# FAST: Fidelity-Adjustable Semantic Transmission over Heterogeneous Wireless
Networks

Peichun Li1,2, Guoliang Cheng1, Jiawen Kang1, Rong Yu1, Liping Qian3, Yuan
Wu2, and Dusit Niyato4  1School of Automation, Guangdong University of
Technology, Guangzhou, China
2State Key Laboratory of Internet of Things for Smart City, University of
Macau, Macau, China
3College of Information Engineering, Zhejiang University of Technology,
Hangzhou, China
4School of Computer Science and Engineering, Nanyang Technological University,
Singapore
Email<EMAIL_ADDRESS><EMAIL_ADDRESS>{kavinkang,
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###### Abstract

In this work, we investigate the challenging problem of on-demand semantic
communication over heterogeneous wireless networks. We propose a fidelity-
adjustable semantic transmission framework (FAST) that empowers wireless
devices to send data efficiently under different application scenarios and
resource conditions. To this end, we first design a dynamic sub-model training
scheme to learn the flexible semantic model, which enables edge devices to
customize the transmission fidelity with different widths of the semantic
model. After that, we focus on the FAST optimization problem to minimize the
system energy consumption with latency and fidelity constraints. Following
that, the optimal transmission strategies including the scaling factor of the
semantic model, computing frequency, and transmitting power are derived for
the devices. Experiment results indicate that, when compared to the baseline
transmission schemes, the proposed framework can reduce up to one order of
magnitude of the system energy consumption and data size for maintaining
reasonable data fidelity.

###### Index Terms:

Semantic communications, dynamic neural networks, on-demand communications,
resource management.

##  I Introduction

By 2030, 17.1 billion wireless devices equipped with versatile sensors will
produce 5 zettabytes of data per month [1]. The explosive growth of edge-
generated data rises challenges on how to achieve efficient information
exchange among massive devices. Semantic communication is an emerging data
transmission paradigm that aims to extract and deliver the explicative meaning
of the data [2]. The semantic-aware communication systems can reveal the
intrinsic information of the raw data by leveraging the knowledge of prior
models [3, 4]. By integrating semantic communication into wireless networks,
the required data traffic will be significantly reduced, leading to a green
and reliable communication pattern [5, 6].

By leveraging the capacity of neural networks, learning-based semantic
communication systems can extract compact and accurate information from the
image and speech [7, 8, 9]. To improve the freshness of status updates, the
age of semantics is incorporated into the semantic communication systems [10].
Recently, system-level methods focus on improving the efficiency of semantic
communication, such as the spectrum-efficient method that assigns the optimal
channel for the wireless devices [11], and adaptive resource scheduling that
maximizes the successful probability of transmission [12]. However, these
methods employ fixed neural networks to accomplish the extraction of semantic
information during the running time, which hinders the flexibility of
semantic-aware transmission over heterogeneous networks.

Using a fixed learning-based semantic model is a stringent limit for the
communication system over heterogeneous wireless networks. This setting
deteriorates the ability of the communication system in handling different
application scenarios. As illustrated in Figure 1, a typical semantic model
may be designed to concurrently support multiple vision-related tasks under
different resource conditions. Compared with image classification that only
identifies the category of the image, object detection needs to additionally
analyze the location of the object of interest [13]. Thus, given limited
computation and communication resources, high-fidelity semantic data with
fine-grained information should be reserved for object detection [14], while
the low-fidelity one is sufficient for image classification. Also, the quality
of semantic communication should be adapted to the energy status of the
battery-powered devices. Employing high-fidelity mode for performance-first
setting and switching to low-fidelity mode for the purpose of energy saving is
an effective way to maintain the transmission quality while prolonging the
battery lifetime [15, 16].

Figure 1: FAST over heterogeneous wireless networks.

In this paper, we propose FAST, a fidelity-adjustable semantic transmission
framework, to improve the flexibility of learning-based semantic communication
systems over heterogeneous wireless networks. We focus on the image
transmission task with autoencoder as the semantic model [17]. Our goal is to
train a flexible semantic model that enables wireless devices to select
different sizes of sub-models at the running time. Then, we propose a
fidelity-aware resource management approach, where the optimal transmission
strategy is designed to meet the quality and efficiency constraints.
Specifically, a full-size model with powerful capacity is preferred in the
high-fidelity scenario, and a small sub-model is adopted in the low-fidelity
scenario to save the system cost [18].

However, determining the optimal transmission strategies for FAST with
personalized constraints is a non-trivial task, as how to train the flexible
semantic models and how the fidelity of semantic data is affected by model
size is unknown. To address these issues, we first design a dynamic sub-model
training scheme to concurrently support flexible encoding and decoding with
different model widths. Meanwhile, the relationship between model size and the
expected fidelity is empirically quantified. Following that, we study the FAST
optimization problem to improve energy efficiency with given latency and
fidelity budgets. Based on the theoretical analysis, the problem is
transformed into a convex problem. Finally, we develop a hierarchical
bisection algorithm to solve the problem, where the size of the semantic sub-
model, CPU computing frequency, and transmitting power are determined
according to the fidelity constraint and resource status.

Our main contributions are summarized as follows.

  * •

We propose a novel semantic communication framework, named FAST that enables
wireless devices to perform fidelity-adjustable data transmission.

  * •

We investigate the fidelity-aware resource management problem for FAST, and
the optimal transmission strategy is devised to minimize the system energy
cost.

  * •

Extensive experiments demonstrate the efficiency and effectiveness of FAST,
which outperforms the existing baselines in terms of resource utilization and
data fidelity.

The remainder of this paper is organized as follows. Section II details the
main components of FAST to fulfill flexible semantic communication. The
problem formulation, theoretical analysis, and the corresponding solution are
provided in Section III. The experiment simulations are presented in Section
IV, and we finally conclude the paper in Section V.

##  II System Model

###  II-A Outline of FAST

We consider the scenario of wireless semantic communication between two
physical entities, i.e., the transmitter and the receiver. As shown in Figure
2, unlike traditional methods that utilize the fixed model during the running
time, FAST employs a flexible semantic model to accomplish the fidelity-
adjustable transmission. Specifically, the semantic model comprises of two
parts, including the encoder and decoder. For the full-size model, we use
$\boldsymbol{\theta}$ and $\boldsymbol{\vartheta}$ to parameterize the weights
of the encoder and decoder, respectively. Here, we introduce a scaling factor
$\pi\in(0,1]$ for the width of each layer in the flexible model. Given a
scaling factor $\pi$, we can derive a pair of small encoder and decoder from
the full-size model, denoted as $\boldsymbol{\theta}_{\pi}$ and
$\boldsymbol{\vartheta}_{\pi}$, respectively. The process of the FAST is
divided into the following three phases.

####  II-A1 Phase I for encoding

With the pre-determined scaling factor $\pi$, the source device switches from
the full-size encoder to a small one parameterized by
$\boldsymbol{\theta}_{\pi}$. Let $\boldsymbol{x}$ denote the raw data, and let
${\boldsymbol{h}}_{\pi}$ represent the corresponding semantic data. The
function of semantic encoding can be expressed as

${\boldsymbol{h}}_{\pi}=\texttt{enc}({\boldsymbol{x}};\boldsymbol{\theta}_{\pi}).$
(1)

####  II-A2 Phase II for transmission

After obtaining the semantic data ${\boldsymbol{h}}_{\pi}$, the source device
transmits it to the destination. Here, we consider that the semantic data is
converted into binary symbols. Thus, the transmission for the semantic
information still follows the Shannon capacity [12].

####  II-A3 Phase III for decoding

After receiving the semantic data, the destination device decodes it to
reconstruct the data $\hat{{\boldsymbol{x}}}$. Specifically, the encoder and
decoder have the symmetry structures, i.e., the decoder shares the same
scaling factor $\pi$ as the encoder does. The process of semantic decoding can
be represented as

$\hat{{\boldsymbol{x}}}=\texttt{dec}({\boldsymbol{h}}_{\pi};\boldsymbol{\vartheta}_{\pi}).$
(2)

Figure 2: FAST with flexible semantic model.

###  II-B Flexible Semantic Model

We aim to train a flexible semantic model that supports nearly continuous
scaling factor $\pi\in(0,1]$. We first focus on deriving a pair of small-size
encoder and decoder
$\\{\boldsymbol{\theta}_{\pi},\boldsymbol{\vartheta}_{\pi}\\}$ from
$\\{\boldsymbol{\theta},\boldsymbol{\vartheta}\\}$. Then, we propose a dynamic
sub-model training scheme to train the flexible semantic model efficiently.

####  II-B1 Sub-model derivation

Given a scaling factor $\pi$, we aim to derive the small-size sub-model
$\\{\boldsymbol{\theta}_{\pi},\boldsymbol{\vartheta}_{\pi}\\}$ from the full-
size one $\\{\boldsymbol{\theta},\boldsymbol{\vartheta}\\}$. Specifically, the
sub-model derivation is performed in a layer-by-layer manner. For a
convolution layer with a number of filters as $C$ (i.e., the width of the
layer), we select the weights of the first $\lfloor\pi C\rfloor$ filters to
construct the layer for a small-size sub-model. Given a scaling factor $\pi$,
the derivations for $\boldsymbol{\theta}_{\pi}$ and
$\boldsymbol{\vartheta}_{\pi}$ are respectively expressed by

$\boldsymbol{\theta}_{\pi}=\texttt{sel}(\boldsymbol{\theta},{\pi})\quad\textrm{and
\quad}\boldsymbol{\vartheta}_{\pi}=\texttt{sel}(\boldsymbol{\vartheta},{\pi}),$
(3)

where $\texttt{sel}(\cdot,\cdot)$ denotes the function to select the weight
from the full-size model to the small one.

####  II-B2 Learning objective

Our goal is to minimize the data reconstruction error for any sub-models
derived from the full-size model. Let
$\ell(\boldsymbol{x},\hat{\boldsymbol{x}})$ be the pre-determined loss
function. The learning objective can be expressed as

$\mathop{\min}\limits_{\boldsymbol{\theta},\boldsymbol{\vartheta}}\int_{\pi_{\min}}^{1}\sum\limits_{\boldsymbol{x}\in\boldsymbol{X}_{\textrm{test}}}{\ell\big{(}{\boldsymbol{x}},F({\boldsymbol{x}};\boldsymbol{\theta},\boldsymbol{\vartheta},\pi)\big{)}}d\pi,$
(4)

where the function
$F({\boldsymbol{x}};\boldsymbol{\theta},\boldsymbol{\vartheta},\pi)=\texttt{dec}\big{(}\texttt{enc}({\boldsymbol{x}};\boldsymbol{\theta}_{\pi});\boldsymbol{\vartheta}_{\pi}\big{)}$
computes the reconstructed data with given
$\boldsymbol{x},\boldsymbol{\theta},\boldsymbol{\vartheta}$ and $\pi$.

####  II-B3 Dynamic sub-model training

Note that the process of optimizing Eqn. (4) requires enumerating all sub-
models, which incurs a prohibitive cost for the model training. Inspired by
the study in [19], an efficient training method via sub-model sampling is
proposed to reduce the computation cost. As presented in Algorithm 1, we
propose to dynamically sample a sub-model at each iteration (i.e., Steps 3).
Specifically, the total training loss is computed as the sum of the losses of
the random sub-model and the full-size model (i.e., Step 5). In this way, we
maintain the performance of different sub-model while reducing the training
overhead.

Input: Training dataset $\boldsymbol{X}_{\textrm{train}}$, and
$\\{\boldsymbol{\theta},\boldsymbol{\vartheta}\\}$.

1 for  _each epoch $i=1,2,\ldots,I$_ do

2 for  _each batch of training data
$\boldsymbol{x}_{\textrm{batch}}\in\boldsymbol{X}_{\textrm{train}}$_ do

3 Randomly sample a scaling factor $\pi$. Perform forward propagation with
$\pi$:
$\hat{\boldsymbol{x}}_{\textrm{sub}}=F(\boldsymbol{x}_{\textrm{batch}};\boldsymbol{\theta},\boldsymbol{\vartheta},\pi)$;

4 Perform forward propagation with full-size model:
$\hat{\boldsymbol{x}}_{\textrm{full}}=F(\boldsymbol{x}_{\textrm{batch}};\boldsymbol{\theta},\boldsymbol{\vartheta},1)$;

5 Compute the total reconstruction loss:
$Loss=\ell(\boldsymbol{x}_{\textrm{batch}},\hat{\boldsymbol{x}}_{\textrm{sub}})+\ell(\boldsymbol{x}_{\textrm{batch}},\hat{\boldsymbol{x}}_{\textrm{full}})$;

6 Apply backward propagation to update
$\\{\boldsymbol{\theta},\boldsymbol{\vartheta}\\}$;

7

8 end for

9

10 end for

Algorithm 1 Dynamic sub-model training

###  II-C Characterizing the Semantic Fidelity

We next investigate how the scaling factor $\pi$ affects the performance of
the corresponding sub-model. We first present the definition of semantic
fidelity of the given semantic model, and then reveal the relationship between
the scaling factor $\pi$ and the semantic fidelity of the sub-model.

####  II-C1 Definition of semantic fidelity

We define the semantic fidelity of the semantic model as the capability of
reconstructing data over the testing dataset. Formally, given a scaling factor
of $\pi$, the semantic fidelity $\phi_{\pi}$ of the corresponding sub-model is
calculated by

$\phi_{\pi}=1-\frac{1}{M|\boldsymbol{X}_{\textrm{test}}|}\sum\limits_{\boldsymbol{x}\in\boldsymbol{X}_{\textrm{test}}}{{\|{\boldsymbol{x}}-F({\boldsymbol{x}};\boldsymbol{\theta},\boldsymbol{\vartheta},\pi)\|}},$
(5)

where $M$ denotes the number of pixels in the image,
$|\boldsymbol{X}_{\textrm{test}}|$ measures the number of samples of the
testing dataset, and $\|\cdot\|$ calculates the L1 norm for the given vector.

