1 Introduction, reminder, and main result
1.1 Random matrices and matrix-valued stochastic processes
The study of moments of random matrices and of their large size limits was initiated by Wigner and Dyson in their analysis of spectra of heavy nuclei (see [Reference Forrester16]). For unitarily-invariant self-adjoint and unitary matrix models, these limits fit the moments of spectral distributions of self-adjoint and unitary operators in a non commutative probability space endowed with a trace (see, e.g., [Reference Hiai and Petz21]). Famous and widely studied examples include Wigner semi-circle, Marchenko-Pastur, Wachter, and Haar distributions. The third one, named after K. W. Wachter who introduced it more than four decades ago, includes up to a variable change the so-called Kesten–McKay distribution arising in the study of symmetric random walks on free groups [Reference Kesten22] and of large random regular graphs [Reference McKay25]. It also arises as the limiting empirical distribution of matrices drawn from the Jacobi unitary ensemble [Reference Capitaine and Casalis4, Reference Collins7] and an elegant expression of its moments may be found in [Reference Carré, Deneufchatel, Luque and Vivo5].
Dynamical versions of random matrices are, when they exist, matrix-valued stochastic processes. For instance, Gaussian ensembles and Wishart matrices correspond to marginals (fixed time) matrix Brownian motions and Wishart processes respectively. While matrices drawn from the classical compact Lie groups, known also as Dyson circular models, show up in the limit as time goes to infinity of Brownian motions on the aforesaid groups. In particular, the limiting empirical distribution of the Brownian motion on the unitary group was determined in [Reference Biane3] relying on the expansion of the heat kernel in the Schur polynomial basis. Subsequently, traces of powers of Brownian motions in classical Lie group were thoroughly studied in [Reference Lévy24] and [Reference Dahlqvist8] making use of algebraic-combinatorics and representation theory.
1.2 The Hermitian and free Jacobi processes
The dynamical version of the Jacobi unitary ensemble is the so-called complex Hermitian Jacobi process
$(J_t)_{t \geq 0}$
whose marginal distribution converges weakly to it as
$t \rightarrow +\infty $
. This matrix-valued stochastic process, introduced in [Reference Doumerc15], is constructed as the radial part of the upper-left corner
$ C_t $
of a Brownian motion in the unitary group. This means
$J_t $
is the Hermitian positive semi-definite matrix derived from
$ C_t $
by the relation
$ J_t = C_t C_t^* $
(see [Reference Deleaval and Demni9] and references therein). The stochastic properties of its eigenvalues process and of their
$\beta $
-extension were studied in [Reference Demni11], while the low and high temperature behaviors of these particle systems were determined in [Reference Voit32] and [Reference Trinh and Trinh30] respectively.
In the large size limit, the Hermitian Jacobi process converges (when suitably normalized) in the sense of mixed moments to the free Jacobi process introduced in [Reference Demni10]. In particular, at any fixed time
$t> 0$
, the limiting moment sequence of the former uniquely determines the spectral distribution of the latter in a non commutative probability space. Denoted
$\mu _t^{(\lambda , \theta )}$
, this compactly-supported probability measure depends on two parameters
$\lambda \in ]0,1], \theta \in ]0,1[,$
which respectively encodes in the large size limit the rectangular shape of the corner out of which
$J_t$
is built and its magnitude regarding the size of the underlying Brownian motion in the unitary group (see below). To the best of our knowledge, the moments of
$\mu _t^{(\lambda , \theta )}$
are explicit only in the particular case corresponding to
$\lambda = 1$
and
$\theta = 1/2$
. However, the Lebesgue decomposition of
$\mu _t^{(\lambda , \theta )}$
was determined in [Reference Hamdi19]. The corresponding absolutely continuous part is encoded by the inverse (in a compositional sense) of the solution to a radial Loewner equation. Specifically, this inverse extends continuously to the unit circle [Reference Hamdi20]. Unfortunately, in general, the explicit expressions for the moments and the density remains elusive.
Motivated by quantum information theory, particularly by dynamical random quantum states, we introduced in [Reference Demni and Hamdi12] a dynamical random density matrix constructed from the sum of
$k \geq 2$
independent unitary Brownian motions. More precisely, we considered the normalized square of this sum, a random matrix inspired by quantum information theory [Reference Zyczkowski, Penson, Nechita and Collins33], whose large-size limit distribution is related to that of the free Jacobi process in a compressed non-commutative probability space through a combinatorial expansion. We provided expressions for the average purity and the entanglement entropy of the large-size limiting density matrix at a fixed time
$t> 0$
. The moments of
$\mu _t^{(1,1/k)}$
, for
$k \geq 2$
, have been further analyzed in [Reference Demni and Hamdi12] using enumeration techniques. This analysis reveals a striking combinatorial structure governed by the Catalan triangle [Reference Shapiro29],
$$ \begin{align} \binom{2n}{n-h}, \quad 0 \leq h \leq n, \end{align} $$
which is independent of k. However, it involves the moments of a non-normal bounded operator that do not seem to have a simple form unless
$k = 2$
. Besides, unless
$h=0$
, the presence of the above binomial coefficients in relation with the moments of the free Jacobi process remains till now unexplained. As to the binomial coefficient
$$ \begin{align*} \binom{2n}{n}, \end{align*} $$
it arises in the moment sequence of the stationary (arcsine) distribution
$\mu _{\infty }^{(1, 1/2)}$
and counts the number of non crossing partitions of type B [Reference Reiner28].
1.3 Moments of the Hermitian Jacobi process
In order to get more insight into both the combinatorial and analytical structures of the moment sequence of
$\mu _t^{(\lambda , \theta )}$
, in particular into the occurrence of the above binomial coefficients, we pursued a matrix approach in [Reference Demni, Hamdi and Souaissi14] where we proved a formula for the moments
$\mathbb {E}\left (\text {tr}[(J_{t})^n]\right ), n \geq 1$
. The latter, recalled below, is given by a sign alternating double sum over hooks of fixed weight n and those contained in. It may be also expressed through terminating and
$1$
-balanced
${}_4F_3$
hypergeometric functions taken at unity [Reference Bailey2]. However, computing the large size limit is still challenging due to the presence of alternating signs and to the complicated form of the summands as well.
At this stage, let us provide more details on the moment formula derived in [Reference Demni, Hamdi and Souaissi14]. To this end, consider a
$d \times d$
Brownian motion on the unitary group and its upper-left corner of shape
$m \times p$
, where without loss of generality we assume
$p \geq m$
. Then
$J_t$
is the
$m \times m$
radial part of this corner. Now, recall from [Reference Demni, Hamdi and Souaissi14] the parameters
$$ \begin{align*} r:= p-m, \quad q:=d-p, \quad s:= q-m, \end{align*} $$
and denote
$$ \begin{align} M_{n}^{(r,s,m)}(t):= \mathbb{E}\left(\text{tr}[(J_{t/d})^n]\right), \end{align} $$
the time-normalized nth moment of
$J_{t/d}$
. Given a hook
$\alpha = (n-k, 1^k)$
, denote further
$|\alpha | = n$
its weight,
$\alpha _1 = n-k$
its largest component, and
$\check{R} (\alpha ) = k+1$
its length. Then, Theorem 1 in [Reference Demni, Hamdi and Souaissi14] reads:
$$ \begin{align} &M_{n}^{(r,s,m)}(t) \\&= M_n^{(r,s,m)}(\infty) + \sum_{\substack{\alpha \, \, \text{hook} \\ |\alpha| = n, l(\alpha) \leq m}} (-1)^{n-\alpha_1} \sum_{\substack{\tau \subseteq \alpha \\ \tau \neq \emptyset}} \frac{e^{-\nu_\tau(r,s,m) t/d} \, {V}_{\alpha_1, \tau_1}^{r,s,m} \, U_{l(\alpha), l(\tau)}^{r,s,m} }{(r+s+\tau_1+2m-l(\tau)) (\tau_1+l(\tau)-1)}.\nonumber \end{align} $$
In (3), the notation
$\tau \subseteq \alpha $
stands for the containment of the corresponding Young diagrams (so
$\tau $
is a hook as well and
$\tau _1$
is its largest component),
$$ \begin{align} \nu_{\tau}(r,s,m) :=\sum_{i=1}^m\tau_i(\tau_i+r+s+1+2(m-i)), \end{align} $$
$$ \begin{align} {V}_{\alpha_1, \tau_1}^{(r,s,m)} := \frac{(d+2\tau_1-1)\Gamma(d+\tau_1)\Gamma(\alpha_1+m)\Gamma(p+\alpha_1)\Gamma(q+\tau_1)}{(\alpha_1-\tau_1)!\Gamma(d+\alpha_1+\tau_1)\Gamma(p+\tau_1)\Gamma(\tau_1)}, \end{align} $$
$$ \begin{align} &U_{l(\alpha), l(\tau)}^{(r,s,m)} \\&\!\quad:= \frac{(d+1-2l(\tau))\Gamma(d-l(\alpha)-l(\tau)+1)\Gamma(p-l(\tau)+1)}{(l(\alpha)- l(\tau))!(l(\tau)-1)!\Gamma(m-l(\alpha)+1)\Gamma(p-l(\alpha)+1)\Gamma(q-l(\tau)+1)\Gamma(d-l(\tau)+1)},\nonumber \end{align} $$
and
$M_n^{(r,s,m)}(\infty )$
is the nth moment of the Jacobi unitary ensemble. We refer the reader to [Reference Krattenthaler23] for an expression of the moment sequence
$(M_n^{(r,s,m)}(\infty ))_{n \geq 0}$
and for its asymptotic analysis based on terminating
${}_4F_3$
hypergeometric functions taken at unity.
