text:
\begin{document}

	\title{A characterization of the Arf property for quadratic quotients of the Rees algebra}

	\author{Alessio Borzì
			\thanks{Dipartimento di Matematica e Informatica, Università degli Studi di Catania.} \thanks{Scuola Superiore di Catania.}}

	\maketitle

	\begin{abstract}
		We provide a characterization of the Arf property in both the numerical duplication of a numerical semigroup and in a member of a family of quotients of the Rees algebra studied in \cite{barucci2015family}.
	\end{abstract}

	\section*{Introduction}

	Let $R$ be a Noetherian one-dimensional local domain, $I$ be an ideal of $R$ and $t$ be an indeterminate. Let $R[It] = \bigoplus_{n \in \mathbb{N}}I^nt^n$ be the Rees algebra associated with $R$ and $I$. In \cite{barucci2015family} the authors, looking for a unified approach to the Nagata's idealization and the amalgamated duplication of a ring (see \cite{d2007amalgamated}), studied the following family of quotients of the Rees algebra for every $a,b \in R$
	\[ R(I)_{a,b} = \frac{R[It]}{(t^2+at+b) \cap R[It]}, \]
	showing that the Nagata's idealization is obtained for $a=b=0$ and the amalgamated duplication for $a=1$, $b=0$. A remarkable fact about this family of rings is that we can always find domains among its members, if the original ring $R$ is itself a domain. In particular, it was shown in \cite{barucci2015family} that this ring construction can be connected to a semigroup construction called numerical duplication (see \cite{d2013numerical}). More precisely let $S$ be a numerical semigroup, $E$ be a semigroup ideal, and $m \in S$ be an odd integer. For any set of integers $A \subseteq \mathbb{Z}$ we set $2 \cdot A = \{ 2a : a \in A \}$. Then we define the numerical duplication $S \Join^m E$ of $S$ with respect to $E$ and $m$ as the numerical semigroup
	\[ S \Join^m E = 2 \cdot S \cup (2 \cdot E + m). \]
	Now, if we start with an algebroid branch $R$ and $b \in R$ with $v(b)$ odd, the member of the family of the type $R(I)_{0,-b}$ has its value semigroup equal to the numerical duplication of $v(R)$ with respect to $v(I)$ and $v(b)$.	In this paper we show that this is true in general for every Noetherian, one-dimensional, analytically irreducible, local domain $R$.

	In \cite{arf1948interpretation} Arf solved the classification problem of singular branches, using their multiplicity sequence. Later, inspired by the work of Arf, Lipman in \cite{lipman1971stable} introduced the notions of Arf ring and Arf closure of a ring. These rings share the same multiplicity sequence. Hence the idea is to calculate the Arf closure of the coordinate ring of a curve and then its value semigroup, which is an Arf numerical semigroup, in order to obtain its multiplicity sequence. \\

	In this paper we provide a characterization of the Arf property in both the numerical duplication and the family of rings $\mathcal{R} = R(I)_{0,-b}$. More precisely, in Section \ref{section 1} we recall all the basic notions on numerical semigroups and Arf rings. In Section \ref{section 2}, we prove the characterization of the Arf property for the numerical duplication (Theorem \ref{characterization on ns}). In Section \ref{section 3}, we show that $\mathcal{R}$ is a Noetherian, one-dimensional, analytically irreducible, local domain and its value semigroup is $v(R) \Join^{v(b)}v(I)$ (Theorem \ref{double valuation}), then we prove a series of technical results for the purpose of proving Theorem \ref{characterization on rings}, that is the extension to $\mathcal{R}$ of the previous numerical characterization. \\

	Several computations are performed by using the GAP system \cite{gap2015gap} and, in
	particular, the NumericalSgps package \cite{delgadonumericalsgps}.

	\section{Preliminaries}\label{section 1}

	A numerical semigroup $S$ is an additive submonoid of $\mathbb{N}$ with finite complement in $\mathbb{N}$. The \emph{multiplicity} of $S$ is $\mu(S) = \min (S \setminus \{0\})$. The \emph{Frobenius number} of $S$ is $F(S) = \max(\mathbb{N} \setminus S)$ and the \emph{conductor} of $S$ is $c(S) = F(S)+1$. A semigroup ideal of $S$ is a subset $E \subseteq S$ such that $E+S \subseteq E$. We call $\tilde{e} = \min E$; then the \emph{integral closure} of $E$ in $S$ is the semigroup ideal $\overline{E} = \{ s \in S: s \geq \tilde{e} \}$, if $E = \overline{E}$ then $E$ is \emph{integrally closed}. We say that $E$ is \emph{stable} if $E+E = E+\tilde{e}$.

	An \emph{Arf numerical semigroup} is a numerical semigroup $S$ in which for every $x,y,z \in S$, such that $x \geq y \geq z$, it results $x+y-z \in S$; equivalently $S$ is Arf if and only if every integrally closed semigroup ideal is stable (see \cite[Theorem 2.2]{lipman1971stable}).

	Given an Arf numerical semigroup $S = \{ 0 = s_0 < s_1 < s_2 < \ldots \}$ the sequence $(e_0,e_1,e_2,\ldots)$, with $e_i = s_{i+1}-s_i$, is the \emph{multiplicity sequence} of $S$. Note that $e_0$ corresponds to the multiplicity of $S$.

	We call an \emph{Arf sequence} a non-increasing sequence of positive integers $(e_0,e_1,e_2,\ldots)$ such that
	\begin{enumerate}
		\item exists $n \in \mathbb{N}$ such that $e_k = 1$ for all $k \geq n$,
		\item for every $i \in \mathbb{N}$ exists $k \geq 1$ such that $e_i = \sum_{j=1}^k e_{i+j}$.
	\end{enumerate}
	A sequence of positive integers $(e_0,e_1,e_2,\ldots)$ is an Arf sequence if and only if it is a multiplicity sequence of an Arf numerical semigroup, that is $S = \{ 0,e_0,e_0+e_1,e_0+e_1+e_2,\ldots \}$, see for instance \cite[Proposition 1]{garcia2017parametrizing}.

	From the Arf numerical semigroup $S$ we can construct a chain of Arf numerical semigroups $S_0 \subseteq S_1 \subseteq S_2 \subseteq \ldots$ with $S=S_0$ and $S_{i+1} = (S_i \setminus \{0\})-e_i$, namely the blow up of $S_i$. The multiplicity of $S_i$ is $e_i$ and $S_n = \mathbb{N}$ for $n$ large enough.

	Let $(R,\mathfrak{m})$ be a Noetherian one-dimensional local domain and $\overline{R}$ its integral closure in its field of fractions $Q(R)$.
	We assume that $R$ is \emph{analytically irreducible}, that is its completion $\hat{R}_{\mathfrak{m}}$ is a domain, or, equivalently, $\overline{R}$ is a discrete valuation ring (DVR) and a finitely generated $R$-module.
	Since the integral closure $\overline{R}$ is a DVR, every non zero element of $R$ has a value as an element of $\overline{R}$. The set of values $v(R)$ is a numerical semigroup. The multiplicity of $R$ is equal to the multiplicity of its value semigroup $\mu(R) = \mu(v(R))$.

	For any two $R$-submodules $E,F$ of $\overline{R}$ set
	\[ (E:F) = \{ x \in \overline{R}: xF \subseteq E \}. \]
	The blow up of $R$ is $L(R) = \bigcup_{n \in \mathbb{N}}(\mathfrak{m}^n : \mathfrak{m}^n)$. If we fix $R = R_0$ and $R_{i+1} = L(R_i)$ then the multiplicity sequence of $R$ is the sequence $(\mu(R_0),\mu(R_1),\ldots)$.

	We will also assume that $R$ is \emph{residually rational}, namely its residue field $k = R/\mathfrak{m}$ is isomorphic to the residue field of $\overline{R}$. With this assumption, for any $x,y \in R$ such that $v(x) = v(y) $ there exists an invertible element $u \in R$ such that $v(x-uy) > v(x) = v(y)$. Furthermore, for any fractional ideals $I,J$ of $R$ such that $J \subseteq I$, it results $\lambda(I/J) = |v(I) \setminus v(J)|$.

	An element $x \in R$ is said to be integral over the ideal $I$ if $x$ satisfies a relation
	\[ x^n + a_1x^{n-1} + \ldots + a_{n-1}x+a_n = 0 \]
	with $a_j \in I^j$ for $j = 1,2,\ldots,n$. The set $\overline{I}$ of all elements of $R$ which are integral over $I$ is an ideal of $R$ (see \cite{lipman1971stable} or \cite{huneke2006integral}), called the integral closure of $I$ in $R$.

	In our setting, the integral closure of an ideal $I$ is equal to
	\[ \overline{I} = I \overline{R} \cap R = \{ x \in R: v(x) \geq \min v(I) \}, \]
	(see \cite[Proposition 1.6.1, Proposition 6.8.1]{huneke2006integral}).
	If $I = \overline{I}$ then $I$ is \emph{integrally closed}. It follows that the ideal $I$ is integrally closed if and only if the semigroup ideal $v(I)$ is integrally closed. The necessity easily follows from the definitions. For the sufficiency we have $\lambda(\overline{I}/I) = |v(\overline{I}) \setminus v(I)| = 0$, then $I = \overline{I}$. Notice that here the assumption that $R$ is residually rational is needed.

	If $I$ and $J$ are two ideals of $R$, then either $\min v(I) \geq \min v(J)$ or $\min v(I) < \min v(J)$. If $I$ and $J$ are integrally closed, from the previous observations it follows that either $I \subseteq J$ or $J \subsetneq I$.

	The \emph{conductor} of $R$ is $C = (R:\overline{R})$. The conductor is an ideal both of $R$ and $\overline{R}$. The conductor of $v(R)$ is $\min v(C)$. Note that $\overline{R}C = C$, so $v(C) = v(\overline{R}C) = v(C)+\mathbb{N} = \min v(C) + \mathbb{N}$, therefore $C$ is integrally closed. In particular, if $J$ is an integrally closed ideal of $R$, then either $J \subseteq C$ or $C \subsetneq J$. We will frequently use this fact in Section \ref{section 3}.

	Let $x \in I$ be such that $v(x) = \min v(I)$. The ideal $I$ is \emph{stable} if $I^2 = xI$, or, equivalently, if $(I:I) = x^{-1}I$. The ring $R$ is an \emph{Arf ring} if every integrally closed ideal is stable. The ring $R$ is Arf if and only if its value semigroup $v(R)$ is an Arf numerical semigroup and the multiplicity sequence of $R$ coincides with the multiplicity sequence of $v(R)$ (see \cite[Theorem II.2.13]{barucci1997maximality}).

	\begin{proposition}\label{stable in conductor}
		Let $J$ be an integrally closed ideal of $R$. If $J \subseteq C$ then $J$ is stable.
	\end{proposition}
	\begin{proof}
		Let $x \in J$ such that $v(x) = \min v(J)$, we show that $J^2 = xJ$. Clearly $xJ \subseteq J^2$. Let $i,j \in J$ then
		\[ v(j) \geq v(x) \Rightarrow v(j)-v(x) = v(jx^{-1}) \geq 0 \Rightarrow jx^{-1} \in \overline{R}. \]
		Since $i \in J \subseteq C$, it results $i(jx^{-1}) \in R$, so
		\[ v(ijx^{-1}) = v(i) + v(jx^{-1}) \geq v(i) \geq v(x) \Rightarrow ijx^{-1} \in J, \]
		therefore $ij \in xJ$.
	\end{proof}

	\begin{proposition}\label{JC c xJ}
		Let $J$ be an integrally closed ideal of $R$. If $C \subseteq J$ and $x \in J$ is an element of minimum value in $J$ then $JC \subseteq xJ$.
	\end{proposition}
	\begin{proof}
		Observe that $x^{-1}J \subseteq \overline{R}$, hence
		\[ JC = x(x^{-1}J)C \subseteq x\overline{R}C = xC \subseteq xJ. \qedhere \]
	\end{proof}

	\section{Arf property in the numerical duplication}\label{section 2}

	In this section $S = \{ 0 = s_0 < s_1 < s_2 < \ldots \}$ will be a numerical semigroup, $E$ a semigroup ideal of $S$ and $m \in S$ an odd integer. Recall that the quotient of $S$ by a positive integer $d$ is
	\[ \frac{S}{d} = \{ x \in \mathbb{N}: dx \in S \}. \]

	\begin{proposition}\label{Arf quotient}
		For every $d>0$, if $S$ is Arf so is $\frac{S}{d}$.
	\end{proposition}
	\begin{proof}
		Let $x,y,z \in \frac{S}{d}$ with $x \geq y \geq z$, then we have $dx,dy,dz \in S$ with $dx \geq dy \geq dz$ and since $S$ is Arf it follows that
		\[ d(x+y-z) = dx+dy-dz \in S, \]
		hence $x+y-z \in \frac{S}{d}$.
	\end{proof}

	By definition of numerical duplication it is clear that $(S \Join^m E) / 2 = S$, hence we immediately get the following

	\begin{corollary}\label{Quotient duplication Arf}
		If $S \Join^m E$ is Arf so is $S$.
	\end{corollary}

	\begin{lemma}\label{E integrally closed}
		If $S \Join^m E$ is Arf then $E$ is integrally closed.
	\end{lemma}
	\begin{proof}
		Suppose by contradiction that $E$ is not integrally closed. Then there exists $i \in \mathbb{N}$ such that $s_i \in E$ and $s_{i+1} \notin E$. Consider $2s_{i+1},2s_i+m, 2s_i \in S \Join^m E$, since $2s_{i+1} \geq 2s_i$, $2s_i+m \geq 2s_i$ and $S \Join^m E$ is Arf, we have
		\[ 2s_{i+1}+2s_i+m - 2s_i = 2s_{i+1}+m \in S \Join^m E, \]
		which means $s_{i+1} \in E$, contradiction.
	\end{proof}

	Let $S$ be an Arf numerical semigroup, and let $(e_0,e_1,e_2,\ldots)$ be its multiplicity sequence. Fix $n \in \mathbb{N}$ to be the smallest integer such that $e_k = 1$ for every $k \geq n$. We recall that $e_0 = \min S \setminus \{0\} = s_1$, and that $s_ {i+1} = e_0+\ldots+e_i$, in particular $s_{n+1} = s_n + 1$ and $s_n = c(S)$ is the conductor of $S$.