####  II-C2 The impact of the scaling factor on semantic fidelity

Intuitively, larger sub-models with strong representation capabilities can
extract more latent information from the raw data, resulting in higher
semantic fidelity. Being consistent with the existing studies in [20], we
employ the parameter fitting method to empirically investigate the
relationship between semantic fidelity $\phi_{\pi}$ and the scaling factor
$\pi$. We adopt the CIFAR-10 and CINIC-10 datasets, and the experiment
settings are provided in Section IV. Then, we sample a subset of sub-models
from the flexible semantic model trained by Algorithm 1, and evaluate their
corresponding performance by Eqn. (5). The relationship between $\phi_{\pi}$
and $\pi$ is formulated as

$\phi_{\pi}=\kappa_{1}\ln(\frac{\kappa_{2}}{\pi}+\kappa_{3})+\kappa_{4},$ (6)

where $\\{\kappa_{1},\kappa_{2},\kappa_{3},\kappa_{4}\\}$ are constant hyper-
parameters that can be experimentally fitted. The experiment results are
provided in Figure 3. As $\pi$ increases, the fidelity of semantic data
increases to carry more detailed information.

Figure 3: Semantic fidelity $\phi_{\pi}$ of sub-model with respect to scaling
factor $\pi$.

##  III Problem and Solution

###  III-A FAST over Wireless Networks

For semantic communication with the full-size model and single image sample,
we use $W_{e}$ and $W_{d}$ to denote the computation workloads for encoding
and decoding, respectively, and the size of the semantic information is $S$.
For FAST with a scaling factor of $\pi$, the encoding workloads, decoding
workloads, and data size to be transmitted are reduced as $\pi^{2}W_{e}$,
$\pi^{2}W_{d}$ and $\pi S$, respectively.

####  III-A1 Computation model

Let $f_{e}$ and $f_{d}$ denote the computing frequency for the encoding and
decoding, respectively. For the semantic-based transmission with $K$ samples,
given the sub-model scaling factor $\pi$, the overall time taken for the model
inference can be measured by

$T_{\text{cmp}}=K\pi^{2}\bigg{(}\frac{{W_{e}}}{f_{e}}+\frac{{W_{d}}}{f_{d}}\bigg{)}.$
(7)

Meanwhile, the overall energy consumption is estimated by

$E_{\text{cmp}}=K\pi^{2}(\epsilon_{e}f_{e}^{2}{W_{e}}+\epsilon_{d}f_{d}^{2}{W_{d}}),$
(8)

where $\epsilon_{e}$ and $\epsilon_{d}$ are the hardware energy coefficients
of the source and destination devices, respectively.

####  III-A2 Communication model

Let $B$ denote the available bandwidth, $P$ the transmitting power of the
source device and $N_{0}$ be the power spectral density of the Gaussian noise.
For the transmission of semantic data from the source to the destination, the
achievable transmitting rate is estimated by

$r={B}{\log_{2}}\Big{(}1+\frac{{|h|^{2}d^{-\eta}P}}{{{N_{0}}{B}}}\Big{)},$ (9)

where $d$ represents the distance between the transmitter and receiver, $\eta$
is the pathloss exponent, and $h$ denotes the Rayleigh channel coefficient.
With given scaling factor of $\pi$, the required time $T_{\text{com}}$ and
energy consumption $E_{\text{com}}$ for the transmission of semantic data can
be respectively calculated by

$T_{\text{com}}=\frac{K\pi S}{r},~{}\text{and
}E_{\text{com}}=PT_{\text{com}}.$ (10)

####  III-A3 Problem formulation

Given a pair of source-destination devices with different local resources, our
goal is to optimize the transmission strategy for these two devices to
minimize the total energy cost with latency and fidelity constraints. To this
end, we formulate the following optimization problem.

$\displaystyle({\text{P1}})$ $\displaystyle\min$
$\displaystyle\;E_{\text{tot}}$ (11) subject to: $\displaystyle
T_{\text{tot}}\leq$ $\displaystyle\;{T^{\max}},$ (11a)
$\displaystyle\phi_{\pi}\geq$ $\displaystyle\;\phi^{{\min}},$ (11b)
$\displaystyle{\pi^{\min}}\leq$ $\displaystyle\;\pi\leq 1,$ (11c)
$\displaystyle 0\leq f_{e}\leq f_{e}^{\max}$ $\displaystyle,\;0\leq f_{d}\leq
f_{d}^{\max},$ (11d) $\displaystyle 0\leq P$ $\displaystyle{}\leq P^{\max},$
(11e) variables: $\displaystyle\pi,\;f_{e}$ $\displaystyle,\;f_{d},\;P,$

where $E_{\text{tot}}=E_{\text{cmp}}+E_{\text{com}}$ and
$T_{\text{tot}}=T_{\text{cmp}}+T_{\text{com}}$ are the total energy cost and
the total system latency, respectively.

###  III-B Problem Simplification

In this subsection, we transform Problem (P1) into a tractable yet equivalent
form via constraint simplification and variable substitution. We first derive
the following lemma.

######  Lemma 1.

The equality always holds for Constraints (11a) and (11b) under the optimal
solution $\\{\pi^{\ast},f_{e}^{\ast},f_{d}^{\ast},P^{\ast}\\}$, namely, we
always have $T_{\text{tot}}^{\ast}={T^{\max}}$ and
$\phi_{\pi}^{\ast}=\phi^{{\min}}$.

  * Proof.

We prove the lemma by showing contradictions. Suppose that there exists an
optimal solution such that $T_{\text{tot}}^{\ast}<T^{\max}$. Then, we
construct a new solution by replacing $f_{e}^{\ast}$ with $f_{e}^{\prime}$ in
the optimal solution such that $f_{e}^{\prime}<f_{e}^{\ast}$ and
$T_{\text{tot}}^{\prime}=T^{\max}$. Let $E_{\text{tot}}^{\prime}$ denote the
corresponding system energy cost of the new solution. Since the system energy
cost decreases with the decrease of $f_{e}$, then we obtain
$E_{\text{tot}}^{\prime}<E_{\text{tot}}^{\ast}$. Similarly, the contradiction
also applies for $\phi_{\pi}>\phi^{\min}$, and thus we complete the proof. ∎

Based on Lemma 1 and Eqn. (6), we can derive the optimal width scaling factor
$\pi^{\ast}$ as

$\pi^{\ast}=\frac{\kappa_{2}}{\exp{\big{(}\frac{\phi^{\min}-\kappa_{4}}{\kappa_{1}}\big{)}}-\kappa_{3}}.$
(12)

Moreover, we introduce three intermediate variables $\alpha>0,\beta>0$, and
$\gamma>0$ such that $\alpha+\beta+\gamma=1$, and they denote the time
splitting factors of the latency for encoding, semantic transmission, and
decoding, respectively. Thus, we obtain the following equations.

$\displaystyle\small\begin{split}\alpha
T^{\max}=K(\pi^{\ast})^{2}\frac{{W_{e}}}{f_{e}}&,\quad\beta
T^{\max}=\frac{K\pi^{\ast}S}{r},\\\ \gamma
T^{\max}=&\,K(\pi^{\ast})^{2}\frac{{W_{d}}}{f_{d}}.\end{split}$ (13)

By combining Eqns. (7)-(10) and (13), the total system energy cost
$E_{\text{tot}}$ can be re-expressed with respect to
$\\{\alpha,\beta,\gamma\\}$ as

$E_{\text{tot}}=\frac{\tau_{1}}{\alpha^{2}}+\tau_{2}\beta\big{(}2^{\frac{\tau_{3}}{\beta}}-1\big{)}+\frac{\tau_{4}}{\gamma^{2}},$
(14)

where the constants $\\{\tau_{1},\tau_{2},\tau_{3},\tau_{4}\\}$ can be
calculated by

$\displaystyle\small\begin{split}\tau_{1}={\epsilon_{e}}\frac{{K^{3}W_{e}^{3}{(\pi^{\ast})^{6}}}}{{{{\left({{T^{\max}}}\right)}^{2}}}}>0,\;\;\tau_{2}=\frac{{B{N_{0}}{T^{\max}}}}{|h|^{2}d^{-\eta}}>0,\\\
\tau_{3}=\frac{{K\pi^{\ast}S}}{{B{T^{\max}}}}>0,\;\;\tau_{4}={\epsilon_{d}}\frac{{K^{3}W_{d}^{3}{(\pi^{\ast})^{6}}}}{{{{\left({{T^{\max}}}\right)}^{2}}}}>0.\end{split}$
(15)

Therefore, given the optimal scaling factor $\pi^{\ast}$, Problem (P1) can be
transformed into the following problem.

$\displaystyle({\text{P2}})$
$\displaystyle\min\;\frac{\tau_{1}}{\alpha^{2}}+\tau_{2}\beta$
$\displaystyle\big{(}2^{\frac{\tau_{3}}{\beta}}-1\big{)}+\frac{\tau_{4}}{\gamma^{2}}$
(16) subject to: $\displaystyle\alpha+\beta\,+$ $\displaystyle\,\gamma=1,$
(16a) $\displaystyle\alpha^{\min}\leq\alpha,\;\beta^{\min}$
$\displaystyle\leq\beta,\;\gamma^{\min}\leq\gamma,$ (16b) variables:
$\displaystyle\alpha,\beta,$ $\displaystyle\;\gamma,$

where the lower limits of $\\{\alpha,\beta,\gamma\\}$ can be acquired by

$\displaystyle\small\begin{split}\alpha^{\min}=\frac{K(\pi^{\ast})^{2}{W_{e}}}{f_{e}^{\max}T^{\max}},\;\gamma^{\min}\,=\frac{K(\pi^{\ast})^{2}{W_{d}}}{f_{d}^{\max}T^{\max}},\\\
\beta^{\min}=\frac{K\pi^{\ast}S}{T^{\max}{B}{\log_{2}}\Big{(}1+\frac{{|h|^{2}d^{-\eta}P^{\max}}}{{{N_{0}}{B}}}\Big{)}}.\end{split}$
(17)

Notably, the optimal solution of Problem (P1) can be obtained directly with
the help of $\\{\alpha^{\ast},\beta^{\ast},\gamma^{\ast}\\}$ according to Eqn.
(13). It can be verified that Problem (P2) is a convex optimization problem,
and we discuss the solution in next subsection.

###  III-C Hierarchical Bisection Search

To solve Problem (P2), we first apply Karush–Kuhn–Tucker (KKT) conditions to
derive necessary equations for achieving the optimality. By utilizing
$\lambda$ as the Lagrange multiplier for the equality Constraint (16a), and
$\\{\mu_{\alpha},\mu_{\beta},\mu_{\gamma}\\}$ as the multipliers for the
inequality Constraint (16b), we obtain

$\displaystyle{\mu_{\alpha}}={\frac{{-2{\tau_{1}}}}{{{\alpha^{3}}}}+\lambda},\,{\mu_{\gamma}}={\frac{{-2{\tau_{4}}}}{{{\gamma^{3}}}}+\lambda},$
(18a)
$\displaystyle{\mu_{\beta}}={\big{(}{{\tau_{2}}-\frac{{{\tau_{2}}{\tau_{3}}\ln
2}}{\beta}}\big{)}{2^{\frac{{{\tau_{3}}}}{\beta}}}-{\tau_{2}}+\lambda},$ (18b)
$\displaystyle{\mu_{\alpha}}(\alpha-{{\alpha^{\min}}})={\mu_{\beta}}(\beta-{{\beta^{\min}}})={\mu_{\gamma}}(\gamma-{{\gamma^{\min}}})=0,$
(18c) $\displaystyle
0\leq{\mu_{\alpha}},0\leq{\mu_{\beta}},0\leq{\mu_{\gamma}},$ (18d)
$\displaystyle\text{Constraints~{}(\ref{eqn:p2-ctr-1}) and
(\ref{eqn:p2-ctr-2})}.$

By substituting $\\{\mu_{\alpha},\mu_{\beta},\mu_{\gamma}\\}$ from Eqns (18a)
and (18b) into Eqn. (18c), we have

$\displaystyle\small\begin{split}&\Big{(}{\frac{{-2{\tau_{1}}}}{{{\alpha^{3}}}}+\lambda}\Big{)}(\alpha-\alpha^{\min})=\Big{(}{\frac{{-2{\tau_{4}}}}{{{\gamma^{3}}}}+\lambda}\Big{)}(\beta-\beta^{\min})=0,\end{split}$
(19)
$\displaystyle\begin{split}\Big{(}\underbrace{{\big{(}{{\tau_{2}}-\frac{{{\tau_{2}}{\tau_{3}}\ln
2}}{\beta}}\big{)}{2^{\frac{{{\tau_{3}}}}{\beta}}}-{\tau_{2}}+\lambda}}_{g_{\lambda}(\beta)}\Big{)}(\gamma-\gamma^{\min})=0.\end{split}$
(20)

According to Constraint (16b), the discussion on the value of $\alpha^{\ast}$
can be divided into two cases, i.e., $\alpha^{\ast}>\alpha^{\min}$ and
$\alpha^{\ast}=\alpha^{\min}$. Based on Eqn. (19), we have
$\alpha^{\ast}=\sqrt[3]{{\frac{{2{\tau_{1}}}}{\lambda}}}$ if
$\alpha^{\ast}>\alpha^{\min}$. Similarly, the optimal values of
$\\{\beta^{\ast},\gamma^{\ast}\\}$ can be analyzed on the same basis.
Therefore, we have

$\displaystyle\small\begin{split}\alpha^{\ast}_{\lambda}=\max\big{\\{}\sqrt[3]{{\frac{{2{\tau_{1}}}}{\lambda}}},\alpha^{\min}\big{\\}}&{},\;\beta^{\ast}_{\lambda}=\max\\{\beta_{\lambda},\beta^{\min}\\},\\\
\gamma^{\ast}_{\lambda}=\max\big{\\{}&{}\sqrt[3]{{\frac{{2{\tau_{4}}}}{\lambda}}},\gamma^{\min}\big{\\}},\end{split}$
(21)

where $\beta_{\lambda}$ is the zero of function $g_{\lambda}(\beta)$ in Eqn.
(20) such that $g_{\lambda}(\beta_{\lambda})=0$. Given a specific $\lambda$,
we define that
$z_{\lambda}=\alpha^{\ast}_{\lambda}+\beta^{\ast}_{\lambda}+\gamma^{\ast}_{\lambda}$.
According to Constraint (16a), the solution of Problem (P2) can be acquired by
searching an optimal Lagrange multiplier $\lambda^{\ast}$ such that
$z_{\lambda^{\ast}}=1$.

