TITLE: Is the equivalent function between isomorphic domains called a dual?
QUESTION [0 upvotes]: Is the equivalent function between isomorphic domains called a dual or something else? If you can also provide citations or examples of mathematicians using the terminology then that would be great.
For example is this the correct use of terminology? Or do I have to change the codomain to a vector space as well?
Let $f(x,y,z)=xe^z+ln(y)$. This has domain $\{(a,b,c)∈R^3 ∶ b≥0\}⊂R^3$. Find the equivalent function, $f^*$ from a subset of a vector space.
Define $f^*(〈x,y,z〉)=xe^z+ln(y)$, with domain $\{〈a,b,c〉∈V_3 ∶ b≥0\}⊂V_3$. This is called the dual of f under the identity isomorphism of $R^3⟼V_3$.

REPLY [1 votes]: Short answer: no. The adjective "dual" has precise definitions in several different mathematical contexts. The example you ask about is not one of them.
In your example the fact that the domain has a vector space structure is irrelevant because the function is not linear where it is defined.


TITLE: max of limit cardinals smaller than a successor cardinal bigger than $\aleph_\omega$
QUESTION [1 upvotes]: This is the statment i want to proove :
if $k$ is a successor cardinal bigger than $\aleph_{\omega}$ then there is a max of $\{\lambda < k | \lambda \text{ is a limit cardinal}\}$.
Assuming (AC), if $k \geq \omega$ then $k^{+}$ is regular, so we can assume that $k$ is regular. I've found this argument but i'm not sure that it's correct :
By contradiction, if there is no max, then the union of all $\lambda$ limit smaller than $k$ is a successor cardinal $\leq k$. But a successor cardinal is regular, so it cannot be reach by this union : a contradiction. So there must be a max. Am I right?

REPLY [1 votes]: Choice has nothing to do with this. The $\aleph$ numbers are defined the same. Of course, for non-$\aleph$ cardinals there is an issue for what would be a successor, or limit, and so on. So let's ignore that issue and focus on $\aleph$ cardinals.
You can easily prove this by induction, noting that successor cardinals are exactly those whose $\aleph$ index is a successor ordinal. So the question is reduced to limit ordinals below infinite successor ordinals, and this question is easier to grasp.


TITLE: Origin of the terminology projective module
QUESTION [17 upvotes]: Projective modules were introduced in 1956 by Cartan and Eilenberg in their book Homological Algebra.  Does anyone know why they chose the word "projective"? Does it have something to do with the notion of projection?

REPLY [24 votes]: The term "projection" has a few possible meanings in linear algebra,
and they are equivalent to the property of being a projective module.

A linear operator $f \colon {\mathbf R}^n \rightarrow {\mathbf R}^n$ is geometrically a projection to a subspace in some coordinate system if and only if $f^2 = f$.
For two vector spaces $V$ and $W$, the function $V \oplus W \rightarrow V$
where $(v,w) \mapsto v$ is called a projection (out of the direct sum).

For a commutative ring $R$ and $R$-module $P$, the following properties that abstract the above two conditions are both equivalent to $P$ being a projective module.

There is a free $R$-module $F$ and an $R$-linear map $f \colon F \rightarrow F$ such that $f^2 = f$ and $f(F) \cong P$ (so $P$ is isomorphic to the image of a "projection").
Any way of making $P$ look like a quotient module essentially amounts to making it the image of the natural projection out of a direct sum module to one of its direct summands: if $f \colon M \twoheadrightarrow P$ is any surjective $R$-linear map then there is an $R$-module isomorphism $h \colon M \cong P \oplus K$ for some $R$-module $K$ such that
$h(m) = (f(m),*)$ for all $m \in M$ (thus $f$ looks like the natural projection map $P \oplus K \rightarrow P$).


TITLE: Proving a subspace of $\mathbb{R}^{2}$ is still a vector space with redefined addition / scalar multiplication operations?
QUESTION [3 upvotes]: Question:
Let $S =$ { $(a,b) \in \mathbb{R}^{2}$ | $(a,b)$ has the form $(1, x)$ }.
Redefine addition and scalar multiplication operations as follows:
$(1, y) + (1, y') = (1, y + y')$
$k(1, y) = (1, ky)$
Can I now say that because $\mathbb{R}^{2}$ is a vector space and $S \subset \mathbb{R}^{2}$, that if additive & scalar multiplicative closures are shown to hold, then $S$ is a vector space?
Or do I first have to show that $\mathbb{R}^{2}$ is still a vector space with the newly defined operations?

REPLY [8 votes]: The operations you've defined do make $S$ into a vector space, but it is not a subspace of $\mathbb{R}^2$. A subspace of a vector space $V$ is a subset $W\subseteq V$ that's a vector space under the same operations as $V$. So, you need to show all of the axioms for a vector space hold for $S$.
Also, I don't know what you mean by "$\mathbb{R}^2$ is still a vector space with the newly defined operations"; how do you intend to extend the definitions $(1, y) + (1, y^\prime) = (1, y + y^\prime)$ and $k(1, y) = (1, ky)$ to all of $\mathbb{R}^2$, in a  way that makes $\mathbb{R}^2$ a vector space?


TITLE: Relationship between volume and surface charge density in the general case
QUESTION [0 upvotes]: Assume we have surface $S$ enclosing a volume $V$ and assume the only charge present in this setting is located on the surface, given by a surface charge density $\sigma$. My question is about the correct way to model such a situation in electrostatics and more precisely, what volume charge density $\rho$ we should take in such a case.
There is a related post (here: Relationship between surface density and volume density) where the problem is discussed in specific examples and it is explained that $\rho$ can in those examples be defined via the Dirac-Delta.
In this question however, I want to know the general "definition" for $\rho$ in such a case, without reference to specific and simple examples like spheres, planes and the like.
My idea is that in the general case of a surface charge, the volume density $\rho$ should be defined as follows: $\rho$ is a distribution with
$$\int_V \varphi\rho \ \text{d}^3x = \int_{S} \varphi\sigma \ \text{da}$$
for all functions $\varphi$ of a suitable function space. Here, "distribution" is understood in the sense of mathematical distribution theory.
My question is now simple: Is this the right way to do this?

REPLY [2 votes]: Sure, though if you want to be super strict, then $\rho$ shouldn't be written inside an integral. You should simply say $\langle\rho,\varphi\rangle:=\int_S\sigma\varphi\,da$, i.e the value of the functional $\rho$ on the test function $\varphi$ is defined to be the integral on the right. This formulation works in any number of dimensions, not just the case of a 2D surface $S$ embedded in $\Bbb{R}^3$.
For instance, let $M$ be a $k$-dimensional embedded submanifold of $\Bbb{R}^n$ ($0\leq k\leq n$), $d\mu_k$ be induced $k$-dimensional volume measure on $M$, and let $\sigma\in L^1(M;d\mu_k)$. We can define a distribution $\delta_{M,\sigma}$ on $\Bbb{R}^n$ by setting for each test function $\varphi$ on $\Bbb{R}^n$ (i.e $\varphi\in C^{\infty}_c(\Bbb{R}^n)$ is smooth compactly supported),
\begin{align}
\langle\delta_{M,\sigma}, \varphi\rangle&:=\int_M\sigma\varphi\,d\mu_k.
\end{align}
We refer to $\delta_{M,\sigma}$ as the single-layer distribution on the submanifold $M$ with density $\sigma$. The higher derivatives of this distribution are known as multiplet-layers (the "layer" terminology comes from classical PDE solving methods of single and double layer potentials for Laplace's/Poisson's equation).
See analysis/PDE texts for more information about this sort of stuff, for example, Dieudonne's Treatise on Analysis Vol III (Chapter 17.10), or Hormander's The Analyis of Linear Partial Differential Operators I (Chapter 6.1), or the classic text by Gelfand and Shilov Generalized Functions Volume I (Chapter 3.1 is all about distributions on $\Bbb{R}^n$ concentrated  lower dimensional manifolds).


text:
\section{Linear Algebra Lemmas}
In this section, we provide some lemmas from linear algebra which were previously used in Appendix.
\begin{lemma}[Re-parametrization of an ellipsoid]\label{ellipse}
	Suppose $x_0\in\mathbb{R}^d$ and $\Sigma\in\mathbb{R}^{d\times d}$ is positive definite. Then
	\[E=\left\{x\in\mathbb{R}^{d}:\,\,\left(x_{0}-x\right)^{\top} \Sigma^{-1}\left(x_{0}-x\right) \leqslant r^{2}\right\}=\left\{x_{0}-r\, \Sigma^{1 / 2}u: \,\, u\in\mathbb{R}^d,\,\, \|u\|\, \leqslant 1\right\}.\]
\end{lemma}
\begin{proof}
	Define
	\begin{equation*}
	u:=\frac{1}{r}\,\Sigma^{-1 / 2}\left(x_{0}-x\right).
	\end{equation*}
	Therefore,
	\begin{align*}
	x\in E \quad &\Leftrightarrow\quad x= x_{0}-r\, \Sigma^{1 / 2} u, \quad r^2\,u^{\top}u \leqslant r^2\\
	&\Leftrightarrow \quad x= x_{0}-r\, \Sigma^{1 / 2} u, \quad \|u\|\, \leqslant 1.
	\end{align*}
\end{proof}

