text:
\begin{figure*}
   \centering
    \includegraphics[width=\textwidth]{comparison_kron.pdf}
   \caption{\footnotesize{Dense core with $R=3$, $N_1=50,N_2=60,N_3=70$, $K_1=10,K_2=20,K_3=15$, and $\alpha_1=\alpha_2=\alpha_3=2$.}}
   \label{fig:kron_comp}
      \vspace{-4mm}
 \end{figure*}

  \begin{figure*}
   \centering
        \includegraphics[width=\textwidth]{comparison_diag.pdf}
   \caption{\footnotesize{Diagonal core with $R=3$ with $N_1=50,N_2=60,N_3=70$, $K_\text{c}=20$, $\beta_1=\beta_2=1$, and $\beta_3=20$.}}
   \label{fig:diag_comp}
   \vspace{-3mm}
 \end{figure*}

\section{Numerical results}\label{sec:numerical_results}

In this section\footnote{The code to reproduce these experiments can be found at https://gitlab.com/gortizji/sparse\_tensor\_sensing.}, we will illustrate the developed framework through several examples. First, we will show some results obtained on synthetic datasets to compare the performance of the different near-optimal algorithms. Then, we will focus on large-scale real-world examples related to (i) graph signal processing: \emph{sampling product graphs for active learning in recommender systems},  and (ii) array processing for wireless communications: \emph{multiuser source separation}, to show the benefits of the developed framework.

\subsection{Synthetic example}
\subsubsection{Dense core}

We compare the performance in terms of the theoretical MSE of our proposed greedy algorithm (henceforth referred to as greedy-FP) to a random sampling scheme based on randomly selecting rows of $\matx{U}_i$ such that the resulting subset of samples also have a Kronecker structure. Only those selections that satisfy the identifiability constraints in \eqref{eq:identifiability_constraint} are considered  valid. We note that the time complexity of evaluating $M$ times the MSE for a Kronecker-structured sampler is $\mathcal{O}(MN_\text{max}^2K_\text{max})$. For this reason, using many realizations (say, a large number $M$)  of random sampling to obtain a good sparse sampler is computationally intense.

To perform this comparison, we draw $M=100$ realizations of three random Gaussian matrices $\{\matx{U}_i\in\R^{N_i\times K_i}\}_{i=1}^{R=3}$ with dimensions $N_1=50$, $N_2=60$, and $N_3=70$. For each of these models, we solve $\eqref{eq:submodular_max_kron_matroid}$ for different number of sensors $L$ using greedy-FP. We also compute $M=100$ realizations of random sampling for each $L$. Fig.~\ref{fig:kron_comp} shows the results of these experiments. The plot on the left shows the performance averaged over the different models against the number of sensors, wherein the blue shaded area represents the 10-90 percentile average interval of the random sampling scheme. The performance values, in dB scale, are normalized by the value of the unsampled MSE. Because the estimation performance is heavily influenced by its related number of samples $\tilde{L}$, and noting the fact that a value of $L$ may lead to different $\tilde{L}$, we also present, in the plot on the right side of Fig.~\ref{fig:kron_comp}, the performance comparison for one model realization against the relative number of samples $\tilde{L}/\tilde{N}$ so that differences in the informative quality of the selections are highlighted.

The plots in Fig.~\ref{fig:kron_comp} illustrate some important features of the proposed sparse sampling method. When comparing the performance against the number of sensors, we see that there are areas where greedy-FP performs as well as random sampling. However, when comparing the same results against the number of samples we see that greedy-FP consistently performs better than random sampling. The reason for this disparity is due to characteristics of greedy-FP that we introduced in Section~\ref{sec:dense_fp}. Namely, the tendency of greedy-FP to produce sampling sets with the minimum number of samples.

On the other hand, the performance curve of greedy-FP shows three bumps (recall that we use $R=3$). Again, this is a consequence of greedy-FP trying to meet the identifiability constraints in \eqref{eq:submodular_max_kron_matroid} with equality. As we increase $L$, the solutions of greedy-FP increase in cardinality by adding more elements to a single domain until the constraints are met, and then proceed to the next domain. The bumps in Fig.~\ref{fig:kron_comp} correspond precisely to these instances.

\subsubsection{Diagonal core}

We perform the same experiment for the diagonal core case. The results are shown in Fig.~\ref{fig:diag_comp}. Again we see that the proposed algorithm outperforms random sampling, especially when collecting just a few samples. Furthermore, as happened in the dense core case, the performance curve of greedy-FP follows a stairway shape.

\subsection{Active learning for recommender systems}

Current recommendation algorithms seek solving an estimation problem of the form: given the past recorded preferences of a set of users, what is the \emph{rating} that these would give to a set of products? In this paper, in contrast, we focus on the data acquisition phase of the recommender system, which is also referred to as active learning/sampling. In particular, we claim that by carefully designing which users to poll and on which items, we can obtain an estimation performance on par with the state-of-the-art methods, but using only a fraction of the data that current methods require, and using a simple least-squares estimator.

We showcase this idea on the MovieLens $100k$ dataset \cite{movielens} that contains partial ratings of $N_1=943$ users over $N_2=1682$ movies which are stored in a second-order tensor $\mathbfcal{X}\in\R^{N_1\times N_2}$. At this point, we emphasize the need for our proposed framework, since it is obvious that designing an unstructured sampling set with about 1.5 million candidate locations is unfeasible with current computing resources.

A model of $\mathbfcal{X}$ in the form of \eqref{eq:tensor_linear_model} can be obtained by viewing $\mathbfcal{X}$ as a signal that lives on a graph. In particular, the first two modes of $\mathbfcal{X}$ can be viewed as a signal defined on the Cartesian product of a user and movie graph, respectively. These two graphs, shown in Fig.~\ref{fig:graphs}, are provided in the dataset and are two 10-nearest-neighbors graphs created based on the user and movie features.

Based on the recent advances in graph signal processing (GSP)~\cite{moura,gsp}, $\mathbfcal{X}$ can be decomposed as $\mathbfcal{X}=\mathbfcal{X}_\text{f}\bullet_1 \matx{V}_1\bullet_2\matx{V}_2$. Here, $\matx{V}_1\in\mathbb{R}^{N_1\times N_1}$ and $\matx{V}_2\in\mathbb{R}^{N_2\times N_2}$ are the eigenbases of the Laplacians of the user and movie graphs, respectively, and $\mathbfcal{X}_\text{f}\in\matx{C}^{N_1\times N_2}$ is the so-called graph spectrum of $\mathbfcal{X}$~\cite{moura,gsp}. Suppose the energy of the spectrum of $\mathbfcal{X}$ is concentrated in the first few $K_1$ and $K_2$ columns of $\matx{V}_1$ and $\matx{V}_2$, respectively, then $\mathbfcal{X}$ admits a low-dimensional representation, or $\mathbfcal{X}$ is said to be smooth or bandlimited with respect to the underlying graph~\cite{gsp}. This property has been exploited in~\cite{kalofolias2014matrix,marques} to impute the missing entries in $\mathbfcal{X}$. In contrast, we propose a scheme for sampling and reconstruction of signals defined on product graphs.

 In our experiments, we set $K_1=K_2=20$, and obtain the decomposition $\mathbfcal{X}=\mathbfcal{G}\bullet_1 \matx{U}_1\bullet_2\matx{U}_2$, where $\matx{U}_1\in\C^{N_1\times K_1}$ and $\matx{U}_2\in\C^{N_2\times K_1}$ consist of the first $K_1$ and $K_2$ columns of $\matx{V}_1$ and $\matx{V}_2$, respectively; and $\mathbfcal{G}=\mathbfcal{X}(1:K_1,1:K_2)$.

\begin{figure}
    \centering
    \begin{subfigure}[b]{0.49\columnwidth}
        \includegraphics[width=\textwidth]{users.png}
        \caption{User graph}
        \label{fig:users}
    \end{subfigure}
    \hfill
    \begin{subfigure}[b]{0.49\columnwidth}
        \includegraphics[width=\textwidth]{movies.png}
        \caption{Movie graph}
        \label{fig:movies}
    \end{subfigure}
    \caption{\footnotesize{User and movie networks. The red (black) dots represent the observed (unobserved) vertices. Visualization obtained using Gephi \cite{gephi}.}}
    \label{fig:graphs}
    \vspace{-2mm}
\end{figure}

For the greedy algorithm we use $L = 100$ and $\alpha_1=\alpha_2=5$, resulting in a selection of $L_1 = 25$ users and $L_2 = 75$ movies, i.e., a total of $1875$ vertices in the product graph. Fig.~\ref{fig:graphs}, shows the sampled users and movies, i.e., users to be probed for movie ratings. The user graph [cf. Fig.~\ref{fig:users}] is made out of small clusters connected in a chain-like structure, resulting in a uniformly spread distribution of observed vertices. On the other hand, the movies graph [cf. Fig.~\ref{fig:movies}] is made out of a few big and small clusters. Hence, the proposed active querying scheme assigns more observations to the bigger clusters and fewer observations to the smaller ones.

To evaluate the performance of our algorithm, we compute the RMSE of the estimated data using the test mask provided by the dataset. Nevertheless, since our active query method requires access to ground truth data (i.e., we need access to the samples at locations suggested by the greedy algorithm) which is not provided in the dataset, we use GRALS \cite{grals} to complete the matrix, and use its estimates when required. A comparison of our algorithm to the performance of the state-of-the-art methods run on the same dataset is shown in Table~\ref{tab:performance}. In light of these results, it is clear that a proper design of the sampling set allows to obtain top performance with significantly fewer ratings, i.e., about an order of magnitude, and using a much simpler non-iterative estimator.

\begin{table}
\centering{}
\begin{tabular}{@{}ccc@{}}
\toprule
\textbf{Method} & \textbf{Number of samples} & \textbf{RMSE} \\ \midrule
GMC \cite{kalofolias2014matrix}            & 80,000                      & 0.996         \\ \midrule
GRALS  \cite{grals}         & 80,000                      & 0.945         \\ \midrule
sRGCNN\cite{monti2017geometric}         & 80,000                      & 0.929         \\ \midrule
GC-MC    \cite{berg2017graph}       & 80,000                      & \textbf{0.905}         \\ \midrule\midrule
Our method      & \textbf{1,875}                       & 0.9347        \\ \bottomrule
\end{tabular}
\caption{\footnotesize{Performance on MovieLens $100k$. Baseline scores are taken from \cite{berg2017graph}.}}
\label{tab:performance}

\end{table}

\subsection{Multiuser source separation}

In multiple-input multiple-output (MIMO) communications \cite{mimo_magazine}, the use of rectangular arrays \cite{ura} allows to separate signals coming from different azimuth and elevation angles, and it is common that users transmit data using different spreading codes to reduce the interference from other sources. Reducing hardware complexity by minimizing the number of antennas and samples to be processed is an important concern in the design of MIMO receivers. This design can be seen as a particular instance of sparse tensor sampling.