Since any hook
$\tau \subseteq \alpha , \tau \neq \emptyset ,$
is a partition of the form
$$ \begin{align*} \tau := (h-j,1^j), \quad 0 \leq j \leq h-1, \quad 1 \leq h \leq n, \end{align*} $$
then the double sum displayed in the RHS of (3) may be written as ([Reference Demni, Hamdi and Souaissi14, Section 4.2]):
$$ \begin{align} & \sum_{h=1}^n \sum_{j=0}^{h-1}\frac{(-1)^{n-h+j}e^{-\nu_{h,j(d)} t/d}}{h(n-h)!(h-j-1)!j!} \frac{(2m+2h-2j-1)(d-2j-1)\Gamma(p-j)\Gamma(q+h-j)\Gamma(d+h-j)}{(d+h-2j-1)\Gamma(q-j)\Gamma(d-j)} \\ & \nonumber \frac{\Gamma(h-j+m)\Gamma(d-n+h-2j-1)}{\Gamma(m+h-n-j)\Gamma(p+h-n-j)\Gamma(d+2h-2j)} {}_4F_3\left(\begin{matrix} -(n-h), m+h-j, p+h-j, d-n+h-2j-1; \\ m-n+h-j, p-n+h-j, d+2h-2j; \end{matrix}, 1\right), \end{align} $$
where
$$ \begin{align} \nu_{h,j}(d) := \nu_{(h-j,1^j)}(r,s,m) = dh + (h-j)(h-j-1) - j(j+1) = dh +h(h-2j-1). \end{align} $$
Here,
${}_4F_3$
is a terminating and
$1$
-balanced (i.e., the sum of the bottom row parameters equals one plus the sum of the top row parameters) hypergeometric functions taken at unity. Various parameter transformations and reduction formulas exist for these functions [Reference Bailey2]. For instance, generalized hypergeometric series whose row and bottom parameters differ by integers admits simpler representations (see [Reference Miller and Paris27] and references therein). Recently, a reduction formula proved in [Reference Chu6, Theorem 2.1] reduces a
$1$
-balanced terminating
${}_4F_3(1)$
to a terminating (non necessarily balanced)
${}_3F_2(1)$
. In particular, it includes the celebrated Carlitz Summation formula for a special choice of the parameter set [Reference Abiodun1]. As we shall see below, the latter allows to simplify considerably (7) when choosing
$p(m) = m+(1/2), d(m)=2m,$
which in turn opens the way to retrieve the moment sequence of
$\mu _t^{(1,1/2)}$
derived in [Reference Demni, Hamdi and Hmidi13]. Note in passing that though this choice does not correspond to the Hermitian Jacobi process with integer parameters, the eigenvalue process still exists as the unique strong solution of a vector-valued stochastic differential equation with a singular drift ([Reference Graczyk and Malecki18, Corollary 6.7.]).
1.4 Main result
More generally, following another route inspired by the seminal paper [Reference Biane3], we shall prove the following theorem which is the main result of our article:
Theorem 1 For any
$\lambda \in ]0,1]$
and
$\theta \in ]0,1[$
, let
$m=(\lambda \theta ) d$
and
$p = \theta d$
. Then, for any
$ n\geq 1$
, there exist real numbers
$\left (c_{n,h,l}^{(\lambda , \theta )}\right )_{0 \leq l \leq h-1 \leq n-1}$
such that
$$ \begin{align*} &\lim_{d \rightarrow +\infty} \frac{1}{m}\sum_{\substack{\alpha \, \, \text{hook} \\ |\alpha| = n, l(\alpha) \leq m}} \sum_{\substack{\tau \subseteq \alpha \\ \tau \neq \emptyset}} \frac{(-1)^{n-\alpha_1} e^{-\nu_\tau(r,s,m) t/d} \, {V}_{\alpha_1, \tau_1}^{r,s,m} \, U_{l(\alpha), l(\tau)}^{r,s,m} }{(r+s+\tau_1+2m-l(\tau)) (\tau_1+l(\tau)-1)} \\&\quad= \sum_{h=1}^n(-1)^{h-1}\frac{e^{-ht}}{h}\sum_{l=0}^{h-1} \frac{(2ht)^{l}}{l!} c_{n,h,l}^{(\lambda,\theta)}. \end{align*} $$
Since
$$ \begin{align*} \lim_{d \rightarrow \infty} \frac{1}{m} M_n^{(r,s,m)}(\infty), \quad n \geq 1, \end{align*} $$
are the moments of the stationary distribution of the free Jacobi process (see [Reference Demni10, Section 5]), then Theorem 1 provides an expression for the moments of the free Jacobi process at any time
$t> 0$
. As to the description of
$c_{n,h,l}^{(\lambda , \theta )}$
, it is easily read off the end of the proof of Theorem 1. Loosely speaking, the proof of this theorem relies on a polynomial long division with respect to the variable d and only the constant term of the quotient polynomial contributes to the large d-limit. Besides, this term is shown to be polynomial in two variables, denoted
$(j,k)$
in the proof of Theorem 1, and
$c_{n,h,l}^{(\lambda , \theta )}$
is the coefficient of
$j^{h-l-1}k^{n-h}$
there.
Using the generating series of elementary symmetric and completely homogeneous polynomials, we shall represent the constant term of the quotient polynomial alluded to above as a doubly symmetric function in two alphabets. The latter are indeed the roots of the dividend and of the divisor in the polynomial long division. However, though this representation is interesting in its own, we are not able to use it to obtain a relatively simple expression of
$c_{n,h,l}^{(\lambda , \theta )}$
. That is why we shall write down a finite and doubly-indexed recurrence equation whose last term is
$c_{n,h,l}^{(\lambda , \theta )}$
. Doing so leads to simple explicit expressions of both
$c_{n,n,l}^{(\lambda , \theta )}$
and
$c_{n,h,h-1}^{(\lambda , \theta )}$
. We also derive the expression
$$ \begin{align*} c_{n,h,h-1}^{(1,\theta)} = (1-\theta)^n \theta^{n-1} \binom{2n}{n-h}, \end{align*} $$
which shows that the binomial coefficient (1) is not restricted to the value
$\theta =1/2$
.
More generally and provided that
$\lambda \neq 1, \theta = 1/2$
, we realize that the binomial coefficient (1) is (by the virtue of the Chu–Vandermonde formula) the special value at
$\lambda = 1$
of the sum
$$ \begin{align*} \sum_{i=0}^{n-h}\binom{n}{i}\binom{n}{n-h-i}\frac{1}{\lambda^i}, \quad 0 \leq h \leq n. \end{align*} $$
Besides, our doubly-indexed recurrence equation takes a simpler form through binomial coefficients and Gauss hypergeometric functions taken at
$\lambda =1$
.
The article is organized as follows. In the next section, we use Carlitz summation formula to give another derivation of the moments of
$\mu _t^{(1,1/2)}$
. Section 3 is entirely devoted to the proof of Theorem 1. In the Section 4, we prove the representation of the constant term of the quotient polynomial through elementary symmetric and completely homogeneous polynomials. In the Section 5, we derive the expressions of
$c_{n,n,l}^{(\lambda , \theta )}$
and
$c_{n,h,h-1}^{(\lambda , \theta )}$
as well as the recurrence equation allowing to compute
$c_{n,h,l}^{(\lambda , \theta )}$
for arbitrary parameters
$(\lambda , \theta )$
. We also revisit there the particular case corresponding to
$\lambda = 1, \theta =1/2$
. In Section 6, we simplify the recurrence equation as much as possible the recurrence equation for arbitrary
$\lambda $
and
$\theta = 1/2$
.
2 Carlitz summation formula
In [Reference Demni, Hamdi and Hmidi13], we proved the following result.
Theorem 2 ([Reference Demni, Hamdi and Hmidi13, Corollary 1])
The moments of
$\mu _t^{(1,1/2)}$
are given by:
$$ \begin{align} \int_0^1 x^n \mu_t^{(1,1/2)}(dx) = \frac{1}{2^{2n}}\binom{2n}{n} + \frac{1}{2^{2n-1}} \sum_{h=1}^n\binom{2n}{n-h} \frac{e^{-ht}}{h}L_{h-1}^{(1)}(2ht), \quad n \geq 1, \end{align} $$
where
$L_{h-1}^{(1)}$
is the
$(h-1)$
th Laguerre polynomial of index one.
In this section, we supply another derivation of (9) starting from (7) and appealing to Carlitz summation formula [Reference Abiodun1]. Of course, we shall restrict ourselves to the time-dependent part of the RHS of (9) since the stationary one is known to be the nth moment of the arcsine distribution
$$ \begin{align*} \mu_{\infty}^{(1,1/2)} := \frac{1}{\pi\sqrt{x(1-x)}} \mathbf{1}_{[0,1]}(x). \end{align*} $$
Another proof of (9)
Substitute
$p(m) = m+(1/2), d=2m,$
in (7) and recall the Pochhammer symbol:
$$ \begin{align*} (a)_k = \frac{\Gamma(a+k)}{\Gamma(a)}, \, a> 0, \quad (-n)_k = (-1)^k\frac{n!}{(n-k)!}, \, n \in \mathbb{N}. \end{align*} $$
Then Carlitz Summation formula entails (see, e.g., [Reference Abiodun1, eq. (1)]):
$$ \begin{align} & {}_4F_3\left(\begin{matrix} -(n-h), m+h-j, m+h-j+1/2, 2m-n+h-2j-1; \\ m-n+h-j, m-n+h-j+1/2, 2m+2h-2j; \end{matrix}, 1\right) \\[5pt] & \quad =\nonumber \frac{(m-n+h-j-1/2)}{m-j-1/2}\frac{(-2n)_{n-h}}{(2m-2n+2h-2j-1)_{n-h}} \\[5pt] & \quad = (-1)^{n-h} \frac{(m-n+h-j-1/2)}{m-j-1/2} \frac{(2n)!}{(n+h)!}\frac{\Gamma(2m-2n+2h-2j-1)}{\Gamma(2m-n+h-2j-1)}.\nonumber \end{align} $$
Taking into account the time change
$t \mapsto t/d = t/(2m)$
in (2) and keeping in mind (7) and (8), we need to compute the limit as
$m \rightarrow +\infty $
of:
$$ \begin{align*} & \sum_{h=1}^n \frac{e^{-ht}}{h}\binom{2n}{n-h}\sum_{j=0}^{h-1} \frac{(-1)^{j}e^{-h(h-2j-1)t/(2m)}}{(h-j-1)!j!} \frac{(2m+2h-2j-1)(m-n+h-j-1/2)}{m(2m+h-2j-1)} \\[5pt] & \frac{\Gamma(m-j+1/2)\Gamma(m+h-j-1/2)\Gamma(m+h-j)\Gamma(2m-2n+2h-2j-1)\Gamma(2m+h-j)}{(m-j-1/2)\Gamma(m-j-1/2)\Gamma(2m+2h-2j)\Gamma(m+h-n-j+1/2)\Gamma(m+h-n-j)\Gamma(2m-j)}. \end{align*} $$
To this end, we use Legendre duplication formula to get:
$$ \begin{align*} \frac{\Gamma(m+h-j-1/2)\Gamma(m+h-j)}{\Gamma(2m+2h-2j)}& = \frac{\sqrt{\pi}}{2^{2m+2h-2j-1}(m+h-j-1/2)}, \\[5pt] \frac{\Gamma(2m-2n+2h-2j-1)}{\Gamma(m+h-n-j+1/2)\Gamma(m+h-n-j)} & = \frac{2^{2m+2h-2j-2n-2}}{\sqrt{\pi}(m-n+h-j-1/2)}. \end{align*} $$
Consequently, we are led after some simplifications to compute:
$$ \begin{align*} & \lim_{m \rightarrow +\infty}\sum_{h=1}^n \frac{e^{-h(h-1)t/(2m)}}{2^{2n}}\frac{e^{-ht}}{h}\binom{2n}{n-h}\sum_{j=0}^{h-1} \frac{(-1)^{j}e^{jht/m}}{(h-j-1)!j!} \frac{(2m-2j-1)}{m(2m+h-2j-1)} \frac{\Gamma(2m+h-j)}{\Gamma(2m-j)} = \\[5pt] & \lim_{m \rightarrow +\infty}\sum_{h=1}^n \frac{e^{-h(h-1)t/(2m)}}{2^{2n}}e^{-ht}\binom{2n}{n-h}\sum_{j=0}^{h-1}e^{(2ht)j/(2m)} \binom{h-1}{j} \frac{ (-1)^{j}(2m-2j-1)}{m(2m+h-2j-1)} \frac{(2m+h-j-1)!}{h!(2m-j-1)!}. \end{align*} $$
Up to the factor
$(2m-2j-1)/(2m+h-2j-1)$
and to the time change
$t \mapsto 2t$
, the inner sum
$$ \begin{align*} \frac{1}{2m}\sum_{j=0}^{h-1} (-1)^{j}e^{(ht)j/(2m)} \binom{h-1}{j} \frac{(2m+h-j-1)!}{h!(2m-j-1)!}, \end{align*} $$
was considered by Biane in [Reference Biane3] (see the proof of Lemma 3 there) who proved the convergence to
$$ \begin{align*} \frac{1}{h}L_{h-1}^{(1)}(ht). \end{align*} $$
Writing
$$ \begin{align*} \frac{2m-2j-1}{2m+h-2j-1} = 1-\frac{h}{2m+h-2j-1} \end{align*} $$
we get
$$ \begin{align*} &\lim_{m \rightarrow \infty} \frac{e^{-h(h-1)t/(2m)}}{2^{2n}}\sum_{h=1}^n e^{-ht}\binom{2n}{n-h}\sum_{j=0}^{h-1} (-1)^{j}e^{(2ht)j/(2m)} \binom{h-1}{j} \\[5pt]&\qquad \times\frac{2(2m-2j-1)}{(2m)(2m+h-2j-1)} \frac{(2m+h-j-1)!}{h!(2m-j-1)!} \\[5pt] & \quad= \frac{2}{2^{2n}}\sum_{h=1}^n e^{-ht}\binom{2n}{n-h}\frac{1}{h}L_{h-1}^{(1)}(2ht), \end{align*} $$
as desired.
