# Towards Open Domain Event Trigger Identification using Adversarial Domain
Adaptation

Aakanksha Naik
Carnegie Mellon University
<EMAIL_ADDRESS>
&Carolyn Rosé
Carnegie Mellon University
<EMAIL_ADDRESS>
###### Abstract

We tackle the task of building supervised event trigger identification models
which can generalize better across domains. Our work leverages the adversarial
domain adaptation (ADA) framework to introduce domain-invariance. ADA uses
adversarial training to construct representations that _are predictive_ for
trigger identification, but _not predictive_ of the example’s domain. It
requires no labeled data from the target domain, making it completely
unsupervised. Experiments with two domains (English literature and news) show
that ADA leads to an average F1 score improvement of 3.9 on out-of-domain
data. Our best performing model (BERT-A) reaches 44-49 F1 across both domains,
using no labeled target data. Preliminary experiments reveal that finetuning
on 1% labeled data, followed by self-training leads to substantial
improvement, reaching 51.5 and 67.2 F1 on literature and news
respectively.111Our system is available at
https://github.com/aakanksha19/ODETTE

##  1 Introduction

Events are a key semantic phenomenon in natural language understanding. They
embody a basic function of language: the ability to report happenings. Events
are a basic building block for narratives across multiple domains such as news
articles, stories and scientific abstracts, and are important for many
downstream tasks such as question answering Saurí et al. (2005) and
summarization Daniel et al. (2003). Despite their utility, event extraction
remains an onerous task. A major reason for this is that the notion of what
counts as an “event” depends heavily on the domain and task at hand. For
example, should a system which extracts events from doctor notes only focus on
medical events (eg: symptoms, treatments), or also annotate lifestyle events
(eg: dietary changes, exercise habits) which may have bearing on the patient’s
illness? To circumvent this, prior work has mainly focused on annotating
specific categories of events Grishman and Sundheim (1996); Doddington et al.
(2004); Kim et al. (2008) or narratives from specific domains Pustejovsky et
al. (2003); Sims et al. (2019). This has an important implication for
supervised event extractors: they do not generalize to data from a different
domain or containing different event types Keith et al. (2017). Conversely,
event extractors that incorporate syntactic rule-based modules Saurí et al.
(2005); Chambers et al. (2014) tend to overgenerate, labeling most verbs and
nouns as events. Achieving a balance between these extremes will help in
building generalizable event extractors, a crucial problem since annotated
training data may be expensive to obtain for every new domain.

Prior work has explored unsupervised Huang et al. (2016); Yuan et al. (2018),
distantly supervised Keith et al. (2017); Chen et al. (2017); Araki and
Mitamura (2018); Zeng et al. (2018) and semi-supervised approaches Liao and
Grishman (2010); Huang and Riloff (2012); Ferguson et al. (2018), which
largely focus on automatically generating in-domain training data. In our
work, we try to leverage annotated training data from other domains. Motivated
by the hypothesis that events, despite being domain/ task-specific, often
occur in similar contextual patterns, we try to inject lexical domain-
invariance into supervised models, improving generalization, while not
overpredicting events.

Concretely, we focus on event trigger identification, which aims to identify
triggers (words) that instantiate an event. For example, in “John was born in
Sussex”, _born_ is a trigger, invoking a BIRTH event. To introduce domain-
invariance, we adopt the adversarial domain adaptation (ADA) framework Ganin
and Lempitsky (2015) which constructs representations that _are predictive_
for trigger identification, but _not predictive_ of the example’s domain,
using adversarial training. This framework requires no labeled target domain
data, making it completely unsupervised. Our experiments with two domains
(English literature and news) show that ADA makes supervised models more
robust on out-of-domain data, with an average F1 score improvement of 3.9, at
no loss of in-domain performance. Our best performing model (BERT-A) reaches
44-49 F1 across both domains using no labeled data from the target domain.
Further, preliminary experiments demonstrate that finetuning on 1% labeled
data, followed by self-training leads to substantial improvement, reaching
51.5 and 67.2 F1 on literature and news respectively.

##  2 Approaching Open Domain Event Trigger Identification

Throughout this work, we treat the task of event trigger identification as a
token-level classification task. For each token in a sequence, we predict
whether it is an event trigger. To ensure that our trigger identification
model can transfer across domains, we leverage the adversarial domain
adaptation (ADA) framework Ganin and Lempitsky (2015), which has been used in
several NLP tasks Ganin et al. (2016); Li et al. (2017); Liu et al. (2017);
Chen et al. (2018); Shah et al. (2018); Yu et al. (2018).

###  2.1 Adversarial Domain Adaptation

Figure 1 gives an overview of the ADA framework for event trigger
identification. It consists of three components: i) representation learner
($R$) ii) event classifier ($E$) and iii) domain predictor ($D$). The
representation learner generates token-level representations, while the event
classifier and domain predictor use these representations to identify event
triggers and predict the domain to which the sequence belongs. The key idea is
to train the representation learner to generate representations which _are
predictive_ for trigger identification but _not predictive_ for domain
prediction, making it more domain-invariant. A notable benefit here is that
the only data we need from the target domain is unlabeled data.

Figure 1: Adversarial Domain Adaptation Framework for Event Trigger
Identification

To ensure domain-invariant representation learning, ADA uses adversarial
training. Assume that we have a labeled source domain dataset $D^{s}$ with
examples $\\{(x^{s}_{1},e^{s}_{1}),...,(x^{s}_{n},e^{s}_{n})\\}$, where
$x^{s}_{i}$ is the token sequence and $e^{s}_{i}$ is the sequence of event
tags. We construct auxiliary dataset $D^{a}$ with examples
$\\{(x^{a}_{1},d^{a}_{1}),...,(x^{a}_{n},d^{a}_{n})\\}$, where $x^{a}_{i}$ is
the token sequence and $d^{a}_{i}$ is the domain label, using token sequences
from $D^{s}$ and unlabeled target domain sentences. The representation learner
$R$ maps a token sequence $x_{i}=(x_{i1},...,x_{ik})$ into token
representations $h_{i}=(h_{i1},...,h_{ik})$. The event classifier $E$ maps
representations $h_{i}=(h_{i1},...,h_{ik})$ to event tags
$e_{i}=(e_{i1},...,e_{ik})$. The domain predictor $D$ creates a pooled
representation $p_{i}=Pool(h_{i1},...,h_{ik})$ and maps it to domain label
$d^{a}_{i}$. Given this setup, we apply an alternating optimization procedure.
In the first step, we train the domain predictor using $D^{a}$, to optimize
the following loss:

$\displaystyle\operatorname*{arg\,min}_{D}\mathcal{L}(D(h^{a}_{i}),d^{a}_{i})$

In the second step, we train the representation learner and event classifier
using $D^{s}$ to optimize the following loss:

$\displaystyle\operatorname*{arg\,min}_{R,E}\Bigg{[}\sum_{k}\Big{(}\mathcal{L}(E(h^{s}_{ik}),e^{s}_{ik})\Big{)}-\lambda\mathcal{L}(D(h^{s}_{i}),d^{s}_{i})\Bigg{]}$

$\mathcal{L}$ refers to the cross-entropy loss and $\lambda$ is a
hyperparameter. In practice, the optimization in the above equation is
performed using a gradient reversal layer (GRL) Ganin and Lempitsky (2015). A
GRL works as follows. During the forward pass, it acts as the identity, but
during the backward pass it scales the gradients flowing through by
$-\lambda$. We apply a GRL $g_{\lambda}$ before mapping the pooled
representation to a domain label using $D$. This changes the optimization to:

$\displaystyle\operatorname*{arg\,min}_{R,E}\Bigg{[}\mathcal{L}(D(g_{\lambda}(p^{s}_{i})),d^{s}_{i})+\sum_{k}\mathcal{L}(E(h^{s}_{ik}),e^{s}_{ik})\Bigg{]}$

In our setup, the event classifier and domain predictors are MLP classifiers.
For the representation learner, we experiment with several architectures.

Statistic | LitBank | TimeBank
---|---|---
#Docs | 100 | 183
#Tokens | 210,532 | 80,281
#Events | 7849 | 8103
Event Density | 3.73% | 10.10%
Table 1: Dataset Statistics

###  2.2 Representation Learner Models

We experiment with the following models:222Complete implementation details in
the appendix
LSTM: A unidirectional LSTM over tokens represented using word embeddings.
BiLSTM: A bidirectional LSTM over word embeddings to incorporate both left and
right context.
POS: A BiLSTM over token representations constructed by concatenating word
embeddings with embeddings corresponding to part-of-speech tags. This model
explicitly introduces syntax.
BERT: A BiLSTM over contextual token representations extracted using BERT
Devlin et al. (2019), similar to the best-performing model on LitBank,
reported by Sims et al. (2019).

Model | In-Domain | Out-of-Domain
---|---|---
| P | R | F1 | P | R | F1
LSTM | 61.9 | 61.5 | 61.7 | 86.1 | 17.1 | 28.5
LSTM-A | 61.1 | 61.6 | 61.3 | 89.0 | 18.9 | 31.2
BiLSTM | 64.5 | 61.7 | 63.1 | 91.8 | 14.4 | 24.9
BiLSTM-A | 66.1 | 62.8 | 64.4 | 92.9 | 18.5 | 30.9
POS | 74.1 | 51.9 | 61.1 | 93.5 | 9.6 | 17.4
POS-A | 69.6 | 57.7 | 63.1 | 92.5 | 15.2 | 26.1
BERT | 73.5 | 72.7 | 73.1 | 88.1 | 28.2 | 42.7
BERT-A | 71.9 | 71.3 | 71.6 | 85.0 | 35.0 | 49.6
Table 2: Model performance on domain transfer experiments from LitBank to
TimeBank. Presence of the -A suffix indicates that the model uses adversarial
training.

##  3 Experiments

###  3.1 Datasets

In our experiments, we use the following datasets:333Unlike prior work, we
cannot use the ACE-2005 dataset since it tags specific categories of events,
whereas we focus on tagging _all_ possible events.

  * •

LitBank Sims et al. (2019): 100 English literary texts with entity and event
annotations.

  * •

TimeBank Pustejovsky et al. (2003): 183 English news articles containing
annotations for events and temporal relations between them.

Both datasets follow similar guidelines for event annotation, with an
important distinction: LitBank does not annotate events which have not
occurred (eg: future, hypothetical or negated events). To overcome this gap,
we remove all such events from TimeBank using available metadata about event
modality and tense. Table 1 provides a brief overview of statistics for both
datasets.

Model | In-Domain | Out-of-Domain
---|---|---
| P | R | F1 | P | R | F1
LSTM | 70.7 | 78.4 | 74.4 | 23.5 | 75.2 | 35.8
LSTM-A | 69.3 | 87.5 | 77.3 | 25.6 | 72.9 | 37.9
BiLSTM | 75.4 | 76.3 | 75.9 | 27.6 | 68.8 | 39.4
BiLSTM-A | 74.2 | 79.4 | 76.7 | 26.3 | 72.0 | 38.6
POS | 77.4 | 81.1 | 79.2 | 26.4 | 79.8 | 39.6
POS-A | 76.4 | 83.0 | 79.6 | 27.3 | 81.9 | 40.9
BERT | 79.6 | 84.3 | 81.9 | 28.1 | 84.8 | 42.2
BERT-A | 79.8 | 85.6 | 82.6 | 30.3 | 80.8 | 44.1
Table 3: Model performance on domain transfer experiments from TimeBank to
LitBank. Presence of the -A suffix indicates that the model uses adversarial
training.

###  3.2 Results and Analysis

Tables 2 and 3 present the results of our experiments. Table 2 shows the
results when transferring from LitBank to TimeBank while Table 3 presents
transfer results in the other direction. From Table 2 (transfer from LitBank
to TimeBank), we see that ADA improves out-of-domain performance for all
models, by 6.08 F1 on average. BERT-A performs best, reaching an F1 score of
49.6, using no labeled news data. Transfer experiments from TimeBank to
LitBank (Table 3) showcase similar trends, with only BiLSTM not showing
improvement with ADA. For other models, ADA results in an average out-of-
domain F1 score improvement of 1.77. BERT-A performs best, reaching an F1
score of 44.1. We also note that models transferred from LitBank to TimeBank
have high precision, while models transferred in the other direction have high
recall. We believe this difference stems from the disparity in event density
across corpora (Table 1). Since event density in LitBank is much lower, models
transferred from LitBank tend to be slightly conservative (high precision),
while models transferred from TimeBank are less so (high recall).

Category | % |  Example
---|---|---
TimeBank Improvements
Finance | 54 |  the accord was unanimously approved
Political | 12 |  the ukrainian parliament has already ratified it
Reporting | 10 |  from member station kqed , auncil martinez reports
Law | 10 |  mr. antar was charged last month in a civil suit
LitBank Improvements
Archaic | 6 |  his countenance became intolerably fervid
Animal Actions | 6 |  the dogs left off barking , and ran about every way
Human Actions | 18 |  a nod was the answer
Literary | 14 |  there strikes the ebony clock
Table 4: Categorization of TimeBank and LitBank examples on which ADA shows
improvement. Words in bold indicate events missed by BERT, but captured by
BERT-A.

When transferring from LitBank to TimeBank, LSTM generalizes better than
BiLSTM, which may be because BiLSTM has twice as many parameters making it
more prone to overfitting. ADA gives a higher F1 boost with BiLSTM, indicating
that it may be acting as a regularizer. Another interesting result is the poor
performance of POS when transferring from LitBank to TimeBank. This might stem
from the Stanford CoreNLP tagger (trained on news data) producing inaccurate
tags for LitBank. Hence using automatically generated POS tags while training
on LitBank does not produce reliable POS embeddings.