We consider a scenario with $K_\text{c}$ users located at different angles of azimuth ($\phi$) and elevation ($\theta$) transmitting using unique spreading sequences of length $N_3$. The receiver consists of a uniform rectangular array (URA) with antennas located on a $N_1\times N_2$ grid. Each time instant, every antenna receives \cite{ura}
\begin{align*}
  x(r,l,m,n) &= \sum_{k=1}^{K_\text{c}} s_k(r) c_k(l) e^{j2\pi n \Delta_x \sin \theta_k}e^{j 2\pi m \Delta_y \sin \phi_k} \\
  &\hspace{1em}+ w(r,l,m,n),
\end{align*}
where $s_k(r)$ the symbol transmitted by user $k$ in the $r$th symbol period; $c_k(l)$ the $l$th sample of the spreading sequence of the $k$th user; $\Delta_x$ and $\Delta_y$ the antenna separations in wavelengths of the URA in the $x$ and $y$ dimensions, respectively; and $\phi_{k}$ and $\theta_k$ the azimuth and elevation coordinates of user $k$, respectively; and where $w(r,l,m,n)$ represents an additive white Gaussian noise term with zero mean and variance $\sigma^2$. For the $r$-th symbol period, all these signals can be collected in a 3rd-order tensor $\mathbfcal{X}(r)\in\C^{N_1\times N_2\times N_3}$ that can be decomposed as
\begin{equation*}
  \mathbfcal{X}(r)=\mathbfcal{S}(r)\bullet_1 \matx{U}_1\bullet_2 \matx{U}_2 \bullet_3 \matx{U}_3+\mathbfcal{W}(r)
\end{equation*}
where $\matx{U}_1\in\C^{N_1\times K_\text{c}}$ and $\matx{U}_2\in\C^{N_2\times K_\text{c}}$ are the array responses for the $x$ and $y$ directions, respectively; $\matx{U}_3\in\C^{N_3\times K_\text{c}}$ contains the spreading sequences of all users in its columns; and $\mathbfcal{S}(r)\in\C^{K_\text{c}\times K_\text{c}\times K_\text{c}}$ is a diagonal tensor that stores the symbols of all users for the $r$th symbol period on its diagonal.

\begin{figure}[t]
  \centering
  \includegraphics[width=\columnwidth]{mse.pdf}
  \caption{\footnotesize{MSE of symbol reconstruction. $N_1=50$, $N_2=60$, $N_3=100$, and $L=15$.}}
  \label{fig:error_prob}
  \vspace{-5mm}
\end{figure}

We simulate this setup using $K_\text{c}=10$ users that transmit BPSK symbols with different random powers and that are equispaced in azimuth and elevation. We use a rectangular array with $N_1=50$ and $N_2=60$ for the ground set locations of the antennas, and binary random spreading sequences of length $N_3=100$. With these parameters, each $\mathbfcal{X}(r)$ has $300,000$ entries. We generate many realizations of these signals for different levels of signal-to-noise ratio (SNR) and sample the resulting tensors using the greedy algorithm for the diagonal core case with $L=15$, resulting in a relative number of samples of $0.048\%$. The results are depicted in Fig.~\ref{fig:error_prob}, where the blue shaded area represents the MSE obtained with the best and worst random samplers. As expected, the MSE of the reconstruction decreases exponentially with the SNR. For a given MSE, achieving maximum compression requires transmitting with a higher SNR of about 30dB than the one needed for no compression. Besides, we see that our proposed greedy algorithm consistently performs as well as the best random sampling scheme.


TITLE: What does the discovery of the Higgs Boson mean for physics?
QUESTION [9 upvotes]: Will this unite some theories, or cause some other change in physics, and perhaps our undertanding of the universe?

REPLY [17 votes]: Physicists have expected the Higgs boson to be discovered for a long time, and in fact a lot of theoretical work has been done under the assumption that the Higgs boson exists. So honestly, the mere fact that the Higgs has been discovered doesn't cause any drastic changes in physics. There are no theories that are suddenly far more or less viable today than they were yesterday, for example.
The real question is what kind of Higgs boson has been discovered. There are a few options here: it could be the standard model Higgs boson, which is the simplest kind. If that turns out to be the case, then we've found exactly the particle that was predicted to exist by electroweak symmetry breaking (the Higgs mechanism). That's the most boring possibility because it doesn't suggest the existence of any sort of new or unknown physical phenomenon - particle physics would kind of hit a dead end in that case. (I'm exaggerating somewhat, it wouldn't really be a dead end.)
Alternatively, the particle that has been discovered could be some different kind of Higgs boson, which doesn't behave in quite the way we'd expect the standard model Higgs to behave. That would be far more interesting, because it would strongly suggest that there is something new out there, beyond the standard model, and it would give theorists their first clue as to what it might be. That's the case in which you could get a bunch of new theories that might start to gain validity.
The way we're going to be able to tell the difference between the various options is by collecting more data at the LHC and analyzing it closely to figure out the behavior of the new particle. This process is going to take several years, so we're not likely to find out definite results on this any time soon.


TITLE: Best algorithm (Time Complexity) to find Minimum spanning tree of an complete, positive weighted, undirected, graph
QUESTION [0 upvotes]: Suppose that we have a complete undirected positive weighted graph G={V,E} and I want to find one MST of it. My problem is what is the best (in time complexity) algorithm for it?
The best prime complexity is O(V^2)
The best Kruskal complexity is O(E.Log(E))=~O(V^2. Log(V))
any idea for lower time complexity?

REPLY [0 votes]: I posted my question in computer science part and this is a resonable answer that I found:


TITLE: How could you find the set of all complex numbers $z$ such that $\lvert z − a \rvert - \lvert z − b \rvert = d$?
QUESTION [0 upvotes]: If I've got 2 points in the complex plane, A and B, of respective affixes a and b,
how could I find the set of all points M, of affix z, such that the difference between the distances AM and BM equals a given constant d, ie such that |a-z|-|b-z|=d ?
It feels like there should be a general solution for this. If d=0, then it would just be the perpedicular bisector of the segment [AB].
But once I give d a different value, whether I work it out in imaginary and real parts, or in modulus and complex argument, I get really clunky equations.
There's got to be an elegant solution out there, right ?
What's more, if you've got such a point M, then the point symetric to M according to the segment [AB] also works out, so there seems to be a general symmetry to this problem.
Any help on this would be appreciated, thanks.

REPLY [0 votes]: Let $2\lambda=|a-b|$, $2c=a+b$, $\theta=\arg (b-a)$ and $2\lambda \cos v=d$  then
$$z=c\pm \lambda\cosh [e^{i\theta}(u+vi)]$$
Also see another posts of mine here, here and here.


TITLE: The Euler's totient function and the product of distinct primes dividing $n$ versus the Heronian means
QUESTION [2 upvotes]: For integers $n\geq 1$ with $$\operatorname{rad}(n)=\prod_{\substack{p\mid n\\p\text{ prime}}}p$$ we denote the squarefree kernel or radical of an integer $n$ (see if you want this Wikipedia). And $\varphi(n)$ denotes the Euler's totient funciton. Then while I was stuying the equation $$\operatorname{rad}(\varphi(2\cdot n))=2$$
in the context of constructible polygons with compass and straightedge, see this section of this Wikipedia, I wondered next variation
$$\operatorname{rad}(\varphi(3\cdot n))=2\cdot 3.\tag{1}$$
Ater I search the resulting sequence of solutions of our equation $(1)$, the first few terms are $3,6,7,9,12,13,14,\ldots$, in The On-Line Encyclopedia of Integer Sequences, I would like to do a comparison to the sequence A135412. I know that there are terms in A135412, but don't satisfy our equation (1). You can read the definition of Heronian mean from this  Wikipedia.

Question. I would like to know* if it is possible to prove or refute that:
If $n$ satisfies $$\operatorname{rad}(\varphi(3 n))=6$$ then our solution $n$ equals three times the Heronian mean of two positive integers.
Many thanks.

I think that this question is curious. If a full answer isn't feasible I'm interested to know what work can be done.
*Additionally if you want to add remarks in your answer about different equations involving the Euler's totient function and/or the squarefree kernel,  showing some remarkable relationship to the sequence of the Heronian means in some way feel free to provide me such feedback.

REPLY [3 votes]: Your conclusion is valid, and for this we only need to assume that $\varphi(3n)$ is divisible by $3$. Indeed, this weaker condition is equivalent to the existence of a prime divisor $p\mid n$ that is either $3$ or congruent to $1$ modulo $3$. It is known that $p$ is the norm of some Eulerian integer $z=x+y\frac{1+i\sqrt{3}}{2}$. Multiplying $z$ with a sixth root of unity, we can achieve that $0<\arg(z)<\pi/3$. Then, $x$ and $y$ are positive integers satisfying $p=x^2+xy+y^2$, and hence $n$ is three times the Heronian mean of $(n/p)x^2$ and $(n/p)y^2$.


TITLE: Exponential Equations and Inequalities and logarithms
QUESTION [0 upvotes]: So this is one kind of task:
$2\times 8^{x}-7\times4^{x}+7\times2^{x}-2=0$
And I don't get any of these tasks, so I am asking for some kind of literature with introduction about this and good explanation if you have to suggest. By the way we did exponential functions so it would be nice to give me some literature about exponential functions too...
By the way we started working with logarithms, basics things I do understand how to calculate:
$\log _{2}4$ =?
, but I don't know how to calculate: $\log _{49}  \sqrt[3]{7^{5}}$=? , so if you also have some nice pdf book with introduction and examples for logarithms too it will be nice.
Thank you.

REPLY [0 votes]: For the second question, there is this identity:
$$\log_b a = \frac{\log a}{\log b}$$
This is derived from: if you want to find out $\log_b a = ?$,
$$\begin{align*}
\log_b a &= \ ?\\
a &= b^?\\
\log a &= \log \left(b^?\right) \tag{1}\\
&= \ ?\cdot\log b\\
? &= \frac{\log a}{\log b}
\end{align*}$$
For your example,
$$\begin{align*}
\log_{49}\sqrt[3]{7^5} &= \frac{\log{\sqrt[3]{7^5}}}{\log 49}\\
&= \frac{\frac53\log 7}{2\log 7}\\
&= \frac56
\end{align*}$$

Note that, on line $(1)$, there is nothing special for taking base-$10$ logarithm. The base can be any base $c > 0$ and $c \ne 1$. The natural logarithm $\ln$ with base-$e$ is equally valid:
$$\begin{align*}
? &= \frac{\ln a}{\ln b}\\
&= \frac{\log_c a}{\log_c b}
\end{align*}$$


TITLE: Does the Skorokhod space with the uniform topology admit a smooth partition of unity?
QUESTION [1 upvotes]: Does the Skorokhod Banach space $D[0,1]$ (cadlag functions equipped with the uniform norm) admit a smooth partition of unity? I found Johanis - Smooth partitions of unity on Banach spaces, which provides several classes of Banach spaces with this property, but don't see how to relate the given criteria to the Skorokhod space.

REPLY [2 votes]: The Skorohod space $D$ has $C([0,1])$ as a Banach subspace that does not have smooth bump functions by results of Bonic and Frampton from 1965. If $D$ had smooth partitions of unity, it would posses smooth bump functions, hence also $C([0,1])$ which is false. So the answer is no.
See Section 14 (pp. 152−158, in particular 14.11(1)) in Kriegl and Michor's The Convenient Setting of Global Analysis (AMS 1997) for an account on bump functions.


TITLE: How does the light source fire a single photon in the double-slit experiment
QUESTION [43 upvotes]: All the youtube videos I have seen on the double slit experiment broadly fall into one of the following three categories:

Documentaries and fan made videos heavy on animation which 'admire' the wave-particle duality of light rather than 'explaining' it.
University lectures where the professors draw diagrams on the board and go into the mathematics (e.g. how to calculate the distance between two bright bands) but never explain how the actual apparatus is built.
Double slit DIY videos on how to do the experiment with sunlight or laser but they seem to leave out the most critical part, how do you make the light source so faint that only one photon is emitted at a time?

As far as I understand, the classical wave theory of light fails to explain the interference only when you do the experiment with one photon at a time. My question here is, how do you make the light source fire a single particle of light at a time when you are trying to determine whether or not light is made of particles in the first place? I am more interested in the engineering of the apparatus rather than the mathematical explanation.
A follow up question might be, how can you be sure that indeed only one photon is coming out of the source at a time? (a single photon is hitting the screen at a time is not the same as a single photon is coming out of the source at a time, especially given the weird wave-particle duality).

