# Towards Flexible Multi-modal Document Models

Naoto Inoue1 Kotaro Kikuchi1 Edgar Simo-Serra2 Mayu Otani1 Kota Yamaguchi1
1CyberAgent, Japan 2Waseda University, Japan
{inoue_naoto<EMAIL_ADDRESS><EMAIL_ADDRESS>
{otani_mayu<EMAIL_ADDRESS>
###### Abstract

Creative workflows for generating graphical documents involve complex inter-
related tasks, such as aligning elements, choosing appropriate fonts, or
employing aesthetically harmonious colors. In this work, we attempt at
building a holistic model that can jointly solve many different design tasks.
Our model, which we denote by _FlexDM_ , treats vector graphic documents as a
set of multi-modal elements, and learns to predict masked fields such as
element type, position, styling attributes, image, or text, using a unified
architecture. Through the use of explicit multi-task learning and in-domain
pre-training, our model can better capture the multi-modal relationships among
the different document fields. Experimental results corroborate that our
single FlexDM is able to successfully solve a multitude of different design
tasks, while achieving performance that is competitive with task-specific and
costly baselines. 111Please find the code and models at:
https://cyberagentailab.github.io/flex-dm.

##  1 Introduction

Vector graphic documents are composed of diverse multi-modal elements such as
text or images and serve as the dominant medium for visual communication
today. The graphical documents are created through many different design
tasks, _e.g_., filling in a background image, changing font and color, adding
a decoration, or aligning texts. While skilled designers perform tasks based
on their design knowledge and expertise, novice designers often struggle to
make decisions to create an effective visual presentation. To assist such
novice designers, interactive frameworks equipped based on models that learn
design knowledge from completed designs have been proposed [38, 12]. Our
present work proposes models that can be used in such systems, with a
particular focus on developing holistic models that can flexibly switch
between design tasks.

Design tasks are characterized by 1) the variety of possible actions and 2)
the complex interaction between multi-modal elements. As discussed above, a
designer can make almost any edit to the appearance of a vector graphic
document, ranging from basic layout to nuanced font styling. While there have
been several studies in solving specific tasks of a single modality, such as
layout generation [30, 13, 3, 26, 23], font recommendation [56], or
colorization [22, 54, 40], in realistic design applications, we believe it is
essential to build a _flexible_ model that can consider multiple design tasks
in a principled manner to make automated decisions on creative workflow.

Figure 1:  Examples of the design tasks that can be solved by our proposed
FlexDM model, which is designed to process a vector graphic document
consisting of an arbitrary number of elements (_e.g_., text). Each element is
composed of multi-modal fields indicating its attribute properties (_e.g_.,
text content, position, font color, etc.).

In this work, we refer to a certain attribute of an element as a _field_ and
formulate the various design tasks as a unified _masked field prediction_ ,
which is inspired by the recent masked autoencoders [9, 15] and multi-task
models [19, 36]. The key idea is to utilize masking patterns to switch among
different design tasks within a single model; _e.g_., element filling can be
formulated as predicting all the fields of the newly added element. Our
flexible document model, denoted by _FlexDM_ , consists of an encoder-decoder
architecture with a multi-modal head dedicated to handling different fields
within a visual element. After pre-training with random masking strategy, we
train FlexDM by explicit multi-task learning where we randomly sample tasks in
the form of masking patterns corresponding to the target design task. We
illustrate in Figs. 1 and 2 an overview of FlexDM, with emphasis on the
correspondence between design tasks and masking patterns.

Through our carefully designed experiments, we show that our proposed FlexDM
performs favorably against baselines in five design tasks using the Rico [7]
and Crello [52] datasets. We also study how different modeling approaches
affect the final task performance in the ablation study. Finally, we apply our
framework to several previously studied design tasks with minimal
modifications and show that the performance matches or even surpasses the
current task-specific approaches.

Our contributions can be summarized in the following.

  * •

We formulate multiple design tasks for vector graphic documents by masked
multi-modal field prediction in a set of visual elements.

  * •

We build a flexible model to solve various design tasks jointly in a single
Transformer-based model via multi-task learning.

  * •

We empirically demonstrate that our model constitutes a strong baseline for
various design tasks.

##  2 Related Work

###  2.1 Vector Graphic Generation

There has been a growing interest in vector graphics to realize
resolution/artifact-free rendering that is easy to interpret and edit, such as
Scalable Vector Graphics (SVG) [8]. Modeling documents in a vector format is
much more complex than the stroke or path level vector graphics [14, 35, 5]
since each element contains multi-modal features such as text and image.
CanvasVAE [52] tackles the document-level unconditional generation of vector
graphics, but is not a multi-task model and cannot solve specific design tasks
such as element filling. Doc2PPT [11] generates slides given a longer and more
detailed multi-modal document, but it is a summarization task and cannot infer
what is missing in the incomplete document.

Obtaining transferable representation for downstream tasks learned from multi-
modal large-scale data is getting popular. Domains closest to our setting are
document understanding [50, 51, 32, 49] and UI understanding [16, 4], where
the data consist of elements with multi-modal attributes. Despite the
generalizable representation, all the methods fine-tune different parameters
for each downstream task (mainly in classification). In contrast, we aim to
solve many essential tasks for design creation in a single model.

###  2.2 Multi-task Learning

Multi-task learning (MTL) [6, 10, 2] aims at solving different tasks at the
same time while sharing information and computation among them, which is
crucial for deployment. MTL methods achieve a good tradeoff between
performance and computational cost by (i) multiple lightweight heads at the
top of shared backbone [55, 25] and (ii) efficient use of task-specific
parameters [43, 44, 33]. On the contrary, our model obtains the task
information from the masking patterns of the input fields and we empirically
show that extra task-specific parameters are not necessary.

Training a single model that generalizes to many different tasks has been a
long-standing goal. 12-in-1 [37] and UniT [17] handle multiple tasks in vision
and language domain with small task-specific parameters. In a more unified
manner, Perceiver [20] and Perceiver IO [19] treat different modalities as the
same data format, OFA [48] and Unified-IO [36] consider similar attempts in
the sequence-to-sequence framework, resulting in a single model or
architecture with no task-specific tuning. We are highly inspired by these
works and explore how to unify the design tasks in vector graphic document
domain.

###  2.3 Computational Assistance for Graphic Design

There is a long history of automatic graphic design [34, 1, 53]. Recent
approaches rely on the learning-based formulation, where the primal focus is
in predicting layouts given label sets [21, 30] or in an unconditional manner
[13, 3], and avoids the manual design of the energy functions seen in the
earlier work [39]. Some works additionally take positional/relational
constraints [31, 27, 24] or textual descriptions [57] for finer design
control, but are not applicable in a more complex scenario. In contrast, our
multi-task approach solves many conditional tasks thanks to the flexible
multi-modal fields in both inputs and targets.

Considering multi-modal features is essential to go beyond layout generation
for intelligent graphic design assistance. Wang  _et al_. [47] retrieve images
from layout information and keywords for each element to obtain visually
pleasing visual design by reinforcement learning. Zhao  _et al_. [56] predict
font properties of a text on a webpage over a background image considering
metadata. Li  _et al_. [29] predict position and size for a single text box
over a background image considering saliency. We demonstrate that we can apply
our flexible model to solve these tasks with almost no modification, and our
model performs favorably against the task-specific well-tuned approaches.

Figure 2:  Top: example of a vector graphic document consisting of five
elements. The array is used to illustrate the data structure of the document.
Each column corresponds to a single visual element. Each row corresponds to an
attribute or a group of attributes consisting the element. Bottom:
Correspondence between design tasks and masking patterns for our masked field
prediction.  Figure 3:  The architecture of FlexDM. E, T, and D are short for
Encoder, Transformer blocks, and Decoder, respectively.

##  3 Approach

We first describe the formal definition of the vector graphic document and
notations in Sec. 3.1. We then introduce the idea of masked field prediction
and a model for it in Sec. 3.2 and Sec. 3.3. Finally, we describe how we train
FlexDM in Sec. 3.4.

###  3.1 Preliminary

Document Structure: In this work, a vector graphic document $X$ consists of a
set of elements $X=(X_{1},X_{2},\ldots,X_{S})$, where $S$ is the number of
elements in $X$. Each element $X_{i}$ consists of a set of multi-modal fields
and denoted by $X_{i}=\\{x_{i}^{k}\;|\;k\in\mathcal{E}\\}$, where
$\mathcal{E}$ indicates the indices for all the attributes. Each field
$x_{i}^{k}$ can be either a categorical or numerical variable such as element
type, position, text content, or image embedding. For ease of explanation, we
illustrate $X$ by a 2D-array as shown in the top of Fig. 2. Note that the
order in the array does not matter because $X$ is a set of sets. Since
processing high-dimensional data such as raw images and texts during
optimization is computationally intensive, we extract a low-dimensional
numerical vector from such data for $x_{i}^{k}$ using pre-trained models.
Special Tokens: In a similar spirit to the masked language model [9], we use a
few special tokens to represent $x_{i}^{k}$.
[NULL]: appears when $x_{i}^{k}$ is inevitably missing (_e.g_., font type for
an image element), or padding variable-length sequence within a mini-batch on
training.
[MASK]: appears when $x_{i}^{k}$ is masked for prediction.

###  3.2 Masked Field Prediction

Given an incomplete document $X$ containing [MASK] as context, our goal is to
predict values for all the fields filled with [MASK] and generate a complete
document $\hat{X}$. We refer to this problem by masked field prediction, where
a model has to predict the masked field considering the different multi-modal
relations between the fields. While the masking approach is similar to the
masked language model [9], there is a key distinction in that we process an
order-less set of multi-modal items (_i.e_., document $X$). For this reason,
we design our architecture to 1) efficiently capture inter-field relationships
of vector graphic attributes, and 2) ensure that the model works without
positional encodings commonly used to model an ordered sequence.

###  3.3 FlexDM Architecture

As shown in Fig. 3, our architecture consists of three modules; encoder,
Transformer blocks, and decoder. Given a document, we first project a set of
partially masked fields (_e.g_., position or font) into embeddings using the
encoder, and then feed the output to the intermediate Transformer blocks. The
final decoder takes the transformed embeddings and projects them back to the
original fields space. The Transformer blocks only process $S$ embeddings,
which is efficient compared to architecture processing $S\times N$ fields with
off-the-shelf Transformer [46] directly, when there are $N$ attributes. In the
following, let us denote all model parameters by $\theta$.