	In the proof of the following result we will use the fact that for an Arf numerical semigroup $S$, if $x,x+1 \in S$ then $x+\mathbb{N} \subseteq S$ (see for instance \cite[Lemma 11]{rosales2004arf}).

	\begin{theorem}\label{characterization on ns}
		The numerical duplication $D = S \Join^m E$ is Arf if and only if $S$ is Arf with multiplicity sequence $(e_0,e_1,\ldots,e_{n-1},1,1,\ldots)$, $e_{n-1} \neq 1$, $E$ is integrally closed and, if $\min(E) < c(S)$, $e_0 = e_1 = \ldots = e_{n-1} = m$.
	\end{theorem}

	\begin{proof}
		\emph{Necessity}. From Corollary \ref{Quotient duplication Arf} and Lemma \ref{E integrally closed}, $S$ is Arf and $E$ is integrally closed.
		Now if $\min(E) < c(S) = s_n$ then $s_{n-1} \in E$. Suppose that $m \geq 2e_{n-1} = 2(s_n - s_{n-1})$, then
		\[ 2s_{n-1} + m \geq 2s_n. \]
		Since $2s_{n-1}+m+1$ is even and $s_n$ is the conductor of $S$ we obtain $2s_{n-1}+m+1 = 2s_k$ for some $k \in \mathbb{N}$. Setting $x = 2s_{n-1}+m$ we have $x,x+1 \in D$ which is Arf, hence $x+\mathbb{N} \subseteq D$, and so $x+2 \in D$; this means
		\begin{gather*}
		2s_{n-1}+m+2 = 2(s_{n-1}+1)+m \in D \Rightarrow \\
		\Rightarrow s_{n-1}+1 \in S \Rightarrow s_n = s_{n-1}+1 \Rightarrow \\
		\Rightarrow e_{n-1} = s_n-s_{n-1} = 1,
		\end{gather*}
		which is a contradiction. Therefore $m < 2e_{n-1}$, hence $2s_{n-1} + m < 2s_{n}$, since $D$ is Arf, this implies
		\[ 2s_n+2s_n-(2s_{n-1}+m) = 2s_n + 2e_{n-1}-m \in D. \]
		Furthermore $2s_n + 2e_{n-1}-m$ is odd and
		\[ 2s_n + 2e_{n-1}-m \geq 2s_{n} > 2s_{n-1}+m. \]
		It follows that
		\begin{gather*}
			2s_n + 2e_{n-1}-m \geq 2s_n+m \\
			\Rightarrow m \leq e_{n-1} \leq e_0.
		\end{gather*}
		Since $m \in S$ and it is odd, we must have $m = e_0 = e_1 = \ldots = e_{n-1}$. \\
		\emph{Sufficiency}. If $\min(E) \geq s_n$, then $E = x+\mathbb{N}$ with $x \geq s_n$, so it results $D = 2S \cup ((2x+m)+\mathbb{N})$ and it is easy to check that $D$ is Arf. \\
		Otherwise if $\min(E) < s_n$ and $m=e_0=e_1=\ldots=e_{n-1}$, then $S = e_0\mathbb{N} \cup (ne_0 + \mathbb{N})$ and $E = \{ ie_0,(i+1)e_0,\ldots,(n-1)e_0 \} \cup (ne_0+\mathbb{N})$ for some $i \leq n$. Hence, if $D = \{ 0 = d_0 < d_1 < \ldots < d_k < \ldots \}$ then after some easy calculations it results
		\[ ( d_{k+1}-d_k : k \in \mathbb{N} ) = ( \underbrace{2e_0,2e_0,\ldots,2e_0}_{i\text{ times}}, \underbrace{e_0,e_0,\ldots,e_0}_{2(n-i) \text{ times}}, \underbrace{2,2,\ldots,2}_{\frac{e_0-1}{2} \text{ times}},1,\ldots ), \]
		which is an Arf sequence, so $D$ is Arf.
	\end{proof}

	\begin{example}
		Let $S = \langle 3,7,8 \rangle = \{ 0,3,6,\rightarrow \}$, $S$ is Arf and its multiplicity sequence is $(3,3,1,\ldots)$, so $n = 2$. Let $E=S \setminus \{0\}$ and $m = 3$, $E$ is integrally closed, $\min(E) = 3 < 6 = s_n$ and $m = e_0 = e_1$. The numerical duplication is
		\[ S \Join^m E = \langle 6,9,14,16,17,19 \rangle = \{ 0,6,9,12,14,\rightarrow \}, \]
		and it is an Arf numerical semigroup.
	\end{example}

	\begin{remark}\label{remark multiples of m}
		In the case $\min(E) < c(S)$ of Theorem \ref{characterization on ns}, the elements of $D= S \Join^m E$ smaller than the conductor of $D$ are multiples of $m$, so they are of the form $k m$ for some $k \in \mathbb{N}$.
	\end{remark}

	Recall that the Arf closure $\operatorname{Arf}(S)$ of a numerical semigroup $S$ is the smallest Arf numerical semigroup that contains $S$ (see \cite{rosales2004arf}). Let $\tilde{E}$ be the integral closure in $\operatorname{Arf}(S)$ of the ideal generated by $E$ in $\operatorname{Arf}(S)$. More explicitly, if $\tilde{e} = \min E$, then $\tilde{E} = \{ s \in \operatorname{Arf}(S) : s \geq \tilde{e} \}$.

	\begin{proposition}\label{Arf S dup int E}
		With the notation introduced above we have
		\[ \operatorname{Arf}(S) \Join^m \tilde{E} \subseteq \operatorname{Arf}(S \Join^m E). \]
	\end{proposition}
	\begin{proof}
		Since $S = (S \Join^m E)/2 \subseteq \operatorname{Arf}(S \Join^m E)/2$ and, by Proposition \ref{Arf quotient}, $\operatorname{Arf}(S \Join^m E)/2$ is Arf, we got
		\[ \operatorname{Arf}(S) \subseteq \frac{\operatorname{Arf}(S \Join^m E)}{2}; \]
		it follows that $2 \cdot \operatorname{Arf}(S) \subseteq \operatorname{Arf}(S \Join^m E)$. Now if $e \in \tilde{E}$, then $2e \geq 2\tilde{e}$ and $2\tilde{e} + m \geq 2\tilde{e}$, and, since $\operatorname{Arf}(S \Join^m E)$ is Arf and $2e,2\tilde{e},2\tilde{e}+m \in \operatorname{Arf}(S \Join^m E)$, we have
		\[ 2e + 2\tilde{e}+m - 2\tilde{e} = 2e+m \in \operatorname{Arf}(S \Join^m E). \]
		So, $2\tilde{E}+m \subseteq \operatorname{Arf}(S \Join^m E)$ and therefore $\operatorname{Arf}(S) \Join^m \tilde{E} \subseteq \operatorname{Arf}(S \Join^m E)$.
	\end{proof}

	Theorem \ref{characterization on ns} gives us sufficient conditions so that the inclusion of Proposition \ref{Arf S dup int E} is an equality. Recall that $\overline{E}$ is the integral closure in $S$ of the semigroup ideal $E$.

	\begin{corollary}
		If one of the following conditions holds
		\begin{enumerate}
			\item $\min(E) \geq c(S)$,
			\item $S$ is Arf and $m = e_0 = e_1 = \ldots = e_{n-1}$,
		\end{enumerate}
		then $\overline{E} = \tilde{E}$ and $\operatorname{Arf}(S) \Join^m \overline{E} = \operatorname{Arf}(S \Join^m E)$.
	\end{corollary}

	It is easy to see that the previous equality is not true in the general case. In particular, the following example shows that neither the equality
	\[ \operatorname{Arf}(S) = \frac{\operatorname{Arf}(S \Join^m E)}{2} \]
	holds true in general.

	\begin{example}
		Let $S= \langle 5,8,11,12,14\rangle = \{ 0,5,8,10,\rightarrow \}$, $E = S \setminus \{0\}$ and $m = 5$. Note that $S$ is Arf, so $S = \operatorname{Arf}(S)$. The numerical duplication of $S$ with respect to $E$ and $m$ is
		\[ S \Join^m E = \{ 0,10,15,16,20,21,22,24,\rightarrow \}. \]
		Since $15,16 \in S \Join^m E$, its Arf closure is $\operatorname{Arf}(S \Join^m E) = \{ 0,10,15\rightarrow \}$; moreover $9 \in \operatorname{Arf}(S \Join^m E)/2$, but $9 \notin \operatorname{Arf}(S) = S$.
	\end{example}

	A couple of questions naturally arise.

	\begin{question}
		Are there any sufficient and necessary conditions so that the inclusion of Proposition \ref{Arf S dup int E} is an equality? Is there a way to express $\operatorname{Arf}(S \Join^m E)$ in terms of $S$, $E$ and $m$?
	\end{question}

	\section{Arf property in $\mathcal{R}$}\label{section 3}

	In this section $(R,\mathfrak{m})$ will be a Noetherian, analytically irreducible, residually rational, one-dimensional, local domain with $\operatorname{char}(R) \neq 2$; $v:Q(R) \rightarrow \mathbb{Z}$ will denote the valuation on $Q(R)$ associated to $\overline{R}$. Let $I$ be an ideal of $R$ and let $t$ be an indeterminate; the Rees algebra (also called Blow-up algebra) associated with $R$ and $I$ is the graded subring of $R[t]$ defined as
	\[ R[It] = \bigoplus_{n \in \mathbb{N}}I^nt^n. \]
	Let $b \in R$ such that $v(b) = m$ is odd, we define
	\[ \mathcal{R} = \frac{R[It]}{(t^2-b) \cap R[It]}. \]
	Then $\mathcal{R}$ is a subring of $R[\alpha]$ with $\alpha = t+(t^2-b)$. Furthermore $\mathcal{R}$ and $R[\alpha]$ have the same integral closure $\overline{\mathcal{R}}$ and the same field of fractions $Q(R)[\alpha]$ (see \cite[Corollary 1.8]{barucci2015family}). Since $\alpha$ is integral over $R$, the integral closure $\overline{R}^{Q(R)[\alpha]}$ of $R$ in $Q(R)[\alpha]$ is the same as the integral closure of $\mathcal{R}$. The extension field $Q(R) \subseteq Q(R)[\alpha]$ is finite and since $\operatorname{char}(R) \neq 2$, it is also separable.

	\begin{theorem}\label{double valuation}
		The ring $\mathcal{R}$ is a Noetherian one-dimensional local domain analytically irreducible and residually rational. If $v':Q(R)[\alpha] \rightarrow \mathbb{Z}$ is the extension on $Q(R)[\alpha]$ of the valuation of $\overline{\mathcal{R}}$, then $v'_{|Q(R)} = 2v$, $v'(\alpha)=m$ and
		\[ v'(\mathcal{R}) = v(R) \Join^{v(b)} v(I). \]
	\end{theorem}
	\begin{proof}
		From \cite{barucci2015family} we know that if $R$ is Noetherian, one-dimensional and local so is $\mathcal{R}$. Moreover, since $v(b)$ is odd, the polynomial $t^2-b$ is irreducible in $Q(R)[t]$, then, from \cite[Corollary 1.3]{d2017new}, $\mathcal{R}$ is a domain. \\
		Now we prove that $\mathcal{R}$ is analytically irreducible. It is enough to prove that $\overline{\mathcal{R}}$ is local and a finitely generated $\mathcal{R}$-module. Let $x \in \overline{R}$ be an element of valuation $1$, $k =\frac{m-1}{2}$ and $\beta = \frac{\alpha}{x^k} \in Q(R)[\alpha]$; then $\beta^2 = \frac{b}{x^{m-1}} \in \overline{R}$ since $v(\frac{b}{x^{m-1}}) = 1 > 0$, so $\beta$ is integral over $\overline{R}$. We prove that $\overline{\mathcal{R}} = \overline{R}+\overline{R}\beta$. The inclusion $\overline{R}+\overline{R}\beta \subseteq \overline{\mathcal{R}}$ follows from the fact that $\overline{R} \subseteq \overline{\mathcal{R}}$ and that $\beta$ is integral over $\overline{R}$. Conversely, let $p+q\alpha \in \overline{\mathcal{R}} = \overline{R}^{Q(R)[\alpha]}$, where $p,q \in Q(R)$, $q \neq 0$. Then, since $Q(R) \subseteq Q(R)[\alpha]$ is algebraic, from \cite[Theorem 2.1.17]{huneke2006integral} the coefficients $2p$ and $p^2-q^2b$ of the minimal polynomial of $p+q\alpha$ over $Q(R)$ are in $\overline{R}$. In addition, $v(p^2) = 2v(p)$ is even and $v(q^2b) = 2v(q)+m$ is odd, then $v(p^2) \neq v(q^2b)$, therefore
		\begin{gather*}
			0 \leq v(p^2-q^2b) = \min\{ 2v(p),2v(q)+m \} \\
			\Rightarrow
			\begin{cases}
				v(p) \geq 0 \Rightarrow p \in \overline{R} \\
				2v(q)+m \geq 0 \Rightarrow v(q) \geq -k \Rightarrow qx^k \in \overline{R}.
			\end{cases}
		\end{gather*}
		Hence $p + q \alpha = p + qx^k \beta \in \overline{R} + \overline{R}\beta$. Now if we denote by $\overline{\mathfrak{m}}$ the maximal ideal of $\overline{R}$, the ring $\overline{R} + \overline{R}\beta$ is local with maximal ideal $\overline{M} = \overline{\mathfrak{m}} + \overline{R} \beta$; indeed the inverse of $p+q\beta \in \overline{R}+\overline{R}\beta$ with $p \in \overline{R} \setminus \overline{\mathfrak{m}}$, is $\frac{p-q\beta}{p^2-q^2\beta^2}$, in fact $p^2-q^2\beta^2$ is invertible since $0 = v(p^2) \neq v(q^2\beta^2) = 2v(q)+1$ and
		\[ v(p^2-q^2\beta^2) = \min\{v(p^2),v\left(q^2\beta^2\right)\} = v(p^2) = 0.  \]
		It follows that $\overline{\mathcal{R}} = \overline{R} + \overline{R}\beta$ is local.

		Now we prove that $\overline{\mathcal{R}}$ is a finitely generated $\mathcal{R}$-module. The field extension $Q(R) \subseteq Q(R)[\alpha]$ is finite and separable, then, by \cite[Theorem 3.1.3]{huneke2006integral}, the integral closure of $\overline{R}$ in $Q(R)[\alpha]$, which is equal to $\overline{R}^{Q(R)[\alpha]} = \overline{\mathcal{R}}$, is a finite module over $\overline{R}$. Now $\overline{R}$ and $\mathcal{R} \simeq R + I \alpha$ are finite modules over $R$, so $\overline{\mathcal{R}}$ is a finite module over $\mathcal{R}$.

		Now let $v'(Q(R)) = d\mathbb{Z}$ for some $d \in \mathbb{N}$. It results $v'(x) = d$ and $b \in (x^m) \setminus (x^{m+1})$, namely $b = ux^m$ with $u \in \overline{R}$ invertible. It follows that $v'(b) = mv'(x)$; in addition from $\alpha^2 = b$ we obtain $2v'(\alpha) = v'(b) = mv'(x) = md$. Since
		\[ \begin{split}
			v'(Q(R)[\alpha]) = v'(Q(R) + Q(R) \alpha) = v'(Q(R)) \cup [v'(Q(R)) + v'(\alpha)] = \\ = d\mathbb{Z} \cup (d\mathbb{Z}+v(\alpha)) = \mathbb{Z},
		\end{split} \]
		it must be $d=2$, so $v'_{|Q(R)} = 2v$ and also $v'(\alpha) = m$. It easily follows that
		\[ v'(\mathcal{R}) = v(R) \Join^{v(b)} v(I) = S \Join^m v(I). \]

		Finally we show that $\mathcal{R}$ is residually rational. Recall that $\overline{\mathfrak{m}}$ is the maximal ideal of $\overline{R}$ and $\overline{M} = \overline{\mathfrak{m}}+\overline{R}\beta$ is the maximal ideal of $\overline{\mathcal{R}}$. From \cite[Proposition 2.1]{barucci2015family} the maximal ideal of $\mathcal{R}$ is $M = \mathfrak{m}+I\alpha$. Thus
		\[ \overline{\mathcal{R}}/\overline{M} = \frac{\overline{R}+\overline{R}\beta}{\overline{\mathfrak{m}}+\overline{R}\beta} \simeq\overline{R}/\overline{\mathfrak{m}} \simeq R/\mathfrak{m} \simeq \frac{R+I\alpha}{\mathfrak{m}+I\alpha} = \mathcal{R}/M. \]
	\end{proof}

	In the following, $v'$ will denote the extension on $Q(R)[\alpha]$ of the valuation of $\overline{\mathcal{R}}$.

	\begin{proposition}\label{R is Arf}
		If $\mathcal{R}$ is Arf, so is $R$.
	\end{proposition}
	\begin{proof}
		Let $J$ be an integrally closed ideal of $R$ and let $x \in J$ such that $v(x) = \min v(J)$. Fix $\tilde{J} = \{ y \in \mathcal{R} : v'(y) \geq v'(x) \}$, $\tilde{J}$ is an integrally closed ideal of $\mathcal{R}$, so it is stable, namely $(\tilde{J}:\tilde{J}) = x^{-1}\tilde{J}$. Furthermore, since $\overline{R} \cap \mathcal{R} = R$, we have $J = \tilde{J} \cap R = \tilde{J} \cap \overline{R}$. It suffices to prove that $(J:J) = x^{-1}J$. It is clear that $(J:J) \subseteq x^{-1}J$, in fact, if $j \in (J:J)$, then by definition $xj \in J$, so $j \in x^{-1}J$. Conversely let $j \in x^{-1}J$ and $j' \in J$, we have
		\[
		\left. \begin{array}{r}
		j \in x^{-1}J \subseteq x^{-1}\tilde{J} = (\tilde{J}:\tilde{J}) \\
		j' \in J \subseteq \tilde{J}
		\end{array} \right\} \Rightarrow jj' \in \tilde{J}.
		\]
		Further
		\[
		\left. \begin{array}{r}
		j \in x^{-1}J \subseteq \overline{R} \\
		j' \in J \subseteq R \subseteq \overline{R}
		\end{array} \right\} \Rightarrow jj' \in \overline{R},
		\]
		it follows that $jj' \in \tilde{J} \cap \overline{R} = J$, so $j \in (J:J)$.
	\end{proof}

	Now our aim is to prove the extension of Theorem \ref{characterization on ns} to $\mathcal{R}$, that is $\mathcal{R}$ is Arf if and only if $R$ is Arf, $I$ is integrally closed and a similar condition on the multiplicity sequence. For the necessity, from Proposition \ref{R is Arf} $R$ is Arf, then the multiplicity sequences of $R$ and $v(R)$ coincides. Therefore we can directly apply Theorem \ref{characterization on ns} (see the proof of Theorem \ref{characterization on rings}). For the sufficiency, from Theorem \ref{characterization on ns} we have that $v(R) \Join^{v(b)} v(I)$ is an Arf numerical semigroup, but this is not enough to prove that $\mathcal{R}$ is Arf. In order to prove necessity, we need a series of technical results.

