TITLE: Parametric version of a simple equation
QUESTION [4 upvotes]: I have a simple relation that I need to plot in a plane.  I could do it, but I believe that I don't get the best way.  A plane curve is defined implicitely by the following equation :
\begin{equation}\tag{1}
\cos{\chi} + \sin{\eta} = \sqrt{f} \; \cos{\eta},
\end{equation}
where $f$ is a positive constant.  I need to plot the curve in the $(\chi, \eta)$ plane.  It's easy to invert the relation above (expressing $\chi$ as a function of $\eta$ is simpler but isn't well adapted to the curve) :
\begin{equation}
\tag{2}
\sin{\eta} = g_f(\chi) \equiv \frac{\sqrt{f^2 + f \, \sin^2{\chi}} - \cos{\chi}}{1 + f},
\end{equation}
so I could define a parametric version of equation (1) above :
\begin{equation}
\tag{3}
(\chi, \; \arcsin{g_f(\chi)}),
\end{equation}
This does the job.  However, I strongly suspect there's a simpler parametrisation of the same curve, represented by equation (1).
So the question is what is the best or simplest parametrisation $(\chi(s), \; \eta(s))$ of the equation (1) above ?
EDIT :  Take note that $\chi$ is only defined on the interval $-\pi$ to $\pi$.  Also : $-\frac{\pi}{2} < \eta < \frac{\pi}{2}$.

REPLY [1 votes]: Using Andrew's answer, I could define the following parametrization, which appears to be working great :
\begin{align}
\big(\, \chi(s), \; \eta(s) \big) = \big(\arccos{s}, \quad \arccos{\frac{s}{\sqrt{1 + f}}} - \arccos{\sqrt{\frac{f}{1 + f}}} \big),
\end{align}
where $-1 < s < 1$ (or $s = \cos{\chi}$).  But I still suspect this could be improved (i.e. simplified).  That parametrization isn't much simpler than the expression (2) I gave in my question.


TITLE: Given a matrix A $\in$ $R^{3 \times 4}$ and a vector b $\in$ $R³$, let Sol(A,b)={$x \in R^n:Ax=b$}.
QUESTION [0 upvotes]: Prove or show that does not exist:
(i) There exist a echelon matrix U $\in R^{3 \times 4}$ and b $\in R³$ such that Sol(U,b) is empty.
(ii) There exist a echelon matrix U $\in R^{3 \times 4}$ and b $\in R³$ such that Sol(U,b) has exactly one element.
(ii) There exist a echelon matrix U $\in R^{3 \times 4}$ and b $\in R³$ such that Sol(U,b) has infinite elements.\
Hello,I need some help with those proofs. Here's what I have so far:
Let $Col(U)$ be the column space generated by the columns-vectors the matrix U.
If $b \notin Col(U)$ then the solution of $U*x=b$ is empty.
And if $b \in Col(U)$ then we either have one possible solution for the system or several possible solutions.\
I read in the book that for every matrix $dim \text{ } Col(U) = dim \text{ } Row(U)$. So, I conclude that the maximum value of $dim \text{ } Col(U)$ would be 3, since the matrix has only 3 rows. Is that correct?
First, if $dim \text{ } Col(U) = 3$ then the columns of U form a basis for $Col(U)$. And:
Ux=b
[$U_{*1},U_{*2},U_{*3},U_{*4}$]*($x_1,x_2,x_3,x_4$)=($b_1,b_2,b_3)$
I have to be honest and say that I'm not sure if this is the appropriate approach to prove. I mean, since $dim \text{ } Col(U) = dim \text{ } Row(U)$, does it mean that $Col(U)=Row(U)$? I think this cannot be true, since the row vectors are 4-vector and the columns vectors are 3-vector, so the vector space is different. Right?
However, if $dim \text{ } Col(U) = 3$ since there are 4 columns, does it mean that one of the columns can always be generated by the other three?
And finally, how can I use those information to prove the statements are true or false?

REPLY [0 votes]: Your anticipation is correct but this kind of approach may not work well since your approach is abstract. To be effective in this approach, your background of linear algebra has to be strong.
Another simple way to approach your exercise is that you can write $Ax = B$ as a system of linear equations. E.g., the first item (i) says that: Find a system of three linear equations of four unknowns so that the system has no solution. It would be easier than your abstract approach.


text:
\begin{document}
\title{$r$-vertex-strongly-distinguishing total coloring of graphs\thanks{This work is supported by National Natural Science Foundation of China (Grant Nos. 11961041, 61802158), and the Foundation of China Scholarship Council (No. 201908620009).}}

\author{{\small Fei Wen$^{1,4,}$\footnote{Corresponding author. {E-mail addresses:}
wenfei@lzjtu.edu.cn(F. Wen).},\ \ \ Zepeng Li$^{2}$,\ \ \ Xiang'en Chen$^{3}$}\\[2mm]
\scriptsize 1.  Institute of Applied Mathematics, Lanzhou Jiaotong
University, Lanzhou 730070, P.R.China\\
\scriptsize 2. School of Information Science and Engineering, Lanzhou University, Lanzhou 730000, P.R.China\\
\scriptsize 3. College of Mathematics and Statistics, Northwest Normal University, Lanzhou 730070, P.R.China\\}
\date{}
\maketitle
{\flushleft\large\bf Abstract.} Inspired by the phenomenon of \emph{co-channel interference} in communication network, a novel graph parameter, called \emph{$r$-vertex-strongly-distinguishing total coloring} (abbreviate as $D(r)$-VSDTC), is proposed in this paper. Given a graph $G$, an $r$-VSDTC is an assignment of $k$ colors to $V(G)\cup E(G)$ such that any two adjacent or incident elements receive different colors and any two vertices with distance at most $r$ have distinct color-set, where the color-set of a vertex $u$ is the set of colors assigned on $u$ and its neighborhoods and incident edges. The \emph{$r$-vertex-strongly-distinguishing total chromatic number} of $G$, denoted by $\chi_{r-vsdt}(G)$, is the minimum integer $k$ for which $G$ admits a $k$-$D(r)$-VSDTC. We show that $\chi_{1-vsdt}(G)\leq 4\Delta(G)$ for every graph $G$ without isolated edges and $\chi_{1-vsdt}(G)\le k\Delta(G)+3$ for a $k$-degenerated graph $G$ without isolated edges, where $1\le k\le 3$.

{\flushleft\large\bf Keywords:} $k$--degenerated graph; $r$-vertex-strongly-distinguishing total coloring; $r$-vertex-strongly-distinguishing total chromatic number
 \vspace{1mm}

\maketitle{\flushleft\large\bf AMS Classifications} 05C15

\section{Introduction}
Let $G=(V,E)$ be a finite, simple and undirected graph with vertex set $V(G)$ and edge set $E(G)$. For any vertex $u,v\in V(G)$, the \emph{distance} between $u$ and $v$, denoted by $d(u,v)$, is defined as the length of a shortest path between them. The \emph{diameter} of $G$, denoted by $\mathbf{d}$, is the maximum distance between any two vertices of $G$. The \emph{incident set} of the vertex $u$, denoted by $N\langle u\rangle$, is a set consisting of $N[u]$ and incident edges of $u$, i.e., $N\langle u\rangle= N[u]\cup\{uv|v\in N(u)\}$, where $N[u]$ and $N(u)$ are the \emph{closed neighbor set} and \emph{open neighbor set} of $u$, respectively. In addition, $d(u)$ represents the \emph{degree} of $u$.

As usual, let $\Delta(G)$ and $\delta(G)$ denote the \emph{maximum degree} and the \emph{minimum degree} of $G$, respectively. If there is no
confusion, sometimes we write $\Delta(G)$ as $\Delta$, and write $\delta(G)$ as $\delta$. And we always denote respectively, $K_{n}$ and $P_{n}$ by the \emph{complete graph} and \emph{path} of order $n$. For a universal set $I$, suppose $A, B\subseteq I$, the symbols $A-B=A\setminus (A\cap B)$ and $|A|$ respectively represent the \emph{minus set} and the \emph{cardinal number} of $A$. For the other terminologies and notations refer to \cite{bondy}.

Graph coloring theory is becoming an increasingly useful
family of mathematical models for a broad range of applications,
such as \emph{register allocation}, \emph{frequency assignment}, \emph{time tabling and scheduling} and so on. A \emph{$\kappa$-vertex coloring} of $G$ is an assignment of $\kappa$ colors to the vertices of $G$ such that any two adjacent vertices have different colors, and the \emph{chromatic number} of $G$, denote by $\chi(G)$, is the minimum $\kappa$ for which $G$ admits a $\kappa$-coloring.

In 1997, Burris and Schelp\cite{Burris} first introduced  \emph{vertex distinguishing edge-coloring} of graphs. After that, \emph{(adjacent-) vertex distinguishing edge coloring} of graphs was proposed and has been drawn a lot of attention of the researchers (see Ref.\cite{zhang-4,hatami,balister}). In 2006, Akbari et al.\cite{limc} proposed \emph{$r$-strong edge coloring of graphs} in the following.

\begin{defin}[\!\!\cite{limc}]\label{def-2}
Let $G$ be a graph, and $f$ a \emph{proper $\kappa$-total coloring} from $E(G)$ to $\{1,2,\ldots,$ $\kappa\}$. If $C_{f}(u) \not=C_{f}(v)$ for any two vertices $u$ and $v$ with distance at most $r$, then $f$ is called \emph{$\kappa$-$r$-strong edge coloring} (denoted by $\kappa$-$D(r)$-SEC for short) where $C_{f}(u)=\{f(uv)|uv\in E(G)\}$, and the \emph{$r$-strong edge chromatic number} $\chi'_{s}(G, r)$ is the minimum number of colors required for such a coloring of $G$.
\end{defin}

Moreover, they gave an upper bound of \emph{$1$-strong edge chromatic number} of a graph, that is, for any graph $G$ with no isolated edges, $\chi'_{s}(G,1)\le 3\Delta(G)$. Moreover, they also obtained \emph{$2$-strong edge chromatic number} and \emph{$3$-strong edge chromatic number} of a tree, which generalized Zhang's result on \emph{$2$-strong edge coloring} of a tree\cite{zhang-4}. In 2005, Zhang et al.\cite{zhang-1} presented a concept of the \emph{adjacent-vertex-distinguishing total coloring of graphs} and, meanwhile, obtained the adjacent-vertex-distinguishing total chromatic number of some graphs such as \emph{cycle, complete graph, complete bipartite graph, fan, wheel and tree}.

After then, Zhang et al.\cite{zhang-2} defined the \emph{$D(\beta)$-vertex-distinguishing total coloring} (abbreviated as $D(\beta)$-VDTC for short) of graph $G$, that is a proper \emph{total coloring} such that for any vertex $u,v\in V(G)$ with  $1\leq d(u,v)\leq \beta$, and $C(u)\neq C(v)$ always holds, where $C(u)$ is the color set of the edges incident to $u$ together with the color on $u$, and $\chi_{\beta-vdt}(G)$ is called the \emph{$D(\beta)$-vertex-distinguishing total chromatic number} for a $D(\beta)$--VDTC of $G$. They subsequently investigated the chromatic number $\chi_{\beta-vdt}(G)$ on some familiar graphs (such as paths, cycles, complete graphs and so on) when $\beta=2,3$.