REPLY [7 votes]: You can use a single photon source, such that there is (provably) only one photon in the system at any one time.  These are quantum mechanically single-photon emitters, as distinct from just turning down the intensity of a source emitting random photons.
One convenient example is the nitrogen-vacancy centre in diamond.  A single NV centre only has one excitation and so can only emit a single photon each time it's excited.
Because the experiment is so inefficient (very few photons actually make it through the slits) finding a nice paper that directly shows it is tricky.  However there are plenty of similar experiments using these sources (example) for duality experiments.


text:
\begin{document}

\begin{abstract}
A classical theorem of Kempner states that the sum of the reciprocals of positive integers
with missing decimal digits converges.
This result is extended to much larger families of ``missing digits'' sets of positive integers
with both convergent and divergent harmonic series.
\end{abstract}

\maketitle

\section{Kempner's theorem}

\begin{quotation}
``It is well known that the series
\[
\sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{ 1 } +  \frac{1}{ 2 } +  \frac{1}{ 3 } + \cdots
\]
diverges.  The object of this Note is to prove that if the denominators do not include all natural
numbers 1, 2, 3, \ldots, but only those which do not contain any figure 9, the series converges.
The method of proof holds unchanged if, instead of 9, any other figure 1, 2, \ldots, 8 is excluded,
but not for the figure 0.''
\begin{flushright}
A. J. Kempner, \emph{Amer. Math. Monthly} 21 (1914), 48--50
\end{flushright}
\end{quotation}

A \emph{harmonic series} is a series of the form $\sum_{a\in A} 1/a$, where $A$ is a set of positive integers.
Mathematicians have long been interested in the
convergence or divergence of harmonic series.
Let $c \in \{1,2,\ldots, 9\}$, and let $A_{10}(c)$ be the set of positive integers
in which the digit $c$ does not occur in the usual decimal representation.
Kempner~\cite{kemp14}  proved in 1914 that $\sum_{a\in A_{10}(c)} 1/a$ converges.
He called this  ``a curious convergent series.''
More generally, for every integer $g \geq 2$,  every positive integer $n$
has a unique  \textit{$g$-adic representation} of the form $n = \sum_{i=0}^k c_i g^i$,
with digits $c_i \in \{0,1,2,\ldots, g-1\}$ for $i = 0,1,\ldots, k$ and $c_k \neq 0$.
If $A_g(c)$ is the set of integers whose $g$-adic representation contains no digit $c$, then
the infinite series $\sum_{a\in A_g(c)} 1/a$ converges.
This includes the case $c=0$, which was not discussed by Kempner.

Kempner's theorem has been studied and extended by
Baillie~\cite{bail79}, Farhi~\cite{fahr08}, Gordon~\cite{gord19}, Irwin~\cite{irwi16},
Lubeck-Ponomarenko~\cite{lube-pono18},
Schmelzer and Baillie~\cite{schm-bail08}, and Wadhwa~\cite{wadh75,wadh78}.
It is Theorem 144 in Hardy and Wright~\cite{hard-wrig08}.

The $g$-adic representation is a special case of a more general method to represent the positive integers.
A \emph{\mcg-adic sequence} is a strictly increasing sequence of positive integers $\mcg = (g_i)_{i=0}^{\infty}$
such that  $g_ 0 = 1$ and $g_{i}$ divides $g_{i+1}$ for all $ i \geq 0$.
The integer quotients
\[
d_{i} = \frac{g_{i+1}}{g_{i}}
\]
satisfy $d_i \geq 2$ and
\beq               \label{MissingDigits:gk-product}
g_{k+1} = g_{k}d_{k} = d_0 d_1 d_2 \cdots d_{k}
\eeq
for all $k \geq 0$.
Every positive integer $n$ has a unique representation in the form
\beq               \label{MissingDigits:n}
n = \sum_{i=0}^{k} c_i g_i
\eeq
where $c_i \in \{0,1,\ldots, d_i-1\}$ for all $i  \in \{ 0,1,\ldots, k \}$ and $c_{k} \neq 0$.
We call~\eqref{MissingDigits:n} the \textit{$\mcg$-adic representation} of $n$.
This is equivalent to de Bruijn's additive system (Nathanson~\cite{nath2014-150, nath2017-172}).

Harmonic series constructed from sets of positive integers
with missing $\mcg$-adic digits do not necessarily  converge.
In Theorem~\ref{MissingDigits:theorem:converges} we construct sets of integers
with missing $\mcg$-adic digits whose harmonic series converge,
and also sets of integers
with missing $\mcg$-adic digits whose harmonic series diverge.

\section{\mcg-adic representations with bounded quotients}

Define the \textit{interval of integers}
\[
[a,b] = \{n \in \Z: a \leq n \leq b\}.
\]

Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence with quotients
$d_i = g_{i+1}/g_i$.
Let $I$ be a set of nonnegative integers, and, for all $i \in I$,
let $U_i$ be a nonempty proper subset of  $[0,d_i-1]$.
For every nonnegative integer $k$, let $A_{k}$ be the set of integers $n \in [g_k,g_{k+1} - 1]$
 whose  $\mcg$-adic representation
 $n = \sum_{i=0}^k c_i g_i $
satisfies the following \textit{missing digits condition}:
\beq                    \label{Digits:MissingDigitsCondition}
c_i \in [0,d_i -1]  \setminus U_i \qquad \text{for all $i \in I \cap [0,k]$.}
\eeq

\begin{lemma}                      \label{MissingDigits:lemma:Ak}
The set $A_k$ satisfies:
\benum
\item[(a)]
$A_k = \emptyset$ if and only if $k \in I$ and $U_k = [1,d_k-1]$.
\item[(b)]
If $A_k \neq \emptyset$, then

\beq               \label{MissingDigits:|Ak|}
|A_k| \leq \prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)
 \prod_{\substack{i=0 \\ i \notin I}}^{k} d_{i} \leq 2|A_k|.
\eeq
\eenum
\end{lemma}

\begin{proof}
We use the inequality
\beq                                             \label{MissingDigits:inequality-2}
x \leq 2(x-1) \qquad \text{for $x \geq 2$.}
\eeq

If $n \in [g_k,g_{k+1} - 1]$, then $n$ has the \mcg-adic representation
\[
n = \sum_{i=0}^{k-1} c_i g_i + c_kg_k
\]
with $c_k \neq 0$ and so $c_k \in [1,d_k-1]$.
It follows that $A_k = \emptyset$
if and only if $k \in I$ and $U_k = [1,d_k-1]$.

For $A_k \neq \emptyset$, there are three cases.
\begin{enumerate}
\item[(i)]
If  $k \in I$ and $0 \in U_k$, then
\[
|A_k| = \prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)  \prod_{\substack{i=0 \\ i \notin I}}^{k} d_{i}
< 2|A_k|.
\]

\item[(ii)]
If $k \in I$ and and $0 \notin U_k$, then $U_k$ is a proper subset of $[1, d_k-1]$
and so $|U_k| \leq d_k-2$.
Inequality~\eqref{MissingDigits:inequality-2} gives
\[
d_k - |U_k| \leq 2\left( d_{k} - |U_k| -1 \right).
\]
We obtain
\begin{align*}
|A_k| & = (d_{k} - |U_k| -1) \prod_{\substack{i=0 \\ i \in I}}^{k-1} (d_{i}-|U_i|)
 \prod_{\substack{i=0 \\ i \notin I}}^{k} d_{i} \\
&  < \prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)  \prod_{\substack{i=0 \\ i \notin I}}^{k} d_{i} \\
& \leq 2 (d_{k} - |U_k| -1) \prod_{\substack{i=0 \\ i \in I}}^{k-1} (d_{i}-|U_i|)
\prod_{\substack{i=0 \\ i \notin I}}^{k} d_{i}  \\
& = 2|A_k|.
\end{align*}

\item[(iii)]
We have $d_k \geq 2$ and so $d_k \leq 2(d_k-1)$ from inequality~\eqref{MissingDigits:inequality-2}.
If $k \notin I$, then
\begin{align*}
|A_k|
& = (d_{k} - 1) \prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)  \prod_{\substack{i=0 \\ i \notin I}}^{k-1} d_{i}
 < \prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)  \prod_{\substack{i=0 \\ i \notin I}}^{k} d_{i} \\
& =  d_{k}  \prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)  \prod_{\substack{i=0 \\ i \notin I}}^{k-1} d_{i}
\leq 2 (d_{k} - 1) \prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)  \prod_{\substack{i=0 \\ i \notin I}}^{k-1} d_{i} \\
& = 2|A_k|.
\end{align*}
\end{enumerate}
This completes the proof.
\end{proof}

Let $A$ be a set of nonnegative integers, and let
$A(n)$ be the number of elements $a \in A$ with $a \leq n$.
The   \textit{upper asymptotic density} of the set $A$ is $d_U(A) = \limsup_{n \rightarrow \infty} A(n)/n$.
If  $\lim_{n \rightarrow \infty} A(n)/n $ exists, then  $d(A) = \lim_{n \rightarrow \infty} A(n)/n$
is called the \textit{asymptotic density} of the set $A$.
The set $A$ has \emph{asymptotic density zero} if $d(A) = d_U(A) = 0$.

\begin{lemma}                                             \label{MissingDigits:lemma:series}
Let $A$ be a set of positive integers.  If $\sum_{a\in A} 1/a < \infty$, then $d(A) = 0$.
\end{lemma}

\begin{proof}
We show that $d_U(A) > 0$ implies $\sum_{a\in A} 1/a = \infty$.

If $d_U(A) = \limsup_{n \rightarrow \infty} A(n)/n = \alpha > 0$,
then, for every $\varepsilon > 0$, we have
\[
\frac{A(n)}{n} < \alpha + \varepsilon \qquad \text{for all integers $n \geq N(\varepsilon)$}
\]
and
\beq                                          \label{MissingDigits:beta}
\frac{A(n_i)}{n_i} > \alpha - \varepsilon \qquad  \text{for infinitely many integers $n_i $.}
\eeq
Let $\varepsilon < \alpha/3$.  There is a sequence of positive integers
$(n_i)_{i=0}^{\infty}$ satisfying inequality~\eqref{MissingDigits:beta}
such that $n_0 \geq N(\varepsilon)$ and
$n_{i} > 2n_{i-1}$ for all $i \geq 1$.  We have
\begin{align*}
A(n_i) - A(n_{i-1}) & > (\alpha -\varepsilon) n_i - (\alpha +\varepsilon) n_{i-1} \\
& > (\alpha -\varepsilon) n_i - \frac{(\alpha +\varepsilon) n_{i}}{2} \\
& = n_i\left( \frac{\alpha - 3\varepsilon}{2}\right)
\end{align*}
and so
\begin{align*}
\sum_{\substack{a\in A \\ n_{i-1} < a \leq n_{i} }} \frac{1}{a}
& \geq \frac{A(n_i) - A(n_{i-1})}{n_i} >  \frac{\alpha  - 3\varepsilon}{2} > 0.
\end{align*}
It follows that
\[
\sum_{\substack{a\in A \\ 1 \leq a \leq n_k}} \frac{1}{a}
\geq \sum_{i=1}^k \sum_{\substack{a\in A \\ n_{i-1} < a \leq n_{i} }} \frac{1}{a}
> k\left(  \frac{\alpha  - 3\varepsilon}{2} \right)
\]
and the infinite series $\sum_{a\in A} 1/a$ diverges.
Equivalently, convergence of the infinite series $\sum_{a\in A} 1/a$ implies  $d(A) = 0$.
This completes the proof.
\end{proof}

The converse of Lemma~\ref{MissingDigits:lemma:series} is false.
The set of prime numbers has asymptotic density zero, but the sum of the reciprocals of the primes diverges.

\begin{theorem}                      \label{MissingDigits:theorem:converges}
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence with bounded quotients, that is,
\beq                \label{Digits:limsup}
d_i = \frac{g_{i+1}}{g_i}  \leq d
\eeq
for some integer $d \geq 2$  and all $i =0,1,2,\ldots$.
Let $I$ be a set of nonnegative integers, and, for all $i \in I$,
let $U_i$ be a nonempty proper subset of  $[0,d_i-1]$.