\begin{lemma}[Sherman Morrison Woodbury Formula]\label{matrix-inversion2}
	Suppose $A$, $B$, and $C$ are matrices such that $A$, $C$, and $A+BCD$ are well-defined and invertible. Then
	\begin{align*}
	(A+B C D)^{-1}=A^{-1}-A^{-1} B\left(C^{-1}+D A^{-1} B\right)^{-1} D A^{-1}.
	\end{align*}
\end{lemma}
\begin{proof}
	Note that if $P$ is a square matrix such that $I+P$ is invertible, then
	\begin{align}\label{50}
	(I+P)\left(I-(I+P)^{-1} P\right)  = I+P-P=I\quad  \Rightarrow \quad(I+P)^{-1}= I-(I+P)^{-1} P.
	\end{align}
	Also, note that if $U$ and $V$ are such that $I+UV$ is invertible, then $I+VU$ is also invertible. To see that, suppose $\left(I+VU\right)x=0$. We need to show that $x=0$. We have
	\begin{align}\label{52}
	\left(I+VU\right)x=0\,\, \Rightarrow \,\,U\left(I+VU\right)x=0\,\,  \Rightarrow \,\, \left(I+UV\right)Ux=0 \,\, \overset{\left(I+UV\right)^{-1}}{\Longrightarrow} \,\, Ux=0.
	\end{align}
	Thus,
	\begin{align*}
	\left(I+VU\right)x=0\,\, \Rightarrow \,\,x = -V(Ux) \overset{\eqref{52}}{=}0.
	\end{align*}
	Hence, if $U$ and $V$ are such that $I+UV$ is invertible, then $I+VU$ is invertible and we can write
	\begin{align}\label{51}
	(I+UV)U = U(I+VU)\quad  \Rightarrow \quad U(I+V U)^{-1}=(I+U V)^{-1} U.
	\end{align}
	Moreover, we can conclude that under these assumptions,
	\begin{align}\label{53}
	(I+U V)^{-1} \overset{\eqref{50}}{=} I-(I+U V)^{-1} U V \overset{\eqref{51}}{=} I-U(I+V U)^{-1} V.
	\end{align}
	Now by replacing $U = A^{-1}B$ and $V=CD$, \eqref{53} results in
	\begin{align*}
	(A+B C D)^{-1} &= \left(I+A^{-1}BCD\right)^{-1}A^{-1}\overset{\eqref{53}}{=}\left(I- A^{-1}B\left(I+CDA^{-1}B\right)^{-1}CD\right)A^{-1}\\
	&=A^{-1}- A^{-1}B\left(I+CDA^{-1}B\right)^{-1}CDA^{-1}\\
	&=A^{-1}- A^{-1}B\left(C^{-1}+DA^{-1}B\right)^{-1}DA^{-1}.
	\end{align*}
\end{proof}

\begin{lemma}\label{matrix-inversion}
	If the right-hand side of \eqref{48} is well-defined, we have
	\begin{align}\label{48}
	\left[\begin{array}{ll}{A} & {B} \\ {C} & {D}\end{array}\right]^{-1}=\left[\begin{array}{cc}{\left(A-B D^{-1} C\right)^{-1}} & {-\left(A-B D^{-1} C\right)^{-1} B D^{-1}} \\ {-D^{-1} C\left(A-B D^{-1} C\right)^{-1}} & {\left(D-C A^{-1} B\right)^{-1}}\end{array}\right].
	\end{align}
\end{lemma}
\begin{proof}
	Simply through a matrix multiplication and using Lemma~\ref{matrix-inversion2}.
\end{proof}

\begin{lemma}\label{9}
	Consider $x\in\mathbb{R}^d$ and the identity matrix $I\in\mathbb{R}^{d\times d}$. Then $\left\|\left[{I}, {x}\right]\right\|=\sqrt{1+\left\|x\right\|^{2}}$.
\end{lemma}
\begin{proof}
	Let $\lambda_{\max}(\cdot)$ denote the maximum eigenvalue of the corresponding matrix. Then
	\begin{align*}
	\left\|\left[{I}, {x}\right]\right\|=\sqrt{\lambda_{\max}\left(\left[I, x\right]\left[\begin{array}{l}{I} \\ {x^{\top}}\end{array}\right] \right)}=\sqrt{\lambda_{\max}\left(I+xx^{\top}\right)}=\sqrt{I+\lambda_{\max} \left(xx^{\top}\right)}=\sqrt{1+\left\|x\right\|^{2}}.
	\end{align*}
\end{proof}

\begin{lemma}\label{matrix-inversion3}
	Suppose $A$ and $B$ are matrices such that $A$ and $A+B$ are well-defined and invertible. Then
	\begin{align*}
	{(A+B)^{-1}=A^{-1}-\left(I+A^{-1} B\right)^{-1} A^{-1} B A^{-1}}.
	\end{align*}
\end{lemma}
\begin{proof}
	Replace $P=A^{-1}B$ in \eqref{50} to get
	\begin{align*}
	(A+B)^{-1}&=\left(I+A^{-1}B\right)^{-1}A^{-1}   \overset{\eqref{50}}{=} \left(I-\left(I+A^{-1}B\right)^{-1} A^{-1}B\right)A^{-1}\\
	&= A^{-1}-\left(I+A^{-1} B\right)^{-1} A^{-1} B A^{-1}.
	\end{align*}
\end{proof}

\begin{lemma}\label{matrix-norm}
	If  $\|A\| \leqslant 1$,  then  $\left\|(I+A)^{-1}\right\| \leqslant \frac{1}{1-\|A\|}$.
\end{lemma}
\begin{proof}
	Note that
	\begin{align*}
	\left\|(I+A)^{-1}\right\|\overset{\eqref{50}}{=}\left\|I-(I+A)^{-1} A\right\| \leqslant 1+\left\|(I+A)^{-1}\right\|\|A \|\,\Rightarrow\,\left\|(I+A)^{-1}\right\|\left(1-\|A\|\right) \leqslant 1.
	\end{align*}
	Since  $\|A\| \leqslant 1$,  we get $\left\|(I+A)^{-1}\right\| \leqslant \frac{1}{1-\|A\|}$.
\end{proof}

config: arxiv


TITLE: Prove that if a space $X$ is locally path connected then each path component of each open subset of $X$ is open.
QUESTION [0 upvotes]: I am having difficulty proving the following statement.

If a space $X$ is locally path connected, then each path component of
  each open subset of $X$ is open.

In particular, while I have been able to argue that each path component of $X$ is open, I am having difficulty extending this to the open subsets of $X$. My proof of the former is as follows.

First, let $X$ be locally path connected, let $A$ be a path component
  of $X$, and let $x\in A$ be an arbitrary point. Because $X$ is locally
  path connected, there exists a path connected open subset $U\subset X$
  containing $x$. Then $x\in A\cap U$, so $A\cup U$ is path connected.
  Since $A$ is maximal, this implies that $U\subset A$, i.e. that $x$ is
  an interior point of $A$. Since the same argument holds for every
  point $x\in A$, we have that $A$ is open. Thus, each path component of
  $X$ is open.

Would anyone here be willing to push me in the right direction? In case there is ambiguity, the text I am following defines a locally path connected space to be one that has a basis of path connected open sets.

REPLY [1 votes]: If $O$ is open in $X$ and $X$ is locally path-connected, so is $O$ (in its subspace topology): let $X \in O$ and $U \subseteq O$ open in $U$, which however just means that $U$ is also open in $X$ (as $U=U'\cap O$ with $U'$ open in $X$ by the definition of the subspace topology on $O$, but then $U$ is a finite intersection of open sets of $X$, so indeed open in $X$!). As $X$ is locally path-connected it has a local base of path-connected neighbourhoods so one of these sits inside $U$ and we're done, as this is still a path-connected neighbourhood of $x$ in $O$ too (as $O$ is open).
So you can apply your argument got $X$ to $O$ too: its path-components are also open (in $O$ and thus in $X$ too).


TITLE: Could an external being predict quantum uncertainty given all the information of our Universe?
QUESTION [2 upvotes]: I'll elaborate and be more specific. I understand that this is almost a metaphysical question but nonetheless I want to give it a try.
Could an external being outside our Universe create 10 simulations of our Universe and get always the same results when the same initial conditions are applied?
Or would quantum mechanics behave unpredictably even in this situation?

REPLY [1 votes]: The point is that "all the information about the universe" could well comprise knowledge about states of the universe in the future as well as knowledge about yet unknown laws of physics. Since to my knowledge nobody knows the future, and the unknown is, by definition, unknown, your question is impossible to answer. But at least if knowledge about the (infinite) future is required to describe the universe, the notion of prediction becomes meaningless.
What you probably meant instead is, that the external being knows everything about the current state of the universe and some time in the past (represented for example by time derivatives of measurable quantities), plus our current knowledge about the laws of physics. If this knowledge would allow a simulation to compute the future, we would live in a deterministic universe.
From what we know, the universe is not deterministic. The path integral formalism of quantum field theory allows a pretty intuitive understanding what the probabilistic nature of the universe means: we can only determine the probability, that the state of the universe evolves over a series of given intermediate states that we provide for testing. In other words, we can conjecture that something that was arbitrarily chosen by us happens in the universe, and after we have made this conjecture, we can compute the probability that this evolution is compatible with what we know about physics. But we will never know for sure. We can only repeat experiments and see if the computed probability matches the experimental histograms.
There is even a very tiny probability that someone who looks like Captain Kirk suddenly materializes next to you. This does not mean that "beaming" is a regular part of physics, but only that the lottery of the universe might throw the dices so that even the weirdest things might actually happen. But again, only with a minute probability that practically means "never".
Whether the laws of physics can be completed to be deterministic again, is an open question in my opinion, although many people try/claim to prove that it cannot.


TITLE: Showing The Expression Behavior as x approach infinity
QUESTION [1 upvotes]: I have an expression
$x^2[log(1+\sqrt x+x)+C], x\in \mathbb{R},>0$
Where C is an arbitrary constant
I want to find whether this expression is asymptotic to $x$ when $x\rightarrow\infty$
So far I tried log inequality properties and taking advantage of choosing C and so on...
But no luck.
Any suggestions or help will be appreciated!
Thank you so much for reading and having interest in my question.

REPLY [1 votes]: No, it is not asymptotic to $x$
$$\lim_{x \to \infty} \frac{x^2(\log(1 + \sqrt{x} + x) + C)}{x} = \infty$$


text:
\section{Introduction}
During the last two decades, there has been a transition from simulation to
coupling of optimization and simulation~\cite{tekin2004simulation}.
Of particular industrial relevance are shape optimization problems, which aim
to optimize the shape of an object subject to physical constraints, typically
described by partial differential equations (PDEs).
Examples of industrial problems that have been modeled are the drag minimization of airplanes and cars
~\cite{mohammadi2010applied,nielsen2006adjoint,reuther1999constrained}, the shape optimization of acoustic horns~\cite{schmidt2016large}, and the optimal design of current carrying multi-cables~\cite{harbrecht2016}.
The success of these applications is driven by efficient optimization algorithms and fast methods for solving PDEs.
More specifically, gradient-based optimization methods have shown to converge quickly and often independent of the number of design variables. The required shape gradients are derived through shape calculus and the adjoint PDE~\cite{delfour2011shapes,schmidt2013three,sokolowski1992introduction}. The Finite Element Method (FEM) is an efficient and flexible method for solving a wide range of PDEs. In the last decades, this method has gained popularity in both the scientific and industrial environment due to its mathematical foundation and geometrical flexibility.