On average, ADA makes supervised models more robust on out-of-domain data,
with an average F1 score improvement of 3.9, at no loss of in-domain
performance.

What cases does ADA improve on? To gain more insight into the improvements
observed on using ADA, we perform a manual analysis of out-of-domain examples
that BERT labels incorrectly, but BERT-A gets right. We carry out this
analysis on 50 examples from TimeBank and LitBank each. We observe that an
overwhelming number of cases from TimeBank use vocabulary in contexts unique
to news (43/50 or 86%). This includes examples of financial events, political
events and reporting events that are rarer in literature, indicating that ADA
manages to reduce event extraction models’ reliance on lexical features. We
make similar observations for LitBank though the proportion of improvement
cases with literature-specific vocabulary is more modest (22/50 or 44%). These
cases include examples with archaic vocabulary, words that have a different
meaning in literary contexts and human/ animal actions, which are not common
in news. Table 4 presents a detailed breakdown of these cases, along with
examples.444This table does not include generic improvement cases (i.e. no
domain-specific vocabulary used), which formed 14% and 56% of improvement
cases in TimeBank and LitBank.

Figure 2: Improvement in model performance when finetuning on labeled training
data from TimeBank Figure 3: Improvement in model performance when finetuning
on labeled training data from LitBank

##  4 Incorporating Minimal Labeled Data

Finetuning on labeled data: We run finetuning experiments to study improvement
in model performance on incorporating small amounts of labeled target domain
data. For both domains, we finetune BERT-A, slowly increasing the percentage
of labeled data used from 1%-5%.555We run these experiments 5 times with
different random subsets and average performance across all runs. We compare
BERT-A with two other models. The first model is naive BERT with no domain
adaptation (BERT-NoDA). The second model is a BERT model trained via
supervised domain adaptation (BERT-FEDA), which we use as an indicator of
ceiling performance. The supervised domain adaptation method we use is the
neural modification of frustratingly easy domain adaptation developed in Kim
et al. (2016). Frustratingly easy domain adaptation Daumé III (2007) uses a
feature augmentation strategy to improve performance when annotated data from
both source and target domains is available. This algorithm simply duplicates
input features 3 times, creating a source-specific, target-specific and
general version of each feature. For source data, only the source-specific and
general features are active, while only the target-specific and general
features are active for target data. The neural modification works by
duplicating the feature extractor module, which is the BiLSTM in our case.

Figures 2 and 3 present the results of these experiments. Performance of all
models steadily improves with more data, but BERT-A starts with a much higher
F1 score than BERT-NoDA, demonstrating that ADA boosts performance when little
annotated training data is available. Performance increase of BERT-NoDA is
suprisingly rapid, especially on LitBank. However, it is worth noting that 5%
of the LitBank training set is $\sim$10,000 tokens, which is a substantial
amount to annotate. Therefore, BERT-A beats BERT-NoDA on sample efficiency. We
can also see that BERT-A does not do much worse than BERT-FEDA, which performs
supervised adaptation.

Using BERT-A to provide weak supervision: We run further experiments to
determine whether finetuned BERT-A can be leveraged for self-training Yarowsky
(1995); Riloff and Wiebe (2003). Self-training creates a teacher model from
labeled data, which is then used to label a large amount of unlabeled data.
Both labeled and unlabeled datasets are jointly used to train a student model.
Algorithm 1 gives a quick overview of our self-training procedure. We use 1%
of the training data as $\mathcal{D}^{l}$, with the remaining 99% used as
$\mathcal{D}^{u}$. BERT-A acts as $\mathcal{T}$, while $\mathcal{S}$ is a
vanilla BERT model. Table 5 shows the results of self-training on both
domains. Self-training improves model performance by nearly 7 F1 points on
average. Increase on TimeBank is much higher which may be due to the high
precision-low recall tendency of the teacher model.

Algorithm 1 SelfTrain($\mathcal{D}^{l},\mathcal{D}^{u},\mathcal{T}$)

1:Teacher Model $\mathcal{T}$, Labeled Data
$\mathcal{D}^{l}=\\{(x^{l}_{1},e^{l}_{1}),...,(x^{l}_{m},x^{l}_{m})\\}$,
Unlabeled Data $\mathcal{D}^{u}=\\{x^{u}_{1},...x^{u}_{n}\\}$,

2:Trained Student Model $\mathcal{S}$

3:Finetune the teacher model $\mathcal{T}$ by minimizing cross-entropy loss on
labeled data
$\displaystyle\frac{1}{m}\sum_{i=1}^{m}\mathcal{L}(\mathcal{T}(x^{l}_{i}),e^{l}_{i})$

4:Generate labels $\\{e^{u}_{1},...,e^{u}_{n}\\}$ for unlabeled data
$\mathcal{D}^{u}$ using $\mathcal{T}$

5:Train a student model $\mathcal{S}$ by minimizing cross-entropy loss on both
datasets $\mathcal{D}^{l},\mathcal{D}^{u}$
$\displaystyle\frac{1}{m}\sum_{i=1}^{m}\mathcal{L}(\mathcal{S}(x^{l}_{i}),e^{l}_{i})+\frac{1}{n}\sum_{i=1}^{n}\mathcal{L}(\mathcal{S}(x^{u}_{i}),e^{u}_{i})$

6:Iterative training: Repeat step 2 using updated student model $\mathcal{S}$

Dataset | P | R | F1
---|---|---|---
TimeBank | 68.9 | 65.5 | 67.2
LitBank | 40.3 | 71.5 | 51.5
Table 5: Model performance on both domains in the self-training paradigm

##  5 Conclusion

In this work, we tackled the task of building generalizable supervised event
trigger identification models using adversarial domain adaptation (ADA) to
introduce domain-invariance. Our experiments with two domains (English
literature and news) showed that ADA made supervised models more robust on
out-of-domain data, with an average F1 score improvement of 3.9. Our best
performing model (BERT-A) was able to reach 44-49 F1 across both domains using
no labeled target domain data. Preliminary experiments showed that finetuning
BERT-A on 1% labeled data, followed by self-training led to substantial
improvement, reaching 51.5 and 67.2 F1 on literature and news respectively.
While these results are encouraging, we are yet to match supervised in-domain
model performance. Future directions to explore include incorporating noise-
robust training procedures Goldberger and Ben-Reuven (2017) and example
weighting Dehghani et al. (2018) during self-training, and exploring lexical
alignment methods from literature on learning cross-lingual embeddings.

## Acknowledgements

This work was supported by the University of Pittsburgh Medical Center (UPMC)
and Abridge AI Inc through the Center for Machine Learning and Health at
Carnegie Mellon University. The authors would like to thank the anonymous
reviewers for their helpful feedback on this work.

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  * Yuan et al. (2018) Quan Yuan, Xiang Ren, Wenqi He, Chao Zhang, Xinhe Geng, Lifu Huang, Heng Ji, Chin-Yew Lin, and Jiawei Han. 2018.  Open-schema event profiling for massive news corpora.  In _Proceedings of the 27th ACM International Conference on Information and Knowledge Management, CIKM 2018, Torino, Italy, October 22-26, 2018_ , pages 587–596. ACM.
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## Appendix

##  Appendix A Implementation Details

All models are implemented in PyTorch. We use 300-dimensional GloVe embeddings
while training on TimeBank and 100-dimensional Word2Vec embeddings trained on
Project Gutenberg texts (similar to Sims et al. (2019)) while training on
LitBank. Both source and target domains share a common vocabulary and
embedding layer which is not finetuned during the training process. All LSTM
models use a hidden size of 100, with an input dropout of 0.5. The POS model
uses 50-dimensional embeddings for POS tags which are randomly initialized and
finetuned during training. The BERT model uses the uncased variant of BERT-
Base for feature extraction. We generate token representations by running
BERT-Base and concatenating the outputs of the model’s last 4 hidden layers.
The event classifier is a single-layer 100-dimensional MLP. For the
adversarial training setup, we experiment with values from
[0.1,0.2,0.5,1.0,2.0,5.0] for the hyperparameter $\lambda$. The domain
predictor (adversary) is a 3-layer MLP with each layer having a dimensionality
of 100 and ReLU activations between layers. We train all models with a batch
size of 16 and use the Adam optimizer with default learning rate settings.
Models are trained for 1000 epochs, with early stopping. For finetuning
experiments, we train for 10 epochs.


# Recurrence equations and their classical orthogonal polynomial solutions on
a quadratic or $q$-quadratic lattice

Daniel Duviol Tcheutia<EMAIL_ADDRESS>Institute of Mathematics,
University of Kassel, Heinrich-Plett Str. 40, 34132 Kassel, Germany

###### Abstract

Every classical orthogonal polynomial system $p_{n}(x)$ satisfies a three-term
recurrence relation of the type

$p_{n+1}(x)=(A_{n}x+B_{n})p_{n}(x)-C_{n}p_{n-1}(x)~{}(n=0,1,2,\ldots,p_{-1}\equiv
0),$

with $C_{n}A_{n}A_{n-1}>0$. Moreover, Favard’s theorem states that the
converse is true. A general method to derive the coefficients $A_{n}$,
$B_{n}$, $C_{n}$ in terms of the polynomial coefficients of the divided-
difference equations satisfied by orthogonal polynomials on a quadratic or
$q$-quadratic lattice is recalled. The Maple implementations rec2ortho of
Koorwinder and Swarttouw or retode of Koepf and Schmersau were developed to
identify classical orthogonal polynomials given by their three-term recurrence
relation as special functions. The two implementations rec2ortho and retode do
not handle classical orthogonal polynomials on a quadratic or $q$-quadratic
lattice. In this manuscript, the Maple implementation retode of Koepf and
Schmersau is extended to cover classical orthogonal polynomials on quadratic
or $q$-quadratic lattices and to answer as application an open problem
submitted by Alhaidari during the 14th International Symposium on Orthogonal
Polynomials, Special Functions and Applications.

###### keywords:

Computer algebra , Maple , Three-term recurrence equation , Divided-difference
equation.

###### MSC:

[2010]
33C45 , 33D45 , 33F99 , 42C05

††journal:

##  1 Introduction

Foupouagnigni showed in [4] that classical orthogonal polynomials on a
quadratic or $q$-quadratic lattice satisfy a second-order divided difference
equation of the form

$\phi(x(s))\mathbb{D}_{x}^{2}p_{n}(x(s))+\psi(x(s))\mathbb{S}_{x}\mathbb{D}_{x}p_{n}(x(s))+\lambda_{n}p_{n}(x(s))=0,$
(1)

where $\phi(x)=ax^{2}+bx+c$, $\psi(x)=dx+e~{}(d\neq 0)$, are polynomials of
degree at most 2 and of degree one, respectively, the operators
$\mathbb{D}_{x}$ and $\mathbb{S}_{x}$ are given by

$\mathbb{D}_{x}f(x(s))=\frac{f(x(s+\frac{1}{2}))-f(x(s-\frac{1}{2}))}{x(s+\frac{1}{2})-x(s-\frac{1}{2})},\quad\mathbb{S}_{x}f(x(s))=\frac{f(x(s+\frac{1}{2}))+f(x(s-\frac{1}{2}))}{2},$

and $x(s)$ is a quadratic or $q$-quadratic lattice defined by [11]

$x(s)=\left\\{\begin{array}[]{lll}c_{1}q^{s}+c_{2}q^{-s}+c_{3}&\textrm{if
}&0<q<1,\\\ c_{4}s^{2}+c_{5}s+c_{6}&\textrm{if
}&q=1,\end{array}\right.c_{1},\ldots,c_{6}\in\mathbb{C}.$

Note that (1) is equivalent to a difference or $q$-difference equation of the
form (see [9, chaps. 9, 14])

$\lambda_{n}y(x(s))=B(s)y(x(s+1))-(B(s)+D(s))y(x(s))+D(s)y(x(s-1)),$ (2)

with

$\phi(x(s))=-\frac{1}{2}\Big{(}x(s+\frac{1}{2})-x(s-\frac{1}{2})\Big{)}\Big{(}(x(s+1)-x(s))B(s)+(x(s)-x(s-1))D(s)\Big{)},$
$\psi(x(s))=(x(s)-x(s+1))B(s)+(x(s)-x(s-1))D(s).$

Following the work by Foupouagnigni [4], Njionou Sadjang et al. [13] proved
that the Wilson and the continuous dual Hahn polynomials are solutions of a
divided-difference equation of the form

$\phi(x)\mathbf{D}_{x}^{2}p_{n}(x)+\psi(x)\mathbf{S}_{x}\mathbf{D}_{x}p_{n}(x)+\lambda_{n}p_{n}(x)=0,$
(3)

where the operators $\mathbf{S}_{x}$ and the Wilson operator (see [3], [6])
$\mathbf{D}_{x}$ are defined by

$\mathbf{D}_{x}f(x)=\frac{f\Big{(}x+\frac{i}{2}\Big{)}-f\Big{(}x-\frac{i}{2}\Big{)}}{2ix},\quad\mathbf{S}_{x}f(x)=\frac{f\Big{(}x+\frac{i}{2}\Big{)}+f\Big{(}x-\frac{i}{2}\Big{)}}{2}.$

Using the same approach, Tcheutia et al. [17] derived a divided–difference
equation of type

$\phi(x)\mathbf{\delta}_{x}^{2}y(x)+\psi(x)\mathbf{S}_{x}\mathbf{\delta}_{x}y(x)+\lambda_{n}y(x)=0,$
(4)

satisfied by the continuous Hahn and the Meixner–Pollaczek polynomials, where
the difference operator $\mathbf{\delta}_{x}$ (see [15, p. 436], compare [9,
p. 201 and 214], [12, 14], [18, Equation (1.15)]) is defined as follows:

$\mathbf{\delta}_{x}f(x)=\frac{f\Big{(}x+\frac{i}{2}\Big{)}-f\Big{(}x-\frac{i}{2}\Big{)}}{i}.$

(3) and (4) are equivalent to the difference equation (see [9, chap. 9])

$\lambda_{n}y(x)=B(x)y(x+i)-(B(x)+D(x))y(x)+D(x)y(x-i),$ (5)

with

$\phi(x)=x((2x+i)B(x)+(2x-i)D(x)),\,\psi(x)=-i((2x+i)B(x)-(2x-i)D(x)),$

and

$\phi(x)=\frac{1}{2}(B(x)+D(x)),\,\psi(x)=-i(B(x)-D(x)),$

respectively.