It can be verified that $\alpha^{\ast}_{\lambda}$, $\beta^{\ast}_{\lambda}$
and $\gamma^{\ast}_{\lambda}$ are monotonically non-increasing with respect to
$\lambda$. Hence, $\lambda^{\ast}$ can be efficiently obtained by the
bisection search as shown in Algorithm 2. Specifically, given a $\lambda$,
$\alpha^{\ast}_{\lambda}$ and $\gamma^{\ast}_{\lambda}$ be directly calculated
while $\beta^{\ast}_{\lambda}$ involves another bisection search (i.e., Steps
8-12). Given the tolerance value $\varepsilon$ and searching range
$\\{\lambda^{\min},\lambda^{\max},\beta^{\min},\beta^{\max}\\}$ and
$J=\max\\{\lambda^{\max}-\lambda^{\min},\beta^{\max}-\beta^{\min}\\}$, the
computational complexity of Algorithm 2 is estimated by ${\cal
O}(log_{2}^{2}J)$.

Input: $\lambda^{\min},\lambda^{\max},\beta^{\min},\beta^{\max}$, and
$\varepsilon$.

Output: The optimal Lagrange multiplier $\lambda^{\ast}$.

1 repeat

2 $\lambda=(\lambda^{\max}+\lambda^{\min})/2$;

3 Compute $\alpha^{\ast}_{\lambda}$ and $\gamma^{\ast}_{\lambda}$ based on
Eqn. (21);

4 Search for $\beta^{\ast}_{\lambda}$, and compute
$z_{\lambda}=\alpha^{\ast}_{\lambda}+\beta^{\ast}_{\lambda}+\gamma^{\ast}_{\lambda}$;

5 if $z_{\lambda}<1$ then $\lambda^{\max}=\lambda$ else
$\lambda^{\min}=\lambda$;

6

7until  _$|\lambda^{\max}-\lambda^{\min}|\leq\varepsilon$_ ;

8return $\lambda^{\ast}$

/* Function for searching $\beta^{\ast}_{\lambda}$. */

9 repeat

10 $\beta=(\beta^{\max}+\beta^{\min})/2$;

11 Compute $g_{\lambda}(\beta)$ based on Eqn. (20);

12 if $g_{\lambda}(\beta)>0$ then $\beta^{\max}=\beta$ else
$\beta^{\min}=\beta$;

13

14until  _$|\beta^{\max}-\beta^{\min}|\leq\varepsilon$_ ;

15return $\beta^{\ast}_{\lambda}$

Algorithm 2 Hierarchical bisection search

##  IV Simulation Results

###  IV-A Experiment Settings

We consider the semantic communication for image transmission with CIFAR10
dataset. For the semantic model, we use two three-layer convolutional neural
networks with kernel size as 4, stride as 2, and padding as 1 for the encoder
and decoder. Specifically, the widths of the encoder and decoder are {12, 24,
32} and {24, 12, 3}, respectively. The semantic data is represented by
$32\times 4\times 4=512$ numbers in 8-bit unsigned integer and $S=4096$ bits.
For training hyper-parameters, the batch size, total epochs and $\pi_{\min}$
are set as 16, 30 and 0.25, respectively. The hyper-parameters for
communication $\\{B,|h|^{2},d,\eta,N_{0}\\}$ are set as {1MHz, 10-3W, 200m,
3.76, $-95$dBm/MHz} by default. The hyper-parameters for computation
$\\{W_{e},W_{d},\epsilon_{e},\epsilon_{d},K\\}$ are empirically set as {0.65
MCycles, 3.25MCycles, $1^{-26}$, $1^{-26}$, 512} by default.

TABLE I: Performance comparison between FAST and baseline methods on image transmission with CIFAR-10 datasets. Method |  Data size (Mbit) |  Comp. Cost (GFLOPs) |  $E_{\text{cmp}}$ (J) |  $E_{\text{com}}$ (J) |  $E_{\text{tot}}$ (J) |  Fidelity
---|---|---|---|---|---|---
Raw |  12.58 (1$\times$) | 0 | 0 | 2.24 | 2.24 | 1
JPEG |  2.76 (4.56$\times$) | — | — | — | — | 0.73
Prune ($\rho$=0.3) |  2.31(5.5$\times$) | 3.31 | 1.65 | 0.53 | 2.18 | 0.80
Quant (3 bits) |  1.18 (10.7$\times$) | 3.31 | 1.44 | 0.26 | 1.70 | 0.80
FAST ($\pi$=0.3) |  0.92 (13.7$\times$) | 0.97 | 0.01 | 0.10 | 0.11 | 0.80
Prune ($\rho$=0.1) |  2.71(4.6$\times$) | 3.31 | 1.73 | 0.64 | 2.37 | 0.85
Quant (4 bits) |  1.57 (8.0$\times$) | 3.31 | 1.51 | 0.35 | 1.86 | 0.85
FAST ($\pi$=0.5) |  1.67 (7.5$\times$) | 1.76 | 0.06 | 0.21 | 0.27 | 0.85

###  IV-B Performance Evalutions

We first show that the hierarchical bisection search algorithm can converge to
the optimal solution of Problem (P2). Figure 4(a) presents the evolution of
the total energy cost with respect to the number of iterations. We observe
that the algorithm can achieve the optimum after about 30 iterations.

We next compare the proposed FAST with the following three baseline methods
under $T^{\max}=8$ seconds. (1) JPEG: we reduce the size of the raw data with
a radical compression ratio of about 4.5. (2) Prune: we employ filter-wise
feature pruning for the semantic data, and $\rho\in[0,1]$ is the pruning rate.
(3) Quant: we quantize the semantic data with fewer bits (from 1 to 8 bits)
before the transmission.

Table I provides the comparison results of FAST against the baseline methods
in terms of the size of semantic data, computation cost, and energy
consumption under two types of fidelity constraints. Compared with Prune and
Quant, FAST can respectively reduce 15 times and 6.9 times the total energy
consumption for realizing semantic communication with the fidelity of 0.80 and
0.85. Particularly, the proposed FAST can reduce 13.7 times the data size
under the low fidelity scenario. Meanwhile, one of the important advantages of
FAST is that the proposed flexible semantic model enables on-demand
computation to mitigate the computation cost.

Figure 4(b) shows the total energy consumption of different methods with
different fidelity constraints. With a given fidelity constraint, the proposed
FAST consistently outperforms the baseline methods to mitigate the total
energy consumption. Specifically, FAST can switch to the small sub-model to
significantly reduce the computation cost in the low-fidelity scenario. Figure
4(c) provides the image plots of reconstructed samples with different fidelity
constraints. Specifically, the proposed FAST can achieve the best semantic
fidelity with the least total energy cost, which strikes the balance between
transmission quality and resource utilization.

Figure 4: The main advantages of FAST. ((a): convergence of the searching
algorithm; (b-c): performance of different methods; (d-f): impact of key
settings.)

Figures 4(d)-(f) show how the energy efficiency is affected by the key system
configurations, including the transmission distance (i.e., channel state),
computational energy coefficient, and latency constraint. It can be observed
that the proposed FAST consistently outperforms the baseline methods under a
variety of system settings. The experiments show that FAST is more resilient
than other baselines to achieve green transmission under heterogeneous
scenarios.

##  V Conclusion

In this paper, we proposed FAST, a fidelity-adjustable semantic transmission
framework for green communication. We presented the dynamic model training
scheme to enable wireless devices to adopt different sub-model on demand. To
improve the energy efficiency of FAST, we focused on minimizing the total
energy consumption under personalized latency and fidelity constraints. By
leveraging the theoretical analysis, we transformed the optimization problem
into a tractable form and designed an algorithm to efficiently search for the
optimal transmission strategy. Experimental results demonstrate the advantage
of FAST in improving energy efficiency and semantic quality against the
baseline methods.

## Acknowledgment

Rong Yu and Yuan Wu are the corresponding authors. This work was supported in
part by National Natural Science Foundation of China under Grants 61971148,
U22A2054 and 62102099, in part by National Key R&D Program of China under
Grants 2020YFB1807802 and 2020YFB1807800, in part by Science and Technology
Development Fund of Macau SAR under Grant 0162/2019/A3, in part by FDCT-MOST
Joint Project under Grant 0066/2019/AMJ, and in part by Research Grant of
University of Macau under Grant MYRG2020-00107-IOTSC. This work was also
supported in part by the National Research Foundation (NRF), Singapore and
Infocomm Media Development Authority under the Future Communications Research
Development Programme (FCP), and DSO National Laboratories under the AI
Singapore Programme (AISG Award No: AISG2-RP-2020-019).

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#  On the positivity of the hypergeometric Veneziano amplitude

###### Abstract

Recently, an infinite one-parameter generalisation of the Veneziano amplitude
was bootstrapped using as input assumptions an integer mass spectrum, crossing
symmetry, high-energy boundedness, and exchange of finite spins. This new
result was dubbed hypergeometric Veneziano amplitude, with the deformation
parameter $r$ being a real number. Using the partial-wave decomposition and
the positivity of said decomposition’s coefficients we are able to bound the
deformation parameter to $r\geq 0$ and, also, to obtain an upper bound on the
number of spacetime dimensions $D\leq 26$, which is the critical dimension of
bosonic string theory.

###### Contents

  1. 1 Prologue
  2. 2 Generalities and setup
  3. 3 The partial wave coefficients in $D=4$ dimensions
    1. 3.1 The leading Regge trajectory
    2. 3.2 More on the Regge trajectories
    3. 3.3 The general coefficients
    4. 3.4 A simpler expression for the general coefficients
  4. 4 The partial wave coefficients in $D$ dimensions
    1. 4.1 The leading Regge trajectory
    2. 4.2 More on the Regge trajectories
    3. 4.3 The general coefficients
  5. 5 Comments on unitarity
  6. 6 Epilogue
  7. A Partial-wave coefficients that are equal to zero
    1. A.1 The effect of a non-vanishing value for the r-parameter
    2. A.2 Vanishing coefficients in $D=4$
    3. A.3 Vanishing coefficients in any $D$
  8. B The polynomials for the Regge trajectories in $D=4$ dimensions

##  1 Prologue

Bootstrapping scattering amplitudes of massless and massive particles, see [1]
for a recent summarized exposition to advances and developments of the
S-matrix bootstrap, is an old theme in theoretical high-energy physics. The
idea behind it is to provide an alternative approach to understanding and
examining physics theories. Concretely, we can formulate specific mathematical
questions about the S-matrix and attempt to answer this kind of questions,
rather than resorting to Lagrangian descriptions and sophisticated geometrical
approaches. This, in turn, implies that we can understand the theories of
interest as being fixed by constraints and conditions that are imposed on
scattering amplitudes.

In order to employ any bootstrap algorithm, we have to choose a set of
assumptions and conditions and then impose them on a landscape of objects. For
the purposes of bootstrapping scattering amplitudes such a set can consist of
crossing symmetry, polynomial residues, and high-energy boundedness.

An explicit four-point amplitude that satisfies the above and is a meromorphic
function, except for its simple poles, was constructed by Veneziano [2] and is
given by:

$\mathcal{M}(s,t)=\frac{\Gamma(-(1+\alpha^{\prime}s))\Gamma(-(1+\alpha^{\prime}t))}{\Gamma(-(1+\alpha^{\prime}s)-(1+\alpha^{\prime}t))}\,.$
(1.1)

Today we know, of course, that it describes the $2\rightarrow 2$ scattering of
open-string tachyons of mass $\alpha^{\prime}m^{2}=-1$111we work with
conventions in which $\alpha^{\prime}=1$ for open string theory..

It is worthwhile stressing that a scattering amplitude violating the
requirement of tame ultraviolet behaviour is an indication for the breakdown
of unitarity and causality of the theory [3, 4]. Very robust expressions have
been derived describing bounds for theories that are gapped; the Regge and the
Froissart bounds.

The Veneziano amplitude has been studied quite extensively, with the recent
works of [5, 6] focusing on the coefficients of the partial-wave decomposition
of the amplitude and discussing its unitarity from tree-level considerations,
as well as the critical dimension of string theory. Questions regarding its
uniqueness were posed since the early days of its discovery.

Along this particular line of investigation, an answer was given in [7, 8, 9]
that is nowadays known as the Coon amplitude. This is another amplitude that
satisfies the criteria of polynomial boundedness, finite-spin exchange, and
meromorphicity. It comes with logarithmic Regge trajectories, and for many
years it was disregarded by virtually everyone. Recently, however, it has
received revived activity [10, 11, 12, 13, 14, 15]. The Coon amplitude is a
deformation of the Veneziano in terms of one-parameter and it exhibits a mass
spectrum with discrete levels converging to an infinite density at an
accumulation point that is followed by a branch cut.

While the work of [16] raises concerns regarding the status of unitarity of
the Coon amplitude, it was realised in [17] that string theory admits
amplitudes behaving like that since they arise from the of open-string
scattering with open strings having their endpoints on a D-brane in AdS.

With the Coon amplitude being able to provide us with an explicit and
consistent generalisation of the Veneziano amplitude, people have been
revisiting the question of constructing more general four-point amplitudes
that are consistent with the principles of the S-matrix bootstrap [18, 19, 20,
21]. This can, also, be phrased as a question to the uniqueness of string
theory. Phrased in simple terms, since string theory amplitudes satisfy
particular constraints, are they the only objects doing so?

In this work we focus our attention on an infinite generalisation of the
Veneziano amplitude, recently derived in [21]. Similarly to the Coon
amplitude, it is, also, written as a one-parameter deformation and has been
dubbed the hypergeometric Veneziano amplitude; this name will become perfectly
clear in the next section. Using the partial-wave decomposition of the
amplitude, we wish to impose the positivity of the coefficients in order
derive bounds on the allowed values for the parameter $r$ and the spacetime
dimensions $D$222The authors of [21] have discussed constraints and bounds
resulting from unitarity using a numerical evaluation of the partial-wave
coefficients. More specifically, starting from the integral representation of
the coefficients, they were able to re-express them as a double-sum and upon
an explicit numerical evaluation they were able to constrain the allowed
values for the deformation parameter, $r$, and the number of allowed spacetime
dimensions, $D$. Our analysis is a more systematic examination of the partial-
wave coefficients..