In \emph{communication network}, if two nodes within certain distance send out the same frequency signals, then it would be appeared the phenomenon of \emph{interference} since the same frequency waves have resonated\cite{Liang}. In some literature\cite{winter}, which is also called the \emph{co-channel interference}. Especially, if the frequency of  the two nodes that send out signals, and the signals as well as the nodes that receive signals are identical, then the phenomenon of interference are more serious. So, combine with the above physical phenomenon and the nice results of \cite{limc,zhang-3}, we defined a new concept entitled \emph{$r$-vertex-strongly-distinguishing total coloring} of graph in the following.
\begin{defin}\label{def-1}
Let $G$ be a graph with no isolated edges as its components, and let $f$ be a \emph{proper $\kappa$-total coloring} from $V(G)\cup E(G)$ to $\{1,2,\ldots, \kappa\}$. We call $f$ an \emph{$\kappa$-$r$-vertex-strongly-distinguishing total coloring} (denoted by $\kappa$-$D(r)$-VSDTC for simplicity) if any two vertices $u$ and $v$ with distance at most $r$ meet $C_{f}\langle u\rangle \not=C_{f}\langle v\rangle$, where $C_{f}\langle u\rangle=\{f(x)|x\in N\langle u\rangle\}$. The $r$-vertex-strongly-distinguishing total chromatic number of $G$ is the minimum $\kappa$ for which $G$ admits an \emph{$\kappa$-$D(r)$-VSDTC}, denote by $\chi_{r-vsdt}(G)$ for short.
\end{defin}
Particularly, if $r=1$ or $r=\mathbf{d}$, it is respectively called to be the \emph{adjacent-vertex-strongly-distinguishing total coloring} and \emph{vertex-strongly-distinguishing total coloring} of $G$, those concepts have been introduced by Zhang et al. in \cite{zhang-3}, but the latter one has no any research till now. Clearly, if $G$ admits an $r$-vertex-strongly-distinguishing total coloring, it is easy to see that $\chi_{1-vsdt}(G)\le \chi_{r-vsdt}(G)\le \chi_{\mathbf{d}-vsdt}(G)$.

\emph{Determining the bound of vertex-distinguishing colorings} of a graph is an important and exciting research topic in mathematics science and theoretical computer science. The \emph{vertex-strongly-distinguishing chromatic numbers} of a graph provide information on its structural properties and also on some relevant topological properties, in particular those related to information science. In this paper, we mainly consider \emph{$r$-vertex-strongly-distinguishing total coloring} of some graphs, especially $k$-degenerated graphs $k=1,2,3$. At first, we by Definition \ref{def-1} give two obvious results of \emph{$r$-vertex-strongly-distinguishing total coloring} of a graph.
\begin{thm}\label{thm-1}
Let $G$ be a connected graph with order no less than $3$, and $r_{i}$ the positive numbers. If $r_{1}<r_{2}$, then $$\chi_{r_{1}-vsdt}(G)\le \chi_{r_{2}-svdt}(G).$$
\end{thm}
\begin{thm}\label{thm-2}
Let $G$ be a graph with $m$ connected components $G_{1},G_{2},\ldots,G_{m}$. Then
$$\chi_{r-vsdt}(G)=\max\{\chi_{r-vsdt}(G_{i})|i=1,2,\ldots,m\}.$$
\end{thm}
\begin{remark}\label{remark-1}
Theorem \ref{thm-2} implies that if one would like to consider \emph{$r$-vertex-strongly-distinguishing total coloring} of a graph $G$, it only needs to verify the connected component with maximum chromatic number of such coloring.
\end{remark}

An \emph{$r$-vertex-strongly-distinguishing total coloring} of a graph can be obtained from  an \emph{$r$-strong edge coloring} and a \emph{proper vertex-coloring} of the graph, so one can get the following result.
\begin{thm}\label{thm-3}
For any graph $G$ with no isolated edges,
$$\chi_{r-vsdt}(G)\le \chi'_{s}(G,r)+\chi(G).$$
\end{thm}

In \cite{limc}, Akbari et al. obtained an upper bound of $1$-strong edge chromatic number of a graph $G$.

\begin{lem}[\!\cite{limc}]\label{lem-0}
For any graph $G$ with no isolated edges, $\chi'_{s}(G,1)=3\Delta(G)$.
\end{lem}

By Lemma \ref{lem-0} and \emph{Brook's Theorem}, we obtain the following result.

\begin{cor}\label{cor-1}
For any graph $G$ with no isolated edges,
$$\chi_{1-vsdt}(G)\le 4\Delta(G).$$
\end{cor}
\begin{remark}\label{remark-2}
Hatami\cite{hatami} proved that $\chi'_{s}(G,1)\le \Delta(G)+300$, provided that  $\Delta(G)>10^{20}$. By Theorem \ref{thm-3} one can also get $\chi_{1-vsdt}(G)\le 2\Delta(G)+300$ when $\Delta(G)>10^{20}$. In fact, this bound for familiar graphs is invalid. In \cite{balister}, Balister et al. proved $\chi'_{s}(G,1)\le 5$ for such graphs with maximum degree $\Delta(G)=3$, and prove $\chi'_{s}(G,1)\le \Delta(G)+2$ for bipartite graphs. Moreover, for $k$-chromatic graphs $G$ without isolated edges they prove a weaker result of the form $\chi'_{s}(G,1)\le \Delta(G)+O(\log k)$. Also by Theorem \ref{thm-3}, $\chi_{1-vsdt}(G)\le 8$ for graphs with maximum degree $\Delta(G)=3$, and $\chi_{1-vsdt}(G)\le \Delta(G)+4$ for bipartite graphs, and for $k$-chromatic graphs $G$ without isolated edges, one can obtain form $\chi_{1-vsdt}(G)\le \Delta(G)+k+O(\log k)$.
\end{remark}

A graph $G$ is \emph{$k$-degenerated} if $\delta(H)\le k$ for every subgraph $H$
of $G$. In literature \cite{zhang-4}, Zhang et al. proved that for any tree $T$, $\chi_{s}'(T,1)\le \Delta (T)+1$. Combine with Theorems \ref{thm-2}, \ref{thm-3} and \emph{Brook's Theorem} we have the following corollary.
\begin{cor}\label{cor-3}
Let $G$ be a $1$-degenerated graph. Then
$$\chi_{1-vsdt}(G)\le \Delta(G)+3.$$
\end{cor}
Next, we give an upper bound of $\chi_{1-vsdt}(G)$ for $k$-degenerated graphs $G$, where $k\ge 2$.

\begin{thm}\label{thm-4}
For $k\ge 2$, let $G$ be a $k$-degenerated graph without isolated edges. Then
$$\chi_{1-vsdt}(G)\le  k\Delta(G)+3.$$
\end{thm}

Together with Theorems \ref{thm-3} and \ref{thm-4}, one can obtain the following corollary.
\begin{cor}\label{cor-2}
For $k\ge 2$, let $G$ be a $k$-degenerated graph without isolated edges. Then
$$\chi_{1-vsdt}(G)\le \min\{k\Delta(G)+3,\ \ 4\Delta(G)\}.$$
\end{cor}

From Corollaries \ref{cor-3} and \ref{cor-2}, we give an improved upper bound of $\chi_{1-vsdt}(G)$ for $k$-degenerated graphs $G$ with $k=1,2,3$.
\begin{thm}\label{thm-5}
Let $G$ be a $k$-degenerated graph with $k=1,2,3$ and without any isolated edge. Then
$$\chi_{1-vsdt}(G)\le  k\Delta(G)+3.$$
\end{thm}

In light of Akbari et al.'s results\cite{limc}  about a tree, we also give an upper bound of \emph{$2$-vertex-strongly-distinguishing total chromatic number} of a tree.
\begin{cor}\label{pro-1}
Let $T$ be any tree with $\Delta(\geq2)$. Then
$\chi_{2-svdt}(T)\leq \Delta(T)+3.$ Moreover, if for any two vertices $u$ and $v$ with maximum degree, and $d(u,v)\ge3$, then $\chi_{2-svdt}(T)=\Delta(T)+2$.
\end{cor}

From Lemma \ref{lem-6} we know that any tree $T$ with $\Delta\ge 3$ admits a $(2\Delta-1)$-$D(3)$-VSDTC, applying the same method as the proof of Corollary \ref{pro-1}, one can obtain the following corollary .
\begin{cor}\label{pro-2}
Let $T$ be any tree with $\Delta\ge 3$. Then
$\chi_{3-vsdt}(T)\leq 2\Delta(T)+1$.
\end{cor}

\section{Preliminaries}
In the section, we give some useful lemmas which will play an important role in the proof of our main results.

\begin{lem}\label{lem-1}
Let $A,B\subseteq S$ and $A\not=B$. Then there exists at most one $c\in S$ such that $A= B\cup \{c\}$ or $B= A\cup \{c\}$.
\end{lem}
\begin{proof}
Without loss of generality, we assume that $|B|\le |A|$, then one can obtain that either $A\not= B\cup \{x\}$ for any $x\in S$, or there is some $c\in S$ such that $A=B\cup \{c\}$. Now we prove that $c$ is unique if the latter one holds. Assume, to the contrary, that there are two distinct $c,c'\in S$ such that the letter one holds. Since $A= B\cup \{c'\}$, we have $c'\in A$ and $c'\not\in B$. Similarly, we have $c\in A$ and $c\not\in B$. This means that $A$ is more than $B$ with at least $2$ elements $c'$ and $c$, it is a contradiction.
\end{proof}

\begin{lem}\label{lem-2}
Suppose $A,B\subseteq S$ and $A\not=B$. Then the element $c$ yielding $A\cup \{c\}= B\cup \{c\}$, has at most one in $S$.