The coefficients of the divided-difference equations given in the forms (1),
(3) or (4) can be used for instance to compute the three–term recurrence
relation or some structure formulae, the inversion coefficients of classical
orthogonal polynomials on a quadratic and $q$-quadratic lattice (see e. g.
[5], [13], [16], [17] and references therein).

Every classical orthogonal polynomial system $p_{n}(x)$ satisfies a three-term
recurrence relation of the type

$p_{n+1}(x)=(A_{n}x+B_{n})p_{n}(x)-C_{n}p_{n-1}(x)~{}(n=0,1,2,\ldots,p_{-1}\equiv
0),$ (6)

with $C_{n}A_{n}A_{n-1}>0$. Moreover, Favard’s theorem states that the
converse is true. In Section 2, a general method to derive such three–term
recurrence relations for classical orthogonal polynomials on a quadratic or
$q$-quadratic lattice in terms of the given polynomials $\phi(x)$ and
$\psi(x)$ will be recalled. Alhaidari [1] submitted (as open problem during
the 14th International Symposium on Orthogonal Polynomials, Special Functions
and Applications) two polynomials defined by their three-term recurrence
relations and initial values. He was interested by the derivation of their
weight functions, generating functions, orthogonality relations, etc.. In
order to solve this problem as suggested in the comments by W. Van Assche in
[2], we use the computer algebra system Maple to identify the polynomials from
their recurrence relations, such as in the Maple implementation rec2ortho of
Koorwinder and Swarttouw [10] or retode of Koepf and Schmersau [8]. The two
implementations rec2ortho and retode do not handle classical orthogonal
polynomials on a quadratic or $q$-quadratic lattice. In Section 3, we extend
the Maple implementation retode of Koepf and Schmersau [8] to cover classical
orthogonal polynomials on a quadratic or $q$-quadratic lattice and to answer
the problem by Alhaidari [1] as application.

##  2 Three-term recurrence equation satisfied by classical orthogonal
polynomials on a quadratic or $q$-quadratic lattices

We recall the definition of the Pochhammer symbol (or shifted factorial) given
by

$(a)_{m}={\Gamma(a+m)\over\Gamma(a)}=a(a+1)(a+2)\cdots(a+m-1),~{}m=0,1,2,\ldots,$

and the $q$-Pochhammer symbol

$(a;q)_{m}=\left\\{\begin{array}[]{ll}\prod\limits_{j=0}^{m-1}(1-aq^{j})&\text{
if }\ m=1,2,3,\ldots\\\ 1&\text{ if }\ m=0.\end{array}\right.$

We set the following polynomial basis:

$B_{n}(a,x)=(aq^{s};q)_{n}(aq^{-s};q)_{n}=\prod_{k=0}^{n-1}(1-2axq^{k}+a^{2}q^{2k}),~{}n\geq
1,~{}B_{0}(a,x)\equiv 1,$ (7)

where $\displaystyle{x=x(s)=\cos\theta={q^{s}+q^{-s}\over
2},~{}q^{s}=e^{i\theta}}$;

$\vartheta_{n}(a,x)=(a+ix)_{n}(a-ix)_{n};$ (8)
$\left\\{\begin{array}[]{ll}\xi_{n}(\gamma,\delta,\mu(x))=(q^{-x};q)_{n}(\gamma\delta
q^{x+1};q)_{n}=\prod\limits_{k=0}^{n-1}(1+\gamma\delta
q^{2k+1}-\mu(x)q^{k}),~{}n\geq 1,\\\ \xi_{0}(\gamma,\delta,\mu(x))\equiv
1,\end{array}\right.$ (9)

with $\mu(x)=q^{-x}+\gamma\delta q^{x+1}$;

${\small\left\\{\begin{array}[]{ll}\chi_{n}(\gamma,\delta,\lambda(x))=(-x)_{n}(x+\gamma+\delta+1)_{n}=\prod\limits_{k=0}^{n-1}\Big{(}k(\gamma+\delta+k+1)-\lambda(x)\Big{)},~{}n\geq
1,\\\ \chi_{0}(\gamma,\delta,\lambda(x))\equiv 1,\end{array}\right.}$ (10)

for $\lambda(x)=x(x+\gamma+\delta+1)$. From the hypergeometric and the basic
hypergeometric representations (see [9, chaps. 9, 14]) of classical orthogonal
polynomials on a quadratic or $q$-quadratic lattice, their natural bases are
$\\{B_{n}(a,x)\\}$, $\\{(a+ix)_{n}\\}$, $\\{\xi_{n}(\gamma,\delta,\mu(x))\\}$
or $\\{\chi_{n}(\gamma,\delta,\lambda(x))\\}$ whose elements are polynomials
of degree $n$ in the variables $x$, $x$, $\mu(x)$ or $\lambda(x)$,
respectively, and the basis $\\{\vartheta_{n}(a,x)\\}$ whose elements are
polynomials of degree $n$ in the variable $x^{2}$. The operator
$\mathbb{D}_{x}$ is appropriate for $B_{n}(a,x)$,
$\xi_{n}(\gamma,\delta,\mu(x))$ and $\chi_{n}(\gamma,\delta,\lambda(x))$,
$\mathbf{\delta}_{x}$ is appropriate for $\\{(a+ix)_{n}\\}$, whereas the
corresponding operator for the basis $\\{\vartheta_{n}(a,x)\\}$ is
$\mathbf{D}_{x}$.

Starting from a difference equation of type (2) or (5) given in [9], we deduce
the divided-difference equation of type (1), (3) or (4) satisfied by each
classical orthogonal polynomial on a quadratic or $q$-quadratic lattice. Some
of them can be found in [4], [13], [17] and we recall all of them here to make
the manuscript self-contained. They will also be recovered using the
algorithms implemented in this manuscript.

###  2.1 Polynomials expanded in the basis $\\{\vartheta_{n}(\alpha,x)\\}$

In this basis are expanded:
1\. the Wilson polynomials

$W_{n}(x^{2};a,b,c,d)=(a+b)_{n}(a+c)_{n}(a+d)_{n}\mbox{${}_{4}F_{3}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}a+b,a+c,a+d\end{array}\hfil\lx@intercol\end{array}\\!\right|\displaystyle{1}\right)$},$

with

$\displaystyle\phi(x)$ $\displaystyle=$
$\displaystyle({x}^{2}-\left(cd+ab+ac+bc+bd+ad\right){x}+abcd),$
$\displaystyle\psi(x)$ $\displaystyle=$ $\displaystyle(a+b+c+d)x-abc-abd-acd-
bcd,~{}\lambda_{n}=-n(n-1+a+b+c+d);$

2\. the continuous dual Hahn polynomials

$S_{n}(x^{2};a,b,c)=(a+b)_{n}(a+c)_{n}\mbox{${}_{3}F_{2}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}a+b,a+c\end{array}\hfil\lx@intercol\end{array}\\!\right|\displaystyle{1}\right)$},$

with

$\phi(x)=-\left(a+b+c\right){x}+abc,~{}\psi(x)={x}-(bc+ab+ac),~{}\lambda_{n}=-n.$

The procedure to find the coefficients of the recurrence equation (6) (with
$x$ substituted by $x^{2}$) in terms of the coefficients
$a,~{}b,~{}c,~{}d,~{}e$ of $\phi(x)$ and $\psi(x)$ is as follows (cf. [5, 7,
8, 13, 17]):

  1. 1.

Substitute

$p_{n}(x):=p_{n}(x^{2})=k_{n}\vartheta_{n}(\alpha,x)+k^{\prime}_{n}\vartheta_{n}(\alpha,x)+k^{\prime\prime}_{n}\vartheta_{n-2}(\alpha,x)+\ldots$
(11)

in the divided-difference equation (3) (with $x$ substituted by $x^{2}$). Next
multiply this equation by $\vartheta_{1}(\alpha,x)$ and use the relations [13]

$\displaystyle\vartheta_{1}(\alpha,x)\mathbf{D}^{2}_{x}\vartheta_{n}(\alpha,x)=\eta(n)\eta(n-1)\vartheta_{n-1}(\alpha,x),$
$\displaystyle\vartheta_{1}(\alpha,x)\mathbf{S}_{x}\mathbf{D}_{x}\vartheta_{n}(\alpha,x)=\eta(n)\left(\beta(\alpha+\frac{1}{2},n-1)\vartheta_{n-1}(\alpha,x)+\vartheta_{n}(\alpha,x)\right),$
$\displaystyle\vartheta_{1}(\alpha,x)\vartheta_{n}(\alpha,x)=\nu(\alpha,n)\vartheta_{n}(\alpha,x)+\vartheta_{n+1}(\alpha,x),$

with

$\eta(n)=n,~{}\beta(\alpha,n)=-n(n+\alpha-\frac{1}{2}),~{}\nu(\alpha,n)=-(n^{2}+2\alpha
n).$

  2. 2.

To eliminate the terms $x^{2}\vartheta_{n}(\alpha,x)$ and
$x^{4}\vartheta_{n}(\alpha,x)$, use twice the relation [13]

$x^{2}\vartheta_{n}(\alpha,x)=-(n+\alpha)^{2}\vartheta_{n}(\alpha,x)+\vartheta_{n+1}(\alpha,x).$
(12)

  3. 3.

Equating the coefficients of $\vartheta_{n+1}(\alpha,x)$ gives
$\lambda_{n}=-n((n-1)a+d)$ in (3). Equating the coefficients of
$\vartheta_{n}(\alpha,x)$ and $\vartheta_{n-1}(\alpha,x)$ gives
$k^{\prime}_{n}/k_{n},~{}k^{\prime\prime}_{n}/k_{n}\in\mathbb{Q}(n)$.

  4. 4.

Substitute the expression of $p_{n}$ given by (11) in the recurrence relation
(6) (with $x$ substituted by $x^{2}$) and use (12). By equating the
coefficients of $\vartheta_{n+1}(\alpha,x)$, $\vartheta_{n}(\alpha,x)$,
$\vartheta_{n-1}(\alpha,x)$, we get $A_{n}$, $B_{n}$ and $C_{n}$,
respectively, in terms of $k_{n}$, $k^{\prime}_{n}$ and
$k^{\prime\prime}_{n}$.

  5. 5.

Substituting the values of $k^{\prime}_{n}$ and $k^{\prime\prime}_{n}$ given
in step 3 in these equations yields the three unknowns in terms of $a$, $b$,
$c$, $d$, $e$, $n$, $k_{n-1}$, $k_{n}$, $k_{n+1}$.

######  Proposition 1.

Let
$p_{n}(x):=p_{n}(x^{2})=k_{n}\vartheta_{n}(\alpha,x)+k^{\prime}_{n}\vartheta_{n}(\alpha,x)+k^{\prime\prime}_{n}\vartheta_{n-2}(\alpha,x)+\ldots~{}(n\in\mathbb{N}_{0})$
be a family of polynomial solutions of the divided-difference equation (3).
Then the recurrence equation (6) (with $x$ substituted by $x^{2}$) holds with

$\displaystyle\frac{k_{n}}{k_{n+1}}A_{n}=1,$
$\displaystyle\frac{k_{n}}{k_{n+1}}B_{n}=-\frac{n\left(n-1\right)a\left(2\,a{n}^{2}-2\,an+4\,nd-2\,b-d\right)-nd\left(2\,b+d-2\,nd\right)+e\left(2\,a-d\right)}{\left(\left(2\,n-2\right)a+d\right)\left(2\,an+d\right)},$
$\displaystyle\frac{k_{n-1}}{k_{n+1}}C_{n}=\frac{n\left(an-2\,a+d\right)}{\left(2\,an-a+d\right)\left(2\,an-3\,a+d\right)\left(2\,an-2\,a+d\right)^{2}}\times\Big{\\{}\left(n-1\right)^{6}{a}^{3}+\left(n-1\right)d{b}^{2}$
$\displaystyle+\left(-2\,\left(n-1\right)^{4}b+3\,\left(n-1\right)^{5}d-4\,c\left(n-1\right)^{2}\right){a}^{2}+\left(-2\,\left(n-1\right)^{2}{d}^{2}+de\right)b+\left(-en-c+e\right){d}^{2}$
$\displaystyle+\left(\left(n-1\right)^{2}{b}^{2}-4\,\left(n-1\right)^{3}db+3\,\left(n-1\right)^{4}{d}^{2}-\left(n-1\right)\left(en+4\,c-e\right)d-{e}^{2}\right)a+\left(n-1\right)^{3}{d}^{3}\Big{\\}}.$