	For this purpose we introduce some more notation. We fix an integrally closed ideal $\tilde{J}$ of $\mathcal{R}$; set $J = \tilde{J} \cap R$ and $\tilde{j_1} = \min v'(\tilde{J})$, $j_1 = \min v'(J)$. We denote the conductor of $R$ with $C = (R:\overline{R})$, and the conductor of $\mathcal{R}$ with $C_{\mathcal{R}} = (\mathcal{R}:\overline{\mathcal{R}})$. Note that the inclusion $C_{\mathcal{R}} \cap R \subseteq C$ may be strict. In addition, we will suppose that $C_{\mathcal{R}} \subsetneq \tilde{J}$. \newpage

	\begin{proposition}\label{J int closed}
		The ideal $J$ is integrally closed in $R$. Further
		\begin{enumerate}
			\item If $x+y\alpha \in \tilde{J}$ then $x \in J$ and $y\alpha \in \tilde{J}$.
			\item If $\tilde{j_1}$ is even then $j_1 = \tilde{j_1}$ and there exists $x \in J$ such that $v'(x) = \tilde{j_1}$.
			\item $v'(J) = v'(\tilde{J}) \cap v'(\overline{R})$.
			\item If $\tilde{j_1} < \min v'(I \alpha)$ then $\tilde{j_1}$ is even.
		\end{enumerate}
	\end{proposition}
	\begin{proof}
		We have
		\[ \tilde{J} = \{ x \in \mathcal{R} : v'(x) \geq \tilde{j}_1 \}, \quad J = \tilde{J} \cap R = \{ x \in R: v'(x) \geq \tilde{j}_1 \}, \]
		then $j_1 \geq \tilde{j_1}$; since for any $x,y \in R$ $v(x) \geq v(y)$ if and only if $v'(x) \geq v'(y)$, $\overline{J} \subseteq J$, i.e. $J$ is integrally closed.

		Let $z = x+y\alpha \in \tilde{J}$, with $x \in R$ and $y \in I$. Now $v'(x) = 2v(x)$ is even and $v'(y\alpha) = 2v(y)+m$ is odd, therefore $v'(x) \neq v'(y\alpha)$ and it results
		\[ v'(x) \geq \min\{v'(x),v'(y\alpha)\} = v'(z) \geq \tilde{j_1}. \]
		It follows that $x \in \tilde{J} \cap R = J$ and consequently $y\alpha = z-x \in \tilde{J}$.

		Now, if $\tilde{j_1}$ is even, let $z = x+y\alpha \in \tilde{J}$ such that $v'(z) = \tilde{j_1}$. From the previous observations it must be $\tilde{j_1} = v'(z) = v'(x) \geq j_1 \geq \tilde{j_1}$, hence $v'(x) = j_1 = \tilde{j_1}$ with $x \in J$.

		Since $J \subseteq \tilde{J}$ and $J \subseteq \overline{R}$, we have $v'(J) \subseteq v'(\tilde{J})$ and $v'(J) \subseteq v'(\overline{R})$, so $v'(J) \subseteq v'(\tilde{J}) \cap v'(\overline{R})$. Conversely let $z = x+y\alpha \in \mathcal{R}$ such that $v'(z) \in v'(\tilde{J}) \cap v'(\overline{R})$; $\tilde{J}$ integrally closed implies $z \in \tilde{J}$, for what we have proved so far $x \in J$. Since $v'(z) \in v'(\overline{R})$, it is even, thus $v'(z) = v'(x) \in v'(J)$.

		Now if $\tilde{j_1} < \min v'(I \alpha)$, assume by contradiction that $\tilde{j_1}$ is odd. Let $z = x+y\alpha \in \tilde{J}$ such that $v'(z) = \tilde{j_1}$, then $\tilde{j_1} = v'(z) = v'(y\alpha) \in v'(I\alpha)$, contradicting $\tilde{j_1} < \min v'(I\alpha)$.
	\end{proof}

	\begin{lemma}\label{I c C}
		If $I \subseteq C$ then $I \alpha \subsetneq C_{\mathcal{R}}$ and $\tilde{j_1}$ is even.
	\end{lemma}
	\begin{proof}
		Recalling that $v'(\mathcal{R}) = v(R) \Join^{v(b)}v(I)$, from \cite[Proposition 2.1]{d2013numerical} we obtain
		\begin{gather*}
			\min v'(C_{\mathcal{R}}) = 2\min v(I) + v(b)-1 = \min v'(I) + v'(\alpha) - 1 = \\
			= \min v'(I \alpha) -1 < \min v'(I \alpha).
		\end{gather*}
		Hence $\min v'(C_{\mathcal{R}}) < \min v'(I \alpha)$, so $I\alpha \subsetneq C_{\mathcal{R}}$ ($C_{\mathcal{R}}$ is integrally closed). Now, since $C_{\mathcal{R}} \subsetneq \tilde{J}$ and both $C_{\mathcal{R}}$ and $\tilde{J}$ are integrally closed, we have $\tilde{j_1} < \min v'(C_{\mathcal{R}}) < \min v'(I\alpha)$, from Proposition \ref{J int closed}, $\tilde{j_1}$ is even.
	\end{proof}

	\begin{lemma}\label{I c J}
		If $I \subseteq J$ then $I\alpha \subsetneq \tilde{J}$ and $\tilde{j_1}$ is even. Further, if $R$ is Arf and $I$ is integrally closed, then $JI\alpha \subseteq x \tilde{J}$, where $x \in J$ is such that $v'(x) = j_1 = \tilde{j_1}$.
	\end{lemma}
	\begin{proof}
		For every $i \in I$ it follows that $v'(i\alpha) = v'(i) + v'(\alpha) > v'(i)$, then $\min v'(I \alpha) > \min v'(I)$. Moreover $I \subseteq J \subseteq \tilde{J}$, hence
		\[ \min v'(I \alpha) > \min v'(I) \geq \min v'(J) \geq \tilde{j_1}. \]
		It follows that $I \alpha \subsetneq \tilde{J}$ ($\tilde{J}$ is integrally closed), and from Proposition \ref{J int closed} $\tilde{j_1}$ is even.

		Now suppose that $R$ is Arf and $I$ is integrally closed. The choice of $x \in J$ is allowed by Proposition \ref{J int closed}. Let $i \in I$ and $j \in J$, then $ij \in IJ \subseteq J^2 = xJ$ ($R$ is Arf, $J$ is integrally closed and $x$ is of minimum value in $J$), so there exists $j' \in J$ such that $ij = xj'$, furthermore
		\[ v'(j') = v'(ijx^{-1}) = v'(i) + v'(j)-v'(x) \geq v'(i), \]
		hence $j' \in I$ ($I$ is integrally closed). Finally $ji\alpha = xj'\alpha \in xI\alpha \subseteq x \tilde{J}$, therefore $JI\alpha \subseteq x\tilde{J}$.
	\end{proof}

	\begin{remark}
		Recall that $\min v(C)$ is the conductor of $v(R)$. Note that if $I$ is integrally closed, the condition $\min(v(I)) < \min v(C) = c(v(R))$ (similar to the one of Theorem \ref{characterization on ns}) is equivalent to $C \subsetneq I$.
	\end{remark}

	Recall that, if $R$ is Arf, then the multiplicity sequences of $R$ and of $v(R)$ coincides. In this case, we will denote the multiplicity sequence of $R$ (and of $v(R)$) with $(e_0,e_1,\ldots,e_{n-1},1,1,\ldots)$, $e_{n-1} \neq 1$.

	\begin{lemma}\label{J c I}
		If $J \subseteq I$, then $J\alpha \subseteq \tilde{J}$. Further, if $R$ is Arf, $I$ is integrally closed and $v(b) = e_0 = e_1 = \ldots = e_{n-1}$, then $J^2 \subseteq x\tilde{J}$ and $J^2\alpha \subseteq x\tilde{J}$, where $x \in \tilde{J}$ is an element of minimum value in $\tilde{J}$.
	\end{lemma}
	\begin{proof}
		If $j \in J$ then $j\alpha \in J\alpha \subseteq I\alpha \subseteq \mathcal{R}$, moreover $v'(j\alpha) \geq v'(j) \geq \tilde{j_1}$. It follows that $j\alpha \in \tilde{J}$, therefore $J\alpha \subseteq \tilde{J}$.

		Now suppose that $R$ is Arf, $I$ is integrally closed and $v(b) = e_0 = e_1 = \ldots = e_{n-1}$. If $\tilde{j_1}$ is even, from Proposition \ref{J int closed} we can choose $x \in J$; it follows that $J^2 = xJ \subseteq x\tilde{J}$ and $J^2\alpha = xJ\alpha \subseteq x \tilde{J}$.

		On the other hand, if $\tilde{j_1}$ is odd we can choose $y \in I$ such that $x = y\alpha \in \tilde{J}$. In this case it must be $C \subsetneq I$, otherwise if $I \subseteq C$ ($I$ is integrally closed), then from Lemma \ref{I c C} $\tilde{j_1}$ can not be odd. Therefore, since $C_{\mathcal{R}} \subsetneq \tilde{J}$ the value of $y\alpha$ is less than $\min v'(C_{\mathcal{R}})$, which is the conductor of $v'(\mathcal{R})$, from Remark \ref{remark multiples of m} we have $v'(y\alpha) = kv(b)$ for some $k \in \mathbb{N}$ odd. From Proposition \ref{J int closed} it follows that the minimum of $v'(J)$ is equal to $(k+1)v(b)$, hence
		\[ (k+1)v(b) = kv(b) + v(b) = v'(y\alpha) + v'(\alpha) = v'(yb) = j_1. \]
		Since $yb \in R$, $yb = x\alpha$ is an element of minimum value in $J$, therefore $J^2 = x\alpha J \subseteq x \tilde{J}$ and $J^2 \alpha = x\alpha^2J = xbJ \subseteq xJ \subseteq x\tilde{J}$.
	\end{proof}

	\begin{lemma}\label{J minus C}
		Suppose that $R$ is Arf, $I$ is integrally closed and $v(b) = e_0 = e_1 = \ldots = e_{n-1}$, and let $i \in I$. If $i\alpha \in \tilde{J} \setminus C_{\mathcal{R}}$ with $v'(i\alpha) > \tilde{j_1}$, then $i \in J$.
	\end{lemma}
	\begin{proof}
		Note that we can assume $C \subsetneq I$, otherwise if $I \subseteq C$ from Lemma \ref{I c C} $I \alpha \subseteq C_{\mathcal{R}}$, so $I \alpha \cap (\tilde{J} \setminus C_{\mathcal{R}}) = \emptyset$.

		In view of Theorem \ref{characterization on ns} and Theorem \ref{double valuation}, since $i\alpha \notin C_{\mathcal{R}}$ we have $v'(i\alpha) = kv(b)$ for some $k \in \mathbb{N}$ odd. Moreover, $v'(i\alpha) > \tilde{j_1}$, so $(k-1)v(b) \in v'(\tilde{J})$, therefore
		\[ v'(i) = v'(i\alpha) - v'(\alpha) = kv(b) - v(b) = (k-1)v(b) \in v'(\tilde{J}). \]
		Hence $v'(i) \geq \tilde{j_1}$ with $i \in I \subseteq R$, so $i \in \tilde{J} \cap R = J$.
	\end{proof}

	Now we are ready to prove Theorem \ref{characterization on rings}, so we no longer hold the assumptions made on $\tilde{J}$.

	\begin{theorem}\label{characterization on rings}
		$\mathcal{R}$ is Arf if and only if $R$ is Arf with multiplicity sequence $(e_0,e_1,\ldots,e_{n-1},1,1,\ldots)$, $e_{n-1} \neq 1$, $I$ is integrally closed and if $C \subsetneq I$ then $v(b) = e_0 = e_1 = \ldots = e_{n-1}$.
	\end{theorem}
	\begin{proof}
		\emph{Necessity}. From Proposition \ref{R is Arf} $R$ is Arf, therefore the multiplicity sequences of $R$ and $v(R)$ coincides. Since $\mathcal{R}$ is Arf, $v(R) \Join^{v(b)}\nolinebreak v(I)$ is an Arf numerical semigroup, so from Theorem \ref{characterization on ns} $v(I)$ is integrally closed, equivalently $I$ is integrally closed. Furthermore, if $C \subsetneq I$, equivalently if $\min v(I) < c(v(R))$, then $v(b) = e_0 = e_1 = \ldots = e_{n-1}$. \\
		\emph{Sufficiency}. Let $\tilde{J}$ be an integrally closed ideal of $\mathcal{R}$. From Proposition \ref{stable in conductor} applied to $\tilde{J}$ and $\mathcal{R}$, if $\tilde{J} \subseteq C_{\mathcal{R}}$ then $\tilde{J}$ is stable, so suppose that $C_{\mathcal{R}} \subsetneq \tilde{J}$. We denote with $J = \tilde{J} \cap R$, from Proposition \ref{J int closed} $J$ is an integrally closed ideal of $R$, so it is stable. Let $x \in \tilde{J}$ be an element of minimum value, we want to prove that $x\tilde{J} = \tilde{J}^2$, the inclusion $x\tilde{J} \subseteq \tilde{J}^2$ is clear, so it is suffice to prove that $\tilde{J}^2 \subseteq x\tilde{J}$. Now the two ideals $I$ and $C$ of $R$ are both integrally closed, so one is contained in the other.

		If $I \subseteq C$ from Proposition \ref{J int closed} and Lemma \ref{I c C} we can choose $x \in J$ and it results $\tilde{J} \subseteq J + I \alpha \subseteq J + C_{\mathcal{R}}$. In the view of Proposition \ref{JC c xJ} applied to $\tilde{J}$ and $\mathcal{R}$ we obtain
		\[ \tilde{J}^2 \subseteq (J+C_{\mathcal{R}})^2 = J^2 + JC_{\mathcal{R}} + C_{\mathcal{R}}^2 \subseteq xJ + \tilde{J}C_{\mathcal{R}} + \tilde{J}C_{\mathcal{R}} \subseteq x\tilde{J}. \]

		If $C \subsetneq I$ in this case we have $v(b) = e_0 = e_1 = \ldots = e_{n-1}$. Again $I$ and $J$ are two integrally closed ideal of $R$ and we distinguish two cases.