Encoder: The encoder takes a document input $X$ and embeds it into
$h^{\text{enc}}=\\{h_{1}^{\text{enc}},h_{2}^{\text{enc}},\ldots,h_{S}^{\text{enc}}\\}$
with element-wise operations. The encoder first maps each field $x_{i}^{k}$ to
a fixed dimensional vector with $f^{\text{enc},k}$, and sums up all the fields
in the element to produce a latent vector for the $i$-th element with:

$h_{i}^{\text{enc}}=\sum_{k\in\mathcal{E}}f^{\text{enc},k}(x_{i}^{k};\theta),$
(1)

where $f^{\text{enc},k}$ is an embedding function that retrieves learnable
dense embeddings for each category id if $x_{i}^{k}$ is a categorical
variable, or a simple linear projection layer if $x_{i}^{k}$ is a numerical
variable. We treat the special tokens (_i.e_., [NULL] and [MASK]) in the same
manner to the categorical variable.
Transformer Blocks: Transformer blocks take $h^{\text{enc}}$ as input and
transform it to
$h^{\text{dec}}=\\{h_{1}^{\text{dec}},h_{2}^{\text{dec}},\ldots,h_{S}^{\text{dec}}\\}$.
We stack these intermediate blocks to process complex inter-element relations.
Our model can stack any off-the-shelf Transformer layer to build up the blocks
$f^{\text{trans}}$:

$h^{\text{dec}}=f^{\text{trans}}(h^{\text{enc}};\theta)$ (2)

Decoder: The final decoder takes $h_{\text{dec}}$ and decodes them back into a
document $\hat{X}=(\hat{X}_{1},\hat{X}_{2},\ldots,\hat{X}_{S})$, where
$\hat{X}_{i}=\\{\hat{x}_{i}^{k}\;|\;k\in\mathcal{E}\\}$. We compute each
$\hat{x}_{i}^{k}$ by a linear layer $f^{\text{dec},k}$ for both categorical
and numerical variables:

$\hat{x}_{i}^{k}=f^{\text{dec},k}(h_{i}^{\text{dec}};\theta).$ (3)

Loss: We train our model using reconstruction losses. Let us denote by $X^{*}$
the ground truth of the incomplete document $X$, and also denote by $M$ a set
of tuples indicating the indices for [MASK] tokens in $X$. We define the loss
function by:

$\mathcal{L}=\sum_{(i,k)\in M}l^{k}(\hat{x}_{i}^{k},{x^{*}}_{i}^{k}),$ (4)

where $l^{k}$ is the loss function for the $k$-th attribute. For each $l^{k}$,
we use softmax cross-entropy loss for categorical variables and mean squared
error for numerical variables.

###  3.4 FlexDM Training

Masked field prediction allows us to represent diverse design tasks having
various input/output formats just by altering the masking pattern. The pattern
can be both deterministic or stochastic. The bottom of Fig. 2 illustrates
example tasks and the corresponding masking patterns. Although we can
formulate arbitrary tasks with masked field prediction, we consider several
subsets of representative design tasks for our evaluation and analyses in Sec.
4.

We describe typical masking patterns in the following. Note that fields
already filled with [NULL] will never be replaced in priority to the masking
operations. _Element masking_ randomly selects elements and masks all the
fields within the element; _i.e_., we can formulate the element filling task
by single element masking. _Attribute masking_ randomly selects attributes and
mask the fields across all the elements; _e.g_., masking position and size of
all the elements becomes layout prediction, and masking fonts becomes font
prediction. _Random masking_ strategy masks fields by some probability without
considering the data structure, which is similar to BERT [9].

Pre-training: To learn the initial model, we employ a pre-training by ordinary
random masking similar to the prevailing pre-training strategy of BERT [9].
One distinction is that our pre-training happens in the same, in-domain
dataset, unlike the common setup where a model is pre-trained on a larger
dataset in a different domain and then fine-tuned on a target task in a target
dataset. We show in Sec. 4 that this in-domain pre-training moderately
improves the final task performance.

Explicit Multi-task Learning: The random masking pre-training above is a solid
baseline for any task. Radford  _et al_. [42] hypothesize that this implicit
multi-task training leads to the astonishingly strong zero-shot performance of
large language models. However, the random masking strategy actually produces
any task with an extraordinarily low probability as the number of attributes
and elements increases. Instead, we employ the explicit masking strategy to
maximize the performance on all the target tasks. During training we randomly
sample a task from the target tasks, sample a complete document $X^{*}$, and
make the triplet ($X$, $X^{*},M$) by using the masking pattern associated with
the task. We repeat this procedure to build each mini-batch when training
FlexDM.

##  4 Experiments

###  4.1 Dataset

We mainly use two datasets containing vector graphic documents, Rico [7] and
Crello [52], to evaluate FlexDM. We basically follow the setting used in [52].
Due to memory limitations, we discard documents having more than fifty
elements. Position, size, and color information are discretized in order to
enhance the implicit alignment of multiple elements. We describe the overview
of each dataset.
Rico [7]: The dataset collects UI designs from mobile apps. We follow previous
works [30, 27] and exclude elements whose labels are not in the most frequent
13 labels. We divide the dataset into 45,012 / 5,565 / 5,674 examples for
train, validation, and test splits.
Crello [52]: The dataset provides design templates from an online design
service. Crello contains various design formats such as social media posts,
banner ads, blog headers, or printed posters. We divide the dataset into
18,738 / 2,313 / 2,271 examples for train, validation, and test splits. Please
refer to the original paper [52] for the definition of each attribute. For
image and text features, we extract 768-dimensional features using CLIP [41].
We also additionally extract categorical font information (called Font). We
group the attributes into some groups based on their property. TYPE denotes
Type attribute. POS denotes Position and Size attributes. IMG denotes Image
attribute. TXT denotes Text attribute. ATTR denotes attributes not listed
above, and these attributes have a large impact on fine-grained appearance.

###  4.2 Tasks

We carefully select tasks to evaluate how our model performs in various design
tasks. We select evaluation tasks such that (i) they are practical, (ii) they
have various combinations of input/output modalities, and (iii) the masking
ratio is modest. We impose the masking ratio requirement because the extreme
masking ratio makes the task too difficult or trivial to solve and makes the
baseline comparison impossible.
Element Filling (ELEM): This task is to predict a new element that can enhance
the document. We mask all the attributes of a single element in a complete
document during training and evaluation.
Attribute Prediction: This task is to predict missing attributes at once in
the document, which is very challenging. We apply attribute masking on a
complete document to make the masked inputs during training and evaluation. We
select an attribute group discussed in Sec. 4.1 and apply the attribute
masking for all the attributes in the group. We consider each group-level
prediction task as an individual task. Note that we do not consider TYPE
prediction since it is too trivial and unrealistic. Therefore, we have two
(POS and ATTR) and four (POS, ATTR, IMG, and TXT) attribute prediction tasks
for Rico and Crello, respectively.

###  4.3 Evaluation Metrics

For each task, we quantitatively evaluate the reconstruction performance. The
score $\mathcal{S}$ for each document is computed by:

$\mathcal{S}=\frac{1}{|M|}\sum_{(i,k)\in
M}s^{k}(\hat{x}_{i}^{k},{x^{*}}_{i}^{k}),$ (5)

where $s^{k}\in[0,1]$ is a scoring function for $k$-th attribute. If the
attribute is categorical, $s^{k}$ is an indicator function that takes 1 if
$\hat{x}_{i}^{k}$ and ${x^{*}}_{i}^{k}$ are identical, otherwise 0. For image
and text features that are the only numerical attributes in our experiments,
we use cosine similarity in $[0,1]$ scale.

Dataset | Rico [7] |  | Crello [52]
---|---|---|---
Model | #par. | ELEM | POS | ATTR |  | #par. | ELEM | POS | ATTR | IMG | TXT
Most-frequent | 0.0x | 0.461 | 0.213 | 0.830 |  | 0.0x | 0.402 | 0.134 | 0.382 | 0.922 | 0.932
BERT [9] | 1.0x | 0.517 | 0.238 | 0.847 |  | 1.0x | 0.524 | 0.155 | 0.632 | 0.935 | 0.949
BART [28] | 1.2x | 0.515 | 0.220 | 0.714 |  | 1.2x | 0.469 | 0.156 | 0.615 | 0.932 | 0.945
CVAE [21, 27] | 1.1x | 0.511 | 0.214 | 0.917 |  | 1.0x | 0.499 | 0.197 | 0.587 | 0.942 | 0.947
CanvasVAE [52] | 1.2x | 0.437 | 0.192 | 0.790 |  | 1.2x | 0.475 | 0.138 | 0.586 | 0.912 | 0.946
Ours-IMP | 1.0x | 0.505 | 0.259 | 0.923 |  | 1.0x | 0.483 | 0.197 | 0.607 | 0.945 | 0.949
Ours-EXP | 1.0x | 0.540 | 0.226 | 0.937 |  | 1.0x | 0.499 | 0.218 | 0.679 | 0.948 | 0.952
Ours-EXP-FT | 1.0x | 0.552 | 0.215 | 0.945 |  | 1.0x | 0.508 | 0.227 | 0.688 | 0.950 | 0.954
Expert | 3.0x | 0.575 | 0.228 | 0.952 |  | 5.0x | 0.534 | 0.255 | 0.703 | 0.948 | 0.955
Table 1:  Quantitative evaluation in two datasets. A higher score indicates
the better performance. Top two results are highlighted in bold and underline,
respectively. LGAN++ is short for LayoutGAN++.

Figure 4:  Results in element filling using Rico dataset. The red dotted box
indicates the target element to be predicted.

###  4.4 Training Details

We use 256-dimensional latent representations within the encoder, Transformer
blocks, and decoder. For the Transformer blocks we use the one from DeepSVG
[5]. We apply a dropout probability of 0.1 to all the dropout layers. We train
the model with a batch size of 256 sequences for 500 epochs in all the
experiments. We use Adam with learning rate of 1e-4, $\beta_{1}=0.9$,
$\beta_{2}=0.99$, and L2 weight decay of 1e-2. In experiments on Rico, we make
FlexDM take positional embedding as the additional input, since otherwise the
model is unable to distinguish elements having a completely similar set of
attributes, which often occurs in POS prediction.