\end{lem}
\begin{proof}
Since $A,B\subseteq S$ and $A\not=B$, the equation of $A\cup \{c\}= B\cup \{c\}$ implies that either $A\subset B$ and $|B|-|A|=1$, or $B\subset A$ and $|A|-|B|=1$. Furthermore, it follows that there at most exists one $c\in B$ but $c\not\in A$, or $c\in A$ but $c\not\in B$. So the proof completes.
\end{proof}

\begin{lem}[\!\cite{zhang-3}]\label{lem-3}
Let $P_{n}$ be a path of order $n(\geq3)$. If $n$ is odd, then $\chi_{1-vsdt}(P_{n})=4$, otherwise, $\chi_{1-vsdt}(P_{n})=5$.
\end{lem}
\begin{lem}[\!\cite{zhang-3}]\label{lem-4}
Let $C_{n}$ be a cycle of order $n(\geq3)$. Then $\chi_{1-vsdt}(C_{n})\le 5$.
\end{lem}

\begin{lem}[\!\cite{limc}]\label{lem-5}
For any tree $T$ with at least three vertices, $\chi'_{s}(T,2)\le \Delta(T)+1$. Moreover, if for any two vertices $u$ and $v$ with maximum degree, and $d(u,v)\ge3$, then $\chi'_{s}(T,2)=\Delta(T)$.
\end{lem}

\begin{lem}[\!\cite{limc}]\label{lem-6}
For any tree $T$ with $\Delta(T)\ge 3$, $\chi'_{s}(T,3)\le 2\Delta(T)-1$.
\end{lem}

From Definition \ref{def-1} we see that the following lemma always holds.
\begin{lem}\label{lem-7}
Let $G$ be a graph. Then $\chi_{r-vsdt}(G)\ge \Delta(G)+1$, and further $\chi_{r-vsdt}(G)\ge \Delta(G)+2$ if $G$ has two vertices $u,v$ with maximum degree and $d(u,v)\le r$.
\end{lem}

\section{Proofs of main results}
In this section, we first give a proof of Theorem \ref{thm-3} in the following.
\begin{proof0}
 \rm For convenience, we denote by $\chi'_{s}(G,r)=p$ and $\chi(G)=q (q\ge p)$. Let $f'$ be a $p$-$D(r)$-SEC of $G$ with colors $\{1,2,\ldots, p\}$, and $f^{*}$ a proper $q$-vertex coloring of $G$ with colors $\{p+1,p+2,\ldots, p+q\}$. We here define a new proper $(p+q)$-total coloring $f$ as follows
\begin{equation*}
f(z)=\begin{cases}
f'(z), & \forall z\in E(G)\\
f^{*}(z),  &\forall z\in V(G)\\
\end{cases}
\end{equation*}
Clearly, $f$ is a proper $(p+q)$-total coloring of $G$, and $\forall y\in V(G)$, $C_{f}\langle y\rangle=C_{f'}(y)\cup \{f^{*}(x)|x\in N[y]\}$ where $C_{f'}(y)\subseteq \{1,2,$ $\ldots, p\}$ and $\{f^{*}(x)|x\in N[y]\}\subseteq \{p+1,p+2,\ldots, p+q\}$.

Let $u,v$ be any two distinct vertices of $G$.
Since $f'$ is a $p$-$D(r)$-SEC of $G$, it follows from Definition \ref{def-2} that $C_{f}(u)\not= C_{f}(v)$. Furthermore, $C_{f'}(u)\cup \{f^{*}(x)|x\in N[u]\}\not= C_{f'}(v)\cup \{f^{*}(x)|x\in N[v]\}$ because $\{1,2,\ldots, p\}\cap \{p+1,p+2,\ldots, p+q\}=\emptyset$. Namely, $C_{f}\langle u\rangle \not =C_{f}\langle v\rangle$. Thus, $f$ is a $(p+q)$-$D(r)$-VSDTC of $G$.

The proof completes.  \hfill$\square$
\end{proof0}

Next, we give some notations before proving Theorem \ref{thm-4}. For a graph $G$, let $A$ be a proper subset of $V(G)\cup E(G)$, and $f$ an \emph{$r$-vertex-strongly-distinguished} mapping from $A$ to $\{1,2,\ldots,\kappa\}$. Then a vertex $u$ of $A$ is called to be `\emph{good}' if all elements of $N\langle u\rangle$ is colored under the current $f$. Otherwise, the vertex of $u$ is called to be `\emph{bad}'. In addition, two vertices $u$ and $v$ are called \emph{strongly distinguished} under $f$ if the two vertices satisfy $C_{f}\langle u\rangle \not=C_{f}\langle v\rangle$.

\begin{proof1}
\rm Let $G$ be a $k$-degenerated graph and $\lambda=k\Delta+3$, where $k\geq 2$. Clearly, $\Delta\geq2$ since $G$ has no isolated edges as its components.
When $\Delta =2$, by Lemmas \ref{lem-3} and \ref{lem-4} we know that the conclusion holds.
When $\Delta \ge 3$, it implies that $|V(G)|\ge 4$, we will prove the results by induction on $|V(G)|$.

If $|V(G)|=4$, the theorem is obvious; suppose otherwise, that $H$ is a vertex-induced subgraph of $G$ with $|V(H)|$ less than $|V(G)|$, and $H$ has a $\lambda$-$D(1)$-VDSTC $f$. We here verify that $G$ also admits a $\lambda$-$D(1)$-VDSTC.

Let $v$ be a vertex of $G$ with $d(v)=\delta$, and $N(v)=\{v_1,v_2,\ldots,v_\delta\}$. We denote by $e_i=vv_i$ and $N(v_i)=\{w_{i1},w_{i2},\ldots, w_{it_i}, v\}$, where $i=1,2,\ldots, \delta$. As is shown in Fig.\ref{figure-1}.

\vspace{-3.0cm}
\begin{figure}[!hbt]
    \centering
    \includegraphics[height=20 cm, width =15 cm]{new.eps}
\vspace{-13.5cm}\caption{\footnotesize The illustration of $G$}\label{figure-1}
\end{figure}

Set $H:=G\backslash \{v\}$. By the induction hypothesis above, suppose $H$ has a $\lambda$-$D(1)$-VDSTC $f$ such that for any $uw\in E(H)$, $C_{f}\langle u\rangle\not =C_{f}\langle w\rangle$. In order to extend $f$ to a $\lambda$-$D(1)$-VDSTC $f'$ of $G$, one should firstly consider the coloring of $v$. Now, a claim is given in the following.

\begin{claim}\label{claim-1}
There is some color $c_{0}\in \{1,2,\ldots,\lambda\}$ such that
\begin{equation}\label{eq-1}
c_0\not\in \{f(v_i)\mid 1\leq i \leq\delta\} \text{~~and~~} C_{f}\langle w_{ij}\rangle\not=C_{f}\langle v_{i}\rangle\cup\{c_{0}\}
\end{equation}
for any $1\le j\le t_{i}$ and $1\le i\le \delta$.
\end{claim}
\begin{proof}
For any neighbor $w_{ij}$ of each $v_{i}$ $(1\le j\le t_{i},1\le i\le \delta)$, we have $C_{f}\langle w_{ij}\rangle\not=C_{f}\langle v_{i}\rangle$ since $f$ is a $D(1)$-VDSTC of $H$. Furthermore, it follows from Lemma \ref{lem-1} that there is at most one $c_{ij} \in \{1,2,\ldots,\lambda\}$ satisfying $C_{f}\langle w_{ij}\rangle =C_{f}\langle v_{i}\rangle\cup\{c_{ij}\}$. Note that $t_{i}\le \Delta-1$ and $|\{f(v_i): 1\leq i \leq\delta\}|\leq \delta$. Thus, there are at most $\delta+\delta(\Delta-1)$ colors $c$ such that $C_{f}\langle w_{ij}\rangle =C_{f}\langle v_{i}\rangle\cup\{c\}$.

Since $|\{f(v_i): 1\leq i \leq\delta\}|\le \delta\leq k$, and $\lambda-(\delta+\delta(\Delta-1))=k\Delta+3-\delta\Delta=(k-\delta)\Delta+3> 1$, and so one can conclude that $c_{0}$ exists.
\end{proof}

Based on Claim \ref{claim-1}, we define a new coloring $f_{0}$ of $H\cup \{v\}$ as follows
\begin{equation*}
f_{0}(z)=\begin{cases}
c_{0}, &\hspace{1cm}z=v\\
f(z),  &\forall z \in V(H)\cup E(H)\\
\end{cases}
\end{equation*}

Next we will successively color the edges $e_{1},e_{2},\ldots, e_{\delta}$ of $v$ and yield that $v_{1},v_{2},\ldots, v_{\delta}$ become `good' vertices. For convenience, we denote $f_{i}$ by the current coloring of $G$ after the $i$-th edge $e_{i}$ of $v$ is colored, and $R_i$ the number of colors on the corresponding $e_i$ such that the current $f_{i}$ is not a $D(1)$-VDSTC of $G$, and also call it to be the \emph{forbidden colors number}, where $1\leq i\leq\delta$. Especially, those colors are called the \emph{forbidden colors}. In addition, let $C_{f_{i}}\langle z\rangle$ denote the color-set of a vertex $z$ consisting of the colors assigned on $N[z]$ and the incident edges of $z$ under the current  $f_{i}$.

For the positive integer of $\delta$, we distinct three cases in the following.

\textbf{Case A.} If $\delta\ge 3$, then we firstly consider the coloring of $e_1$ to obtain $f_{1}$. Obviously, for the coloring of $f_{1}$, there are two types forbidden colors on $e_{1}$ below.
\begin{enumerate}
\item [(a)] To ensure $f_{1}$ is a local proper coloring, the forbidden colors include the following: $f_{0}(v),f_{0}(v_{1})(=f(v_{1}))$ and $f_{0}(v_{1}w_{1j})(=f(v_{1}w_{1j}))$ for $j=1,2,\ldots,t_{1}$;
\item [(b)] To ensure $v_{1}$ and $w_{1j}$ are strongly distinguished, we notice that $C_{f_{0}}\langle w_{1j}\rangle\not =C_{f_{0}}\langle v_{1}\rangle$ for $j=1,2,\ldots,t_{1}$. Then by Lemma \ref{lem-1}, there is at most one color (say) $c_{1j}$ to yield $C_{f_{0}}\langle w_{1j}\rangle=C_{f_{0}}\langle v_{1}\rangle\cup\{c_{1j}\}$ for each $j$, where $j=1,2,\ldots,t_{1}$.
\end{enumerate}
From the above, it has at most $t_{1}+2$ forbidden colors in the subcase (a), and at most $t_{1}$ forbidden colors in the subcase (b). Thus, $R_{1}\le 2+2t_{1}\le 2\Delta$. Clearly, $\lambda-2\Delta>0$. So one can choose one color from $\{1,2,\ldots,\lambda\}$ except for the above $2\Delta$ forbidden colors to dye $e_{1}$, and the reminders remain unchange. At this point, we write the revised coloring function as $f_{1}$. Then under the coloring $f_{1}$, the vertex $v_{1}$ becomes a `good' vertex, and other vertices $v$ and $v_{i}(i=2,3,\ldots, \delta)$ are also `bad'.