A by-product of the previous proof is given in the following corollary:
Corollary 1 For any
$n \geq 1$
, the nth moment of
$J_t$
corresponding to the parameter set
$$ \begin{align*} p = m+1/2, \quad d= 2m \quad \Leftrightarrow \quad r= 1/2, \quad s=-1/2, \end{align*} $$
reads:
$$ \begin{align*} & M_{n}^{(1/2,-1/2,m)}(t) = \frac{1}{n!}\sum_{h=0}^{n-1}(-1)^h\binom{n-1}{h}\frac{(m-h)_n(3/2 +m-h-1)_n}{(2m-i)_n} + \\ & \quad \sum_{h=1}^n \frac{e^{-h(h-1)t/(2m)}}{2^{2n}}e^{-ht}\binom{2n}{n-h}\sum_{j=0}^{h-1}e^{(2ht)j/(2m)} \binom{h-1}{j} \frac{ (-1)^{j}(2m-2j-1)}{2m+h-2j-1} \frac{(2m+h-j-1)!}{h!(2m-j-1)!}. \end{align*} $$
Proof The stationary distribution of the eigenvalues of the Hermitian Jacobi process is given by the Selberg weight:
$$ \begin{align*} \prod_{i=1}^m \lambda_i^r(1-\lambda_i)^s \prod_{1 \leq i < v \leq m}(\lambda_i - \lambda_v)^2 \mathbf{1}_{\{0 < \lambda_m < \cdots < \lambda_1 < 1\}}, \end{align*} $$
up to a normalizing constant. Its nth moment was computed in [Reference Carré, Deneufchatel, Luque and Vivo5, Corollary II. 3.], and our sought expression follows after substituting in that corollary
$a=3/2, b= 1/2, N = m, k=n$
.
Remark In [Reference Chu6, Theorem 2.1], the following reduction formula was proved for terminating
$1$
-balanced
${}_4F_3$
at unity:
$$ \begin{align*} &{}_4F_3\left(\begin{matrix} -N, a-c+N, (c/2), (c+1)/2; \\ 1+a-e, (e/2), (e+1)/2; \end{matrix}, 1\right) \\&\quad= \frac{(1+a-c-e)_N(e-c)_N}{(1+a-e)_N(e)_N}{}_3F_2\left(\begin{matrix} -N, a-c+N, c; \\ c+e-a-N, e+N; \end{matrix}, 1\right). \end{align*} $$
Specializing it with
$$ \begin{align*} N = n-h, \quad p = p(m) = m + \frac{1}{2}, \quad c = 2(m+h-j), \quad e = 2(m+h-j-n), \end{align*} $$
we get
$$ \begin{align*} & {}_4F_3\left(\begin{matrix} -(n-h), m+h-j, m+h-j+1/2, d-n+h-2j-1; \\ m-n+h-j, m-n+h-j+1/2, d+2h-2j; \end{matrix}, 1\right) \\&\quad= \frac{(d-2m)_{n-h}(-2n)_{n-h}}{(d+2h-2j)_{n-h}(2m+2h-2j-2n)_{n-h}}\\ & \qquad {}_3F_2\left(\begin{matrix} -(n-h), d-n+h-2j-1, 2m+2h-2j; \\ 2m - d +h+1-n, 2m-2j+h-n; \end{matrix}, 1\right). \end{align*} $$
The RHS of the last equation may be further written as ([Reference Chu6, Corollary 2.2]):
$$ \begin{align*} \frac{(-2n)_{n-h}}{(2m+2h-2j-2n)_{n-h}}{}_3F_2\left(\begin{matrix} -(n-h), 2m-d+1, 2m+2h-2j; \\ 2m-2j+h-n, d+2h-2j; \end{matrix}, 1\right), \end{align*} $$
which reduces when
$d=2m$
to Cariltz summation formula (10):
$$ \begin{align*} &\frac{(-2n)_{n-h}}{(2m+2h-2j-2n)_{n-h}}{}_2F_1\left(\begin{matrix} -(n-h), 1, \\ 2m-2j+h-n; \end{matrix}, 1\right) \\&\quad= \frac{(m-n+h-j-1/2)(-2n)_{n-h}}{(m-j-1/2)(2m-2n+2h-2j-1)_{n-h}}, \end{align*} $$
by the virtue of the Gauss hypergeometric Theorem:
$$ \begin{align*} {}_2F_1(-N, b;c, 1) = \frac{(c-b)_{N}}{(c)_N}. \end{align*} $$
However, we do not know whether a simple expression of the above terminating
${}_3F_2$
hypergeometric function exists for
$d \neq 2m$
.
3 The general case: proof of Theorem 1
We now proceed with the general case and prove Theorem 1. The proof is long and for ease of reading, we divide it into three steps:
-
(1) Express the summands of the second term in the RHS of (3) as a rational function in the variable d.
-
(2) Inspired by the proof of Lemma 2 in [Reference Biane3], we expand the exponential factor
$e^{(2thj)/d}$
into powers of
$1/d$
, truncate the obtained series and perform there a polynomial long division. -
(3) Obtain cancellations and determine exactly the terms having non zero contributions in the large d-limit. The expression displayed in Theorem 1 follows after straightforward computations.
3.1 First step
For any
$n \geq 1$
, denote
$$ \begin{align*} \tilde{M}_n^{(r,s,m)}(t) := \frac{1}{m} \sum_{\substack{\alpha \, \, \text{hook} \\ |\alpha| = n, l(\alpha) \leq m}} (-1)^{n-\alpha_1} \sum_{\substack{\tau \subseteq \alpha \\ \tau \neq \emptyset}} \frac{e^{-\nu_\tau(r,s,m) t/d} \, V_{\alpha_1, \tau_1}^{(r,s,m)} \, U_{l(\alpha), l(\tau)}^{(r,s,m)} }{(r+s+\tau_1+2m-l(\tau)) (\tau_1+l(\tau)-1)}, \end{align*} $$
where we recall that
$\nu _\tau (r,s,m)$
,
$V_{\alpha _1, \tau _1}^{(r,s,m)}$
and
$U_{l(\alpha ), l(\tau )}^{(r,s,m)}$
were defined in (4), (5), and (6), respectively. Recall also the parameterization of the hooks
$\tau \subseteq \alpha $
:
$$ \begin{align*} \alpha_1=n-k, \, \tau_1=h-j, \quad l(\alpha)=k+1, \, l(\tau)=j+1. \end{align*} $$
Consequently, we may rewrite
$\tilde {M}_n^{(r,s,m)}(t)$
as
$$ \begin{align*} \tilde{M}_n^{(r,s,m)}(t) &= \frac{1}{m} \sum_{h=1}^n\sum_{j=0}^{h-1}\sum_{k=j}^{j+n-h}(-1)^kF(m,p,d,n,h,j,k)e^{-ht-\frac{ht}{d}(h-(2j+1))}\end{align*} $$
where
$$ \begin{align*} F(m,p,d,n,h,j,k) &= \frac{\Gamma(d+h-j)\Gamma(n-k+m)\Gamma(d-j-k-1)\Gamma(p+n-k)\Gamma(q+h-j)}{(n-k-h+j)!(k-j)!j!\Gamma(d+n-k+h-j)\Gamma(p+h-j)\Gamma(h-j)} \\&\quad \frac{(d+2h-2j-1)(d-2j-1)\Gamma(p-j)}{h(d+h-2j-1)\Gamma(m-k)\Gamma(p-k)\Gamma(q-j)\Gamma(d-j)}. \end{align*} $$
Performing the index changes
$k\mapsto n-h+j-k$
, we further obtain
$$ \begin{align} &\tilde{M}_n^{(r,s,m)}(t) \\&\quad= \sum_{h=1}^n\frac{e^{-ht - th(h-1)/d}}{n!}\binom{n}{h}\sum_{k=0}^{n-h}\sum_{j=0}^{h-1}(-1)^{n-h-k+j}\binom{n-h}{k}\binom{h-1}{j}\tilde{F}(m,p,d,n,h,j,k)e^{\frac{2thj}{d}}\nonumber \end{align} $$
where
$$ \begin{align*} &\tilde{F}(m,p,d,n,h,j,k)\\&\quad=\frac{(d-2j-1)(d+2h-2j-1)(d-j)_h(d-p-j)_h(m+h-n-j+k)_n(p+h-n-j+k)_n}{m(d+h-2j-1)(p-j)_h(d-n+h-2j-1+k)_{n+h+1}}. \end{align*} $$
Write
$m=(\lambda \theta ) d, p = \theta d$
, then
$$ \begin{align*} &\tilde{F}(m,p, d,n,h,j,k)= \tilde{F}((\lambda \theta)d,\theta d,d,n,h,j,k) \\&\quad =\frac{(d-2j-1)(d+2h-2j-1)(d-j)_h((1-\theta)d-j)_h((\lambda \theta)d+h-n-j+k)_n(\theta d+h-n-j+k)_n}{(\lambda \theta)d(d+h-2j-1)(\theta d-j)_h(d-n+h-2j-1+k)_{n+h+1}}. \end{align*} $$
3.2 Second step
Since the Pochhammer symbol
$(a)_k$
has exactly k factors, then a quick inspection shows that
$$ \begin{align} \tilde{F}((\lambda \theta)d,\theta d,d,n,h,j,k) \sim \frac{(1-\theta)^h(\lambda \theta^2)^n}{(\lambda \theta)\theta^h} d^{n-1}, \quad d = d(m) \rightarrow +\infty, \end{align} $$
so that taking the limit as
$m \rightarrow +\infty $
in (11) yields an indeterminate form unless
${n = 1}$
. Note that the situation was similar in [Reference Biane3] when dealing with the moments of the unitary Brownian motion. In that paper, the author succeeded to obtain cancellations making use of the following elementary, yet useful, lemma:
Lemma 1 ([Reference Biane3, Lemma 2])
Let
$0 \leq a \leq b$
be integers. Then
$$ \begin{align*} \sum_{j=0}^b (-1)^j j^b \binom{b}{j} = \left\{\begin{array}{lcr} 0 & \text{if} & a < b, \\ (-1)^a a! & \text{if} & a =b. \end{array} \right. \end{align*} $$
Mimicking the proof of Lemma 3 in [Reference Biane3], we expand the exponential term