###  4.5 Quantitative Evaluation

We test three models based on our proposed framework to clarify the
contribution of both explicit multi-task learning and pre-training.
Ours-IMP: As in the standard masked language modeling such as BERT [9], we
randomly mask 15% of the fields during training. Since this randomized
training is called implicit multi-task learning [42], we call it Ours-IMP.
Ours-EXP: All the tasks are explicitly and jointly trained in a single model
by sampling the masking patterns corresponding to each task. For simplicity,
$T$ tasks introduced in Sec. 4.2 are uniformly sampled in a mini-batch.
Ours-EXP-FT: This is our entire model. We use weights of the model trained on
IMP, and fine-tune the model. The rest of the training is the same as Ours-
EXP.

We compare these models with the following baselines, some of which are
adapted from existing task-specific models to our multi-task, multi-attribute,
and arbitrary masking setting with minimal modification.
Expert: We train the network individually for each task. Note that the number
of the parameters used in this variant is $T$ times larger than our models.
Most-frequent: We calculate the statistics of the training dataset. For a
categorical attribute, we count the occurrences and pick the most frequent
category. For a numerical attribute, we compute the average because the
numerical attributes that we use are only image and text features.
BERT [9]: We convert all the fields into a single sequence and process them
with Transformer blocks. This evaluates the effect of element-wise embedding
discussed in Sec. 3.3.
BART [28]: BART employs an encoder-decoder-based sequence-to-sequence model
for pre-training text generation models by masked language modeling. We
replace our Transformer blocks with the blocks from BART.
CVAE [21, 27]: Recent methods for conditional layout generation such as
LayoutVAE [21] and NDN [27] employ Conditional VAE [45] in an auto-regressive
manner. We replace our Transformer block and decoder parts with CVAE variants
used in [21, 27] and predict the fields in an element-by-element manner. Note
that the full version of NDN contains relation prediction and layout
refinement modules in addition to CVAE modules. We omit the full NDN pipeline
evaluation due to their specific approach.
CanvasVAE [52]: CanvasVAE is for an unconditional generation. Although direct
comparison is impossible, we adapt CanvasVAE to our setting, similar to other
baselines.

Table 1 summarizes the performance of all the models. Our full model (Ours-
EXP-FT) is almost comparable to Expert model while being much more efficient
in the number of parameters. Ours-IMP exhibits moderate performance, resulting
in a better initial weight for fine-tuning in Ours-EXP-FT. We can see that
most of the compared baselines perform clearly worse compared to Ours-EXP. The
result suggests that applying existing Transformer models for sequence
modeling or conditional layout generation models is not enough in our
challenging setting. POS-prediction in Rico is the exceptional case, where
most of the methods fail because of the larger number of elements compared to
the benchmark setup in the literature [24] (nine at maximum).

ATTR prediction | TXT prediction | IMG prediction
---|---|---
Input | Output | Input | Output | Input | Output
|  |  |  |  |
POS prediction | Element filling
Output (bbox.) | Output (img.) | Output (bbox.) | Output (img.) | Input | Output
|  |  |  |  |
Figure 5:  Prediction of FlexDM (Ours-EXP-FT trained on Crello). FlexDM
jointly handles a large variety of design tasks with a single Transformer-
based model. In the input of ATTR/TXT/IMG prediction, the target fields
assigned [MASK] are visualized using fixed default values (_i.e_., black for
text color, gray for image and solid fill, ‘TEXT’ for text). In POS
prediction, we additionally show the layout of the elements. The
correspondence between the color and type of the element is as follows: green
= vector shape, magenta = image, purple = text, yellow = solid fill. Best
viewed with zoom and color.

###  4.6 Qualitative Evaluation

We show the prediction quality of our full FlexDM (Ours-EXP-FT) for Rico
dataset in the element-filling task in Fig. 4. For Rico, we show a color map
indicating the position and type information. In Fig. 5, we show the
prediction of our full FlexDM (Ours-EXP-FT) on all the target design tasks.
For visualizing predicted low-dimensional image and text features, we conduct
a nearest neighbor search to retrieve actual images and texts using the assets
in the test subset, following CanvasVAE [52].

###  4.7 Ablation Study

In this section, we perform several ablation experiments in the Crello
dataset, as shown in Tab. 2. We demonstrate that our design choices non-
trivially affect the final performance of FlexDM.
Task-specific Embedding: The previous work [17] on unifying multiple tasks in
a single Transformer uses small task-specific learnable query embedding to
feed information of the current task explicitly. We append the query as
$h_{0}^{\text{enc}}$ at the beginning of
$h^{\text{enc}}=\\{h_{1}^{\text{enc}},h_{2}^{\text{enc}},\ldots,h_{S}^{\text{enc}}\\}$
and train the model. The result suggests the benefit of the embedding is
marginal. We conjecture that the model implicitly captures the task
information from the masked inputs in our setting.
Attention: Here we study the importance of self-attention to model the inter-
element relationship by training a model without self-attention. We increase
the number of layers to eight to roughly match the total number of parameters
with Ours-EXP. As expected, the result clearly suggests the importance of
modeling the inter-element relationship.
Additional Loss: Our objective function in Eq. 4 only considers
reconstruction. One may argue that incorporating adversarial losses such as
those used in LayoutGAN++ [24] could improve the model. While we tried our
best in implementing and tuning the additional adversarial loss, we did not
find a clear benefit in adversarial training.

Table 2:  Ablation study results in Crello dataset. Top two results are highlighted in bold and underline, respectively.  | Model | ELEM | POS | ATTR | IMG | TXT
---|---|---|---|---|---|---
| Ours-EXP | 0.499 | 0.218 | 0.679 | 0.948 | 0.952
(i) | w/ task-ID | 0.496 | 0.222 | 0.674 | 0.949 | 0.953
(ii) | w/o attention | 0.446 | 0.208 | 0.605 | 0.939 | 0.947
(iii) | w/ adv. | 0.499 | 0.215 | 0.677 | 0.948 | 0.952

###  4.8 Comparison with Task-specific Baselines

In this section, we show that our data-driven masked field prediction model
can match or even surpasses task-specific approaches. We perform experiments
in two tasks: 1) single text styling and 2) single text box placement. Since
each task uses partially overlapping set of attributes, we train our model for
each single task for fair comparison. Note that we are unable to compare to
contextual images filling [47] discussed in Sec. 2.3 due to their task setup
where they retrieve an image only from pre-defined sets used during training.

Input (background) | SmartText+ [29] | Ours | Ground truth
---|---|---|---
|  |  |
|  |  |
|  |  |
Figure 6:  Qualitative comparison of single text box placement with SmartText+
[29]. Best viewed with zoom and color.

####  4.8.1 Single Text Styling

Zhao  _et al_. [56] propose an MLP-based model to predict desirable font
properties for a single text box (_i.e_., font emb., color, and size), given
context in web designs. We consider that each design is a document with one
text and two image elements, and regard all the context information as
attributes in the elements so that we can just apply FlexDM. We implement Zhao
_et al_. [56] with the following minor difference, since the code is not
publicly available. We quantize the color and size into 16 bins and 64 bins,
respectively. We did not apply data augmentation using the external dataset,
since the dataset used for the augmentation is not available. We show the
results in Tab. 3. The metrics are accuracy for font color and size, and
cosine similarity for font type, which is represented by a low-dimensional
embedding. We can clearly see that our model is comparable to the task-
specific model.

Table 3:  Comparison of models for font properties prediction in CTXFont dataset [56]. The average and standard deviation of three runs are reported. The values are multiplied by 100x for visibility.  Model | Color | Size | Emb. | Avg.
---|---|---|---|---
Zhao  _et al_. [56] | 45.8$\pm$2.9 | 19.9$\pm$3.1 |  79.2$\pm$0.5 | 48.2$\pm$1.2
Ours |  54.2$\pm$0.7 |  24.2$\pm$0.1 | 77.7$\pm$1.3 |  52.0$\pm$0.5

####  4.8.2 Single Text Box Placement

Table 4:  Quantitative evaluation of models for single text box placement in
Crello dataset. The samples are divided into two groups: no other text box
available (Single) and some text boxes available as the context (Multiple).
| Single | Multiple
---|---|---
| IoU $\uparrow$ | BDE $\downarrow$ | IoU $\uparrow$ | BDE $\downarrow$
SmartText+ [29] | 0.047 | 0.262 | 0.023 | 0.300
Ours | 0.357 | 0.098 | 0.110 | 0.141
w/o image | 0.355 | 0.100 | 0.103 | 0.156
w/o text | 0.350 | 0.106 | 0.086 | 0.178

Li  _et al_. [29] propose to predict the size and position of a single text
box given a natural image and aspect ratio of the text box. We perform
comparison in Crello dataset, since the dataset used for their model training
and evaluation is not publicly available. We evaluate the performance in terms
of the intersection over union (IoU) and boundary displacement error (BDE)
[29]. As shown in the upper half of Tab. 4, our model clearly outperforms Li
_et al_. [29]’s model. To measure the contribution of multi-modal features to
the prediction, we exclude each of them and train the model. The results in
the lower half of Tab. 4 suggest that those features contribute to the better
performance. Some results are shown in Fig. 6.

##  5 Limitation and Discussion

As image and text generation quality is astonishingly improving, one may want
to generate images and texts directly. However, retrieval-based generation is
still a practical option. For instance, due to clients’ requests, designers
often need to use images from private collections or public photo stock
services such as Adobe Stock or Shutterstock. Moreover, some people avoid
using generated images or text as there are controversies about the legal and
ethical issues of AI-generated images.

Our model does not support design tasks that cannot be framed as masked field
prediction. We do not consider unconditional generation; _i.e_., generating a
complete document without input. Extending FlexDM to an unconditional scenario
requires us to apply a generative formulation instead of BERT-style masked
modeling, and we leave such formulation as future work. However, we believe
that our model nicely fits in a common application scenario where there exist
initial design materials to start with.

The model’s performance decreases when the input document has more elements.
Whether bigger models or datasets alleviate the issue is worth investigating.
Developing other evaluation metrics would be helpful for further analysis
since current metrics simply evaluate reconstruction performance. In
conditional generation, the input context may correspond to multiple possible
outputs, especially when the input context is sparse (_e.g_., label sets).
Modeling such variability as in layout generation models [21, 24, 18] would be
an exciting direction.