		If $I \subseteq J$ then from Proposition \ref{J int closed} and Lemma \ref{I c J} we can choose $x \in J$ and we have $JI\alpha \subseteq x \tilde{J}$. It follows
		\[ \tilde{J}^2 \subseteq (J+I\alpha)^2 = J^2 + JI\alpha + I^2b \subseteq J^2 + x\tilde{J} + J^2 = xJ + x\tilde{J} \subseteq x\tilde{J}. \]
		If $J \subseteq I$ then let $x_1+y_1\alpha, x_2+y_2\alpha \in \tilde{J}$. For $k=1,2$ we distinguish three cases.
		\begin{enumerate}
		\item $v'(y_k \alpha) = \tilde{j_1}$. In this case set $y_k \alpha = x$.
		\item $y_k \alpha \in \tilde{J} \setminus C_{\mathcal{R}}$ with $v'(y_k \alpha) > \tilde{j_1}$. In this case, from Lemma \ref{J minus C}, $y_k \in J$.
		\item $y_k \alpha \in C_{\mathcal{R}}$.
		\end{enumerate}
		In view of Lemma \ref{J c I} and Proposition \ref{JC c xJ} we have to verify six different cases.
		\begin{itemize}
			\item[$(1,1)$]
			Set $y_1 \alpha = x$. Since $v'(y_1 \alpha) = v'(y_2 \alpha) = \tilde{j_1}$ and $\mathcal{R}$ is residually rational, there exists a unit $u \in \mathcal{R}$ such that $v'((y_2-uy_1)\alpha) > \tilde{j_1}$, thus for the element $(y_2-uy_1)\alpha$ we can proceed as in the cases $2$ or $3$:
			\begin{gather*}
			(x_1 +y_1 \alpha)(x_2 +y_2 \alpha) = x_1x_2 + x_1(y_2-uy_1)\alpha + y_1 \alpha(ux_1+x_2+y_2\alpha) \in \\
			\begin{split}
				\in J^2 + (y_2-uy_1) \alpha J + x \tilde{J} & \subseteq x J + (y_2-uy_1) \alpha J + x\tilde{J} \subseteq \\
				& \subseteq (y_2-uy_1) \alpha J + x\tilde{J}.
			\end{split}
			\end{gather*}
			Now if $(y_2-uy_1) \alpha \in \tilde{J} \setminus C_{\mathcal{R}}$, then from Lemma \ref{J minus C} $(y_2-uy_1) \in J$ so $(y_2-uy_1) \alpha J \subseteq \alpha J^2 \subseteq x \tilde{J}$. Otherwise if $(y_2-uy_1) \alpha \in C_{\mathcal{R}}$, then $(y_2-uy_1) \alpha J \subseteq C_{\mathcal{R}} \tilde{J} \subseteq x \tilde{J}$.

			\item[$(1,2)$] \begin{gather*}
			(x_1 +y_1 \alpha)(x_2 +y_2 \alpha) = x_1(x_2 +y_2 \alpha) + y_1 \alpha (x_2 +y_2 \alpha) \in \\
			\in J(J+J\alpha) + x \tilde{J} \subseteq J^2 + J^2 \alpha + x\tilde{J} \subseteq x\tilde{J}.
			\end{gather*}

			\item[$(1,3)$] \begin{gather*}
			(x_1 +y_1 \alpha)(x_2 +y_2 \alpha) = x_1(x_2 +y_2 \alpha) + y_1 \alpha (x_2 +y_2 \alpha) \in \\
			J(J+C_{\mathcal{R}}) + x\tilde{J} = J^2 + JC_{\mathcal{R}} + x\tilde{J} \subseteq x J + \tilde{J}C_{\mathcal{R}} + x\tilde{J} \subseteq x \tilde{J}.
			\end{gather*}

			\item[$(2,2)$] \begin{gather*}
			(x_1 +y_1 \alpha)(x_2 +y_2 \alpha) \in (J+J\alpha)^2 = J^2 + J^2\alpha + J^2b \subseteq \\
			\subseteq x\tilde{J} + x\tilde{J} + J^2\subseteq x\tilde{J}.
			\end{gather*}

			\item[$(2,3)$] \begin{gather*}
			(x_1 +y_1 \alpha)(x_2 +y_2 \alpha) \in (J+J\alpha)(J+C_{\mathcal{R}}) = \\
			= J^2 + JC_{\mathcal{R}} + J^2\alpha + J\alpha C_{\mathcal{R}} \subseteq x\tilde{J} + \tilde{J}C_{\mathcal{R}} + x\tilde{J} + \tilde{J}C_{\mathcal{R}} \subseteq x \tilde{J}.
			\end{gather*}

			\item[$(3,3)$] \begin{gather*}
			(x_1 +y_1 \alpha)(x_2 +y_2 \alpha) \in (J+C_{\mathcal{R}})^2 = J^2 + J C_{\mathcal{R}} + C_{\mathcal{R}}^2 \subseteq \\
			\subseteq x\tilde{J} + \tilde{J} C_{\mathcal{R}} + \tilde{J} C_{\mathcal{R}} \subseteq x\tilde{J}.
			\end{gather*}
		\end{itemize}
		This proves that $\tilde{J}^2 \subseteq x\tilde{J}$.
	\end{proof}
	\noindent
	\textbf{Acknowledgments}. I would like to thank Marco D'Anna for his constant help during the drafting of the paper and for his useful suggestions.

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	\bibliography{Reference}

\end{document}


TITLE: Derivative of $l_1$ norm
QUESTION [6 upvotes]: I want to compute the following derivative with respect to $n\times1$ vector $\mathbf x$.
$$g = \left\lVert \mathbf x - A \mathbf x  \right\rVert_1 $$
My work:
$$g = \left\lVert \mathbf x - A \mathbf x  \right\rVert_1  = \sum_{i=1}^{n} \lvert x_i - (A\mathbf x)_i\rvert = \sum_{i=1}^{n} \lvert x_i - A_i \cdot \mathbf x \rvert = \sum_{i=1}^{n} \lvert x_i - \sum_{j=1}^n a_{ij} x_j\rvert$$
So the $k$th element of derivative is:
$$\frac{\partial g}{\partial x_k} = \frac{\partial }{\partial x_k}\sum_{i=1}^n \lvert x_i - \sum_{j=1}^n a_{ij} x_j\rvert $$
$$= \frac{\partial }{\partial x_k}\bigg(\lvert x_1 - \sum_{j=1}^n a_{1j} x_j\rvert +\cdots+ \lvert x_k - \sum_{j=1}^n a_{kj} x_j\rvert + \cdots\lvert x_n - \sum_{j=1}^n a_{nj} x_j\rvert  \bigg)$$
$$ =-a_{1k}sign(x_1 - \sum_{j=1}^n a_{1j} x_j)-\cdots+(1-a_{kk})sign(x_k - \sum_{j=1}^n a_{kj} x_j)-\cdots -a_{nk}sign(x_n - \sum_{j=1}^n a_{nj} x_j)$$
And my questions:

Is this derivation correct?
How I can represent the answer compactly?
Can you introduce me a source to master this material?

Thanks.

REPLY [1 votes]: $\ell_1$ norm does not have a derivative. It is a nonsmooth function. It has subdifferential which is the set of subgradients. A vector $s$ is a subgradient of a function $f$ at a point $x$ if for all $y$, $s$ satisfies
\begin{equation}
f(x+y)\geq f(x)+y^*s.
\end{equation}
The subdifferential of $\ell_1$ norm is connected to nonzero entries of the vector $x$. In particular, let $sign(x)$ return $+1,-1,0$ for $x_i>0$, $x_i<0$ and $x_i=0$ respectively. Let $S$ be the set of nonzero coordinates of $x$. Then, you can verify that the $\ell_1$ subdifferential $\partial \|x\|_1$ is given by
\begin{equation}
\partial \|x\|_1=\{s~\big|~ s_i=sign(x_i)~\forall~i\in S,~\|s\|_{\infty}\leq 1\}
\end{equation}
that is the vectors $s$ which are equal to $sign(x)$ over $S$ and that have infinity norm (maximum absolute value) less than $1$.
Regarding your question: You should calculate the subdifferential of your function which will involve the nonzero pattern of the vector $(I-A)x$. For more info: https://en.wikipedia.org/wiki/Subderivative


text:
/-
Copyright (c) 2021 OpenAI. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Kunhao Zheng, Stanislas Polu, David Renshaw, OpenAI GPT-f
-/
import mathzoo.imports.miniF2F

open_locale nat rat real big_operators topological_space

theorem mathd_numbertheory_551 :
  1529 % 6 = 5 :=
begin
  norm_num,
end

subset_name: curated


TITLE: How to solve for the parameters of the Gamma distribution given x for the 50th and 90th percentiles?
QUESTION [1 upvotes]: How can I solve for the parameters alpha and beta given that x = 20 is the 50th percentile, and x = 300 is the 90th percentile?

REPLY [2 votes]: I'll use the notation of the wikipedia article on the gamma distribution throughout.
Given that there's no simple closed form for the median of a gamma distribution, you'll want to do this numerically.
You can be smart about this, though.  First, note that the ratio of the 90th to 50th percentiles of a gamma distribution depends only on the shape parameter $k$, not the scale parameter $\theta$.  So we can ask: what is the shape parameter $k$ for which the 90th percentile divided by the 50th percentile is $300/20 = 15$?  This is a couple lines in R:
f = function(k){qgamma(.9, k, 1)/qgamma(.5, k, 1)}
k0 = uniroot(function(k){f(k)-15}, c(0.1, 1))$root

The first line defines a function f which returns the 90th percentile of the $gamma(x,1)$ distribution divided by its 50th percentile.    The second line finds a root of $f(x) - 15$ in the interval $[0.1, 1]$ and assigns it to k0.
This returns the root $\theta \approx  0.2672395$.
Now the median of a $Gamma(k_0, 1)$ random variable is given by qgamma(.5, k0, 1) in R and is equal to $m_0 = 0.05319006$.  So to get the median equal to 20, you need a scale parameter $\theta_0 = 20/m_0 \approx 376.0101$.


TITLE: If $A=[0,1)$ then show that its interior is $(0,1)$.
QUESTION [0 upvotes]: I want to show that if $A=[0,1)$ then its interior is $(0,1)$. I know that $int(A) \subset A$, and that $\forall x \in int(A)$ $\exists R>0 $ such that $B(x,R) \subset A$. Thus immediately we see that $0 \notin int(A)$ because $\not \exists R>0$ such that $B(0,R)\subset A$.
What I struggle to do is to show that the final set is equal to $(0,1)$.

REPLY [3 votes]: Draw a picture : let $x \in (0,1)$ and $R=\frac{1}{2} \min\{x,1-x\}$.
Now show: $B(x, R)\subseteq (0,1)$


TITLE: Help with a symmetric random walk problems
QUESTION [2 upvotes]: Given a simple symmetric random walk $X_1,X_2,…,X_n,…$ where $X_t,t=1,2,…$ are i.i.d random variables distributed according to $\Bbb P(X_t=1)=1/2$ and $\Bbb P(X_t=-1)=1/2$. Define partial sum $S_t=∑_{k=1}^tX_k$ . Define $T_x=\min\{t∈N:S_t=x\}$. Use $\Bbb P_0 (T_x=k)$ to denote the probability of a random walk first reaching position $x$ at time $k$ (the subscript $0$ of $\Bbb P_0$ means the random walk is assumed to start at position $0$).
It can be shown by reflection that $\Bbb P{_0}\left( {{T_1} = 2k - 1} \right) = \frac{{\left( {\begin{array}{*{20}{c}}
{2k}\\
k
\end{array}} \right)}}{{{2^{2k}}\left( {2k - 1} \right)}}$. The problem is to show

$\Bbb P_0 (T_x≥N)\simeq cN^{-\frac{1}{2}}$ when N→∞ where $\simeq$ means infinitesimal of the same order.

The textbook offers the following hint:

Use induction. For $x=a+b$, $\Bbb P_0 (T_x≥N)$ is the sum of $\Bbb P_0 (T_a≥i)\Bbb P_0 (T_b≥j)$ over $i+j≥N$. Also, bulk of the sum comes from $i≤L,j≥N-i$ and $j≤L,i≥N-j$ for some large number $L$.

I tried to follow the hint. It is not hard to show $\Bbb P_0 (T_1≥N) \simeq  N^{-1/2}$. Also Suppose $\Bbb P_0 (T_x≥N) \simeq N^{-\frac{1}{2}}$ holds for $x$.
Then for $x+1$, first breaks $\Bbb P_0 (T_{x+1}≥N)=∑_{i+j≥N}\Bbb P_0 (T_x=i) \Bbb P_0 (T_1=j)=p_1+p_2+p_3$ where
$p_1=∑_{i=1}^L∑_{j≥N-i}\Bbb P_0 (T_x=i) \Bbb P_0 (T_1=j)=∑_{i=1}^L \Bbb P_0 (T_x=i) ∑_{j≥N-i}\Bbb P_0 (T_1=j)$
$p_2=∑_{j=1}^L∑_{i≥N-j}\Bbb P_0 (T_1=j) \Bbb P_0 (T_x=i)=∑_{j=1}^L \Bbb P_0 (T_1=j) ∑_{i≥N-j}\Bbb P_0 (T_x=i)$
$p_3=∑_{i=L}^∞ ∑_{j≥\max(L+1,N-i) }P_0 (T_x=i) P_0 (T_1=j)$
The following graph illustrates the breakdown.

For $p_1$, we have
$\mathop {\lim }\limits_{N \to \infty } \frac{{{p_1}}}{{{N^{ - \frac{1}{2}}}}} = \mathop \sum \limits_{i = 1}^L {\Bbb P_0}\left( {{T_x} = i} \right)\mathop {{\rm{lim}}}\limits_{N \to \infty } \frac{{\mathop \sum \limits_{j \ge N - i} {\Bbb P_0}\left( {{T_1} = j} \right)}}{{{N^{ - \frac{1}{2}}}}} = \mathop \sum \limits_{i = 1}^L {\Bbb P_0}\left( {{T_x} = i} \right){c_i} \Rightarrow {p_1} \simeq {N^{ - \frac{1}{2}}}$
The same thing for $p_2$. However, I encountered difficulty to show $p_3$ is an infinitesimal of order not lower than $N^-{\frac{1}{2}}$. By hint, I think $p_3$ should decrease faster than $N^{-\frac{1}{2}}$.

REPLY [1 votes]: Hint: For a symmetric random walk with $p=q=1/2$, you can use the reflection principle directly to show that, if $x+N$ is odd, then $\mathbb{P}_0(T_x>N)=\mathbb{P}_0(-x<S_N<x).$


TITLE: On the space of continuous complex-valued functions with compact support
QUESTION [1 upvotes]: Let $X$ be a topological space. Recall that $C_c (X)$ is the space of continuous complex-valued functions with compact support on $X$.
Is $C_c (X)$ (with the supremum norm) a Banach space?

REPLY [2 votes]: It is obviously a normed space but it is not complete. [ In fact, when $X=\mathbb R$ its completion is the space $C_0 (\mathbb R)$ of continuous functions that vanish at $\infty$ and $-\infty$. To show that the space is not complete when $X=\mathbb R$ take any function g in $C_0 (\mathbb R)$. There exists a sequence $\{f_n\}$ in $C_c (\mathbb R)$ such that $f_n =1$ on $[-n,n]$ and $0 \leq f_n \leq 1$ everywhere. The sequence $\{gf_n\}$ is a Cauchy sequence which is not convergent.


