• n=0

Mn tn (1 + α0t) · · · (1 + αnt) = L

• 1

1 − xt

• =

∞

• k=0

L(xk)tk =

∞

• k=0

µktk.

Let’s look at an example. First, the standard notation: (α)n = α(α + 1) · · · (α + n − 1)

pFq

a1, . . . , ap b1, . . . , bp

• z
• =

∞

• n=0

(a1)n · · · (ap)n n! (b1)n · · · (bq)n zn The Hahn polynomials are defined by Qn(x; α, β, N) = 3F2 −n, n + α + β + 1, −x α + 1, −N

• 1
• They are orthogonal with respect to the linear functional L given

by L

• p(x)
• =

N

• x=0

α + x x β + N − x N − x

• p(x),

By applying Vandermonde’s theorem we find that L

• (x + α + 1)k
• = (α + β + 2)N

N! (α + 1)k(α + β + 2 + N)k (α + β + 2)k .

Since we want L(1) to be 1, we normalize this by dividing L by (α + β + 2)N/N!. (This is independent of k.) Then with our new normalization, L

• (x + α + 1)k
• = (α + 1)k(α + β + 2 + N)k

(α + β + 2)k . Since the right side is a rational function of α, β, and N, we don’t need N to be a nonnegative integer and we can therefore make a change of variables, α = A − 1, β = C − A − 1, N = B − C. Then L

• (x +A)n
• = L
• (x +A)(x +A+n) · · · (x +A+n−1)
• = (A)n(B)n

(C)n . Now we can apply our lemma to get the ordinary generating function for the moments µn of the Hahn polynomials:

∞

• n=0

µntn =

∞

• n=0

(A)n(B)n (C)n tn n

j=0(1 + (A + j)t).

This can be written as a rather strange-looking hypergeometric series 1 1 + At

∞

• n=0

(A)n (Bn) (C)n(1 + A + t−1)n = 1 1 + At 3F2

• A, B, 1

C, 1 + A + t−1

• 1
• .

This can be written as a rather strange-looking hypergeometric series 1 1 + At

∞

• n=0

(A)n (Bn) (C)n(1 + A + t−1)n = 1 1 + At 3F2

• A, B, 1

C, 1 + A + t−1

• 1
• .

How do we get an exponential generating function for the moments?

This can be written as a rather strange-looking hypergeometric series 1 1 + At

∞

• n=0

(A)n (Bn) (C)n(1 + A + t−1)n = 1 1 + At 3F2

• A, B, 1

C, 1 + A + t−1

• 1
• .

How do we get an exponential generating function for the moments? We define a linear operator ε on formal power series by ε ∞

• n=0

untn

• =

∞

• n=0

un tn n!.

This can be written as a rather strange-looking hypergeometric series 1 1 + At

∞

• n=0

(A)n (Bn) (C)n(1 + A + t−1)n = 1 1 + At 3F2

• A, B, 1

C, 1 + A + t−1

• 1
• .

How do we get an exponential generating function for the moments? We define a linear operator ε on formal power series by ε ∞

• n=0

untn

• =

∞

• n=0

un tn n!. Then we apply ε to ∞

n=0 µntn.

Lemma. ε

• tn

(1 + At)(1 + (A + 1)t) · · · (1 + (A + n)t)

• = e−At (1 − e−t)n

n! . Proof #1. Expand the left side by partial fractions and the right side by the binomial theorem. Proof #2. Without explicitly computing the partial fraction on the left, we can see that it will be a linear combination of e−At, e−(A+1)t, . . . , e−(A+n)t and the first nonzero term is tn/n!. The right side is the only possibility.

Lemma. ε

• tn

(1 + At)(1 + (A + 1)t) · · · (1 + (A + n)t)

• = e−At (1 − e−t)n

n! . Proof #1. Expand the left side by partial fractions and the right side by the binomial theorem. Proof #2. Without explicitly computing the partial fraction on the left, we can see that it will be a linear combination of e−At, e−(A+1)t, . . . , e−(A+n)t and the first nonzero term is tn/n!. The right side is the only possibility. Now we apply the lemma to

∞

• n=0

µntn =

∞

• n=0

(A)n(B)n (C)n tn n

j=0(1 + (A + j)t).

We have

∞

• n=0

µn tn n! =

∞

• n=0

(A)n(B)n (C)n e−At (1 − e−t)n n! = e−At

2F1

A, B C

• 1 − e−t
• .