A critical part in shape optimization algorithms is handling of geometry changes during each optimization iteration. For FEM based models this means that the computational mesh must be updated to a new target geometry at low cost while maintaining a high mesh quality.
Two commonly used mesh updating strategies are mesh deformation and re-meshing.
Mesh deformation methods often involve the solution of an auxiliary PDE. However, the mesh quality may degrade or even degenerate for large deformations. Several deformation schemes have therefore been proposed to handle large deformations~\cite{schmidt2014two,stein2003mesh} of the expense of a high computational cost.
In contrast, a re-meshing strategy produces meshes that are guaranteed to be regular for any geometrical change. However, drawbacks are that the geometry must be reconstructed from the mesh to allow for re-meshing of the boundary elements, and the high computational cost of the meshing algorithms~\cite{boggs2005dart}.

To overcome these limitations, we propose a shape optimization algorithm based on the idea to represent the domain by multiple, non-matching meshes, as illustrated in \cref{fig:Ball}. Because our method is highly embedded in the finite element setting, we call it MultiMesh as opposed to existing approaches like Chimera and Overset methods. Each mesh can be freely rotated, scaled or translated at a low computational cost without impacting the mesh quality.
In an optimization setting, this means that mesh updates can be completely eliminated in cases where the goal is to optimally rotate, scale or translate of objects within a larger geometry. Further, as we will show in the numerical examples, the MultiMesh approach is beneficial for general shape optimization. Applying mesh deformation or re-meshing to a MultiMesh is computationally cheaper and more robust than the traditional single mesh approach, since it is only applied on submeshes. To the best of the authors' knowledge, this is the first instance of a FEM with multiple overlapping meshes in the setting of shape optimization.

\begin{figure}[!ht]
  \centering
  \begin{tikzpicture}
    \node (sm) [above right, inner sep=0pt]{
      \includegraphics[width=\linewidth]{Falling_Ball.pdf}};
    \node (A) at (0.27\linewidth,-0.2) {(a)};
    \node (B) at (0.775\linewidth,-0.2) {(b)};
  \end{tikzpicture}

  \caption{A comparison of a moving object described with a standard mesh and
    with multiple meshes. In (a), the mesh is deformed with an
    Eikonal convection equation, combined with a centroidal Voronoi
    tessellation (CVT) smoothing~\cite{schmidt2014two}. The mesh quality,
    quantified by the minimum radius ratio decreases from $0.75$
    to $6\cdot 10^{-4}$, and the mesh is degenerated.
    In (b), a mesh describing the ball is
    introduced, and can be translated independent of the background mesh. Here
    the minimum radius ratio is constant at $0.72$.
    }\label{fig:Ball}
\end{figure}

The use of multiple meshes dates back to solving the problem of structure mesh generation for finite difference or structured finite volume schemes~\cite{benek1983flexible,henshaw2002overture,starius1977composite,volkov1970method}.
These many-mesh techniques (also known as Chimera or Overset techniques) overcome several limitations of structured grids, such as multiple holes and moving domains, making them  particularly popular for aerodynamic applications~\cite{stangl1996euler}.

These schemes have also been used in an optimization setting, with similar data-transfer for the adjoint equation~\cite{liao2006aerodynamic}.

A recent method for non-matching meshes for FEM is the
Cut Finite Element Method (CutFEM)~\cite{burman2015cutfem}.
This method uses a Nitsche based formulation to weakly enforce boundary
conditions at the interface.
CutFEM has been used for a wide range of shape and topology optimization problems, such as acoustics~\cite{bernlandacoustic}, elasticity~\cite{bandara2016shape,burman2018shape} and incompressible flow~\cite{villanueva2017cutfem}. The MultiMesh FEM~\cite{MMFEMarbit} is a generalization of the CutFEM, where the computational domain is described by arbitrary many overlapping non-matching meshes,
coupled with Nitsche's method. The MultiMesh FEM has been explored for the Poisson and Stokes-equations~\cite{hansen2016simulation, johansson2013high, johansson2015high}.
Other methods used for shape optimization of complex computational domains are available, see for example~\cite{NAJAFI20151, noel2017shape, van2005generalized} and the references therein.

The MultiMesh FEM introduces several interesting
aspects when applied to a shape optimization problem.
When creating a solution algorithm for the optimization problem, a choice has to
be made, namely, should one use the \emph{first optimize, then discretize approach} or
the first discretize, then optimize approach. An initial analysis of these approaches has been investigated with respect to shape optimization problems~\cite{berggren}. However, there is no general recipe for
which method is to be preferred~\cite{hinze2008optimization}.

Due to generality and independence of the FEM framework, we have chosen the optimize then discretize approach. However, the authors plan to investigate the effects of a discretize then optimize strategy in a later publication.
Numerical examples show that the numerical inconsistency in the shape gradient is insignificant for fine meshes.

The remainder of this paper is organized as follows. \cref{sec:math} introduces the mathematical notation and presents the new algorithm for solving shape optimization on multiple domains.
\cref{sec:MMFEM} gives a brief introduction solving PDEs on multiple meshes with the MultiMesh FEM.
In \cref{sec:shape}, we then derive shape derivatives using shape calculus, and the associated adjoint equations.
\cref{sec:updatemesh} discusses the optimization step and the mesh updating strategies.
Thereafter, we present several numerical examples in \cref{sec:numerical_examples} and compare the new approach to a traditional FEM when feasible. Finally, we summarize and draw conclusions in \cref{sec:conc}.


text:
\begin{document}

\title{Reconfigurable Intelligent Surfaces for EnergyEfficiency in D2D Communication Network}

\author{Shuaiqi Jia, Xiaojun Yuan, \IEEEmembership{Senior Member, IEEE}, and Ying-Chang Liang, \IEEEmembership{Fellow, IEEE}
\vspace{-.2cm}
\thanks{The authors are with the National Key Laboratory of Science and Technology on Communications and also with the Center for Intelligent Networking and Communications, University of Electronic Science and Technology of China, Chengdu 611731, China (e-mail: sqjia@std.uestc.edu.cn;xjyuan@uestc.edu.cn;liangyc@ieee.org).}
}

\maketitle

\begin{abstract}
In this letter, the joint power control of D2D users and the passive beamforming of reconfigurable intelligent surfaces (RIS) for a RIS-aided device-to-device (D2D) communication network is investigated to maximize energy efficiency. This non-convex optimization problem is divided into two subproblems, which are passive beamforming and power control. The two subproblems are optimized alternately. We first decouple the passive beamforming at RIS based on the Lagrangian dual transform. This problem is solved by using fractional programming. Then we optimize the power control by using the Dinkelbach method. By iteratively solving the two subproblems, we obtain a suboptimal solution for the joint optimization problem. Numerical results have verified the effectiveness of the proposed algorithm, which can significantly improve the energy efficiency of the D2D network.
\end{abstract}

\begin{IEEEkeywords}
D2D communication, power control, reconfigurable intelligent surface, passive beamforming. \vspace{-3mm}
\end{IEEEkeywords}

\section{Introduction}
\IEEEPARstart{E}{nergy} efficiency has emerged as a key performance indicator in the design of wireless communication systems, due to operational, economic, and environmental concerns \cite{buzzi2016survey}. Device-to-device (D2D) communication has been proposed as a promising technology to increase energy efficiency (EE), which allows new peer-to-peer communication service by leveraging the proximity and reuse gains \cite{jiang2015energy}.
\par
Reconfigurable intelligent surface (RIS) is designed to improve the propagation environment and enhance wireless communications, and has attracted much attention recently \cite{8959174,8941126,huang2019reconfigurable}. Specifically, RIS is a meta-surface with a large number of passive reflecting elements. Each element can induce a phase shift to the incident signal independently. All elements cooperatively perform the reflect beamforming. Unlike the conventional amplify-and-forward (AF) relay, the RIS forwards the incident signals using passive reflection beamforming, which reduces the energy consumption of the system. These characteristics make the RIS technology attractive from an energy efficiency standpoint \cite{huang2019reconfigurable}.
\par
In this letter, we propose a RIS-assisted D2D communication network with a full-frequency reuse to support D2D users, which is assisted by some RISs. The EE maximization problem is formulated to optimize the RIS phase shifts and the transmit powers under phase-shift, maximum power and minimum transmission rate constraints. We divide the non-convex problem into two subproblems, which are passive beamforming and power control. The two subproblems can be optimized alternately. We first rewrite the passive beamforming at RIS based on the Lagrangian dual transform \cite{shen2018fractional}, and solve this subproblem based on the fractional programming \cite{shen2018fractional}. Then, we transform the power control subproblem in a fractional form into an equivalent optimization problem in a subtractive form, which is solvable by the Dinkelbach method \cite{dinkelbach1967nonlinear}. The two subproblems are solved iteratively until convergence, yielding a suboptimal solution for the joint optimization problem. Numerical results have verified the effectiveness of the proposed algorithm, which can significantly improve the energy efficiency of the D2D network.