For the coloring of $f_{i}$, we successively promote the following procedure on $e_{i}$, where $i=2,3,\ldots,\delta-2$.
\begin{enumerate}
\item [(a')] To ensure $f_{i}$ is a local proper coloring, the forbidden colors include the following: $f_{i-1}(v)(=f_{0}(v))$, $f_{i-1}(v_{i})(=f(v_{i}))$, $f_{i-1}(e_{1}), \ldots, f_{i-1}(e_{i-1})$ and $f_{i-1}(v_{i}w_{ij})$ $(=f(v_{i}w_{ij}))$ for $j=1,2,\ldots,t_{i}$;
\item [(b')] Ensuring $v_{i}$ and $w_{ij}$ are strongly distinguished, we notice that $C_{f_{i-1}}\langle w_{ij}\rangle\not=C_{f_{i-1}}\langle v_{i}\rangle$ for $j=1,2,\ldots,t_{i}$. Then by Lemma \ref{lem-1}, there is at most one color (say) $c_{ij}$ to yield $C_{f_{i-1}}\langle w_{ij}\rangle=C_{f_{i-1}}\langle v_{i}\rangle\cup\{c_{ij}\}$ for each $j$, where $j=1,2,\ldots,t_{i}$.
\end{enumerate}
According to the above, two subcases have at most $t_{i}+i+1$ and $t_{i}$ forbidden colors, respectively, so we have $R_{i}\le 2t_{i}+i+1\le 2\Delta+i-1$. Clearly, $\lambda-R_{i}\ge \lambda-(2\Delta+i-1)=(k-2)\Delta+4-i\ge (\delta-1.5)^{2}+3.75>1$ for $2\le i\le \delta-2$ (Note that in this case $k\geq \delta\geq i+2\geq 4$). Thus, for each $i$, one can orderly choose one color from $\{1,2,\ldots,\lambda\}$ except for the $2\Delta+i-1$ forbidden colors to dye $e_{i}$, and the reminders remain unchange. We here write the resulting coloring function as $f_{i}$ for $i=2,3,\ldots,\delta-2$. Under each $f_{i}$ by order, $v_{1},\ldots,v_{i-1}$ and $v_{i}$ become `good' vertices, and $v_{i+1},\ldots, v_{\delta}$ and $v$ are also `bad' vertices.

For the coloring $f_{\delta-1}$, there are three types of forbidden colors on $e_{\delta-1}$ to be considered below.
\begin{enumerate}
\item [(i)] To ensure $f_{\delta-1}$ is a local proper coloring, the forbidden colors include the following: $f_{\delta-2}(v)(=f_{0}(v)),f_{\delta-2}(v_{\delta-1})(=f(v_{\delta-1}))$, $f_{\delta-2}(e_{1}), \ldots, f_{\delta-2}(e_{\delta-2})$ and $f_{\delta-2}(v_{\delta-1}$ $w_{\delta-1,j})(=f(v_{\delta-1}w_{\delta-1,j}))$ for $j=1,2,\ldots,t_{\delta-1}$;
\item [(ii)] Ensuring $v_{\delta-1}$ and $w_{\delta-1,j}$ are strongly distinguished.  Note that $C_{f_{\delta-2}}\langle w_{\delta-1,j}\rangle=C_{f_{\delta-2}}\langle v_{\delta-1}\rangle$. Then by Lemma \ref{lem-1}, there is at most one color (say) $c_{\delta-1,j}$ to yield $C_{f_{\delta-2}}\langle w_{\delta-1,j}\rangle=C_{f_{\delta-2}}\langle v_{\delta-1}\rangle\cup\{c_{\delta-1,j}\}$ for each $j$, where $j=1,2,\ldots,t_{\delta-1}$;
\item [(iii)] There are at most $\delta$ forbidden colors of $f_{\delta-1}$ under the following restrictions: $C_{f_{\delta-1}}\langle v\rangle\not =C_{f_{\delta-1}}\langle v_{\delta}\rangle$ if $C_{f_{\delta-2}}\langle v\rangle=C_{f_{\delta-2}}\langle v_{\delta}\rangle$, otherwise $|C_{f_{\delta-1}}\langle v\rangle\oplus C_{f_{\delta-1}}\langle v_{\delta}\rangle|\ge 2$ where $\oplus$ is the symmetric minus.
\end{enumerate}
\begin{claim}\label{claim-2}
Under the above assumptions of \emph{(iii)}, if $f_{\delta-1}$ is a local proper coloring, and $C_{f_{\delta-1}}\langle w_{\delta-1,j}\rangle$ $\not=C_{f_{\delta-1}}\langle v_{\delta-1}\rangle$ for $j=1,2,\ldots,$ $t_{\delta-1}$. There are at most $\delta$ forbidden colors for the following restrictions:
\begin{enumerate}
 \item [$\bullet$] if $C_{f_{\delta-2}}\langle v\rangle=C_{f_{\delta-2}}\langle v_{\delta}\rangle$, then $C_{f_{\delta-1}}\langle v\rangle\not =C_{f_{\delta-1}}\langle v_{\delta}\rangle$, otherwise,
  \item [$\bullet$] $|C_{f_{\delta-1}}\langle v\rangle\oplus C_{f_{\delta-1}}\langle v_{\delta}\rangle|\ge 2$.
\end{enumerate}
\end{claim}
\begin{proof}
We here distinct two cases to discuss the number of forbidden colors for the above two restrictions.

\textbf{Case 1.} $C_{f_{\delta-1}}\langle v\rangle\not =C_{f_{\delta-1}}\langle v_{\delta}\rangle$ if $C_{f_{\delta-2}}\langle v\rangle=C_{f_{\delta-2}}\langle v_{\delta}\rangle$.

From the definition of $f_{i}$ we know that $C_{f_{\delta-2}}\langle v\rangle=\{f_{0}(v),f(v_{\delta}),f(v_{\delta-1}), f(v_{i}),f_{\delta-2}(e_{i})|$  $i=1,2,\ldots, \delta-2\}$ and $C_{f_{\delta-2}}\langle v_{\delta}\rangle=C_{f}\langle v_{\delta}\rangle\cup \{f_{0}(v)\}$, but $C_{f_{\delta-1}}\langle v\rangle = C_{f_{\delta-2}}\langle v\rangle \cup \{f_{\delta-1}(e_{\delta-1})\}$ and $C_{f_{\delta-1}}\langle v_{\delta}\rangle=C_{f_{\delta-2}}\langle v_{\delta}\rangle$.
Thus, $C_{f_{\delta-1}}\langle v\rangle =C_{f_{\delta-1}}\langle v_{\delta}\rangle$ implies that $f_{\delta-1}$ should be taken one color from $\{f(v_{\delta}), f(v_{i})|$  $i=1,2,\ldots, \delta-2\}$. Therefore, there are at most $\delta-1$ forbidden colors.

\textbf{Case 2.} $|C_{f_{\delta-1}}\langle v\rangle\oplus C_{f_{\delta-1}}\langle v_{\delta}\rangle|\ge 2$. By the analysis of $|C_{f_{\delta-2}}\langle v\rangle\oplus C_{f_{\delta-2}}\langle v_{\delta}\rangle|$ to study the forbidden colors, there are three subcases to be considered in the following.

\textbf{Case 2.1.} $|C_{f_{\delta-2}}\langle v\rangle\oplus C_{f_{\delta-2}}\langle v_{\delta}\rangle|=1$. It is easy to find that the number of forbidden colors is maximum if $C_{f_{\delta-2}}\langle v\rangle\subset C_{f_{\delta-2}}\langle v_{\delta}\rangle$ and $|C_{f_{\delta-2}}\langle v_{\delta}\rangle|-|C_{f_{\delta-2}}\langle v\rangle|=1$. In this case, $C_{f_{\delta-2}}\langle v_{\delta}\rangle$ has just one color different from that of $C_{f_{\delta-2}}\langle v\rangle$, and the other colors is the same as $C_{f_{\delta-2}}\langle v\rangle$. From the proof of Case1. we know that $f_{\delta-1}$ has at most $\delta-1$ forbidden colors, plus the one color different from that of $C_{f_{\delta-2}}\langle v\rangle$. Consequently, there are at most $\delta$ forbidden colors.

\textbf{Case 2.2.} $|C_{f_{\delta-2}}\langle v\rangle\oplus C_{f_{\delta-2}}\langle v_{\delta}\rangle|=2$. Then the forbidden colors should be only taken form $C_{f_{\delta-2}}\langle v\rangle\oplus C_{f_{\delta-2}}\langle v_{\delta}\rangle$, thus, there are $2$ forbidden colors.

\textbf{Case 2.3.} $|C_{f_{\delta-2}}\langle v\rangle\oplus C_{f_{\delta-2}}\langle v_{\delta}\rangle|\ge 3$. Clearly, taking one color from $\{1,2,\ldots,\lambda\}$ to dye $e_{\delta-1}$ yields $|C_{f_{\delta-2}}\langle v\rangle\oplus C_{f_{\delta-2}}\langle v_{\delta}\rangle|$ at most descend $1$. Hence, it has no forbidden colors in the case.

From the above discussion, there are at most $\delta$ forbidden colors for the statements.
\end{proof}

In light of the above, there is at most $\delta+\Delta-1$ forbidden colors in (i), and $\Delta-1$ forbidden colors in (ii), and $\delta$ forbidden colors in (iii). Therefore, $R_{\delta-1}\le (\delta+\Delta-1)+ (\Delta-1)+\delta=2\Delta+2\delta-2$. Notice that $\lambda-R_{\delta-1}=k\Delta+3-(2\Delta+2\delta-2)\ge(\delta-2)\Delta-2\delta+5
\ge(\delta-2)\delta-2\delta+5=(\delta-2)^2+1> 1$. So one can also choose one color from $\{1,2,\ldots,\lambda\}$ except for the $2\Delta+2\delta-2$ forbidden colors to dye $e_{i}$, and similarly obtain the new coloring $f_{\delta-1}$. Under the coloring $f_{\delta-1}$, it is not difficult to find that $v$ and $v_{\delta}$ are also `bad' vertices.

 At last, we consider the coloring $f_{\delta}$ on $e_{\delta}$. According to the analysis of the current coloring, there are four types forbidden colors in the following.
\begin{enumerate}
  \item [(i')] To ensure $f_{\delta}$ is a proper total coloring of $G$, the forbidden colors includes the following: $f_{\delta-1}(=f_{0}(v)),f_{\delta-1}(v_{\delta})(=f(v_{\delta}))$, $f_{\delta-1}(e_{i})$ for $i=1,2,\ldots,\delta-1$ and $f_{\delta-1}(v_{\delta}w_{\delta j})$ $(=f(v_{\delta}w_{\delta j}))$ for $j=1,2,\ldots,t_{\delta}$;
  \item [(ii')] To ensure $v_{\delta}$ and $w_{\delta j}$ are strongly distinguished, it follows from Lemma \ref{lem-1} that there is at most one color $c_{\delta j}$ to yield $C_{f_{\delta-1}}\langle w_{\delta j}\rangle=C_{f_{\delta-1}}\langle v_{\delta}\rangle\cup\{c_{\delta j}\}$ for each $j$, where $j=1,2,\ldots,t_{\delta}$;
  \item [(iii')] To ensure $v$ and $v_{i}(i=1,2,\ldots,\delta-1)$ are strongly distinguished, there will present two possibilities:
      \begin{enumerate}
        \item [(1)] when $C_{f_{\delta-1}}\langle v\rangle = C_{f_{\delta-1}}\langle v_{i}\rangle$ for some $i\in\{1,2,\ldots, \delta-1\}$, the forbidden color of $e_{\delta}$ should be taken from one of $\{f_{\delta-1}(v_{j})|j=1,2,\ldots,\delta-1\}$. Except for those forbidden colors and that of (i'), one can take any one color $c$ such that $C_{f_{\delta-1}}\langle v\rangle\cup \{c\}\not= C_{f_{\delta-1}}\langle v_{i}\rangle$.
      \item [(2)] when $C_{f_{\delta-1}}\langle v\rangle\not=C_{f_{\delta-1}}\langle v_{i}\rangle $for some $i\in\{1,2,\ldots, \delta-1\}$, then by Lemma \ref{lem-1},  $e_{\delta}$ has at most one forbidden color $c^{i}$ (say).
      \end{enumerate}
      \quad \quad If $(1)$ appears $0$ time, and $(2)$ appears $\delta-1$ times, in order to ensure the strongly distinguished property of $v$ and $v_{1},v_{2}, \ldots, v_{\delta-1}$, then the number of the forbidden colors of $e_{\delta}$ is at most $\delta-1$, say $c^{1},c^{2},\ldots, c^{\delta-1}$; if $(1)$ appears $t~(1\le t <\delta-1)$ times, and $(2)$ appears $\delta-1-t$ times, in order to ensure the strongly distinguished property of $v$ and $v_{i}(i=1,2,\ldots,\delta-1)$, then the number of the forbidden colors of $e_{\delta}$ is at most $\delta-1+\delta-1-t\le 2\delta-3$; if $(1)$ appears $\delta-1$ times, and $(2)$ appears $0$ time, then the number of the forbidden colors of $e_{\delta}$ is at most $\delta-1$. To sum up above, $e_{\delta}$ has at most $\max\{\delta-1,2\delta-3\}=2\delta-3$ forbidden colors in (iii').