Let’s look at a few examples. First let’s see how this generalizes the Meixner polynomials, which are the most general Sheffer

• rthogonal polynomials. To get (one form of) the Meixner

polynomials, we take B = uC and then take the limit as C → ∞, getting as the exponential generating function for the moments

• (1 − u)et + u

−A

We have

∞

• n=0

µn tn n! =

∞

• n=0

(A)n(B)n (C)n e−At (1 − e−t)n n! = e−At

2F1

A, B C

• 1 − e−t
• .

As another example, if we take A = B = 1, C = 2, we get the Bernoulli numbers:

∞

• n=0

µn tn n! = t et − 1, so µn = Bn.

We have

∞

• n=0

µn tn n! =

∞

• n=0

(A)n(B)n (C)n e−At (1 − e−t)n n! = e−At

2F1

A, B C

• 1 − e−t
• .

Similarly, if we take A = 1, B = 2, C = 3, we get the Bernoulli numbers shifted by 1:

∞

• n=0

µn tn n! = −2d dt t et − 1, so µn = −2Bn+1.

We can use the same approach on the Wilson polynomials, which are the most general “classical” orthogonal polynomials (with q = 1). They are defined by Wn(x2) = 4F3 −n, n + a + c + d − 1, a + ix, a − ix a + b, a + c, a + d

• 1
• ,

(in this form they are not monic). The corresponding linear functional is L

• p(x)
• =

1 2π ∞

• Γ(a + ix)Γ(b + ix)Γ(c + ix)Γ(d + ix)

Γ(2ix)

• 2

p(x2) dx

It follows easily from known facts that L

• (x + a2)(x + (a + 1)2) · · · (x + (a + n − 1)2)
• = (a + b)n(a + c)n(a + d)n

(a + b + c + d)n so from our first lemma,

∞

• n=0

µntn =

∞

• n=0

(a + b)n(a + c)n(a + d)ntn (a + b + c + d)n n

j=0

• 1 + (a + j)2t

It follows easily from known facts that L

• (x + a2)(x + (a + 1)2) · · · (x + (a + n − 1)2)
• = (a + b)n(a + c)n(a + d)n

(a + b + c + d)n so from our first lemma,

∞

• n=0

µntn =

∞

• n=0

(a + b)n(a + c)n(a + d)ntn (a + b + c + d)n n

j=0

• 1 + (a + j)2t
• What about an exponential generating function?

It follows easily from known facts that L

• (x + a2)(x + (a + 1)2) · · · (x + (a + n − 1)2)
• = (a + b)n(a + c)n(a + d)n

(a + b + c + d)n so from our first lemma,

∞

• n=0

µntn =

∞

• n=0

(a + b)n(a + c)n(a + d)ntn (a + b + c + d)n n

j=0

• 1 + (a + j)2t
• What about an exponential generating function?

If we replace t with −t2, then the denominator factors into linear factors:

n

• j=0
• 1 − (a + j)2t2

=

• 1 − (a + n)t
• · · ·
• 1 − (a + 1)t
• 1 − at
• ×
• 1 + at
• 1 + (a + 1
• t · · ·
• 1 + (a + n)t
• .

We want to apply our lemma ε

• tm

(1 + At)(1 + (A + 1)t) · · · (1 + (A + m)t)

• = e−At (1 − e−t)m

m! . to t2n

• 1 − (a + n)t
• · · ·
• 1 − (a + 1)t
• 1 − at
• ×
• 1 + at
• 1 + (a + 1
• t · · ·
• 1 + (a + n)t
• .

We want to apply our lemma ε

• tm

(1 + At)(1 + (A + 1)t) · · · (1 + (A + m)t)

• = e−At (1 − e−t)m

m! . to t2n

• 1 − (a + n)t
• · · ·
• 1 − (a + 1)t
• 1 − at
• ×
• 1 + at
• 1 + (a + 1
• t · · ·
• 1 + (a + n)t
• .

We can do this if a = 0 or if a = 1/2. (If a = 1/2, we must multiply by t.) Our conclusion is

For a = 0,

∞

• n=0

µn t2n (2n)! = 3F2

• b, c, d

b + c + d, 1

2

• sin2 t

2

• and for a = 1/2,

∞

• n=0

µn t2n+1 (2n + 1)! = 2 sin t 2 3F2

• b + 1

2, c + 1 2, d + 1 2

b + c + d + 1

2, 3 2

• sin2 t

2

If we take the limit as d → ∞ in the Wilson polynomials Wn(x2) = 4F3 −n, n + a + c + d − 1, a + ix, a − ix a + b, a + c, a + d

• 1
• ,

we get the continuous dual Hahn polynomials pn(x2) = 3F2 −n, a + ix, a − ix a + b, a + c

• 1
• .