\section{System Model}

In this letter, we consider the RIS-aided D2D communication network with full-frequency reuse to support D2D users. Without loss of generality, we assume that there are $L$ D2D links in the network, assisted by geographically separated $M$ RISs, where the $m$-th RIS, denoted by $\textrm{RIS}_{m}$, has $N_{m}$ elements. We assume that the power of the signals reflected by the RIS for two or more times can be ignored due to significant path loss. Besides, we assume that all the channels experience quasi-static flat-fading, and the channel state information (CSI) can be estimated by using the existing channel estimation methods, such as in \cite{8879620}.
\begin{comment}
\begin{figure}[H]
    \centering
    \includegraphics[scale=0.6]{1.png}
    \caption{RIS aided D2D communication}
    \label{4}

\end{figure}
\end{comment}
\par
Denote by $h_{ij}\in \mathbb{C}$ the channel coefficient from transmitter $i$ to receiver $j$. The channel coefficient vectors from transmitter $i$ to $\textrm{RIS}_{m}$ and from receiver $j$ to $\textrm{RIS}_{m}$ are denoted by $\mbf{g}_{i}^{m}\in \mathbb{C}^{ N_{m}\times 1}$ and $\mbf{f}_{j}^{m}\in \mathbb{C}^{ N_{m}\times 1}$, respectively. Let $N=\sum_{m=1}^{M}N_m$ be the total number of reflecting elements. The channel coefficient vectors from transmitter $i$ to RISs and from receiever $j$ to RISs are denoted by $\mbf{g}_{i}\in \mathbb{C}^{ N\times 1}$ and $\mbf{f}_{j}\in \mathbb{C}^{ N\times 1}$, respectively. A simple scenario with one RIS and two D2D pairs is shown in Fig. \ref{f1}.
\begin{figure}[h]
    \centering
    \includegraphics[scale=0.7]{1.pdf}
    \caption{The RIS-aided D2D communication network with one RIS and two D2D pairs.}
    \label{f1}

\end{figure}
\par
The phase-shift matrix of the RISs is denoted by a diagonal matrix $\mbf{\Theta}=\sqrt{\eta} \operatorname{diag}\left(\boldsymbol{\theta}\right)$, where $\boldsymbol{\theta}=[\theta_{1}, \cdots, \theta_{N}]^{H}$, $|\theta_{n}|=1$ and $\eta \in (0,1)$.
\par
Denote by $x_{i}$ the transmit data symbol to user $i$. The received signal of receiver $l$ is given by
\begin{equation}
    y_{l}=\sum_{i=1}^{L} (h_{il}+f_{l}^{T}\mathbf{\Theta}g_{i}) x_{i}+w_{l},
\end{equation}
where $w_{l} \sim \mathcal{C} \mathcal{N}\left(0, \sigma^{2}\right)$ is the additive white Gaussian noise (AWGN) at the $l$-th receiver. Denote by $p_{l}$ as the transmit power of D2D link $l$. The signal-to-interference-plus-noise ratio (SINR) of link $l$ is expressed as
\begin{equation}
z_{l}=\frac{|h_{l l}+ f_{l}^{T}\mathbf{\Theta}g_{l}|^{2}p_{l}} {\sum_{i \neq l,i=1}^{L} |h_{i l}+f_{l}^{T}\mathbf{\Theta}g_{i}|^{2} p_{i}+\sigma^{2}}.
\end{equation}
Then, the achievable sum rate of the system is given by
\begin{equation}
\mathcal{R} = \sum_{l=1}^{L} \log _{2}\left(1+z_{l}\right).
\end{equation}
\par
The total power consumption contains D2D users' transmit power $\mbf{p}$, circuit power $P_{c}$ and the RISs' power. The RISs' power consumption depends on the type and the resolution of its individual elements \cite{huang2019reconfigurable}. Elements with $b$-bit resolution can perform $b$-bit phase shifting on the impinging signal. The power consumption of the RISs with $N$ reflecting elements can be written as
\begin{equation}
    P_{\mathrm{RIS}}=N P(b),
\end{equation}
where $P(b)$ denotes the power consumption of each element having $b$-bit resolution. Thus, the toal power consumption can be expressed as
\begin{equation}
\mathcal{P}_{\text {total }}=\sum_{l=1}^{L}\left( p_{l}+2P_{c}\right)+NP(b).
\end{equation}
\par
We define the ratio between the achievable sum rate and the total power consumption as the energy efficiency (EE). The energy efficiency is maximized by jointly optimizing the transmit power $\mbf{p}$ and the phase-shift matirx $\mbf{\Theta}$, subject to RIS’s phase-shift, maximum power and minimum transmission rate constraints. Thus, the problem can be formulated as
\begin{subequations} \label{P1}
\begin{align}
 \max_{\mbf{p}, \mbf{\Theta}} \quad & {\mathrm{EE}}=\frac{ \sum_{l=1}^{L} \log _{2}\left(1+z_{l}\right)}{\sum_{l=1}^{L} p_{l}+2L P_{c}+N P(b)} & \\
\mbox{s.t.}\quad
&\log _{2}\left(1+z_{l}\right) \geq R_{\min , l} \quad \forall l=1,2, \ldots, L  & \\
&0 \leq p_{l} \leq P_{\max } \quad \forall l=1,2, \ldots, L\\
& {\left|\theta_{n}\right|=1 \quad \forall n=1,2, \ldots, N}.&
\end{align}
\end{subequations}
It is difficult to obtain the optimal solution to this problem due to its nonconvex objective function and constant-modulus constraints. In the next section, an efficient algorithm is proposed to obtain a suboptimal solution to this problem.

\section{Energy Efficiency Maximization} \label{section3}
In this section, we adress the energy efficiency maximization problem of the RIS-aided D2D system. We divide the problem (6) into two subproblems, i.e., the passive beamforming and the power control, respectively. Then, the power control and passive beamforming can be optimized alternately.
\subsection{Optimizing $\boldsymbol{\Theta}$ for Given $\mbf{p}$}
For a fixed $\mbf{p}$, the problem (\ref{P1}) reduces to
\begin{align} \label{P2}
\max_{ \boldsymbol{\Theta}} \quad &~  \sum_{l=1}^{L} \log _{2}\left(1+z_{l}\right) \notag \\
\mbox{s.t.}\quad &(6b),(6d).
\end{align}
\par
We apply the the Lagrangian dual transform \cite{guo2019weighted} to (\ref{P2}), with the new objective function given by
\begin{equation}
\begin{aligned}
f(\boldsymbol{\Theta},\boldsymbol{\beta})=\sum_{l=1}^{L} \log _{2}&~\left(1+\beta_{l}\right) -\sum_{l=1}^{L} \beta_{l} \\
&+\sum_{l=1}^{L} \frac{(1+\beta_{l})|h_{l l}+ f_{l}^{T}\mathbf{\Theta}g_{l}|^{2}p_{l}}{\sum_{i=1}^{L} |h_{i l}+f_{l}^{T}\mathbf{\Theta}g_{i}|^{2} p_{i}+\sigma^{2}},
\end{aligned}
\end{equation}
where $\boldsymbol{\beta}$ refers to a set of auxiliary variables. We propose to optimize $\boldsymbol{\beta}$ and $\boldsymbol{\Theta}$ alternately. For a fixed $\boldsymbol{\Theta}$, the optimal $\beta_{l}$ is
\begin{equation}
    \beta_{l}^{\circ}=\frac{|h_{l l}+ f_{l}^{T}\mathbf{\Theta}g_{l}|^{2}p_{l}} {\sum_{i \neq l,i=1}^{L} |h_{i l}+f_{l}^{T}\mathbf{\Theta}g_{i}|^{2} p_{i}+\sigma^{2}}.
\end{equation}
Then, for a fixed $\boldsymbol{\beta}$, optimizing $\boldsymbol{\Theta}$ is reduced to
\begin{align} \label{P3}
\max_{ \boldsymbol{\Theta}} \quad &  f_{1}(\boldsymbol{\Theta})=\sum_{l=1}^{L}\frac{(1+\beta_{l})|h_{l l}+ f_{l}^{T}\mathbf{\Theta}g_{l}|^{2}p_{l}}{\sum_{i=1}^{L} |h_{i l}+f_{l}^{T}\mathbf{\Theta}g_{i}|^{2} p_{i}+\sigma^{2}}& \notag \\
\mbox{s.t.}\quad
&(6b),(6d).&
\end{align}

\par
The problem (\ref{P3}) can be solved by using the FP method. Define $\mathbf{A}_{il}=diag(f_{l})g_{i}\sqrt{p_{i}}$ and $b_{il}=h_{il}\sqrt{p_{i}}$. Using the quadratic transform proposed in \cite{shen2018fractional}, $f_{1}(\boldsymbol{\Theta})$ is reformulated as
\begin{equation} \label{obj1}
\begin{aligned}
f_{2}(\boldsymbol{\theta},\boldsymbol{\varepsilon})=\sum_{l=1}^{L} 2\sqrt{1+\beta_{l}}&~\mathbf{Re}\{
\varepsilon_{l}^{*}\boldsymbol{\theta^{\mathbf{H}}}\mathbf{A}_{ll}+\varepsilon_{l}^{*}b_{ll}
\}\\
&-\sum_{l=1}^{L}|\varepsilon_{l}|^{2} (\sum_{i=1}^{L}|b_{il}+\boldsymbol{\theta^{\mathbf{H}}}\mathbf{A}_{il}|^{2}+\sigma^{2}).
\end{aligned}
\end{equation}
\par
We optimize $\boldsymbol{\theta}$ and $\boldsymbol{\varepsilon}$ alternately. The optimal $\varepsilon_{l}$ can be obtained by letting $\partial f_{2}/ \partial \varepsilon_{l}=0$, yielding
\begin{equation}
    \varepsilon_{l}^{\circ}=\frac{\sqrt{1+\beta_{l}}(b_{ll}+\boldsymbol{\theta^{\mathbf{H}}}\mathbf{A}_{ll})}{\sum_{i=1}^{L}|b_{il}+\boldsymbol{\theta^{\mathbf{H}}}\mathbf{A}_{il}|^{2}+\sigma^{2}}.
\end{equation}
\par
Then the remaining problem is to optimize $\boldsymbol{\theta}$ for a given $\boldsymbol{\varepsilon}$. By simplifying (\ref{obj1}), the optimization problem for $\boldsymbol{\theta}$ is represented as
\begin{align} \label{P4}
 \max_{ \boldsymbol{\theta}} \quad &  f_{3}(\boldsymbol{\theta})= -\boldsymbol{\theta}^\mathbf{H}\mathbf{U}\boldsymbol{\theta}+2\textbf{Re}\{\boldsymbol{\theta}^\mathbf{H}\mbf{v} \}+C \notag \\
\mbox{s.t.}\quad
&(6b),(6d),&
\end{align}