\item [(iv')] To ensure $v$ and $v_{\delta}$ are strongly distinguished. According to (iii), there are two cases to discuss.

      \textbf{Case 1.} If $C_{f_{\delta-1}}\langle v\rangle\not =C_{f_{\delta-1}}\langle v_{\delta}\rangle$ while
      $C_{f_{\delta-2}}\langle v\rangle=C_{f_{\delta-2}}\langle v_{\delta}\rangle$. It can be checked that $e_{\delta-1}$ has no
      forbidden color as $f_{\delta}$ is a proper total coloring of $G$.

      \textbf{Case 2.} If $|C_{f_{\delta-1}}\langle v\rangle\oplus C_{f_{\delta-1}}\langle v_{\delta}\rangle|\ge 2$, then the coloring of $e_{\delta}$ at most yields the symmetric minus decline $1$, so the number of forbidden color is also zero.
\end{enumerate}

From the above we known that there are at most $\delta+\Delta$ forbidden colors in (i'), and $\Delta-1$ forbidden colors in (ii'), and $2\delta-3$ forbidden colors in (iii'), respectively. Note that (iv') has no forbidden color. So we have
$$R_{\delta}\le (\delta+\Delta) +(\Delta-1)+ 2\delta-3 =2\Delta+3\delta-4.$$

Hence, $\lambda-R_{\delta}\ge k\Delta+3-(2\Delta+3\delta-4)\ge (\delta-2)\Delta-3\delta+7\ge (\delta-2)\delta-3\delta+7=(\delta-2.5)^2+0.75\geq 1$ since $\delta$ is an integer. Thus, $f_{\delta}$ is a $\lambda$-$D(1)$-VDSTC of $G$.

\textbf{Case B.} If $\delta=2$, then one should consider three types of the forbidden colors for the coloring $f_{1}$ on $e_{1}$: subcases (a), (b) and (c) respectively. The first two subcases see \textbf{Case A}, the repetitions will not be made here. Now we only give subcase (c) as follows:
\begin{enumerate}

\item [(c)] To avoid $C_{f_{1}}\langle v\rangle =C_{f_{1}}\langle v_{2}\rangle$, there is at most one forbidden color for $e_{1}$  since if either $C_{f_{0}}\langle v\rangle\not =C_{f_{0}}\langle v_{2}\rangle$, by Lemma \ref{lem-1} there is a possible color for $e_{1}$ to yield $C_{f_{1}}\langle v\rangle =C_{f_{1}}\langle v_{2}\rangle$, or $C_{f_{0}}\langle v\rangle =C_{f_{0}}\langle v_{2}\rangle$ then, the forbidden color of $e_{1}$ is only taken by $f(v_{2})$.

\end{enumerate}
From \textbf{Case A.} we know that there are at most $2\Delta$ forbidden colors in (a) and (b), and there is one forbidden color in (c). Therefore, $R_{1}\le (\Delta+1)+ (\Delta-1)+1=2\Delta+1$. Note that $\lambda-R_{1}=k\Delta+3-(2\Delta+1)\ge(k-2)\Delta+2\ge 2$. So one can also choose one color from $\{1,2,\ldots,\lambda\}$ except for the above $2\Delta+1$ forbidden colors to dye $e_{1}$, and obtain the new coloring $f_{1}$. Under the coloring $f_{1}$,  it can be seen that $v$ and $v_{2}$ are also `bad' vertices.

For the coloring $f_{2}$ on $e_{2}$, we also consider three types of the forbidden colors: subcases (a'), (b'), (c') and (d') respectively. Subcases (a') and (b') shown in \textbf{Case A} while $i=2$, here is no longer repeated it. Now we give subcases (c') and (d') in the following.
\begin{enumerate}

\item [(c')] To obtain $C_{f_{2}}\langle v\rangle \not=C_{f_{2}}\langle v_{2}\rangle$, $e_{2}$ has no any forbidden color since $|C_{f_{1}}\langle v\rangle \oplus C_{f_{1}}\langle v_{2}\rangle|\ge 2$, which can be easily deduced from the above subcase (c) and Claim \ref{claim-2}, here we omit its proof.
\item [(d')] To ensure that $C_{f_{2}}\langle v\rangle\not \not=C_{f_{2}}\langle v_{1}\rangle$, there is at most one forbidden color for $e_{2}$ whether $C_{f_{1}}\langle v\rangle$ and $C_{f_{1}}\langle v_{1}\rangle$ are identical or not.
\end{enumerate}
Analogous to the above of $R_{1}$, one can get $R_{2}\le (\Delta+2)+ (\Delta-1)+0+1=2\Delta+2$. Thus, there is at least one color from $\{1,2,\ldots,\lambda\}$ except for the above $2\Delta+2$ forbidden colors to dye $e_{2}$. Consequently, $f_{2}$ is a $\lambda$-$D(1)$-VDSTC of $G$.

\textbf{Case C.} If $\delta=1$, then $f_{1}$ has two types of the forbidden colors on $e_{1}$: subcases (a) and (b) respectively, see \textbf{Case A}. Clearly, there are at most $2\Delta$ forbidden colors in those situations. On the other hand, one should consider the strongly distinguished property of $v$ and $v_{1}$. Clearly, $C_{f_{0}}\langle v\rangle \not =C_{f_{0}}\langle v_{1}\rangle$ since if not, $|C_{f_{0}}\langle v\rangle|=2$ but $|C_{f_{0}}\langle v_{1}\rangle|\ge 3$, a contradiction. So it follows from Lemma \ref{lem-1} that there is at most one forbidden color here. Therefore, $R_{1}\le 2\Delta+1$, and so, it can be deduced that $f_{1}$ is a $\lambda$-$D(1)$-VDSTC of $G$.

Sum up the above, the proof is completed.  \hfill$\square$
\end{proof1}

\begin{proof2}
\rm Since $T$ is a bipartite graph, $T$ admits a $2$-vertex coloring, namely $\chi(T)=2$. From Lemma \ref{lem-5}, one can obtain $\chi'_{s}(T,2)\le \Delta(T)+1$. Thus, it follows from Theorem \ref{thm-3} that $\chi_{2-vsdt}(T)\le \Delta(T)+3$.

Similarly, if $d(u,v)\ge3$ for any two vertices $u,v\in V(T)$ with maximum degree, one can also obtain that $\chi_{2-vsdt}(T)\le \Delta(T)+2$.

The proof follows. \hfill$\square$
\end{proof2}

\section{Further works and problems }

As far as we know, for the \emph{(adjacent-) vertex-distinguishing colorings}, there exist some proper subgraphs $H$ of $G$ such that its \emph{(adjacent-) vertex-distinguishing
chromatic number} is greater than that of $G$, see \cite{zhang-4,zhang-1,wen} for instance.
Thus, we propose the following problem as a further work.
\begin{prob}\label{problem-1}
For any positive integer $r$, if $H$ be a proper subgraph of $G$, what does $G$ satisfy to yield $\chi_{r-vsdt}(G)<\chi_{r-vsdt}(H)$ ?
\end{prob}
From Theorem 6 of literature \cite{zhang-3}, one can obtain  $\chi_{1-vsdt}(K_{n})\ge n+\lceil\log_{2}^{n}\rceil$, so too is $\chi_{r-vsdt}(K_{n})$ for $r\ge 2$ since $\mathbf{d}(K_{n})=1$. In fact, if $G$ admits an $r$-VDSTC, then the restriction of the color-set on each vertex of $G$, that is, any two vertices of $G$ with distance at most $r$ are \emph{vertex-strongly-distinguishing}, can be viewed as two adjacent vertices with \emph{$1$-vertex-strongly-distinguishing property}. Moreover, Zhang et al. in \cite{zhang-3} also obtained the accurate value of  $\chi_{1-vsdt}(K_{n})$ for $3\le n\le 18$, so we can get the following table.

\begin{table}[htbp]
 \caption{$\chi_{r-vsdt}(K_{n})$ for $3\le n\le 18$}\label{tab-1}
 \begin{tabular}{c|cccccccccccccccccc}
  \toprule
$n$ &3 & 4 &5&6&7&8&9&10&11&12&13&14&15&16&17&18  \\
  \midrule
$\chi_{r-vsdt}(K_{n})$ &5 & 6 &8&10&10&11&13&14&15&16&17&19&19&20&22&23 \\
  \bottomrule
 \end{tabular}
\end{table}
Hence, combining the above we have the following conjecture.
\begin{conj}\label{conj-1}
Let $G$ be a graph on $n$ vertices and without isolated edges. Then
$$\chi_{r-vsdt}(G)\le n+\lceil\log_{2}n\rceil+1$$
where the equality holds if $n=2^{k}-2$.
\end{conj}

From Hatami's results\cite{hatami} and Theorem \ref{thm-3} we know that if $\Delta(G)$ is very large, then $\chi_{1-vsdt}(G)\le 2\Delta(G)+300$. So we give the following conjecture.
\begin{conj}\label{conj-2}
For any graph $G$ without no isolated edges, there is a positive constant $c$ such that
$$\chi_{1-vsdt}(G)\le 2\Delta(G)+c.$$
\end{conj}

config: arxiv


TITLE: Epsilon-Delta proof of $\lim_{x\to 2} x^2=4$
QUESTION [6 upvotes]: I have seen and understand the delta-epsilon proof of the limit of $x^2$ for $x\to2$, such as explained here: https://www.youtube.com/watch?v=gLpQgWWXgMM
Now I am wondering, is there also another way? How about this:
Verify that $\lim x^2=4$ (for $x\to2$)
STEP A: Express epsilon in terms of $x$:
\begin{align}
|x^2-4| &< \varepsilon\\
-\varepsilon &< x^2-4 < \varepsilon\\
4-\varepsilon &< x^2 < 4+\varepsilon\\
\sqrt{4-\varepsilon} &< x < \sqrt{4+\varepsilon}
\end{align}
STEP B: Express delta in terms of $x$
\begin{align}
|x-2| &< \delta\\
-\delta &< x-2 < \delta\\
2-\delta &< x < 2+\delta
\end{align}
STEP C: Now we can express $\delta$ in terms of $\varepsilon$ hence proving the limit.
If we take $\delta=\min\{-2+\sqrt{4+\varepsilon},2-\sqrt{4-\varepsilon}\}$ then the limit is proven
Did I make any mistake? Thanks! Cheers!