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# Absolutely monotonic functions related to the asymptotic formula for the
complete elliptic integral of the first kind

Tiehong Zhao and Zhen-Hang Yang  Tiehong Zhao
School of mathematics, Hangzhou Normal University, Hangzhou, Zhejiang, China,
311121<EMAIL_ADDRESS>https://orcid.org/0000-0002-6394-1049 Zhen-
Hang Yang
State Grid Zhejiang Electric Power Company Research Institute, Hangzhou,
Zhejiang, China, 310014<EMAIL_ADDRESS>https://orcid.org/0000-0002-2719-4728

###### Abstract.

Let $\mathcal{K}\left(x\right)$ be the complete elliptic integral of the first
kind and

$\text{\
}\mathcal{G}_{p}\left(x\right)=e^{\mathcal{K}\left(\sqrt{x}\right)}-\frac{p}{\sqrt{1-x}}$

for $p\in\mathbb{R}$ and $x\in\left(0,1\right)$. In this paper we find the
necessary and sufficient conditions for the functions
$\pm\mathcal{G}_{p}^{\left(k\right)}\left(x\right)$
($k\in\mathbb{N\cup}\left\\{0\right\\}$) to be absolutely monotonic on
$\left(0,1\right)$, which extend previous known results and yield several new
functional inequalities involving the complete elliptic integral of the first
kind. More importantly, we provide a new method to deal with those absolute
monotonicity problem by proving the monotonicity of a sequence generated by
the coefficients of the power series of $\mathcal{G}_{p}\left(x\right)$.

###### Key words and phrases:

Complete elliptic integral of the first kind, absolute monotonicity,
hypergeometric function, recurrence formula, functional inequality

###### 2000 Mathematics Subject Classification:

Primary 33E05, 26A48 Secondary 40A05, 41A10

Corresponding Author: Zhen-Hang Yang*

##  1\. Introduction

The complete elliptic integral of the first kind $\mathcal{K}\left(r\right)$
is defined on $\left(0,1\right)$ by

$\mathcal{K}\left(r\right)=\int_{0}^{\pi/2}\frac{1}{\sqrt{1-r^{2}\sin^{2}t}}dt,$

which can also be expressed by the Gaussian hypergeometric function

(1.1)
$\mathcal{K}\left(r\right)=\frac{\pi}{2}F\left(\frac{1}{2},\frac{1}{2};1;r^{2}\right)=\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{\left(1/2\right)_{n}^{2}}{\left(n!\right)^{2}}r^{2n},$

where

$F\left(a,b;c;x\right)=\sum_{n=0}^{\infty}\frac{\left(a\right)_{n}\left(b\right)_{n}}{\left(c\right)_{n}}\frac{x^{n}}{n!}$

is the Gaussian hypergeometric function defined on $x\in\left(-1,1\right)$,
here $-c\notin\mathbb{N\cup}\left\\{0\right\\}$ and
$\left(a\right)_{n}=\Gamma\left(n+a\right)/\Gamma\left(a\right)$ in which
$\Gamma(x)=\int_{0}^{\infty}t^{x-1}e^{-t}dt$ $(x>0)$ is the gamma function.
The complete elliptic integral of the first kind $\mathcal{K}\left(r\right)$
has the asymptotic formula

(1.2) $\mathcal{K}\left(r\right)\thicksim\ln\frac{4}{r^{\prime}}\text{ \ as
}r\rightarrow 1^{-},$

where and in what follows $r^{\prime}=\sqrt{1-r^{2}}$ (see [1]).

There are many known properties of $\mathcal{K}\left(r\right)$, including
monotonicity, convexity, absolute monotonicity and inequalities. Here we
mention several properties related to the asymptotic formula (1.2), which are
more interesting. Let

(1.3)
$\mathcal{F}_{p}\left(x\right)=\left(1-x\right)^{p}e^{\mathcal{K}\left(\sqrt{x}\right)}\text{
\ and \
}\mathcal{G}_{p}\left(x\right)=e^{\mathcal{K}\left(\sqrt{x}\right)}-\frac{p}{\sqrt{1-x}}.$

In 1996, Qiu and Vamanamurthy [2, Theorem 1.2] showed that the function
$r\mapsto\mathcal{F}_{1/2}\left(r^{2}\right)$ is strictly decreasing and
concave from $\left(0,1\right)$ onto $\left(4,e^{\pi/2}\right)$. Yang and Chu
[3, Theorem 3.1] proved that the function
$r\mapsto\mathcal{F}_{p/2}\left(r^{2}\right)$ is strictly increasing on
$\left(0,1\right)$ if and only if $p\leq\pi/4$ and strictly decreasing on
$\left(0,1\right)$ if and only if $p\geq 1$ using a different way. Further,
Yang and Tian [4] proved that $-\mathcal{F}_{p}^{\prime}$ is absolutely
monotonic on $\left(0,1\right)$ if and only if $1/2\leq
p\leq\left(\pi+4+\sqrt{16-\pi}\right)/8=1.340...$, and $\mathcal{F}_{p}$ is
absolutely monotonic on $\left(0,1\right)$ if and only if
$p\leq\pi/8=0.392...$. Yang and Chu [3, Theorem 3.3] proved that
$\mathcal{G}_{p}$ is strictly decreasing on $(0,1)$ if and only if $p\geq 4$
and strictly increasing on $(0,1)$ if and only if $p\leq\pi e^{\pi/2}/4$.

For the function
$f_{p}\left(x\right)=\mathcal{K}(\sqrt{x})/\ln\left(p/\sqrt{1-x}\right)$
($p\geq 1$), Anderson, Vamanamurthy and Vuorinen [5, Theorem 2.2] proved that
the function $r\mapsto f_{c}\left(r^{2}\right)$ is strictly decreasing if and
only if $1\leq c\leq 4$ and increasing if and only if $c\geq e^{2}$. Qiu and
Vamanamurthy [6] further verified that $r\mapsto f_{4}\left(r^{2}\right)$ is
strictly concave on $\left(0,1\right)$. Yand and Tian [7] showed that $f_{p}$
is concave on $\left(0,1\right)$ if and only if $p=e^{4/3}$. After three
years, Tian and Yang [8, Theorem 1] proved further that $p=e^{4/3}$ is also
the necessary and sufficient condition for the function
$-f_{p}^{\prime\prime}$ to be absolutely monotonic on $\left(0,1\right)$ by
using recurrence technique. Recently, Alzer and Richards [9] found that the
function $x\mapsto 1/f_{p}\left(x\right)$ is strictly concave on
$\left(0,1\right)$ if and only if $p\geq e^{8/5}$. This condition was proved
in [8, Theorem 3] to be also necessary and sufficient such that the function
$x\mapsto-\left[1/f_{p}\left(x\right)\right]^{\prime\prime}$ is absolutely
monotonic on $\left(0,1\right)$. Moreover, Tian and Yang [10, Theorem 1]
discovered the function $x\mapsto 1/f_{p}\left(x\right)$ is strictly convex on
$\left(0,1\right)$ if and only if $0<p\leq 4$. More information about
monotonicity, convexity and absolute monotonicity for the functions
$f_{p}\left(x\right)$ and its reciprocal can be found in [11, 12, 8, 10], also
refer to [13, 14, 15, 16, 17, 18, 19, 20, 21] for their generalizations.
Besides, several remarkable inequalities related to the asymptotic formula
(1.2) can be seen in [22, 23, 24, 25, 26, 27, 28, 29].

Let us return to (1.3). The absolute monotonicity of the functions
$\mathcal{F}_{p}$ in [4] and the monotonicity of $\mathcal{G}_{p}$ in [3]
encourage us to investigate the absolute monotonicity of the function
$\mathcal{G}_{p}$ on $\left(0,1\right)$. This seems to be not difficult,
however, it is actually quite challenging. Fortunately, we succeed in solving
the problem using the following new idea:

Step 1: Let

(1.4) $e^{\mathcal{K}\left(\sqrt{x}\right)}=\sum_{n=0}^{n}b_{n}x^{n}\text{ \
and \ }\frac{1}{\sqrt{1-x}}=\sum_{n=0}^{n}W_{n}x^{n},$

where

(1.5)
$W_{n}=\frac{\left(2n-1\right)!!}{\left(2n\right)!!}=\frac{\Gamma\left(n+1/2\right)}{\Gamma\left(1/2\right)\Gamma\left(n+1\right)}$

which satisfies the recurrence relation

(1.6) $W_{n+1}=\frac{n+1/2}{n+1}W_{n}.$

Then

(1.7)
$G\left(x\right)=\left[\frac{\pi}{8}F\left(\frac{1}{2},\frac{1}{2};2;x\right)-\frac{1}{2}\right]e^{\mathcal{K}\left(\sqrt{x}\right)}=\sum_{n=0}^{\infty}\left[\left(n+1\right)b_{n+1}-\left(n+\frac{1}{2}\right)b_{n}\right]x^{n}.$

Differentiation yields

$G^{\prime}\left(x\right)=\frac{\pi}{64}e^{\mathcal{K}\left(\sqrt{x}\right)}g\left(x\right),$

where

(1.8) $g\left(x\right)=F\left(\frac{3}{2},\frac{3}{2};3;x\right)+\pi
F\left(\frac{3}{2},\frac{3}{2};2;x\right)F\left(\frac{1}{2},\frac{1}{2};2;x\right)-4F\left(\frac{3}{2},\frac{3}{2};2;x\right).$

Step 2: We prove that $g\left(x\right)$ is absolutely monotonic on
$\left(0,1\right)$. Since $e^{\mathcal{K}\left(\sqrt{x}\right)}$ is obviously
absolutely monotonic on $\left(0,1\right)$, so is $G^{\prime}\left(x\right)$.

Step 3: Due to the absolute monotonicity of $G^{\prime}\left(x\right)$, we
deduce the positivity of those coefficients of power series of
$G\left(x\right)$ (1.7), except for the constant term, that is,

(1.9) $\left(n+1\right)b_{n+1}-\left(n+\frac{1}{2}\right)b_{n}>0\text{ \ \ or
equivalently, \ \ }\frac{b_{n+1}}{W_{n+1}}>\frac{b_{n}}{W_{n}}$

for $n\geq 1$. Moreover, we also need to prove that
$\lim_{n\rightarrow\infty}\left(b_{n}/W_{n}\right)=4$.

Step 4: From Step 3, we easily arrive at the absolute monotonicity results for
the functions $\pm\mathcal{G}_{p}^{\left(k\right)}\left(x\right)$ on
$\left(0,1\right)$ for $k\geq 0$, which are contained in the following
theorem.