The approach we take here, in order to derive bounds on the allowed values of
the deformation parameter $r$ and the spacetime dimension $D$, is to examine
the positivity of the coefficients of the partial-wave decomposition of the
amplitude. To do so, we compute the residues of the amplitude at the location
of its poles. Then, we proceed to decompose the residues in a basis spanned by
Gegenbauer polynomials, which is valid for any number of spacetime dimensions,
$D\geq 3$333In the special cases of $D=4$ and $D=5$ the Gegenbauer polynomials
reduce to the Legendre polynomials and the Chebyshev polynomials of the second
kind.. As we have already mentioned, the task at hand is to find the numbers
that multiply these polynomials, since their non-negativity is tied to the
unitarity of the underlying theory. We proceed by utilizing the orthogonality
relations that these polynomials obey, and we obtain a relation for the
partial-wave coefficients as an integral of those special polynomials and some
non-trivial function. Using the generating functions for the special
polynomials we can define a “pseudo-generating function”. After some algebraic
manipulations, which consist of writing our expressions as power series
expansions we, effectively, have two polynomial expansions for the original
equation of the partial-wave coefficients. From that we can read off the terms
of appropriate scaling in order to derive the coefficients.

In addition to the above, we also resort to some experimental guess-work, in
order to derive additional analytic expressions for the partial-wave
coefficients of sub-leading Regge trajectories. This means that, starting from
the original expression for the partial-wave coefficients given as an integral
of the special polynomials times some non-trivial function, we compute the
integral for some values, we make a guess for the general form of the answer
and we proceed to verify our claim by checking explicitly some non-trivial
values444In this context, by non-trivial we mean some values that were not
used in order to claim the answer of the partial-wave coefficients..

The structure of this work is the following: in section 2 we set-up our
notation and conventions. We move on to section 3, where we specialise the
discussion in the $D=4$ case and we derive the partial-wave coefficients for
the leading Regge trajectory, $a_{n,n+1}$. We, also, provide expressions for
the partial-wave coefficients of the sub-leading Regge trajectories,
$a_{n,n-\gamma}$, with $\gamma=\\{0,1,\dots,10\\}$. Finally, we re-write the
original integral representation of the partial-wave coefficients as multiple
sums. In section 4 we analyse the partial-wave coefficients for general
dimensions. We proceed to analyse the positivity constraints of those
coefficients in section 5 and derive bounds on the parameter $r$ and the
spacetime dimensions $D$. We conclude in section 6.

##  2 Generalities and setup

The new infinite family of hypergeometric amplitudes is given by [21]

$\mathcal{A}(s,t)=\frac{\Gamma(-s-1)\Gamma(-t-1)}{\Gamma(-s-t-2)}{}_{3}F_{2}\left(-s-1,-t-1,r;-s-t-2,1+r;1\right)\,,$
(2.1)

where in the above ${}_{3}F_{2}(a;b;z)$ is the generalised hypergeometric
function555Note that relative to [21] we have a shift
$\\{s,t\\}\rightarrow\\{s+1,t+1\\}$..

It is obvious that the amplitude, $\mathcal{A}(s,t)$, has poles in $s$ at
$s=n=-1,0,1\ldots$ and of course the same is true for $t$.

We start by calculating the residues at the $s$ poles. We have

$\mathop{\mathrm{Res}}_{s=n}\mathcal{A}(s,t)=-\frac{r!}{(n+r+1)!}(t+2+r)_{n+1}\,.$
(2.2)

From the Gegenbauer expansion for arbitrary dimensions $D$ we know

$\mathop{\mathrm{Res}}_{s=n}\mathcal{A}(s,t)=-\sum_{l=0}^{n+1}a_{n,l}C_{l}^{(\alpha)}\left(1+\frac{2t}{n+4}\right)\,,$
(2.3)

where in the above the parameter $\alpha$ is related to the spacetime
dimensions $D$ via $\alpha=\tfrac{D-3}{2}$. Now we can just equate equations
2.2 and 2.3 to get

$\frac{r!}{(n+r+1)!}(t+2+r)_{n+1}=\sum_{l=0}^{n+1}a_{n,l}C_{l}^{(\alpha)}\left(1+\frac{2t}{n+4}\right)\,.$
(2.4)

We can use the fact that the Gegenbauer polynomials satisfy the following
orthogonality condition

$\int_{-1}^{+1}dxC_{\ell}^{(\alpha)}(x)C_{\ell^{\prime}}^{(\alpha)}(x)(1-x^{2})^{\alpha-\frac{1}{2}}=2\mathcal{K}(\ell,\alpha)\delta_{\ell\ell^{\prime}}\,,$
(2.5)

where in the above we have defined

$\mathcal{K}(\ell,\alpha)=\frac{\pi\Gamma(\ell+2\alpha)}{2^{2\alpha}\ell!(\ell+\alpha)\Gamma^{2}(\alpha)}\,,$
(2.6)

in order to derive the following expression for the partial-wave coefficients
of equation 2.4

$\displaystyle a_{n,\ell}=$
$\displaystyle\frac{r!}{(n+1+r)!}\frac{1}{\mathcal{K}(\ell,\alpha)}\left[\frac{4}{(n+4)^{2}}\right]^{\alpha-\tfrac{1}{2}}$
(2.7)
$\displaystyle\int^{n+4}_{0}dtC_{\ell}^{(\alpha)}\left(1-\frac{2t}{n+4}\right)(t(n+4-t))^{\alpha-\tfrac{1}{2}}(-t+2+r)_{n+1}\,.$

By examining equation 2.7 we make the following observations:

  * •

Unlike the case of the Veneziano amplitude where $a_{n,\ell}=0$ for $n+\ell$
equal to an even number, here we do not have that. This is due to the presence
of the deformation parameter $r$. In the special case $r=0$ the coefficients
are equal to $0$ as they should.

  * •

We have, however, that $a_{n,\ell}=0$ for $\ell\geq n+2$, as is the case for
the Veneziano amplitude as well.

We conclude this section here and discuss more the above two points in
appendix A.

##  3 The partial wave coefficients in $D=4$ dimensions

Here we specialize the discussion in the $D=4$, or equivalently
$\alpha=\tfrac{1}{2}$, case. Equation 2.7 simplifies to the following
expression

$a_{n,\ell}=\frac{r!}{(n+1+r)!}\frac{1+2\ell}{n+4}\int^{n+4}_{0}dtP_{\ell}\left(1-\frac{2t}{n+4}\right)(-t+2+r)_{n+1}\,,$
(3.1)

where in the above $P_{\ell}(x)$ denotes the Legenedre polynomials.

We can proceed by utilizing the generating function of the Legendre
polynomials

$\sum^{\infty}_{\ell=0}P_{\ell}(x)t^{-\ell-1}=\frac{1}{\sqrt{1-2xt+t^{2}}}\,.$
(3.2)

and the representation of the Pochhammer symbol in terms of the Stirling
number of the first kind, $s^{(b)}_{a}$,666in some places in the literature it
is written as $s(a,b)$ but we opt for the one that is closer to the
Mathematica implementation.

$(x)_{n}=\sum^{n}_{k=0}(-1)^{n-k}s^{(k)}_{n}x^{k}\,,$ (3.3)

in order to obtain

$\sum^{\infty}_{j=0}\frac{1}{2j+1}\frac{a_{n,j}}{h^{j+1}}=\frac{1}{(n+4)(n+1)!}\sum^{n+1}_{k=0}(-1)^{n+1-k}s^{(k)}_{n+1}\underbrace{\int^{n+4}_{0}dt\frac{(-t+2+r)^{k}}{\sqrt{(h-1)^{2}+\tfrac{4ht}{n+4}}}}_{\mathcal{G}^{(r)}_{n,k}(h)}\,.$
(3.4)

In the above, equation 3.4, we have defined a “pseudo generating function”
$\mathcal{G}^{(r)}_{n,k}(h)$ which we can evaluate explicitly and is given by:

$\mathcal{G}^{(r)}_{n,k}(h)=\frac{1}{2^{2k+1}}\frac{(n+4)^{k+1}}{h^{k+1}}\left[(h-1)^{2}+\frac{4h}{n+4}(2+r)\right]^{k}\Big{(}\mathcal{E}^{+}-\mathcal{E}^{-}\Big{)}\,,$
(3.5)

with the shorthands

$\displaystyle\mathcal{E}^{\pm}$ $\displaystyle=(h\pm
1)~{}{}_{2}F_{1}(\tfrac{1}{2},-k,\tfrac{3}{2};f^{(\pm)})\,,$ (3.6)
$\displaystyle f^{(\pm)}$ $\displaystyle=\frac{(h\pm
1)^{2}}{(h-1)^{2}+\tfrac{4h}{n+4}(2+r)}\,.$

Using the formal power-series definition for the hypergeometric function
${}_{2}F_{1}(a,b,c;z)$

${}_{2}F_{1}(a,b,c;x)=\sum^{\infty}_{y=0}\frac{1}{y!}\frac{(a)_{y}(b)_{y}}{(c)_{y}}x^{y}\,,$
(3.7)

we can re-write equation 3.5 as777Note that the $\mathfrak{p}$-sum is
terminated, however, this is natural since $(-a)_{b}=0$ holds $\forall b>a$.

$\displaystyle\mathcal{G}^{(r)}_{n,k}(h)=$
$\displaystyle\frac{1}{2^{2k+1}}\frac{(n+4)^{k+1}}{h^{k+1}}$ (3.8)
$\displaystyle\sum^{k}_{\mathfrak{p}=0}\frac{(-k)_{\mathfrak{p}}}{(2\mathfrak{p}+1)\mathfrak{p}!}\left((h-1)^{2}+\frac{4h}{n+4}(2+r)\right)^{k-\mathfrak{p}}\sum^{2\mathfrak{p}+1}_{\mathfrak{m}=0}\binom{2\mathfrak{p}+1}{\mathfrak{m}}h^{\mathfrak{m}}(1+(-1)^{\mathfrak{m}})\,.$

###  3.1 The leading Regge trajectory

In equation 3.8 the function $\mathcal{G}^{(r)}_{n,k}(h)$ has in total $k+1$
terms that scale according to
$\tfrac{1}{h},\tfrac{1}{h^{2}},\dots,\tfrac{1}{h^{k+1}}$. It is a rather
straightforward exercise to extract the $\tfrac{1}{h^{k+1}}$ coefficient,
which is given by

$\mathcal{D}^{(r)}_{n,k}=\frac{(n+4)^{k+1}}{2^{2k+1}}\sqrt{\pi}\frac{k!}{(k+\tfrac{1}{2})!}\,.$
(3.9)

Let us recall at this point, that the point of the exercise is to extract the
partial-wave coefficient for the leading Regge trajectory, $a_{n,n+1}$. This
is the term that scales like $\tfrac{1}{h^{n+2}}$ in equation 3.4. We have

$\frac{1}{2n+3}a_{n,n+1}=\frac{1}{n+4}\frac{r!}{(n+r+1)!}s^{(n+1)}_{n+1}\mathcal{D}^{(r)}_{n,n+1}\,,$
(3.10)

which yields

$a_{n,n+1}=\frac{1}{4^{n+1}}\sqrt{\pi}\frac{(n+4)^{n+1}}{(n+\tfrac{1}{2})!}\frac{(n+r+1)!}{r!}\,,$
(3.11)

and equation 3.11 holds for any integer $n\geq-1$. Note, also, that for $r=0$
since the amplitude reduces to the Veneziano amplitude, the partial-wave
coefficient of the leading Regge trajectory, given by equation 3.11, should
reduce to the result derived for that case. We have checked against the result
derived in [5] and this is indeed the case.

###  3.2 More on the Regge trajectories

In this section we wish to provide some, hopefully, useful expressions for the
partial-wave coefficients of the Regge trajectories. We start the discussion
by considering the case $a_{n,n}$. We work in an experimental manner. That is,
we compute the coefficients of interest using equation 3.1 for some low-lying
values of $n$ and manage to spot the pattern. Subsequently, we perform
numerous non-trivial tests of our expression. By non-trivial we mean against
values for the quantum number $n$ that lie outside the data-range used to
obtain the original expression. We find the following

$a_{n,n}=a_{n,n+1}~{}\frac{2n+1}{n+4}~{}2r\,.$ (3.12)

Let us make the comment that in the $r=0$ case the coefficients given by
equation 3.12 should coincide with those of the undeformed Veneziano
amplitude. It is useful at this point to remind ourselves of the fact that for
the undeformed Veneziano amplitude the partial-wave coefficients,
$a_{n,\ell}$, are always equal to zero when $n+\ell$ is an even number. This
agreement is manifest in equation 3.12.

Working in a similar vein for a couple more Regge trajectories, we observe
that there is a striking pattern. All the coefficients can be written as

$a_{n,n-\gamma}=a_{n,n+1}~{}\frac{2(n-\gamma)+1}{(n+4)^{\gamma+1}}~{}\frac{1}{2}\left((-1)^{\gamma}(r-1)+r+1\right)~{}\mathcal{P}_{\gamma}(n,r)$
(3.13)

where in the above $\mathcal{P}_{\gamma}(n,r)$ is a polynomial in $n$ and $r$.
The $r$-degree of the polynomial is
$\tfrac{1}{2}(-1)^{\gamma}\left((-1)^{\gamma}(2\gamma+1)-1\right)$ and only
the even powers in $r$ appear. The degree in $n$ is
$\tfrac{1}{4}(-1)^{\gamma}\left((-1)^{\gamma}(6\gamma+1)-1\right)$.
Unfortunately, we do not have a closed-form expression for all Regge
trajectories, however, we have closed-form expressions for
$\gamma=\\{0,1,\ldots,10\\}$. Since the polynomials become quite unwieldy,
below we provide the expressions for those that we will need in order to
constrain the parameter $r$ and more expressions can be found in appendix B.