REPLY [2 votes]: This is the general solution for a problem like this but the final steps only work in $\mathbb{R}$.

Problem: $\forall\epsilon>0$ $\exists\delta>0$ such that $|x-2|<\delta$ implies $|x^2-4|<\epsilon$

This may seem a bit over-complicated but it is saying, $$\lim_{x \to 2} x^2=4$$ if you can derive $|x^2-4|<\epsilon$ from $|x-2|<\delta$.
Solution:
$|x^2-4|=|x-2||x+2|$ therefore $|x-2|<\frac{\epsilon}{|x+2|}$. We can't have that since delta must be a function of just epsilon. The theorem only applies when $x$ is close to 2 we can set, $|x-2|<1$ (this is the step that only works in the reals) this implies $-1<x-2<1$
 which implies $x<3$.
Therefore, $\frac{\epsilon}{|x+2|}<\frac{\epsilon}{5}$ therefore the $\delta=\min \left\{1,\frac{\epsilon}{5}\right\}$.
Post Script: I believe we can go so far as to set $\delta=\frac{1}{2}\min \left\{1,\frac{\epsilon}{5}\right\}$.


TITLE: How many straight lines can be made between 10 points such that 4 of them are colinear?
QUESTION [4 upvotes]: So i know how to get the answer. We just have to find $C(10,2)$ and subtract $C(4,2)$ and add 1.
We are basically counting all the points between co-linear points as 1.
So the question is why we are doing that.
Lets consider 4 points to be $A,B,C,D$ and the lines that could be made from them are $AB,BC,CD,DA,AC,BD$.
Clearly these lines are different from each other unless it is specified that distance between them are equal.
So Why we are counting all of them as 1 ? This makes no sense to me.

REPLY [3 votes]: You have to assume that there are no lines through more than 2 points except for the line $L$ containing the 4. So for each of the points on $L$ we have 6 lines to the other points. That is 24 lines plus $L$, giving 25. We then have lines between the other 6 points. There are ${6\choose 2}=15$ pairs of points each of which gives a line, so a grand total of 40. As a check note that there are ${10\choose 2}=45$ pairs of points in all. But ${4\choose 2}=6$ give the same line $L$, in other words 5 duplicates, so a total of $45-5=40$ distinct lines.


TITLE: The set that only contains itself
QUESTION [12 upvotes]: Ignoring the axiom of regularity (and therefore the implication of "no set can contain itself"), would it be correct to state that the set that contains only itself is unique?
My argument is that if $x$ is said set, then
$$ x = \{x\} = \{\{x\}\} = \{\{\{\cdot\cdot\cdot\}\}\} $$
Ad infinitum, which seems to be unique.

REPLY [14 votes]: It need not be unique; in fact, you can weaken the axiom of foundation to allow either well-founded sets or sets of the form $x=\{x\}$. Sets of the latter form are called Quine atoms, and play the role of urelements. These are useful in set theory with atoms because they allow you to formulate a theory of sets-and-atoms without resorting to multi-sorted logic.
For more, see The Axiom of Choice by T. J. Jech.


TITLE: Is there an algebraic number that cannot be expressed using only elementary functions?
QUESTION [15 upvotes]: (this is basically a repost of a question I asked at M.SE last year)
Is there an explicit real algebraic number (such that we can write its minimal polynomial and a rational isolating interval) that cannot be expressed as a combination of the constants $\pi, e,$ integers and elementary functions (rational functions, powers, logarithms, direct and inverse trigonometric functions)?
In case it is an open question, can you give an example of an algebraic number such that we do not know how to express it in elementary functions?

REPLY [23 votes]: I addressed this exact question in my American Mathematical Monthly paper,
What is a closed-form number? Corollary 1 in that paper states that if Schanuel's conjecture holds, then the EL numbers (i.e., the numbers expressible according to your list of rules) that are algebraic are precisely the numbers expressible using radicals.  So any algebraic number that is not expressible in radicals would be an answer to your question.  However, Schanuel's conjecture is not known to be true.  I believe that your question is still open.


TITLE: Probability of 'at least' event
QUESTION [1 upvotes]: In a roulette of letters A and B, the probability of choosing A is 0.92, B is 0.33, and both is 0.29. What is the probability of choosing at least 1 of the letters?

P(at least 1) = P(A) + P(B) - P(A and B) = 0.92 + 0.33 - 0.29 = 0.96

I am confused if this is the right way to solve P (at least 1), that is, similar to P(A U B). I have consulted other references but they do not give examples with given probabilities, particularly P(both).

REPLY [1 votes]: The $P(\text{at least }1)$ is exactly $P(A \cup B)$. To calculate $P(A \cup B)$, when you add up $P(A)$ and $P(B)$, the part they share, i.e, $P(A \text{ and }B)$, has been added for two times so we subtract $P(A \text{ and }B)$ from the sum.


text:
{\bf Problem.} The median of a set of consecutive odd integers is 138. If the greatest integer in the set is 145, what is the smallest integer in the set?
{\bf Level.} Level 3
{\bf Type.} Prealgebra
{\bf Solution.} The median of a set of consecutive integers is the middle value of that set. Since the median is an even number, but there are only odd integers in this set, there must be an even number of integers in the set. The set must be $$\{131, 133, 135, 137, 139, 141, 143, 145\},$$ and $\boxed{131}$ is the smallest integer in the set.

set_name: MATH


TITLE: Tangents to a parabola that go through the same point
QUESTION [1 upvotes]: The question is: The two lines tangent to f (x) = $x^2$ + 4x + 2 through the point (2, -12)have equations y = ax + b and y = cx + d, respectively. What is the value of a + b + c + d?
What I did to solve it:
f '(x) = 2x + 4. The point is (2 -12) so I plugged in two to get f '(2) = 8. I used this in point-slope form to get y + 12 = 8(x-2) => y = 8x -28. Since it is two equations, I added them up and multiplied by 2 and got -40, which is the answer. I feel like this is a fluke and it doesn't make sense to me why this worked. Can someone explain either why it works or a way that will always work? Thanks in advance!

REPLY [1 votes]: This is no coincidence, or if it is, it’s a happy one.
Let $P_0=(x_0,y_0)$ be the given external point and $x_1$ and $x_2$ the ordinates of the points of tangency. The equations of the two tangent lines are $y=f'(x_1)(x-x_0)+y_0$ and $y=f'(x_2)(x-x_0)+y_0$. On the other hand, the equation of the line that you found is $y=f'(x_0)(x-x_0)+y_0$. So, for the given parabola, we have $$f'(x_1)+y_0-f'(x_1)x_0+f'(x_2)+y_0-f'(x_2)x_0 = 2(f'(x_0)+y_0-f'(x_0)x_0)$$ or $$(1-x_0)(f'(x_1)+f'(x_2))+2y_0 = 2(1-x_0)f'(x_0)+2y_0$$ which reduces to $$f'(x_1)+f'(x_2)=2f'(x_0).\tag{1}$$
As it turns out, this holds for any parabola given by the general formula $f(x)=ax^2+bx+c$. The ordinates of the points of tangency are the solutions to $$f(x)-y_0=f'(x)(x-x_0),$$ which after expanding, rearranging and simplifying becomes $$x^2-2x_0x+{y_0-bx_0-c\over a}=0.\tag{2}$$ On the other hand, substituting the general formula into (1) yields $x_1+x_2=2x_0$, which can be verified by a glance at (2).
Observe that this last equation tells us that $x_0$ is midway between $x_1$ and $x_2$, which means that the chord connecting the two tangent points is parallel to the tangent at $x=x_0$. This follows from general properties of parabolas, but can be verified directly: $$\begin{align} f(x_2)-f(x_1) &= f'(x_2)(x_2-x_0)-f'(x_1)(x_1-x_0) \\ &= (2ax_2+b)(x_2-x_0)-(2ax_1+b)(x_1-x_0) \\ &= (2a(x_1+x_2-x_0)+b)(x_2-x_1) \\ &= (2ax_0+b)(x_2-x_1) \end{align}$$ therefore $${f(x_2)-f(x_1)\over x_2-x_1}=2ax_0+b=f'(x_0).$$ A similar derivation demonstrates that for an arbitrary chord, the tangents at its endpoints intersect at a point with ordinate halfway between the ordinates of those endpoints. Combining this with (1), we find that the slope of a chord is equal to the average of the slopes of the tangents at its endpoints.


TITLE: Update: Matrix-Diagonal Multiplication Proof Verification
QUESTION [0 upvotes]: Define the following:

$\mathbf{T}\in\mathbb{R^{n\times n}}$ an $n\times n$ matrix.
$\mathbf{c}\in\mathbb{R^{n}}$ a column vector.
$\Psi=\operatorname{diag}(c)$ outputs an $n\times n$ diagonal matrix.

I am asked to show when I multiply $\mathbf{\Psi}\mathbf{T}$, I am essentially multiplying each row $i$ of $\mathbf{T}$ by $c_{i}$.

I proceed to show this as follow:
A matrix-matrix multiplication is defined by the following mathematical operation:
$$
\mathbf{AB}=\left[\sum_{{k=1}}^{n}\mathbf{A}(i,k)\mathbf{B}(k,j)\right]_{i,j}=\left[\sum_{{k=1}}^{n}a_{i,k}b_{k,j}\right]_{i,j}
$$
Our case, the matrix $\mathbf{\Psi}\mathbf{T}$ is non-zero whenever $i=k$ this means that:
\begin{align*}
\mathbf{\Psi T}&=\left[\sum_{k=1}^{n}\mathbf{\Psi}(i,k)\mathbf{T}(k,j)\right]_{i,j}\\\\
&=\left[\mathbf{\Psi}(i,i)\mathbf{T}(i,j)\right]_{i,j}\\\\
&=\left[\operatorname{diag}(\mathbf{c})(i,i)\mathbf{T}(i,j)\right]_{i,j}\\\\
&=c_{i}\mathbf{T}(i,:)
\end{align*}
where $1\leqslant i,j\leqslant n$.
I am unsure if I solved it correctly or not.
[Note]: I may need to state that $\mathbf{T}(i,:)$ is MATLAB's way of looping over the columns of $\mathbf{T}$ for a fixed row $i$.
[Update]: I have taken @0XLR remarks into account and have update my derivation. I hope someone confirms if this is a complete proof.