######  Theorem 1.

Let $\mathcal{G}_{p}$ be defined on $\left(0,1\right)$ in (1.3). Then the
following statements are valid:

(i) $\mathcal{G}_{p}$ is absolutely monotonic on $\left(0,1\right)$ if and
only if $p\leq\pi e^{\pi/2}/4=3.778...$, and $-\mathcal{G}_{p}$ is absolutely
monotonic on $\left(0,1\right)$ if and only if $p\geq e^{\pi/2}$.

(ii) For $k\in\mathbb{N}$, $\mathcal{G}_{p}^{\left(k\right)}$ is absolutely
monotonic on $\left(0,1\right)$ if and only if $p\leq b_{k}/W_{k}$, and
$-\mathcal{G}_{p}^{\left(k\right)}$ is absolutely monotonic on
$\left(0,1\right)$ if and only if $p\geq 4$.

######  Remark 1.

The conclusion in Step 3 follows from the following obvious fact:

A real power series $S\left(x\right)=\sum_{n=0}^{\infty}a_{n}x^{n}$ converging
on $\left(-R,R\right)$ ($R>0$) is absolutely monotonic on $\left(0,R\right)$
if and only if $a_{k}\geq 0$ for all $k\geq 0$.

In fact, if $S\left(x\right)$ is absolutely monotonic on $\left(0,R\right)$,
then $S^{\left(k\right)}\left(0\right)\geq 0$ for all $k\geq 0$, which means
that $a_{k}\geq 0$ for all $k\geq 0$. Conversely, if $a_{k}\geq 0$ for all
$k\geq 0$, then $S^{\left(k\right)}\left(x\right)\geq 0$ for
$x\in\left(0,R\right)$ and all $k\geq 0$.

##  2\. Preliminaries and proof of Theorem 1

###  2.1. Basic knowledge

In order to prove our results, we need several basic properties of the
hypergeometric function for real $a,b,c$, where
$-c\notin\mathbb{N\cup}\left\\{0\right\\}$.

(i) Differentiation formula

(2.1)
$F^{\prime}\left(a,b;c;x\right)=\frac{ab}{c}F\left(a+1,b+1;c+1;x\right).$

(ii) The value of $x=1$:

$F\left(a,b;c;1\right)=\dfrac{\Gamma\left(c\right)\Gamma\left(c-a-b\right)}{\Gamma\left(c-a\right)\Gamma\left(c-b\right)},$

where $-c\notin\mathbb{N\cup}\left\\{0\right\\}$, and $c>a+b$. In particular,
we have

(2.2) $F\left(\frac{1}{2},\frac{1}{2};2;1\right)=\frac{4}{\pi}.$

(iii) Linear transformation formula

(2.3) $F\left(a,b;c;x\right)=(1-x)^{c-a-b}F(c-a,c-b;c;x).$

(iv) Asymptotic formula

(2.4)
$F\left(a,b;a+b;x\right)=\dfrac{R\left(a,b\right)-\ln\left(1-x\right)}{B\left(a,b\right)}+O\left(\left(1-x\right)\ln\left(1-x\right)\right)\text{
as }x\rightarrow 1^{-},$

where

$B\left(a,b\right)=\frac{\Gamma\left(a\right)\Gamma\left(b\right)}{\Gamma\left(a+v\right)}\text{,
\ }a,b>0$

is the classical beta function,

(2.5) $R\equiv
R\left(a,b\right)=-2\gamma-\psi\left(a\right)-\psi\left(b\right)\text{,}$

here $\psi\left(x\right)=\Gamma^{\prime}\left(x\right)/\Gamma\left(x\right)$,
$x>0$ is the psi function and $\gamma$ is the Euler-Mascheroni constant. In
particular, we have

(2.6)
$F\left(\frac{3}{2},\frac{3}{2};3;x;\right)\thicksim\frac{8}{\pi}\left[4\ln
2-4-\ln\left(1-x\right)\right]\rightarrow\infty\text{ as }x\rightarrow 1^{-}.$

Moreover, using the notation $W_{n}$, certain specific hypergeometric
functions can be simply represented as

(2.7)
$\begin{array}[]{ll}F\left(\dfrac{1}{2},\dfrac{1}{2};1;x\right)=\mathop{\displaystyle\sum}\limits_{n=0}^{\infty}W_{n}^{2}x^{n},&F\left(\dfrac{3}{2},\dfrac{3}{2};2;x\right)=4\mathop{\displaystyle\sum}\limits_{n=0}^{\infty}\left(n+1\right)W_{n+1}^{2}x^{n},\vskip
12.0pt plus 4.0pt minus 4.0pt\\\
F\left(\dfrac{1}{2},\dfrac{1}{2};2;x\right)=\mathop{\displaystyle\sum}\limits_{n=0}^{\infty}\dfrac{W_{n}^{2}}{n+1}x^{n},&F\left(\dfrac{3}{2},\dfrac{3}{2};3;x\right)=8\mathop{\displaystyle\sum}\limits_{n=0}^{\infty}\dfrac{n+1}{n+2}W_{n+1}^{2}x^{n},\end{array}$

###  2.2. The first three steps

In this subsection, we complete the proofs of Step 1–3 introduced in Section
1.

######  Lemma 1 (Step 1).

Let $b_{n}$ be defined as in (1.4) Then the formula (1.7) is valid. Moreover,
$b_{n}$ satisfies the recurrence formula: $b_{0}=e^{\pi/2}$ and for $n\geq 0$,

(2.8)
$b_{n+1}=\frac{n}{n+1}b_{n}+\frac{\pi}{8\left(n+1\right)}\sum_{k=0}^{n}\frac{W_{k}^{2}}{k+1}b_{n-k}.$

In particular, the first few coefficients are:

$b_{1}=\frac{1}{8}\pi e^{\pi/2}\text{, \
}b_{2}=\frac{1}{128}\pi\left(\pi+9\right)e^{\pi/2}\text{, \ and \
}b_{3}=\frac{\pi}{3072}\left(\pi^{2}+27\pi+150\right)e^{\pi/2}.$

###### Proof.

Differentiation yields

$\left[e^{\mathcal{K}\left(\sqrt{x}\right)}\right]^{\prime}=\frac{\pi}{8}F\left(\frac{3}{2},\frac{3}{2};2;x\right)e^{\mathcal{K}\left(\sqrt{x}\right)}=\sum_{n=0}^{\infty}nb_{n}x^{n-1},$

which, by (2.3), implies that

(2.9)
$\frac{\pi}{8}F\left(\frac{1}{2},\frac{1}{2};2;x\right)e^{\mathcal{K}\left(\sqrt{x}\right)}=\left(1-x\right)\sum_{n=0}^{\infty}nb_{n}x^{n-1}=\sum_{n=0}^{\infty}\left[\left(n+1\right)b_{n+1}-nb_{n}\right]x^{n}.$

Hence,

$G\left(x\right)=\frac{\pi}{8}F\left(\frac{1}{2},\frac{1}{2};2;x\right)e^{\mathcal{K}\left(\sqrt{x}\right)}-\frac{1}{2}e^{\mathcal{K}\left(\sqrt{x}\right)}=\sum_{n=0}^{\infty}\left[\left(n+1\right)b_{n+1}-\left(n+\frac{1}{2}\right)b_{n}\right]x^{n},$

which proves the formula (1.7).

On the other hand, applying the third formula of (2.7) and Cauchy product
formula to (2.9) gives

$\frac{\pi}{8}\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{W_{k}^{2}}{k+1}b_{n-k}\right)x^{n}=\sum_{n=0}^{\infty}\left[\left(n+1\right)b_{n+1}-nb_{n}\right]x^{n},$

then comparing coefficients of $x^{n}$ leads to

$\frac{\pi}{8}\sum_{k=0}^{n}\frac{W_{k}^{2}}{k+1}b_{n-k}=\left(n+1\right)b_{n+1}-nb_{n},$

which implies (2.8). Letting $x\rightarrow 0^{+}$ leads us to
$b_{0}=e^{\pi/2}$, and the first few coefficients follows from the recurrence
formula (2.8), which completes the proof.

######  Lemma 2 (Step 2).

Let $G\left(x\right)$ be defined by (1.7). Then $G^{\prime}\left(x\right)$ is
absolutely monotonic on $\left(0,1\right)$.

###### Proof.

Differentiation yields

$G^{\prime}\left(x\right)=\frac{\pi}{64}e^{\mathcal{K}\left(\sqrt{x}\right)}g\left(x\right),$

where $g\left(x\right)$ is defined by (1.8). Since
$e^{\mathcal{K}\left(\sqrt{x}\right)}$ is obviously absolutely monotonic on
$\left(0,1\right)$, if we prove that $g\left(x\right)$ is absolutely monotonic
on $\left(0,1\right)$, then by Remark 1, so is $G^{\prime}\left(x\right)$.

Let

$g_{0}\left(x\right)=\left(1-x\right)F\left(\frac{3}{2},\frac{3}{2};3;x\right)+\pi
F^{2}\left(\frac{1}{2},\frac{1}{2};2;x\right)-4F\left(\frac{1}{2},\frac{1}{2};2;x\right).$

By the third and fourth formulas of (2.7), $g_{0}\left(x\right)$ can be
expressed as

$g_{0}\left(x\right)=\sum_{n=0}^{\infty}u_{n}x^{n},$

where

(2.10)
$u_{n}=\pi\sum_{k=0}^{n}\frac{W_{k}^{2}W_{n-k}^{2}}{\left(k+1\right)\left(n-k+1\right)}-6\frac{2n+1}{\left(n+2\right)\left(n+1\right)}W_{n}^{2}.$

It was proved in [3, Lemma 2.5] that $u_{n}<0$ for all $n\geq 8$. This
together with [3, Eqs. (3.6)–(3.8)] gives that $u_{0}=\pi-3>0$,
$u_{1}=\left(\pi-3\right)/4>0$ and $u_{n}<0$ for all $n\geq 2$. Moreover, by
(2.2) and (2.6), we have

$g_{0}\left(1^{-}\right)=\sum_{n=0}^{\infty}u_{n}=0.$

On the other hand, applying the linear transformation formula (2.3) and the
Cauchy product formula gives

$\displaystyle g\left(x\right)$ $\displaystyle=$ $\displaystyle
F\left(\frac{3}{2},\frac{3}{2};3;x\right)+\pi
F\left(\frac{3}{2},\frac{3}{2};2;x\right)F\left(\frac{1}{2},\frac{1}{2};2;x\right)-4F\left(\frac{3}{2},\frac{3}{2};2;x\right)$
$\displaystyle=$
$\displaystyle\frac{g_{0}\left(x\right)}{1-x}=\frac{1}{1-x}\sum_{n=0}^{\infty}u_{n}x^{n}=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}u_{k}\right)x^{n}:=\sum_{n=0}^{\infty}v_{n}x^{n}.$

It is verified that $v_{0}=u_{0}=\pi-3>0$,
$v_{1}=u_{0}+u_{1}=5\left(\pi-3\right)/4>0$. For $n\geq 2$, since

$v_{n}-v_{n-1}=\sum_{k=0}^{n}u_{k}-\sum_{k=0}^{n-1}u_{k}=u_{n}<0,$

it is readily deduced that

$v_{n}>\lim_{n\rightarrow\infty}v_{n}=\sum_{n=0}^{\infty}u_{n}=0.$

This proves the absolute monotonicity of $g\left(x\right)$ on
$\left(0,1\right)$, thereby completing the proof.