$\displaystyle\mathcal{P}_{0}(n,r)$ $\displaystyle=2\,,$ (3.14)
$\displaystyle\mathcal{P}_{1}(n,r)$
$\displaystyle=(4n+2)r^{2}+\frac{n^{2}}{6}+\frac{19n}{6}+\frac{23}{3}\,,$
$\displaystyle\mathcal{P}_{2}(n,r)$
$\displaystyle=\left(\frac{16n^{2}}{3}-\frac{4}{3}\right)r^{2}+\frac{2n^{3}}{3}+\frac{43n^{2}}{3}+\frac{121n}{3}+\frac{50}{3}\,,$

###  3.3 The general coefficients

It is possible, with the relations that we have derived so far, to re-write
the general partial-wave coefficients given by equation 3.1. Algorithmically
we will work in the same way as above for the leading Regge trajectory,
$a_{n,n+1}$. The task at hand, now, is to compute the general term that scales
like $h^{-\ell-1}$ from both sides of the $h$-expansion. We re-state below
what we have obtained, for the reader’s convenience. The relation that we need
to use in order to get the $a_{n,\ell}$ is given by:

$\displaystyle\sum^{\infty}_{j=0}\frac{1}{2j+1}\frac{1}{h^{j+1}}a_{n,j}=\frac{r!}{(n+r+1)!}\sum^{n+1}_{k=0}(-1)^{n+1-k}s^{(k)}_{n+1}\frac{1}{2^{2k+1}}(n+4)^{k}\frac{1}{h^{k+1}}$
(3.15)
$\displaystyle\sum^{k}_{\mathfrak{p}=0}\frac{(-k)_{\mathfrak{p}}}{\mathfrak{p}!(2\mathfrak{p}+1)}\left[(h-1)^{2}+\frac{4h}{n+4}(r+2)\right]^{k-\mathfrak{p}}\sum^{2\mathfrak{p}+1}_{\mathfrak{m}=0}\binom{2\mathfrak{p}+1}{\mathfrak{m}}h^{\mathfrak{m}}(1+(-1)^{\mathfrak{m}})\,.$

We will use the binomial expansion

$(a+b)^{c}=\sum^{c}_{d=0}\binom{c}{d}a^{c-d}b^{d}\,,$ (3.16)

in order to re-write

$\left[(h-1)^{2}+\frac{4h}{n+4}(2+r)\right]^{k-\mathfrak{p}}=\sum^{k-\mathfrak{p}}_{\mathfrak{r}=0}\binom{k-\mathfrak{p}}{\mathfrak{r}}\left(\frac{4}{n+4}(2+r)\right)^{k-\mathfrak{p}-\mathfrak{r}}h^{k-\mathfrak{p}-\mathfrak{r}}(h-1)^{2\mathfrak{r}}\,,$
(3.17)

and subsequently

$(h-1)^{2\mathfrak{r}}=\sum^{2\mathfrak{r}}_{\mathfrak{z}=0}\binom{2\mathfrak{r}}{\mathfrak{z}}(-1)^{2\mathfrak{r}-\mathfrak{z}}h^{\mathfrak{z}}\,,$
(3.18)

thus arriving at

$\displaystyle\sum^{\infty}_{j=0}\frac{1}{2j+1}\frac{1}{h^{j+1}}a_{n,j}=$
(3.19)
$\displaystyle\frac{r!}{(n+r+1)!}\sum^{n+1}_{k=0}(-1)^{n+1-k}s^{(k)}_{n+1}\frac{1}{2^{2k+1}}(n+4)^{k}~{}\frac{1}{h^{k+1}}~{}\times~{}$
$\displaystyle\sum^{k}_{\mathfrak{p}=0}\sum^{2\mathfrak{p}+1}_{\mathfrak{m}=0}\sum^{k-\mathfrak{p}}_{\mathfrak{r}=0}\sum^{2\mathfrak{r}}_{\mathfrak{z}=0}\binom{2\mathfrak{p}+1}{\mathfrak{m}}\binom{k-\mathfrak{p}}{\mathfrak{r}}\binom{2\mathfrak{r}}{\mathfrak{z}}\left(\frac{4}{n+4}(r+2)\right)^{k-\mathfrak{p}-\mathfrak{r}}$
$\displaystyle\frac{(-k)_{\mathfrak{p}}}{\mathfrak{p}!(2\mathfrak{p}+1)}(-1)^{\mathfrak{z}}h^{\mathfrak{z}+\mathfrak{m}+k-\mathfrak{p}-\mathfrak{r}}(1+(-1)^{\mathfrak{m}})\,.$

From the above we can read-off the terms proportional to $h^{-\ell-1}$ from
both sides and the result yields888We have checked that our result, equation
3.20, matches the expression for the partial-wave coefficients in [21], Note
that there is a shift by one in the definitions of $n$ between the two works;
the $n$ here has to be shifted as $n\rightarrow n+1$ to match the result in
[21].:

$\displaystyle a_{n,\ell}=$
$\displaystyle\frac{\left(\ell+1\right)r!}{(n+r+1)!}\sum^{n+1}_{k=0}(-1)^{n+1-k}s^{(k)}_{n+1}\frac{1}{2^{2k+1}}(n+4)^{k}$
(3.20)
$\displaystyle\sum^{k}_{\mathfrak{p}=0}\sum^{2\mathfrak{p}+1}_{\mathfrak{m}=0}\sum^{k-\mathfrak{p}}_{\mathfrak{r}=0}\binom{2\mathfrak{p}+1}{\mathfrak{m}}\binom{k-\mathfrak{p}}{\mathfrak{r}}\binom{2\mathfrak{r}}{\mathfrak{p}+\mathfrak{r}-\ell-\mathfrak{m}}\frac{(-k)_{\mathfrak{p}}}{\mathfrak{p}!(2\mathfrak{p}+1)}$
$\displaystyle\left(\frac{4}{n+4}(r+2)\right)^{k-\mathfrak{p}-\mathfrak{r}}(-1)^{\mathfrak{p}+\mathfrak{r}-\ell-\mathfrak{m}}(1+(-1)^{\mathfrak{m}})$

###  3.4 A simpler expression for the general coefficients

We will derive a simpler expression for the $a_{n,\ell}$ coefficients compared
to equation 3.20. To do so, we remind ourselves of the definition of the
“pseudo generating function” $\mathcal{G}^{(r)}_{n,k}(h)$ in $D=4$ which is
given by:

$\mathcal{G}^{(r)}_{n,k}(h)=\int^{n+4}_{0}dt\frac{(-t+2+r)^{k}}{\sqrt{(h-1)^{2}+\tfrac{4ht}{n+4}}}\,.$
(3.21)

As we have seen, the integral in equation 3.21 can be performed analytically
in terms of the ordinary Gauss hypergeometrics functions,
${}_{2}F_{1}(a,b,c;x)$. There exists another way of obtaining the integral in
terms of the Appell hypergeometrics function, $F_{1}(a;b,c;d;x,y)$. The answer
is given by the following:

$\mathcal{G}^{(r)}_{n,k}(h)=2^{k}\frac{n+4}{h-1}F_{1}\left(1;-k,\tfrac{1}{2};2;\tfrac{n+4}{2+r},-\tfrac{4h}{(h-1)^{2}}\right)\,.$
(3.22)

We recall now that the formal definition of the Appell hypergeometrics
function as a power-series in the following manner

$F_{1}(a;b,c;d;x,y)=\sum^{\infty}_{e=0}\sum^{\infty}_{f=0}\frac{1}{e!}\frac{1}{f!}\frac{(a)_{e+f}(b)_{e}(c)_{f}}{(d)_{e+d}}x^{e}y^{f}\,,$
(3.23)

and we also make note of the simplification:

$\frac{(1)_{x+y}}{(2)_{x+y}}=1+x+y\ ,$ (3.24)

in order to obtain:

$\mathcal{G}^{(r)}_{n,k}(h)=(r+2)^{k}\frac{n+4}{h-1}\sum^{k}_{\mathfrak{u}=0}\sum^{\infty}_{\mathfrak{v}=0}\frac{1}{\mathfrak{u}!}\frac{1}{\mathfrak{v}!}\frac{(-k)_{\mathfrak{u}}(\tfrac{1}{2})_{\mathfrak{v}}}{1+\mathfrak{u}+\mathfrak{v}}\left(\frac{n+4}{r+2}\right)^{\mathfrak{u}}\left(-\frac{4h}{(h-1)^{2}}\right)^{\mathfrak{v}}\,.$
(3.25)

To proceed we perform the $\mathfrak{u}$-sum in the above relation and we
obtain

$\mathcal{G}^{(r)}_{n,k}(h)=(r+2)^{k}(n+4)\sum^{k}_{\mathfrak{u}=0}\frac{(\tfrac{1}{2})_{\mathfrak{v}}}{(\mathfrak{v}+1)!}{}_{2}F_{1}\left(-k,1+\mathfrak{v},2+\mathfrak{v};\tfrac{n+4}{r+2}\right)(-4)^{\mathfrak{v}}\frac{h^{\mathfrak{v}}}{(h-1)^{2\mathfrak{v}+1}}\,.$
(3.26)

Now, we can use, once more, the binomial expansion

$\frac{1}{(1+x)^{n}}=\sum^{\infty}_{y=0}\binom{n+y-1}{y}(-1)^{y}x^{y}\,,$
(3.27)

as well as the relation between the binomial coefficients and the Pochhammer
symbol

$\binom{x+y-1}{y}=\frac{(x)_{y}}{y!}\,,$ (3.28)

in order to re-write

$\frac{1}{(h-1)^{2\mathfrak{v}+1}}=(-1)^{-2\mathfrak{v}-1}\sum^{\infty}_{\mathfrak{r}=0}\frac{(2\mathfrak{v}+1)_{\mathfrak{r}}}{\mathfrak{r}!}(-1)^{2\mathfrak{r}}h^{\mathfrak{r}}\,,$
(3.29)

and thus the expression for $\mathcal{G}^{(r)}_{n,k}(h)$ becomes

$\displaystyle\mathcal{G}_{n,k}(h)=(r+2)^{k}(n+4)$
$\displaystyle\sum^{k}_{\mathfrak{v}=0}\frac{(\tfrac{1}{2})_{\mathfrak{v}}}{(\mathfrak{v}+1)!}{}_{2}F_{1}\left(-k,1+\mathfrak{v},2+\mathfrak{v};\tfrac{n+4}{r+2}\right)(-4)^{\mathfrak{v}}(-1)^{-2\mathfrak{v}-1}~{}h^{\mathfrak{v}}~{}$
(3.30)
$\displaystyle\sum^{\infty}_{\mathfrak{r}=0}\frac{(2\mathfrak{v}+1)_{\mathfrak{r}}}{\mathfrak{r}!}(-1)^{2\mathfrak{r}}h^{\mathfrak{r}}\,.$

Having obtained an explicit form as power series expansion for the “pseudo
generating function”, $\mathcal{G}_{n,k}(h)$, in terms of $h$, equation 3.30,
it is quite straightforward to extract the coefficient $a_{n,\ell}$. It is
given by999We have checked in this case as well that, equation 3.31, matches
the expression for the partial-wave coefficients in [21] as we did
previously.:

$\displaystyle a_{n,\ell}=$
$\displaystyle\frac{(2\ell+1)r!}{(n+4)(n+r+1)!}\sum_{k=0}^{n+1}(-1)^{n+1-k}s^{(k)}_{n+1}(r+2)^{k}(n+4)$
(3.31)
$\displaystyle\sum_{\mathfrak{v}=0}^{\ell}\frac{(\frac{1}{2})_{\mathfrak{v}}}{(\mathfrak{v}+1)!}{}_{2}F_{1}\left(-k,1+\mathfrak{v};2+\mathfrak{v};\tfrac{n+4}{r+2}\right)(-4)^{\mathfrak{v}}\frac{(2\mathfrak{v}+1)_{\ell-\mathfrak{v}}}{(\ell-\mathfrak{v})!}\,.$

Note that, while equation 3.31, is completely equivalent to equation 3.20 they
are distinct parametrisations of the partial-wave coefficients. Equation 3.31
comes with only two sums, instead of the four ones that appear in equation
3.20. An argument can be made in favour of both those equations in comparison
to equation 3.1 as they are just sums, rather than integration of the Legendre
polynomials times non-trivial functions.

Also, note that for $\ell=0$ we have just a single sum

$\displaystyle a_{n,0}=$
$\displaystyle\frac{r!}{(n+4)(n+r+1)!}\sum_{k=0}^{n+1}(-1)^{n+1-k}s^{(k)}_{n+1}(r+2)^{k}$
(3.32)
$\displaystyle\left(\frac{r+2}{k+1}+(-1)^{k}\frac{(r+2)^{-k}(n-r+2)^{k+1}}{k+1}\right)\,.$

##  4 The partial wave coefficients in $D$ dimensions

In this section, we wish to derive similar expressions as we did previously in
$D=4$ dimensions, section 3, but without specifying the number of spacetime
dimensions. The steps and basic relations that we need in order to manipulate
the expressions have already appeared in the previous section, and hence we
will proceed with a faster pace here.

For the reader’s convenience we record here again, the basic relation, as an
integral over the Gegenbauers, that gives the partial-wave coefficients
$a_{n,\ell}$

$\displaystyle a_{n,\ell}=$
$\displaystyle\frac{r!}{(n+r+1)!}\frac{1}{\mathcal{K}(\ell,\alpha)}\frac{1}{n+4}\left(\frac{4}{(n+4)^{2}}\right)^{\alpha-\tfrac{1}{2}}$
(4.1)
$\displaystyle\int^{n+4}_{0}dtC^{(\alpha)}_{\ell}\left(1-\frac{2t}{n+4}\right)(t(n+4-t))^{\alpha-\tfrac{1}{2}}(-t+2+r)_{n+1}\,.$

In order to proceed, we want to make use of the generating function of the
Gegenbauer polynomials

$\sum^{\infty}_{\ell=0}C^{(\alpha)}_{\ell}(x)t^{\ell}=\frac{1}{(1-2xt+t^{2})^{\alpha}}\,,$
(4.2)

the representation of the Pochhammer symbol in terms of the Stirling number of
the first kind, see equation 3.3, and also

$(n+4-t)^{\alpha-\tfrac{1}{2}}=\sum^{\infty}_{p=0}\binom{\alpha-\tfrac{1}{2}}{p}(-1)^{p}(n+4)^{\alpha-\tfrac{1}{2}-p}t^{p}\,.$
(4.3)

After using the above, we obtain the following:

$\displaystyle\sum^{\infty}_{j=0}\mathcal{K}(j,\alpha)a_{n,j}h^{j}$ (4.4)
$\displaystyle=\frac{r!}{(n+r+1)!}\frac{1}{n+4}\left[\frac{4}{(n+4)^{2}}\right]^{\alpha-\tfrac{1}{2}}\sum^{n+1}_{k=0}(-1)^{n+1-k}s^{(k)}_{n+1}\sum^{\infty}_{p=0}\binom{\alpha-\tfrac{1}{2}}{p}(-1)^{p}(n+4)^{\alpha-\tfrac{1}{2}-p}\times$
$\displaystyle\underbrace{\int^{n+4}_{0}dt~{}\frac{(-t+2+r)^{k}t^{p+\alpha-\tfrac{1}{2}}}{\left[(h-1)^{2}+\frac{4ht}{n+4}\right]^{\alpha}}}_{\mathcal{G}^{(\alpha)(r)}_{n,k,p}(h)}\,,$

where in the above relation, equation 4.4, we have defined the “pseudo
generating function” $\mathcal{G}^{(\alpha)(r)}_{n,k,p}(h)$. The integral can
be performed analytically and we obtain