where
\begin{equation}
\mathbf{U}=\sum_{l=1}^{l}\left|\varepsilon_{l}\right|^{2} \sum_{i=1}^{L} \mathbf{A}_{il} \mathbf{A}_{il}^{\mathrm{H}}
\end{equation}
\begin{equation}
\mbf{v}=\sum_{l=1}^{L}\left(\sqrt{\beta_{l}+1} \varepsilon_{l}^{*} \mathbf{A}_{ll}-\left|\varepsilon_{l}\right|^{2} \sum_{i=1}^{l} b_{il}^{*} \mathbf{A}_{il}\right),
\end{equation}
\begin{comment}
\begin{equation}
C=\sum_{l=1}^{L}\left(2 \sqrt{\beta_{l}+1} \operatorname{Re}\left\{\varepsilon_{l}^{*} b_{ll}\right\}-\left|\varepsilon_{l}\right|^{2}\left(\sigma^{2}+\sum_{i=1}^{L}\left|b_{il}\right|^{2}\right)\right)
\end{equation}
\end{comment}
and $C$ is a constant. Since $\mbf{A}_{il}\mbf{A}_{il}^{\mathrm{H}}$ for all $i$ and $l$ are positive-definite matrices, $\mathbf{U}$ is a positive-definite matrix and $f_{3}(\boldsymbol{\theta})$ is a quadratic concave functions of $\boldsymbol{\theta}$. Problem (\ref{P4}) is non-convex and inhomogeneous due to the non-convexity of the quadratic constraints and the unit modulus constraints. Letting $\boldsymbol{\bar{\theta}^{\mbf{H}}}=[\boldsymbol{\theta^\mbf{H}}, 1]$ and $\mbf{Q}=\boldsymbol{\bar{\theta}}\boldsymbol{\bar{\theta}^\mbf{H}}$, and dropping the rank-one constraint of $\mathbf{Q}$, problem (\ref{P4}) can be rewritten as
\begin{subequations} \label{P5}
\begin{align}
\max_{ \mathbf{Q}} \quad &  \operatorname{Tr}(\mathbf{\bar{U}} \mathbf{Q}) & \\
\mbox{s.t.}\quad
&\gamma_{l}^{\min }\left(\sum_{i \neq l}^{L}\left(\operatorname{Tr}\left(\mbf{R}_{i, l} \mathbf{Q}\right)+\left|b_{il}\right|^{2}\right)+\sigma^{2}\right) &\notag\\
&\leq \operatorname{Tr}\left(\mbf{R}_{l, l} \mathbf{Q}\right)+\left|b_{ll}\right|^{2}, \forall l, l=1, \ldots, L& \\
& \mathbf{Q}_{n, n}=1, \forall n=1, \ldots, N+1&\\
&\mathbf{Q} \succcurlyeq 0,&
\end{align}
\end{subequations}
where,
\par
$\mbf{R}_{i, l}=\left[\begin{array}{cc}
\mbf{A}_{il} \mbf{A}_{il}^{\mathrm{H}} & \mbf{A}_{il} b_{il}^{*} \\
b_{il} \mbf{A}_{il}^{\mathrm{H}} & 0
\end{array}\right]$, $\mathbf{\bar{U}}=\left[\begin{array}{cc}
 \mathbf{U}& \mbf{-v}\\
\mbf{-v}^{\mathrm{H}} & 0
\end{array}\right]$.
\par
We can solve problem (\ref{P5}) via CVX \cite{cvx}. Then the standard Gaussian randomization can be used to obtain a feasible rank-one solution. We summarize the optimization method for $\mathbf{\Theta}$ with a fixed $\mbf{p}$ in Algorithm 1.
\begin{algorithm}[H]
\vspace{0.2mm}
\caption{\textbf{\!:} \vspace{0.15mm} Optimizing $\boldsymbol{\Theta}$ for Given $\mbf{p}$ }
\textbf{\,Initialization:} Initialize $\boldsymbol{\Theta}^{(0)}$ to a feasible value, $i=0$.
\begin{algorithmic}[1]

\State Update the auxiliary variable $\beta_{l}^{i}$ by (9).
\\ Update the auxiliary variable $\epsilon_{l}^{i}$ by (12).
\\Update $\boldsymbol{\Theta}^{(i)}$ by solving (\ref{P5}) together with Gaussian randomization.
\\Repeat setps 1-3 until the value of $\boldsymbol{\Theta}$ converges.
\end{algorithmic}
\textbf{\,Output:} $\boldsymbol{\Theta}^{(i)}$
\end{algorithm}

\subsection{Optimizing $\mbf{p}$ for Given $\boldsymbol{\Theta}$}
For a fixed $\boldsymbol{\theta}$, the problem (\ref{P1}) reduces to
\begin{align} \label{P6}
\max_{\mbf{p}} \quad&\frac{ \sum_{l=1}^{L} \log _{2}\left(1+z_{l}\right)}{\sum_{l=1}^{L} p_{l}+2L P_{c}+N P(b)} & \notag \\
\mbox{s.t.}\quad
&(6b),(6c).&
\end{align}
Problem (\ref{P6}) is a non-concave FP problem. From \cite{dinkelbach1967nonlinear}, we define a new optimization problem as
\begin{align} \label{P7}
\max_{\mbf{p}} \quad&~F(\lambda) & \notag \\
\mbox{s.t.}\quad
&(6b),(6c),&
\end{align}
where $\lambda$ is a non-negative parameter, and
\begin{equation} \label{obj2}
  F(\lambda)={\sum_{l=1}^{L} \log _{2}\left(1+z_{l}\right)}-\lambda({\sum_{l=1}^{L} p_{l}+2L P_{c}+N P(b)}).
\end{equation}
$F(\lambda)$ is continuous and strictly monotonically decreasing in $\lambda$ and has a unique root $\lambda^{*}$. The optimal solution $\mbf{p^{*}}$ of problem (\ref{P6}) is the same as that of problem (\ref{P7}) with $\lambda=\lambda^{*}$, where $\lambda^{*}$ can be obtained by using the Dinkelbach method \cite{dinkelbach1967nonlinear}.
\par
The key of the method is to solve problem (\ref{P7}) for a given $\lambda$. To be specific, (\ref{obj2}) can be rewritten as
\begin{equation}
    F(\lambda)=f_{1}(\mbf{p})-f_{2}(\mbf{p})
\end{equation}
where
\begin{equation}
    \begin{aligned}
    f_{1}(\mbf{p})=\sum_{l=1}^{L}\log_{2}(\sum_{i=1}^{L}&~|h_{il}+f_{l}\boldsymbol{\Theta}g_{i}|^{2}p_{i}+\sigma^{2})\\
    &-\lambda \sum_{i=1}^{L}p_{i}-\lambda(2LP_{c}+NP(b))
    \end{aligned}
\end{equation}
\begin{equation}
    f_{2}(\mbf{p})=\sum_{l=1}^{L}\log_{2}(\sum_{i=1,i\neq l}^{L}|h_{il}+f_{l}\boldsymbol{\Theta}g_{i}|^{2}p_{i}+\sigma^{2}).
\end{equation}
Clearly, $f_{1}(\mbf{p})$ and $f_{2}(\mbf{p})$ are concave. Hence, the objective function of (\ref{P7}) is the difference of two concave functions. Constraints in (6b) are also the difference of two concave functions, which are
\begin{equation}
    c_{1,l}(\mbf{p})-c_{2,l}(\mbf{p}) \geq R_{\min , l} \quad \forall l=1,2, \ldots, L
\end{equation}
where
\begin{align}
   c_{1,l}(\mbf{p})=\log_{2}(\sum_{i=1}^{L}|h_{il}+f_{l}\boldsymbol{\Theta}g_{i}|^{2}p_{i}+\sigma^{2})
\end{align}
\begin{align}
     c_{2,l}(\mbf{p})=\log_{2}(\sum_{i=1,i\neq l}^{L}|h_{il}+f_{l}\boldsymbol{\Theta}g_{i}|^{2}p_{i}+\sigma^{2}).
\end{align}
\par
Consequently, problem (\ref{P7}) is a difference of convex (DC) programming problem. We apply the DC algorithm \cite{tao1997convex} to obtain a suboptimal solution of problem (18).
\begin{comment}
We summarize the optimization method for $\mbf{\Theta}$ with a fixed $\mbf{p}$ in Algorithm 2.
\begin{algorithm}[H]
\vspace{0.2mm}
\caption{\textbf{\!:} \vspace{0.15mm} Optimizing $\mbf{p}$ for Given $\boldsymbol{\theta}$}
\textbf{\,Initialization:} Set tolerance $\epsilon$, $s=1$, $\lambda_{s}=0$.
\begin{algorithmic}[1]

\State Obtain $\mbf{p}_{s}$ by solving problem (18) with DCA.\\
Update $\lambda_{s}$ by (6a).\\
If $|F(\lambda_{s})| \geq \epsilon$, set $s=s+1$.\\
Repeat steps 1-3 until $F(q_{s})| < \epsilon$.
\end{algorithmic}
\textbf{\,Output:} $\mbf{p}_{s}$
\end{algorithm}
\end{comment}

\section{Numerical Result}
In this section, numerical results are provided to demonstrate the validity of the proposed algorithm. We consider a RIS-aided D2D network with multiple single-antenna D2D users, which are assumed randomly and uniformly placed in the $200m \times 200m$ rectangular. And there are four RISs in the rectangular. The distance between D2D users is 20-40m. The baseband channels of the RIS-user link, user-RIS link, and user-user link are modeled as Rician fading channels with Rician factor $\beta =2$. The background noise at the receivers is $\sigma^{2}=-117$ dBm. We set the path-loss constant $k=10^{-3}$, the path-loss exponent $\chi=4$. Circuit power at each user $P_{C}=15$ dBm. RIS reflection efficiency $\eta=0.8$. All presented results are obtained by averaging over 1000 independent channel realizations.
\par
We compare the performance of the proposed algorithm with two baselines. Baseline 1 is the energy efficiency optimized by power control without the aid of RIS. Baseline 2 is the energy efficiency optimized by optimal power control with the aid of RISs. We adopt the Branch and Bound method to obtain the optimal power control \cite{yang2016energy}. Fig. 2 illustrates the energy efficiency of different power parameters $P(b)$ with respect to the size of $N$ of RIS. As we can see, significant performance gains are achieved by joint power control and RISs' passive beamforming optimization. Typical power consumption values of each RIS element are 1.5, 4.5, 6, and 7.8 mW for 3-, 4-, 5-, and 6-bit resolution \cite{ribeiro2018energy}. As the number of quantized bits increases, the energy efficiency decreases gradually. Although high-bit RIS will improve the spectral efficiency of the network, its high power consumption leads to the decrease of energy efficiency.
\begin{figure}[h]
    \centering
    \includegraphics[scale=0.5]{4.pdf}
    \caption{The energy efficiency versus the number of RIS elements for  $L=10$, $P_{max}=0.1$ W, $R_{min}=0$ bps/Hz}
    \label{4}

\end{figure}
\par
Particularly, EE performance increases as $N$ increases. However, for a large $N$, EE starts decreasing. This is because for a large $N$, RISs' improvement in spectrum efficiency is no longer enough to compensate for the decrease in system energy efficiency caused by high power consumption. So, there exists an optimal number $N$ of RIS elements. Besides, the energy efficiency tends to decrease firstly and then increase. This is because when $N$ is too small, there will be little improvement in spectral efficiency with the assistance of the RISs. However, the power consumption of the RISs results in the decrease of energy efficiency.