######  Lemma 3 (Step 3).

Let $b_{n}$ be defined by (2.8). Then the sequence
$\left\\{b_{n}/W_{n}\right\\}_{n\geq 1}$ is increasing with

(2.11) $\lim_{n\rightarrow\infty}\frac{b_{n}}{W_{n}}=4.$

###### Proof.

To prove the increasing property of the sequence
$\left\\{b_{n}/W_{n}\right\\}_{n\geq 1}$, it suffices to prove the inequality

$b_{n+1}>\frac{W_{n+1}}{W_{n}}b_{n}=\frac{n+1/2}{n+1}b_{n},$

or equivalently,

(2.12) $\left(n+1\right)b_{n+1}-\left(n+\frac{1}{2}\right)b_{n}>0$

for $n\in\mathbb{N}$. By Lemma 2, $G^{\prime}\left(x\right)$ is absolutely
monotonic on $\left(0,1\right)$, which together with (1.7) and Remark 1
implies (2.12).

Since the sequence $\left\\{b_{n}/W_{n}\right\\}_{n\geq 1}$ is increasing and
$b_{1}/W_{1}>0$, the $\lim\limits_{n\rightarrow\infty}(b_{n}/W_{n})=p_{0}>0$
or tends to $\infty$. We first prove that
$\lim_{n\rightarrow\infty}\left(b_{n}/W_{n}\right)\leq 4$. If not, that is,
$\lim_{n\rightarrow\infty}\left(b_{n}/W_{n}\right)>4$, then there exists
$n_{0}>1$ such that $b_{n}/W_{n}>4$ for $n\geq n_{0}$. Let us denote by
$p^{\ast}=b_{n_{0}}/W_{n_{0}}$. The increasing property of
$\\{b_{n}/W_{n}\\}_{n\geq 1}$ makes us to deduce

$\frac{b_{n}}{W_{n}}-p^{\ast}=\frac{b_{n}}{W_{n}}-\frac{b_{n_{0}}}{W_{n_{0}}}\geq
0\text{, that is,}\quad b_{n}-p^{\ast}W_{n}\geq 0\text{ for }n\geq
n_{0}\text{,}$

which in conjunction with (1.4) yields

$\mathcal{G}_{p^{\ast}}(x)=\sum_{n=0}^{\infty}\left(b_{n}-p^{\ast}W_{n}\right)x^{n}\geq\sum_{n=0}^{n_{0}-1}(b_{n}-p^{\ast}W_{n})x^{n},$

which contradicts with

$\mathcal{G}_{p^{\ast}}(1^{-})=\lim_{x\rightarrow
1^{-}}\left[e^{\ln(4/\sqrt{1-x})}-\frac{p^{\ast}}{\sqrt{1-x}}\right]=-\infty.$

If $\lim\limits_{n\rightarrow\infty}(b_{n}/W_{n})=p_{0}<4$, that is to say,
$\sup_{n\geq 1}\\{b_{n}/W_{n}\\}=p_{0}$, then $b_{n}/W_{n}\leq p_{0}$ for all
$n\geq 1$. Similarly,

$\mathcal{G}_{p_{0}}(x)=\sum_{n=0}^{\infty}\left(b_{n}-p_{0}W_{n}\right)x^{n}\leq
b_{0}-p_{0}W_{0}$

which is a contradiction with $\mathcal{G}_{p_{0}}(1^{-})=\infty$ due to
$p_{0}<4$. This proves $\lim\limits_{n\rightarrow\infty}(b_{n}/W_{n})=4$, and
the proof is done.

###  2.3. Step 4: Proof of Theorem 1

By means of Lemma (Step 3) 3, we can easily complete the proof of our main
result.

###### Proof of Theorem 1.

From (1.4), $\mathcal{G}_{p}\left(x\right)$ can be expressed in power series:

(2.13)
$\mathcal{G}_{p}\left(x\right)=\sum_{n=0}^{\infty}b_{n}x^{n}-p\sum_{n=0}^{\infty}W_{n}x^{n}=\sum_{n=0}^{\infty}\left(b_{n}-pW_{n}\right)x^{n}:=\sum_{n=0}^{\infty}c_{n}\left(p\right)x^{n}.$

As a matter to remember, Lemma 3 states that the sequence $\\{b_{n}/W_{n}\\}$
is strictly increasing for $n\geq 1$ with
$\lim\limits_{n\rightarrow\infty}(b_{n}/W_{n})=4$.

(i) By (2.13), $\mathcal{G}_{p}$ is absolutely monotonic on $\left(0,1\right)$
if and only if

$c_{n}(p)=b_{n}-pW_{n}\geq 0\text{ for}\quad n\geq 0,$

which implies that

$p\leq\inf_{n\geq
0}\left\\{\frac{b_{n}}{W_{n}}\right\\}=\min\left\\{\frac{b_{0}}{W_{0}},\frac{b_{1}}{W_{1}}\right\\}=\min\left\\{e^{\pi/2},\frac{\pi}{4}e^{\pi/2}\right\\}=\frac{\pi}{4}e^{\pi/2}.$

Similarly, $-\mathcal{G}_{p}$ is absolutely monotonic on $\left(0,1\right)$ if
and only if

$p\geq\sup_{n\geq
0}\left\\{\frac{b_{n}}{W_{n}}\right\\}=\max\left\\{\frac{b_{0}}{W_{0}},4\right\\}=\max\left\\{e^{\pi/2},4\right\\}=e^{\pi/2}.$

(ii) In a fashion to (i), $\mathcal{G}_{p}^{(k)}$ for $k\geq 1$ is absolutely
monotonic on $(0,1)$ if and only if

$p\leq\inf_{n\geq k}\left\\{\frac{b_{n}}{W_{n}}\right\\}=\frac{b_{k}}{W_{k}};$

and $-\mathcal{G}_{p}^{(k)}$ is absolutely monotonic on $(0,1)$ if and only if

$p\geq\sup_{n\geq k}\left\\{\frac{b_{n}}{W_{n}}\right\\}=4.$

This completes the proof.

##  3\. New functional inequalities related to $\mathcal{K}(r)$

In this Section, Theorem 1 will be applied to establish several functional
inequalities involving the complete elliptic integral of the first kind
$\mathcal{K}(r)$.

######  Proposition 1.

Let $b_{n}$ be defined in (1.4) with $b_{0}=e^{\pi/2}$ and
$c_{n}(p)=b_{n}-pW_{n}$ for $n\geq 0$. The double inequality

(3.1)
$\ln\left(\frac{p}{r^{\prime}}+\sum_{n=0}^{m}c_{n}(p)r^{2n}\right)<\mathcal{K}(r)<\ln\left(\frac{q}{r^{\prime}}+\sum_{n=0}^{m}c_{n}(q)r^{2n}\right)$

holds for $r\in(0,1)$ if and only if $p\leq b_{m+1}/W_{m+1}$ and $q\geq 4$.

###### Proof.

As shown in (2.13), we denote by

(3.2)
$\mathcal{R}_{p,m}(x):=e^{\mathcal{K}(\sqrt{x})}-\frac{p}{\sqrt{1-x}}-\sum_{n=0}^{m}c_{n}(p)x^{n}=\sum_{n=m+1}^{\infty}c_{n}(p)x^{n}.$

Theorem 1 (ii) enables us to know that $c_{n}(p)\geq 0$ for $n\geq m+1$ if
$p\leq b_{m+1}/W_{m+1}$ and $c_{n}(q)\leq 0$ for $n\geq m+1$ if $q\geq 4$.
This together with (3.2) gives the desired inequality (3.1).

Conversely, the necessary conditions that (3.1) holds for all $r\in(0,1)$
require to satisfy $\mathcal{R}_{p,m}(0^{+})\geq 0$ and
$\mathcal{R}_{q,m}(1^{-})\leq 0$, which imply $c_{m+1}(p)=b_{m+1}-pW_{m+1}\geq
0$, (i.e. $p\leq b_{m+1}/W_{m+1}$) and

$\displaystyle\mathcal{R}_{q,m}(1^{-})$ $\displaystyle=\lim_{x\rightarrow
1^{-}}\frac{\sqrt{1-x}e^{\mathcal{K}(\sqrt{x})}-q-\sqrt{1-x}\sum\limits_{n=0}^{m}c_{n}(p)x^{n}}{\sqrt{1-x}}$
$\displaystyle=\lim_{x\rightarrow 1^{-}}\frac{4-q}{\sqrt{1-x}}\leq 0,$

namely, $q\geq 4$.

######  Remark 2.

Taking $m=0$ in the double inequality (3.1) leads to

(3.3)
$\ln\left(e^{\pi/2}-p+\frac{p}{r^{\prime}}\right)<\mathcal{K}(r)<\ln\left(e^{\pi/2}-q+\frac{q}{r^{\prime}}\right)$

for all $r\in(0,1)$ with the best constants $p=b_{1}/W_{1}=\pi e^{\pi/2}/4$
and $q=4$. This result was proven in [3, Corollary 3.7]. Clearly, Proposition
1 is a generalization of [3, Corollary 3.7].