The generating function for the moments of the continuous dual Hahn polynomials is

∞

• n=0

µn(a, b, c)tn =

∞

• n=0

(a + b)n(a + c)ntn n

j=0

• 1 + (a + j)2t
• Then it is clear that the moments µn(a, b, c) are polynomials in

a, b, and c, and they have nonnegative coefficients, as is clear from the corresponding continued fraction.

In fact µn(1, 1, 1) = G2n+4 and µn(0, 1, 1) = G2n+2, where the Genocchi numbers are given by x tan 1

2x = ∞ n=2 Gnxn/n!, and

the polynomials µn(a, b, c) are polynomials studied by Dumont and Foata as refinements of the Genocchi numbers.

In fact µn(1, 1, 1) = G2n+4 and µn(0, 1, 1) = G2n+2, where the Genocchi numbers are given by x tan 1

2x = ∞ n=2 Gnxn/n!, and

the polynomials µn(a, b, c) are polynomials studied by Dumont and Foata as refinements of the Genocchi numbers. For a = 0 we have the generating function

∞

• n=0

µn(0, b, c) t2n (2n)! = 2F1 b, c

1 2

• sin2 t

2

• ,

In fact µn(1, 1, 1) = G2n+4 and µn(0, 1, 1) = G2n+2, where the Genocchi numbers are given by x tan 1

2x = ∞ n=2 Gnxn/n!, and

the polynomials µn(a, b, c) are polynomials studied by Dumont and Foata as refinements of the Genocchi numbers. For a = 0 we have the generating function

∞

• n=0

µn(0, b, c) t2n (2n)! = 2F1 b, c

1 2

• sin2 t

2

• ,

and it’s not hard to check that indeed,

2F1

1, 1

1 2

• sin2 t

2

• =

∞

• n=0

Gn+2 xn n! = d2 dx2 x tan x 2 using

2F1

1, 1

1 2

• y2
• =
• 1 − y2 + y sin−1 y

(1 − y2)3/2 .

Another application of the continuous dual Hahn polynomials

We have

∞

• n=0

µn(1

2, −1 2γ, 1 2γ) t2n

(2n)! = 2F1 1

2(1 − γ), 1 2(1 + γ) 3 2

• sin2 t

2

• = sin 1

2γt

γ sin 1

2t

In the case γ = 1/3, the corresponding Hankel determinant counts alternating sign matrices (Colomo and Pronko).

Hankel Determinants and Ternary Tree Numbers

Let an = 1 2n + 1 3n n

• ,

so that an is the number of trees with n vertices. Michael Somos conjectured that det(ai+j)0≤i,j≤n−1 is the number of cyclically symmetric transpose complement plane partitions whose Ferrers diagrams fit in an n × n × n box. This number is known to be

n

• i=1

(3i + 1) (6i)! (2i)! (4i + 1)! (4i)! . Somos had similar conjectures for the Hankel determinants det(ai+j)0≤i,j≤n−1 and det(a(i+j+1)/2)0≤i,j≤n−1, relating them to alternating sign matrices invariant under vertical reflection and alternating sign matrices invariant under both vertical and horizontal reflection.

Let g(x) =

∞

• n=0

1 2n + 1 3n n

• xn =

∞

• n=0

anxn. Then g(x) satisfies g(x) = 1 + xg(x)3. If we compute the continued fraction for g(x), we find empirically that it has a simple formula, which implies the determinant evaluation. I found a (complicated) proof of this continued fraction. Then Guoce Xin found a much nicer proof, using Gauss’s continued fraction.

Let g(x) =

∞

• n=0

1 2n + 1 3n n

• xn =

∞

• n=0

anxn. Then g(x) satisfies g(x) = 1 + xg(x)3. If we compute the continued fraction for g(x), we find empirically that it has a simple formula, which implies the determinant evaluation. I found a (complicated) proof of this continued fraction. Then Guoce Xin found a much nicer proof, using Gauss’s continued fraction. However, the same method had already been used by Ulrich Tamm to evaluate these determinants, before Somos stated his conjectures. (But there is no known combinatorial connection between these Hankel determinants and plane partitions or alternating sign matrices.)