$e^{\frac {2thj}{d}}$
in (11):
$$ \begin{align*} \sum_{k=0}^{n-h}(-1)^{n-h-k} \sum_{j=0}^{h-1}(-1)^{j}\binom{h-1}{j}\binom{n-h}{k}\tilde{F}((\lambda \theta) d,\theta d,d,n,h,j,k)\left(\sum_{l=0}^{n-1}\frac{(2thj)^l}{l!d^l}+O(d^{-n})\right). \end{align*} $$
By the virtue of (12), we only need to analyse on the following triple sum:
$$ \begin{align*} \sum_{l=0}^{n-1}\frac{(-2th)^l}{l!}\sum_{k=0}^{n-h}\sum_{j=0}^{h-1} (-1)^{n-h-k+j}\binom{h-1}{j}j^l\binom{n-h}{k}\frac{\tilde{F}((\lambda \theta) d,\theta d,d,n,h,j,k)}{d^l}. \end{align*} $$
More precisely, we shall use Lemma 1 to obtain cancellations in the
$(j,k)$
-double sum. To this end, letFootnote
1
$$ \begin{align*} &P_{n,h}(j,k,d)\\&\quad:=(d-2j-1)(d+2h-2j-1)(d-j)_h((1-\theta)d-j)_h((\lambda \theta) d+h-n-j+k)_n(\theta d +h-n-j+k)_n \end{align*} $$
denote the numerator of
$\tilde {F}((\lambda \theta ) d,\theta d,d,n,h,j,k)/d^l$
, and for fixed
$0 \leq l \leq n-1$
, let
$$ \begin{align*} D_{n,h,l}(j,k,d):= (\lambda \theta) d^{l+1}(d+h-2j-1)(\theta d-j)_h(d-n+h-2j-1+k)_{n+h+1}, \end{align*} $$
be its denominator. Performing polynomial long division with respect to the variable d for fixed
$(n,h,l)$
, there exists a polynomial (quotient)
$Q_{n,h,l}(j,k,d)$
of degree
${n-l-1}$
such that
$$ \begin{align*} P_{n,h}(j,k,d) = D_{n,h,l}(j,k,d)Q_{n,h,l}(j,k,d) + \, \text{Remainder}. \end{align*} $$
Consequently, the asymptotic behaviour of
$\tilde {M}_n^{(r,s,m)}(t)$
as
$d \rightarrow +\infty $
is governed by:
$$ \begin{align} \sum_{l=0}^{n-1} \frac{(2th)^l}{l!} \sum_{j=0}^{h-1}(-1)^{j}\binom{h-1}{j}j^l\sum_{k=0}^{n-h}(-1)^{n-h-k}\binom{n-h}{k}Q_{n,h,l}(j,k,d). \end{align} $$
3.3 Third step
Write
$$ \begin{align} Q_{n,h,l}(j,k,d) := \sum_{i=0}^{n-l-1}q_{n-l-1-i}^{(n,h)}(j,k)d^i, \end{align} $$
and note that its coefficients
$(q_{i}^{(n,h)}(j,k))_{i=0}^{n-l-1}$
do not depend on the index l since the latter only appears in the factor
$d^l$
of
$D_{n,h,l}(j,k,d)$
. The following lemma describes the polynomial structure of these coefficients in the variables
$(j,k)$
.
Lemma 2 For any
$0\le i\le n-l-1$
, the coefficient
$q_i^{(n,h)}(j,k)$
is a polynomial of degree at most i in both variables j and k. Moreover, the coefficient of
$j^s$
there (resp.
$k^s$
) is a polynomial of degree at most
$i-s$
in k (resp. j).
Proof Let
$(a_i^{(n,h)})_{0\le i\le 2n+2h+2}$
and
$(b_i^{(n,h)})_{0\le i\le n+2h+l+3}$
be respectively the coefficients of the polynomials
$P_{n,h}(j,k,\cdot )$
and
$D_{n,h,l}(j,k,\cdot )$
so that:
$$ \begin{align*} P_{n,h}(j,k,d) =\sum_{i=0}^{2n+2h+2}a_{i}^{(n,h)}(j,k)d^i \quad \mathrm{and\ } D_{n,h,l}(j,k,d) =\sum_{i=l+1}^{n+2h+l+3}b^{(n,h)}_{i}(j,k)d^i. \end{align*} $$
Since the degree of the remainder of the polynomial long division is strictly less than B, then
$(q_{i}^{(n,h)})_{i=0}^{n-l-1}$
satisfy the following relations:
$$ \begin{align} q_{i}^{(n,h)}(j,k) = \tilde{a}_{2n+2h+2-i}^{(n,h)}(j,k) - \sum_{v=0}^{i-1}\tilde{b}_{n+2h+l+3-i+v}^{(n,h)}(j,k)q_{v}^{(n,h)}(j,k), \quad 0 \leq i \leq n-l-1 \end{align} $$
where
$$ \begin{align*} \tilde{a}_{2n+2h+2-i}^{(n,h)}(j,k)=\frac{a_{2n+2h+2-i}^{(n,h)}(j,k)}{(\lambda\theta)\theta^h}, \quad \tilde{b}_{n+2h+l+3-i+v}^{(n,h)}(j,k)=\frac{b_{n+2h+l+3-i+v}^{(n,h)}(j,k)}{(\lambda\theta)\theta^h}. \end{align*} $$
Consequently, we readily infer from (15) by induction and from the product forms of
$P_{n,h}(j,k,d)$
and
$D_{n,h,l}(j,k,d)$
that
$q_{i}^{(n,h)}(j,k)$
is a polynomial of degree at most i in both j and k. The last statements follows similarly from (15) by induction and a close (yet straightforward) inspection of the coefficients of
$j^s$
and of
$k^s$
in
$q_{i}^{(n,h)}(j,k)$
.
Remark Later, we shall express the coefficients
$(q_{i}^{(n,h)}(j,k))$
through elementary and completely homogeneous symmetric polynomials in the roots of
$P_{n,h}(j,k,d)$
and of
$D_{n,h}(j,k,d)$
respectively. These expressions may be used to write another proof of the previous lemma.
With the help of Lemmas 1 and 2, we are now able to obtain all the cancellations occurring in (13) and determine the terms having non zero contributions in the large d-limit.
Proposition 1 All the terms in (13) vanish except those corresponding to the
$(k,j)$
-monomials
$k^{n-h}j^{h-1-l}$
in the constant term coefficient
$q_{n-l-1}^{(n,h)}(j,k)$
of the polynomial
$Q_{n,h,l}(j,k,d)$
.
Proof By the virtue of (13) and of (14), we need firstly to show that
$$ \begin{align*} \sum_{j=0}^{h-1}(-1)^{j}\binom{h-1}{j}j^l\sum_{k=0}^{n-h}(-1)^{n-h-k}\binom{n-h}{k}q_{n-l-1-i}^{(n,h)}(j,k) \end{align*} $$
is zero for any
$1\le i\le n-l-1$
. According to Lemma 1 and since
$q_{n-l-1-i}^{(n,h)}$
is a polynomial of degree at most
$n-l-1-i$
in j and k by Lemma 2, we notice that the terms corresponding to
$l> h-1$
in (13) give zero contribution. Indeed,
$$ \begin{align*} n-l-1-i=n-(l+1)-i < n-h-i < n-h \end{align*} $$
so that the inner sum over k is zero for all
$1\le i\le n-l-1$
.
Now consider a j-monomial of degree s in
$q_{n-l-1-i}^{(n,h)}$
. If
$s+l < h-1$
then the sum over j vanishes. Otherwise
$s+l\ge h-1$
and the sum over k is zero for any
$1\le i\le n-l-1$
since
$$ \begin{align*} n-l-1-i-s=n-1-i-(l+s)\le n-1-i-(h-1)=n-h-i < n-h. \end{align*} $$
Secondly, let
$i=0$
and recall from Lemma 2 that the coefficient of
$j^s$
in
$q_{n-l-1}^{(n,h)}$
is a polynomial in k of degree at most
$n-l-1-s$
. Reasoning as above, if
$s+l < h-1$
then the sum over j vanishes while the sum over k does so if
$s+l> h-1$
since
$$ \begin{align*} n-l-1-s=n-1-(l+s) < n-1-(h-1)=n-h. \end{align*} $$
Hence, we are only left with the monomials
$k^{v}j^{h-1-l}, v \leq n-h,$
in
$q_{n-l-1}^{(n,h)}(j,k)$
. But once more, the sum over k vanishes unless
$v=n-h$
, the proposition is proved.
The end of the proof of Theorem 1 goes as follows. The sum (13) reduces to
$$ \begin{align*} \sum_{l=0}^{h-1} \frac{(2th)^l}{l!} \sum_{j=0}^{h-1}(-1)^{j}\binom{h-1}{j}j^l\sum_{k=0}^{n-h}(-1)^{n-h-k}\binom{n-h}{k}c_{n,h,l}^{(\lambda,\theta)}k^{n-h}j^{h-l-1}, \end{align*} $$
where
$c_{n,h,l}^{(\lambda ,\theta )}$
is the coefficient of the monomial
$k^{n-h}j^{h-l-1}$
in
$q_{n-l-1}^{(n,h)}(j,k)$
. Appealing again to Lemma 1, we may simplify further this triple sum to get
$$ \begin{align*} \sum_{l=0}^{h-1} \frac{(2th)^l}{l!} (-1)^{h-1}(h-1)!(n-h)!c_{n,h,l}^{(\lambda,\theta)}, \end{align*} $$
so that
$$ \begin{align*} \lim_{d \rightarrow \infty}\tilde{M}_n^{(r,s,m)}(t) &= \sum_{h=1}^n\frac{e^{-ht}}{n!}\binom{n}{h}(-1)^{h-1}(n-h)!(h-1)!\sum_{l=0}^{h-1} \frac{(2th)^l}{l!} c_{n,h,l}^{(\lambda,\theta)} \\ & = \sum_{h=1}^n(-1)^{h-1}\frac{e^{-ht}}{h}\sum_{l=0}^{h-1} \frac{(2ht)^{l}}{l!} c_{n,h,l}^{(\lambda,\theta)}, \end{align*} $$
as stated in Theorem 1.