\begin{figure}[t]

   \centering
    \includegraphics[width=4.5cm,height=4.5cm]{5.pdf}\,\includegraphics[width=4.5cm,height=4.5cm]{6.pdf}\\
    \vspace{-2.5mm}
    \caption{(a) The energy efficiency versus $P_{max}$ for $L=10$, $N=80$; (b) The failure rate versus $P_{max}$ for $L=10$, $N=80$.}
    \label{fig1}

\end{figure}
\par
The effect of the different values of $R_{min}$ and EE performances versus $P_{max}$ in dBm is depicted in Fig. 3. For the cases where the design problems turned out to be infeasible, EE is set to zero, corresponding to a communication failure. Fig. 3(a) compares the EE with the different minimum rate $R_{min}$. When $R_{min}=0$, EE increases to a fixed value as the power limit $P_{max}$ increases. For the higher $R_{min}$, EE will increase to a lower fixed value. Fig. 3(b) compares the failure rate with the different minimum rate $R_{min}$. Similarly, for the higher $R_{min}$, the failure rate will reduce to a higher fixed value. At this moment, increasing $P_{max}$ does not improve efficiency or reduce failure rate.
\section{Conclusion}
In this letter, the RIS technique was applied to enhance the energy efficiency of D2D communication network. To maximize the energy efficiency, we proposed a joint power control and RISs' passive beamforming optimization algorithm for obtaining the high-quality suboptimal solution. Simulation results demonstrated that the assistance of the RISs is beneficial to substantially improve the energy efficiency of the D2D communication network.

\renewcommand\refname{References}
\bibliographystyle{IEEEtran}
\bibliography{bib-he}

\begin{comment}


text:
\begin{document}

\title{A Sauer-Shelah-Perles Lemma for Sumsets}

\author{Zeev Dvir\thanks{Department of Computer Science and Department of Mathematics,
Princeton University.
Email: \texttt{zeev.dvir@gmail.com}. Research supported by NSF CAREER award DMS-1451191 and NSF grant CCF-1523816 } \and
Shay Moran\thanks{School of Mathematics, IAS. Email: \texttt{shaymoran1@gmail.com}.
Research supported by the National Science Foundation under agreement No. CCF-1412958 and by the Simons Foundations.}}

\date{}
\maketitle

\begin{abstract}
We show that any family of subsets  $A\subseteq 2^{[n]}$
satisfies $\lvert A\rvert \leq O\bigl(n^{\lceil{d}/{2}\rceil}\bigr)$,
where $d$ is the VC dimension of $\{S\triangle T \,\vert\, S,T\in A\}$, and $\triangle$ is the symmetric difference operator.
We also observe that replacing $\symdif$ by either $\cup$ or $\cap$ fails to satisfy an analogous statement.
Our proof is based on the polynomial method; specifically, on an argument due to [Croot, Lev, Pach '17].
\end{abstract}

\section{Introduction}

Let $A \subset 2^{[n]}$ be a family of subsets of an $n$ element set ($[n]$ w.l.o.g).
The VC dimension of $A$, denoted by $\VC(A)$, is the size of the largest $Y\subseteq [n]$
such that $\{S\cap Y \,\vert\, S\in A\} = 2^{Y}$.
One of the most useful facts about the VC dimension is given by the Sauer-Shelah-Perles Lemma.
\begin{thm}[Sauer-Shelah-Perles Lemma \citep{sauer72density,shelah72comb}]\label{thm:sauer}
Let $d\leq n\in\N$.
Suppose $A \subset 2^{[n]}$ satisfies $\VC(A) \leq d$. Then $\lvert A\rvert \leq {n \choose \leq d}$.
\end{thm}
The Sauer-Shelah-Perles Lemma has numerous applications ranging from model theory, probability theory,
geometry, combinatorics, and various fields in computer science.
A simple-yet-useful corollary of this lemma is that if $\VC(A) \leq d$, and $\star$ is any binary set-operation (e.g.\ $\star\in\{\cap,\cup,\symdif\}$)
then
\[\bigl\lvert\{ S\star T \,\vert\, S,T\in A\}\bigr\rvert \leq  {n \choose \leq d}\cdot{n \choose \leq d} = O(n^{2d}).\]
This corollary is used, for example, by~\cite{blumer89VC} to derive closure properties for {\it PAC learnability}.
Let~$A{\circled{$\star$}} A$ denote the family $\{ S\star T \,\vert\, S,T\in A\}$.
In this work we explore the converse direction:
Does an upper bound on the VC-dimension $\VC(A\circled{$\star$} A)$ imply an upper bound on~$\lvert A\rvert$?
It is not hard to see that  $\VC(A)\leq\VC(A\circled{$\star$} A)$ for $\star\in\{\cup,\cap,\symdif\}$,
and therefore, by \Cref{thm:sauer}: $\VC(A\circled{$\star$} A) < d\implies \lvert A\rvert\leq O(n^d)$.

Our main result quadratically improves this naive bound when $\star$ is symmetric difference:
\begin{thm}\label{thm-main}
Let $d\leq n\in\N$.
Suppose $A \subset 2^{[n]}$ satisfies $\VC(A \circled{$\symdif$} A) \leq d$.
Then
\[\lvert A\rvert \leq 2 {n \choose \leq \lfloor {d}/{2}\rfloor}.\]
\end{thm}

We note that \Cref{thm-main} does not hold when $\star\in\{\cup,\cap\}$: pick $d\geq 2$, and set
\[ A = \{ S \subseteq [n] \,\vert\, \lvert S\rvert \leq d\}.\]
Note that $A = A\circled{$\cap$} A $ and therefore $d =\VC(A)= \VC(A\circled{$\cap$} A)$.
However $\lvert A \rvert = {n \choose \leq d} = \Theta(n^d)$, which is not upper bounded by $O(n^{\lceil {d}/{2}\rceil})$.
Picking $A = \{ S \subseteq [n] \,\vert\, \lvert S\rvert \geq n-d\}$ shows that $\cup$ behaves similarly like $\cap$
in this context.

The above examples rules out the analog of \Cref{thm-main} for exactly one of $\cup,\cap$.
This suggests the following open question:
\begin{question}
Let $d\leq n\in\N$.
Suppose $A \subset 2^{[n]}$ satisfies $\VC(A\circled{$\cap$} A) \leq d$ and $\VC(A\circled{$\cup$} A) \leq d$.
Is it necessarily the case that $\lvert A\rvert  \leq n^{{d}/{2} + O(1)}$?
\end{question}

Another natural question is whether this phenomenon extends to several applications of the symmetric difference operator, for example:
\begin{question}
Does there exist an $\eps < 1/2$ such that for every $d\leq n$ and every $A\subset 2^{[n]}$:
\[ \VC\bigl(A\circled{$\triangle$} A \circled{$\symdif$} A\bigr)\leq d \implies \lvert A\rvert \leq n^{\eps \cdot d + O(1)}?\]
\end{question}
{In Section~\ref{sec:modp} we derive a related statement when $\symdif$ is replaced by addition modulo $p$ for a prime~$p$,
and the VC dimension is replaced by the interpolation degree (which is defined in the next section).}

\subsection{Interpolation degree}
Since our proof method is algebraic, it is convenient to view $A\subset 2^{[n]}$ as a subset of the $n$-dimensional vector space $\F_2^n$ over the field of two elements.  In this setting $A \circled{$\symdif$}A$ is the {\em sumset} of $A$, denoted~$A+A$.

Theorem~\ref{thm-main} will  follow from a stronger statement involving a quantity referred to in some places as the {\em regularity} (as a special case of Castelnuovo-Mumford regularity from algebraic geometry)~\citep{Remscrim16} and in other as the {\em interpolation-degree} \citep{moran16shattered}. We will use the more descriptive interpolation-degree for the rest of this paper. We begin with some preliminary notations and definitions.

Let $A \subset \F_2^n$.
It is a basic fact that for each function $f : \F_2^n \mapsto \F_2$
there exists a unique multilinear polynomial $P_f \in \F_2[x_1,\ldots,x_n]$
such that $f(a) = P_f(a)$ for all $a\in \F_2^n$ (existence is via simple interpolation and uniqueness follows from dimension counting).
For a partial function $f : A \mapsto \F_2$ there are  many  (precisely $2^{{2^n - |A|}}$)
multilinear polynomials whose restriction to $A$ computes $f$.
Let $\deg_A(f)$ denote the minimal degree of any polynomial whose restriction to $A$ computes $f$.
We define the {\em interpolation-degree} of $A$,
denoted $\intdeg(A)$ to be the maximum of $\deg_A(f)$ taken over all functions $f : A \mapsto \F_2$.
In other words, $\intdeg(A)$ is the smallest $d$ such that
any function from $A$ to $\F_2$ can be realized by a polynomial of degree at most $d$.
Clearly, $\intdeg(A)$ is an integer between $0$ and $n$.
It is also not hard to see that, if $A$ is a proper subset of ${\F_2^n}$ then $\intdeg(A) < n$.
Our interest in $\intdeg(A)$ comes from the following connection to VC-dimension.
\begin{lem}[\cite{babai92book,gurvits97linear,smolensky97vc,moran16shattered}]\label{thm:intdegvc}
For $A \subset \F_2^n$ we have $\intdeg(A) \leq \VC(A)$.
\end{lem}
This Lemma, under various formulations, was proved in several works.
The formulation that appears here can be found in \citep{moran16shattered}.
For completeness, we next sketch the proof:
since the set of all multilinear monomials (also those of degree larger than $\VC(A)$)
span the set of functions $f: A\to\F_2$,
it suffices to show that any monomial (when seen as an $A\to \F_2$ function)
can be represented a polynomial of degree at most $d=\VC(A)$.
The crucial observation is that if $x_S = \pi_{i\in S} x_i$ is a monomial of degree larger than $d$,
then $S$ is not shattered by $A$.
This means that there is a pattern $v:S\to\{0,1\}$ that does not appear
in any of the vectors in $A$ and therefore
\[ \Pi_{i\in S}(x_i + v_i + 1) =_{A} 0,\]
where ``$=_A$'' means equality as functions over $A$.
Now, expanding this product and rearranging the equation
yields a representation of $x_S$ as sum of monomials~$x_{S'}$,
where $S' \subset S$, which by induction can also be represented
by polynomials of degree at most $d$.