###  2.2 Polynomials expanded in the basis $\\{(\alpha+ix)_{n}\\}$

The polynomials expanded in this basis are:
1\. the continuous Hahn polynomials

$p_{n}(x;a,b,c,d)=i^{n}{(a+c)_{n}(a+d)_{n}\over
n!}\mbox{${}_{3}F_{2}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}a+c,a+d\end{array}\hfil\lx@intercol\end{array}\\!\right|\displaystyle{1}\right)$},$

with

$\phi(x)=2\,{x}^{2}-i\left(a+b-c-d\right)x-cd-
ab,~{}\psi(x)=2\left(a+b+c+d\right)x-2\,i\left(ab-cd\right);$

2\. the Meixner-Pollaczek polynomials

$P_{n}^{(\lambda)}(x;\theta)={(2\lambda)_{n}\over
n!}e^{in\theta}\mbox{${}_{2}F_{1}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}2\lambda\end{array}\hfil\lx@intercol\end{array}\\!\right|\displaystyle{1-e^{-2i\theta}}\right)$},$

with

$\phi(x)=x\cos\left(\theta\right)-\lambda\,\sin\left(\theta\right),~{}\psi(x)=2\left(x\sin\left(\theta\right)+\lambda\cos\left(\theta\right)\right).$

The action of the operators $\mathbf{\delta}_{x}$ and $\mathbf{S}_{x}$ on the
basis $(\alpha+ix)_{n}$ is given by [17]

$\displaystyle(\alpha+ix)\mathbf{\delta}_{x}^{2}(\alpha+ix)_{n}$
$\displaystyle=-n(n-1)(\alpha+ix)_{n-1};$
$\displaystyle(\alpha+ix)\mathbf{S}_{x}\mathbf{\delta}_{x}(\alpha+ix)_{n}$
$\displaystyle=ni(\alpha+ix)_{n}-\frac{n(n-1)}{2}i(\alpha+ix)_{n-1};$
$\displaystyle(\alpha+ix)(\alpha+ix)_{n}$
$\displaystyle=(\alpha+ix)_{n+1}-n(\alpha+ix)_{n};$ $\displaystyle
x(\alpha+ix)_{n}$
$\displaystyle=-i(\alpha+ix)_{n+1}+i(n+\alpha)(\alpha+ix)_{n}.$

We suppose that

$p_{n}(x)=k_{n}(\alpha+ix)_{n}+k^{\prime}_{n}(\alpha+ix)_{n-1}+k^{\prime\prime}_{n}(\alpha+ix)_{n-2}+\ldots.$

Using the same approach as in section 2.1, it follows that
$\lambda_{n}=-n((n-1)a+d)$ in (4) and the following result holds.

######  Proposition 2.

Let
$p_{n}(x)=k_{n}(\alpha+ix)_{n}+k^{\prime}_{n}(\alpha+ix)_{n-1}+k^{\prime\prime}_{n}(\alpha+ix)_{n-2}+\ldots~{}(n\in\mathbb{N}_{0})$
be a family of polynomial solutions of the divided-difference equation (4).
Then the recurrence equation (6) is valid with

$\displaystyle\frac{k_{n}}{k_{n+1}}A_{n}=i,$
$\displaystyle\frac{k_{n}}{k_{n+1}}B_{n}=i\frac{\left(2\,b{n}^{2}-2\,bn-2\,e\right)a+d\left(2\,bn+e\right)}{\left(2\,an-2\,a+d\right)\left(2\,an+d\right)},$
$\displaystyle\frac{k_{n-1}}{k_{n+1}}C_{n}=n\Big{\\{}-8\,n\left(n-2\right)\left(n-1\right)^{4}{a}^{5}+\left(-4\,\left(7\,{n}^{2}-13\,n+2\right)\left(n-1\right)^{3}d+32\,cn\left(n-2\right)\left(n-1\right)^{2}\right){a}^{4}$
$\displaystyle+\Big{(}-8\,n\left(n-2\right)\left(n-1\right)^{2}{b}^{2}-2\,\left(19\,{n}^{2}-34\,n+10\right)\left(n-1\right)^{2}{d}^{2}+16\,c\left(n-1\right)\left(5\,{n}^{2}-9\,n+2\right)d$
$\displaystyle+8\,{e}^{2}n\left(n-2\right)\Big{)}{a}^{3}+\Big{(}-4\,\left(n-1\right)\left(5\,{n}^{2}-9\,n+2\right)d{b}^{2}-8\,en\left(n-2\right)db+\left(72\,{n}^{2}-128\,n+48\right){d}^{2}c$
$\displaystyle-\left(n-1\right)\left(5\,n-3\right)\left(5\,n-6\right){d}^{3}+\left(12\,n-8\right){e}^{2}d\Big{)}{a}^{2}+\Big{(}-4\,\left(4\,n-3\right)\left(n-1\right){d}^{2}{b}^{2}+\left(-12\,n+8\right)e{d}^{2}b$
$\displaystyle+\left(28\,n-24\right){d}^{3}c-\left(8\,n-7\right)\left(n-1\right){d}^{4}+4\,{e}^{2}{d}^{2}\Big{)}a+\left(-4\,n+4\right){d}^{3}{b}^{2}-4\,b{d}^{3}e+\left(1-n\right){d}^{5}+4\,c{d}^{4}\Big{\\}}$
$\displaystyle\Big{/}\Big{\\{}4\left(2\,an-2\,a+d\right)^{2}\left(2\,an-3\,a+d\right)\left(2\,an+d\right)\left(2\,an-a+d\right)\Big{\\}}.$

###  2.3 Polynomials expanded in the basis $\\{\
\chi_{n}(\gamma,\delta,\lambda(x))\\}$

In this basis, we have:
1\. the Racah polynomials

$R_{n}(\lambda(x);\alpha,\beta,\gamma,\delta)=\mbox{${}_{4}F_{3}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}\alpha+1,\beta+\delta+1,\gamma+1\end{array}\hfil\lx@intercol\end{array}\\!\right|\displaystyle{1}\right)$},~{}n=0,1,\ldots,N,$

with

$\displaystyle\phi(\lambda(x))$ $\displaystyle=$
$\displaystyle\lambda(x)^{2}+\frac{1}{2}\Big{(}2\,\gamma+\gamma\,\alpha+2\,\gamma\,\delta+\gamma\,\beta+3\,\alpha+2\,\delta+2\,\alpha\,\beta+$
$\displaystyle\delta\,\alpha+3\,\beta+4-\delta\,\beta\Big{)}\lambda(x)+1/2\,\left(\gamma+1\right)\left(1+\gamma+\delta\right)\left(1+\delta+\beta\right)\left(1+\alpha\right),$
$\displaystyle\psi(\lambda(x))$ $\displaystyle=$
$\displaystyle\left(2+\alpha+\beta\right)\lambda(x)+\left(1+\delta+\beta\right)\left(1+\alpha\right)\left(\gamma+1\right);$

2\. the Dual Hahn polynomials

$R_{n}(\lambda(x);\gamma,\delta,N)=\mbox{${}_{3}F_{2}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}\gamma+1,-N\end{array}\hfil\lx@intercol\end{array}\\!\right|\displaystyle{1}\right)$},~{}n=0,1,\ldots,N,$

with

$\displaystyle\phi(\lambda(x))=\frac{1}{2}\left(-\gamma-1+2\,N+\delta\right)\lambda(x)+\frac{1}{2}N\left(\gamma+1\right)\left(1+\gamma+\delta\right),\,\psi(\lambda(x))=-\lambda(x)+N\left(\gamma+1\right);$

We get by direct computations that the action of the operators
$\mathbb{D}_{x}$ and $\mathbb{S}_{x}$ on $\chi_{n}(\gamma,\delta,\lambda(x))$
is given by

$\displaystyle\chi_{1}(\gamma,\delta,\lambda(x))\ \mathbb{D}_{x}^{2}\
\chi_{n}(\gamma,\delta,\lambda(x))=\eta(n)\eta(n-1)\
\chi_{n-1}(\gamma,\delta,\lambda(x)),$
$\displaystyle\chi_{1}(\gamma,\delta,\lambda(x))\
\mathbb{S}_{x}\mathbb{D}_{x}\
\chi_{n}(\gamma,\delta,\lambda(x))=\eta(n)\left(\beta(\frac{1}{2},\gamma,\delta,n-1)\
\chi_{n-1}(\gamma,\delta,\lambda(x))+\chi_{n}(\gamma,\delta,\lambda(x))\right),$
$\displaystyle\chi_{1}(\gamma,\delta,\lambda(x))\
\chi_{n}(\gamma,\delta,\lambda(x))=\nu(\gamma,\delta,n)\
\chi_{n}(\gamma,\delta,\lambda(x))+\chi_{n+1}(\gamma,\delta,\lambda(x)),$
$\displaystyle\lambda(x)\
\chi_{n}(\gamma,\delta,\lambda(x))=\mu(\gamma,\delta,n)\
\chi_{n}(\gamma,\delta,\lambda(x))-\chi_{n+1}(\gamma,\delta,\lambda(x))$

where

$\displaystyle\mu(\gamma,\delta,n)=n(n+\gamma+\delta+1),\
\nu(\gamma,\delta,n)=-n(n+\gamma+\delta+1),$
$\displaystyle\beta(a,\gamma,\delta,n)=\frac{-n(2n+\gamma+\delta+2a)}{2},\ \
\eta(n)=-n.$

We set

$p_{n}(x):=p_{n}(\lambda(x))=k_{n}\
\chi_{n}(\gamma,\delta,\lambda(x))+k^{\prime}_{n}\
\chi_{n-1}(\gamma,\delta,\lambda(x))+k^{\prime\prime}_{n}\
\chi_{n-2}(\gamma,\delta,\lambda(x))+\ldots,$

use the latter structure relations satisfied by
$\chi_{n}(\gamma,\delta,\lambda(x))$, and proceed as in section 2.1 to get
$\lambda_{n}=-n((n-1)a+d)$ in (1) and the following result.

######  Proposition 3.

Let $p_{n}(x):=p_{n}(\lambda(x))=k_{n}\
\chi_{n}(\gamma,\delta,\lambda(x))+k^{\prime}_{n}\
\chi_{n-1}(\gamma,\delta,\lambda(x))+k^{\prime\prime}_{n}\
\chi_{n-2}(\gamma,\delta,\lambda(x))+\ldots~{}(n\in\mathbb{N}_{0})$ be a
family of polynomial solutions of the divided-difference equation (1) (where
$\phi(x)\leftarrow\phi(\lambda(x))$ and $\psi(x)\leftarrow\psi(\lambda(x))$).
Then the recurrence equation (6) (with $x\leftarrow\lambda(x)$) is valid with

$\displaystyle\frac{k_{n}}{k_{n+1}}A_{n}=-1,$
$\displaystyle\frac{k_{n}}{k_{n+1}}B_{n}=-\frac{an\left(n-1\right)\left(2\,a{n}^{2}-2\,an+4\,dn+2\,b-d\right)+2\,bdn+n\left(2\,n-1\right){d}^{2}+de-2\,ae}{\left(2\,an-2\,a+d\right)\left(2\,an+d\right)},$
$\displaystyle\frac{k_{n-1}}{k_{n+1}}C_{n}=-\frac{n\left(an-2\,a+d\right)}{4\,\left(2\,an-a+d\right)\left(2\,an-3\,a+d\right)\left(2\,an-2\,a+d\right)^{2}}\times\Big{\\{}\left(4\,c+\left(-4\,n+4\right)e\right){d}^{2}$
$\displaystyle+\left(-8\,\left(n-1\right)^{4}b+4\,\left(n-1\right)^{3}\left(2\,{\delta}^{2}+4\,\delta\,\gamma+2\,{\gamma}^{2}-3\,{n}^{2}+4\,\delta+4\,\gamma+6\,n-1\right)d+16\,c\left(n-1\right)^{2}\right){a}^{2}$
$\displaystyle+\Big{(}-4\,\left(n-1\right)^{2}{b}^{2}-16\,\left(n-1\right)^{3}db+\left(n-1\right)^{2}\left(5\,{\delta}^{2}+10\,\delta\,\gamma+5\,{\gamma}^{2}-12\,{n}^{2}+10\,\delta+10\,\gamma+24\,n-7\right){d}^{2}$
$\displaystyle+\left(\left(16\,n-16\right)c-4\,e\left(n-1\right)^{2}\right)d+4\,{e}^{2}\Big{)}a+4\,\left(n-1\right)^{4}\left(n+\delta+\gamma\right)\left(-n+2+\delta+\gamma\right){a}^{3}+\left(-4\,n+4\right)d{b}^{2}$
$\displaystyle+\left(-8\,\left(n-1\right)^{2}{d}^{2}-4\,de\right)b+\left(n-1\right)\left(-2\,n+3+\delta+\gamma\right)\left(2\,n-1+\delta+\gamma\right){d}^{3}\Big{\\}}.$

###  2.4 Polynomials expanded in the basis $\\{B_{n}(\alpha,x)\\}$

The following polynomial families are expanded in the basis
$\\{B_{n}(\alpha,x)\\}$:
1\. the Askey-Wilson polynomials defined by

$p_{n}(x;a,b,c,d|q)={(ab,ac,ad;q)_{n}\over
a^{n}}\mbox{${}_{4}\phi_{3}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}ab,ac,ad\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos\theta,$

for which

$\displaystyle\phi(x)$ $\displaystyle=$
$\displaystyle\frac{\left(q-1\right)^{2}}{4{q}^{3/2}}\Big{(}2(abcd+1)x^{2}-\left(a+b+c+d+acd+abc+abd+bcd\right)x$
$\displaystyle+ab+cd+bd+bc+ac+ad-abcd-1\Big{)},$ $\displaystyle\psi(x)$
$\displaystyle=$
$\displaystyle\frac{\left(q-1\right)}{2q}\Big{(}2\left(abcd-1\right)x+a+b+c+d-abc-
abd-acd-bcd)\Big{)};$