4 Representations of the constant term coefficient
$q_{n-l-1}(j,k)$
So far, we proved that the large size asymptotics of the moments of
$J_t$
are governed by the constant term coefficient
$q_{n-l-1}(j,k)$
of the polynomial
$Q_{n,h,l}(j,k,\cdot )$
. Besides, this coefficient is the last term
$i=n-l-1$
of the inductive relation (15). In this section, we derive another representation of
$q_{n-l-1}(j,k)$
and of all the coefficients of
$Q_{n,h,l}(j,k,\cdot )$
through elementary and complete homogeneous symmetric polynomials taken at the roots of
$P_{n,h}(j,k,\cdot )$
and of
$D_{n,h}(j,k,\cdot )$
. Though we believe that this fact is very likely known, we were not able to find any reference to it and as such, we state and prove it below. Using a result of Goulden and Greene [Reference Goulden and Greene17], the obtained representation may be further written as a complete symmetric polynomial in some set of variables (alphabet) formed out of the the roots of
$P_{n,h}(j,k,\cdot )$
and of
$D_{n,h}(j,k,\cdot )$
.
For sake of simplicity, set
$$ \begin{align*} A := 2n+2h+2, \quad B := n+2h+l+3, \quad C := A-B = n-l-1, \end{align*} $$
for fixed
$n,h,l,$
and denote
$(y_1,\dots ,y_A)$
and
$(z_1, \dots , z_B)$
the roots of the polynomials
$P_{n,h}(j,k,\cdot )$
and
$D_{n,h,l}(j,k,\cdot )$
, respectively (which depend of course of
$(n,h,j,k)$
). Then,
Proposition 2 The polynomial
$Q_{n,h,l}(j,k, d)$
admits the following expansion:
$$ \begin{align} Q_{n,h,l}(j,k,d) &= d^{A-B}\frac{a_A}{b_B}\sum_{i=0}^{A-B} \frac{(-1)^i}{d^i}\sum_{v=0}^i(-1)^ve_{i-v}(y_1,\dots, y_A) h_{v}(z_1,\dots, z_B) \nonumber \\& = \frac{a_A}{b_B}\sum_{i=0}^{C}d^i (-1)^{C-i}\sum_{v=0}^{C-i}(-1)^v e_{C-i-v}(y_1,\dots, y_A) h_{v}(z_1,\dots, z_B), \end{align} $$
where
$a_A$
and
$b_B$
are the leading terms
$a_A^{(n,h)}$
and
$b_B^{(n,h)}$
of
$P_{n,h}(j,k,d)$
and
$D_{n,h,l}(j,k,d)$
respectively and
$e_i(y_1,\dots , y_A)$
and
$h_i(z_1,\dots , z_B)$
are the ith elementary and complete homogeneous symmetric polynomials respectively. In particular, the constant term coefficient of
$Q_{n,h,l}(j,k,\cdot )$
is given by
$$ \begin{align} q_C^{(n,h)}(j,k) = (-1)^C \frac{(1-\theta)^h(\lambda \theta^2)^n}{\theta^h(\lambda\theta)} \sum_{v=0}^C (-1)^v e_{C-v}(y_1,\dots, y_A) h_v(z_1,\dots, z_B). \end{align} $$
Proof One obviously has:
$$ \begin{align*} P_{n,h}(j,k,d) = a_Ad^A \prod_{v=1}^A(1-y_v/d) = a^Ad^A\sum_{v=0}^{\infty}(-1)^ve_{v}(y_1,\dots, y_A)\frac{1}{d^v}, \end{align*} $$
where we set
$e_{v}(y_1,\dots , y_A) = 0$
for any
$v> A$
, and (for large enough d):
$$ \begin{align*} \frac{1}{D_{n,h}(j,k,d)} = \frac{1}{b_Bd^{B}}\prod_{v=1}^B(1-z_v/d) = \frac{1}{b_Bd^{B}}\sum_{v=0}^{+\infty} h_{v}(z_1,\dots, z_B)\frac{1}{d^v}. \end{align*} $$
As a result,
$$ \begin{align*} \frac{P_{n,h}(j,k,d)}{D_{n,h}(j,k,d)} = d^{A-B}\frac{a_A}{b_B}\sum_{i=0}^{+\infty} \frac{(-1)^i}{d^i}\sum_{v=0}^i (-1)^ve_{i-v}(y_1,\dots, y_A) h_{v}(z_1,\dots, z_B). \end{align*} $$
But the quotient
$Q_{n,h,l}(j,k,d)$
of the polynomial division in the LHS is the polynomial part of the RHS in the variable d whence the first expression of
$Q_{n,h,l}(j,k,d)$
follows. Performing the index change
$i \mapsto A-B-i=C-i$
, one gets the second one.
Finally, the leading coefficients of
$P_{n,h}(j,k,d)$
and
$D_{n,h}(j,k,d)$
are given by
$$ \begin{align*} a_A = \theta^n (1-\theta)^h(\lambda \theta)^n, \quad b_B = (\lambda \theta) \theta^h, \end{align*} $$
respectively. Extracting the constant term coefficient in (16), the lemma is proved.
Remark Recalling the representation (14)
$$ \begin{align*} Q_{n,h,l}(j,k,d) = \sum_{i=0}^{n-l-1}q_{n-l-1-i}^{(n,h)}(j,k)d^i = \sum_{i=0}^{C}q_{C-i}^{(n,h)}(j,k)d^i, \end{align*} $$
we readily infer from (16) that
$$ \begin{align*} q_{C-i}^{(n,h)}(j,k)= (-1)^{C-i}\sum_{v=0}^{C-i}(-1)^ve_{C-i-v}(y_1,\dots, y_A) h_{v}(z_1,\dots, z_B), \quad 0 \leq i \leq C. \end{align*} $$
On the other hand, the roots
$(y_1, \dots , y_A)$
are given by:
$$ \begin{align} y_{i}= \begin{cases} 2j-2h+1, & i=1\\ 2j+1, & i=2\\ j-i+3, & 3\le i\le h+2\\ (j+h-i+3)/(1-\theta), & h+3\le i\le 2h+2\\ (j-k+n+h-i+3)/(\lambda \theta), & 2h+3\le i\le n+2h+2\\ (j-k+2n+h-i+3)/\theta, & n+2h+3\le i\le 2n+2h+2=A \end{cases}, \end{align} $$
while
$$ \begin{align} z_{i}= \begin{cases} 0, & 1\le i\le l+1\\ 2j-h+1, & i=l+2\\ (j-i+l+3)/\theta, & l+3\le i\le h+l+2\\ 2j-k+n+l-i+4, & h+l+3\le i\le n+2h+l+3 = B. \end{cases}. \end{align} $$
As a matter fact, the simple monomial j is present in all the roots except
$z_i, 1\le i\le l+1$
. Since
$h_v$
and
$e_{C-i-v}$
are homogenous polynomials (in their variables) of orders v and
$C-i-v$
respectively, then (16) shows that
$q_{C-i}^{(n,h)}(j,k)$
is a polynomial of degree at most
$C-i$
in the variable j. As to the simple monomial k, it is present in more than C roots. Consequently, the same conclusion holds regarding the degree of
$q_{C-i}^{(n,h)}(j,k)$
in the variable k and we get another proof of Lemma 2.
The formula (17) is clearly symmetric in each of the alphabets
$\mathcal {Y}:= (y_1, \dots , y_A)$
and
$\mathcal {Z}:= (z_1, \dots , z_B)$
. In the realm of algebraic combinatorics, a function enjoying this property is referred to as doubly symmetric. Moreover,
$q_C$
may be seen as a completely homogeneous function in an another alphabet formed out of
$\mathcal {Y}$
and
$\mathcal {Z}$
. Indeed, appealing again to the homogeneity of
$h_i$
, we write
$$ \begin{align*} q_C &= (-1)^C \frac{(1-\theta)^h(\lambda \theta^2)^n}{\theta^h(\lambda\theta)} \sum_{i=0}^C h_i(-z_1,\dots, -z_B) e_{C-i}(y_1,\dots, y_A) \\[5pt]& = (-1)^C \frac{(1-\theta)^h(\lambda \theta^2)^n}{\theta^h(\lambda\theta)} \sum_{i=0}^C h_i(-z_1,\dots, -z_B, \underbrace{0, \dots, 0}_{C \text{times}}) e_{C-i}(y_1,\dots, y_A). \end{align*} $$
Moreover, Lemma 2.1 and formula (18) in [Reference Goulden and Greene17] entail:
Corollary 2 The constant term coefficient
$q_C$
admits the following representation:
$$ \begin{align*} q_C^{(n,h)}(j,k) = (-1)^C \frac{(1-\theta)^h(\lambda \theta^2)^n}{\theta^h(\lambda\theta)} \sum_{1 \leq \tau_1 \leq \tau_2 \leq \dots \leq \tau_C \leq A} \prod_{i=1}^C(y_{\tau_i + i-1} - z_{\tau_i}), \end{align*} $$
where it is understood that
$y_i = 0, i> A$
and
$z_i = 0, i> B$
.
5 On the coefficient
$c_{n,h,l}^{(\lambda ,\theta )}$
The limiting moment formula stated in Theorem 1 depends only on the coefficient
$c_{n,h,l}^{(\lambda ,\theta )}$
of the monomial
$k^{n-h}j^{h-l-1}$
in the
$(j,k)$
-polynomial
$q_{n-l-1}^{(n,h)}(j,k)$
. The extraction of this coefficient may be obtained from the above representations of
$q_{n-l-1}^{(n,h)}(j,k)$
by successive differentiations with respect to
$(j,k)$
(viewed as real variables). In this section, we shall rather use (15) to compute inductively
$c_{n,h,l}^{(\lambda ,\theta )}$
for any
$(\lambda , \theta )$
. As a warm up, we firstly investigate the two special (univariate) cases corresponding to
$n=h$
and to
$l=h-1$
, for which the coefficients of
$j^{n-l-1}$
and of
$k^{n-l-1}$
in
$q_{n-l-1}^{(n,h)}(j,k)$
are constant (by the virtue of the second statement of Lemma 2). Doing so reveals further the role played by the special values
$\lambda = 1$
and
$\theta = 1/2$
. For instance, we determine explicitly
$c_{n,h,h-1}^{(\lambda ,\theta )}$
whose expression considerably simplifies when
$\lambda =1$
without any restriction on
$\theta $
. The same holds for
$c_{n,n,l}^{(\lambda ,1/2)}$
while
$c_{n,n,l}^{(\lambda ,\theta )}$
is expressed through complete ordinary Bell polynomials.
