text:
\section{Divergence Measures}
\label{sec:divMeasures}
Divergence measures are used to quantify the dissimilarity between distributions.
We recall the definitions of the Kullback-Leibler divergence, the (differential) Shannon cross-entropy, and the Pearson-Vajda and \renyi \ divergences.
We also present the definition of the (differential) \renyi \ cross-entropy and examine some of its properties.
We also present the Jensen-\renyi \ divergence, which is a natural extension of the Jensen-Shannon divergence by virtue of being a mixture of two \renyi \ divergences.
This Jensen-\renyi \ divergence was recently introduced in~\cite{kluza19} for discrete distributions

and studied in the context of generalized (\renyi-type) $f$-divergences. It differs from the identically named divergence studied in \cite{jenRenyi} and~\cite{jenRenyi2}, an earlier extension of the Jensen-Shannon divergence consisting of the difference between the \renyi \ entropy of a mixture of multiple probability distributions and the mixture of the \renyi \ entropies of the individual distributions.
Other recent (but different) extensions of the Jensen-Shannon divergence can be found in~\cite{nielsen19} and the references therein.

Let $p$ and $q$ be two probability densities
with common support $\mathcal{R} \subset \real$
on the Lebesgue measurable space $(\real, \mathcal{B}(\real), \mu)$, and let
\begin{equation}\KL(p \Vert q) := \int_{\mathcal{R}} p \log \frac{p}{q} \, \der \mu

\qquad \text{and} \qquad

h(p;q):=-\int_{\mathcal{R}} p \log q \, \der \mu \label{kl-ce}\end{equation}
denote the KL divergence and the differential Shannon cross-entropy
between $p$ and $q$, respectively, where both information measures are assumed to be finite.
In~\eqref{kl-ce} and throughout, we use the short form $\int_{\mathcal{R}} f \, \der \mu
:= \int_{x\in\mathcal{R}} f(x) \, \der \mu(x)$ for any measurable function $f$.
When  $p = q$ almost everywhere (a.e.), then $h(p;q)$ reduces to the differential
Shannon entropy of $p$, denoted by $h(p)$.
We next describe the Pearson-Vajda divergence, which is itself an $f$-divergence \cite{nielsen}.
\begin{definition}
        The \textbf{Pearson-Vajda divergence of order $\bm{k}$}, $\PV(p \Vert q)$,
        between $p$ and $q$, where $k\ge 1$, is given by

        \begin{eqnarray}
                \PV (p \Vert q) := \int_{\mathcal{R}} \frac{\vert q - p \vert^k}{p^{k-1}} \der \mu.
        \end{eqnarray}
\end{definition}
Note that $\PV (p \Vert q) \ge 0$ with equality if and only if (iff) $p = q$ (a.e.).
Also when $k = 2$, $\PV (\cdot \Vert \cdot)$ reduces to the {\bf Pearson $\bm{\chi^2}$ divergence}:
\begin{eqnarray}
                \chi^2(p \Vert q) := \int_{\mathcal{R}} \frac{(q - p)^2}{p} \der \mu.
        \end{eqnarray}

\begin{definition}
         The \textbf{\renyi \ divergence of order} $\bm{\alpha}$ between $p$ and $q$, where $\alpha > 0$, $\alpha \neq 1$, is given by