$\displaystyle\mathcal{G}^{(\alpha)(r)}_{n,k,p}(h)=$
$\displaystyle\frac{2(r+2)^{k}}{2\alpha+2p+1}(n+4)^{\alpha+p+\tfrac{1}{2}}\frac{1}{(h-1)^{2\alpha}}$
(4.5) $\displaystyle
F_{1}\left(\alpha+p+\tfrac{1}{2};-k,\alpha;\alpha+p+\tfrac{3}{2};\tfrac{n+4}{r+2},-\tfrac{4h}{(h-1)^{2}}\right)\,.$

Now, we can proceed as we did in section 3.4, in order to re-write equation
4.5 in a form that is appropriate for our manipulations. Namely, we can use
the definition of the Appell hypergeometrics function as a power-series, which
is given by equation 3.23, alongside with the simplification

$\frac{(\alpha+p+\tfrac{1}{2})_{x+y}}{(\alpha+p+\tfrac{3}{2})_{x+y}}=\frac{1+2\alpha+2p}{1+2\alpha+2p+2x+2y}\,,$
(4.6)

and then analytically perform the $\mathfrak{u}$-sum, after which we need to
use the binomial expansion, equation 3.27, and the relation between the
binomial coefficients and the Pochhammer symbol given by equation 3.28 in
order to obtain

$\displaystyle\mathcal{G}^{(\alpha)(r)}_{n,k,p}(h)=2(r+2)^{k}(n+4)^{\alpha+p+\tfrac{1}{2}}$
$\displaystyle\sum^{k}_{\mathfrak{v}=0}\frac{(\alpha)_{\mathfrak{v}}}{\mathfrak{v}!}{}_{2}F_{1}\left(-k,\alpha+p+\tfrac{1}{2}+\mathfrak{v},\alpha+p+\tfrac{3}{2}+\mathfrak{v};\tfrac{n+4}{r+2}\right)$
(4.7)
$\displaystyle(-4)^{\mathfrak{v}}(-1)^{-2\mathfrak{v}-2\alpha}~{}h^{\mathfrak{v}}~{}\sum^{\infty}_{\mathfrak{r}=0}\frac{(2\mathfrak{v}+2\alpha)_{\mathfrak{r}}}{\mathfrak{r}!}(-1)^{2\mathfrak{r}}h^{\mathfrak{r}}\,.$

After the above simplifications, the equation we need to consider in order to
extract the partial-wave coefficients is given by:

$\displaystyle\sum^{\infty}_{j=0}\mathcal{K}(j,\alpha)a_{n,j}h^{j}=\frac{r!}{(n+r+1)!}4^{\alpha-\tfrac{1}{2}}\sum^{n+1}_{k=0}(-1)^{n+1-k}s^{(k)}_{n+1}\sum^{\infty}_{p=0}\binom{\alpha-\tfrac{1}{2}}{p}(-1)^{p}2(r+)2^{k}$
(4.8)
$\displaystyle\sum^{k}_{\mathfrak{v}=0}\frac{(\alpha)_{\mathfrak{v}}}{\mathfrak{v}!}\frac{1}{1+2\alpha+2p+2\mathfrak{v}}(-4)^{\mathfrak{v}}{}_{2}F_{1}\left(-k,\alpha+p+\tfrac{1}{2}+\mathfrak{v},\alpha+p+\tfrac{3}{2}+\mathfrak{v};\tfrac{n+4}{r+2}\right)$
$\displaystyle
h^{\mathfrak{v}}\sum^{\infty}_{\mathfrak{r}=0}\frac{(2\mathfrak{v}+2\alpha)_{\mathfrak{r}}}{\mathfrak{r}!}(-1)^{2\mathfrak{r}}h^{\mathfrak{r}}\,.$

###  4.1 The leading Regge trajectory

Using equation 4.8 we can read-off from both sides the terms that scale as
$h^{n+1}$ in order to extract the expression for the partial-wave coefficients
on the leading Regge trajectory, $a_{n,n+1}$. We have that:

$\displaystyle a_{n,n+1}=$
$\displaystyle\frac{r!}{(n+r+1)!}\frac{2(r+2)^{n+1}}{\mathcal{K}(n+1,\alpha)}4^{\alpha-\tfrac{1}{2}}\sum^{\infty}_{p=0}\sum^{n+1}_{\mathfrak{v}=0}\binom{\alpha-\tfrac{1}{2}}{p}\frac{(\alpha)_{\mathfrak{v}}}{\mathfrak{v}!}\frac{(-1)^{p}(-4)^{\mathfrak{v}}}{1+2\alpha+2p+2\mathfrak{v}}$
(4.9)
$\displaystyle{}_{2}F_{1}\left(-n-1,\alpha+p+\tfrac{1}{2}+\mathfrak{v},\alpha+p+\tfrac{3}{2}+\mathfrak{v};\tfrac{n+4}{r+2}\right)\frac{(2\mathfrak{v}+2\alpha)_{n+1-\mathfrak{v}}}{(n+1-\mathfrak{v})!}\,.$

###  4.2 More on the Regge trajectories

While, equation 4.9 is a formal derivation for the partial-wave coefficients
of the leading Regge trajectories in arbitrary $D$-dimensions, it looks more
like a simplified re-writing of the integral, rather than a helpful
expression. Again, we can work experimentally in order to derive

$a_{n,n+1}=\frac{(n+4)^{n+1}(n+1)!}{4^{n}}\frac{1}{(D-3)(D-1)}\frac{r!}{(n+r+1)!}\frac{1}{\left(\frac{D+2}{2}\right)_{n-1}}\,,$
(4.10)

for the leading Regge trajectory. It turns out that a similar behaviour to the
one in the $D=4$ case is observed here as well. The sub-leading trajectories
can be expressed as

$a_{n,n-\gamma}=a_{n,n+1}~{}\frac{1}{(n+4)^{\gamma+1}}~{}\frac{1}{2}\left(1+(-1)^{\gamma}(r-1)+r\right)~{}\left(D-3+2(n-\gamma)\right)~{}\mathbb{P}_{\gamma}(n,D,r)\,,$
(4.11)

where in the above $\mathbb{P}_{\gamma}(n,d,r)$ is a polynomial, which we do
not have in a closed-form for all levels. For the first few we have the
expressions:

$\displaystyle\mathbb{P}_{0}(n,D,r)$ $\displaystyle=2\,,$ (4.12)
$\displaystyle\mathbb{P}_{1}(n,D,r)$
$\displaystyle=\frac{12r^{2}(D+2n-3)-Dn-2D+n^{2}+23n+54}{6(n+4)^{2}}\,,$
$\displaystyle\mathbb{P}_{2}(n,D,r)$
$\displaystyle=\frac{(D+2n-3)\left(-Dn-2D+n^{2}+25n+58\right)+4r^{2}(D+2n-5)(D+2n-3)}{3(n+4)^{3}}\,.$

###  4.3 The general coefficients

Finally, it is a straightforward exercise to extract the $h^{\ell}$ term from
both sides of equation 4.8 in order to obtain the expression for all partial-
wave coefficients. It is given by101010We have checked in this case also that,
equation 4.13, matches the expression for the partial-wave coefficients in
[21] as we did in the previous cases.:

$\displaystyle a_{n,\ell}=$
$\displaystyle\frac{r!}{(n+r+1)!}\frac{1}{\mathcal{K}(\ell,\alpha)}4^{\alpha-\tfrac{1}{2}}\sum_{k=0}^{n+1}(-1)^{n+1-k}s^{(k)}_{n+1}(r+2)^{k}\sum^{\infty}_{p=0}\binom{\alpha-\tfrac{1}{2}}{p}(-1)^{p}$
(4.13)
$\displaystyle\sum_{\mathfrak{v}=0}^{\ell}\frac{(\alpha)_{\mathfrak{v}}}{\mathfrak{v}!}\frac{1}{1+2\alpha+2p+2\mathfrak{v}}{}_{2}F_{1}\left(-k,\alpha+p+\tfrac{1}{2}+\mathfrak{v},\alpha+p+\tfrac{3}{2}+\mathfrak{v};\tfrac{n+4}{r+2}\right)$
$\displaystyle(-4)^{\mathfrak{v}}\frac{(2\mathfrak{v}+2\alpha)_{\ell-\mathfrak{v}}}{(\ell-\mathfrak{v})!}\,.$

##  5 Comments on unitarity

The unitarity of the underlying theory, no matter what the theory, is directly
related to the positivity of the partial-wave coefficients that we derived in
the previous sections. The reason is that negative partial-wave coefficients
indicate the exchange of ghost states. Hence, requiring that the partial-wave
coefficients are non-negative numbers, is equivalent to the requirement that
the theory is ghost-free.

The deformation parameter $r$ that enters equation 2.1 can be any real number.
With that in mind, we start from the expressions we derived for the $D=4$
case. From equation 3.11 and for $n=0$ we obtain

$a_{0,1}=2\frac{1}{r+1}\,,$ (5.1)

from which we conclude that $r>-1$ and this is our first unitarity bound.

A second more stringent bound comes from the study of $a_{n,n}$ given by
equation 3.12 already at $n=0$. We obtain

$a_{0,0}=r\frac{1}{r+1}\,,$ (5.2)

and requiring positivity of the above yields

$r<-1\lor r\geq 0\,.$ (5.3)

From the above, we conclude that $r\geq 0$, since the $r<-1$ solution does not
allow the deformation parameter $r$ to become $0$ and thus undo the
deformation of the Veneziano amplitude. The study of coefficients derived from
the sub-leading trajectories, does not lead to further bounds on the allowed
values of the parameter $r$.

We continue to the general $D$ dimensions and check if we can derive any
bounds on the allowed value of $D$. Here, we have already derived $r\geq 0$
and we will use this as an input. Note, also, that we are interested in
integer spacetime dimensions $D$. The first non-trivial constraints in this
case come from the $a_{n,n-1}$ and $n=1$. We have the expression

$a_{1,0}=\frac{12(D-1)r^{2}-3D+78}{12(D-1)(r+1)(r+2)}\,,$ (5.4)

the positivity of which leads to

$\left(r>0\land D\leq 4(D-1)r^{2}+26\right)\lor 2r\geq 1\lor D\leq 26\,.$
(5.5)

Clearly, the first part, which reads $\left(r>0\land D\leq
4(D-1)r^{2}+26\right)$ is inconsistent as it does not allow the deformation
parameter to return to the value $r=0$ and thus to obtain the un-deformed
Veneziano amplitude. For the same reason we can exclude the second solution,
$2r\geq 1$, and we are only left with $D\leq 26$. Namely the underlying theory
has to live below the critical dimension of string theory. Since, $r\geq 0$
one could imagine that the partial-wave coefficients of the Veneziano
amplitude going to negative values above the critical dimensions of string
theory could become positive for some appropriate value of $r$, however, such
is not the case.

Before concluding this section we would like to add some clarifying comments.

As we have already mentioned, in [21] the authors provided an expression for
all partial-wave coefficients as a double-sum. Upon explicit evaluations of
the relation, they were able to derive bounds on the allowed values for the
deformation parameter, $r$ and the number of spacetime dimensions, $D$.

The results we have obtained here are in agreement with those presented in
[21] upon the appropriate shift that we have already mentioned in the previous
section and for the special case of $m^{2}_{0}=-1$. While our formulae given
by equations 3.20 and 3.31 for the special case of $D=4$ and equation 4.13 for
general-$D$ dimensions are equivalent to the result obtained in [21] they are
distinct parametrisations of the partial-wave coefficients.

Furthermore, the general expressions for all partial-wave coefficients that
appear both here and in [21] are not expressed in terms of simple analytic
functions, but rather as sums. This is more of a formal re-writing of the
coefficients, rather than a straightforward expression that allows the non-
negativity of the said coefficients to be manifest. Taking that into
consideration, the formulae describing the partial-wave coefficients on the
Regge trajectories, while not formally derived, appear to be more useful in a
practical sense.

##  6 Epilogue

In this work, we focused on examining the following hypergeometric deformation
of the Veneziano amplitude

$\mathcal{A}(s,t)=\frac{\Gamma(-s-1)\Gamma(-t-1)}{\Gamma(-s-t-2)}{}_{3}F_{2}\left(-s-1,-t-1,r;-s-t-2,1+r;1\right)\,,$
(6.1)

that was derived in [21]. Using the decomposition into partial waves, we were
able to derive bounds on the deformation parameter $r$ and the number of
spacetime dimensions $D$, namely

$r\geq 0\,,\qquad\text{and}\qquad D\leq 26\,,$ (6.2)

based on the requirement that the coefficients in the partial-wave
decomposition are non-negative numbers. This requirement is a consequence of
the unitarity of the underlying theory.

We find it quite remarkable that even though the deformation parameter $r$
could be any real number at the beginning, and naively one could expect that
this would not allow to derive any bounds on the dimensions of the underlying
theory, the positivity of the coefficients in the expansion requires that $D$
is bounded from above. More extraordinary is the fact that this upper bound
matches precisely the critical dimensions of the bosonic string.

As we have mentioned already in the introduction, generalising the Veneziano
amplitude is motivated from many different points of view, one of which is a
question on the uniqueness of string theory. Being able to derive the critical
dimension of the string from the examination of the hypergeometric Veneziano
amplitude, is not a proof that string theory is unique, however, it can be
seen as suggestive evidence. Of course, as was explained in [20], one can
argue that the input assumptions used as constraints to bootstrap equation 6.1
were neither strong nor restrictive enough, and hence it is not a big surprise
that new mathematical functions were found that satisfy the bootstrap
conditions.

We hope that this work is a first step towards the more systematic study of
these new and exciting hypergeometric amplitudes, supplementing and extending
the unitarity analysis of [21].

There are many exciting and interesting avenues for future work.

The first and most straightforward path, would be to consider performing a
similar analysis for different values of $m^{2}_{0}$. There is already
numerical evidence from [21] that for different values of $m^{2}_{0}$ the
deformation parameter can be negative without violating the unitarity of the
underlying theory111111We are grateful to Grant Remmen for stressing this
possibility to us..