2\. the continuous dual $q$-Hahn polnomials

$p_{n}(x;a,b,c|q)={(ab,ac;q)_{n}\over
a^{n}}\mbox{${}_{3}\phi_{2}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}ab,ac\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos\theta,$

with

$\displaystyle\phi(x)$ $\displaystyle=$
$\displaystyle\frac{\left(q-1\right)^{2}}{4{q}^{3/2}}\Big{(}2x^{2}-\left(a+b+c+abc\right)x+bc+ac+ab-1\Big{)},$
$\displaystyle\psi(x)$ $\displaystyle=$
$\displaystyle\frac{\left(q-1\right)}{2q}\Big{(}-2x+a+b+c-abc\Big{)};$

3\. the continuous $q$-Hahn polynomials

$p_{n}(x;a,b,c,d;q)={(abe^{2i\hat{\theta}},ac,ad;q)_{n}\over(ae^{i\hat{\theta}})^{n}}\mbox{${}_{4}\phi_{3}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}abe^{2i\hat{\theta}},ac,ad\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos(\theta+\hat{\theta}),$

with

$\displaystyle\phi(x)$ $\displaystyle=$
$\displaystyle\frac{\left(q-1\right)^{2}}{4q^{3/2}}\Big{(}2\,\left(abcd+1\right)x^{2}-\frac{\left(d+bcd+a{t}^{2}+abd{t}^{2}+abc{t}^{2}+c+acd+b{t}^{2}\right)}{t}x$
$\displaystyle+\frac{ac{t}^{2}+bd{t}^{2}-abcd{t}^{2}+bc{t}^{2}+cd-{t}^{2}+ab{t}^{4}+ad{t}^{2}}{{t}^{2}}\Big{)},~{}t=e^{i\hat{\theta}},$
$\displaystyle\psi(x)$ $\displaystyle=$
$\displaystyle\frac{\left(q-1\right)}{2q}\Big{(}2\left(abcd-1\right)x+\frac{\left(c+d-acd+b{t}^{2}+a{t}^{2}-bcd-
abc{t}^{2}-abd{t}^{2}\right)}{t}\Big{)};$

4\. the Al-Salam-Chihara polynomials

$Q_{n}(x;a,b|q)={(ab;q)_{n}\over
a^{n}}\mbox{${}_{3}\phi_{2}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}ab,0\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos\theta,$

with

$\phi(x)=\frac{\left(q-1\right)^{2}}{4{q}^{3/2}}\Big{(}2x^{2}-\left(b+a\right)x+ab-1\Big{)},\
\psi(x)=\frac{\left(q-1\right)}{2q}\Big{(}-2x+b+a\Big{)};$

5\. the $q$-Meixner-Pollaczek polynomials

$P_{n}(x;a|q)=a^{-n}e^{-in\hat{\theta}}{(a^{2};q)_{n}\over(q;q)_{n}}\mbox{${}_{3}\phi_{2}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}a^{2},0\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos(\theta+\hat{\theta}),$

with $(t=e^{i\hat{\theta}})$

$\phi(x)=\frac{(q-1)^{2}}{4q^{3\over
2}}\Big{(}2x^{2}-\frac{a(1+t^{2})}{t}x+a^{2}-1\Big{)},\,\psi(x)=\frac{q-1}{2q}\Big{(}-2x+\frac{a(1+t^{2})}{t}\Big{)};$

6\. the continuous $q$-Jacobi polynomials

$P_{n}^{(\alpha,\beta)}(x|q)={(q^{\alpha+1};q)_{n}\over(q;q)_{n}}\mbox{${}_{4}\phi_{3}\left(\\!\left.\begin{array}[]{c}\lx@intercol\hfil\begin{array}[]{c}q^{-n},q^{n+\alpha+\beta+1},q^{{\alpha\over
2}+{1\over 4}}e^{i\theta},q^{{\alpha\over 2}+{1\over
<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}q^{\alpha+1},-q^{\alpha+\beta+1\over
2},-q^{\alpha+\beta+2\over
2}\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos\theta,$

with ($p^{2}=q$)

$\displaystyle\phi(x)=\frac{(p^{2}-1)^{2}}{4p^{3}}\Big{(}2(p^{2\alpha+2\beta+4}+1)x^{2}+\sqrt{p}(p+1)(p^{\alpha}-p^{\beta})(p^{\alpha+\beta+2}-1)x$
$\displaystyle+p^{2\alpha+2}+p^{2\beta+2}-p^{\alpha+\beta+1}-p^{\alpha+\beta+3}-2p^{\alpha+\beta+2}-p^{2\alpha+2\beta+4}-1\Big{)},$
$\displaystyle\psi(x)=\frac{p^{2}-1}{2p^{2}}\Big{(}2(p^{2\alpha+2\beta+4}-1)x+\sqrt{p}(p+1)(p^{\alpha}-p^{\beta})(p^{\alpha+\beta+2}+1)\Big{)};$

7\. the continuous $q$-ultraspherical/Rogers polynomials

$C_{n}(x;\beta|q)={(\beta^{2};q)_{n}\over(q;q)_{n}}\beta^{-{n\over
2}}\mbox{${}_{4}\phi_{3}\left(\\!\left.\begin{array}[]{c}\lx@intercol\hfil\begin{array}[]{c}q^{-n},\beta^{2}q^{n},\beta^{1\over
2}e^{i\theta},\beta^{1\over
<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}\beta q^{1\over 2},-\beta,-\beta q^{1\over
2}\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos\theta,$

with

$\phi(x)=\left(q-1\right)^{2}\left(\left(2\,{\beta}^{2}q+2\right){x}^{2}-\left(\beta+1\right)\left(\beta
q+1\right)\right),~{}\psi(x)=4\,\left(q-1\right)x\left({\beta}^{2}q-1\right)\sqrt{q};$

8\. the continuous big $q$-Hermite polynomials

$H_{n}(x;a|q)=a^{-n}\mbox{${}_{3}\phi_{2}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}0,0\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos\theta,$

with

$\displaystyle\phi(x)$ $\displaystyle=$
$\displaystyle\frac{(q-1)^{2}}{4q^{3\over 2}}\Big{(}2x^{2}-ax-1\Big{)},\
\psi(x)=\frac{q-1}{2q}\Big{(}-2x+a\Big{)};$

9\. the continuous $q$-Laguerre polynomials

$P_{n}^{(\alpha)}(x|q)={(q^{\alpha+1};q)_{n}\over(q;q)_{n}}\mbox{${}_{3}\phi_{2}\left(\\!\left.\begin{array}[]{c}\lx@intercol\hfil\begin{array}[]{c}q^{-n},q^{{\alpha\over
2}+{1\over 4}}e^{i\theta},q^{{\alpha\over 2}+{1\over
<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}q^{\alpha+1},0\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}x=\cos\theta,$

with

$\displaystyle\phi(x)=\frac{(p^{2}-1)^{2}}{4p^{3}}\Big{(}2x^{2}-p^{\alpha+{1\over
2}}(p+1)x+p^{2\alpha+2}-1\Big{)},$
$\displaystyle\psi(x)=\frac{p^{2}-1}{2p^{2}}\Big{(}-2x+p^{\alpha+{1\over
2}}(p+1)\Big{)}.$

The procedure to find the coefficients of the recurrence equation (6) in terms
of the coefficients $a,~{}b,~{}c,~{}d,~{}e$ of $\phi(x)$ and $\psi(x)$ is as
in section 2.1:

  1. 1.

Substitute

$p_{n}(x)=k_{n}B_{n}(\alpha,x)+k^{\prime}_{n}B_{n-1}(\alpha,x)+k^{\prime\prime}_{n}B_{n-2}(\alpha,x)+\ldots$
(13)

in the divided-difference equation (1). Next multiply this equation by
$B_{1}(\alpha,x)$ and use the relations [5]

$\displaystyle B_{1}(\alpha,x)\mathbb{D}_{x}^{2}B_{n}(\alpha,x)$
$\displaystyle=\eta(\alpha,n)\eta(\alpha\sqrt{q},n-1)B_{n-1}(\alpha,x),$ (14)
$\displaystyle B_{1}(\alpha,x)\mathbb{S}_{x}\mathbb{D}_{x}B_{n}(\alpha,x)$
$\displaystyle=\eta(\alpha,n)\left(\beta_{1}(\alpha\sqrt{q},n-1)B_{n-1}(\alpha,x)+\beta_{2}(n-1)B_{n}(\alpha,x)\right),$
(15) $\displaystyle B_{1}(\alpha,x)B_{n}(\alpha,x)$
$\displaystyle=\nu_{1}(\alpha,n)B_{n}(\alpha,x)+\nu_{2}(n)B_{n+1}(\alpha,x),$
(16)

with

$\displaystyle\eta(\alpha,n)=\frac{2\alpha(1-q^{n})}{q-1},\ \
\beta_{1}(\alpha,n)=\frac{1}{2}(1-\alpha^{2}q^{2n-1})(1-q^{-n}),$
$\displaystyle\beta_{2}(n)=\frac{1}{2}+\frac{1}{2q^{n}},\
\nu_{1}(\alpha,n)=(1-q^{-n})(1-\alpha^{2}q^{n}),\ \ \nu_{2}(n)=q^{-n}.$

  2. 2.

To eliminate the terms $xB_{n}(\alpha,x)$ and $x^{2}B_{n}(\alpha,x)$, use the
relations [5]

$\displaystyle xB_{n}(\alpha,x)$
$\displaystyle=\mu_{1}(\alpha,n)B_{n}(\alpha,x)+\mu_{2}(\alpha,n)B_{n+1}(\alpha,x),$
(17) $\displaystyle x^{2}B_{n}(\alpha,x)$
$\displaystyle=\mu_{1}^{2}(\alpha,n)B_{n}(\alpha,x)+\mu_{2}(\alpha,n)(\mu_{1}(\alpha,n)+\mu_{1}(\alpha,n+1))B_{n+1}(\alpha,x)$
$\displaystyle+\mu_{2}(\alpha,n)\mu_{2}(\alpha,n+1)B_{n+2}(\alpha,x),$ (18)

with

$\mu_{1}(\alpha,n)=\frac{1+\alpha^{2}q^{2n}}{2\alpha q^{n}},\ \
\mu_{2}(\alpha,n)=\frac{-1}{2\alpha q^{n}}.$

  3. 3.

Equating the coefficients of $B_{n+1}(\alpha,x)$ gives

$\lambda_{n}=-\frac{1}{2}\,{\frac{\left({q}^{n}-1\right)\left(2\,\sqrt{q}\left({q}^{n}-q\right)a+\left(q-1\right)\left({q}^{n}+q\right)d\right)}{{q}^{n}\left(q-1\right)^{2}}}.$
(19)

Equating the coefficients of $B_{n}(\alpha,x)$ and $B_{n-1}(\alpha,x)$ gives
$k^{\prime}_{n}/k_{n},~{}k^{\prime\prime}_{n}/k_{n}\in\mathbb{Q}(q^{n},\sqrt{q})$.

  4. 4.

Substitute the expression of $p_{n}$ given by (13) in the recurrence relation
(6) and use (17). By equating the coefficients of $B_{n+1}(\alpha,x)$,
$B_{n}(\alpha,x)$, $B_{n-1}(\alpha,x)$, we get $A_{n}$, $B_{n}$ and $C_{n}$,
respectively, given as

$\displaystyle A_{n}\frac{k_{n}}{k_{n+1}}=$
$\displaystyle\frac{1}{\mu_{2}(\alpha,n)},~{}B_{n}\frac{k_{n}}{k_{n+1}}=-\frac{\mu_{2}(\alpha,n-1)k^{\prime}_{n}}{\mu_{2}(\alpha,n)k_{n}}+\frac{k^{\prime}_{n+1}}{k_{n+1}}-\frac{\mu_{1}(\alpha,n)}{\mu_{2}(\alpha,n)},$
(20) $\displaystyle C_{n}\frac{k_{n-1}}{k_{n+1}}=$
$\displaystyle-\frac{\mu_{2}(\alpha,n-1)(k^{\prime}_{n})^{2}}{\mu_{2}(\alpha,n)k_{n}^{2}}+\frac{\mu_{2}(\alpha,n-2)k^{\prime\prime}_{n}}{\mu_{2}(\alpha,n)k_{n}}-\frac{k^{\prime\prime}_{n+1}}{k_{n+1}}+\frac{k^{\prime}_{n}k^{\prime}_{n+1}}{k_{n}k_{n+1}}-\frac{(\mu_{1}(\alpha,n)-\mu_{1}(\alpha,n-1))k^{\prime}_{n}}{\mu_{2}(\alpha,n)k_{n}}.$

  5. 5.