5.1 The case
$n=h$
For any
$0 \leq i \leq n-l-1$
, let
$X_i^{(n,\lambda ,\theta )}$
denote the coefficient of
$j^i$
in
$q_{i}^{(n,n)}(j,k)$
, so that
$$ \begin{align*} c_{n,n,l}^{(\lambda,\theta)}= X_{n-l-1}^{(n,\lambda,\theta)}. \end{align*} $$
Then
Proposition 3 We have,
$X_0^{(n,\lambda ,\theta )}=(\lambda \theta )^{n-1}(1-\theta )^n$
and for any
$1\le i \le n-l-1$
,
$X_i^{(n,\lambda ,\theta )}$
satisfies
$$ \begin{align} X_i^{(n,\lambda,\theta)}=&\alpha_i -\sum _{v=0}^{i-1} \beta_{i-v}^{(n,\theta)} X_v^{(n,\lambda,\theta)}, \end{align} $$
where
$$ \begin{align*} \alpha_i^{(n,\lambda,\theta)} = (-1)^i\sum _{i_1+i_2+i_3+i_4+i_5=i} \binom{2}{i_1} \binom{n}{i_2} \binom{n}{i_3} \binom{n}{i_4} \binom{n}{i_5}2^{i_1}(\lambda\theta)^{n-1-i_3} (1-\theta)^{n-i_4} \theta^{-i_5} \end{align*} $$
and
$$ \begin{align*} \beta_{i-v}^{(n,\theta)}=(-1)^{i-v} \sum _{i_1+i_2+i_3=i-v} \binom{1}{i_1} \binom{n}{i_2} \binom{2 n+1}{i_3} 2^{i_1+i_3} \theta^{-i_2}, \quad 0 \leq v \leq i-1. \end{align*} $$
Proof From Lemma 2,
$X_0^{(n,\lambda ,\theta )}$
is the highest degree coefficient
$q_0^{(n,h)}$
in
$Q_{n,h,l}(j.k,d)$
and we infer from (15) together with the definition of
$P_{n,h,l}(j, k,d)$
that:
$$ \begin{align*} q_{0}^{(n,h)}(j,k) = \tilde{a}_{2n+2h+2}^{(n,h)}(j,k) = (\lambda\theta)^{n-1}\theta^{n-h}(1-\theta)^n \end{align*} $$
so that
$X_0^{(n,\lambda ,\theta )} = q_{0}^{(n,h)}(j,k) = (\lambda \theta )^{n-1}(1-\theta )^n$
.
Appealing again to (15), we get (20) where
$\alpha _i^{(n,\lambda ,\theta )}$
and
$\beta _{i-v}^{(n,\lambda ,\theta )}, 0 \leq v \leq i-1,$
are the coefficients of
$j^i$
and of
$j^{i-v}$
in
$\tilde {a}_{2n+2h+l+2-i}^{(n,h)}(j,k)$
and in
$\tilde {b}_{n+2h+l+3-i+v}^{(n,h)}(j,k)$
, respectively. Indeed, the first statement of Lemma 2 shows that
$q_v^{(n,h)}(j,k), 1\le v \le i-1,$
is a polynomial of degree at most v in the variable j. Moreover,
$$ \begin{align} &\tilde{a}_{2n+2h+2-i}^{(n,h)}(j,k) \\&\quad= (-1)^i(\lambda\theta)^{n-1}\theta^{n-h}(1-\theta)^h\sum_{1\le j_1< \dots j_i \le 2n+2h+2}y_{j_1}\ldots y_{j_i}, \quad 1 \leq i \leq n-l-1,\nonumber \end{align} $$
and
$$ \begin{align} \tilde{b}_{n+2h+l+3-i+v}^{(n,h)}(j,k)=(-1)^{i-v} \sum_{1\le j_1 < \dots < j_{i-v}\le n+2h+l+3}z_{j_1}\ldots z_{j_{i-v}}, \quad 0 \leq v \leq i-1, \end{align} $$
are elementary symmetric polynomials in the roots
$(y_1,\dots , y_{2n+2h+1})$
and
$(z_1, \dots , z_{n+2h+l+3})$
, respectively. As such, they are polynomials of degrees i and
$i-v$
in each of the variables j and k.
Finally, the expressions of
$\alpha _i$
and of
$\beta _{i-v}$
readily follow from (18) and (19), keeping in mind
$n=h$
. The proposition is proved.
5.2 The case
$\lambda =1, \theta =1/2$
If
$\theta =1/2$
then the generalized Chu–Vandermonde identity entails:
$$ \begin{align*} \beta_{i-v}^{(n,1/2)} & =(-1)^{i-v} \sum _{i_1+i_2+i_3=i-v} \binom{1}{i_1} \binom{n}{i_2} \binom{2 n+1}{i_3} 2^{i_1+i_2+i_3} \\& = (-2)^{i-v} \sum _{i_1+i_2+i_3=i-v} \binom{1}{i_1} \binom{n}{i_2} \binom{2 n+1}{i_3} \\& = (-2)^{i-v}\binom{3n+2}{i-v}. \end{align*} $$
Since
$\beta _0=1$
then the recurrence relation (20) may be then written as:
$$ \begin{align*} \sum _{v=0}^{i} (-2)^{i-v}\binom{3n+2}{i-v} X_v^{(n,\lambda,1/2)} = \alpha_i^{(n,\lambda,1/2)} , \end{align*} $$
and is inverted as:
$$ \begin{align*} X_i^{(n,\lambda,1/2)} = \sum_{v=0}^i \frac{(3n+2)_{i-v}}{(i-v)!}\alpha_i^{(n,\lambda,1/2)} , \quad 0 \leq i \leq n-l-1. \end{align*} $$
In particular,
$$ \begin{align*} X_{n-l-1}^{(n,\lambda,1/2)}= \sum_{v=0}^{n-l-1} \frac{(3n+2)_{n-l-1-v}}{(n-l-1-v)!}\alpha_v^{(n,\lambda,\theta)}. \end{align*} $$
If further
$\lambda = 1$
then
$$ \begin{align*} \alpha_i^{(n,1,1/2)} &= \frac{(-1)^i}{2^{2n-1}}\sum _{i_1+i_2+i_3+i_4+i_5=i} \binom{2}{i_1} \binom{n}{i_2} \binom{n}{i_3} \binom{n}{i_4} \binom{n}{i_5}2^{i_1+i_3+i_4+i_5} \\& = \frac{(-1)^i}{2^{2n-1}}\sum_{i_2=0}^i\binom{n}{i_2} 2^{i-i_2} \sum _{i_1+i_3+i_4+i_5=i-i_2} \binom{2}{i_1} \binom{n}{i_3} \binom{n}{i_4} \binom{n}{i_5} \\& = \frac{(-1)^i}{2^{2n-1}}\sum_{i_2=0}^i\binom{n}{i_2} \binom{3n+2}{i-i_2} 2^{i-i_2}, \end{align*} $$
where the last equality follows again from the generalized Chu–Vandermonde identity. In this case, (20) simplifies further to
$$ \begin{align*} \sum _{v=0}^{i} (-1)^{v}\binom{3n+2}{i-v} 2^{i-v}X_v^{(n,1,1/2)} = \frac{1}{2^{2n-1}}\sum_{v=0}^i\binom{n}{v} \binom{3n+2}{i-i_2} 2^{i-v}. \end{align*} $$
and the uniqueness of its solution (subject to the initial condition
$X_0= 1/2^{2n-1}$
) clearly yields
$$ \begin{align*} X_v^{(n,1,1/2)} = \frac{(-1)^v}{2^{2n-1}} \binom{n}{v}. \end{align*} $$
As a result,
$$ \begin{align*} c_{n,n,l}^{(1,1/2)}= X_{n-l-1}^{(n,1,1/2)} = \frac{(-1)^{n-l-1}}{2^{2n-1}} \binom{n}{n-l-1} = \frac{(-1)^{n-l-1}}{2^{2n-1}} \binom{n}{l+1}, \end{align*} $$
whence we deduce that the
$h=n$
term of the sum displayed in the RHS of Theorem 1 reads
$$ \begin{align*} \frac{1}{2^{2n-1}}\frac{e^{-nt}}{n}\sum_{l=0}^{h-1} \frac{(-2nt)^{l}}{l!}\binom{n}{l+1} = \frac{e^{-nt}}{2^{2n-1}n}L_{n-1}^{(1)}(2nt), \end{align*} $$
which agrees with (9). More generally,
Proposition 4 For all
$0\le l\le n-1$
, we have
$$ \begin{align*} c_{n,n,l}^{(\lambda,\theta)}=\sum_{v=0}^{n-l-1}\alpha_v^{(n,\lambda,\theta)} \gamma_{n-l-1-v}^{(n,\theta)} \end{align*} $$
where
$$ \begin{align*} \gamma_a^{(n,\theta)} =\sum _{\substack{v_1 \geq 0, \dots, v_a \geq 0 \\ v_1+2s_2\ldots+av_a=a}} \frac{(v_1+\cdots + v_a)!}{v_1!\ldots v_a!}(-\beta_1^{(n,\theta)})^{v_1}\ldots (-\beta_a^{(n,\theta)})^{v_a}, \quad 0 \leq a \leq n-l-1. \end{align*} $$
Proof The system of equations (20) may be written as
$$ \begin{align*} \left(\begin{array}{c} \alpha_0^{(n,\lambda,\theta)} \\ \alpha_1^{(n,\lambda,\theta)} \\ \vdots \\ \alpha_{n-l-1}^{(n,\lambda,\theta)} \end{array}\right) = \underbrace{\left(\begin{array}{cccc} 1 & 0 & \dots & 0 \\ \beta_{1}^{(n,\theta)} & 1 & \dots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ \beta_{n-l-1}^{(n,\theta)} &\dots & \dots & 1 \end{array} \right)}_{\mathcal{B}}\left(\begin{array}{c} X_0^{(n,\lambda,\theta)} \\ X_1^{(n,\lambda,\theta)} \\ \vdots \\ X_{n-l-1}^{(n,\lambda,\theta)} \end{array}\right), \end{align*} $$
where
$\mathcal {B}$
is a lower triangular invertible Toeplitz matrix. Appealing to [Reference Merca26, Corollary 1], we obtain the expression of the inverse matrix
$\mathcal {B}^{-1}$
whence the formula for
$c_{n,n,l}^{(\lambda ,\theta )}$
follows.
Remark For any
$1 \leq a \leq n-l-1$
,
$\gamma _a^{(n,\theta )}$
is a complete ordinary Bell polynomial in the variables
$(-\beta _1^{(n,\theta )}, \dots , -\beta _a^{(n,\theta )})$
.
5.3 The case
$l=h-1$
This case seems a bit easier than the previous one and the proof of the following proposition is similar to that of Proposition 20.