We still do not know the underlying theory of the amplitude given by equation
6.1, if any. Taking our findings into consideration, as well as the fact that
this four-point amplitude has an integral representation in terms of the Koba-
Nielsen formula [21], perhaps the first and more natural place to try and look
for an answer would be some $2\rightarrow 2$ scattering process within string
theory itself.

Furthermore, it is well-known that the Veneziano and the Virasoro-Shapiro
amplitudes given by121212We are considering the scattering of four tachyons of
mass $\alpha^{\prime}m^{2}=-1$ and $\alpha^{\prime}m^{2}=-4$ in open and
closed string theory respectively, and we are following conventions in which
$\alpha^{\prime}=1$ for open strings and $\alpha^{\prime}=4$ for closed-string
theory. We have used $s+t+u=4m^{2}$.:

$\displaystyle\mathcal{A}_{\text{ven}}$
$\displaystyle=\frac{\Gamma(-s-1)\Gamma(-t-1)}{\Gamma(-s-t-2)}\,,$ (6.3)
$\displaystyle\mathcal{A}_{\text{vs}}$
$\displaystyle=\frac{\Gamma(-s-1)\Gamma(-t-1)\Gamma(s+t+3)}{\Gamma(s+2)\Gamma(t+2)\Gamma(-s-t-2)}\,,$

are related via the KLT relation [22]

$\mathcal{A}_{\text{vs}}=\underbrace{\frac{\sin(\pi s)\sin(\pi
t)}{\pi\sin(\pi(-s-t))}}_{\text{kernel}}\mathcal{A}^{2}_{\text{ven}}\,.$ (6.4)

It would be very interesting to examine if a similar relation holds true for
the hypergeometric deformations of the Veneziano and Virasoro-Shapiro
amplitudes, to derive the kernel in this generalised context, and thus the
generalised KLT relation.

Finally, a straightforward path is to consider the hypergeometric Coon
amplitude that was, also, derived in [21] and attempt to obtain the
corresponding unitarity bounds for the deformation parameters, $q$ and $r$,
and perhaps the spacetime dimensions $D$ in that case. Note that this is the
most general construction in terms of the hypergeometric deformations that
were discussed in that article and certain limits can be taken in order to
derive equation 6.1 from that. We believe that the bounds derived here can be
used as useful input in order to derive bounds on the allowed region of values
that the parameters can have in the case of the hypergeometric Coon amplitude.

## Acknowledgments

We are grateful to James M. Drummond for bringing [21] to our attention and
suggesting to carry out the unitarity analysis. We have greatly benefited from
discussions with James M. Drummond, Pronobesh Maity, and Theodoros Nakas
throughout the various stages of this project. We are, also, indebted to Grant
Remmen, and Xinan Zhou for reading a draft of this work and offering their
valuable insight and comments. Finally, we would like to acknowledge the
hospitality of ShanghaiTech University where parts of this work were
completed. The work of KCR is supported by starting funds from University of
Chinese Academy of Sciences (UCAS), the Kavli Institute for Theoretical
Sciences (KITS), and the Fundamental Research Funds for the Central
Universities.

##  Appendix A Partial-wave coefficients that are equal to zero

###  A.1 The effect of a non-vanishing value for the r-parameter

Let us discuss the first point of section 2 at a bit more depth in order to
showcase our argument. To do so, we remind ourselves that in the case of the
Veneziano amplitude, in order to prove that $a_{n,\ell}=0$ when $n+\ell$ is
equal to an even number, we have to consider the shift $t=n+4-t^{\prime}$ [5].
Using the properties $(-x)_{y}=(-1)^{y}(x-y+1)_{y}$ and
$C^{(\alpha)}_{\ell}(-x)=(-1)^{\ell}C^{(\alpha)}_{\ell}(x)$ and the integral
of equation 2.7 becomes

$\displaystyle\int^{n+4}_{0}dtC_{\ell}^{(\alpha)}\left(1-\frac{2t}{n+4}\right)(t(n+4-t))^{\alpha-\tfrac{1}{2}}(-t+2+r)_{n+1}=$
(A.1) $\displaystyle-(-1)^{n+\ell}$
$\displaystyle\int^{n+4}_{0}dtC_{\ell}^{(\alpha)}\left(1-\frac{2t}{n+4}\right)(t(n+4-t))^{\alpha-\tfrac{1}{2}}(-t+2-r)_{n+1}\,,$

where in the above we renamed $t^{\prime}$ as $t$ after using the properties.
Now, in the Veneziano case, which is the case $r=0$, the original integral is
just re-written as $-(-1)^{n+\ell}$ times itself, and hence for $n+\ell$ any
even number the result is zero. It is clear, that due to the presence of a
non-zero $r$ this is no longer the case.

###  A.2 Vanishing coefficients in $D=4$

Here we will specify the discussion to the case $D=4$. In this case, we have
seen that the expression for the partial-wave coefficient simplifies
drastically, see equation 3.1. Let us focus on the integral of that
expression, given by

$\int^{n+4}_{0}dtP_{\ell}\left(1-\frac{2t}{n+4}\right)(-t+2+r)_{n+1}\,.$ (A.2)

We will use that the Legendre polynomials satisfy

$P_{\ell}\left(1-\frac{2t}{n+4}\right)=\sum^{\ell}_{k=0}\binom{\ell}{k}\binom{\ell+k}{k}\left(-\frac{t}{n+4}\right)^{k}\,,$
(A.3)

in order to re-write equation A.2 as:

$\sum^{\ell}_{k=0}\binom{\ell}{k}\binom{\ell+k}{k}\left(-\frac{1}{n+4}\right)^{k}\int^{n+4}_{0}dtt^{k}\underbrace{\left(-t+2\right)\left(-t+3\right)\ldots\left(-t+2+n+r\right)}_{(n+1)-\text{terms}}\,.$
(A.4)

Let us consider that $\mathfrak{a}_{j}$ is the coefficient of the term $t^{j}$
in the above and hence we have that equation A.4 becomes

$\sum^{n+1}_{j=0}\mathfrak{a}_{j}(n+4)^{j+1}\mathcal{T}\left(\ell,j\right)\,,$
(A.5)

where in the above we have defined:

$\mathcal{T}(\ell,j)=\sum^{\ell}_{k=0}(-1)^{k}\binom{\ell}{k}\binom{\ell+k}{k}\frac{1}{k+j+1}\,.$
(A.6)

Notice that equation A.6 can be written as:

$\mathcal{T}(\ell,j)=\int^{1}_{0}dz~{}z^{j}\widetilde{P}_{\ell}(z)\,,$ (A.7)

with $\widetilde{P}_{\ell}(z)$ being the shifted Legendre polynomial that
satisfy

$\widetilde{P}_{\ell}(z)=P_{\ell}(1-2z)=\sum^{\ell}_{k=0}(-1)^{k}\binom{\ell}{k}\binom{\ell+k}{k}z^{k}\,.$
(A.8)

Recall that the task at hand was to evaluate $\mathcal{T}(\ell,j)$. The
integral in equation A.7 can be evaluated to be

$\mathcal{T}(\ell,j)=\frac{(-j)_{\ell}}{(j+1)(j+2)_{\ell}}\,.$ (A.9)

From equation A.9 we conclude that $\mathcal{T}(\ell\geq n+2,j)$ for any
$j=0,1,\ldots,n+1$.

###  A.3 Vanishing coefficients in any $D$

Now, we proceed to compute the integral and show the vanishing of the partial-
wave coefficients in any dimensions for $\ell\geq n+2$. We begin by
considering the following representation of the Gegenbauer polynomials

$C^{(\alpha)}_{\ell}(x)=\frac{\left(2\alpha\right)_{\ell}}{\ell!}{}_{2}F_{1}\left(-n,2\alpha+n,\alpha+\tfrac{1}{2};\tfrac{1-x}{2}\right)\,,$
(A.10)

which can be written in the more convenient, for our purposes, form

$C^{(\alpha)}_{\ell}(x)=\frac{\left(2\alpha\right)_{\ell}}{\ell!}\sum^{\ell}_{j=0}\binom{n}{j}\frac{\left(2\alpha+n\right)_{j}}{\left(\alpha+\tfrac{1}{2}\right)_{j}}\left(\frac{x-1}{2}\right)^{j}\,.$
(A.11)

Using equation A.11, the integral appearing in equation 2.7 becomes

$\int^{n+4}_{0}dt\frac{\left(2\alpha\right)_{\ell}}{\ell!}\sum^{\ell}_{k=0}\binom{\ell}{k}\frac{\left(2\alpha+\ell\right)_{k}}{\left(\alpha+\tfrac{1}{2}\right)_{k}}\left(-\frac{1}{n+4}\right)^{k}t^{k}t^{\alpha-\tfrac{1}{2}}\left(-t+n+4\right)^{\alpha-\tfrac{1}{2}}\left(-t+2+r\right)_{n+1}\,.$
(A.12)

In the above, $\left(-t+2+r\right)_{n+1}$ has in total $(n+1)$-terms of the
form:
$\left(-t+2+r\right)\left(-t+3+r\right)\ldots\left(-t+2+n+r\right)$.
Additionally, we can use the binomial theorem to express
$\left(-t+n+4\right)^{\alpha-\tfrac{1}{2}}$ as:

$\left(-t+n+4\right)^{\alpha-\tfrac{1}{2}}=\sum^{\infty}_{p=0}\binom{\alpha-\tfrac{1}{2}}{p}(-1)^{p}(n+4)^{\alpha-\tfrac{1}{2}-p}t^{p}\,.$
(A.13)

Now, we consider that $a_{j}$ is the coefficient of $t^{j}$ in
$\left(-t+2+r\right)\left(-t+3+r\right)\ldots\left(-t+2+n+r\right)$ and
equation A.12 becomes

$\frac{\left(2\alpha\right)_{\ell}}{\ell!}\sum^{n+1}_{j=0}a_{j}(n+4)^{j+2\alpha}\mathcal{T}(\ell,j,\alpha)\,,$
(A.14)

where in the above

$\mathcal{T}(\ell,j,\alpha)=\sum^{\ell}_{k=0}\binom{\ell}{k}\frac{\left(2\alpha+\ell\right)_{k}}{\left(\alpha+\tfrac{1}{2}\right)_{k}}(-1)^{k}\sum^{\infty}_{p=0}\binom{\alpha-\tfrac{1}{2}}{p}(-1)^{p}\frac{2}{1+2\alpha+2j+2k+2p}\,.$
(A.15)

The sums in equation A.15 can be performed analytically and we obtain

$\mathcal{T}(\ell,j,\alpha)=\left[\Gamma\left(\alpha+\tfrac{1}{2}\right)\right]^{2}\Gamma\left(\alpha+j+\tfrac{1}{2}\right){}_{3}\mathcal{F}_{2}\left(\\{\alpha+j+\tfrac{1}{2},-\ell,\ell+2\alpha\\};\\{\alpha+\tfrac{1}{2},2\alpha+j+1\\};1\right)\,,$
(A.16)

where in the above we have used
${}_{p}\mathcal{F}_{q}(\\{a_{1},a_{2},\ldots,a_{p}\\};\\{b_{1},b_{2},\ldots,b_{q}\\};z)$
to denote the regularised hypergeometric function, which is given in terms of
the generalised hypergeometric function as:

${}_{p}\mathcal{F}_{q}(\\{a_{1},a_{2},\ldots,a_{p}\\};\\{b_{1},b_{2},\ldots,b_{q}\\};z)=\frac{1}{\Gamma(b_{1})\Gamma(b_{2})\ldots\Gamma(b_{q})}{}_{p}F_{q}(\\{a_{1},a_{2},\ldots,a_{p}\\};\\{b_{1},b_{2},\ldots,b_{q}\\};z)\,,$
(A.17)

From equation A.16 we can conclude that $\mathcal{T}(\ell\geq n+2,j,\alpha)=0$
for any number of spacetime dimensions $D$ and any $j=0,1,\ldots,n+1$.

##  Appendix B The polynomials for the Regge trajectories in $D=4$ dimensions

In this appendix we provide some additional examples for the polynomials
governing the Regge trajectories in $D=4$ dimensions from section 3.2.

$\displaystyle\mathcal{P}_{3}(n,r)=$
$\displaystyle\left(\frac{16n^{3}}{3}-8n^{2}-\frac{4n}{3}+2\right)r^{4}+\left(\frac{4n^{4}}{3}+\frac{92n^{3}}{3}+\frac{215n^{2}}{3}-\frac{23n}{3}-18\right)r^{2}+$
(B.1)
$\displaystyle\frac{n^{5}}{36}+\frac{401n^{4}}{360}+\frac{1367n^{3}}{90}+\frac{23503n^{2}}{360}+\frac{5923n}{60}+\frac{523}{15}\,,$
$\displaystyle\mathcal{P}_{4}(n,r)=$
$\displaystyle\left(\frac{64n^{4}}{15}-\frac{256n^{3}}{15}+\frac{224n^{2}}{15}+\frac{64n}{15}-4\right)r^{4}+$
$\displaystyle\left(\frac{16n^{5}}{9}+\frac{376n^{4}}{9}+36n^{3}-\frac{1486n^{2}}{9}-\frac{82n}{9}+\frac{116}{3}\right)r^{2}+$
$\displaystyle\frac{n^{6}}{9}+\frac{218n^{5}}{45}+\frac{4157n^{4}}{60}+\frac{24743n^{3}}{90}+\frac{55921n^{2}}{180}-\frac{2071n}{30}-82\,,$
$\displaystyle\mathcal{P}_{5}(n,r)=$
$\displaystyle\left(\frac{128n^{5}}{45}-\frac{64n^{4}}{3}+\frac{448n^{3}}{9}-32n^{2}-\frac{568n}{45}+\frac{28}{3}\right)r^{6}+$
$\displaystyle\left(\frac{16n^{6}}{9}+\frac{368n^{5}}{9}-\frac{680n^{4}}{9}-\frac{2440n^{3}}{9}+\frac{3889n^{2}}{9}+\frac{587n}{9}-\frac{310}{3}\right)r^{4}+$
$\displaystyle\left(\frac{2n^{7}}{9}+\frac{461n^{6}}{45}+\frac{13139n^{5}}{90}+\frac{14515n^{4}}{36}-\frac{2221n^{3}}{9}-\frac{224179n^{2}}{180}+\frac{1577n}{30}+286\right)r^{2}+$
$\displaystyle\frac{n^{8}}{324}+\frac{103n^{7}}{540}+\frac{72011n^{6}}{15120}+\frac{31553n^{5}}{560}+\frac{285899n^{4}}{1008}+\frac{1033789n^{3}}{1680}+$
$\displaystyle\frac{4928053n^{2}}{11340}-\frac{594829n}{3780}-\frac{1138}{9}\,,$
$\displaystyle\mathcal{P}_{6}(n,r)=$
$\displaystyle\left(\frac{512n^{6}}{315}-\frac{2048n^{5}}{105}+\frac{5248n^{4}}{63}-\frac{1024n^{3}}{7}+\frac{23648n^{2}}{315}+\frac{3968n}{105}-24\right)r^{6}+$
$\displaystyle\left(\frac{64n^{7}}{45}+\frac{1376n^{6}}{45}-\frac{8576n^{5}}{45}+\frac{16n^{4}}{9}+\frac{52756n^{3}}{45}-\frac{55546n^{2}}{45}-\frac{1406n}{5}+308\right)r^{4}+$
$\displaystyle\left(\frac{8n^{8}}{27}+\frac{1904n^{7}}{135}+\frac{5108n^{6}}{27}+\frac{8144n^{5}}{135}-\frac{105671n^{4}}{54}-\frac{126079n^{3}}{135}+\right.$
$\displaystyle\left.\frac{87601n^{2}}{18}+\frac{10327n}{45}-\frac{3292}{3}\right)r^{2}+$
$\displaystyle\frac{n^{9}}{81}+\frac{221n^{8}}{270}+\frac{16141n^{7}}{756}+\frac{69851n^{6}}{280}+\frac{2588081n^{5}}{2520}+$
$\displaystyle\frac{272901n^{4}}{280}-\frac{11443967n^{3}}{4536}-\frac{7777853n^{2}}{1890}+\frac{356711n}{630}+964\,,$