TITLE: Help needed with axiomatic deductions
QUESTION [3 upvotes]: Proof axiomatically:
$\vdash \forall x (\neg (Ax \rightarrow Bx) \rightarrow (\neg Ax \rightarrow \neg Bx))$
You can use in your deduction as step the equivalence of $\varphi$ with $\neg\neg\varphi$:
$\varphi\vdash\neg\neg\varphi$ and $\neg\neg \varphi \vdash \varphi$
Also used are the following axioms:
1 $(∀x(φ→ψ)) \rightarrow (\forall x \varphi \rightarrow \forall x \psi)$
2 $\varphi \rightarrow \forall x \varphi$ ($x$ not free in $\varphi$)
3 $\forall x \varphi \rightarrow [t/x]\varphi$
a $\varphi \rightarrow (\psi \rightarrow \varphi)$
b $\varphi \rightarrow (\psi \rightarrow \chi) \rightarrow ((\varphi \rightarrow \psi)\rightarrow(\varphi \rightarrow \chi))$
c $(\neg \varphi \rightarrow \neg \psi) \rightarrow (\psi \rightarrow \varphi)$
So starting out on this question, I had hopes of delivering an answer. But alas, my plans were foiled, I could not get a deduction. So, I decided to have a peek at the answer, and it baffled me completely:

$\vdash Bx \rightarrow (Ax \rightarrow Bx)$ (a)
$\neg\neg Bx\vdash Bx$ ($\neg\neg\varphi\vdash\varphi$)
$\neg\neg Bx \vdash Ax \rightarrow Bx$ (MP, 1, 2)
$Ax \rightarrow Bx \vdash \neg\neg(Ax \rightarrow Bx)$ ($\varphi\vdash\neg\neg\varphi$)
$\vdash (Ax \rightarrow Bx) \rightarrow \neg\neg(Ax \rightarrow Bx)$ (deduction theorem on 4)
$\neg\neg Bx \vdash \neg\neg(Ax \rightarrow Bx)$ (MP, 3, 5)
$\vdash \neg\neg Bx \rightarrow \neg\neg(Ax \rightarrow Bx)$ (deduction theorem on 6)
$\vdash (\neg\neg Bx \rightarrow \neg\neg(Ax \rightarrow Bx)) \rightarrow (\neg(Ax \rightarrow Bx) \rightarrow \neg Bx)$ (c)
$\vdash \neg(Ax \rightarrow Bx) \rightarrow \neg Bx$ (MP, 7, 8)
$\neg(Ax \rightarrow Bx) \vdash \neg(Ax \rightarrow Bx)$ (assumption)
$\neg(Ax \rightarrow Bx) \vdash \neg Bx$ (MP, 9, 10)
$\vdash \neg Bx \rightarrow (\neg Ax \rightarrow \neg Bx)$ (a)
$\neg(Ax \rightarrow Bx) \vdash (\neg Ax \rightarrow \neg Bx)$ (MP, 11, 12)
$\vdash \neg(Ax \rightarrow Bx) \rightarrow (\neg Ax \rightarrow \neg Bx)$ (deduction theorem on 13)
$\vdash \forall x (\neg(Ax \rightarrow Bx) \rightarrow (\neg Ax \rightarrow \neg Bx))$ (universal generalization on 14)

Starting from 1, I immediately run into a wall. I have absolutely no idea how it got here. Where does $Bx$ come from? So, I start from the bottom. This goes a bit easier, but backtracking it is still not easy. This goes well up until 10/9-ish, again, I run into a wall. No ideas on how to get to the following statement, let alone to 1.
My question is this: can someone guide me through this deduction, with the simplest of explanation?

REPLY [2 votes]: This is a long comment ...
The above proof is "complicated" and difficult to grasp because the axiomatic (or Hilbert-style) proof-system is very "opaque", in terms of euristic (finding how to build up the required proof).
Natural deduction is far better for this.
A possible way to "simplify" the search of a proof in a case like that above, which is basically a "propositional" proof (the only "quantifier" rule is invoked in the last step), is to "enlarge" the set of rules available.
How ? Starting from the propositional axioms a)-c) and proving their completeness for the tautologies, i.e.proving the meta-theorem which states that all tautologies are provable from the above axioms with modus ponens.
In this way, we may use in our proofs every instance of a tautology.
Assuming having done this, we may build another proof of the theorem using the following tautologies :

$\vDash_{TAUT} (\varphi \rightarrow \psi) \lor (\psi \rightarrow \varphi)$ --- $(TAUT_1)$

and :

$\vDash_{TAUT} \lnot \sigma \rightarrow [ (\sigma \lor \tau) \rightarrow \tau) ]$ --- $(TAUT_2)$

and :

$\vDash_{TAUT} (\varphi \rightarrow \psi) \rightarrow (\lnot \psi \rightarrow \lnot \varphi)$ --- $(TAUT_3)$.

Now for the proof.
(1) Instantiate $TAUT_1$ with $Ax$ in place of $\varphi$ and $Bx$ in place of $\psi$ :

$\vdash (Ax \rightarrow Bx) \lor (Bx \rightarrow Ax)$

(2) Instantiate $TAUT_2$ with $(Ax \rightarrow Bx)$ in place of $\sigma$ and $(Bx \rightarrow Ax)$ in place of $\tau$ :

$\vdash \lnot (Ax \rightarrow Bx) \rightarrow ( [ (Ax \rightarrow Bx)\lor (Bx \rightarrow Ax) ] \rightarrow (Bx \rightarrow Ax) )$

(3) Assume :

$ \lnot (Ax \rightarrow Bx)$

(4) $\lnot (Ax \rightarrow Bx) \vdash [ (Ax \rightarrow Bx)\lor (Bx \rightarrow Ax) ] \rightarrow (Bx \rightarrow Ax)$ --- from 2) and 3) by modus ponens
(5) $\lnot (Ax \rightarrow Bx) \vdash (Bx \rightarrow Ax)$ --- from 1) and 4) by modus ponens
(6) Instantiate $TAUT_3$ with $Bx$ in place of $\varphi$ and $Ax$ in place of $\psi$ :

$ \vdash (Bx \rightarrow Ax) \rightarrow (\lnot Ax \rightarrow \lnot Bx)$

(7) $\lnot (Ax \rightarrow Bx) \vdash (\lnot Ax \rightarrow \lnot Bx)$ --- from 5) and 6) by modus ponens
(8) $\vdash \lnot (Ax \rightarrow Bx) \rightarrow (\lnot Ax \rightarrow \lnot Bx)$ --- from 7) by Deduction Theorem.
Now the "hard" work is done.
We have only to apply the Generalization Theorem :

if $\Gamma \vdash \varphi$ and $x$ does not occur free in any formula in $\Gamma$, then $\Gamma \vdash \forall x \varphi$,

with $\Gamma = \emptyset$, to conclude from 8) :

$\vdash \forall x [\lnot (Ax \rightarrow Bx) \rightarrow (\lnot Ax \rightarrow \lnot Bx) ]$.


text:
(*******************************************************************************

  Project: Refining Authenticated Key Agreement with Strong Adversaries

  Module:  dhlvl3_symmetric.thy (Isabelle/HOL 2016-1)
  ID:      $Id: dhlvl3_symmetric.thy 133183 2017-01-31 13:55:43Z csprenge $
  Author:  Joseph Lallemand, INRIA Nancy <joseph.lallemand@loria.fr>
           Christoph Sprenger, ETH Zurich <sprenger@inf.ethz.ch>

  Level-3 Diffie-Hellman protocol; symmetric instantiation of
  version with generic channel implementation.

  Copyright (c) 2015-2016 Joseph Lallemand and Christoph Sprenger
  Licence: LGPL

*******************************************************************************)

section \<open>Authenticated Diffie-Hellman Protocol (L3, symmetric)\<close>

theory dhlvl3_symmetric
imports dhlvl3 Implem_symmetric
begin

interpretation dhlvl3_sym: dhlvl3 implem_sym
by (unfold_locales)

end


TITLE: Seismic science questions
QUESTION [2 upvotes]: We just (20mn ago) suffered from another strong aftershock (of the initial March 11th earthquake) in Japan.
I have some questions related to seismic science.

Why is there so much time between the initial quake and the following strong aftershocks (initial was March 11, the stronger aftershock happened 20mn ago, April 7)?
Why does it seem that progress in seismic science (and quakes predictions) is so poor? It seems people have only raw probabilities, based on historical data.
Would anybody have any graphical / imagery data of the physical fault move on March 11?

Thank you.

REPLY [2 votes]: Why does it seem that progress in seismic science (and quakes predictions) is so poor? It seems people have only raw probabilities, based on historical data.

It's because the actual earthquake depends on tiny initial events.
Imagine standing on a glass so it breaks. We can calculate how much force it takes for a glass to break. But it would be very hard to predict in advance at which microscopic atomic scale flaw in the material the crack will start and at exactly what nano-second.
An earthquake is just the same on a larger scale.


TITLE: Why does different part of a spring having mass expand proportional to their distance while the spring has some mass hanged in the bottom?
QUESTION [3 upvotes]: Why do different parts of a massive spring expand proportional to their distance while the spring has some mass hung on the bottom which is comparably very less than the mass of the spring? If we take a massless (or practically negligible mass) spring then the elongation of the spring is uniform. But why does it make an exception when the spring has mass? A theoretical analytical answer will be appreciated. Now you might say it's an approximation but the approximation should be nearly true.
My book, The Physics of Waves and Oscillations by N. K. Bajaj, says:

Let us assume that the various parts of the spring undergo displacements proportional to their distance from the fixed end as indicated in the diagram.
  We can now compute the total kinetic energy of the spring at an instant when the mass M has a displacement $\phi$.
  Now the displacement per unit length of the spring is $\phi/l$.

(from page 40)

REPLY [1 votes]: I think it might be because the helical springs when compressed are not actually compressed, they are twisted.The spring as a whole might be acted upon by an axial compressive load but the wire is actually twisted.With twisting, the deformation at the far end(away from fixed) will be more than the fixed end.But Iam not sure about this.


TITLE: How to find the derivative of this?
QUESTION [0 upvotes]: $$f(z) = \frac{\exp(z)}{1+\exp(z)}$$
My thought was to apply quotient rule with the denominator needing chain rule. But I feel like my answer is off.
Here's my working:
$$\frac{exp(z)}{1+exp(z)} = \frac{(1 + exp(z))^{-1}exp(z) - exp(z)(-1)(1 + exp(z))^{-2}(exp(z))}{(1 + exp(z))^2}$$
$$= \frac{exp(z)}{(1+exp(z))^3} + \frac{exp(z)^2}{(1+exp(z))^4}$$
$$= \frac{exp(z) + exp(z)^2 + exp(z)^2}{(1 + exp(z))^4}$$
$$= \frac{exp(z)(1 + 2exp(z))}{(1 + exp(z))^4}$$

REPLY [0 votes]: Quotient rule approach:  $(\dfrac{e^z}{1+e^z})'=\dfrac {e^z(1+e^z)-e^z(e^z)}{(1+e^z)^2}=\dfrac {e^z}{(1+e^z)^2}$.


TITLE: Techniques to prove that solutions of ODE are defined for all time.
QUESTION [2 upvotes]: What are the main techniques for proving that the solutions of ODE are defined for all times? Obviously I refer to qualitative study of $$y'=f(t,y)$$ with $f:\mathbb{R}\times\mathbb{R}^n\to\mathbb{R}^n $ at least $C^0$. The standard way is to use classical reults (e.g. prove that $f$ is sublinear or globally lipschitz) but these rusults are often not applicable. In this case I try to show that the solutions are bounded (it is known that if the orbits are bounded then the solutions are defined for all time). To do this I usually try one of the following techniques:

I look for a constant of the motion and I study the level sets to understand the shape of the orbits (and see if they are bounded);
I find a particular solution that is a closed and curve (e.g. a circle): in fact for the uniqueness the orbits can not intersect and therefore all orbits that start from inside this closed curve will have to remain there and will therefore be limited;
I study the function $h(t): = x^2(t) + y^2(t)$: I make the derivative and I try to show that it is limited;
I try a change of variables (e.g. polar coordinates) that simplifies the equations.

In conclusion:
1) Do you know other general techniques to show that the orbits are bounded?
2) However, boundedness is only a sufficient condition to prove that solutions are defined for all time! In case the solutions are unbounded how can I prove that they are defined for all time? For example consider $$\begin{cases}\dot x=-y^2\\
\dot y = x^2\end{cases}$$ or $$\begin{cases}\dot x=y^2\\
\dot y = x^2\end{cases}$$
The level sets of the constants of motion ($E:=x^3\pm y^3$) of these system are open and unbounded curve. In this case the only techniques that I know is to cumpute $t=\int^{+\infty}_{x_0}...dx$ and understand if this integral converges or diverges (for other details see the end of this page: Qualitative study of $ \dot x = - y^2$, $\dot y= x^2 $).
3) Advice on books and readings about this problem are welcome.

REPLY [3 votes]: An answer to 1):
In mathematical population biology, a totally competitive system  of ODEs is defined as
$$
x'_i = x_i f_i(x_1, \dots, x_n), \quad i = 1, \dots, n,
$$
where
$$
\frac{\partial f_i}{\partial x_j} < 0
$$
(of course $f_i$ are assumed to be $C^1$, at least).  The natural phase space is the nonnegative orthant $C = \{\, x = (x_1, \dots, x_n) : x_i \ge 0$ for all $i = 1, \dots, n \,\}$.
As $x_i$ represents the population density of the $i$-th species, only solutions that are bounded for positive time have biological significance.  There is a very simple, but extremely useful, sufficient condition for such boundedness:

For each $i = 1, \dots, n\ $  there is $x_i^* > 0$ such that $f_i(0, \dots, 0, x_i^*, 0, \dots, 0) = 0$.

Indeed, let $x \in C$ be such that for some $j$, $x_j > x_j^*$.  Then
$$
f_j(x) = f_j(x_1, \dots, x_{j-1}, x_j, x_{j+1}, \dots, x_n) \le  f_j(0, \dots, 0, x_j, 0, \dots, 0) < f_j(0, \dots, 0, x_j^*, 0, \dots, 0) = 0.
$$
From this it follows that all solutions are attracted toward the ($n$-dimensional) parallelepiped $[0, x_1^*] \times \ldots \times [0, x_n^*]$. Consequently, each solution in $C$ is defined for all positive times $t$.
For (much) more on such systems, see, e.g., Zhao's book Dynamical Systems in Population Biology.  See also the classical paper by M. W. Hirsch Systems of differential equations which are competitive or cooperative: III. Competing species.


text:
\section{The foliation}
\label{sec:foliation}
In this section we show that the surfaces $\Sigma_r$ indeed are a
foliation of a pointed neighborhood of $p\in M$. The method used is
very close to the arguments in \cite[pp. 390--391]{Ye:1991}. We start
with the following observation.
\begin{lemma}
  \label{thm:Phi-phi-r-r-is-even}
  The operator $\Phi_{\varphi r r} (0,\tau,0,\lambda)$ maps even
  functions to even functions.
\end{lemma}
\begin{proof}
  Note that  $\Phi_{\varphi r r} (0,\tau,0,\lambda)= \drr W_{r,\tau,\lambda}$
  where $W_{r,\tau,\lambda}$ is given by the expression in
  equation~\eqref{eq:8}. To prove the claim we check this expression
  term by term as in the proof of Lemma~\ref{thm:derivativePhi1}.

  We start by quoting from~\cite[Lemma 1.3]{Ye:1991} that $\drr
  L_{r,\tau}$ is an even operator. Hence, the claim follows from the
  facts that $\drr Q_{r,\tau}$ is an even function and in conjunction
  with equation~\eqref{eq:23} from the fact that the operator
  \begin{equation}
    \label{eq:14}
    f\mapsto
    -\tfrac12 H_{r,\tau}^2 \Delta_{r,\tau} f
    + 2 \Acirc_{r,\tau}(\nabla H_{r,\tau},\nabla_{r,\tau} f)
    + 2 H_{r,\tau} \langle \Acirc_{r,\tau},\nabla_{r,\tau}^2 f\rangle
    + 2 H_{r,\tau}\omega_{r,\tau}(\nabla_{r,\tau} f)
  \end{equation}
  maps even functions to even functions.

  To show this, we quote from the proof of~\cite[Lemma 1.3]{Ye:1991}
  that $\drr \Delta_{r,\tau}|_{r=0}$ is an even operator,
  $\drr\Ric_{r,\tau}(\nu_{r,\tau},\nu_{r,\tau})$ is an even
  function, $\drr |A_{r,\tau}|^2$ is an even function. Note that
  \cite[Equation (1.17)]{Ye:1991} implies that $\drr A_{r,\tau}$ is
  even, so that also $\drr H_{r,\tau}$ and $\drr \Acirc_{r,\tau}$ are even.