Let $n = \sum_{i=0}^k c_i g_i$ be the $\mcg$-adic representation of the positive integer $n$.
Let $A$ be the set of positive integers $n$  that satisfy the missing digits
condition~\eqref{Digits:MissingDigitsCondition}.
If
\beq                \label{Digits:I(x)-1}
I(k) \geq \frac{(1+\delta)\log k}{\log (d/(d-1))}
\eeq
for some $\delta > 0$ and all $k \geq k_0 = k_0(\delta) $,
then the set $A$ has asymptotic density zero and the harmonic series $\sum_{a\in A} 1/a$ converges.

If
\beq                \label{Digits:I(x)-2}
I(k) \leq \frac{(1-\delta) \log k}{\log d}
\eeq
for some $\delta> 0$ and all $k \geq k_1 = k_1(\delta) $,
then the harmonic series $\sum_{a\in A} 1/a$ diverges.
\end{theorem}

Kempner's theorem is the special case  $g_i = 10^i$, $d_i = 10$,
and $U_i = \{9\}$ for all $i \in I = \mathbf{N}_0$.

\begin{proof}
For all $k \in \mathbf{N}_0$,  the finite sets
\[
A_{k} = A \cap [g_k,g_{k+1} - 1]
\]
are pairwise disjoint and
$
A = \bigcup_{k=0}^{\infty} A_k.
$

For all $i \in I$, we have
\[
1 \leq |U_i| \leq d_i -1
\]
and
\[
 \frac{1}{d}  \leq \frac{1}{d_i}  \leq \frac{d_i-|U_i|}{d_i} = 1 -  \frac{|U_i|}{d_i} \leq 1 - \frac{1}{d} < 1.
\]

Let $I(k)$ satisfy inequality~\eqref{Digits:I(x)-1}.  We obtain
\beq                                                \label{MissingDigits:UpperBound}
\left(1 - \frac{1}{d} \right)^{I(k)} \leq
\left( \frac{d-1}{d} \right)^{ \frac{(1+\delta)\log k}{\log (d/(d-1))}} = \frac{1}{k^{1+\delta}}.
\eeq
If $a \in A_k$, then $a \geq g_k = d_0d_1\cdots d_{k-1}$.  By Lemma~\ref{MissingDigits:lemma:Ak},
\begin{align*}
\sum_{ \substack{a\in A \\ a \geq g_{k_0}}} \frac{1}{a}
& =  \sum_{k= k_0 }^{\infty} \sum_{a\in A_k} \frac{1}{a}
\leq \sum_{k= k_0}^{\infty} \frac{|A_k|}{g_{k}} \\
& \leq  \sum_{k= k_0}^{\infty} \frac{d_k}{\prod_{i=0}^{k} d_{i} }
\prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)   \prod_{\substack{i=0 \\ i \notin I}}^{k} d_{i} \\
& \leq d  \sum_{k= k_0}^{\infty} \prod_{\substack{i=0 \\ i \in I}}^{k} \frac{d_i -|U_i|}{d_i} \\
& \leq d  \sum_{k= k_0}^{\infty} \left(1 - \frac{1}{d} \right)^{I(k)} \\
& \leq d  \sum_{k= k_0}^{\infty} \frac{1}{k^{1+\delta}} < \infty.
\end{align*}
Thus, the harmonic series converges.  By Lemma~\ref{MissingDigits:lemma:series},
the set $A$ has  asymptotic density zero.

Let $I(k)$ satisfy inequality~\eqref{Digits:I(x)-2}.  We obtain
\beq                                                \label{MissingDigits:LowerBound}
\left(\frac{1}{d} \right)^{I(k)} \geq
\left( \frac{1}{d} \right)^{ \frac{(1-\delta)\log k}{\log d}} = \frac{1}{k^{1-\delta}}.
\eeq
If $a \in A_k$, then $a < g_{k+1} = d_0d_1\cdots d_{k-1}d_k$.
By Lemma~\ref{MissingDigits:lemma:Ak},
\begin{align*}
\sum_{\substack{a\in A \\ a \geq g_{k_1}} }\frac{1}{a}
& =  \sum_{k=k_1}^{\infty} \sum_{a\in A_k} \frac{1}{a}
\geq \sum_{k=k_1}^{\infty} \frac{|A_k|}{g_{k+1}} \\
& \geq \frac{1}{2}  \sum_{k=k_1}^{\infty} \frac{1}{\prod_{i=0}^{k} d_{i} }
\prod_{\substack{i=0 \\ i \in I}}^{k} (d_{i}-|U_i|)   \prod_{\substack{i=0 \\ i \notin I}}^{k} d_{i} \\
&  = \frac{1}{2}  \sum_{k=k_1}^{\infty} \prod_{\substack{i=0 \\ i \in I}}^{k} \frac{d_i -|U_i|}{d_i}
\geq  \frac{1}{2}  \sum_{k=k_1}^{\infty} \prod_{\substack{i=0 \\ i \in I}}^{k} \frac{1}{d_i} \\
& \geq  \frac{1}{2}   \sum_{k=k_1}^{\infty} \left(\frac{1}{d} \right)^{I(k)} \\
& \geq  \frac{1}{2}  \sum_{k=k_1}^{\infty} \frac{1}{k^{1-\delta}}
\end{align*}
and the harmonic series diverges.
This completes the proof.
\end{proof}

\begin{corollary}
Let $I$ be a set of nonnegative integers, and let $(v_i)_{i \in I}$ be a sequence of 0s and 1s.
Let $A$ be the set of integers $n$ such that,
if $n \in \left[2^k, 2^{k+1}-1 \right]$ has the $2$-adic representation $n = \sum_{i=0}^k c_i 2^i$,
then $c_i = v_i$ for all $i \in I \cap [0,k]$.
If
\[
I(k) \geq (1+\delta)\log_2 k
\]
for some $\delta > 0$ and all $k \geq k_0(\delta) $,
then the harmonic series $\sum_{a\in A} 1/a$ converges.
If
\[
I(k) \leq (1-\delta) \log_2 k
\]
for some $\delta> 0$ and all $k \geq k_1(\delta) $,
then the harmonic series $\sum_{a\in A} 1/a$ diverges.
\end{corollary}

\begin{proof}
For all $i \in I$, let $u_i = 1-v_i$ and $U_i = \{u_i\}$.
Apply Theorem~\ref{MissingDigits:theorem:converges}.
\end{proof}

It is an open problem to determine the convergence or divergence of $\sum_{a\in A} 1/a$
if $I(k) \sim \log_2 k$.

\section{\mcg-adic representations with unbounded quotients}
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence
with quotients $d_i = g_{i+1}/g_i$.
Let $I$ be an infinite  set of nonnegative integers, and,
for all $i \in I$, let $U_i$ be a nonempty proper subset of  $[0,d_i-1]$.
If the sequence  $\mcg = (g_i)_{i=0}^{\infty}$ has bounded quotients $d_i \leq d$,
then
\[
\frac{|U_i|}{d_i} \geq \frac{1}{d}
\]
for all $i \in I$ and the infinite series $\sum_{i\in I} \frac{|U_i|}{d_i}$ diverges.
Equivalently, the convergence of this series implies that \mcg\
has unbounded quotients.

Let $n = \sum_{i=0}^k c_i g_i$ be the $\mcg$-adic representation of the positive integer $n$.
Let  $A$ be the set of positive integers whose  $\mcg$-adic representations
satisfy the missing digits condition~\eqref{Digits:MissingDigitsCondition}.
The missing digits set $A$ is finite if and only if $I$ is a cofinite set of nonnegative
integers and $U_i = [1,d_i-1]$ for all sufficiently large $i$.
The harmonic series of a finite set of positive integers converges.

Theorem~\ref{MissingDigits:theorem:converges} shows that harmonic series constructed
from infinite sets of integers with missing $\mcg$-adic digits do not always converge.
It follows from Theorem~\ref{MissingDigits:theorem:converges} that if
\[
I(k) \geq (\log k)^{1+\delta}
\]
for some $\delta > 0$ and all $k \geq k_0(\delta) $,
and if $\sum_{a \in A} 1/a$ diverges,
then the sequence \mcg\ must have \emph{unbounded quotients}, that is,
\[
\limsup d_i = \infty.
\]

Theorem~\ref{MissingDigits:theorem:diverges}  gives a sufficient condition
for the divergence of harmonic series of sets of positive integers
constructed from \mcg-adic sequences with unbounded  quotients.
We use the following inequality,
which is easily proved by induction:
If $0 \leq x_i < 1$ for $i=1,\ldots, n$, then
\beq               \label{MissingDigits:InfiniteProduct}
\prod_{i=1}^n (1-x_i) \geq 1-\sum_{i=1}^n x_i.
\eeq

\begin{theorem}              \label{MissingDigits:theorem:diverges}
Let $\mcg = (g_i)_{i=0}^{\infty}$ be a \mcg-adic sequence,
and let $n = \sum_{i=0}^k c_i g_i$ be the $\mcg$-adic representation of the positive integer $n$.
Let $I$ be an infinite  set of nonnegative integers, and,
for all $i \in I$, let $U_i$ be a nonempty proper subset of  $[0,d_i-1]$.
Let  $A$ be the set of positive integers whose  $\mcg$-adic representations
satisfy the missing digits condition~\eqref{Digits:MissingDigitsCondition}.
If the set $A$ is infinite and if
\beq                                      \label{MissingDigits:ConvergentSeries}
\sum_{i\in I} \frac{|U_i|}{d_i} < \infty
\eeq
then  the sequence $\mcg = (g_i)_{i=0}^{\infty}$
has unbounded  quotients and the harmonic series $\sum_{a\in A} 1/a$ diverges.
\end{theorem}

For example, the ``missing digits'' set constructed from
$\mcg = (g_i)_{i=0}^{\infty}$ with  $g_i = 2^{i(i+1)/2}$ and $d_i = 2^{i+1}$
and with $I = \mathbf{N}_0$
and $U_i = \{0\}$ for all $i \in I$ has a divergent harmonic series.

\begin{proof}
Because the infinite series~\eqref{MissingDigits:ConvergentSeries} converges,
there is an integer $i_0 \in I$ such that
\[
\sum_{\substack{i=i_0\\i \in I}}^{\infty} \ \frac{|U_i|}{d_i} < \frac{1}{2}.
\]
Inequality~\eqref{MissingDigits:InfiniteProduct} implies that, for all $k \in \mathbf{N}_0$,
\[
\prod_{\substack{i=i_0\\i \in I}}^{k} \ \left( 1 - \frac{|U_i|}{d_i} \right)
\geq 1 - \sum_{\substack{i=i_0\\i \in I}}^{k}  \frac{|U_i|}{d_i} > \frac{1}{2}
\]
and so
\begin{align*}
\prod_{\substack{i=0\\i \in I}}^{k} \left( 1 - \frac{|U_i|}{d_i} \right)
& = \prod_{\substack{i=0\\i \in I}}^{i_0 -1} \left( 1 - \frac{|U_i|}{d_i} \right)
 \prod_{\substack{i=i_0 \\i \in I}}^{k} \left( 1 - \frac{|U_i|}{d_i} \right) \\
& > \frac{1}{2} \prod_{\substack{i=0\\i \in I}}^{i_0 -1} \left( 1 - \frac{|U_i|}{d_i} \right) = \delta > 0.
\end{align*}