$\displaystyle\mathcal{P}_{7}(n,r)=$
$\displaystyle\left(\frac{256n^{7}}{315}-\frac{128n^{6}}{9}+\frac{4288n^{5}}{45}-\frac{2720n^{4}}{9}+\frac{19792n^{3}}{45}-\frac{1688n^{2}}{9}-\frac{12172n}{105}+66\right)r^{8}+$
(B.2)
$\displaystyle\left(\frac{128n^{8}}{135}+\frac{2432n^{7}}{135}-\frac{32096n^{6}}{135}+\frac{16864n^{5}}{27}+\right.$
$\displaystyle\left.\frac{107992n^{4}}{135}-\frac{620152n^{3}}{135}+\frac{504206n^{2}}{135}+\frac{9982n}{9}-980\right)r^{6}+$
$\displaystyle\left(\frac{8n^{9}}{27}+\frac{1924n^{8}}{135}+\frac{22508n^{7}}{135}-\frac{94526n^{6}}{135}-\frac{677699n^{5}}{270}+\right.$
$\displaystyle\left.\frac{3716929n^{4}}{540}+\frac{1240327n^{3}}{135}-\frac{3456947n^{2}}{180}-\frac{64279n}{30}+4382\right)r^{4}+$
$\displaystyle\left(\frac{2n^{10}}{81}+\frac{698n^{9}}{405}+\frac{8641n^{8}}{189}+\frac{468323n^{7}}{945}+\right.$
$\displaystyle\left.\frac{75233n^{6}}{72}-\frac{1451983n^{5}}{360}-\frac{8123005n^{4}}{648}+\frac{18345539n^{3}}{3240}+\right.$
$\displaystyle\left.\frac{10273985n^{2}}{378}-\frac{737869n}{630}-6028\right)r^{2}+$
$\displaystyle\frac{n^{11}}{3888}+\frac{841n^{10}}{38880}+\frac{2152081n^{9}}{2721600}+\frac{5665843n^{8}}{362880}+\frac{5406743n^{7}}{32400}+\frac{30769397n^{6}}{36288}+$
$\displaystyle\frac{972735997n^{5}}{544320}-\frac{38975383n^{4}}{1088640}-\frac{7255356281n^{3}}{1360800}-\frac{460545103n^{2}}{90720}+\frac{15354599n}{12600}+\frac{18871}{15}\,,$

$\displaystyle\mathcal{P}_{8}(n,r)=$
$\displaystyle\left(\frac{1024n^{8}}{2835}-\frac{8192n^{7}}{945}+\frac{11264n^{6}}{135}-\frac{2048n^{5}}{5}+\frac{144256n^{4}}{135}-\frac{60928n^{3}}{45}+\right.$
(B.3)
$\displaystyle\left.\frac{1390016n^{2}}{2835}+\frac{344192n}{945}-\frac{572}{3}\right)r^{8}+$
$\displaystyle\left(\frac{512n^{9}}{945}+\frac{7936n^{8}}{945}-\frac{13184n^{7}}{63}+\frac{53696n^{6}}{45}-\frac{74464n^{5}}{45}-\frac{241136n^{4}}{45}+\right.$
$\displaystyle\left.\frac{3306440n^{3}}{189}-\frac{11112316n^{2}}{945}-\frac{1344076n}{315}+3256\right)r^{6}+$
$\displaystyle\left(\frac{32n^{10}}{135}+\frac{7616n^{9}}{675}+\frac{7592n^{8}}{75}-\frac{303824n^{7}}{225}+\frac{134866n^{6}}{225}+\right.$
$\displaystyle\left.\frac{3756572n^{5}}{225}-\frac{27364001n^{4}}{1350}-\frac{36813233n^{3}}{675}+\frac{34407883n^{2}}{450}+\frac{945907n}{75}-\frac{89292}{5}\right)r^{4}+$
$\displaystyle\left(\frac{8n^{11}}{243}+\frac{964n^{10}}{405}+\frac{531536n^{9}}{8505}+\frac{1619614n^{8}}{2835}-\frac{1250071n^{7}}{1134}-\frac{2318551n^{6}}{180}+\right.$
$\displaystyle\left.\frac{12720941n^{5}}{4860}+\frac{133287227n^{4}}{1620}+\frac{77483591n^{3}}{6804}-\frac{449387851n^{2}}{2835}-\frac{2828951n}{945}+\frac{104056}{3}\right)r^{2}+$
$\displaystyle\frac{n^{12}}{972}+\frac{112n^{11}}{1215}+\frac{1196773n^{10}}{340200}+\frac{11905489n^{9}}{170100}+\frac{70047469n^{8}}{100800}+$
$\displaystyle\frac{548246269n^{7}}{226800}-\frac{66696559n^{6}}{38880}-\frac{48406423n^{5}}{1944}-\frac{79199036753n^{4}}{2721600}+\frac{31022789003n^{3}}{680400}+$
$\displaystyle\frac{17353447943n^{2}}{226800}-\frac{20749049n}{2100}-\frac{259286}{15}\,,$

$\displaystyle\mathcal{P}_{9}(n,r)=$
$\displaystyle\left(\frac{2048n^{9}}{14175}-\frac{1024n^{8}}{225}+\frac{280576n^{7}}{4725}-\frac{93184n^{6}}{225}+\right.$
(B.4)
$\displaystyle\left.\frac{1117952n^{5}}{675}-\frac{843136n^{4}}{225}+\frac{60356992n^{3}}{14175}-\frac{99392n^{2}}{75}-\frac{368216n}{315}+572\right)r^{10}+$
$\displaystyle\left(\frac{256n^{10}}{945}+\frac{2816n^{9}}{945}-\frac{15104n^{8}}{105}+\frac{433408n^{7}}{315}-\frac{228128n^{6}}{45}+\frac{138272n^{5}}{45}+\right.$
$\displaystyle\left.\frac{25833776n^{4}}{945}-\frac{62429264n^{3}}{945}+\frac{12001739n^{2}}{315}+\frac{1711723n}{105}-11154\right)r^{8}+$
$\displaystyle\left(\frac{64n^{11}}{405}+\frac{14752n^{10}}{2025}+\frac{24752n^{9}}{675}-\frac{1002248n^{8}}{675}+\frac{4159412n^{7}}{675}+\right.$
$\displaystyle\left.\frac{6835234n^{6}}{675}-\frac{172516181n^{5}}{2025}+\frac{181332943n^{4}}{4050}+\right.$
$\displaystyle\left.\frac{189778778n^{3}}{675}-\frac{413439001n^{2}}{1350}-\frac{2927713n}{45}+73612\right)r^{6}+$
$\displaystyle\left(\frac{8n^{12}}{243}+\frac{328n^{11}}{135}+\frac{521078n^{10}}{8505}+\frac{1100114n^{9}}{2835}-\frac{25659041n^{8}}{5670}-\frac{19007911n^{7}}{1890}+\right.$
$\displaystyle\left.\frac{145457699n^{6}}{1944}+\frac{93677327n^{5}}{1080}-\frac{30023039233n^{4}}{68040}-\frac{5731391107n^{3}}{22680}+\right.$
$\displaystyle\left.\frac{1636399621n^{2}}{1890}+\frac{36484051n}{630}-190028\right)r^{4}+$
$\displaystyle\left(\frac{n^{13}}{486}+\frac{941n^{12}}{4860}+\frac{1288333n^{11}}{170100}+\frac{50049121n^{10}}{340200}+\frac{557418061n^{9}}{453600}+\right.$
$\displaystyle\left.\frac{137847077n^{8}}{907200}-\frac{933350689n^{7}}{34020}-\frac{2027689027n^{6}}{38880}+\frac{212491115287n^{5}}{1360800}+\right.$
$\displaystyle\left.\frac{1087556641399n^{4}}{2721600}-\frac{50810550289n^{3}}{226800}-\frac{2094388529n^{2}}{2800}+\frac{11761469n}{252}+162838\right)r^{2}+$
$\displaystyle\frac{n^{14}}{58320}+\frac{107n^{13}}{58320}+\frac{362693n^{12}}{4082400}+\frac{10121317n^{11}}{4082400}+\frac{1667002441n^{10}}{39916800}+\frac{3207436571n^{9}}{7983360}+$
$\displaystyle\frac{163569574211n^{8}}{89812800}+\frac{68609574763n^{7}}{35925120}-\frac{4576464369611n^{6}}{359251200}-\frac{14953602874033n^{5}}{359251200}-$
$\displaystyle\frac{927033772163n^{4}}{59875200}+\frac{2596960946917n^{3}}{29937600}+\frac{441409885271n^{2}}{4989600}-\frac{2649524531n}{138600}-\frac{314354}{15}\,,$

$\displaystyle\mathcal{P}_{10}(n,r)=$
$\displaystyle\left(\frac{8192n^{10}}{155925}-\frac{65536n^{9}}{31185}+\frac{370688n^{8}}{10395}-\frac{3473408n^{7}}{10395}+\frac{13976576n^{6}}{7425}-\frac{1925120n^{5}}{297}+\right.$
(B.5)
$\displaystyle\left.\frac{409485056n^{4}}{31185}-\frac{425455616n^{3}}{31185}+\frac{63601568n^{2}}{17325}+\frac{13240064n}{3465}-1768\right)r^{10}+$
$\displaystyle\left(\frac{1024n^{11}}{8505}+\frac{5632n^{10}}{8505}-\frac{137728n^{9}}{1701}+\frac{666112n^{8}}{567}-\frac{2309504n^{7}}{315}+\frac{7955776n^{6}}{405}+\right.$
$\displaystyle\left.\frac{2012032n^{5}}{1701}-\frac{214121696n^{4}}{1701}+\frac{2118827684n^{3}}{8505}-\frac{119186042n^{2}}{945}-\frac{11763526n}{189}+\frac{117260}{3}\right)r^{8}+$
$\displaystyle\left(\frac{256n^{12}}{2835}+\frac{55808n^{11}}{14175}-\frac{24064n^{10}}{14175}-\frac{1096192n^{9}}{945}+\frac{1090592n^{8}}{105}-\frac{92613952n^{7}}{4725}-\right.$
$\displaystyle\left.\frac{1380237872n^{6}}{14175}+\frac{1098647552n^{5}}{2835}-\frac{8138435n^{4}}{567}-\frac{2140478078n^{3}}{1575}+\frac{646434493n^{2}}{525}+\right.$
$\displaystyle\left.\frac{33164102n}{105}-305448\right)r^{6}+$
$\displaystyle\left(\frac{32n^{13}}{1215}+\frac{11888n^{12}}{6075}+\frac{1942184n^{11}}{42525}+\frac{3789124n^{10}}{42525}-\frac{6254986n^{9}}{945}+\frac{200483701n^{8}}{14175}+\right.$
$\displaystyle\left.\frac{10897731679n^{7}}{85050}-\frac{7691537891n^{6}}{24300}-\frac{30358969699n^{5}}{34020}+\frac{359628196349n^{4}}{170100}+\right.$
$\displaystyle\left.\frac{122628506327n^{3}}{56700}-\frac{21427008719n^{2}}{4725}-\frac{153371047n}{315}+1006456\right)r^{4}+$
$\displaystyle\left(\frac{2n^{14}}{729}+\frac{976n^{13}}{3645}+\frac{897097n^{12}}{85050}+\frac{981730n^{11}}{5103}+\frac{1102788149n^{10}}{1020600}-\frac{15745694n^{9}}{1701}-\right.$
$\displaystyle\left.\frac{207637186087n^{8}}{4082400}+\frac{25174204033n^{7}}{204120}+\frac{157940397331n^{6}}{226800}-\frac{10725272695n^{5}}{20412}-\right.$
$\displaystyle\left.\frac{14608972386719n^{4}}{4082400}+\frac{10269635113n^{3}}{22680}+\frac{681990515411n^{2}}{113400}-\frac{17273033n}{210}-1290532\right)r^{2}+$
$\displaystyle\frac{n^{15}}{14580}+\frac{227n^{14}}{29160}+\frac{11153n^{13}}{28350}+\frac{22796371n^{12}}{2041200}+\frac{16378424261n^{11}}{89812800}+\frac{89710139017n^{10}}{59875200}+$
$\displaystyle\frac{488830522129n^{9}}{179625600}-\frac{1146948964289n^{8}}{44906400}-\frac{1733919836167n^{7}}{14968800}+\frac{1953314586743n^{6}}{179625600}+$
$\displaystyle\frac{137895712720613n^{5}}{179625600}+\frac{8473021434481n^{4}}{9979200}-\frac{2101852716961n^{3}}{1663200}-\frac{1626950129029n^{2}}{831600}+$
$\displaystyle\frac{747757867n}{2772}+435964\,.$

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