REPLY [4 votes]: $\def\D#1{{\rm Diag}\left(#1\right)}\def\e{\varepsilon}$Suppose
you have two vectors and their corresponding diagonal matrices
$$\eqalign{
A = \D{a}\qquad B = \D{b} \\
}$$
The product of these diagonal matrices with an arbitrary matrix
$X\in{\mathbb R}^{n\times m}$ is equivalent to the Hadamard product
of the matrix with the two vectors
$$AXB = ab^T\odot X$$
Since the matrix can be written as a sum of its columns $\{c_i\}$ or rows $\{r_j^T\}$ multiplied by the cartesian basis vectors, i.e.
$$\eqalign{
X &= \sum_{i=1}^m c_i\e_i^T = \sum_{j=1}^n e_jr_j^T \\
}$$
Therefore the product with the diagonal matrices can be expanded as
$$\eqalign{
AXB
 &= \sum_{i=1}^m ab^T\odot c_i\e_i^T
 &= \sum_{i=1}^m (a\odot c_i)(b\odot\e_i)^T \\
 &= \sum_{j=1}^n ab^T\odot e_jr_j^T
 &= \sum_{j=1}^n (a\odot e_j)(b\odot r_j)^T \\
}$$
So, as you suspected, the columns get multiplied (element-wise) by $a$ and the rows by $b$.
In your particular problem $\;\,B=I,\;b={\tt1},\;(b\odot\e_i)=\e_i$


TITLE: quantum states and observables for the non-commutative torus
QUESTION [3 upvotes]: The noncommutative torus $A_\theta$ is a $C^*$-algebra corresponding to an irrational foliation of the torus $\mathbb{T}^2$ by lines of slope $\theta \notin \mathbb{Q}$.
As far as I am reading it is generated by two bounded linear operators $U,V$ acting on $L^2(\mathbb{T})$.
$$ U(f)(z) = z f(z) \hspace{0.5in} V(f)(z) = f(e^{-2\pi i \theta}z) $$
This is cool... but what quantum mechanical system does this correspond to?
In particular,

can we find a hamiltonian $\mathcal{H}$ corresponding to this system?
Or is this a quantization of a different Poisson structure than the free particle?

In order to address this question, I tried writing the two formulas in bra-ket notation:
$$ \langle f |U | z \rangle = z \langle f| z\rangle \hspace{0.5in} \langle f |V| z \rangle =  \langle f| e^{-2\pi i \theta} z\rangle $$
Here $\langle f | \in L^2 (\mathbb{T})$ and $|z\rangle \in \mathbb{T}$.  In a sense it didn't matter which function $f$ we picked we can just write:
$$ U | z \rangle = z | z\rangle \hspace{0.5in} V| z \rangle =  | e^{-2\pi i \theta} z\rangle $$
These look like the position and momentum operators and we get an representation of the Heisenberg algebra, since $UV = e^{2\pi i \theta} VU$.

Basically... $C^\ast$ algebras supposed to be the "same" as quantum mechanical systems.  So what is the quantum mechanical system corresponding to the non-commutative torus $A_\theta$ ?
The article says define Poisson bracket on the phase space $C^\infty(\mathbb{T}^2)$ by
$$\{ f,g \} = \theta \left( \frac{\partial f}{\partial x_1} \frac{\partial g}{\partial x_2} - \frac{\partial f}{\partial x_2} \frac{\partial g}{\partial x_1} \right)$$
Then build this deformed multiplication $\ast_\hbar$ such that $$f \ast_\hbar g - g \ast_\hbar f = i\hbar\{ f,g\} + O(\hbar^2) $$
Then quantizing we get $[ x_1, x_2] = i\theta\hbar$, which looks like $\theta$ could be absorbed into $\hbar$ ? This might be semiclassical analysis or something.
I don't understand this:

When $n=2$ the skew-symmetric matrix $\theta$ is just determined by a real number, again denoted $\theta$, and $A_\theta$ will be isomorphic to the crossed-product $C^\ast$-algebra for the action of $\mathbb{Z}$ on the circle $\mathbb{T}$ coming from rotation by angle $2\pi \theta$.  For this reason the algebras are often called rotation algebras or irrational rotation algebras if $\theta \notin \mathbb{Q}$.

REPLY [2 votes]: I am not completely sure what the question is here but I'all attempt an answer anyway, hopefully with some reasonable informations.
Point 1 - Dependence on $\theta$ How does the non commutative algebra $A_\theta$ depends on $\theta$? Much. The first obvious difference is between rational and irrational, If $\theta=\frac p q $ something similar to what happens on quantum groups at roots of unity shows up: a big center, in fact $Z(A_{\frac p q})={\cal C}(\mathbb T^2)$ and $A_{\frac p q}$ turns out to be isomorphic to the algebra of continous sections of a vector bundle over $\mathbb T^2$ whose fibers are $q\times q$ matrix algebras. What if $\theta$ is irrational? Well, well, things are not so easy in that case as well. Consider, for example, that if $\theta,\theta'\in ]0,\frac 1 2 [$ are two irrationals then $A_\theta$ and $A_{\theta'}$ are not isomorphic as $C^*$--algebras. But we know that isomorphism as $C^*$--algebras is pretty difficult to obtain, right? Let us relax things and ask for a weaker condtion, Morita equivalence. Then you can prove that two quantum torus algebras for two different values of the parameter are Morita equivalent if and only if
$$
\exists\, A=\left(\begin{array}{cc}a&b\\ c& d\end{array}\right)\in GL(2;\mathbb Z)\, : \theta'=\frac{a\theta+b}{c\theta+d}
$$
thus showing that dependence on the parameter is quite subtle. In this sense it is not easily "rescaled"... (best reference on this, imho, chapter 12 of Elements of NC Geometry, Gracia-Bondia, Varilly, Figueroa).
Point 2 - Relation with Poisson The formula as a $\star_\hbar=\star_{\hbar,\theta}$--product shows that there is a relation with a specific Poisson structure on the torus. In fact that $\theta$--depending Poisson brackets you wrote exhausts all possible left invariant Poisson structures on $\mathbb T^2$: they are all symplectic. From each fixed $A_\theta$ you can recover the corresponding Poisson bracket as a semiclassical limit, basically as
$$
\{f,g\}_\theta=\frac{f\star_{\hbar,\theta}g-g\star_{\hbar,\theta}f}{\hbar}\, \mathrm{mod}\,\hbar
$$
which can be made more precise in the context of continous fields of $C^*$--algebras. Remark that the semiclassical limit requires an identification of ${\cal C}^\infty(\mathbb T^2)$ with a subalgebra of $A_\theta$ and this identification is $\theta$--depending. For this part papers by Rieffel keep being the best possible source of information. Let me add this reference http://arxiv.org/pdf/math/0305413v1.pdf as source of very interesting consideration on the relation between classification of Poisson structures (as Dirac ones) and Morita equivalence of $C^*$--algebras. Always keep in mind that quantization, here, is not a functor...
Point 3 - Poisson geometry+irrational rotation algebra which seems to puzzle you somehow. Either you take it as a fact: life is hard and NC torus are isomorphic as $C^*$--algebras to certain crossed products, or you try to give it a geometric meaning which, as often happens, brings in a certain amount of complication... One possible way to quantize Poisson manifolds is the following: if the Poisson manifold is integrable, integrate it to a symplectic groupoid and then perform geometric quantization on this symplectic groupoid. Here the symplectic groupoid, being $\mathbb T^2$ symplectic itself, is $T^*\mathbb T^2$. You have two possible reasonable choices of polarization: one is the vertical one, like in every vector bundle: since fibers are planes you have no Bohr-Sommerfeld conditions and you basically end up with an identification of ${\cal C}^\infty(\mathbb T^2)$ with a subalgebra of $A_\theta$ when you identify functions on the base with polarized sections. Or you can take a cylindrical polarization in which leaves are of the form $\mathbb T\times\mathbb R$; then you have a Bohr-Sommerfeld conditions on the first factor so that your space of quantizable leaves is of the form $\mathbb Z\times\mathbb R$ and when you carry through the whole procedure you end up with the irrational rotation algebra. There are too many details to be explained here. This is described originally in Weinstein, A., Symplectic groupoids, geometric quantization, and irrational rotation algebras, in Symplectic geometry, groupoids, and integrable systems, Seminaire sud-Rhodanien de geometrie a Berkeley (1989), P. Dazord and A. Weinstein, eds., Springer-MSRI Series (1991), 281–290; but you can also profit by much later work of Eli Hawkins in http://arxiv.org/abs/math/0612363v3
Hope this answers some of your questions...


TITLE: Solutions of $x\ln x=u$ is called $x(u)$, show $\frac{u}{\ln u}<x\leq (1+1/e)\frac{u}{\ln u}$. Here $u>e$
QUESTION [0 upvotes]: Solutions of $x\ln x=u$ is called $x(u)$, show $\frac{u}{\ln u}<x\leq (1+1/e)\frac{u}{\ln u}$. Here $u>e$.
I could just see $x\ln x$ is strictly increasing, so $\lim_{u\to\infty}x(u)=+\infty$. How to prove this exact control?
Next, how to show $x(u)\sim \frac{u}{\ln u}$ as $u\to+\infty$?

REPLY [2 votes]: $f(x) = x \ln(x)$ is strictly increasing on $[1, \infty)$ with $f(e) = e$, so that $u > e$ implies $x > e$. Then $u = x \ln(x) > x$ and therefore
$$
\frac{u}{\ln(u)} < \frac{u}{\ln(x)} = x \, .
$$
For the other direction we use the convexity of
$$
g(u) = e^u \ge g(1) + (u-1)g'(1) = e u
$$
so that
$$
 x^{1/e} = e^{\ln(x) / e} \ge \ln(x)
$$
and therefore
$$
\left(1 + \frac 1e \right) \frac{u}{\ln(u)} = \left(1 + \frac 1e\right) \frac{x \ln(x)}{\ln(x \ln(x))}
= x \frac{\ln(x) + \ln(x^{1/e})}{\ln(x) + \ln(\ln(x))} \ge x \, .
$$
Equality holds if and only if $x=e^e$, $u=e^{e+1}$.
In order to show that $x(u)\sim \frac{u}{\ln (u)}$ for $u \to \infty$ we need to show that, given $\epsilon > 0$,
$$
 x(u) < (1+\epsilon) \frac{u}{\ln (u)}
$$
for sufficiently large $u$. A simple calculation shows that this is equivalent to
$$
 \frac{\ln(\ln(x(u)))}{\ln(x(u))} < \epsilon
$$
and that is true for large $u$ because the left-hand side converges to zero if $u \to \infty$.


TITLE: The topology of a "closed" universe - is it really closed?
QUESTION [2 upvotes]: The spatial part of the positive curvature FRW metric has the form
\begin{equation}
ds^2=\frac{dr^2}{1-(r/R)^2}+r^2d\Omega^2
\end{equation}
or
\begin{equation}
ds^2=R^2(d\chi^2+\sin{\chi}^2d\Omega^2)
\end{equation}
This is described as "closed", as it has the metric of a three-sphere, but I want to know what this actually means. Is $\chi$ not limited to the interval $[0,\pi/2)$, since beyond that we are simply reproducing the same $r$? In which case, is the universe not really one "hemi-threesphere" and not closed in the same way as a full sphere is?
(Edit: I note that Carroll states that the "only possible global structure is the complete three-sphere", but he doesn't go into why.)