Substituting the values of $k^{\prime}_{n}$ and $k^{\prime\prime}_{n}$ given
in step 3 in these equations yields the three unknowns in terms of $\alpha$,
$a$, $b$, $c$, $d$, $e$, $n$, $k_{n-1}$, $k_{n}$, $k_{n+1}$ given by
$(N=q^{n})$:

$\displaystyle\frac{k_{n}}{k_{n+1}}A_{n}=-2\alpha N,$
$\displaystyle\frac{k_{n}}{k_{n+1}}B_{n}=2\,{N}^{2}\alpha\,\Big{\\{}\left(-4\,{q}^{34}\left(q+1\right)\left(q-1\right)^{6}\left(N-1\right)\left(N-q\right)b+2\,{q}^{{\frac{67}{2}}}\left(q-1\right)^{8}\left(N+1\right)\left(N+q\right)e\right)a$
$\displaystyle-2\,{q}^{{\frac{67}{2}}}\left(q+1\right)\left(q-1\right)^{7}\left(N-1\right)\left(N+q\right)db+{q}^{33}\left(q-1\right)^{8}\left({N}^{2}q-N{q}^{2}-{N}^{2}-2\,Nq-{q}^{2}-N+q\right)ed\Big{\\}}$
$\displaystyle\Big{/}\Big{\\{}4\,{q}^{34}\left(q-1\right)^{6}\left({N}^{2}-1\right)\left({N}^{2}-{q}^{2}\right){a}^{2}+4\,{q}^{{\frac{67}{2}}}\left(q-1\right)^{7}\left({N}^{4}-{q}^{2}\right)da+{q}^{33}\left(q-1\right)^{8}\left({N}^{2}+1\right)\left({N}^{2}+{q}^{2}\right){d}^{2}\Big{\\}}.$

The expression of $\frac{k_{n-1}}{k_{n+1}}C_{n}$ is very huge and will not be
displayed here. However, it can be found in the Maple file associated to this
manuscript at http://www.mathematik.uni-kassel.de/~tcheutia/.

###  2.5 Polynomials expanded in the basis
$\\{\xi_{n}(\gamma,\delta,\mu(x))\\}$

The polynomials represented in this basis are:
1\. the $q$–Racah polynomials

$R_{n}(\mu(x);\alpha,\beta,\gamma,\delta|q)=\mbox{${}_{4}\phi_{3}\left(\\!\left.\begin{array}[]{c}\lx@intercol\hfil\begin{array}[]{c}q^{-n},\alpha\beta
q^{n+1},q^{-x},\gamma\delta<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}\alpha q,\beta\delta q,\gamma
q\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}n=0,1,\ldots,N,$

with

$\displaystyle\phi(\mu(x))$ $\displaystyle=$
$\displaystyle\frac{\left(q-1\right)^{2}}{2{q}^{3/2}}\Big{(}\left({q}^{2}\alpha\,\beta+1\right)\mu(x)^{2}-\left(q\gamma\,\delta\,\beta+q\delta\,\beta\,\alpha+q\gamma\,\alpha+q\beta\,\alpha+\gamma+\gamma\,\delta+\delta\,\beta+\alpha\right)\mu(x)$
$\displaystyle+2\left(q\gamma\,\delta\,\alpha+q\delta\,\beta\,\alpha+q\gamma\,\alpha+q{\delta}^{2}\beta\,\gamma+q{\gamma}^{2}\delta+q\gamma\,\delta\,\beta-\gamma\,\delta\,{q}^{2}\alpha\,\beta-\gamma\,\delta\right)\Big{)},$
$\displaystyle\psi(\mu(x))$ $\displaystyle=$
$\displaystyle\frac{\left(q-1\right)}{q}\Big{(}\left({q}^{2}\alpha\,\beta-1\right)\mu(x)-q\left(q\gamma\,\delta\,\beta+q\delta\,\beta\,\alpha+q\gamma\,\alpha+q\beta\,\alpha-\gamma-\gamma\,\delta-\delta\,\beta-\alpha\right)\Big{)};$

2\. the dual $q$–Hahn polynomials

$R_{n}(\mu(x);\gamma,\delta,N|q)=\mbox{${}_{3}\phi_{2}\left(\\!\left.\begin{array}[]{c}\lx@intercol\hfil\begin{array}[]{c}q^{-n},q^{-x},\gamma\delta
<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}\gamma
q,q^{-N}\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}n=0,1,\ldots,N,$

with

$\displaystyle\phi(\mu(x))$ $\displaystyle=$
$\displaystyle\frac{\left(q-1\right)^{2}}{2{q}^{3/2}}\Big{(}\mu(x)^{2}-\frac{\gamma\,q+{q}^{N}\gamma\,q+\gamma\,\delta\,{q}^{N}q+1}{q^{N}}\mu(x)+2\frac{\gamma
q\,\left(\gamma\,\delta\,{q}^{N}q+1+\delta-{q}^{N}\delta\right)}{{q}^{N}}\Big{)},$
$\displaystyle\psi(\mu(x))$ $\displaystyle=$
$\displaystyle{\frac{\left(q-1\right)}{q}}\Big{(}-\mu(x)+\frac{-\gamma\,q+{q}^{N}\gamma\,q+\gamma\,\delta\,{q}^{N}q+1}{{q}^{N}}\Big{)};$

3\. the dual $q$–Krawtchouk polynomials

$K_{n}(\mu(x);c,N|q)=\mbox{${}_{3}\phi_{2}\left(\\!\left.\begin{array}[]{c}\lx<EMAIL_ADDRESS>\lx@intercol\hfil\begin{array}[]{c}q^{-N},0\end{array}\hfil\lx@intercol\end{array}\right|\displaystyle{q;q}\right)$},~{}n=0,1,\ldots,N,$

with

$\displaystyle\phi(\mu(x))$ $\displaystyle=$
$\displaystyle\frac{(q-1)^{2}}{2q^{3\over
2}}\Big{(}\mu(x)^{2}-\frac{c+1}{q^{N}}\mu(x)+2c\frac{1-q^{N}}{q^{2N}}\Big{)},\
\psi(\mu(x))=\frac{q-1}{q}\Big{(}-\mu(x)+\frac{c+1}{q^{N}}\Big{)};$

By direct computations, we have the structure relations

$\displaystyle\xi_{1}(\gamma,\delta,\mu(x))\mathbb{D}_{x}^{2}\xi_{n}(\gamma,\delta,\mu(x))$
$\displaystyle=\eta(1,n)\eta(\sqrt{q},n-1)\xi_{n-1}(\gamma,\delta,\mu(x));$
$\displaystyle\xi_{1}(\gamma,\delta,\mu(x))\mathbb{S}_{x}\mathbb{D}_{x}\xi_{n}(\gamma,\delta,\mu(x))$
$\displaystyle=\eta(1,n)\left(\beta_{1}(\sqrt{q},\gamma,\delta,n-1)\xi_{n-1}(\gamma,\delta,\mu(x))+\beta_{2}(n-1)\xi_{n}(\gamma,\delta,\mu(x))\right);$
$\displaystyle\xi_{1}(\gamma,\delta,\mu(x))\xi_{n}(\gamma,\delta,\mu(x))$
$\displaystyle=\nu_{1}(\gamma,\delta,n)\xi_{n}(\gamma,\delta,\mu(x))+\nu_{2}(n)\xi_{n+1}(\gamma,\delta,\mu(x));$
$\displaystyle\mu(x)\xi_{n}(\gamma,\delta,\mu(x))$
$\displaystyle=\mu_{1}(\gamma,\delta,n)\xi_{n}(\gamma,\delta,\mu(x))+\mu_{2}(n)\xi_{n+1}(\gamma,\delta,\mu(x));$

where

$\displaystyle\mu_{1}(\gamma,\delta,n)=\frac{1+\gamma\delta q^{2n+1}}{q^{n}},\
\ \mu_{2}(n)=\frac{-1}{q^{n}},\
\nu_{1}(\gamma,\delta,n)=(1-q^{-n})(1-\gamma\delta q^{n+1}),$
$\displaystyle\nu_{2}(n)=q^{-n},~{}\beta_{1}(a,\gamma,\delta,n)=\frac{1}{2}(1-a^{2}\gamma\delta
q^{2n})(1-q^{-n}),\ \ \beta_{2}(n)=\frac{1+q^{n}}{2q^{n}},\ \
\eta(a,n)=\frac{a(1-q^{n})}{q-1}.$

We suppose now that

$p_{n}(x):=p_{n}(\mu(x))=k_{n}\xi_{n}(\gamma,\delta,\mu(x))+k^{\prime}_{n}\xi_{n-1}(\gamma,\delta,\mu(x))+k^{\prime\prime}_{n}\xi_{n-2}(\gamma,\delta,\mu(x))+\ldots.$

To get the coefficients $A_{n}$, $B_{n}$ and $C_{n}$ of (6), we proceed as in
section 2.4 and obtain $\lambda_{n}$ given by (19) and for $N=q^{n}$,

$\displaystyle\frac{k_{n}}{k_{n+1}}A_{n}=-N,$
$\displaystyle\frac{k_{n}}{k_{n+1}}B_{n}=\Big{\\{}\left(-4\,{N}^{2}{q}^{4}\left(q+1\right)\left(q-1\right)^{2}\left(N-1\right)\left(N-q\right)b+2\,{N}^{2}{q}^{7/2}\left(q-1\right)^{4}\left(N+1\right)\left(N+q\right)e\right)a$
$\displaystyle-2\,{N}^{2}{q}^{7/2}\left(q+1\right)\left(q-1\right)^{3}\left(N-1\right)\left(N+q\right)db+{N}^{2}{q}^{3}\left(q-1\right)^{4}\left({N}^{2}q-N{q}^{2}-{N}^{2}-2\,Nq-{q}^{2}-N+q\right)ed\Big{\\}}$
$\displaystyle\Big{/}\Big{\\{}4\,{q}^{4}\left({N}^{2}-1\right)\left({N}^{2}-{q}^{2}\right)\left(q-1\right)^{2}{a}^{2}+4\,{q}^{7/2}\left(q-1\right)^{3}\left({N}^{4}-{q}^{2}\right)da+{q}^{3}\left(q-1\right)^{4}\left({N}^{2}+1\right)\left({N}^{2}+{q}^{2}\right){d}^{2}\Big{\\}}.$

As in section 2.4, the coefficients $\frac{k_{n-1}}{k_{n+1}}C_{n}$ is huge and
can be found in the Maple file associated to this manuscript at
http://www.mathematik.uni-kassel.de/~tcheutia/.

##  3 Extension of the algorithms implemented in the Maple package retode

As shown in section 2, the classical orthogonal polynomials on a quadratic or
a $q$-quadratic lattice satisfy a recurrence equation

$p_{n}(x)=(A_{n}x+B_{n})p_{n}(x)-C_{n}p_{n-1}(x),$

with $A_{n}$, $B_{n}$, $C_{n}$ given explicitly. Koepf and Schmersau [8]
implemented algorithms to test wether or not a given holonomic recurrence
equation (i. e. linear, homogeneous with polynomial coefficients)

$q_{n}(x)p_{n+2}(x)+r_{n}(x)p_{n+1}(x)+s_{n}(x)p_{n}(x)=0~{}(q_{n}(x),\
r_{n}(x),\ s_{n}(x)\in\mathbb{Q}[n,x]),$ (21)

has classical orthogonal polynomial solutions of a continuous, a discrete or a
$q$-discrete variable. In fact, it is less obvious to find out whether there
is a polynomial system satisfying (21), being a linear transform of one of the
classical system (of a continuous, a discrete or a $q$-discrete variable), and
to identify the system in the affirmative case. In this section, we aim to
implement, using the same approach, algorithms to test wether a given
holonomic recurrence equation has classical orthogonal polynomial solutions of
a quadratic or a $q$-quadratic lattice. The algorithms were explicitly given
and explained in [8] for classical orthogonal polynomials of a continuous, a
discrete or a $q$-discrete variable and we will adapt them here for classical
orthogonal polynomials on a quadratic or a $q$-quadratic lattice according to
the basis in which the polynomials are expanded.

###  3.1 Polynomials expanded in the basis $\\{\vartheta_{n}(\alpha,x)\\}$

Algorithm 1 (see [8, Algorithms 1 and 2]). This algorithm decides whether a
given holonomic three-term recurrence equation has classical orthogonal
polynomial solutions expanded in the basis $\\{\vartheta_{n}(\alpha,x)\\}$,
and returns their data if applicable.

  1. 1.

Input: A holonomic three-term recurrence equation

$q_{n}(x)p_{n+2}(x)+r_{n}(x)p_{n+1}(x)+s_{n}(x)p_{n}(x)=0~{}(q_{n}(x),\
r_{n}(x),\ s_{n}(x)\in\mathbb{Q}[n,x]).$

  2. 2.

Shift: Shift by $\max\\{n\in\mathbb{N}_{0}\ |\ n\text{ is zero of either
}q_{n-1}(x)\text{ or }s_{n}(x)\\}+1$ if necessary.

  3. 3.

Rewriting: Rewrite the recurrence equation in the form

$p_{n+1}(x)=t_{n}(x)p_{n}(x)+u_{n}(x)p_{n-1}(x)~{}~{}(t_{n}(x),u_{n}(x)\in\mathbb{Q}(n,x)).$

If either $t_{n}(x)$ is not a polynomial of degree one in $x$ or $u_{n}(x)$ is
not a constant with respect to $x$, return “no classical orthogonal polynomial
solution exists”; exit.

  4. 4.

Linear transformation: Rewrite the recurrence equation by the linear
transformation $x\mapsto(x-g)/f$ with unknowns $f$ and $g$.

  5. 5.

Standardization: Given now $A_{n}$, $B_{n}$ and $C_{n}$ by

$p_{n+1}(x)=(A_{n}x+B_{n})p_{n}(x)-C_{n}p_{n-1}(x)~{}(A_{n},\ B_{n},\ C_{n}\
\in\mathbb{Q}(n),\ A_{n}\neq 0),$

define

$\frac{k_{n+1}}{k_{n}}:=A_{n}=\frac{v_{n}}{w_{n}}~{}(v_{n},\ w_{n}\
\in\mathbb{Q}[n])$

according to proposition 1.

  6. 6.