\Cref{thm:intdegvc} reduces \Cref{thm-main} to the following stronger statement that is proved in the next section.
\begin{thm}\label{thm-intdeg}
Let $d\leq n\in\N$, and let $A \subset \F_2^n$ satisfy $|A| > 2{n \choose \leq \lfloor{d}/{2} \rfloor}$. Then $\intdeg(A+A) > d$.
\end{thm}

\section{Proof of Theorem~\ref{thm-intdeg}}

The main technical tool will be a lemma of Croot-Lev-Pach \citep{CLP17} that was the main ingredient in the recent solution of the cap-set problem \citep{EG17} and has found many other applications since then (e.g., \citep{Green16, Soly18, DB17, Fox17} to name a few).

\begin{lem}[CLP lemma \citep{CLP17}]\label{lem-clp}
	Let $P \in \F_q[x_1,\ldots,x_n]$ be a polynomial of degree at most $d$ over  any finite field $\F_q$,
	and let $M$ denote the $q^n \times q^n$ matrix with entries $M_{x,y} = P(x+y)$ for $x,y \in \F_q^n$.
	Then $\rank(M) \leq 2\cdot m_{\lfloor d/2 \rfloor}(q,n),$
	where $m_{k}(q,n)$ denotes the number of monomials in $n$ variables $x_1,\ldots,x_n$
	such that each variable appears with individual degree at most $q-1$ and the total degree of the monomial is at most $k$.
\end{lem}

Specializing to our setting of $\F_2$ multilinear polynomials, we see that $m_k(2,n) = {n \choose \leq k}$ and so we conclude:

\begin{cor}\label{cor-clp}
	Let $P \in \F_2[x_1,\ldots,x_n]$ be a polynomial of degree at most $d$  and let $M$ be as in Lemma~\ref{lem-clp}. Then $\rank(M) \leq 2 {n \choose \leq \lfloor d/2 \rfloor}$.
\end{cor}

We are now ready to prove Theorem~\ref{thm-intdeg}.

\begin{proof}[Proof of Theorem~\ref{thm-intdeg}]
Suppose $A \subset \F_2^n$ is such that $|A| \geq 2{n \choose \leq \lfloor{d}/{2}\rfloor}$. Let $f : A+A \mapsto \F_2$ be such that $f(\bar 0)=1$,
where $\bar 0$ is the all zero vector in $\F_2^n$, and $f(a)=0$ for all non-zero $a \in A+A$.
It suffices to show that $deg_{A+A}(f)\geq \lfloor{d}/{2}\rfloor$
({notice that since $A\neq\emptyset$} it follows that $\bar 0\in A+A$ and so~$f$ is not constantly $0$ on $A+A$).
Let $M$ be the $2^n \times 2^n$ matrix whose rows and columns are indexed by $\F_2^n$ and with entries $M_{x,y} = f(x+y)$.
By our definition of $f$ we have that the sub-matrix of $M$ whose rows and columns are indexed by $A$ is just the $|A|\times |A|$ identity matrix. This implies $$ \rank(M) \geq |A|.$$ Let $d_f = \deg_{A+A}(f)$ denote the smallest degree of a polynomial whose restriction to $A+A$ computes~$f$. Applying Corollary~\ref{cor-clp} we get that
\[ \rank(M) \leq 2{ n \choose \leq \lfloor d_f/2 \rfloor}.\]
Combining the two inequalities on $\rank(M)$ and using the bound on the size of $A$ we get that
\[2{n \choose \leq \lfloor{d}/{2}\rfloor} < |A| \leq \rank(M) \leq 2{ n \choose \leq \lfloor d_f/2 \rfloor}, \]
which implies $\lfloor d/2 \rfloor < \lfloor d_f/2 \rfloor$. This means that $d_f > d$ and so $\intdeg(A+A) > d$.
\end{proof}

\section{Generalization to sums modulo $p$}\label{sec:modp}

In this section we observe that our proof can be generalized to give stronger bounds in the case when we take $p$-fold sums of boolean vectors over $\F_p$. The case proved in the last section corresponds to (two fold) sums modulo 2. For a subset $A \subset \F_p^n$ and a positive integer $k$, we denote by $$k\cdot A = \{ a_1 + \ldots + a_k \,|\, a_i \in A \}$$ the $k$-fold sumset of $A$. To formally define the interpolation degree over $\F_p$ we need to consider, instead of multilinear polynomials, polynomials in which each variable has degree at most $p-1$. We call such polynomials {\em $p$-reduced} polynomials. The space of all $p$-reduced polynomials has dimension $p^n$ and  can uniquely represent any function $f: \F_p^n \mapsto \F_p$. The degree of such a function is defined to be the total degree of the unique $p$-reduced polynomial representing it and can range between 0 and $(p-1)n$. The interpolation degree of a set $A \subset \F_q^n$ is the minimum $d$ such that any function $f : A \mapsto \F_p$ can be represented by a $p$-reduced polynomial of degree at most $d$. To avoid confusion we will denote the interpolation degree over $\F_p^n$ as $\intdeg_p(A)$.

We denote by $\cM_d(p,n)$ the set of monomials in $n$ variables $x_1,\ldots,x_n$ in which each variables has degree at most $p-1$ and the total degree is at most $d$. When $p=2$ we have the closed formula $|\cM_d(2,n)| = {n \choose \leq d}$. When $p >2$ the quantity $|\cM_d(p,n)|$ is a bit more tricky to compute but is known to satisfy certain asymptotic inequalities (e.g., large deviations \citep{LDP15} showing that $\cM_{\delta n}(p,n) \leq 2^{\eps n}$ with $\eps(\delta)$ going to zero with $\delta$).

The following theorem generalizes Theorem~\ref{thm-main} when $p > 2$.
\begin{thm}\label{thm-psums}
Let $p$ be any prime number and let $A \subset \{0,1\}^n \subset \F_p^n$	be such that $|A| > p\cdot |\cM_{\lfloor d/p \rfloor}(p,n)|$. Then $\intdeg_p(p \cdot A) > d$.
\end{thm}

The proof of the theorem requires the notion of {\em slice-rank} of a tensor which was introduced by Tao in his symmetric interpretation of the proof of the cap-set conjecture \citep{Tao-capset}. By a $k$-fold tensor of dimension $D$ over a field $\F$ we mean a function $T$ mapping  ordered tuples $(j_1,\ldots,j_k) \in [D]^n$ to $\F$.  The {\em slice-rank} of a $k$-fold tensor $T$ is a the smallest integer $R$ such that $T$ can be written as a sum $T = \sum_{i=1}^R T_i$ such that, for every $i \in [R]$ there is some $j_i \in [k]$ so that $T_i(j_1,\ldots,j_k)= A(j_i)B(j_1,\ldots,j_{i-1},j_{i+1},\ldots,j_k)$. In other words, we define the `rank one' tensors to be those in which the dependence on one of the variables is multiplicative (by a function $A(j_i)$) and the rank of a tensor is the smallest number of rank one tensors needed to describe it. For 2-fold tensors (or matrices) this notion coincides with the usual definition of matrix rank.

The proof of Theorem~\ref{thm-psums} will follow from a combination of two lemmas regarding slice rank. The first lemma generalizes the Croot-Lev-Pach lemma (and proved in an a similar way).

\begin{lem}\label{lem-clpmodp}
Let $f : \F_p^n \mapsto \F_p$ be of degree $d$. Then the $p$-fold $p^n$ dimensional tensor $T: (\F_p^n)^k \mapsto \F_p$ defined by $T(X^1,\ldots,X^p) = f(X^1 + \ldots +X^p)$ has slice rank at most $p\cdot \cM_{\lfloor d/p \rfloor}(p,n)$.
\end{lem}
\begin{proof}
Consider $T$ as a polynomial in $p$ groups of variables $X^i = (x^i_1,\ldots,x^i_n)$ with $i = 1,2,\ldots,p$. Since the degree of $f$ is $d$, the degree of $T$ as a polynomial will also be at most $d$. This means that, in each monomial of $T(X^1,\ldots,X^p) = f(X^1 + \ldots +X^p)$, the degree of at least one group of variables will be at most $\lfloor d/p \rfloor$. Grouping together  monomials according to which group has low degree (if there is more than one group take the one with lowest index) we can represent $T$ as a sum of $p$ tensors, each having rank at most $\cM_{\lfloor d/p \rfloor}(p,n)$. This completes the proof.
\end{proof}

The second lemma needed to prove Theorem~\ref{thm-psums} is due to Tao and shows that the 'diagonal' tensor has full rank.
\begin{lem}[\cite{Tao-capset}]\label{lem-taorank}
Let $\delta(j_1,\ldots,j_k): [D]^k \mapsto \F$ be defined as $\delta(j,j,\ldots,j) = 1$ for all $j$ and is zero otherwise. Then the slice rank of $\delta$ is equal to $D$.
\end{lem}

\begin{proof}[Proof of Theorem~\ref{thm-psums}]
To prove the bound on $\intdeg_p(p\cdot A)$ we describe a function $f : p\cdot A \mapsto \F_p$ that cannot be represented by a low degree polynomial. We take $f$ to be equal to $1$ on the zero vector and zero otherwise. We now consider the tensor $T(X^1,\ldots,X^p) = f(X^1 + \ldots +X^p)$ defined on $A^p$. Notice that, since $A \subset \{0,1\}^n$, the sum of $p$ of them is equal to zero iff all $p$ summands are identical. This implies that $T$ is the diagonal tensor $\delta$ of Lemma~\ref{lem-taorank} and hence has rank equal to $|A|$. On the other hand, if the degree of $f$ (over $p\cdot A$) is at most $d$ then, by Lemma~\ref{lem-clpmodp}, the tensor $T$ has rank at most $p\cdot \cM_{\lfloor d/p \rfloor}(p,n)$. Since we assume that $|A| > p\cdot \cM_{\lfloor d/p \rfloor}(p,n)$ this cannot happen and so $\intdeg_p(pA) > d$.
\end{proof}

\bibliographystyle{plainnat}
\bibliography{refs}

\end{document}


TITLE: Intuition behind the coefficients in Taylor expansion of a function
QUESTION [1 upvotes]: " For example, the best linear approximation for f(x) is
$$ f(x) \approx f(a) + f'(a) (x-a)$$

The linear approximation fits f(x) (shown in green below) with a line (shown in blue) through x=a that matches the slope of 'f' at a.
And, we could add higher order terms for a better approximation, as shown below,
$$ f(x) = f(a) + \frac{ f'(a)}{1} (x-a) + \frac{ f''(a) }{2} (x-a)^2 + \frac{ f''(a)}{6}..$$
What is the intuitive understanding of coefficients in the Taylor expansion?