######  Remark 3.

Taking $m=1$ in (3.1) yields

(3.4)
$\ln\left[e^{\pi/2}-p-\left(\frac{p}{2}-\frac{\pi}{8}e^{\pi/2}\right)r^{2}+\frac{p}{r^{\prime}}\right]<\mathcal{K}(r)<\ln\left[e^{\pi/2}-q-\left(\frac{q}{2}-\frac{\pi}{8}e^{\pi/2}\right)r^{2}+\frac{q}{r^{\prime}}\right]$

for all $r\in(0,1)$ with the best constants
$p=b_{2}/W_{2}=\left[\pi(\pi+9)e^{\pi/2}\right]/48$ and $q=4$. Since

$\displaystyle\left[e^{\pi/2}-\frac{b_{2}}{W_{2}}-\left(\frac{b_{2}}{2W_{2}}-\frac{\pi}{8}e^{\pi/2}\right)r^{2}+\frac{b_{2}}{W_{2}r^{\prime}}\right]-\left(e^{\pi/2}-\frac{b_{1}}{W_{1}}+\frac{b_{1}}{W_{1}r^{\prime}}\right)$
$\displaystyle=\frac{\pi(\pi-3)e^{\pi/2}}{96}\frac{r^{4}(r^{2}+3)}{r^{\prime}\left[2+(r^{2}+2)r^{\prime}\right]}>0$

and $2-\pi e^{\pi/2}/8=0.110\cdots>0$, we conclude that the optimal inequality
of (3.4) improves the one of (3.3).

For $n\geq 0$, Theorem 1 (ii) demonstrates that the function
$x\mapsto\mathcal{R}_{4,n}(x)/x^{n+1}$ is strictly decreasing on $(0,1)$
together with

$\lim_{x\rightarrow
0^{+}}\frac{\mathcal{R}_{4,n}(x)}{x^{n+1}}=c_{n+1}(4)\quad\text{and}\quad\lim_{x\rightarrow
1^{-}}\frac{\mathcal{R}_{4,n}(x)}{x^{n+1}}=-\sum_{k=0}^{n}c_{k}(4).$

This yields

$-\left(\sum_{n=0}^{m}c_{n}(4)\right)x^{m+1}<e^{\mathcal{K}(\sqrt{x})}-\frac{4}{\sqrt{1-x}}-\sum_{n=0}^{m}c_{n}(4)x^{n}<c_{m+1}(4)x^{m+1}$

for $x\in(0,1)$. By replacing $x=r^{2}$, we obtain therefore the following
proposition.

######  Proposition 2.

Let $b_{n}$ be defined by (1.4) and $c_{n}=b_{n}-4W_{n}$ for $n\geq 0$. The
double inequality

(3.5)
$\ln\left[\frac{4}{r^{\prime}}+\sum_{n=0}^{m}c_{n}r^{2n}-\left(\sum_{n=0}^{m}c_{n}\right)r^{2m+2}\right]<\mathcal{K}(r)<\ln\left[\frac{4}{r^{\prime}}+\sum_{n=0}^{m+1}c_{n}r^{2n}\right]$

holds for $r\in(0,1)$, where the lower and upper bounds are sharp.

######  Remark 4.

Taking $m=0$ in Proposition 2 yields

(3.6) $\ln\left[\frac{4}{r^{\prime}}+\left(e^{\pi/2}-4\right)r^{\prime
2}\right]<\mathcal{K}(r)<\ln\left[\frac{4}{r^{\prime}}+\left(e^{\pi/2}-4\right)-\left(2-\frac{\pi}{8}e^{\pi/2}\right)r^{2}\right]$

for $r\in(0,1)$. It was proved in [24, Corollary 3.5] that

(3.7)
$\ln\left[\frac{4}{r^{\prime}}+\left(e^{\pi/2}-4\right)r^{\prime}\right]<\mathcal{K}(r)<\ln\left(\frac{4}{r^{\prime}}+e^{\pi/2}-4\right)$

for $r\in\left(0,1\right)$. In [3, Eq. (3.5)], it was found that

(3.8)
$\ln\left[\frac{4}{r^{\prime}}+\left(e^{\pi/2}-4\right)r^{\prime}\right]<\mathcal{K}(r)<\ln\left[\frac{(\pi+4)e^{\pi/2}}{8r^{\prime}}+\left(\frac{1}{2}-\frac{\pi}{8}\right)e^{\pi/2}r^{\prime}\right]$

for $r\in(0,1)$. The left side inequality of (3.6) looks like the one of (3.7)
but a bit worse than the left side of (3.6), while the right side of (3.6) is
better than the ones of (3.7) and of (3.8). This can be confirmed by

$\left[\frac{4}{r^{\prime}}+\left(e^{\pi/2}-4\right)-\left(2-\frac{\pi}{8}e^{\pi/2}\right)r^{2}\right]-\left[\frac{4}{r^{\prime}}+\left(e^{\pi/2}-4\right)\right]<0$

and

$\displaystyle\left[\frac{4}{r^{\prime}}+e^{\pi/2}-4-\left(2-\frac{\pi}{8}e^{\pi/2}\right)r^{2}\right]-\left[\frac{(\pi+4)e^{\pi/2}}{8r^{\prime}}+\left(\frac{1}{2}-\frac{\pi}{8}\right)e^{\pi/2}r^{\prime}\right]$
$\displaystyle=-\frac{\left(1-r^{\prime}\right)^{2}}{8r^{\prime}}\left[\left(\left(\pi+4\right)e^{\pi/2}-32\right)-\left(16-\pi
e^{\pi/2}\right)r^{\prime}\right]$
$\displaystyle<-\frac{\left(\pi+2\right)e^{\pi/2}-24}{4}\frac{\left(1-r^{\prime}\right)^{2}}{r^{\prime}}<0$

for $r\in(0,1)$.

It was shown in [30] that every convex function
$f:[0,\infty)\mapsto\mathbb{R}$ satisfies a functional inequality

(3.9) $f(x)+f(y)\leq f(0)+f(x+y)$

for $x,y\in[0,\infty)$. In particular, $f(x)$ is superadditive if $f(0)\leq
0$. As a matter to remind, a function $f(x)$ defined on
$\mathbb{R}_{+}\backslash(0,a]$ is said to be superadditive if

(3.10) $f(x)+f(y)\leq f(x+y)$

for $x,y\in(a,\infty)$, while it is called subadditive if the inequality
(3.10) reverses.

Let $\mathcal{R}_{p,m}$ be defined as in (3.2). Then it is clear that
$\mathcal{R}_{p,m}(0^{+})=0$ for $m\geq 0$. Since

$\mathcal{R}_{p,m}^{\prime\prime}(x)=\left[\sum_{n=m+1}^{\infty}c_{n}(p)x^{n}\right]^{\prime\prime}=\left\\{\begin{array}[]{cc}\sum\limits_{n=2}^{\infty}n\left(n-1\right)c_{n}\left(p\right)x^{n-2}&\text{if
}m=0,\\\
\sum\limits_{n=m+1}^{\infty}n\left(n-1\right)c_{n}\left(p\right)x^{n-2}&\text{if
}m\geq 1,\end{array}\right.$

it follows from Theorem 1 (ii) that $c_{n}(p)\geq 0$ for $n\geq 2$ if $p\leq
b_{2}/W_{2}=[\pi(\pi+9)e^{\pi/2}]/48$, while $c_{n}(p)\leq 0$ for $n\geq 2$ if
$p\geq 4$. This together with (3.9) deduces that the double inequality

$2\mathcal{R}_{p,m}\left(\frac{x+y}{2}\right)<\mathcal{R}_{p,m}(x)+\mathcal{R}_{p,m}(y)<\mathcal{R}_{p,m}(x+y)$

holds for $x,y,x+y\in(0,1)$ if $p\leq[\pi(\pi+9)e^{\pi/2}]/48$ and it reverses
if $p\geq 4$.

######  Proposition 3.

Let $b_{n}$ be defined in (1.4) and $c_{n}(p)=b_{n}-pW_{n}$. If
$p\leq[\pi(\pi+9)e^{\pi/2}]/48$, then the double inequality

$\displaystyle
2e^{\mathcal{K}\left(\sqrt{\frac{x+y}{2}}\right)}+p\left(\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1-y}}-\frac{2}{\sqrt{1-\frac{x+y}{2}}}\right)+\sum_{n=0}^{m}c_{n}(p)\left[x^{n}+y^{n}-2\left(\frac{x+y}{2}\right)^{n}\right]$
$\displaystyle<e^{\mathcal{K}\left(\sqrt{x}\right)}+e^{\mathcal{K}\left(\sqrt{y}\right)}$
$\displaystyle<e^{\mathcal{K}\left(\sqrt{x+y}\right)}+p\left(\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1-y}}-\frac{1}{\sqrt{1-x-y}}\right)+\sum_{n=0}^{m}c_{n}(p)\left[x^{n}+y^{n}-\left(x+y\right)^{n}\right]$

holds for $x,y,x+y\in(0,1)$, and it reverses if $p\geq 4$.

Taking $m=0$ into Proposition 3, we obtain

######  Corollary 1.

If $p\leq[\pi(\pi+9)e^{\pi/2}]/48$, then the double inequality

$\displaystyle
2e^{\mathcal{K}\left(\sqrt{\frac{x+y}{2}}\right)}+p\left(\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1-y}}-\frac{2}{\sqrt{1-\frac{x+y}{2}}}\right)<e^{\mathcal{K}\left(\sqrt{x}\right)}+e^{\mathcal{K}\left(\sqrt{y}\right)}$
$\displaystyle\hskip
85.35826pt<e^{\mathcal{K}\left(\sqrt{x+y}\right)}+p\left(\frac{1}{\sqrt{1-x}}+\frac{1}{\sqrt{1-y}}-\frac{1}{\sqrt{1-x-y}}-1\right)+e^{\pi/2}$

holds for $x,y,x+y\in(0,1)$, and it is reversed if $p\geq 4$.

######  Remark 5.

Let $p=4$, $y=1-r^{2}$ and $x\rightarrow r^{2}$ in Corollary 1. Then we obtain

$e^{\pi/2}-4<e^{\mathcal{K}(r)}+e^{\mathcal{K}(r^{\prime})}-4\left(\frac{1}{r}+\frac{1}{r^{\prime}}\right)\leq
2e^{\mathcal{K}(1/\sqrt{2})}-8\sqrt{2}$

for $r\in(0,1)$, where

$\mathcal{K}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{2}F\left(\frac{1}{2},\frac{1}{2};1;\frac{1}{2}\right)=\frac{\Gamma(1/4)^{2}}{4\sqrt{\pi}},$

which can be computed by the formula (see [31, Eq. (15.1.26)])

(3.11)
$F\left(a,1-a;b;\frac{1}{2}\right)=\frac{2^{1-b}\sqrt{\pi}\Gamma\left(b\right)}{\Gamma\left(a/2+b/2\right)\Gamma\left(1/2+b/2-a/2\right)}.$

Clearly, the lower and upper bounds are sharp by (1.2).