Proposition 5 For all
$1\le h\le n$
, we have
$$ \begin{align*} &c_{n,h,h-1}^{(\lambda,\theta)}\\&\quad= (1-\theta)^h(\lambda\theta)^{n-1}(-\theta)^{n-h}\binom{2n}{n-h}\sum_{v=0}^{n-h} \frac{(h-n)_v(-n)_v}{(-2n)_v v!\theta^{v}}{}_2F_1(-v,-n,n-v+1; 1/\lambda), \end{align*} $$
where
${}_2F_1$
is the Gauss hypergeometric function. In particular, for any
$n\ge 1$
, we have
$$ \begin{align*} c_{n,h,h-1}^{(1,\theta)} = (1-\theta)^n \theta^{n-1} \binom{2n}{n-h}. \end{align*} $$
Proof Let
$Y_i^{(n, h, \lambda , \theta )}$
denote the coefficient
$k^i$
in
$q_{i}^{(n,h)}(j,k)$
so that
$c_{n,h,h-1}^{(\lambda ,\theta )}= Y_{n-h}^{(n, h, \lambda , \theta )}$
. Then
$$ \begin{align*} Y_0=(\lambda\theta)^{n-1}\theta^{n-h}(1-\theta)^h \end{align*} $$
and (15) shows that the sequence
$(Y_v^{(n, h, \lambda , \theta )})_{0 \leq v \leq i}$
satisfies
$$ \begin{align} Y_i^{(n, h, \lambda, \theta)} = \alpha_i^{(n, h, \lambda, \theta)}-\sum _{v=0}^{i-1} Y_v^{(n, h, \lambda, \theta)}\beta_{i-v}^{(n, h)}, \quad 0\le i\le n-h. \end{align} $$
where (by the virtue of (18), (19), (21), and (22)):
$$ \begin{align*} \alpha_i^{(n, h, \lambda, \theta)} &= (1-\theta)^h\sum _{v=0}^i \binom{n}{v} \binom{n}{i-v}(\lambda\theta)^{n-1-v} \theta^{n-h-(i-v)} \\& = (1-\theta)^h(\lambda\theta)^{n-1}\theta^{n-h-i}\sum _{v=0}^i \binom{n}{v} \binom{n}{i-v}(\lambda)^{-v} \end{align*} $$
and
$$ \begin{align*} \beta_i^{(n, h)} = \binom{h+n+1}{i}. \end{align*} $$
The recurrence relation (23) may be then written as:
$$ \begin{align*} \sum_{v=0}^i \binom{h+n+1}{i-v}Y_v = (1-\theta)^h(\lambda\theta)^{n-1}\theta^{n-h-i}\sum _{v=0}^i \binom{n}{v} \binom{n}{i-v}(\lambda)^{-v}, \end{align*} $$
and is inverted as follows: for any
$0 \leq i \leq n-h$
, one has:
$$ \begin{align*} Y_{i}^{(n, h, \lambda, \theta)} = (1-\theta)^h(\lambda\theta)^{n-1}\theta^{n-h}\sum_{v=0}^i (-1)^{i-v}\frac{(n+h+1)_{i-v}}{(i-v)!} \theta^{-v} \sum _{a=0}^v \binom{n}{a} \binom{n}{v-a}(\lambda)^{-a}. \end{align*} $$
Consequently,
$$ \begin{align*} &Y_{n-h}^{(n, h, \lambda, \theta)} = \frac{(1-\theta)^h(\lambda\theta)^{n-1}\theta^{n-h}}{(n+h)!}\sum_{v=0}^{n-h} (-1)^{n-h-v}\frac{(2n-v)!}{(n-h-v)!} \theta^{-v} \sum _{a=0}^v \binom{n}{a} \binom{n}{v-a}(\lambda)^{-a} \\&\quad = (1-\theta)^h(\lambda\theta)^{n-1}(-\theta)^{n-h}\binom{2n}{n-h}\sum_{v=0}^{n-h} (-1)^v\frac{(h-n)_v}{(-2n)_v} \frac{\theta^{-v}n!}{v!} \sum _{a=0}^v\binom{n}{a} \frac{v!}{(v-a)!(n-(v-a))!} (\lambda)^{-a} \\&\quad = (1-\theta)^h(\lambda\theta)^{n-1}(-\theta)^{n-h}\binom{2n}{n-h}\sum_{v=0}^{n-h} \frac{(h-n)_v}{(-2n)_v} \frac{\theta^{-v}(-n)_v}{v!}\sum _{a=0}^v\frac{(-n)_a(-v)_a}{a!(n-v+1)_a} (\lambda)^{-a} \\&\quad = (1-\theta)^h(\lambda\theta)^{n-1}(-\theta)^{n-h}\binom{2n}{n-h}\sum_{v=0}^{n-h} \frac{(h-n)_v(-n)_v}{(-2n)_v v!\theta^{v}} {}_2F_1(-v,-n,n-v+1; 1/\lambda). \end{align*} $$
Finally, if
$\lambda = 1$
then
$$ \begin{align*} {}_2F_1(-v,-n,n-v+1; 1) = \frac{(2n-v+1)_v}{(n-v+1)_v} = \frac{(2n!)(n-v)!}{n!(2n-v)!}, \end{align*} $$
by the Gauss hypergeometric Theorem, whence
$$ \begin{align*} Y_{n-h}^{(n, h, 1, \theta)} &= (1-\theta)^h(\theta)^{n-1}(-\theta)^{n-h}\binom{2n}{n-h}\sum_{v=0}^{n-h} \frac{(h-n)_v}{v!\theta^{v}} \\& = (1-\theta)^h(\theta)^{n-1}(-\theta)^{n-h}\binom{2n}{n-h}\left(1-\frac{1}{\theta}\right)^{n-h} \\& = (1-\theta)^n \theta^{n-1} \binom{2n}{n-h}.\\[-40pt] \end{align*} $$
5.4 The general case
Denote
$Z_{a,b}^{(n, h, \lambda , \theta )}$
the coefficient of
$j^ak^b$
in
$q_{a+b}^{(n,h)}(j,k)$
. Recall also that
$c_{n,h,l}^{(\lambda ,\theta )}$
is the coefficient of
$j^{h-l-1}k^{n-h}$
in
$$ \begin{align*} q_{n-l-1}^{(n,h)}(j,k) =q_{(h-l-1)+(n-h)}^{(n,h)}(j,k), \end{align*} $$
in other words,
$$ \begin{align*} c_{n,h,l}^{(\lambda,\theta)}=Z_{h-l-1,n-h}^{(n, h, \lambda, \theta)}. \end{align*} $$
Appealing once more to (15), we prove the following.
Proposition 6
$Z_{0,0}^{(n, h, \lambda , \theta )} = (\lambda \theta )^{n-1}\theta ^{n-h}(1-\theta )^h$
and for any
$0\le a\le h-l-1$
and
$0\le b\le n-h, (a,b) \neq (0,0)$
, one has:
$$ \begin{align} Z_{a,b}^{(n, h, \lambda, \theta)}=A_{a,b}^{(n, h, \lambda, \theta)} -\sum_{\substack{0\le u\le a,\ 0\le v\le b\\0\le u+v\le a+b-1}} Z_{u,v}\ B_{a-u,b-v}^{(n, h, \theta)}, \end{align} $$
where
$$ \begin{align*} &A_{a,b}^{(n, h, \lambda, \theta)} =(-1)^a\sum _{i=0}^b \binom{n}{i} \binom{n}{b-i} \\ &\quad \!\sum _{i_1+i_2+i_3+i_4+i_5=a} \binom{2}{i_1} \binom{h}{i_2} \binom{h}{i_3}\binom{n-i}{i_4}\binom{n -(b-i)}{i_5}2^{i_1} (1-\theta )^{h-i_3} (\theta \lambda )^{n-1-i-i_4} \theta ^{(n-h) - (b-i)-i_5}, \end{align*} $$
and
$$ \begin{align*} B_{a,b}^{(n, h, \theta)}=(-1)^a\binom{n+h+1}{b} \sum _{j_1+j_2+j_3=a} \binom{1}{j_1}\binom{h}{j_2}\binom{n+h+1-b}{j_3} \frac{2^{j_1+j_3}}{\theta^{j_2}}. \end{align*} $$
Proof We follow the lines of the proof of Propositions 3 and (23):
$Z_{0,0}^{(n, h, \lambda , \theta )}$
is the highest degree coefficient
$q_0^{(n,h)}(j,k)$
in
$Q_{n,h,l}(j.k,d)$
and its expression follows readily from (21). As to
$A_{a,b}^{(n, h, \lambda , \theta )}$
and
$B_{a,b}^{(n, h, \theta )}$
are nothing else but the coefficients of
$j^a k^b$
in
$\tilde {a}_{2n+2h+2-a-b}^{(n,h)}$
and
$\tilde {b}_{n+2h+l+3-a-b}^{(n,h)}$
, respectively, viewed as bivariate polynomials in
$(j,k)$
. In this respect, note from (21) and (22) that their total degrees do not exceed
$a+b$
since the roots
$(y_1, \dots , y_{2n+2h+2})$
and
$(z_1, \dots , z_{n+2h+l+3})$
are linear factors in
$(j,k)$
. The double recurrence relation (24) follows then from (15) and it only remains to derive the expressions of
$A_{a,b}^{(n, h, \lambda , \theta )}$
and
$B_{a,b}^{(n, h, \theta )}$
.
To derive the former, we consider the
$2n= n +n$
roots
$$ \begin{align*} y_{i}= \begin{cases} (j-k+n+h-i+3)/(\lambda \theta), & 2h+3\le i\le n+2h+2\\ (j-k+2n+h-i+3)/\theta, & n+2h+3\le i\le 2n+2h+2 \end{cases}, \end{align*} $$
from which we choose b elements. Obviously, there are
$$ \begin{align*} \binom{n}{i} \binom{n}{b-i}, 0 \leq i \leq b, \end{align*} $$
possible choices, each contributing with a factor
$(-1)^bk^b/[(\lambda \theta )^i\theta ^{b-i}]$
. Once a choice is fixed, we choose further a elements from the remaining
$2n+2h+2-b$
roots in (18). The contribution of any choice is of the form
$$ \begin{align*} \frac{2^{i_1} 1^{i_2}}{(1-\theta)^{i_3}(\lambda \theta)^{i_4}\theta^{i_5}}, \end{align*} $$
where
$i_1+i_2+i_3+i_4+i_5 = a$
and
$0 \leq i_1 \leq 2$
and there are
$$ \begin{align*} \binom{2}{i_1} \binom{h}{i_2} \binom{h}{i_3}\binom{n-i}{i_4}\binom{n-(b-i)}{i_5} \end{align*} $$
possible choices. Keeping in mind (21), the expression of
$A_{a,b}^{(n, h, \lambda , \theta )}$
follows. The derivation of the expression of
$B_{a,b}^{(n, h, \theta )}$
is similar, we are done.