Make monic: Set

$\tilde{B}_{n}:=\frac{B_{n}}{A_{n}}\in\mathbb{Q}(n)\ \text{ and }\
\tilde{C}_{n}:=\frac{C_{n}}{A_{n}A_{n-1}}\in\mathbb{Q}(n)$

and bring these rational functions in lowest terms. If the degree of either
the numerator $\tilde{B}_{n}$ is larger than 4, if the degree of the
denominator of $\tilde{B}_{n}$ is larger than 2, if the degree of the
numerator of $\tilde{C}_{n}$ is larger than 8, or if the degree of the
denominator of $\tilde{C}_{n}$ is larger than 4, then return “no classical
orthogonal polynomial solution exists”.

  7. 7.

Polynomial identities: Set

$\tilde{B}_{n}=\frac{k_{n}}{k_{n+1}}B_{n},~{}\tilde{C}_{n}=\frac{k_{n-1}}{k_{n+1}}C_{n}$

according to proposition 1, in terms of the unknowns $a,\ b,\ c,\ d,\ e,\ f$
and $g$. Multiply these identities by their common denominators, and bring
them therefore in polynomial form.

  8. 8.

Equating coefficients: Equate the coefficients of the powers of $n$ in the two
resulting equations. This results in a nonlinear system in the unknowns $a,\
b,\ c,\ d,\ e,\ f$ and $g$. Solve this system by Gröbner bases methods. If the
system has no solution or only one with $a=d=0$, then return ‘no classical
orthogonal polynomial solution exists”; exit.

  9. 9.

Output: Return the solution vectors $(a,b,c,d,f,g)$ of the last step, the
divided-difference equation (3) together with the information
$\frac{k_{n+1}}{k_{n}}$ and $y=fx+g$.

######  Example 4.

For the first example, we consider the three-term recurrence equation
satisfied by the Wilson polynomials

$\tilde{W}_{n}(x^{2}):=\tilde{W}_{n}(x^{2};a,b,c,d)=\frac{W_{n}(x^{2};a,b,c,d)}{(a+b)_{n}(a+c)_{n}(a+d)_{n}}$

given by [9, eq. (9.1.4)]

$-(a^{2}+x^{2})\tilde{W}_{n}(x^{2})=A_{n}\tilde{W}_{n+1}(x^{2})-(A_{n}+C_{n})\tilde{W}_{n}(x^{2})+C_{n}\tilde{W}_{n-1}(x^{2}),$

where

$\displaystyle A_{n}=$
$\displaystyle\frac{\left(n+a+b+c+d-1\right)\left(n+a+b\right)\left(n+a+c\right)\left(n+a+d\right)}{\left(2\,n+a+b+c+d-1\right)\left(2\,n+a+b+c+d\right)},$
$\displaystyle C_{n}=$
$\displaystyle\frac{n\left(n+b+c-1\right)\left(n+b+d-1\right)\left(n+c+d-1\right)}{\left(2\,n+a+b+c+d-2\right)\left(2\,n+a+b+c+d-1\right)}.$

Using our implementation, the result is obtained by

A[n]:=(n+a+b+c+d-1)*(n+a+b)*(n+a+c)*(n+a+d)/((2*n+a+b+c+d-1)*(2*n+a+b+c+d))

$\displaystyle
A_{{n}}\,:=\,{\frac{\left(n+a+b+c+d-1\right)\left(n+a+b\right)\left(n+a+c\right)\left(n+a+d\right)}{\left(2\,n+a+b+c+d-1\right)\\\
\mbox{}\left(2\,n+a+b+c+d\right)}}$

C[n]:=n*(n+b+c-1)*(n+b+d-1)*(n+c+d-1)/((2*n+a+b+c+d-2)*(2*n+a+b+c+d-1))

$\displaystyle
C_{{n}}\,:=\,{\frac{n\left(n+b+c-1\right)\left(n+b+d-1\right)\left(n+c+d-1\right)}{\left(2\,n+a+b+c+d-2\right)\left(2\,n+a+b+c+d-1\right)\\\
\mbox{}}}$

RecWilson:= -(a^2+x)*p(n)=A[n]*p(n+1)-(A[n]+C[n])*p(n)+C[n]*p(n-1):

strict:=true:

REtoWilsonDE(subs(n=n+1, RecWilson),p(n),x)

$\displaystyle\mbox{{\tt`Warning: parameters have the
values`}},\,\Big{\\{}a=a,b=-a\left(ab+ac+ad+cb+bd+cd\right),$

$\displaystyle
c={a}^{2}bcd,d={a}^{2}+ab+ac+ad,e=-{a}^{2}bc-{a}^{2}bd-{a}^{2}cd-
abcd\Big{\\}}$

$\displaystyle\Big{[}\left(abcd-abx-acx-adx-cbx-bdx-cdx+{x}^{2}\right){\it
DD}\left({\it DD}\left(p\left(n,x\right),x\right),x\right)$

$\displaystyle-\left(abc+abd+acd+bcd-ax-bx-cx-dx\right){\it SS}\left({\it
DD}\left(p\left(n,x\right),x\right),x\right)$

$\displaystyle-n\left(n+a+b+c+d-1\right)p\left(n,x\right)=0,$

$\displaystyle{\frac{k_{{n+1}}}{k_{{n}}}}=-{\frac{\left(2\,n+a+b+c+d-1\right)\left(2\,n+a+b+c+d\right)\\\
\mbox{}}{\left(n+a+b+c+d-1\right)\left(n+a+b\right)\left(n+a+c\right)\left(n+a+d\right)}}\Big{]}$

which gives the divided-difference equation of the Wilson polynomials (see
[13, theo. 2.5]), as well as the term ratio $k_{n+1}/k_{n}$. Here SS and DD
stand for $\mathbf{S}_{x}$ and $\mathbf{D}_{x}$, respectively.

######  Example 5.

Abdulaziz D. Alhaidari [1] encountered two families of orthogonal polynomials
on the real line defined by their three-term recurrence relations and initial
values. The first system is given by

$\displaystyle\cos\theta
H_{n}^{(\mu,\nu)}(z;\alpha,\theta)=\left(z\sin\theta\left[\left(n+\frac{\mu+\nu+1}{2}\right)^{2}+\alpha\right]+\frac{\nu^{2}-\mu^{2}}{(2n+\mu+\nu)(2n+\mu+\nu+2)}\right)H_{n}^{(\mu,\nu)}(z;\alpha,\theta)$
$\displaystyle+\frac{2(n+\mu)(n+\nu)}{(2n+\mu+\nu)(2n+\mu+\nu+1)}H_{n-1}^{(\mu,\nu)}(z;\alpha,\theta)+\frac{2(n+1)(n+\mu+\nu+1)}{(2n+\mu+\nu+1)(2n+\mu+\nu+2)}H_{n+1}^{(\mu,\nu)}(z;\alpha,\theta),$
(22)

with $0\leq\theta\leq\pi$, $\mu,\nu>-1$, $\alpha\in\mathbb{R}$ and initial
values $H_{0}^{(\mu,\nu)}(z;\alpha,\theta)=1$,
$H_{-1}^{(\mu,\nu)}(z;\alpha,\theta)=0$.
The second system is

$\displaystyle zG_{n}^{(\mu,\nu)}(z;\sigma)$
$\displaystyle=\left((\sigma+B_{n}^{2})\left[\frac{\mu^{2}-\nu^{2}}{(2n+\mu+\nu)(2n+\mu+\nu+2)}+1\right]-\frac{2n(n+\nu)}{2n+\mu+\nu}-\frac{(\mu+1)^{2}}{2}\right)G_{n}^{(\mu,\nu)}(z;\sigma)$
$\displaystyle-(\sigma+B_{n-1}^{2})\frac{2(n+\mu)(n+\nu)}{(2n+\mu+\nu)(2n+\mu+\nu+1)}G_{n-1}^{(\mu,\nu)}(z;\sigma)$
$\displaystyle-(\sigma+B_{n}^{2})\frac{2(n+1)(n+\mu+\nu+1)}{(2n+\mu+\nu+1)(2n+\mu+\nu+2)}G_{n+1}^{(\mu,\nu)}(z;\sigma),$
(23)

with $B_{n}=n+1+\frac{\mu+\nu}{2}$, $\mu,\nu>-1$ and $\sigma\in\mathbb{R}$ and
initial values $G_{0}^{(\mu,\nu)}(z;\sigma)=1$,
$G_{-1}^{(\mu,\nu)}(z;\sigma)=0$.

Our implementation with

BB:=n->n+1+(mu+nu)/2:

recOpen2:=z*p(n)=((sigma+BB(n)^2)*((mu^2-nu^2)/((2*n+mu+nu)*(2*n+mu+nu+2))+1)

-2*n*(n+nu)/(2*n+mu+nu)-(mu+1)^2/2)*p(n)-(sigma+BB(n-1)^2)*2*(n+mu)*(n+nu)
/((2*n+mu+nu)*(2*n+mu+nu+1))*p(n-1)

-(sigma+BB(n)^2)*2*(n+1)*(n+mu+nu+1)/((2*n+mu+nu+1)*(2*n+mu+nu+2))*p(n+1):

strict:=false:

REtoWilsonDE(subs(n=n+1, recOpen2),p(n),z)
returns six divided-difference equations (see the Maple file associated to
this manuscript) for the second recurrence equation (23):

$\displaystyle\left({z}^{2}+\left({\mu}^{2}-2\,\sigma-3\right)z+\frac{1}{4}\,\left(\mu-1\right)^{2}\left({\mu}^{2}+2\,\mu+4\,\sigma+1\right)\right)\mathbf{D}_{z}^{2}P_{n}(z/2)$
$\displaystyle+(4\,z+2\,\left(\mu-1\right)\left(\mu+2\,\sigma+1\right))\mathbf{S}_{z}\mathbf{D}_{z}P_{n}(z/2)-4n(n+1)P_{n}(z/2)=0,$
(24)
$\displaystyle\left({z}^{2}+\left({\mu}^{2}+2\,\mu-2\,\sigma+1\right)z+\frac{1}{4}\left(\mu+1\right)^{2}\left({\mu}^{2}+2\,\mu+4\,\sigma+1\right)\right)\mathbf{D}_{z}^{2}P_{n}(z/2)$
$\displaystyle-4\,\sigma\,\left(\mu+1\right)\mathbf{S}_{z}\mathbf{D}_{z}P_{n}(z/2)-4n(n-1)P_{n}(z/2)=0,$
(25)
$\displaystyle\Big{(}{a}^{2}{z}^{2}+\left(\left({\mu}^{2}+2\,\mu-2\,\sigma+1\right){a}^{2}-d\left(\mu+1\right)a-\frac{1}{2}\,{d}^{2}\right)z$
$\displaystyle+\frac{1}{4}\,\left(\mu+1\right)^{2}\left(\left({\mu}^{2}+2\,\mu+4\,\sigma+1\right){a}^{2}-2\,d\left(\mu+1\right)a+{d}^{2}\right)\Big{)}\mathbf{D}_{z}^{2}P_{n}(z/2)$
$\displaystyle+\left(2\,adz-\left(\mu+1\right)\left(4\,{a}^{2}\sigma-d\left(\mu+1\right)a+{d}^{2}\right)\right)\mathbf{S}_{z}\mathbf{D}_{z}P_{n}(z/2)$
$\displaystyle-4na((n-1)a+d)P_{n}(z/2)=0~{}(a\neq 0,~{}d\neq 0),$ (26)
$\displaystyle(\left(4\,\nu+1\right)z-\frac{1}{2}\,{\nu}^{2})\mathbf{D}_{z}^{2}P_{n}(2z)+(\nu^{2}+\nu-2z)\mathbf{S}_{z}\mathbf{D}_{z}P_{n}(2z)+nP_{n}(2z)=0~{}(\nu\neq-\frac{1}{4}),$
(27)

with $\sigma=0$, $\mu=\nu-1$,

$\displaystyle\left(\left(4\,\nu+3\right)z-\frac{1}{2}\,\nu\,\left(\nu+1\right)\right)\mathbf{D}_{z}^{2}P_{n}(2z)+(\nu^{2}+2\nu-2z+\frac{1}{2})\mathbf{S}_{z}\mathbf{D}_{z}P_{n}(2z)+nP_{n}(2z)=0~{}(\nu\neq-\frac{3}{4}),$
(28)

with $\sigma=-\frac{1}{4}$, $\mu=\nu$,

$\displaystyle\left(\left(4\,\nu+5\right)z-\frac{1}{2}\,\left(\nu+1\right)^{2}\right)\mathbf{D}_{z}^{2}P_{n}(2z)+(\nu^{2}+3\nu-2z+2)\mathbf{S}_{z}\mathbf{D}_{z}P_{n}(2z)+nP_{n}(2z)=0~{}(\nu\neq-\frac{5}{4}),$
(29)

with $\sigma=0$, $\mu=\nu+1$. By comparison with the Wilson divided-difference
equation, we deduce from the first three divided-difference equations
(24)–(26) that

$G_{n}^{(\mu,\nu)}(z;\sigma)=constant\times W_{n}(z/2;a,b,c,d)$

where $a,b,c,d$ are permutations of elements of the set
$\\{\frac{1}{2}(-\mu+1),\frac{1}{2}(-\mu+1),\frac{1}{2}(\mu+1)+\sqrt{-\sigma},\frac{1}{2}(\mu+1)-\sqrt{-\sigma}\\}$,
$\\{\frac{1}{2}(\mu+1),\frac{1}{2}(\mu+1),-\frac{1}{2}(\mu+1)+\sqrt{-\sigma},-\frac{1}{2}(\mu+1)-\sqrt{-\sigma}\\}$,
$\\{\frac{1}{2}(\mu+1),\frac{1}{2}(\mu+1),\frac{1}{2}(\delta-\mu-1)+\sqrt{-\sigma},\frac{1}{2}(\delta-\mu-1)-\sqrt{-\sigma}\\}$
for the first equation (24), the second equation (25), and the third equation
(26) in which the parameters $a=1$, $d=\delta$, respectively. This brings then
a new parameter $\delta$ in the definition of the polynomial
$G_{n}^{(\mu,\nu)}(z;\sigma)$ and we also remark that for $d=0$ and $a=1$ in
the third equation (26), we recovered the second equation (25). For the value
$d=\delta=n+\mu+2$, we recover (from the identification of (26) with the
Wilson polynomials) the solution

$G_{n}^{(\mu,\nu)}(z;\sigma)=\frac{W_{n}(z/2;a,b,c,d)}{(a+b)_{n}(a+d)_{n}}$

given in [2] where $a=c=\frac{\mu+1}{2}$, $b=\frac{n+1}{2}+\sqrt{-\sigma}$ and
$d=\frac{n+1}{2}-\sqrt{-\sigma}$.
Comparing the last three equations (27)–(29) with the divided-difference
equation of the continuous dual Hahn polynomials $S_{n}(2z;a,b,c)$, we deduce
that

$G_{n}^{(\nu-1,\nu)}(z;0)=constant\times S_{n}(2z;a,b,c),$

with $a=c=\nu,~{}b=\frac{1}{2}$, or $a=b=\nu,c=\frac{1}{2}$ or
$b=c=\nu,a=\frac{1}{2}$, i. e., $a,b,c$ are permutations of elements of the
set $\\{\nu,\nu,\frac{1}{2}\\}$;

$G_{n}^{(\nu,\nu)}(z;-1/4)=constant\times S_{n}(2z;a,b,c),$

where $a,b,c$ are permutations of elements of the set
$\\{\nu+1,~{}\nu,~{}\frac{1}{2}\\}$;

$G_{n}^{(\nu+1,\nu)}(z;0)=constant\times S_{n}(2z;a,b,c),$

where $a,b,c$ are permutations of the elements of the set
$\\{\nu+1,\nu+1,\frac{1}{2}\\}$.