Let $A_k = A \cap [g_k, g_{k+1}-1]$. The set $A$ is infinite if and only if
$A_k \neq \emptyset$ for infinitely many $k$.
Applying inequality~\eqref{MissingDigits:|Ak|}   of Lemma~\ref{MissingDigits:lemma:Ak},
we obtain
\begin{align*}
\sum_{a\in A} \frac{1}{a}
& =  \sum_{ \substack{k=0 \\ A_k \neq \emptyset} }^{\infty} \sum_{a\in A_k} \frac{1}{a}
\geq \sum_{  \substack{k=0 \\ A_k \neq \emptyset}  }^{\infty} \frac{|A_k|}{g_{k+1}} \\
& \geq  \frac{1}{2} \sum_{ \substack{k=0 \\ A_k \neq \emptyset} }^{\infty} \frac{1}{\prod_{i=0}^{k} d_{i} } \prod_{\substack{i=0\\i\in I}}^{k} (d_i - |U_i|)
\prod_{\substack{i=0\\ i \notin I}}^{k} d_i \\
& =  \frac{1}{2} \sum_{ \substack{k=0 \\ A_k \neq \emptyset} }^{\infty}
\prod_{ \substack{ i=0 \\ i \in I }}^{k}  \left( 1 - \frac{|U_i|}{d_i} \right)
\end{align*}
and so the harmonic series $\sum_{a\in A} \frac{1}{a}$ diverges.
This completes the proof.
\end{proof}

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\providecommand{\MR}{\relax\ifhmode\unskip\space\fi MR }

\providecommand{\MRhref}[2]{
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config: arxiv


text:
\begin{document}
\title{Uplink Cooperative Interference Cancellation for Cellular-Connected UAV: A Quantize-and-Forward Approach}
\author{Weidong Mei and Rui Zhang, \IEEEmembership{Fellow, IEEE}\vspace{-6pt}
\thanks{W. Mei is with the NUS Graduate School for Integrative Sciences and Engineering, National University of Singapore, Singapore 119077, and also with the Department of Electrical and Computer Engineering, National University of Singapore, Singapore 117583 (e-mail: wmei@u.nus.edu).}
\thanks{R. Zhang is with the Department of Electrical and Computer Engineering, National University of Singapore, Singapore 117583 (e-mail: elezhang@nus.edu.sg).}}
\maketitle

\begin{abstract}
Aerial-ground interference is the main obstacle to achieving high spectral efficiency in cellular networks with both traditional terrestrial users and new unmanned aerial vehicle (UAV) users. Due to their strong line-of-sight (LoS) channels with the ground, UAVs could cause/suffer severe interference to/from a large number of non-associated but co-channel base stations (BSs) in their uplink/downlink communications. In this letter, we propose a new cooperative interference cancellation (CIC) scheme for the UAV's uplink communication to mitigate its strong interference to co-channel BSs, which requires only local cooperation between each co-channel BS and its adjacent helping BSs. Specifically, the helping BSs without serving any users in the UAV's communication channel quantize and forward their received signals from the UAV to their aided co-channel BS, which then processes the quantized signals jointly with its own received signal to decode its served terrestrial user's message via canceling the UAV's signal by linear/nonlinear interference cancellation techniques (ICTs). We derive the achievable rates of the proposed CIC scheme with different ICTs as a function of the UAV's transmit power and rate, and thereby unveil the conditions under which the proposed CIC scheme outperforms the existing CIC scheme based on the decode-and-forward (DF) operation of the helping BSs.
\end{abstract}
\begin{IEEEkeywords}
Cellular-connected UAV, interference cancellation, quantize-and-forward, decode-and-forward.\vspace{-6pt}
\end{IEEEkeywords}

\section{Introduction}
Unmanned aerial vehicle (UAV) (or drone) is an aircraft without requiring any human pilot on board. While UAVs originated from military applications, their use in civilian/commercial applications has been rapidly growing in the last decade, such as for packet delivery, aerial photography, airborne communication, etc\cite{zeng2019accessing}. However, most of today's commercial UAVs are controlled via hand-held radio, which limits their flights to the visual or radio line-of-sight (LoS) range only. To enable beyond visible and radio LoS (BVRLoS) UAV communications, both academia and industry have looked into a new paradigm called \emph{cellular-connected UAV}, by reusing the existing cellular base stations (BSs) and spectrum to serve UAVs and traditional terrestrial user equipments (UEs) at the same time\cite{zeng2019cellular,lin2019mobile,garcia2019essential}. Thanks to the advanced cellular technologies, preliminary field trials\cite{3GPP36777} have demonstrated that the current fourth-generation (4G) long-term evolution (LTE) network is able to meet some basic requirements of UAV-ground communications. However, a more severe aerial-ground interference issue than its counterpart in the existing terrestrial networks also arises\cite{3GPP36777}. Specifically, due to the LoS-dominant channels between UAVs and ground BSs, UAVs may impose/suffer overwhelming interference to/from a large number of BSs in their uplink/downlink communications. As such, effective interference mitigation techniques are required to resolve this new and challenging issue.

\begin{figure}[!t]
\centering
\includegraphics[width=3.0in]{UAV_Upcic.pdf}
\DeclareGraphicsExtensions.\vspace{-3pt}
\caption{Uplink UAV communication in a cellular network with cooperative air-to-ground interference cancellation.}\label{UAV_Upcic}
\vspace{-12pt}
\end{figure}
Motivated by the above, this letter aims to investigate the uplink interference mitigation techniques for a cellular network with co-existing UAV and terrestrial UEs, for cancelling the UAV’s interference to a large number of non-associated but co-channel BSs within its wide signal coverage region, denoted by $D_u$ and shown in Fig.\,\ref{UAV_Upcic}. It is worth noting that there have been recent works\cite{cellular2019mei,mei2020uav,liu2019multi,mei2019uplink} on addressing similar problems. In \cite{cellular2019mei} and \cite{mei2020uav}, two new aerial-ground inter-cell interference coordination (ICIC) schemes were proposed for avoiding interference among the UAV and terrestrial UEs by treating the interference as noise. To further improve the rate performance, a decode-and-forward (DF) based cooperative interference cancellation (CIC) scheme was proposed in \cite{liu2019multi} and \cite{mei2019uplink}, by exploiting the existing backhaul links among neighboring BSs. As shown in Fig.\,\ref{UAV_Upcic}, for any co-channel BS (say, BS1) that uses the same communication channel as the UAV for serving its terrestrial UE (say, UE1) and thus suffers the UAV's uplink interference, its adjacent BSs in a local region $D_c$ (termed as {\it helping} BSs) without using this channel (due to fixed/dynamic frequency reuse in existing cellular networks) can decode the UAV's signal and then forward the decoded signal to it for interference cancellation, thus helping improve UE1's achievable rate.

Despite that the above DF-based CIC is effective in general, its performance critically depends on the UAV's transmit power and rate. If the UAV's transmit power is limited as a terminal while its transmit rate is high, which is typical for the high-rate uplink payload communication such as video streaming, none of BS1 and its helping BSs in $D_c$ might be able to decode the UAV's signal, especially when they are far from the UAV (e.g., at the boundary of $D_u$). As a result, UE1 would still suffer severe interference from the UAV at BS1. To deal with the above issue, in this letter we propose a new quantize-and-forward (QF) based CIC scheme. Specifically, the helping BSs in $D_c$ only quantize and forward their received UAV signals to BS1 without decoding them. Then, BS1 can properly combine these quantized signals with its own received signal to either maximize the signal-to-interference-plus-noise ratio (SINR) for decoding the UAV's signal first (by treating UE1's signal as interference) and then cancelling it before decoding UE1's signal (referred to as {\it successive/nonlinear} interference cancellation), or maximize UE1's signal SINR for decoding it directly while linearly suppressing the UAV's interference (refereed to as {\it linear} interference cancellation). With given UAV's transmit power and rate, we derive the optimal combining weights for different interference cancellation techniques (ICTs) and the corresponding UE1's SINRs/achievable rates in closed form. Finally, both analytical and numerical results are provided to unveil the conditions under which the proposed QF-based CIC with different ICTs outperforms the existing DF-based CIC.

{\it Notations:} Bold symbols in capital letter and small letter denote matrices and vectors, respectively. The conjugate transpose and inverse of a matrix are denoted as ${(\cdot)}^{H}$ and ${(\cdot)}^{-1}$, respectively. $\left[ \cdot \right]_i$ denotes the $i$-th diagonal element of a square matrix. ${\text{diag}}(\mv a)$ denotes a diagonal matrix with its main diagonal given by the vector $\mv a$. ${\mathbb{C}}^n$ denotes the set of complex vectors of length $n$. For a complex number $s$, $\lvert s \rvert$ denotes its norm. ${\mathbb E}[\cdot]$ denotes the expected value of random variables. $x \sim \mathcal{CN}(\mu,\sigma^2)$ means that $x$ is a circularly symmetric complex Gaussian (CSCG) random variable with mean $\mu$ and variance $\sigma^2$. ${\bf 0}_n$ denotes a zero vector of length $n$.

\section{System Model}
As shown in Fig.\,\ref{UAV_Upcic}, we consider the uplink UAV communication in a subregion $D_u$ of the cellular network, where a UAV UE is associated with a ground BS (named BS0) over a given resource block (RB)\footnote{The proposed scheme can be extended to the case of multiple UAVs if they are assigned with orthogonal RBs to avoid mutual interference.}. Assume that the UAV's transmit power and rate are given as $p_u$ and $r_u$ in its assigned RB, respectively, with $p_u > 0$ and $r_u > 0$, under which its associated BS (BS0) can decode its message reliably. Due to the dense frequency reuse in the cellular network, the UAV may cause strong interference to a large number of non-associated but co-channel BSs over its LoS-dominant channels to them. Without loss of generality, we consider one of such co-channel BSs and its served terrestrial UE (i.e., BS1 and UE1 shown in Fig.\,\ref{UAV_Upcic}), while the results are applicable to other co-channel BSs/UEs similarly.

Let $h_1$ be the complex-valued baseband equivalent channel from UE1 to BS1, and $f_1$ be that from the UAV to BS1. In general, these channels depend on the BS1/UAV/UE1 antenna gains, path-loss, shadowing and small-scale fading. Note that in the current 4G LTE network, the ground BS antennas are usually tilted downwards for covering the terrestrial UEs and mitigating their inter-cell interference (ICI)\cite{3GPP36777}. As such, we assume that each BS employs an antenna array with fixed directional gain pattern in 3D, while the UAV and terrestrial UEs are assumed to be equipped with a single omnidirectional antenna for simplicity. Let $p_1$ denote the transmit power of UE1. Then, the received signal at its serving BS (BS1) is expressed as
\begin{equation}\label{sum1}
y_1=\sqrt{p_1}h_1x_1+\sqrt{p_u}f_1x_u+z_1,
\end{equation}
where $x_1$ and $x_u$ denote the complex-valued data symbols for UE1 and the UAV with ${\mathbb E}[{\lvert{x_1}\rvert}^2] = 1$ and ${\mathbb E}[{\lvert{x_u}\rvert}^2] = 1$, respectively, and $z_1 \sim \mathcal{CN}(0,\sigma_1^2)$ comprises the background noise and any terrestrial ICI at BS1\footnote{In this letter, we assume that the terrestrial ICI has been well mitigated below the background noise by the existing ICIC techniques in cellular networks, such as cooperative dynamic RB allocation and power control.}, both of which are assumed to be independently Gaussian distributed with $\sigma_1^2$ denoting their total power.