REPLY [3 votes]: The first metric form uses coordinates that are valid only in a patch of the whole space.  This is a fairly common situation.  Consider the example of a 2-sphere projected onto a tangent plane from the center.  This is the gnomonic projection which maps all great circles into straight lines.  The $(x,y)$ coordinate system on the tangent plane define a coordinate system on a patch of the sphere but not the whole sphere.  The sphere is still all there though.  Projecting from infinity onto the tangent plane maps the the sphere onto the unit disk, and $(r,\theta)$ on this disk are the equivalent to the first metric form.  But each point on the disk represents two points on the sphere, so you restrict the mapping to a patch which is a hemisphere.


TITLE: Aren't these linear transformations only orthogonal with respect to *some* inner product?
QUESTION [2 upvotes]: On p.154 in Husemoller's Fibre Bundles, during his introduction of Clifford algebras, I found a claim which seems questionable to me (highlighted in red):

You can click here for some context (e.g. the definition of "orthogonal multiplication") but assuming I've expanded everything correctly, here's what I think the remark is saying:

A module $M$ over this Clifford algebra is the data of a real vector space $M$ with a choice of $u_1,\ldots,u_{k-1}\in\mathrm{GL}(M)$ satisfying $u_i^2=-1$ and $u_iu_j+u_ju_i=0$ for $i\neq j$.
Given such an $M$, there is an inner product $(x|y)$ on $M$ and an isomorphism of inner product spaces $\varphi:M\to\mathbb{R}^n$, where $\mathbb{R}^n$ has the Euclidean inner product, such that the $\varphi\circ u_i\circ \varphi^{-1}$ are elements of $\mathrm{O}(n)$.

My confusion is that while the inner product $(x|y)$ on $M$ is certainly constructed so as to ensure that the linear transformations $u_1,\ldots,u_{k-1}$ are orthogonal for that inner product, I see no reason for that inner product to be equivalent to the/an Euclidean inner product on $M$.
In other words, if the inner product $(x|y)$ has signature (Wikipedia link) equal to $(p,q)$, then I agree we can find an isomorphism $\varphi:M\to\mathbb{R}^n$ such that the $u_i$ get sent to elements of $\mathrm{O}(p,q)$ (Wikipedia link), but I see no reason we can ensure $(x|y)$ has signature $(n,0)$.
Now, the definition of $(x|y)$ depends on an original choice of inner product $\langle x|y\rangle$ on $M$, so I suppose it's possible we can always find some $\langle x|y\rangle$ such that $(x|y)$ has the right signature.
If that's the case, can someone provide a proof? Or is the remark incorrect? Or have I misinterpreted the remark?
If any of the terms are unclear, I'm happy to provide more screenshots from the book.
Thanks in advance for any help.

REPLY [0 votes]: I think I was just being overly cautious here.

The starting inner product $\langle x|y\rangle$ on $M$ is positive definite, i.e. of signature $(n,0)$, simply by the definition of "inner product".
For each $\sigma\in \Gamma_{kl}$, the form $\langle \sigma(x)|\sigma(y)\rangle$ on $M$ is positive definite, again by simple observation: for any $x\in M$ $\sigma(x)$ is again an element of $M$, so $\langle \sigma(x)|\sigma(x)\rangle>0$. This is I think what was the sticking point for me before - for whatever reason I was convinced there was no guarantee $\langle \sigma(x)|\sigma(y)\rangle$ would necessarily be positive definite.
A sum of positive definite matrices is positive definite, so since $\langle \sigma(x)|\sigma(y)\rangle$ is positive definite (and symmetric) for any $\sigma\in\Gamma_{kl}$, their sum $(x|y)$ is as well.
All inner products on a finite-dimensional $\mathbb{R}$-vector space are equivalent, via the obvious argument (take an orthonormal basis for one inner product and define a map by sending it to an orthonormal basis for another inner product)


text:
\subsection{Notion of rationalizability}
\label{sec:Rationalizability}

Let $k$ be any field.
Consider the polynomial ring $R:=k[X_1,...,X_n]$, and denote its field of fractions by $Q:=k(X_1,...,X_n)=\Frac R$.
If we let $f$ and $g$ be two polynomials in $R$ with $g$ non-zero, then $f/g\in Q$.
Fix an algebraic closure $\overline{Q}$ of $Q$, and consider the quantity $\sqrt{f/g}\in \Qbar$.

\begin{definition}\label{d:rationalizability}
We call the square root $\sqrt{f/g}$ {\it rationalizable}
if there is a homomorphism of $k$-algebras $\phi\colon Q \to Q$ such that $\phi(f/g)$ is a square in $Q$.
Otherwise, we say that $\sqrt{f/g}$ is {\it not rationalizable}.
\end{definition}

\begin{remark}
Since $\phi\colon Q\to Q$ is a homomorphism of $k$-algebras and since $Q$ is a field, $\phi$ is in particular a homomorphism of fields and preserves the zero element and the unit.
This implies that $\phi$ is automatically non-constant and, in particular, injective.
\end{remark}

\begin{remark}
Definition~\ref{d:rationalizability} is motivated by the fact that physicists are looking for a {\em change of variables} that turns $f/g$ into a square while preserving the number of variables, i.e., the number of newly introduced variables should be equal to the number of original variables.
Such a change of variables is a non-constant homomorphism of $k$-algebras from $Q$ to $Q$ and vice versa.
\end{remark}

\begin{example}\label{e:defrationalizability}
    The square root $\sqrt{1-X^2}$ is rationalizable.
    If $k$ has characteristic $2$, then
    $$
    1-X^2=X^2+1=(X+1)^2
    $$
    and hence $\sqrt{1-X^2}=X+1$
    (the rationalizing map being the identity).
    If $\mathrm{char}\, k\neq 2$,
    consider $\phi\colon k(X) \to k(X)$
    defined by
    $$
    X \mapsto \frac{2X}{X^2+1},
    $$
    that is, the map that sends the polynomial $g(X)$ to $g\left(\frac{2X}{X^2+1}\right)$.
    Then, we have
    $$\phi (1-X^2)= \left( \frac{X^2-1}{X^2+1}\right)^2.$$
    A square root that is \textit{not} rationalizable is, for example, $\sqrt{1-X^3}$.
    We will explain how to prove its non-rationalizability in Section \ref{sec:criteria} (cf. Corollary~\ref{c:DegreeCriterion}).
\end{example}

\begin{remark}
  Let us stress that our notion of non-rationalizability only implies that there is no \textit{rational} substitution that rationalizes the square root.
  For example, the substitution
  \begin{equation*}\label{eq:FakeMap}
      X\mapsto \sqrt[3]{-X^2+2X}
  \end{equation*}
  turns $\sqrt{1-X^3}$ into a square in $Q$.
  However, it does not define a homomorphism $Q\to Q$ since $\sqrt[3]{-X^2+2X}\notin Q$.
  Physicists dealing with Feynman integrals will mainly be interested in the existence of \textit{rational} substitutions since these do not introduce new square roots in other parts of their computation.
  Nevertheless, the existence of non-rational (called \textit{radical}) parametrizations is studied and applied to other contexts in~\cite{SS11, SS13} and~\cite{SSV17}.
\end{remark}

\begin{remark}
If $\A^n_k$ denotes the affine space with $R$ as its coordinate ring,
then homomorphisms of $k$-algebras $Q\to Q$ are in one-to-one correspondence with dominant rational maps $\A^n_k\to \A^n_k$  \cite[Theorem I.4.4]{Har77}.
This correspondence is crucial since it will allow us to switch between the algebraic and the geometric point of view, see Proposition~\ref{p:criterion}.
\end{remark}

Definition \ref{d:rationalizability} raises an obvious question:
given a square root $\sqrt{f/g}$, can we determine whether or not it is rationalizable?
To answer this question, we will use tools from algebraic geometry.
In particular, we will relate the rationalizability of a square root to the unirationality of a certain variety associated to it.
Before delving into this, let us prove the following elementary lemma.

\begin{lemma}\label{l:eqvrat}
If $p$ and $q$ denote two non-zero elements of $Q$, then $\sqrt{p}$ is rationalizable if and only if $\sqrt{pq^2}$ is rationalizable.
\end{lemma}
\begin{proof}
Assume that $\sqrt{p}$ is rationalizable.
Then, there exists a non-constant homomorphism of $k$-algebras $\phi\colon Q\to Q$ such that $\phi(p)=r^2$ with $r\in Q$.
But this means that
\begin{equation*}
    \phi(pq^2)=\phi(p)\phi(q)^2=r^2\phi(q)^2=(r\phi(q))^2\in Q.
\end{equation*}
Hence, $\sqrt{pq^2}$ is rationalizable.
\par\vspace{\baselineskip}
Conversely, assume that $\sqrt{pq^2}$ is rationalizable.
Then, there is a non-constant homomorphism of $k$-algebras $\phi\colon Q\to Q$ such that $\phi(pq^2)$ is a square in $Q$.
As $\phi$ is a morphism of $k$-algebras, it follows that $\phi(pq^2)=\phi(p)\phi(q^2)=\phi (p)\phi(q)^2$ is a square in $Q$.
As $\phi$ is a non-constant morphism of $k$-algebras that are fields, it is injective,
and so $\phi (q)$ is non-zero, and hence it follows that $\phi (p)=\phi (pq^2)/\phi (q)^2$ is a square.
Hence, $\sqrt{p}$ is rationalizable.
\end{proof}

\begin{corollary}\label{c:SqFreePol}
Let $p$ and $q$ be two polynomials in $R$ with $q$ non-zero and consider the fraction $p/q\in Q$.
Then there exists a squarefree polynomial $f\in R$ such that $\sqrt{p/q}$ is rationalizable if and only if $\sqrt{f}$ is.
\end{corollary}
\begin{proof}
By Lemma~\ref{l:eqvrat},
the square root $\sqrt{p/q}$ is
rationalizable if and only if the square root
$\sqrt{q^2\cdot p/q}=\sqrt{p q}$ is.
Write $g:=p\cdot q$ and notice that $g\in R$.
As $R$ is a unique factorization domain, we can write $g=fh^2$, with $f,h\in R$ and $f$ squarefree.
Again by Lemma~\ref{l:eqvrat}, $\sqrt{g}$ is rationalizable if and only if $\sqrt{f}$ is,
proving the statement.
\end{proof}

\begin{remark}\label{r:Squarefree}
Let us stress that ignoring squares in the argument is, in many cases, even mandatory for our criteria to be applicable.
In other words, one often \textit{should} ignore square factors in a square root argument:
in Subsection~\ref{sec:TwoVariables}, we will present rationalizability criteria that require the associated variety of the square root to have at most rational simple singularities.
Simple singularities are, in particular, isolated singularities.
The problem is that, if the square root argument contained a square, the associated varieties
(see Definition~\ref{def:AssHyp} and~\ref{def:AssDoubleCover})
would have non-isolated singularities, i.e., a singular locus of positive dimension, and our criteria would not be applicable.
\end{remark}