        \begin{eqnarray}
                \D_{\alpha}(p \Vert q) := \frac{1}{\alpha - 1} \log \left (
                                \int_{\mathcal{R}} p^{\alpha}q^{1 - \alpha}\der \mu
                        \right ).
        \end{eqnarray}
\end{definition}
Note that $\D_{\alpha} (p \Vert q) \ge 0$ with equality iff $p = q$ (a.e.).
Furthermore, if $\D_{\gamma}(p \Vert q) < \infty$ for some $\gamma >1$,
then, as shown in~\cite{renyiDivergence}, we have that
\begin{eqnarray} \label{renyi-KL}
\lim_{\alpha \rightarrow 1} \D_{\alpha}(p \Vert q) &=& \KL(p \Vert q).
\end{eqnarray}
For simplicity of analysis, we assume in what follows the
finiteness of $\D_{\gamma}(\cdot \Vert \cdot)$
for some $\gamma >1$ so that convergence of~\eqref{renyi-KL} holds.
Being a function of an $f$-divergence, useful properties and bounds
on the \renyi \ divergence can be elucidated from the study of $f$-divergences,
see~\cite{sason18} and related references.
\begin{definition}\label{r-cross-entropy}
        The \textbf{differential \renyi \ cross-entropy of order} $\bm{\alpha}$ between
        $p$ and $q$,
        where $\alpha > 0$, $\alpha \neq 1$, is given by
        \begin{eqnarray}
                 h_{\alpha} (p;q) &:=& \frac{1}{1-\alpha} \log \left (
                                \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu
                        \right ).
        \end{eqnarray}
\end{definition}
When $p = q$ (a.e.),  $h_{\alpha} (p;q)$ reduces to the {\bf \renyi \ entropy}, $h_{\alpha} (p):= \frac{1}{1-\alpha}\log \left (     \int_{\mathcal{R}} p^{\alpha} \der \mu \right )$.
We next show that $\lim_{\alpha \rightarrow 1} h_{\alpha} (p;q) = h(p;q)$ under some finiteness conditions.
\begin{theorem}
\label{theorem:renyi-crossentropy-limit}

        If $h(p;q) < \infty$, then
        \[\lim_{\alpha \downarrow 1} h_{\alpha} (p;q) = h(p;q).\]
        Moreover, if $\E{A \sim p}\left( \frac{1}{q(A)} \right) < \infty$, then
        \[\lim_{\alpha \uparrow 1} h_{\alpha} (p;q) = h(p;q).\]
\end{theorem}
The proof of the theorem requires the following result.
\begin{lemma} \cite{renyiDivergence} \label{erwen}
        For any $x > \frac{1}{2}$,
        \begin{eqnarray*}
                (x - 1) \left ( 1 + \frac{1-x}{2} \right )\leq \log(x) \leq x - 1.
        \end{eqnarray*}
\end{lemma}
{\bf Proof of Theorem~\ref{theorem:renyi-crossentropy-limit}}
Working in the measure space $(\mathcal{R}, \mathcal{B}(\mathcal{R}), \mu)$, where $\mathcal{B}(\mathcal{R})$ is the Borel
$\sigma$-algebra and $\mu$ is the Lebesgue measure on $\real$, we first note that
	by setting $$x_{\alpha} = \E{\mathbf{A} \sim p}(q(\mathbf{A})^{\alpha - 1}) = \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu$$
        we have that
        $\lim_{\alpha \downarrow 1} x_{\alpha} = 1$.
        Also, Lemma~\ref{erwen} yields that
        $$\lim_{\alpha \downarrow 1}
        \frac{\log\left (x_{\alpha} \right)}{x_{\alpha} - 1} = 1.$$
We then can write
        \begin{eqnarray*}
 \lim_{\alpha \downarrow 1} h_{\alpha}(p;q)

                &=& \lim_{\alpha \downarrow 1} \frac{1}{1 - \alpha} \log\left (x_{\alpha} \right) \\
                &=& \lim_{\alpha \downarrow 1} \frac{x_{\alpha} - 1}{1 - \alpha}
                        \frac{\log\left (x_{\alpha} \right)}{x_{\alpha} - 1} \\
                &=& \lim_{\alpha \downarrow 1} \left( \frac{x_{\alpha} - 1}{1 - \alpha} \right)
                    \lim_{\alpha \downarrow 1} \left( \frac{\log\left (x_{\alpha} \right)}{x_{\alpha} - 1} \right)  \\
		&=& \lim_{\alpha \downarrow 1} \frac{x_{\alpha} - 1}{1 - \alpha}
        \end{eqnarray*}
iff
        $\lim_{\alpha \downarrow 1} \frac{x_{\alpha} - 1}{1 - \alpha}$
        exists. We next show the existence of this limit
        and verify that it is indeed equal to $h(p;q)$.
        Consider
        \begin{eqnarray*}
                \lim_{\alpha \downarrow 1} \frac{x_{\alpha} - 1}{1 - \alpha}
		&=& \lim_{\alpha \downarrow 1} \int_{\mathcal{R}} \frac{p\times q^{\alpha - 1} - p }{1 - \alpha}
                        \der \mu.
        \end{eqnarray*}
        In order to invoke the monotone convergence theorem,
        we prove that the integrand is non-decreasing and bounded below as $\alpha \downarrow 1$.