Based on (\ref{sum1}), UE1's SINR at BS1 is given by
\begin{equation}\label{sinr1}
{\gamma}_1 = \frac{p_1\lvert h_1 \rvert^2}{p_u\lvert f_1 \rvert^2+\sigma_1^2}.
\end{equation}
Note that due to the LoS-dominant UAV-BS channel $f_1$, UE1's received signal power, $p_1\lvert h_1 \rvert^2$, may be comparable to or even lower than the UAV's interference power, $p_u\lvert f_1 \rvert^2$, in (\ref{sinr1}). As a result, its SINR/achievable rate may be very low in practice.

\section{Existing DF-Based CIC}
One straightforward method to improve ${\gamma}_1$ is by performing successive/nonlinear interference cancellation (SIC) at BS1 to first decode the UAV's signal, i.e., $\sqrt{p_u}f_1x_u$, and then cancel it in (\ref{sinr1}) for decoding UE1's signal. To this end, the UAV's transmit rate $r_u$ should be no greater than its achievable rate at BS1 by treating UE1's signal as interference/noise, which is given by
\begin{equation}\label{rate1}
R_1 = {\log_2}\left(1 + \frac{p_u\lvert f_1 \rvert^2}{p_1\lvert h_1 \rvert^2+\sigma_1^2}\right),
\end{equation}
in bits per second per Hertz (bps/Hz). However, due to the comparable power of $p_1\lvert h_1 \rvert^2$ and $p_u\lvert f_1 \rvert^2$, $R_1$ can be lower than $r_u$, making this direct SIC at BS1 infeasible. To overcome this issue, a DF-based CIC scheme was proposed in \cite{liu2019multi} and \cite{mei2019uplink}, where the adjacent helping BSs of BS1 in $D_c$ without serving any UEs in the UAV's assigned RB are exploited to help decoding the UAV's signal. By this means, if one or more helping BSs in $D_c$ are able to decode the UAV's signal and forward it to BS1, BS1 can completely cancel the interference from the UAV before decoding UE1's signal. Note that due to the fixed/dynamic RB allocation in terrestrial ICIC, the adjacent BSs of BS1 usually do not reuse any RB already being used by BS1 to avoid ICI. Hence, they are more likely to be able to decode the UAV's signal as compared to BS1 due to weaker interference from UE1 in general.

In this letter, we assume that BS1 can cooperate with all the BSs in its first $M$-tier neighborhood (i.e., $D_c$), which are given by the set ${\cal J}=\{2,3,4,\cdots,J+1\}$ with $J$ denoting the total number of helping BSs in $\cal J$. Let $f_i$ be the channel from the UAV to BS$i$ with $i \in \cal J$. Then, the received signal at each helping BS$i$ is given by
\begin{equation}\label{sum2}
y_i=\sqrt{p_u}f_ix_u+z_i,
\end{equation}
where $z_i \sim \mathcal{CN}(0,\sigma_i^2)$ comprises the background noise and any terrestrial ICI at BS$i$ (including that from UE1). Thus, the UAV's achievable rate at BS$i$ is given by
\begin{equation}\label{rate2}
R_i = {\log_2}\left(1 + \frac{p_u\lvert f_i \rvert^2}{\sigma_i^2}\right), i \in {\cal J}.
\end{equation}

Define $R_{u,\text{DF}} \triangleq \mathop {\max}\nolimits_{i \in {\cal J} \cup \{1\}} R_i$ as the maximum of the UAV's achievable rates at BS1 and all its helping BSs in $\cal J$. Obviously, the UAV interference can be completely cancelled at BS1 if $R_{u,\text{DF}} \ge r_u$. In this case, the corresponding SINR for decoding UE1's signal at BS1 is expressed as
\begin{equation}\label{sinr2}
{\gamma}_{1,\text{DF}}=
\begin{cases}
\frac{p_1\lvert h_1 \rvert^2}{\sigma_1^2}, &{\text{if}}\;R_{u,\text{DF}} \ge r_u\\
\gamma_1, &\text{otherwise.}
\end{cases}
\end{equation}

By comparing (\ref{sinr2}) with (\ref{sinr1}), it is noted that UE1's SINR can be improved by the DF-based CIC if $R_{u,\text{DF}} \ge r_u$. However, the condition $R_{u,\text{DF}} \ge r_u$ may not hold in practice, e.g., when $r_u$ is high, $p_u$ is low, and/or the UAV is far from BS1 (and thus its helping BSs). In this case, UE1 would still suffer severe SINR loss due to the UAV's uplink transmission. To overcome this limitation, we propose a new QF-based CIC in the next section.\vspace{-6pt}

\section{Proposed QF-Based CIC}
Note that the fundamental limitation of the DF-based CIC lies in the fact that $R_{u,\text{DF}}$ may be limited as compared to $r_u$ due to the individual UAV signal decoding at BS1 or its helping BSs in $\cal J$. In view of this limitation, we propose the following QF-based CIC scheme, where each helping BS$i, i \in \cal J$ only quantizes and forwards $y_i$ to BS1 for joint processing, without the need of individually decoding the UAV's signal.

Mathematically, the quantized signals by all the helping BSs in $\cal J$ are expressed as
\begin{equation}\label{sum3}
\tilde y_i = y_i + e_i = \sqrt{p_u}f_ix_u+z_i+e_i, i \in \cal J,
\end{equation}
where $e_i$ denotes the quantization noise for the received signal at BS$i$ with zero mean and variance $q_i$. We assume that $q_i$’s are independent over $i$ due to independent quantization among the helping BSs. In practice, the quantization noise powers $q_i$'s depend on specific quantization scheme and desired quantization accuracy. If the quantization level is set to be sufficiently high, we have $q_i \approx 0, \forall i \in \cal J$. However, this requires extremely high transmission rate for each helping BS to reliably forward the quantized signals to BS1, which may overwhelm the backhaul links, especially when the UAV UE density is high and the number of helping BSs is large. For example, assume that a simple uniform scalar quantization method is applied by BS$i$ to independently quantize the in-phase and quadrature (I/Q) components of $y_i, i \in \cal J$. Let $D_i$ denote the number of bits that BS$i$ uses to quantize each I/Q component of $y_i, i \in \cal J$. Then, the quantization noise powers for the above quantization method are given by\cite{liu2015joint}
\begin{equation}\label{qerror}
q_i=
\begin{cases}
3(p_u\lvert f_i \rvert^2+\sigma_i^2)2^{-2D_i}, &{\text{if}}\;D_i > 0\\
\infty, &{\text{if}}\;D_i = 0
\end{cases}, i \in \cal J.
\end{equation}
After the quantization, each helping BS$i, i \in \cal J$ forwards $\tilde y_i$ (along with $f_i$, $\sigma_i^2$ and $q_i$) to BS1 for its joint processing via the backhaul link between them. Here, we assume that the transmission rate of each helping BS$i, i \in \cal J$, i.e., $2D_i$ with the above uniform scalar quantization\cite{liu2015joint}, is chosen not to exceed its backhaul capacity. Next, we consider the following nonlinear and linear ICTs (named Scheme 1 and Scheme 2, respectively) for BS1 to cancel the UAV's interference, based on the quantized signals $\tilde y_i, i \in \cal J$ and its received signal $y_1$.\vspace{-6pt}

\subsection{Scheme 1: Nonlinear Interference Cancellation}\label{scheme1}
First, BS1 can properly combine $\tilde y_i, i \in \cal J$ with $y_1$ such that the SINR for decoding the UAV's signal is maximized by treating UE1's signal as noise, so as to facilitate its SIC before decoding UE1's signal. Let ${\mv w}_u \in {\mathbb C}^{J+1}$ be the combining weights applied by BS1. Then, the combined signal is expressed as
\begin{equation}\label{sum4}
{\mv w}^H_u{\mv y} = {\mv w}^H_u{\mv h}x_1 + {\mv w}^H_u{\mv f}x_u+{\mv w}^H_u{\mv z}+{\mv w}^H_u\mv{e},
\end{equation}
where ${\mv y}=[y_1,\tilde y_2,\cdots,\tilde y_{J+1}]^H$, ${\mv h}=[h_1, {\bf 0}_J]^H$, ${\mv f}=[f_1,f_2,\cdots,f_{J+1}]^H$, ${\mv z}=[z_1,z_2,\cdots,z_{J+1}]^H$, and $\mv{e}=[0,e_2,e_3,\cdots,e_{J+1}]^H$.

Based on (\ref{sum4}), the UAV SINR at BS1 is given by
\begin{equation}\label{sinr3}
\gamma_{u,\text{QF1}} = \frac{p_u\lvert {\mv w}^H_u{\mv f}\rvert^2}{{\mv w}^H_u\left(p_1{\mv h}{\mv h}^H+{\text{diag}}({\mv{\Sigma}})+{\text{diag}}(\mv{q})\right){\mv w}_u},
\end{equation}
where ${\mv{\Sigma}}=[\sigma^2_1,\sigma^2_2,\cdots,\sigma^2_{J+1}]$ and $\mv{q}=[0,q_2,\cdots,q_{J+1}]$.

It can be shown that the optimal combining weights that maximize (\ref{sinr3}) are obtained from the well-known minimum mean square error (MMSE) combining\cite{kay1993fundamentals}, i.e.,
\begin{equation}\label{qf.mrc}
{\mv w}_u = (p_1{\mv h}{\mv h}^H+{\mv{\Gamma}})^{-1}{\mv f},
\end{equation}
where ${\mv{\Gamma}} \triangleq {\text{diag}}({\mv{\Sigma}})+{\text{diag}}(\mv{q})$. Correspondingly, the maximum UAV SINR is $\gamma_{u,\text{QF1}}=p_u{\mv f}^H(p_1{\mv h}{\mv h}^H+{\mv{\Gamma}})^{-1}{\mv f}$. Since $p_1{\mv h}{\mv h}^H+{\mv{\Gamma}}$ is a diagonal matrix, it is easy to obtain
\begin{equation}\label{sinr4}
\gamma_{u,\text{QF1}}=\frac{p_u\lvert f_1 \rvert^2}{p_1\lvert h_1 \rvert^2+\sigma_1^2} + \sum\limits_{i \in \cal J}{\frac{p_u\lvert f_i \rvert^2}{\sigma_i^2+q_i}},
\end{equation}
which is the sum of the UAV SINRs in $y_1$ and $\tilde y_i, i \in \cal J$. As a result, with Scheme 1, the UAV's achievable rate at BS1 is given by
\begin{equation}\label{rate3}
R_{u,\text{QF1}} = \log_2\left(1+\frac{p_u\lvert f_1 \rvert^2}{p_1\lvert h_1 \rvert^2+\sigma_1^2} + \sum\limits_{i \in \cal J}{\frac{p_u\lvert f_i \rvert^2}{\sigma_i^2+q_i}}\right).
\end{equation}
and the corresponding SINR for decoding UE1's signal at BS1 is given by
\begin{equation}\label{sinr5}
{\gamma}_{1,\text{QF1}}=
\begin{cases}
\frac{p_1\lvert h_1 \rvert^2}{\sigma_1^2}, &{\text{if}}\;R_{u,\text{QF1}} \ge r_u\\
\gamma_1, &\text{otherwise}
\end{cases}.
\end{equation}

By comparing $R_{u,\text{QF1}}$ in (\ref{rate3}) with $R_{u,\text{DF}}$, it follows that the MMSE combining enables BS1 to jointly decode the UAV's signal based on those received at its helping BSs and itself, instead of decoding it individually under the DF-based CIC. Hence, $R_{u,\text{QF1}}$ and ${\gamma}_{1,\text{QF1}}$ can be improved over $R_{u,\text{DF}}$ and ${\gamma}_{1,\text{DF}}$, respectively, provided that the quantization noise powers $q_i$'s in (\ref{rate3}) are sufficiently small. In particular, if the uniform scalar quantization method is applied, it is easy to verify that $R_{u,\text{QF1}}$ will monotonically increase with $p_u$ and $D_i$'s, by substituting (\ref{qerror}) into (\ref{rate3}). Thus, BS1 is more likely to be able to cancel the interference from the UAV with increasing $p_u$ and $D_i$'s. However, if $D_i$'s are too small, e.g., $D_i=0, \forall i \in \cal J$, we have $R_{u,\text{QF1}}=R_1$, i.e., Scheme 1 reduces to the direct SIC by BS1, i.e., there is no need for QF by the helping BSs.\vspace{-6pt}