  Using these, it is easy to check that
  \begin{equation}
    \label{eq:22}
    \begin{aligned}
      &\drr\big(
      H_{r,\tau}\Delta_{r,\tau} H_{r,\tau}
      + 2\langle \nabla_{r,\tau}^2 H_{r,\tau} ,\Acirc_{r,\tau} \rangle
      +2H_{r,\tau}\langle \Acirc_{r,\tau},T_{r,\tau}\rangle
      \\
      &\ \ \ \
      - H_{r,\tau} \nabla_{r,\tau} \Ric_{r,\tau}(\nu_{r,\tau},\nu_{r,\tau},\nu_{r,\tau})
      + H_{r,\tau}^2|\Acirc_{r,\tau}|^2
      + \frac12 H_{r,\tau}^2 \Ric_{r,\tau}(\nu_{r,\tau},\nu_{r,\tau})
      \big)
    \end{aligned}
  \end{equation}
  is even. For example consider (we omit the subscript $_{r,\tau}$ for
  clarity in the notation:
  \begin{equation*}
    \begin{aligned}
      \drr\!\! H\Delta H
      &=\left(\drr H\right)\Delta H_{0,\tau} + \left((\dd{r} H) (\dd{r} \Delta H)\right)|_{r=0} + H_{0,\tau} \drr (\Delta H)
      \\
      &= H_{0,\tau} \left( 2(\dr\Delta) (\dr H) +  (\drr \Delta) H +
      \Delta_{0,\tau} \drr H \right)
      \\
      &= H_{0,\tau} (\drr \Delta_{r,\tau}) H_{0,\tau} + H_{0,\tau}
      \Delta_{0,\tau} \big( \drr H_{r,\tau} \big).
    \end{aligned}
  \end{equation*}
  In the second and third equality we used from the proof of
  Lemma~\ref{thm:derivativePhi1} that $\dr H_{r,\tau}=0$ and
  the fact that $H_{0,\tau}$ is constant. The right hand side is even,
  since $H_{0,\tau}$ is constant and thus even, since
  $\drr \Delta_{r,\tau}$ and $\Delta_{0,\tau}$ map even functions to
  even functions and since the product of even functions is even. The
  calculation for the other terms in~\eqref{eq:22} is similar. To
  treat the term
  $H_{r,\tau} \nabla_{r,\tau}
  \Ric_{r,\tau}(\nu_{r,\tau},\nu_{r,\tau},\nu_{r,\tau})$
  use that
  $\nabla_{r,\tau}\Ric_{r,\tau}(\nu_{r,\tau},\nu_{r,\tau},\nu_{r,\tau})=O(r^3)$.

  For the remaining terms in $Q_{r,\tau}$ note that the
  $\dr(\nabla_{r,\tau})$ is a first order differential operator that
  vanishes on constant functions. Hence,
  \begin{equation*}
    \begin{aligned}
      &\tfrac{1}{2}\drr |\nabla_{r,\tau} H_{r,\tau}|^2
      \\
      &
      =
      \left\langle\! \drr\!\!(\nabla_{r,\tau} H_{r,\tau}) ,\nabla_{0,\tau} H_{0,\tau} \right\rangle
      +
      \left|(\dr\!\nabla_{r,\tau}) H_{0,\tau} + \nabla_{0,\tau} (\dr H_{r,\tau})\right|^2 = 0.
    \end{aligned}
  \end{equation*}
  This follows, since $H_{0,\tau}$ is constant,
  $\dr(\nabla_{r,\tau})= H_{0,\tau}$ and $\nabla_{0,\tau} H_{0,\tau}
  =0$ and since $\dr H_{r,\tau} =0$ as in
  Lemma~\ref{thm:derivativePhi1}. A similar computation yields that $\drr
  \omega_{r,\tau}(\nabla_{r,\tau} H) = 0$. We established that $\drr
  Q_{r,\tau}$ is an even function.

  It remains to consider the expression in~\eqref{eq:14}. Note that
  the first term is even by reasoning as above. The second term
  satisfies
  \begin{equation*}
    \drr \left( \Acirc_{r,\tau}(\nabla H_{r,\tau},\nabla_{r,\tau}
      f)\right) = 0.
  \end{equation*}
  To treat the third term, use $\Acirc_{0,\tau}=0$ and
  $\dr\Acirc_{0,\tau} =0$ to compute
  \begin{equation*}
    \drr H_{r,\tau} \langle \Acirc_{r,\tau},\nabla_{r,\tau}^2 f\rangle
    =
    H_{0,\tau} \langle \drr \Acirc_{r,\tau} , \nabla_{0,\tau}^2 f\rangle.
  \end{equation*}
  Note that $\drr \Acirc_{r,\tau}$ is even and $\nabla_{0,\tau}^2$
  maps even functions to even functions so that this operator also has
  the desired property.

  For the last term from~\eqref{eq:14} we compute using
  $\omega_{0,\tau} =0$, $\dr\omega_{r,\tau}=0$ and $\dr H_{r,\tau} =0$
  that:
  \begin{equation*}
    \drr H_{r,\tau} \omega_{r,\tau} (\nabla_{r,\tau} f)
      =
      H_{0,\tau} \left(\drr\omega_{r,\tau}\right) \nabla_{0,\tau} f.
  \end{equation*}
  Note that $\nabla_{0,\tau}$ is the tangential gradient on $S^2$ and
  maps even functions to odd vector fields. Furthermore, by
  equation~\eqref{eq:12} and the fact that $\nu_{r,\tau} = x + O(r^2)$
  we have that
  \begin{equation*}
    \omega_{r,\tau} = r^2 \Ric_{c(\tau)} x + O(r^3)
  \end{equation*}
  so that
  \begin{equation*}
    \drr \omega_{r,\tau} = \Ric_{c(\tau)} x
  \end{equation*}
  and hence $\drr \omega_{r,\tau}$ is an odd one form. Consequently
  $H_{0,\tau} \left(\drr\omega_{r,\tau}\right) \nabla_{0,\tau} f$ is
  an even function whenever $f$ is even. This concludes the proof.
\end{proof}
\begin{lemma}
  \label{thm:tau-is-o-r2}
  For $r\in(0,r_1)$ let
  $\Sigma_r := S(r,\tau(r,\varphi(r)),\varphi(r))$ be as in
  Lemma~\ref{thm:solution-final}.  Then  $\tau(r)=O(r^2)$ as $r\to 0$.
\end{lemma}
\begin{proof}
  It follows from the implicit function theorem that $\tau'(r)=O(r)$
  if and only if
  \begin{equation}
    \label{eq:21}
    \tilde P_1 (  \Phi_{\varphi \varphi}(0,\tau,0,\lambda)\varphi_0
    \varphi_0)=0
    \quad\text{and}\quad
    \tilde P_1 (\Phi_{\varphi r r} (0,\tau,0,\lambda)\varphi_0)=0.
  \end{equation}
  To establish the first identity, note that by the fact that
  equation~\eqref{eq:13} has unique solutions and since
  $W_{0,\tau,\lambda}$ is invariant under the reflection at the
  origin, it follows that $\varphi_0$ is an even function.

  Furthermore, for every $t$ in a neighborhood of $0$ the euclidean
  Willmore operator $\Phi(0,\tau,t\varphi_0,\lambda)$ evaluates to an
  even function. Hence also
  \begin{equation*}
    \Phi_{\varphi \varphi}(0,\tau,0,\lambda)\varphi_0 \varphi_0
    =
    \left.\frac{\partial^2}{\partial t^2}\right|_{t=0} \Phi(0,\tau,t\varphi_0,\lambda)
  \end{equation*}
  is even. Since $\tilde P_1$ vanishes on even functions, the first
  claim from \eqref{eq:21} follows.

  To prove the second identity, note that by
  Lemma~\ref{thm:Phi-phi-r-r-is-even} the operator
  $\Phi_{\varphi r r} (0,\tau,0,\lambda)$ maps even function to even
  functions and the claim follows in a similar manner.
\end{proof}

This lemma implies in particular that we can reparameterize the
solutions that we found in section~\ref{sec:equation} by their area.
\begin{corollary}
  \label{thm:reparametrize}
  For $r\in(0,r_1)$ let
  $\Sigma_r := S(r,\tau(r,\varphi(r)),r^2\varphi(r))$ be as in
  Lemma~\ref{thm:solution-final}. Consider the area of $\Sigma_r$ in
  $(M,g)$ as a function of $r$:
  \begin{equation*}
    a :(0,r_1) \to (0,\infty) : r \mapsto \int_{\Sigma_r} 1 \dmu_g.
  \end{equation*}
  Then there exists $r_2\in(0,r_1]$ so that $a$ is strictly increasing
  on $(0,r_2)$. In particular:
  \begin{equation*}
    a(r) = 4\pi r^2 + O(r^4)
    \quad\text{and}\quad
    a'(r) = 8\pi r + O(r^3).
  \end{equation*}
\end{corollary}

\begin{proof}
  Note that $a$ extends as a smooth function to $r=0$ so that $a(0)=0$
  and hence the first claim follows from the second. We first note that
  \begin{equation*}
    a'(r) = - \int_{\Sigma_r} g(\vec{H}, X) \dmu_g\ ,
  \end{equation*}
  where $X$ is the variation vector-field along this family. Note that
  $X$ is not unique, whereas $X^\perp$ is well defined. Recall that
  from Lemma \ref{thm:solution-final} we have that $\Sigma_r$ is an
  exponential normal graph over $S_r(\tau(r))$ with height function
  $r^3\varphi(r)$ such that $\varphi(r) \rightarrow \varphi_0$ as
  $r \rightarrow 0$. Furthermore, by Lemma \ref{thm:tau-is-o-r2} we
  have that $\tau(r) = O(r^2)$ as $r \to 0$. This implies that
  $$ X^\perp\big|_{\Sigma_r} = \frac{\partial}{\partial r_\tau} + O(r^2)$$
  where $r_\tau = d_g(\tau(r), \cdot)$. Furthermore, by the above and
  \eqref{eq:2.8} we have that
  $$ H_{\Sigma_r} = H_{S_r(\tau(r))} + O(r^2) = \frac{2}{r} + O(r)\, ,$$
  as well as
  $$ \nu_{\Sigma_r} = \nu_{S_r(\tau(r))} + O(r^3)\, .$$
  Also note that from \eqref{eq:2.4} we have
$$ \int_{S_r(\tau(r))} 1 \dmu_g = 4\pi r^2 + O(r^4)\ .$$
This implies
  \begin{equation*}
    a'(r) = - \int_{\Sigma_r} g(\vec{H}, X) \dmu_g =  8\pi r + O(r^3)\, .
  \end{equation*}
\end{proof}
Due to Corollary~\ref{thm:reparametrize} there exists $a_0\in
(0,\infty)$ and a map $\tilde r : (0,a_0) \to (0,r_2)$ such that
$|\Sigma_{\tilde r(a)}| = a$. We slightly abuse notation by letting
\begin{equation}
  \label{eq:24}
  \Sigma_a := \Sigma_{\tilde r(a)} \quad\text{for}\quad a\in(0,a_0)
\end{equation}
This finishes the existence part of the proof of
Theorem~\ref{thm:main}. To complete the proof, it remains to show the
following:
\begin{proposition}
  \label{thm:is-foliation}
  For $r\in(0,r_1)$ let
  $\Sigma_r := S(r,\tau(r,\varphi(r)),\varphi(r))$ be the surfaces from
  Lemma~\ref{thm:solution-final}. Then there exist $r_2\in(0,r_1]$ so
  that the family $\{\Sigma_r\}_{r\in(0,r_1)}$ is a foliation of a
  pointed neighborhood of $p$.
\end{proposition}
\begin{proof}
  Define the maps
  \begin{align*}
    \Psi^r &:= \exp_p^{-1} \exp_{c(\tau(r))},\\
    \Psi(r,x) &:= \Psi^r(r(x+r^2\varphi(r)(x)))	\quad \text{and}\\
    \beta(r,x)&:=\frac{\Psi(r,x)}{|\Psi(r,x)|}.
  \end{align*}
  We claim that there exists $\tilde r \in(0,r_1]$ such that
  $|\Psi(r,x)|\not=0$ every $x\in S^2$ and such that
  $\beta(r,\cdot) :S^2 \to S^2 $ is a family of diffeomorphisms which
  can be smoothly extended to $r=0$ by the identity.

  Indeed, this follows from the facts that
  \begin{equation*}
    \frac{\partial \Psi}{\partial r}
    = (d_x\Psi^r)\big(x+r^2\varphi(r)(x)+r(r^2\varphi(r)(x))_r\big)
    +\left(\frac{\partial \Psi^r}{\partial r}\right)\big(r(x+r^2\varphi(r)(x))\big)
  \end{equation*}
  and
  \begin{equation*}
    \left.\frac{\partial \Psi^r}{\partial r}\right|_{r=0}
    = \left.\frac{\partial}{\partial\tau^i}\big(\exp_p^{-1} \exp_{c(\tau(r))}\big)\right|_{\tau=0}
    \cdot \left.\frac{\partial \tau^i}{\partial r}\right|_{r=0}=0
  \end{equation*}
  where we used Lemma~\ref{thm:tau-is-o-r2} in the last equality. In combination
  \begin{equation*}
    \frac{\partial \Psi}{\partial r}(0,x)=x.
  \end{equation*}
  Hence
  \begin{equation*}
    \Psi(r,x)=rx+O(r^2)\quad\text{as}\quad r\to 0.
  \end{equation*}
  In particular, $\Psi(r,x)\not=0$ for $r$ small enough and
  \begin{equation*}
    \beta(r,x)=\frac{x+O(r)}{|x+O(r)|}.
  \end{equation*}
  This establishes the claim.

  Let $\eta(r,x):= |\Psi(r,\beta^{-1}(r,x))|$ and calculate
  \begin{equation*}
    \frac{\partial \eta}{\partial r}
    = \frac{\Psi}{|\Psi|} \left(\frac{\partial \Psi}{\partial r}+(d_x\Psi)\left(\frac{\partial}{\partial r}\beta^{-1}\right)\right)
    = \frac{x+O(r)}{|x+O(r)|}\, (x+O(r)).
  \end{equation*}
  This yields
  \begin{equation*}
    \left.\frac{\partial \eta}{\partial r}\right|_{r=0}=1.
  \end{equation*}
  Consequently $\eta$ is strictly increasing for $r$ small enough
  which shows that all the surfaces are disjoint.
\end{proof}

config: arxiv


TITLE: How to find the range of this specific function?
QUESTION [0 upvotes]: $$f(x) = (x-1)e^{x+2} - \frac {x^2}2 $$
if x is in the domain of $[-2,1]$
Image of the function
Thank you.

REPLY [0 votes]: By computing $f'(x)$ you may easily check that $f(x)$ is decreasing on $[-2,0]$ and increasing on $[0,1]$. Since $f(-2)\in (f(0),f(1))$, it follows that the range of $f$ is $\color{red}{[f(0),f(1)]}$.


TITLE: Isometry without injection and surjection
QUESTION [5 upvotes]: Suppose that $B_1$ and $B_2$ are bilinear form on space $V_1$ and $V_2$. An isometry relative to $B_1$ and $B_2$ is an linear map $\sigma:V_1 \rightarrow V_2$ satisfying $B_2(\sigma(v),\sigma(u))=B_1(v,u)$
Does this definition imply that $\sigma$ is one to one and onto?