Remark Back to the case λ = 1, θ = 1/2
When
$\lambda =1, \theta = 1/2$
, the expressions of
$A_{a,b}^{(n, h, 1, 1/2)}$
and of
$B_{a,b}^{(n, h, 1/2)}$
reduce to:
$$ \begin{align*} A_{a,b}^{(n, h, 1, 1/2)} & = \frac{(-1)^a2^b}{2^{2n-1}} \binom{2n}{b}\sum_{i=0}^a2^{a-i}\binom{h}{i}\binom{2n+h+2-b}{a-i}, \\ B_{a,b}^{(n, h, 1/2)} &= (-1)^a 2^a \binom{n+h+1}{b}\binom{n+2h+2-b}{a}. \end{align*} $$
Consequently, (24) reads:
$$ \begin{align} &\sum_{\substack{0\le u\le a,\ 0\le v\le b\\ u+v\le a+b}} 2^{a-u}(-1)^u\binom{n+h+1}{b-v} \binom{n+2h+2-b+v}{a-u} Z_{u,v}^{(n, h, 1,1/2)} \\ & \nonumber \quad= \frac{2^b}{2^{2n-1}}\binom{2n}{b}\sum_{i=0}^{a}2^{a-i}\binom{h}{i}\binom{2n+h+2-b}{a-i}. \end{align} $$
On the other hand, comparing both (9) and Theorem 1, and using the expansion of the Laguerre polynomial:
$$ \begin{align*} L_{h-1}^{(1)}(2ht) = \sum_{l=0}^{h-1}\frac{(-2ht)^l}{l!}\binom{h}{l+1} = \sum_{l=0}^{h-1}\frac{(-2ht)^l}{l!}\binom{h}{h-l-1}, \end{align*} $$
it follows that
$$ \begin{align} Z_{h-l-1,n-h}^{(n, h, 1,1/2)} = \frac{(-1)^{h-l-1}}{2^{2n-1}} \binom{h}{h-l-1}\binom{2n}{n-h}. \end{align} $$
This result may be proved by considering generation functions of both sides of (25) over
$a \geq 0$
for fixed
$0 \leq b \leq n-h$
. More precisely, set
$$ \begin{align*} \tilde{Z}_{u,v}^{(n, h, 1,1/2)}:= \frac{(-1)^u}{2^{2n-1}} Z_{u,v}^{(n, h, 1,1/2)}, \end{align*} $$
then for any small enough
$T \geq 0$
,
$$ \begin{align*} \sum_{v=0}^b \binom{n+h+1}{b-v} \left(\sum_{u \geq 0} \tilde{Z}_{u,v}^{(n, h, 1,1/2)}T^u\right)(1+2T)^{n+2h+2-b+v} = 2^b\binom{2n}{b}(1+T)^h(1+2T)^{2n+h+2-b}. \end{align*} $$
Equivalently, for any
$0 \leq b \leq n-h, 1 \leq h \leq n$
,
$$ \begin{align*} \sum_{v=0}^b \binom{n+h+1}{b-v} \left(\sum_{u \geq 0} \tilde{Z}_{u,v}^{(n, h, 1,1/2)}T^u\right)(1+2T)^{v} = 2^b\binom{2n}{b}(1+T)^h(1+2T)^{n-h}. \end{align*} $$
Using induction on b, we infer that for any
$$ \begin{align*} \left(\sum_{u \geq 0} \tilde{Z}_{u,v}^{(n, h, 1,1/2)}T^u\right) = \phi_{v}(n,h)(1+T)^{h}(1+2T)^{n-h-v} \end{align*} $$
so that we are only left with the equation
$$ \begin{align*} \sum_{v=0}^b \binom{n+h+1}{b-v} \phi_v(n,h) = 2^b\binom{2n}{b}, \end{align*} $$
whose inverse yields
$$ \begin{align*} \phi_b(n,h) = \sum_{v=0}^b 2^v\binom{2n}{v} \frac{(n+h+1)_{b-v}}{(b-v)!}(-1)^{b-v}, \quad 0 \leq b \leq n-h. \end{align*} $$
Specializing to
$b = n-h$
gives
$$ \begin{align*} \phi_{n-h}(n,h) = \binom{2n}{n-h} \end{align*} $$
and finally
$$ \begin{align*} \left(\sum_{u \geq 0} \tilde{Z}_{u,n-h}^{(n, h, 1,1/2)}T^u\right) = \binom{2n}{n-h}(1+T)^{h}. \end{align*} $$
Consequently,
$$ \begin{align*} \tilde{Z}_{u,n-h}^{(n, h, 1,1/2)} = \binom{2n}{n-h}\binom{h}{u}, \quad 0 \leq u \leq h. \end{align*} $$
Keeping in mind the definition of
$\tilde {Z}_{u,v}^{(n, h, 1,1/2)}$
, we recover (26).
6 Further perspectives:
$\theta = 1/2$
The expressions of
$A_{a,b}^{(n, h, \lambda , \theta )}$
and of
$B_{a,b}^{(n, h, \theta )}$
in Proposition 6 suggest considering the value
$\theta =1/2$
for arbitrary
$\lambda $
. In this respect, we still have:
$$ \begin{align*} B_{a,b}^{(n, h, 1/2)} = (-1)^a 2^a \binom{n+h+1}{b}\binom{n+2h+2-b}{a}, \end{align*} $$
while
$A_{a,b}^{(n, h, \lambda , 1/2)}$
reduces to:
$$ \begin{align*} &A_{a,b}^{(n, h, \lambda, 1/2)} \\&\quad=\frac{(-1)^a \lambda^{n-1}2^b}{2^{2n-1}}\sum _{i=0}^b \binom{n}{i} \binom{n}{b-i} \frac{1}{\lambda^i} \sum _{i_1+i_2+i_3+i_4+i_5=a} \binom{2}{i_1} \binom{h}{i_2} \binom{h}{i_3}\binom{n-i}{i_4}\binom{n -(b-i)}{i_5}\frac{2^{a-i_2}}{\lambda^{i_4}}. \end{align*} $$
We can write this last expression as follows.
Proposition 7 For any
$0 \leq a \leq h-l-1, 0 \leq b \leq n-h$
,
$$ \begin{align*} &A_{a,b}^{(n, h, \lambda, 1/2)} =\frac{(-1)^a \lambda^{n-1}2^b}{2^{2n-1}} \sum_{i_2=0}^{a}\frac{2^{a-i_2}}{\lambda^{a-i_2}} \binom{h}{i_2} \binom{2n+h+2-b}{a-i_2} \\ &\quad \sum _{i=0}^b \binom{n}{i} \binom{n}{b-i} \frac{1}{\lambda^i}{}_2F_1\left(i_2-a, b-i-n-h-2, b-2n-h-2; 1-\lambda\right). \end{align*} $$
Proof We split the inner sum as
$$ \begin{align*} \sum_{i_2+i_4 = 0}^a\binom{h}{i_2} \binom{n-i}{i_4} \frac{2^{a-i_2}}{\lambda^{i_4}} \sum_{i_1+i_3+i_5 = a-i_2-i_4} \binom{2}{i_1} \binom{h}{i_3}\binom{n -(b-i)}{i_5} \end{align*} $$
and use the Chu–Vandermonde identity to get:
$$ \begin{align*} A_{a,b}^{(n, h, \lambda, 1/2)} =\frac{(-1)^a \lambda^{n-1}2^b}{2^{2n-1}}\sum _{i=0}^b \binom{n}{i} \binom{n}{b-i} \frac{1}{\lambda^i} \sum_{i_2+i_4 = 0}^a\binom{h}{i_2} \binom{n-i}{i_4} \binom{n+h+2-(b-i)}{a-i_2-i_4}\frac{2^{a-i_2}}{\lambda^{i_4}}. \end{align*} $$
Next we focus on the sum
$$ \begin{align*} \sum_{i_4 = 0}^{a-i_2}\binom{n-i}{i_4} \binom{n+h+2-(b-i)}{a-i_2-i_4}\frac{1}{\lambda^{i_4}} \end{align*} $$
for fixed
$0 \leq i_2 \leq a, 0 \leq i \leq b$
. Using the identity,
$$ \begin{align*} \binom{N}{j} = (-1)^j \frac{(-N)_j}{j!}, \quad 0 \leq j \leq N, \end{align*} $$
we get
$$ \begin{align*} &\sum_{i_4 = 0}^{a-i_2}\binom{n-i}{i_4} \binom{n+h+2-(b-i)}{a-i_2-i_4}\frac{1}{\lambda^{i_4}} \\&\quad= \binom{n+h+2-(b-i)}{a-i_2}{}_2F_1\left(i_2-a, i-n, n+h+3-(b-i)+i_2-a; \frac{1}{\lambda}\right). \end{align*} $$
Appealing to the argument transformation for terminating Gauss hypergeometric functions (see, e.g., [Reference Vidunas31, eq. (22)]):
$$ \begin{align*} {}_2F_1(-N, b,c; z) = \frac{(c-b)_N}{(c)_N} z^N{}_2F_1 \left(-N, 1-c-N, b-c-N+1; 1-\frac{1}{z}\right), \end{align*} $$
we can further write
$$ \begin{align*} &\sum_{i_4 = 0}^{a-i_2}\binom{n-i}{i_4} \binom{n+h+2-(b-i)}{a-i_2-i_4}\frac{1}{\lambda^{i_4}} = \binom{2n+h+2-b}{a-i_2}\frac{1}{\lambda^{a-i_2}} \\ &\quad {}_2F_1\left(i_2-a, b-i-n-h-2, b-2n-h-2; 1-\lambda\right). \end{align*} $$
As a result,
$$ \begin{align*} &A_{a,b}^{(n, h, \lambda, 1/2)} = \frac{(-1)^a \lambda^{n-1}2^b}{2^{2n-1}}\sum _{i=0}^b \binom{n}{i} \binom{n}{b-i} \frac{1}{\lambda^i} \\ &\quad \sum_{i_2=0}^{a}\frac{2^{a-i_2}}{\lambda^{a-i_2}} \binom{h}{i_2} \binom{2n+h+2-b}{a-i_2}{}_2F_1\left(i_2-a, b-i-n-h-2, b-2n-h-2; 1-\lambda\right) \end{align*} $$
whence the proposition follows after inverting the order of summation.
By the virtue of this proposition, the two-parameters relation (24)
$$ \begin{align*} Z_{a,b}^{(n, h, \lambda, \theta)}=A_{a,b}^{(n, h, \lambda, \theta)} -\sum_{\substack{0\le u\le a,\ 0\le v\le b\\0\le u+v\le a+b-1}} Z_{u,v}^{(n, h, \lambda, \theta)} B_{a-u,b-v}^{(n, h, \theta)}, \end{align*} $$
is given explicitly by
$$ \begin{align} &\sum_{u=0}^a 2^{a-u}(-1)^u\sum_{v=0}^{b}\binom{n+h+1}{b-v} \binom{n+2h+2-b+v}{a-u} Z_{u,v}^{(n, h, \lambda,1/2)}\\&\quad = \frac{\lambda^{n-1}2^b}{2^{2n-1}} \sum_{u=0}^{a}\frac{2^{a-u}}{\lambda^{a-u}} \binom{h}{u} \binom{2n+h+2-b}{a-u}\nonumber \\ & \qquad \sum _{i=0}^b \binom{n}{i} \binom{n}{b-i} \frac{1}{\lambda^i}{}_2F_1\left(u-a, b-i-n-h-2, b-2n-h-2; 1-\lambda\right).\nonumber \end{align} $$