The key fact is that the continued fraction for g(x) is a special case of Gauss’s continued fraction:

2F1

a, b + 1 c + 1

• x

2F1

a, b c

• x
• = S(x; λ1, λ2, . . .),

(1) where λ2n−1 = (a + n − 1)(c − b + n − 1) (c + 2n − 2)(c + 2n − 1) , n = 1, 2, . . . , λ2n = (b + n)(c − a + n) (c + 2n − 1)(c + 2n), n = 1, 2, . . . , and S(x; λ1, λ2, λ3, . . .) denotes the continued fraction S(x; λ1, λ2, λ3, . . .) = 1 1 − λ1x 1 − λ2x 1 − λ3x ... (2)

In fact we have g = 2F1

• 2

3, 4 3; 3 2

• 27

4 x 2F1

• 2

3, 1 3; 1 2

• 27

4 x

• .

and this isn’t too hard to prove; if we set f = g − 1 by Lagrange inversion one can show that the numerator is (1 + f)2/(1 − 2f) and the denominator is (1 + f)/(1 − 2f). Xin and I tried to find all related cases where an instance of Gauss’s continued fraction is a polynomial in f (and thus the coefficients have an explicit formula).

In fact we have g = 2F1

• 2

3, 4 3; 3 2

• 27

4 x 2F1

• 2

3, 1 3; 1 2

• 27

4 x

• .

and this isn’t too hard to prove; if we set f = g − 1 by Lagrange inversion one can show that the numerator is (1 + f)2/(1 − 2f) and the denominator is (1 + f)/(1 − 2f). Xin and I tried to find all related cases where an instance of Gauss’s continued fraction is a polynomial in f (and thus the coefficients have an explicit formula). We found exactly 10, but we couldn’t prove that there aren’t any more.

Five of them are 1 + f = 2F1

• 2

3, 4 3; 3 2

• 27

4 x 2F1

• 2

3, 1 3; 1 2

• 27

4 x

• (1 + f)2 = 2F1
• 4

3, 5 3; 5 2

• 27

4 x 2F1

• 4

3, 2 3; 3 2

• 27

4 x

• (1 + f)(1 + 1

2f) = 2F1

• 5

3, 7 3; 7 2

• 27

4 x 2F1

• 5

3, 4 3; 5 2

• 27

4 x

• (1 + f)(1 − 1

2f) = 2F1

• 5

3, 7 3; 5 2

• 27

4 x 2F1

• 5

3, 4 3; 3 2

• 27

4 x

• (1 + f)(1 + 2

5f) = 2F1

• 2

3, 4 3; 5 2

• 27

4 x 2F1

• 2

3, 1 3; 3 2

• 27

4 x

• Why do they factor?

Five of them are 1 + f = 2F1

• 2

3, 4 3; 3 2

• 27

4 x 2F1

• 2

3, 1 3; 1 2

• 27

4 x

• (1 + f)2 = 2F1
• 4

3, 5 3; 5 2

• 27

4 x 2F1

• 4

3, 2 3; 3 2

• 27

4 x

• (1 + f)(1 + 1

2f) = 2F1

• 5

3, 7 3; 7 2

• 27

4 x 2F1

• 5

3, 4 3; 5 2

• 27

4 x

• (1 + f)(1 − 1

2f) = 2F1

• 5

3, 7 3; 5 2

• 27

4 x 2F1

• 5

3, 4 3; 3 2

• 27

4 x

• (1 + f)(1 + 2

5f) = 2F1

• 2

3, 4 3; 5 2

• 27

4 x 2F1

• 2

3, 1 3; 3 2

• 27

4 x

• Why do they factor? We don’t know.

Five of them are 1 + f = 2F1

• 2

3, 4 3; 3 2

• 27

4 x 2F1

• 2

3, 1 3; 1 2

• 27

4 x

• (1 + f)2 = 2F1
• 4

3, 5 3; 5 2

• 27

4 x 2F1

• 4

3, 2 3; 3 2

• 27

4 x

• (1 + f)(1 + 1

2f) = 2F1

• 5

3, 7 3; 7 2

• 27

4 x 2F1

• 5

3, 4 3; 5 2

• 27

4 x

• (1 + f)(1 − 1

2f) = 2F1

• 5

3, 7 3; 5 2

• 27

4 x 2F1

• 5

3, 4 3; 3 2

• 27

4 x

• (1 + f)(1 + 2

5f) = 2F1

• 2

3, 4 3; 5 2

• 27

4 x 2F1

• 2

3, 1 3; 3 2

• 27

4 x

• Why do they factor? We don’t know.

The other five are the same as these, but with different constant terms.