\subsection{Scheme 2: Linear Interference Cancellation}\label{scheme2}
Alternatively, BS1 can design ${\mv w}_u$ such that UE1's signal SINR is maximized by linearly suppressing the UAV's interference, so as to decode UE1's signal directly. By this means, Scheme 2 avoids decoding the UAV's signal and thus its performance will not be subjected to the UAV's transmit rate. In addition, it requires a lower-cost linear operation at BS1 as compared to the non-linear SIC in both the DF-based CIC and the QF-based Scheme 1. Based on (\ref{sum4}), the SINR for decoding UE1's signal at BS1 is expressed as
\begin{equation}\label{sinr6}
\gamma_{1,\text{QF2}} = \frac{p_1\lvert {\mv w}^H_u{\mv h}\rvert^2}{{\mv w}^H_u\left(p_u{\mv f}{\mv f}^H+{\mv{\Gamma}}\right){\mv w}_u}.
\end{equation}

Similarly as Scheme 1, the optimal combining weights that maximize (\ref{sinr6}) are MMSE combining, which is given by
\begin{equation}\label{qf.mmse}
{\mv w}_u = (p_u{\mv f}{\mv f}^H+{\mv{\Gamma}})^{-1}{\mv h},
\end{equation}
and correspondingly, the SINR in (\ref{sinr6}) becomes
\begin{equation}\label{sinr7}
\gamma_{1,\text{QF2}}=p_1{\mv h}^H(p_u{\mv f}{\mv f}^H+{\mv{\Gamma}})^{-1}{\mv h}.
\end{equation}
Unlike Scheme 1, the matrix $p_u{\mv f}{\mv f}^H+{\mv{\Gamma}}$ is not diagonal in (\ref{sinr7}), thus it is generally difficult to express $\gamma_{1,\text{QF2}}$ in a simplified scalar form as in (\ref{sinr4}). However, this can be achieved in our considered scenario due to the special structure of ${\mv h}=[h_1, {\bf 0}_J]^H$, as shown in the following proposition.

\begin{proposition}
The SINR in (\ref{sinr7}) can be simplified as
\begin{equation}\label{sinr8}
\gamma_{1,\text{QF2}} = \frac{p_1\lvert h_1 \rvert^2}{\sigma_1^2}\left(1-\frac{p_u\lvert f_1 \rvert^2}{\sigma^2_1+p_u\lvert f_1 \rvert^2+p_u\sigma_1^2\sum\limits_{i \in \cal J}{\frac{\lvert f_i \rvert^2}{\sigma_i^2+q_i}}}\right).
\end{equation}
\end{proposition}
\begin{IEEEproof}
It follows from (\ref{sinr7}) that
\begin{align}
\gamma_{1,\text{QF2}} &\mathop = \limits^{(a)} p_1\lvert h_1 \rvert^2[(p_u{\mv f}{\mv f}^H+{\mv{\Gamma}})^{-1}]_{1}\nonumber\\
&\mathop = \limits^{(b)} p_1\lvert h_1 \rvert^2\left[{\mv{\Gamma}}^{-1}-\frac{p_u}{1+p_u{\mv f}^H{\mv{\Gamma}}^{-1}{\mv f}}{\mv{\Gamma}}^{-1}{\mv f}{\mv f}^H{\mv{\Gamma}}^{-1}\right]_1\nonumber\\
&= p_1\lvert h_1 \rvert^2\left(\lbrack \mv{\Gamma}^{-1}\rbrack_1-\frac{p_u\lbrack {\mv{\Gamma}}^{-1}{\mv f}{\mv f}^H{\mv{\Gamma}}^{-1} \rbrack_1}{1+p_u{\mv f}^H{\mv{\Gamma}}^{-1}{\mv f}}\right),\label{eq1}
\end{align}
where equality $(a)$ is due to ${\mv h}=[h_1, {\bf 0}_J]^H$, while equality $(b)$ is derived by invoking the following matrix inverse lemma, i.e., $(\mv A + \mv{BCD})^{-1}=\mv A^{-1}-\mv A^{-1}\mv B(\mv C^{-1}+\mv D\mv A^{-1}\mv B)^{-1}\mv D\mv A^{-1}$ for invertible $\mv A$ and $\mv C$, with setting $\mv A=\mv{\Gamma}$, $\mv B={\mv f}$, $\mv C=p_u$, and $\mv D={\mv f}^H$. Next, it is easy to verify $\lbrack \mv{\Gamma}^{-1}\rbrack_1=\frac{1}{\sigma^2_1}$, $\lbrack {\mv{\Gamma}}^{-1}{\mv f}{\mv f}^H{\mv{\Gamma}}^{-1} \rbrack_1=\frac{\lvert f_1 \rvert^2}{\sigma^4_1}$, and ${\mv f}^H{\mv{\Gamma}}^{-1}{\mv f}=\frac{\lvert f_1 \rvert^2}{\sigma_1^2}+\sum\nolimits_{i \in \cal J}{\frac{\lvert f_i \rvert^2}{\sigma_i^2+q_i}}$. By substituting the above equalities into (\ref{eq1}), we can obtain the SINR expression given in (\ref{sinr8}). The proof is thus completed.
\end{IEEEproof}

It is noted from (\ref{sinr8}) that $\frac{p_1\lvert h_1 \rvert^2}{p_u\lvert f_1 \rvert^2+\sigma_1^2}< \gamma_{1,\text{QF2}}<\frac{p_1\lvert h_1 \rvert^2}{\sigma_1^2}$, i.e., UE1 improves its SINR but still suffers residual interference with the linear interference cancellation. Nonetheless, it can perform better than the QF-based Scheme 1 if $\sum\nolimits_{i \in \cal J}\frac{\lvert f_i \rvert^2}{\sigma_i^2+q_i} \gg \frac{\lvert f_1 \rvert^2}{\sigma_1^2}$, e.g., when the number of helping BSs, $J$, is large enough. However, this comes at the cost of higher cooperation complexity and longer decoding delay. In particular, if the uniform scalar quantization method in (\ref{qerror}) is applied, $\gamma_{1,\text{QF2}}$ can be verified to monotonically increase with $D_i$'s but decrease with $p_u$. In the extreme case of $D_i=0, \forall i \in \cal J$, we have $\gamma_{1,\text{QF2}}={\gamma}_1$ given in (\ref{sinr1}), i.e., there is no need for QF by the helping BSs, similar to Scheme 1.\vspace{-9pt}

\subsection{Performance and Complexity Comparison}\label{comp}
Next, we compare the SINRs/achievable rates for decoding UE1's signal by the considered CIC schemes, namely, the DF-based CIC and the two proposed QF-based CIC schemes (Schemes 1 and 2). The results are summarized as follows.
\begin{itemize}
  \item {\it Scheme 1 versus DF-based CIC}: By comparing (\ref{sinr5}) with (\ref{sinr2}), it follows that the former outperforms the latter if and only if (iff) the UAV's signal can be decoded by the former instead of the latter, i.e., $R_{u,\text{QF1}} \ge r_u>R_{u,\text{DF}}$. Similarly, the latter outperforms the former iff $R_{u,\text{DF}} \ge r_u>R_{u,\text{QF1}}$ (e.g., when the quantization noise powers $q_i, i \in \cal J$ are too high).
  \item {\it Scheme 2 versus DF-based CIC}: Since $\frac{p_1\lvert h_1 \rvert^2}{p_u\lvert f_1 \rvert^2+\sigma_1^2}< \gamma_{1,\text{QF2}}<\frac{p_1\lvert h_1 \rvert^2}{\sigma_1^2}$, the former yields better performance than the latter iff the UAV's signal cannot be decoded by the latter, i.e., $r_u>R_{u,\text{DF}}$. Otherwise, the latter outperforms the former.
  \item {\it Scheme 2 versus Scheme 1}: Similarly as the above, the former outperforms the latter iff the UAV's signal cannot be decoded by the latter, i.e., $r_u>R_{u,\text{QF1}}$. Otherwise, the latter outperforms the former.
\end{itemize}

In practical implementation, Scheme 2 has the lowest decoding complexity among the three considered schemes due to the linear ICT applied at BS1. However, both Schemes 1 and 2 yield higher signaling overhead than the DF-based CIC, since the quantized signals by all helping BSs need to be forwarded to BS1; while in the latter case, only the decoded UAV signal at one of the helping BSs needs to be forwarded.\vspace{-6pt}
\begin{figure*}[hbtp]
\centering
\subfigure[]{\includegraphics[width=0.32\textwidth]{Thres_Rate.pdf}}
\subfigure[]{\includegraphics[width=0.32\textwidth]{Thres_Pw.pdf}}
\subfigure[]{\includegraphics[width=0.32\textwidth]{Thres_dist.pdf}}\vspace{-3pt}
\caption{UE1's achievable rate versus (a) the UAV's transmit rate; (b) the UAV's transmit power; and (c) the UAV's horizontal distance with BS1.}\label{sim}\vspace{-12pt}
\end{figure*}

\section{Numerical Results}
In this section, numerical results are provided to evaluate the performance of the proposed QF-based CIC (Schemes 1 and 2) as compared to the existing DF-based CIC. Unless otherwise specified, the simulation settings are as follows. The shape of each cell is assumed to be hexagonal. The helping BSs of BS1 constitute its nearest adjacent BSs in all considered CIC schemes, i.e., $M=1$ and $J=6$. The bandwidth of the UAV's assigned RB is set to 180 kHz. The carrier frequency $f_c$ is $2$ GHz, and the noise power spectrum density at BS receiver is $-164$ dBm/Hz. For the terrestrial channels, their path-loss and shadowing are modeled based on the urban macro (UMa) scenario in the 3GPP report\cite{3GPP36777}, while the small-scale fading is modeled as Rayleigh fading. The cell radius is set to 800 in meter (m), and the heights of all BSs are 25 m. The altitude of the UAV is fixed as 200 m, while its horizontal distance with BS1 is 3 km. The BS antenna pattern is assumed to be directional in the vertical plane with 10-degree downtilt angle. The total number of non-associated but co-channel BSs in $D_u$ is 8 (including BS1), and their served terrestrial UEs' locations are randomly generated in their located cells. All BS-UAV channels follow the probabilistic LoS channel model in the UMa scenario in \cite{3GPP36777}. The transmit power for all terrestrial UEs is assumed to be identical as $p_1=15$ dBm. We consider that all helping BSs adopt 6-bit uniform scalar quantization to quantize their received UAV signals, i.e., $D_i=6, \forall i \in \cal J$ in (\ref{qerror}). All the results shown below are averaged over 1000 random channel and location realizations of terrestrial UEs.