The following lemma was proved in [8, Lemma 4], where $h$ should be infinitely
differentiable instead of “thrice differentiable”.

######  Lemma 4.

Let the function $h$ be defined on $\mathbb{R}$ which is infinitely
differentiable. If $h^{\left(2k+1\right)}\left(x\right)>\left(<\right)0$ for
$x\in\mathbb{R}$ and $k\in\mathbb{N}$ then the function

$\phi\left(x\right)=\frac{h\left(2c-x\right)-h\left(x\right)}{2c-2x}\text{ if
}x\neq c\text{ \ and \ }\phi\left(c\right)=h^{\prime}\left(c\right)$

satisfies that $\phi^{\left(2k-2\right)}\left(x\right)$ is strictly convex
(concave) on $\mathbb{R}$, and is decreasing (increasing) on
$\left(-\infty,c\right)$ and increasing (decreasing) on
$\left(c,\infty\right)$. Furthermore, if $h^{\prime}\left(x\right)$ is
absolutely monotonic on $\mathbb{R}$ then $\phi\left(x\right)$ is completely
monotonic on $\left(-\infty,c\right)$ and absolutely monotonic on
$\left(c,\infty\right)$.

Applying Lemma 4 and the absolute monotonicity of
$-\mathcal{G}_{4}^{\prime}\left(x\right)$, we have the following statement.

######  Proposition 4.

The function

$\mathcal{H}\left(x\right)=\frac{\mathcal{G}_{4}\left(x\right)-\mathcal{G}_{4}\left(1-x\right)}{1-2x}=\frac{1}{1-2x}\left[e^{\mathcal{K}\left(\sqrt{x}\right)}-e^{\mathcal{K}\left(\sqrt{1-x}\right)}+\frac{4}{\sqrt{x}}-\frac{4}{\sqrt{1-x}}\right]$

is completely monotonic on $\left(0,1/2\right)$ and absolutely monotonic on
$\left(1/2,1\right)$. Consequently, the double inequality

(3.12)
$\beta\left(1-2r^{2}\right)<e^{\mathcal{K}\left(r\right)}-e^{\mathcal{K}\left(r^{\prime}\right)}+\frac{4}{r}-\frac{4}{r^{\prime}}<\alpha\left(1-2r^{2}\right)$

holds for $r\in\left(0,1/\sqrt{2}\right)$ with the best constants
$\alpha=e^{\pi/2}-4\approx 0.810$ and

$\beta=4\sqrt{2}-\frac{1}{\sqrt{\pi}}\Gamma^{2}\left(\frac{3}{4}\right)\exp\left(\frac{\Gamma\left(1/4\right)^{2}}{4\sqrt{\pi}}\right)\approx
0.246.$

The double inequality (3.12) reverses for $r\in\left(1/\sqrt{2},1\right)$.

###### Proof.

Taking $h\left(x\right)=-\mathcal{G}_{4}\left(x\right)$ and $c=1/2$ in Lemma 4
yields the complete and absolute monotonicity of $\mathcal{H}\left(x\right)$
on $\left(0,1/2\right)$ and $\left(1/2,1\right)$, respectively. The decreasing
property of $\mathcal{H}\left(x\right)$ on $\left(0,1/2\right)$ implies that

(3.13)
$-\mathcal{G}_{4}^{\prime}\left(\frac{1}{2}\right)=\mathcal{H}\left(\frac{1}{2}\right)<\mathcal{H}\left(x\right)=\mathcal{H}\left(0\right)=e^{\pi/2}-4$

for $x\in\left(0,1/2\right)$. Using the formula (3.11) we have

$\displaystyle F\left(\frac{1}{2},\frac{1}{2};1;\frac{1}{2}\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2\pi^{3/2}}\Gamma^{2}\left(\frac{1}{4}\right),$
$\displaystyle F\left(\frac{1}{2},\frac{1}{2};2;\frac{1}{2}\right)$
$\displaystyle=$
$\displaystyle\frac{4}{\pi^{3/2}}\Gamma^{2}\left(\frac{3}{4}\right).$

Then

$\displaystyle-\mathcal{G}_{4}^{\prime}\left(\frac{1}{2}\right)$
$\displaystyle=$
$\displaystyle\left[\frac{2}{\left(1-x\right)^{3/2}}-\frac{\pi}{8\left(1-x\right)}F\left(\frac{1}{2},\frac{1}{2};2;x\right)e^{\mathcal{K}\left(\sqrt{x}\right)}\right]_{x=1/2}$
$\displaystyle=$ $\displaystyle
4\sqrt{2}-\frac{1}{\sqrt{\pi}}\Gamma^{2}\left(\frac{3}{4}\right)\exp\left(\frac{\Gamma^{2}\left(1/4\right)}{4\sqrt{\pi}}\right).$

Let $x=r^{2}$. Then (3.13) implies (3.12). In a similar way, we can prove that
(3.12) reverses for $r\in\left(1/\sqrt{2},1\right)$.

##  4\. Concluding remarks

In this paper, we investigated the absolute monotonicity of the functions
$\pm\mathcal{G}_{p}\left(x\right)$ and
$\pm\mathcal{G}_{p}^{\left(k\right)}\left(x\right)$ for $k\in\mathbb{N}$ on
$\left(0,1\right)$, where

$\mathcal{G}_{p}\left(x\right)=e^{\mathcal{K}\left(\sqrt{x}\right)}-\frac{p}{\sqrt{1-x}}=\sum_{n=0}^{\infty}b_{n}x^{n}-p\sum_{n=0}^{\infty}W_{n}x^{n}.$

These absolute monotonicity results yield several new functional inequality
related to $\mathcal{K}\left(r\right)$. It is worth empathizing our approach
used in this paper is very novel and interesing. Follow the usual method, ones
should determine the signs of those coefficients of power series of
$\pm\mathcal{G}_{p}\left(x\right)$ and
$\pm\mathcal{G}_{p}^{\left(k\right)}\left(x\right)$ by directly using the
recurrence formula (1.9) and $W_{n}$ defined by (1.5), which is very
complicated and hard. By our new idea (Steps 1 and 2), we found a surprising
property about $b_{n}$ and $W_{n}$, that is, the sequence
$\left\\{b_{n}/W_{n}\right\\}_{n\geq 1}$ is increasing with
$\lim_{n\rightarrow\infty}\left(b_{n}/W_{n}\right)=4$. It is with this
property that we can conveniently prove the desired absolute monotonicity
results. Such new idea may contribute to deal with certain problems on
monotonicity, convexity and absolute monotonicity occured in special
functions.

Finally, we present two remarks and a problem.

######  Remark 6.

It was proved in [4] that

$\mathcal{F}_{1/2}\left(x\right)=\sqrt{1-x}e^{\mathcal{K}\left(\sqrt{x}\right)}=\sum_{n=0}^{\infty}a_{n}x^{n},$

where $a_{0}=e^{\pi/2}$ and $a_{n}<0$ for all $n\geq 1$. Differentiation
yields

$\sqrt{1-x}\mathcal{F}_{1/2}^{\prime}\left(x\right)=\left[\frac{\pi}{8}F\left(\frac{1}{2},\frac{1}{2};2;x\right)-\frac{1}{2}\right]e^{\mathcal{K}\left(\sqrt{x}\right)}=G\left(x\right).$

Can prove that $G^{\prime}\left(x\right)$ is absolutely monotonic on
$\left(0,1\right)$ by the absolute monotonicity of
$-\mathcal{F}_{1/2}^{\prime}\left(x\right)$? Or vice versa?

######  Remark 7.

Since $b_{0}/W_{0}>b_{1}/W_{1}$, we see that the sequence
$\left\\{b_{n}/W_{n}\right\\}_{n\geq 0}$ is decreasing for $0\leq n\leq 1$ and
increasing for $n\geq 1$. Making use of the unimodal type monotonicity rule of
the ratio of power series given in [32, Theorem 2.1], we can only deduce that

$\mathcal{F}_{1/2}\left(x\right)=\frac{e^{\mathcal{K}\left(\sqrt{x}\right)}}{1/\sqrt{1-x}}=\frac{\sum_{n=0}^{\infty}b_{n}x^{n}}{\sum_{n=0}^{\infty}W_{n}x^{n}}$

is decreasing on $\left(0,1\right)$. Employing the recurrence technique,
however, Yang and Tian have shown that
$-\mathcal{F}_{1/2}^{\prime}\left(x\right)$ is absolutely monotonic on
$\left(0,1\right)$. In a similar way, by the monotonicity rule of the ratio of
power series given in [33], we can draw a conclusion that

$x\mapsto\mathcal{J}\left(x\right)=\frac{e^{\mathcal{K}\left(\sqrt{x}\right)}-e^{\pi/2}}{1/\sqrt{1-x}-1}=\frac{\sum_{n=1}^{\infty}b_{n}x^{n}}{\sum_{n=1}^{\infty}W_{n}x^{n}}$

is increasing on $\left(0,1\right)$. However, by the increasing property of
$\left\\{b_{n}/W_{n}\right\\}_{n\geq 1}$ and recurrence technique, we can
prove that this function $\mathcal{J}\left(x\right)$ is absolutely monotonic
on $\left(0,1\right)$, and the proof of which is left to the reader.

######  Remark 8.

So far, several interesting questions remain unresolved involving the
absolutely monotonic functions related to the asymptotic formula (1.2), for
example,

(i) In [8, Conjecture 1], it was conjectured that the function
$-f_{4}^{\prime\prime\prime}\left(x\right)$ is absolutely monotonic on
$\left(0,1\right)$, where $f_{p}\left(x\right)$ is defined in Introduction of
this paper.

(ii) In [8, Conjecture 1], it was conjectured that the function
$1/f_{4}\left(x\right)$ is absolutely monotonic on $\left(0,1\right)$.

(iii) Is the function $\left(f_{4}\left(x\right)-1\right)/\left(1-x\right)$
absolutely monotonic on $\left(0,1\right)$? This is a special of the open
problem proposed in [34, Appendix G, Open Problems (2)], which is still open.

Acknowledgements

We would like to thank the three referees for the careful review on the
manuscript. This research was supported by the National Natural Science
Foundation of China (11971142) and the Natural Science Foundation of Zhejiang
Province (LY24A010011).

Data availability
Data sharing not applicable to this article as no datasets were generated or
analysed during the current study.

Declarations

Conflict of interest The authors declare that they have no conflict of
interest.

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