REPLY [1 votes]: Right, so suppose you have a polynomial $P(x)$ of degree 'n', where
$$ P(x) = a_o + a_1 x + a_2 x^2 + a_3 x^3 ..a_n x^n = \sum a_k x^k$$
Now, I can rewrite this same polynomial 'centred' around a point x=a, The only catch is  I have to change the coefficients of the polynomial accordingly
$$ P(x) = b_o + b_1 (x-a) + b_2 (x-a)^2 ... b_n (x-a)^n = \sum b_k x^k$$
Now, the idea here is that the 'coefficients' of polynomials are completely controlled by the values of derivative at the point which it is centred around. More specifically speaking,
$$ P(a) = 0! b_o$$
$$ P'(a) =1! b_1$$
--
$$ P^{k} (a) = k! b_k$$
Or, we can write generally,
$$ b_k = \frac{ P^{k} (a)} {k!}$$
plugging this into our summation,
$$ P(x) = \sum \frac{ P^{k} (a) }{k!}  (x-a)^k$$
Now, as I have demonstrated, if you centre a polynomial around a point, then the expression for polynomial around that point has it's coefficients based on the local rates of changes near that point.
So, now, we think in reverse, if a polynomial's coefficient when expressed around a point x=a is reliant on it's derivative, if a function had such a polynomial for it, then the coefficients of terms in polynomial would be reliant on the value of function near that point.
$$ f(x) =  \sum \frac{ f^{k} (a) }{k!}  (x-a)^k$$
And I think this is a common principle to understand, if you think of a concept in a reverse way, then you can extend that concept to be applied on other things.


TITLE: Does there exist an onto map from $[0,1]$ to $\mathbb Q$?
QUESTION [1 upvotes]: Does there exist an onto map from $[0,1]$ to $\mathbb Q$?
I am unable to come up with a proof or a counter example.
If we take $f$ to be an onto mapping $[0,1]$ to $Q$, then $f([0,1])=\mathbb Q$.
I am unsure if it will give any contradiction. If not then what will be the map?
Can someone please help me out?

REPLY [0 votes]: Yes:
Do an enumeration $E_1$ of the Rationals in $[0,1] \cap \mathbb Q$ and an enumeration $E_2$ of the "Stand alone" $ \mathbb Q$. Then map $E_1(j) \rightarrow E_2(j); j=1,2,..$
and map all Irrationals to a constant Rational, such as $1/2$.


TITLE: Power Series -- Convergence, Divergence, and Absolute Convergence
QUESTION [0 upvotes]: Suppose that the power series $$\sum a_nx^n$$ is convergent at $x=-3$ and divergent at $x=5$. What can be said about the following:

convergence at $x=-2$ ?
absolute convergence at $x=2$ ?
convergence at $x=-6$ ?
convergence at $x=3$ ?
divergence at $x=-5$ ?

How can I go about getting the answers to each of these questions?

REPLY [0 votes]: Hint: The radius of convergence is some $R$ for $3 < R < 5$. This should settle questions 1,2,3 for you. For questions 4 and 5, consider $a_n = 1/(n3^n)$ vs. $a_n = 1/(n 4^n)$ and $a_n = 1/(n4^n)$ vs. $a_n = 1/(n5^n)$.


TITLE: How thin a filter to filter particles out of salt water to make salt for human consumption?
QUESTION [1 upvotes]: I have gathered some salt in the Death Valley and for the fun of it I would like to clean it and make it edible. The process I plan to follow is simple:

Dissolve the salt in water.
Boil water for 30 min to kill bacteria.
Filter water to remove foreign particles.
Evaporate water to get clean salt.

At step 3, I'm planning to buy paper filters with pores of 3 µm to filter fine particles. Is it sufficient for the purpose of human consumption?
I'm mostly wondering if there is a key pore size threshold beyond which it is significantly less useful to filter, or if it's just a matter of degree, the smaller the better at an consistent size/benefit ratio.

REPLY [1 votes]: There are 3 kinds of mixtures in liquid...

True Solution
Colloids
Suspension

These three vary in between because of the size of the particle in them. see wiki. Now, the salt solution you were talking about comes under category "true solution" i.e. particle size less than 1 $nm$.
Now we don't have sieve to filter out this particles of this dimension. Even bacterias and microbes are orders of magnitude greater than this dimension.
As per classical textbooks, it's only suspension that can be filtered out using sieve or sedimentation. Even for colloids one needs ultrafiltration methods, with superfine pores in the filter.


TITLE: Definition of positive definite matrix
QUESTION [3 upvotes]: Definitions in the book by Friedberg.
A linear operator $T$ on a finite-dimensional inner product space is called positive definite if $T$ is self-adjoint and $\langle T(x),x\rangle >0$  for all $ \neq 0$. An $n\times n$ matrix $A$ with entries from $\mathbb R $ or $\mathbb C$ is called positive definite if $L_A$ is positive definite.
Question

Is the definition of positive definite matrix defined with respect to the dot product of $\mathbb R $ or $\mathbb C$?
Do the definition of positive definite matrix depends on what inner product of $\mathbb R $ or $\mathbb C$ is given?

Thanks.

REPLY [3 votes]: Yes and no. A matrix $A \in M_n(\mathbb{F})$ (where $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$) is called positive definite if the operator $L_A \colon \mathbb{F}^n \rightarrow \mathbb{F}^n$ is a positive definite operator when $\mathbb{F}^n$ is endowed with the standard inner product. It might be the case that the operator $L_A$ is positive definite as an operator with respect to a different inner product on $\mathbb{F}^n$ but then you won't say that the matrix $A$ is positive definite (unless it is also positive definite as an operator with respect to the standard inner product).


TITLE: Multiplying out a quadratic equation of differential factors acting on a function
QUESTION [3 upvotes]: I am having trouble understanding a multiplying out of a quadratic equation with differential factors:
from the following video:
$$
(D + A(x))(D + B(x))y(x) = 0\\
(D^2 + AD + AB + B^{'} + BD)y = 0 \\
$$
What I don't get is his remark on the last multiplication two terms. It looks like he is using the multiplication rule for differentiation, but in the end, the multiplied out parenthesis is again acting on y, so for me it looks like a double use.
I see that $B^{'} = DB$ but I still do not get why two terms are produced?

REPLY [2 votes]: Omitting the argument $(x)$, we have $$(D+b)y=Dy+by$$
and
$$(D+a)(D+b)y=(D+a)(Dy+by)=D^2y+D(by)+aDy+aby
\\=D^2y+bDy+b'y+aDy+aby
\\=(D^2+(a+b)D+(b'+ab))y.$$


TITLE: Given a field extension $L : K : F$, if $L:F$ and $K:F$ are finitely generated then so is $L:K$.
QUESTION [1 upvotes]: I'm trying to find a source for the proposition from the title: if $L : K : F$ is a tower of field extensions, and furthermore both $L$ and $K$ are finitely generated over $F$, then $L$ is finitely generated over $K$.
This statement does not seem to be in Dummit and Foote, and I have had no luck finding a proof online. Does anybody have a reference, source, or proof for it?

REPLY [0 votes]: Well, I figured out a proof. It turns out it's way simpler that I thought it would be, and I didn't think about it enough before posting the question.
If $L : F$ is finitely generated, then by definition that means that there are some $\alpha_1, \ldots, \alpha_s \in L$ such that $L$ is the smallest field (by inclusion) containing $F$ all the $\alpha_i$. Now consider some intermediate field $K$ (so $L : K : F$ is a tower of extensions). Certainly $L$ contains $K$, and $L$ contains all the $\alpha_i$, and if $L$ was not the smallest field containing $K$ and the $\alpha_i$, then it would also not be the smallest field containing $F$ and the $\alpha_i$, since $F \leq K$, which is absurd since $L$ is finitely generated over $F$. Therefore $L$ is finitely generated over $K$.


TITLE: Will a closed set which is a subset of $\mathbb{R}$ always be only a union of finite number of closed intervals?
QUESTION [0 upvotes]: A closed set is defined as follows:

A subset $X$ of $\mathbb{R}$ is closed if every convergent sequence in $X$ converges to a point in $X$ itself. It is a generalization of closed intervals.

While I realize that it is a useful generalization of closed intervals for spaces other than $\mathbb{R}$, I want to know that for subsets of $\mathbb{R}$, is there an example of a closed set which cannot be described as a union of finite number of closed intervals (including single points, i.e. $[a,a]$)?
EDIT: Earlier, the question had not mentioned that it has to be a union of finite number of sets, but later I edited it to include that condition

REPLY [1 votes]: The set$$\{0\}\cup\bigcup_{n\in\mathbb{n}}\left\{\frac1n\right\}$$is a closed susbset of $\mathbb R$ which cannot be expressed as the union of a finite family of closed intervals.


TITLE: Knowing $\prod_{i=1}^n (c_1 - a_i) = d_1$ can we determine $\prod_{i=1}^n (c_2 - a_i)$?
QUESTION [1 upvotes]: For known set of real values $\{a_i\}$ and $c_1$, if we know
$$\prod_{i=1}^n (c_1 - a_i) = d_1$$
is there a way to determine
$$\prod_{i=1}^n (c_2 - a_i)$$
without having to redo the multiplications?
(Forgot to include additional relevant assumptions):
All $a_i$ are strictly positive, and no $a_i$ is equal to $c_1$ or $c_2$; in fact both ${c_1,c_2}$ are greater than the $\max \{a_i\}$. The product is strictly positive. Thank you.

REPLY [1 votes]: Here is a way to rephrase your question: given a monic polynomial $p$ of degree $n$ with roots $a_1,\ldots,a_n$, and such that $p(c_1)=d_1$, find $p(c_2)$.
Of course this can always be done, as you can recover the coefficients for $p$ from the symmetric polynomials. Namely, you have
$$
p(x)=\sum_{k=0}^n (-1)^{n}e_n(a_1,\ldots,a_n)\,x^{n-k}.
$$
Other than that, I don't think you can expect any kind of general answer. Consider for instance the case where $a_1=\ldots=a_n=0$. Then you are asking something like

If I now that $\pi^n=d_1$, find $e^n$.

Which can be answer, but not in a "polynomial" way.