First, Fig.\,\ref{sim}(a) shows UE1's achievable rate (defined as $\log_2(1+\text{SINR})$ in bps/Hz, where SINR denotes the UE1's achievable SINR in each scheme) versus the UAV's transmit rate, $r_u$. The UAV's transmit power is 15 dBm. It is observed that UE1 can attain its maximum achievable rate in both the DF-based CIC and the QF-based CIC Scheme 1 when $r_u$ is low. This is because the first conditions in (\ref{sinr2}) and (\ref{sinr5}) are both met in this case. Thus, BS1 can completely remove the interference from the UAV. However, as $r_u$ increases, these conditions no longer hold in more channel realizations; thus, UE1's achievable rates by both schemes are observed to decrease and finally attain the minimum given by $\log_2(1+\gamma_1)$. Nonetheless, it is observed that the QF-based CIC Scheme 1 yields a better performance than the DF-based CIC when $r_u$ is sufficiently high, i.e., it is more unlikely for the helping BSs to individually decode the UAV's signal. In contrast, without the need of decoding the UAV's signal, UE1's achievable rate is observed to be unaffected by $r_u$ in the QF-based CIC Scheme 2. It is also observed that this scheme significantly outperforms the other two schemes when $r_u$ is sufficiently high, in accordance with the conditions given in Section \ref{comp}.

Next, in Fig.\,\ref{sim}(b), we plot UE1's achievable rate versus the UAV's transmit power, $p_u$, with fixed $r_u=5$ bps/Hz. It is observed that with increasing $p_u$, UE1's achievable rate by Scheme 2 decreases, which is consistent with our discussion at the end of Section \ref{scheme2}. Whereas for the other two schemes, UE1's achievable rates are observed to first decrease and then increase with $p_u$. This is because when $p_u$ is small, the first conditions in (\ref{sinr2}) and (\ref{sinr5}) cannot be met. As a result, UE1's achievable rates are degraded by increasing $p_u$ due to higher interference power. However, as $p_u$ further increases, the above conditions are more likely to be met, and thus UE1's achievable rates start to increase since a larger $p_u$ facilitates decoding the UAV's signal in both schemes. It is also observed that the proposed QF-based CIC Schemes 1 and 2 outperform the DF-based CIC when $p_u$ is high and low, respectively, which are consistent with the conditions given in Section \ref{comp}.

Last, we plot UE1's achievable rate versus the UAV's horizontal distance with BS1 in Fig.\,\ref{sim}(c), with $r_u=5$ bps/Hz and $p_u=15$ dBm. It is observed that UE1's achievable rate by Scheme 2 increases when the UAV is further away from BS1. This is expected as the UAV's interference to UE1 at BS1 is weakened due to its increased distance with BS1. However, UE1's achievable rates by the other two schemes are observed to first decrease and then increase. This is because increasing the distance between the UAV and BS1 (and its helping BS$i, i \in \cal J$) also degrades BS1's interference cancellation capability (thus, $R_{u,\text{DF}}$ and $R_{u,\text{QF1}}$) due to the reduced channel power gains, $\lvert f_i \rvert^2, i \in \cal J$. It is observed that when the UAV-BS1 horizontal distance is smaller than 3.8 km (4 km) in Scheme 2 (DF-based CIC), BS1's degraded interference cancellation capability dominates over the reduced UAV interference at BS1. However, if this distance continues to increase, the latter will become dominant, thus resulting in the improvement of UE1's achievable rates. \vspace{-6pt}

\section{Conclusions}
This letter proposes a new QF-based CIC scheme with different ICTs to mitigate the strong uplink interference from the UAV to a large number of co-channel BSs serving terrestrial UEs, by exploiting the quantized signals shared by their adjacent helping BSs. Analytical and numerical results both reveal that the proposed QF-based scheme can yield significant performance gains over the existing DF-based CIC scheme for practical UAV's transmit rate/power as well as channels with BSs, while the use of linear or nonlinear ICT for the QF-based CIC should be decided based on the actual condition.\vspace{-3pt}

\begin{comment}
\begin{figure}[htbp]
\centering
\includegraphics[width=2.8in]{Thres_Rate.eps}
\DeclareGraphicsExtensions.
\caption{UE1's achievable rate versus the UAV's transmit rate.}\label{Thres_Rate}
\vspace{-8pt}
\end{figure}
\begin{figure}[htbp]
\centering
\includegraphics[width=2.8in]{Thres_dist.eps}
\DeclareGraphicsExtensions.
\caption{UE1's achievable rate versus the UAV's horizontal distance with BS1.}\label{Thres_dist}
\vspace{-8pt}
\end{figure}
As predicted by the Federal Aviation Administration, there will be over 2.75 million small UAVs flown in the United States by 2021.
\begin{figure}[htbp]
\centering
\includegraphics[width=2.8in]{Thres_bit.eps}
\DeclareGraphicsExtensions.
\caption{UE1's achievable rate versus number of bits for quantization.}\label{Thres_bit}
\vspace{-8pt}
\end{figure}
\begin{figure}[htbp]
\centering
\includegraphics[width=2.8in]{Thres_Pw.eps}
\DeclareGraphicsExtensions.
\caption{UE1's achievable rate versus the UAV's transmit power.}\label{Thres_Pw}
\vspace{-8pt}
\end{figure}

In Fig.\,\ref{Thres_bit}, we plot UE1's achievable rate versus the number of bits for quantization, $D$, with $r_u=5$ bps/Hz and $p_u=15$ dBm. It is observed that UE1's achievable rates by both QF schemes improves as $D$ increases, consistent with our discussion at the end of Sections \ref{scheme1} and \ref{scheme2}. It is also observed that with insufficient bits for quantization, both of the proposed schemes can yield a worse performance than the DF-CIC.

As such, in practice, BS1 can optimally choose between the nonlinear (Scheme 1) and linear (Scheme 2) ICT based on the above conditions for the proposed QF-based CIC.
\end{comment}

\bibliography{UAV_QF}
\bibliographystyle{IEEEtran}
\end{document}


TITLE: Why isn't the free expansion of a gas in an adiabatic container isentropic?
QUESTION [6 upvotes]: If you expand a gas adiabatically using a piston, the process is isentropic. However, if you simply remove the piston and let the gas expand freely, the process is now not isentropic. What makes these processes different? Is it simply because the free expansion is not a reversible process?
But imagine that the two gases are ideal and that they expand from the same initial volume and temperature, and into the same final volume. Then, using the ideal gas law these states are identical, and so should be their entropies. So why aren't they? Why is one process isentropic while the other is producing entropy?
Perhaps because the free expansion does not have any internal energy loss?

REPLY [0 votes]: Another way is see the difference is to look at the problem in terms microscopics. When you look close enough, you can imagine your gas as a number of particles distributed over some set of energy levels $\{ E_n\}$. The exact position of these levels is determined by means of Quantum Mechanics in the geometry of the container.
Now when you change the geometry very slowly, or as people in quantum mechanics say ${\it adiabatically}$, that is, the characteristic timescales $\tau$ over which the deformation occurs, is much longer than the inverse of the smallest inter-level energy difference $min\{E_{n+1}-E_n\}$, then at the end of the process each particle will remain in the same level $n$ that it occupied before deformation. The total energy of the system $E_{tot}=\sum_n N(n) E_n$ ($N(n)$ is the number of particles in the level $n$) will have changed in the process because the energy levels have changed $E'_{tot}=\sum_n N(n) E'_n$. The change in total energy achieved this way is referred to as ${\it work}$: $d W = E'_{tot}-E_{tot}$. Since, this is the only way we have supplied the system with extra energy, such process, - adiabatic in QM sense, - also happens to be adiabatic in thermodynamics sense (of course this is not a coinsidence. It is just the etymological origin of "adiabaticity" in QM).
Now what happens when the adiabaticity condition (in QM sense) is violated? The time dependent perturbation that changing of the container geometry is, generally speaking will cause particles to transition between different energy levels. The new energy of the system is now $E'_{tot}=\sum_n N'(n) E'_n$. The part of the energy difference $\delta E_{tot}\approx \sum_n N(n) \delta E_n + \sum_n \delta N(n) E_n$ coming from change in energy levels (the first term) is again interpreted as work, whereas the part coming from re-distribution of population (the second term) is referred to as ${\it heat}$ $\delta Q$. It is intuitively clear that the change in distribution $\{ \delta N(n) \}$ should lead to a change in entropy $S$ (in Boltzman sense) $\delta Q=TdS$.
Now coming to your original question, in the first process if you move your piston slowly, the particles stick to their "original" levels, no redistribution $\delta N(n)$ occurs and hence to change in entropy. In the second case, the piston moves abruptly. Quantum mechanically speaking you are actually in the sudden change approximation which is the opposite of adiabatic (the particles most likely end up in the new energy levels that are closet in value to their original ones). The population distribution changes is non-zero, hence the process is not adiabatic in both QM and thermodynamic senses.


TITLE: phase transformation of liquid under motion
QUESTION [0 upvotes]: I was testing in my lab with water and found that it starts to solidify when it is stationery at 0 degree centigrade.but when I move the liquid with some velocity it dosent change its phase.my interpretation of this happening because when I move liquid it has kinetic energy which dosent allow it to crystallize.but at certain velocity and further lowered temperature it should start crystallising again.is there any graph which shows at what velocity and temperature it crystallize with atmospheric pressure?as in phase transition of water under temperature and pressure?

REPLY [4 votes]: When I worked at Unilever we investigated the freezing of water being pumped through pipes as part of the study into making ice cream (the pipes in question were part of the plant making ice cream).
We found exactly what you describe. It's well known that very pure stationary water can be supercooled to well below zero, but even for not especially pure water rushing through pipes it could be cooled well below zero without nucleating any ice crystals. The only downside was that as soon as the flow stopped the whole thing immediately froze solid! I don't remember what our record low temperature was, but it was startlingly low - I think, though I wouldn't swear to it, that it was around -20°C.
I'm afraid we never looked into why the water could be supercooled. Our guess was that the turbulence of the flow prevented the nuclei from growing large enough to cause macroscopic freezing. At any temperature above around -50°C there is a kinetic barrier to the formation of ice crystals because it costs a lot of energy to nucleate an ice crystal.
Ig you consider the process of making a sperical ice crystal of radius $r$ the free energy change is given by something like:
$$ \Delta G (r) = 4\pi r^2 \gamma_{sl} - \frac{4}{3}\pi r^3 \rho \Delta G_{sl} $$
The first, positive, term is the increase in energy due to the formation of the ice/water boundary with an interfacial tension $\gamma_{sl}$, and the second negative term is the energy released by converting a volume of water to ice. At small $r$ the first term dominates, and forming an ice crystal actually increases the energy instead of decreasing it. You need to grow the crystal to a certain size before the free energy becomes negative and freezing procedes spontaneously. If you can put in enough mechanical energy to break up the nuclei before they get big enough to grow then you can stop the water freezing.


TITLE: Probability of drawing a specific number out of a pool of numbers with multiple draws
QUESTION [0 upvotes]: I have a pool of 100 different numbers, I draw 5 different numbers without putting them back into the pool. I want to calculate the chance of any single number being drawn.
So for example, what's the chance of the number 13 being drawn during those 5 draws?
I figured, the way to calculate this would be
$$\frac{1}{100}+(\frac{1}{99}*\frac{99}{100})+(\frac{1}{98}*\frac{98}{99})+(\frac{1}{97}*\frac{96}{97})+(\frac{1}{96}*\frac{95}{96})=0.0506$$
Which would mean the chance for that would be ~5.1%
Is this calculation correct? What is this calculation called and whats the formula for it?

REPLY [0 votes]: I know it s late but maybe somebody will find it helpful, you need to calculate probability of successful draw as well as probability of getting into that specific draw, which would be on first 100/100 (since it is your first draw), for second (100-1 draw you already did)/100 of total pool, so basically you would have formula like 100/100 * 1/100 + 99/100 * 1/99 + 98/100 * 1/98....etc.=0,05=5%. That is (when you think about it) the same as if you picked five numbers with one draw which is 100/100(probability that you draw first draw) * 5/100 (chance that you ll get specific number out of 100) which should equal exactly 5% too. If you wanted to calculate chances of getting it at 5th draw exactly that would be different however. Then youwouldstart of chancesof getting through draw starting by 99/100 chance to get through first draw where you actually cannot pull the desired number.