REPLY [11 votes]: It is natural to assume $B_1$ and $B_2$ are both non-degenerate, which means if $B_1(x,y) = 0$ for all $y$ then $x = 0$ and if $B_1(x,y) = 0$ for all $x$ then $y = 0$, and likewise with $B_2$. This condition was suggested in a comment by wxu to countinghaus's answer, and it is a good one. For real vector spaces, a positive-definite bilinear form is non-degenerate (after all, if $B(x,y) = 0$ for all $y$, then just use $y = x$ to see $B(x,x) = 0$, so $x = 0$ by positive-definiteness).  While positive-definite bilinear forms are non-degenerate, not by a long shot are "most" non-degenerate bilinear forms on a real vector space positive definite. In fact, physics gives us a great example: the bilinear form associated to $x_1^2 + x_2^2 + x_3^2 - x_4^2$ in (special) relativity is not positive definite but it is non-degenerate. Also, non-degeneracy is a condition on a bilinear form that makes sense when you're not working over the real or complex numbers (like those exotic fields of positive characteristic) and positive definiteness loses its meaning.
Let's assume $\sigma \colon V_1 \rightarrow V_2$ is a function between two vector spaces over a field $F$ and that $B_1(v,w) = B_2(\sigma(v),\sigma(w))$ for all $v$ and $w$ in $V_1$, where $B_i$ is a non-degenerate bilinear form on $V_i$.
What can we say about $\sigma$?
1) It is injective. Indeed, if $\sigma(v) = \sigma(w)$, then for all $u$ in $V_1$ we have
$$
B_1(v,u) = B_2(\sigma(v),\sigma(u)) = B_2(\sigma(w),\sigma(u)) = B_1(w,u),$$
so $B_1(v-w,u) = 0$. This holds for all $u$, so $v-w = 0$ by non-degeneracy of $B_1$. Thus $v = w$.
2) It preserves linear independence. Suppose $v_1,\dots,v_n$ are linearly independent in $V_1$. We will show $\sigma(v_1),\dots,\sigma(v_n)$ are linearly independent in $V_2$. Assume $\sum_{i=1}^n c_i\sigma(v_i) = 0$ in $V_2$, where $c_i \in F$. For all $u$ in $V_1$,
$$
B_1\left(\sum_{i=1}^n c_iv_i,u\right) = \sum_{i=1}^n c_iB_1(v_i,u) = \sum_{i=1}^n c_iB_2(\sigma(v_i),\sigma(u)) = B_2\left(\sum_{i=1}^n c_i\sigma(v_i),\sigma(u)\right),
$$
and the last term is $B_2(0,u) = 0$. Thus $B_1(\sum_{i=1}^n c_iv_i,u) = 0$ for all $u$ in $V_1$. By non-degeneracy of $B_1$, $\sum_{i=1}^n c_iv_i = 0$, so each $c_i$ is $0$ by linear independence of $v_1,\dots,v_n$.
Notice the non-degeneracy of $B_2$ hasn't been used so far.
3) If $V_1$ and $V_2$ are finite-dimensional and $\dim(V_1) = \dim(V_2)$ then $\sigma$ is linear, so necessarily surjective since linear + injective implies surjective when the vector spaces have the same (finite) dimension. In case linearity surprises you as a consequence, rather than a hypothesis, check that we didn't actually use linearity of $\sigma$ yet. It is a consequence of the isometry condition provided the bilinear forms are non-degenerate and the spaces have the same (finite) dimension.  I will give two proofs that $\sigma$ is linear.
Proof #1. Pick a basis $e_1,\dots,e_n$ of $V_1$. Define $L \colon V_1 \rightarrow V_2$ to be the linear map where $L(e_i) = \sigma(e_i)$ for all $i$. That is,
$$
L(c_1e_1+\cdots+c_ne_n) := c_1L(e_1) + \cdots + c_nL(e_n)
$$
for all scalars $c_i$.  Since $L$ is linear and $B_1$ and $B_2$ are bilinear, it follows from $B_2(L(e_i),L(e_j)) = B_1(e_i,e_j)$ for all $i$ and $j$ that $B_2(L(v),L(w)) = B_1(v,w)$ for all $v$ and $w$ in $V_1$. Then $L$ is injective by the same argument used in part 1 for $\sigma$. Since $L$ is 1-1 and linear between vector spaces of the same (finite) dimension, it is invertible and $B_1(L^{-1}(x),L^{-1}(y)) = B_1(x,y)$ for all $x$ and $y$ in $V_2$. Set $k = L^{-1} \circ \sigma \colon V_1 \rightarrow V_1$. Then $k(e_i) = e_i$ for all $i$ and $B_1(k(v),k(w)) = B_1(v,w)$ for all $v$ and $w$ in $V_1$.
For all $v$ in $V_1$, $B_1(k(v),e_i) = B_1(k(v),k(e_i)) = B_1(v,e_i)$, so the linearity of $B_1$ in its 2nd component implies $B_1(k(v),w) = B_1(v,w)$ for all $w$ in $V_1$. Then nondegeneracy of $B_1$ implies $k(v) = v$ for all $v$ in $V_1$. Thus $k$ is the identity, so $\sigma = L$. This proves $\sigma$ is an invertible linear map. (We didn't use the 2nd item here.)
Proof #2. Pick a basis $e_1,\dots,e_n$ of $V_1$. Then $\sigma(e_1),\dots,\sigma(e_n)$ is linearly independent in $V_2$ by the 2nd item above, so it is a basis of $V_2$. For any scalars $a_1,\dots,a_n$, set $v' = \sum_{i=1}^n a_i\sigma(e_i)$. (All elements of $V_2$ arise in this way.) We expect that $v' = \sigma(\sum_{i=1}^n a_ie_i)$. Write $v' - \sigma(\sum_{i=1}^n a_ie_i)$ as $\sum_{i=1}^n b_i\sigma(e_i)$ for some $b_i$ in $F$. For any $u$ in $V_1$, we will compute $B_2(\sum_{i=1}^n b_i\sigma(e_i),\sigma(u))$ in two ways.
First it is
$$
\sum_{i=1}^n b_iB_2(\sigma(e_i),\sigma(u)) = \sum_{i=1}^n b_iB_1(e_i,u) = B_1\left(\sum_{i=1}^n b_ie_i,u\right).
$$
Second, it is
$$
B_2\left(\sum_{i=1}^n a_i\sigma(e_i)-\sigma\left(\sum_{i=1}^n a_ie_i\right),\sigma(u)\right) = \sum_{i=1}^na_iB_1(e_i,u) - B_1\left(\sum_{i=1}^n a_ie_i,u\right)=0.
$$
Thus $B_1(\sum_{i=1}^n b_ie_i,u)=0$ for all $u$, so $\sum_{i=1}^n b_ie_i = 0$ in $V_1$ by non-degeneracy of $B_1$, and that implies each $b_i$ is 0.  We have shown
$$
\sum_{i=1}^n a_i\sigma(v_i) = \sigma\left(\sum_{i=1}^n a_iv_i\right)
$$
for all $a_i$ in $F$. Thus $\sigma$ is linear.
The linearity of $\sigma$ is due to Andrew Vogt. See Theorem 2.4 in A. Vogt, "On the linearity of form isometries", SIAM Journal on Applied Mathematics 22 (1972), 553--560. The second proof above is Vogt's proof. (He also has a linearity proof in Lemma 1.5 if $B_2$ is non-degenerate but non-degeneracy is not assumed for $B_1$.)
To summarize, if $V_1$ and $V_2$ are finite-dimensional vector spaces over a field of the same dimension and they are equipped with non-degenerate bilinear forms $B_1$ and $B_2$, any function $\sigma \colon V_1 \rightarrow V_2$ such that $B_1(v,w) = B_2(\sigma(v),\sigma(w))$ for all $v$ and $w$ in $V_1$ must be an isomorphism of vector spaces (linear, injective, and surjective). The proof never used non-degeneracy of $B_2$ on $V_2$ in the argument, and that now actually follows from non-degeneracy of $B_1$ on $V_1$.
Some of the comments on this question have pointed out that if you don't assume the spaces have the same dimension then there is no reason to expect surjectivity, e.g., an embedding of one Euclidean space into another with larger dimension. Moreover, Vogt's linearity theorem breaks down too. Consider $\sigma \colon {\mathbf R}^n \rightarrow {\mathbf R}^{n+1,1}$ (here ${\mathbf R}^{p,q}$ is $(p+q)$-dimensional space with the quadratic form having signature $(p,q)$) by
$$
\sigma(x_1,\dots,x_n) = (x_1,\dots,x_n,x_n^3,x_n^3).
$$
This map $\sigma$ is not linear but it preserves the bilinear forms on ${\mathbf R}^n$ and ${\mathbf R}^{n+1,1}$.
If there were no theorem like Vogt's, nothing would change, since we want isometries of bilinear spaces to be linear maps anyway. For instance, if $(V,Q)$ is a non-degenerate quadratic space, you don't define an isometry of $(V,Q)$ to be any function $\sigma \colon V \rightarrow V$ satisfying $Q(\sigma(v)) = Q(v)$ for all $v \in V$, because such a function can be completely wild: think about ${\mathbf R}^3$ with its usual sum of squares quadratic form and let $\sigma$ be any function from ${\mathbf R}^3$ to itself that preserves all spheres around the origin and is a crazy mapping on each sphere. An isometry of quadratic spaces is defined to be a linear map preserving the quadratic forms.


TITLE: Trigonometry and finding max value
QUESTION [0 upvotes]: Is it possible to calculate $\max \{\sin(\alpha) + \cos(\alpha)\} $by JUST using the inequality between arithmetic and quadratic mean?
Assume that $ \alpha   $ is  $ [0, \pi/2] $
What about $\min \{\sin(\alpha) + \cos(\alpha)\}$ for $ \alpha   $ is  $ [0, \pi/2] $

REPLY [0 votes]: Let $x = \sin \alpha, y = \cos \alpha$.
We have $x^2 + y^2 = 1$, so by AM-QM,
$$\frac{1}{2} |x + y| \leq \frac{|x| + |y|}{2} \leq \sqrt{\frac{x^2 + y^2}{2}} = \frac{1}{2}\sqrt{2}.$$
Therefore $-\sqrt{2} \leq x + y \leq \sqrt{2}$.
Equality is attained for $\alpha = \pi/4, 5\pi/4$.
For the minimum  on $[0,\pi/2]$,
$$1 = x^2 + y^2 \leq x^2 + y^2 + 2xy = (x + y)^2,$$
so $x + y \geq 1$. The minimum is attained for $\alpha = 0, \pi/2$.

REPLY [0 votes]: We use this inequality:
$\frac 1n(a^m+b^m+ . . .)<\big(\frac{a+b+ . . }n\big)^m$
where a and b are positive numbers and m is positive  proper fraction. In interval you mentioned both $\sin x$ and $\cos x$ are positive numbers, so we have:
$\frac12[(\sin^2 x)^{\frac 12} +(\cos^2 x)^{\frac 12} )<\big(\frac{\sin^2 x+\cos^2 x}2\big)^{\frac12}$
$A=\sin x+ \cos x<\sqrt 2$
Clearly $\sqrt 2$ can not be maximum of A that is maximum of $A=\sin x+\cos x$ can not be greater than  $\sqrt 2$ in interval $0<x<\frac{\pi}2$. Minimum of A is $\approx 1$ which is at $0+\epsilon$ or at $\frac{\pi}2-\epsilon$


text:
We show that constructing fixed points of locally continuous functors is a functorial operation.
Typically, the ``canonical'' fixed point of a locally continuous functor $F$ on an \O-category is given by the colimit of the $\omega$-chain $\bot \to F\bot \to F^2\bot \to \dotsb$.
Other fixed points can be constructed using a different ``first link'', \ie, by taking the colimit of a chain $K \to FK \to F^2K \to \dotsb$ generated by a link $k : K \to FK$.

Fix an \O-category $\mb{K}$.
Links form a category $\Links_{\mb{K}}$ where
\begin{itemize}
\item objects are triples $(K, k, F)$ called ``links'', where $K$ is an object of $\mb{K}$, $F : \mb{K} \to \mb{K}$ is locally continuous, and $k : K \to FK$ is an embedding;
\item morphisms $\lk{K}{k}{F} \to \lk{L}{l}{G}$ are pairs $(f, \eta)$ where $f : K \to L$ is a morphism of $\mb{K}$, $\eta : F \nto G$ is a natural transformation in $\mb{K}$, and $f$ and $\eta$ satisfy $l \circ f = (\eta \ast f) \circ k : K \to GL$;
\item composition is given component-wise: $(g, \rho) \circ (f, \eta) = (g \circ f, \rho \circ \eta)$.
\end{itemize}

There exists a locally continuous functor $\Omega : \Links_{\mb{K}} \to [ \omega \to \mb{K}^e ]$ that, given a link $\lk{K}{k}{F}$, generates the $\omega$-chain ${\Omega\lk{K}{k}{F} : K \xrightarrow{k} FK \xrightarrow{Fk} F^2K \xrightarrow{F^2k} \dotsb}$.
The action of $\Omega$ on morphisms uses the horizontal iteration of natural transformations.
Consider functors $H, G : \mb{C} \to \mb{C}$ and a natural transformation $\eta : H \nto G$.
We define the family of \defin{horizontal iterates} $\eta^{(i)} : H^i \nto G^i$, $i \in \N$, by recursion on $i$.
When $i = 0$, $H^0 = G^0 = \ms{id}_{\mb{C}}$ and we define $\eta^{(0)}$ to be the identity natural transformation on $\ms{id}_{\mb{C}}$.
Given $\eta^{(i)}$, we set $\eta^{(i+1)} = \eta \ast \eta^{(i)}$.

We define the functor $\Omega : \Links_{\mb{K}} \to [ \omega \to \mb{K} ]$.
The action of $\Omega\lk{K}{k}{F}$ on morphisms $n \to n + k$ is defined by induction on $k$.
\begin{align}
  \Omega\lk{K}{k}{F}(n) &= F^nK\label{eq:main:1}\\
  \Omega\lk{K}{k}{F}(n \to n) &= \ms{id}_{F^{n} K}\label{eq:main:4}\\
  \Omega\lk{K}{k}{F}(n \to n + k + 1) &= F^{n + k}k \circ \Omega\lk{K}{k}{F}(n \to n + k)\label{eq:main:2}\\
  \Omega(f : K \to L, \eta : F \nto G)_n &= \eta^{(n)} \ast f : F^nK \to G^nL\label{eq:main:3}
\end{align}

\Cref{prop:main:1} generalizes the functor $S : [\mb{C} \to \mb{C}] \to [\mb{\omega} \to \mb{C}]$ of \cites[\S~3]{lehmann_smyth_1977:_data_types}[Lemma~2 (p.~118)]{lehmann_smyth_1981:_algeb_specif_data_types} to form chains with an arbitrary initial link in a locally continuous manner.

\begin{restatable}{proposition}{propMainB}
  \label{prop:main:1}
  \Cref{eq:main:1,eq:main:4,eq:main:2,eq:main:3} define a locally continuous functor $\Omega : \Links_{\mb{K}} \to [ \omega \to \mb{K} ]$.
  For all links $\lk{K}{k}{F}$, $\Omega\lk{K}{k}{F} : \omega \to \mb{K}^e$.
  The natural transformation $\Omega(f, \eta)$ lies in $\mb{K}^e$ whenever $f$ and $\eta$ do.
\end{restatable}

\subsection{General Fixed Points}
\label{sec:general-fixed-points}

We can use $\Omega$ to show that taking fixed points of locally continuous functors is itself functorial and locally continuous.
To do so, we use an $\omega$-colimit functor (\cref{prop:main:10}).
Let $[ \omega \to \mb{K}^e \hookrightarrow \mb{K} ]$ be the subcategory of $[ \omega \to \mb{K} ]$ whose objects are functors $\omega \to \mb{K}^e$ and whose morphisms are natural transformations in $\mb{K}$.

\begin{restatable}{proposition}{propMainBA}
  \label{prop:main:10}
  Let $\mb{K}$ be an \O-cocomplete \O-category.
  A choice of \O-colimit in $\mb{K}$ for each diagram $\omega \to \mb{K}^e \hookrightarrow \mb{K}$ defines the action on objects of a locally continuous functor $\colim_\omega : [ \omega \to \mb{K}^e \hookrightarrow \mb{K} ] \to \mb{K}^e$.
  Where $\phi : \Phi \nto \colim_\omega \Phi$ and $\gamma : \Gamma \nto \colim_\omega \Gamma$ are the chosen \O-colimits in $\mb{K}$, its action on morphisms $\eta : \Phi \nto \Gamma$ is given by
  \[
    \colim_\omega(\eta : \Phi \nto \Gamma) = \dirsup_{n \in \N} \gamma_n \circ \eta_n \circ \phi_n^p.
  \]
\end{restatable}

We assume below that, whenever $\mb{K}$ is \O-cocomplete, a suitable choice of \O-colimits has been made so that the functor $\colim_\omega$ given by \cref{prop:main:10}~exists.