###  3.2 Polynomials expanded in the basis $\\{(\alpha+ix)_{n}\\}$

The steps of the algorithm in this case agree with those given in section 3.1.
In steps 5 and 7, we use proposition 2 whereas in step 6, the algorithm will
return “no classical orthogonal polynomial solution exists” if the degree of
either the numerator or the denominator of $\tilde{B}_{n}$ is larger than 2,
if the degree of the numerator of $\tilde{C}_{n}$ is larger than 7, or if the
degree of the denominator of $\tilde{C}_{n}$ is larger than 5.

######  Example 6.

As example here, starting from the three-term recurrence equations (RE) [9,
eq. (9.4.3)] and [9, eq. (9.7.3)] satisfied by the continuous Hahn and the
Meixner-Pollaczek polynomials, respectively, and using our implementation with
REtoContHahnDE(subs(n=n+1, RE), p(n), x), we recover the divided-difference
equations of type (4) satisfied by both families (see [17, prop. 4]). In the
output in this case, SS and DD stand for $\mathbf{S}_{x}$ and
$\mathbf{\delta}_{x}$, respectively.

###  3.3 Polynomials expanded in the basis $\\{\
\chi_{n}(\gamma,\delta,\lambda(x))\\}$

We proceed as in the algorithm of section 3.1. Here, in steps 5 and 7, we use
proposition 3 whereas in step 6, the algorithm will return “no classical
orthogonal polynomial solution exists” if the degree of either the numerator
of $\tilde{B}_{n}$ is larger than 4, the degree of the denominator of
$\tilde{B}_{n}$ is larger than 2, if the degree of the numerator of
$\tilde{C}_{n}$ is larger than 8, or if the degree of the denominator of
$\tilde{C}_{n}$ is larger than 4.

######  Example 7.

If we call RE the three-term recurrence equation of the Racah or the dual Hahn
polynomials given, respectively, by [9, eq. (9.2.3)] and [9, eq. (9.6.3)],
then with our implementation REtoRacahDE(subs(n=n+1, RE), p(n),x), we get the
divided-difference equation satisfied by both families with $\phi$ and $\psi$
given as in section 2.3.

######  Remark 8.

From our implementations of sections 3.1. 3.2 and 3.3, we get for the
recurrence equation (22) the following:

RE:=1/2*(y+y^(-1))*p(n)=(z/(2*I)*(y-y^(-1))*((n+(mu+nu+1)/2)^2+alpha)+
(nu^2-mu^2)/((2*n+mu+nu)*(2*n+mu+nu+2)))*p(n)
+2*(n+mu)*(n+nu)/((2*n+mu+nu)*(2*n+mu+nu+1))*p(n-1)
+2*(n+1)*(n+mu+nu+1)/((2*n+mu+nu+1)*(2*n+mu+nu+2))*p(n+1):

strict:=false:

REtoWilsonDE(subs(n=n+1, RE), p(n), z)

$\displaystyle\mbox{{\tt`Warning: parameters have the
values`}},\,\left\\{a=a,\alpha=\alpha,b=b,c=c,d=d,\right\\}$

$\displaystyle\,\left\\{e=e,f=0,g=g,\mu=\mu,\nu=\nu,y=1\right\\}$

$\displaystyle\mbox{{\tt`Warning: parameters have the
values`}},\,\left\\{a=a,\alpha=\alpha,b=b,c=c,d=d,\right\\}$

$\displaystyle\,\left\\{e=e,f=0,g=g,\mu=\mu,\nu=\nu,y=1\right\\}$

$\displaystyle[[{\it DD}\left({\it
DD}\left(p\left(n,g\right),z\right),z\right)+{\frac{\left(dg+e\right){\it
SS}\left({\it DD}\left(p\left(n,g\right),z\right),z\right)}{a{g}^{2}+bg+c}}\\\
\mbox{}-{\frac{n\left(an-a+d\right)p\left(n,g\right)}{a{g}^{2}+bg+c}}=0],]$

We get the same answer using

REtoRacahDE(subs(n=n+1, RE),p(n),z) and

REtoContHahnDE(subs(n=n+1, RE),p(n),z)
We deduce from our implementations (which return the solution
$H_{n}^{(\mu,\nu)}(z;\alpha,\theta)=0$) that the polynomial family
$H_{n}^{(\mu,\nu)}(z;\alpha,\theta)$ satisfying the recurrence equation (22)
is not related to a known classical orthogonal polynomial sequence on a
quadratic lattice expanded in the basis $\\{\vartheta_{n}(\alpha,x)\\}$,
$\\{(\alpha+ix)_{n}\\}$ or $\\{\ \chi_{n}(\gamma,\delta,\lambda(x))\\}$. This
recurrence equation may lead to a new family of orthogonal polynomial
sequence.

###  3.4 Polynomials expanded in the basis $\\{B_{n}(\alpha,x)\\}$

Algorithm 2 (see [8, Algorithm 3]). This algorithm decides whether a given
holonomic three-term recurrence equation has classical orthogonal polynomial
solutions expanded in the basis $\\{B_{n}(\alpha,x)\\}$, and returns their
data if applicable.

  1. 1.

Input: A holonomic three-term recurrence equation

$q_{n}(x)p_{n+2}(x)+r_{n}(x)p_{n+1}(x)+s_{n}(x)p_{n}(x)=0~{}\left(q_{n}(x),\
r_{n}(x),\ s_{n}(x)\in\mathbb{Q}[q^{n},\sqrt{q},x]\right).$

  2. 2.

Shift: Shift by $\max\\{n\in\mathbb{N}_{0}\ |\ n\text{ is zero of either
}q_{n-1}(x)\text{ or }s_{n}(x)\\}+1$ if necessary.

  3. 3.

Rewriting: Rewrite the recurrence equation in the form

$p_{n+1}(x)=t_{n}(x)p_{n}(x)+u_{n}(x)p_{n-1}(x)~{}~{}\left(t_{n}(x),u_{n}(x)\in\mathbb{Q}(q^{n},\sqrt{q},x)\right).$

If either $t_{n}(x)$ is not a polynomial of degree one in $x$ or $u_{n}(x)$ is
not a constant with respect to $x$, return “no classical orthogonal polynomial
solution exists”; exit.

  4. 4.

Linear transformation: Rewrite the recurrence equation by the linear
transformation $x\mapsto(x-g)/f$ with unknowns $f$ and $g$.

  5. 5.

Standardization: Given now $A_{n}$, $B_{n}$ and $C_{n}$ by

$p_{n+1}(x)=(A_{n}x+B_{n})p_{n}(x)-C_{n}p_{n-1}(x)~{}\left(A_{n},\ B_{n},\
C_{n}\ \in\mathbb{Q}(q^{n},q),\ A_{n}\neq 0\right),$

define

$\frac{k_{n+1}}{k_{n}}:=-\frac{1}{2\alpha
q^{n}}A_{n}=\frac{v_{n}}{w_{n}}~{}\left(v_{n},\ w_{n}\
\in\mathbb{Q}[q^{n},q]\right).$

  6. 6.

Make monic: Set

$\tilde{B}_{n}:=\frac{B_{n}}{A_{n}}\in\mathbb{Q}(q^{n},\sqrt{q})\ \text{ and
}\ \tilde{C}_{n}:=\frac{C_{n}}{A_{n}A_{n-1}}\in\mathbb{Q}(q^{n},\sqrt{q})$

and bring these rational functions in lowest terms. If the degree (w. r. t.
$N:=q^{n}$) of either the numerator $\tilde{B}_{n}$ is larger than 3, if the
degree of the denominator of $\tilde{B}_{n}$ is larger than 4, if the degree
of the numerator or the denominator of $\tilde{C}_{n}$ is larger than 14, then
return “no classical orthogonal polynomial solution exists”.

  7. 7.

Polynomial identities: Set

$\tilde{B}_{n}=\frac{k_{n}}{k_{n+1}}B_{n},~{}\tilde{C}_{n}=\frac{k_{n-1}}{k_{n+1}}C_{n}$

in terms of the unknowns $a,\ b,\ c,\ d,\ e,\ f$ and $g$. Multiply these
identities by their common denominators, and bring them therefore in
polynomial form.

  8. 8.

Equating coefficients: Equate the coefficients of the powers of $N=q^{n}$ in
the two resulting equations. This results in a nonlinear system in the
unknowns $a,\ b,\ c,\ d,\ e,\ f$ and $g$. Solve this system by Gröbner bases
methods. If the system has no solution or only one with $a=d=0$, then return
‘no classical orthogonal polynomial solution exists”; exit.

  9. 9.

Output: Return the solution vectors $(a,b,c,d,f,g)$ of the last step, the
divided-difference equation (1) together with the information
$\frac{k_{n+1}}{k_{n}}$ and $y=fx+g$.

######  Example 9.

As illustrative example, we use our implementation to find the divided-
difference equation of type (1) satisfied by the continuous Hermite
polynomials.

recContinuousqHermite:=2*x*p(n)=p(n+1)+(1-q^n)*p(n-1)

$\displaystyle{\it
recContinuousqHermite}\,:=\,2\,xp\left(n\right)=p\left(n+1\right)+\left(1-{q}^{n}\right)p\left(n-1\right)$

strict:=true:

REtoAskeyWilsonDE(subs(n=n+1, recContinuousqHermite), p(n),q, x)

$\displaystyle[1/2\,\left(2\,{x}^{2}-1\right){\it DD}\left({\it
DD}\left(p\left(n,x\right),x\right),x\right)-2\,{\frac{x{\it SS}\left({\it
DD}\left(p\left(n,x\right),x\right),x\right)\sqrt{q}}{q-1}}$

$\displaystyle\mbox{}+2\,{\frac{{q}^{3/2}\left(-1+{q}^{n}\right)p\left(n,x\right)}{{q}^{n}\left(q-1\right)^{2}}}=0,{\frac{k_{{n+1}}}{k_{{n}}}}=2]$

In the result, SS and DD stand for $\mathbb{S}_{x}$ and $\mathbb{D}_{x}$,
respectively. The results for the other families can be found in the
accompanying Maple file of this manuscript.

###  3.5 Polynomials expanded in the basis
$\\{\xi_{n}(\gamma,\delta,\mu(x))\\}$

The steps of the algorithm in this case agree with those given in section 3.4.
In steps 5 and 7, we use the results from section 2.5 whereas in step 6, the
algorithm will return “no classical orthogonal polynomial solution exists” if
the degree of either the numerator of $\tilde{B}_{n}$ is larger than 3, if the
denominator of $\tilde{B}_{n}$ is larger than 4, if the degree of the
numerator or the denominator of $\tilde{C}_{n}$ is larger than 12.

######  Example 10.

If we consider for example the recurrence equation RE for the $q$-Racah, dual
$q$-Hahn, dual $q$-Krawtchouk given, respectively, by [9, eq. (14.2.3)], [9,
eq. (14.7.3)], [9, eq. (14.17.3)], we use our implementation
REtoqRacahDE(subs(n=n+1,RE), p(n), x, q) to get the divided-difference
equations satisfied by the three families of polynomials and the product
$\gamma\delta$.

Note: The Maple implementation retode by Koepf and Schmersau has been updated
with our extension to classical orthogonal polynomials on a quadratic or a
$q$-quadratic lattice. The package retode.mpl and a worksheet retodedemo.mw
containing the three-term recurrence equations of section 2 and the examples
for all the classical orthogonal polynomials on a quadratic or a $q$-quadratic
lattice can be obtained from http://www.mathematik.uni-kassel.de/~tcheutia/.

##  4 Acknowledgments

This work has been supported by the Institute of Mathematics of the University
of Kassel, Germany. The author thanks Wolfram Koepf for bringing his attention
to this problem and for helpful discussions to improve his Maple
implementation.

## References

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