\begin{restatable}{proposition}{propMainI}
  \label{prop:main:8}
  Let $\mb{K}$ be an \O-cocomplete \O-category.
  There exists a locally continuous functor $\GFIX = {\colim_\omega} \circ \Omega : \Links_{\mb{K}} \to \mb{K}^e$.
  Its action on morphisms $(f, \eta) : \lk{K}{k}{F} \to \lk{L}{l}{G}$ is given by
  \[
    \GFIX(f, \eta) = \dirsup_{n \in \N} \gamma_n \circ \Omega(f, \eta)_n \circ \phi^p_n
  \]
  where $\phi : \Omega\lk{K}{k}{F} \nto \GFIX\lk{K}{k}{F}$ and $\gamma : \Omega\lk{L}{l}{G} \nto \GFIX\lk{L}{l}{G}$ are the chosen \O-colimits in $\mb{K}$.
\end{restatable}

To show that this indeed gives fixed points, we define the ``unfolding'' functor:

\begin{restatable}{proposition}{propMainC}
  \label{prop:main:2}

  Let $\mb{K}$ be an \O-cocomplete \O-category.
  The following defines a locally continuous functor $\UNF : \Links_{\mb{K}} \to \mb{K}^e$.
  \begin{itemize}
  \item On objects: $\UNF\lk{K}{k}{F} = F(\GFIX\lk{K}{k}{F})$,
  \item on morphisms: $\UNF(f, \eta) = \dirsup_{n \in \N} G\gamma_n \circ \Omega(f,\eta)_{n+1} \circ F\phi_n^p  : \UNF\lk{K}{k}{F} \to \UNF\lk{L}{l}{G}$, where $\phi : \Omega\lk{K}{k}{F} \nto \GFIX\lk{K}{k}{F}$ and $\gamma : \Omega\lk{L}{l}{G} \nto \GFIX\lk{L}{l}{G}$ are the chosen \O-colimits in $\mb{K}$.
  \end{itemize}
\end{restatable}

\Cref{prop:main:3} then tells us that $\GFIX$ produces fixed points.
To the best of our knowledge, the naturality result is new.

\begin{proposition}
  \label{prop:main:3}

  Let $\mb{K}$ be an \O-cocomplete \O-category.
  There exists a natural isomorphism $\ms{fold} : \UNF \nto \GFIX$ with inverse $\ms{unfold} : \GFIX \nto \UNF$.
  They are explicitly given as follows.
  Let $\lk{K}{k}{F}$ be an object of $\Links_{\mb{K}}$ and let $\kappa : \Omega\lk{K}{k}{F} \nto \GFIX\lk{K}{k}{F}$ be the chosen \O-colimit.
  The components are:
  \begin{align*}
    \ms{fold}_{\lk{K}{k}{F}} &= \dirsup_{n \in \N} \kappa_{n+1} \circ F\kappa_n^p :  F(\GFIX\lk{K}{k}{F}) \to \GFIX\lk{K}{k}{F}\\
    \ms{unfold}_{\lk{K}{k}{F}} &= \dirsup_{n \in \N} F\kappa_n \circ \kappa_{n+1}^p : \GFIX\lk{K}{k}{F} \to F(\GFIX\lk{K}{k}{F}).
  \end{align*}
  Naturality means that for every $(f, \eta) : \lk{K}{k}{F} \to \lk{L}{l}{G}$, the following diagram commutes:
  \begin{equation}
    \label[diagram]{eq:main:9}
    \begin{tikzcd}[column sep=4em]
      F(\GFIX\lk{K}{k}{F}) \ar[r, shift left=0.5ex, "{\ms{fold}_{\lk{K}{k}{F}}}"] \ar[d, swap, "{\UNF(f, \eta)}"]  & \GFIX\lk{K}{k}{F} \ar[l, shift left=0.5ex, "{\ms{unfold}_{\lk{K}{k}{F}}}"]  \ar[d, "{\GFIX(f, \eta)}"]\\%
      G(\GFIX\lk{L}{l}{G}) \ar[r, shift left=0.5ex, "{\ms{fold}_{\lk{L}{l}{G}}}"] & \GFIX\lk{L}{l}{G} \ar[l, shift left=0.5ex, "{\ms{unfold}_{\lk{L}{l}{G}}}"].
    \end{tikzcd}
  \end{equation}
\end{proposition}

We can specialize these constructions to produce fixed points that are, in a sense made clear in \cref{sec:canon-fixed-points}, canonical.
Assume $\mb{K}^e$ has an initial object.
We first observe that $[ \mb{K} \lcto \mb{K} ]$ embeds fully and faithfully into $\Links_{\mb{K}}$ via the locally continuous functor that maps objects $F : \mb{K} \lcto \mb{K}$ to the link $\lk{\bot}{\bot}{F}$ and natural transformations $\eta : F \nto G$ to the morphism $(\ms{id}_\bot, \eta)$.

We say that an \O-category $\mb{K}$ has \defin{strict morphisms} if it has zero morphisms and $0_{AB}$ is the least element of $\mb{K}(A, B)$ for all object objects $A$ and $B$.
We say that $\mb{K}$ \defin{supports canonical fixed points} if it has an initial object, strict morphisms, and is \O-cocomplete.
Assume $\mb{K}$ supports canonical fixed points.
Then $\bot$ is also the initial object of $\mb{K}^e$.
We define the \defin{canonical-fixed-point functor} $\FIX : [ \mb{K} \lcto \mb{K} ] \to \mb{K}^e$ as the composition $[ \mb{K} \lcto \mb{K} ] \hookrightarrow \Links_{\mb{K}} \xrightarrow{\GFIX} \mb{K}^e$.
This functor is locally continuous by \cref{prop:main:8}.

\subsection{Parametrized Fixed Points}
\label{sec:param-fixed-points}

In this section, we show that taking parametrized fixed points of functors is functorial.
In particular, we show in \cref{prop:main:4} that for appropriate \O-categories $\mb{D}$ and $\mb{E}$, there exists a locally continuous functor $[\mb{D} \times \mb{E} \lcto \mb{E} ] \to [\mb{D} \lcto \mb{E}^e ]$ that induces an external dagger operation.
This proposition generalizes the mapping $F \mapsto \sfix{F}$ of \cite[Proposition~5.2.7]{abramsky_jung_1995:_domain_theor} from functors on categories of pointed domains closed under bilimits to locally continuous functors on $\mb{O}$-categories.
By also defining the action on natural transformation, it makes the mapping's functorial structure evident.
It specializes the construction of \cite[\S~3]{lehmann_smyth_1977:_data_types} to \O-categories.

Given a functor $F : \mb{D} \times \mb{E} \to \mb{E}$ and an object $D$ of $\mb{D}$, we write $F_D$ for the partial application $\Lambda F D = F(D, {-}) : \mb{E} \to \mb{E}$.

\begin{restatable}{proposition}{PropMainE}
  \label{prop:main:4}

  Let $\mb{D}$ and $\mb{E}$ be $\mb{O}$-categories.
  Assume $\mb{E}$ supports canonical fixed points.
  The following defines a locally continuous functor:
  \[
    \sfix{({\cdot})} = [\ms{id}_{\mb{D}} \to \FIX] \circ \Lambda : [\mb{D} \times \mb{E} \lcto \mb{E} ] \to [\mb{D} \lcto \mb{E}^e ].
  \]
  Explicitly, $\sfix{F}D = \FIX(F_D)$ is the canonical fixed point of $F(D, -)$.
  Given a natural transformation $\eta : F \nto G$, $\left(\sfix{\eta}\right)_D = \FIX\left((\Lambda \eta)_D\right)$.
\end{restatable}

Let {\CFP} be the full subcategory of {\O} whose objects are \O-categories that support canonical fixed points.
It is 2-cartesian closed~\cite[Theorem~7.3.11]{fiore_1994:_axiom_domain_theor}.

\begin{corollary}
  \label{cor:main:1}
  \Cref{prop:main:4} defines an external dagger operator on {\CFP}.
\end{corollary}

The weak fixed-point identity gives us an analog of \cref{prop:main:3} for solutions to parametrized equations, \ie, to equations of the form $E \cong F(D,E)$.
The fixed point identity $\sfix{F}D = F(D, \sfix{F}D)$ typically does not hold for daggers of functors because the two functors are not equal on the nose.
However, it holds up to natural isomorphism, giving the weak fixed-point identity:

\begin{proposition}[Weak fixed-point identity]
  \label{prop:32}
  Let $\mb{E}$ and $\mb{D}$ be $\mb{O}$-categories.
  Assume $\mb{E}$ supports canonical fixed points.
  Let $F : \mb{D} \times \mb{E} \to \mb{E}$ be a locally continuous functor.
  There exist natural transformations
  \begin{align*}
    \ms{Unfold}^F &: \sfix F \nto F \circ \langle \ms{id}_{\mb{D}}, \sfix F \rangle\\
    \ms{Fold}^F &: F \circ \langle \ms{id}_{\mb{D}}, \sfix F \rangle \nto \sfix F
  \end{align*}
  that form a natural isomorphism $\sfix F \cong F \circ \langle \ms{id}_{\mb{D}}, \sfix F \rangle$.
  Let $D$ be an object of $\mb{D}$.
  The $D$-components for these natural transformations are given by
  \begin{align*}
    \ms{Unfold}^F_D &= \ms{unfold}_{\lk{\bot}{\bot}{F_D}} : \GFIX\lk{\bot}{\bot}{F_D} \to \UNF\lk{\bot}{\bot}{F_D}\\
    \ms{Fold}^F_D &= \ms{fold}_{\lk{\bot}{\bot}{F_D}} : \UNF\lk{\bot}{\bot}{F_D} \to \GFIX\lk{\bot}{\bot}{F_D}
  \end{align*}
  where $\ms{unfold}$ and $\ms{fold}$ are the natural isomorphisms given by \cref{prop:main:3}.
  The definitions of $\ms{Fold}^F$ and $\ms{Unfold}^F$ are natural in $F$.
  Given any natural transformation $\eta : F \nto G$, the following two squares commute:
  \[
    \begin{tikzcd}[column sep=4em, arrows=Rightarrow]
      \sfix{F}
      \ar[r, "{\ms{Unfold}^F}"]
      \ar[d, swap, "{\sfix\eta}"]
      & F\circ\langle\ms{id}, \sfix{F} \rangle
      \ar[d, "{\eta \ast \langle \ms{id}, \sfix{\eta} \rangle}"]
      &F\circ\langle\ms{id}, \sfix{F} \rangle
      \ar[r, "{\ms{Fold}^F}"]
      \ar[d, swap, "{\eta \ast \langle \ms{id}, \sfix{\eta} \rangle}"]
      & \sfix{F}
      \ar[d, "{\sfix\eta}"]\\
      \sfix{G} \ar[r, "{\ms{Unfold}^G}"]
      & G\circ\langle\ms{id}, \sfix{G} \rangle
      & G\circ\langle\ms{id}, \sfix{G} \rangle
      \ar[r, "{\ms{Fold}^G}"]
      & \sfix{G}
    \end{tikzcd}
  \]
\end{proposition}

To the best of our knowledge, the fact that $\ms{Fold}^F$ and $\ms{Unfold}^F$ are natural in $F$ is new.
It will be key to defining the interpretations of recursive session types in \cref{sec:appl-semant}.

\Cref{prop:main:13} follows easily from \cref{prop:main:4,prop:32}.
It illustrates the action of $\sfix{(\cdot)}$ on natural transformations and gives identities that will be useful in \cref{sec:appl-semant}.
It affirmatively answers the first question of the introduction: the definition of $\sfix{(\cdot)} : [\mb{D} \times \mb{E} \lcto \mb{E} ] \to [\mb{D} \lcto \mb{E}^e ]$ is natural in $\mb{D}$.

\begin{restatable}[Parameter Identity]{proposition}{propMainBD}
  \label{prop:main:13}
  Let $\mb{C}$, $\mb{D}$ and $\mb{E}$ be \O-categories and assume $\mb{E}$ supports canonical fixed points.
  Let $F, H : \mb{D} \times \mb{E} \to \mb{E}$ and $G, I : \mb{C} \to \mb{D}$ be locally continuous.
  Set $F_G = F \circ (G \times \ms{id}_E) : \mb{C} \times \mb{E} \to \mb{E}$, and analogously for $H_I$.
  Let $\phi : F \nto H$ and $\gamma : G \nto I$ be natural transformations.~Then
  \begin{align}
    \sfix{F_G} &= \sfix{F} \circ G : \mb{C} \to \mb{E},\label{eq:main:10}\\
    F_G \circ \langle \ms{id}_{\mb{C}}, \sfix{F_G} \rangle &= F \circ \langle \ms{id}_{\mb{D}}, \sfix{F} \rangle \circ G : \mb{C} \to \mb{E},\label{eq:main:11}\\
    \sfix{\left(\phi \ast (\gamma \times \ms{id})\right)} &= \sfix{\phi} \ast \gamma : \sfix{F_G} \nto \sfix{H_I},\label{eq:main:14}\\
    \ms{Fold}^{F_G} &= \ms{Fold}^FG : F_G \circ \langle \ms{id}_{\mb{C}}, \sfix{F_G} \rangle \nto \sfix{F_G},\label{eq:main:12}\\
    \ms{Unfold}^{F_G} &= \ms{Unfold}^FG : \sfix{F_G} \nto F_G \circ \langle \ms{id}_{\mb{C}}, \sfix{F_G} \rangle.\label{eq:main:13}
  \end{align}
\end{restatable}


TITLE: What is the next perfect square of the form 14444... in decimal notation?
QUESTION [22 upvotes]: We know that $12^2 = 144$ and that $38^2 = 1444$. Are there any other perfect squares in the form of $\frac{13}{9} (10^n - 1) + 1$ (i.e. $1$ followed by $n$ $4$'s), and how would we prove it?

REPLY [7 votes]: Brute force answer:
If $x^2 = 1\cdots4444$ then we can consider this mod $10000$ to see that we must have $x^2 \equiv 4444 \pmod{10000}$.  To see if such $x$ exists, it is sufficient to consider $0 \le x < 10000$.  The following C program terminates with no output, showing that no such $x$ exists.
#include <stdio.h>

int main(void) {
  int x;
  for (x = 0; x < 10000; x++) {
    if ((x * x) % 10000 == 4444) {
      printf("%d\n", x);
    }
  }
  return 0;
}

Note that you should run this on a system where int is at least 32 bits.


TITLE: On the subfield of $\mathbb C(x)$ fixed by the inversion of the variable
QUESTION [2 upvotes]: Let $\mathbb C(x)$ be the field of rational functions with complex coefficients i.e. it is the fraction field of $\mathbb C[x]$ .
I am trying to explicitly determine the subfield of $\mathbb C(x)$ given by $\{f(x) \in \mathbb C(x) : f(x)=f(1/x) \}$ .
I can show that any element in the subfield of the form $p(x)/q(x)$ with $p(x), q(x)$ are polynomial of degree $1$ is actually a member of $\mathbb C$ . I'm not sure what would happen with other elements.
Please help.

REPLY [3 votes]: Note that the element $x + 1/x$ is invariant under $x \mapsto 1/x$. Consider the field extension $K := \mathbb{C}(x + 1/x) \subset L := \mathbb{C}(x)$, it is an easy exercise to see that $K \not= L$. Consider the polynomial
$$
p(t) = t^2 -at+1
$$
where $a = x + 1/x \in K[t]$, then, since $p(x) = 0$, this is the minimal polynomial of $x$ over $K$ and this extension has degree 2 and is Galois (every degree 2 extension is Galois), then its Galois group has order two and contains $x \mapsto 1/x$ and the identity, the only possibility is that $Gal(L/K) = \{ x \mapsto 1/x, Id\}$. Therefore $K = L^{Gal(L/K)}$, i.e. $K$ is the fixed field of $Gal(L/K)$ in L.
Edit: The Fundamental theorem_of Galois theory was used.


TITLE: Distributional equality and independence
QUESTION [0 upvotes]: Let $X$ and $Y$ be random variables. Assume that
\begin{align}
P(X \in A, Y \in B) = P(X \in A) \, P(X \in B).
\end{align}
Claim: Then, $X$ and $Y$ are equal in distribution and independent of each other.
I agree with this intuitively but would be interested in a rigorous explanation.

REPLY [2 votes]: Take $A =\mathbb R$ to see that $P(Y \in B)=P(X \in B)$. Hence $X$ and $Y$ have the same distribution. We now get $P(X \in A, Y \in B)=P(X\in A) P(Y \in B)$ so $X$ and $Y$ are independent.