        Noting that
        \begin{eqnarray*}
                \frac{\der}{\der \alpha} \frac{p\times q^{\alpha - 1} - p}{1 - \alpha}
		&=& \frac{p  q^{\alpha - 1} (1- (\alpha - 1) \log (q)) - p}{(\alpha - 1)^2},
        \end{eqnarray*}
        it is enough to show that
        \[p  q^{\alpha - 1} (1- (\alpha - 1) \log (q)) - p \leq 0.
        \]

        Indeed, we have
        \begin{equation}

		p [q^{\alpha - 1} (1 + \log (q^{1 - \alpha})) - 1]
                \leq p [q^{\alpha - 1} (1 + q^{1 - \alpha} - 1) - 1]
                = p (1 - 1)
                = 0,
\label{eqn2}	\end{equation}
 where \eqref{eqn2} holds since $\log(x) \leq x - 1$, for $x>0$.

        We next show that the integrand is bounded from below.
	The lower bound can be obtained by letting $\alpha \rightarrow \infty$:
        \begin{eqnarray*}
                \lim_{\alpha \rightarrow \infty }\frac{p \times q^{\alpha - 1} - p}{1 - \alpha}
                &=& 0.
        \end{eqnarray*}
        Hence, by the monotone convergence theorem, we have that
        \begin{eqnarray*}
		\lim_{\alpha \downarrow 1} \int_{\mathcal{R}} \frac{p \times q^{\alpha - 1} - p}{1 - \alpha}
                        \der \mu
		 = \int_{\mathcal{R}} \lim_{\alpha \downarrow 1} \frac{p \times q^{\alpha - 1} - p}{1 - \alpha}
                        \der \mu.
        \end{eqnarray*}
        Finally, by L'H\^{o}pital's rule, we obtain that
        \begin{equation*}
                \lim_{\alpha \downarrow 1} \frac{x_{\alpha} - 1}{1 - \alpha}
		= - \int_{\mathcal{R}} \frac{p \log(q)}{1}
                        \der \mu
		= h(p;q).
        \end{equation*}

        Next, we shall prove taking $\alpha \uparrow 1$.
        For $\alpha < 1$,
        $
            x_{\alpha} = \E{\mathbf{A} \sim p}(q(\mathbf{A})^{\alpha - 1}) < \infty
        $
        by the fact that $ x_{\alpha}$
        is non-increasing in $\alpha > 0$ and by the assumption
        that $\E{\mathbf{A} \sim p}\left(\frac{1}{q(\mathbf{A})} \right) < \infty$.
        Hence, $\lim_{\alpha \uparrow 1} x_{\alpha} = 1$. Using Lemma~\ref{erwen}, we can
        apply the same steps as above with the alteration of the following argument.
        Consider
        \begin{eqnarray*}
                \lim_{\alpha \uparrow 1} \frac{x_{\alpha} - 1}{1 - \alpha}
                &=& \lim_{\alpha \uparrow 1} \int_{\X} \frac{p q^{\alpha - 1} - p}{1 - \alpha}
                        \der \mu.
        \end{eqnarray*}
        We know that the integrand is non-increasing in $\alpha > 0$.
        To use the monotone convergence theorem, we need to show
        that the integrand is bounded above. Note that
        \begin{equation*}
                \lim_{\alpha \rightarrow 0 }\frac{p \times q^{\alpha - 1} - p}{1 - \alpha}
                \ = \ \frac{p}{q} - p
		\ \leq \ \frac{p}{q}.
        \end{equation*}
        Since we assumed $\E{\mathbf{A} \sim p}\left(\frac{1}{q(\mathbf{A})} \right) < \infty$,
	we have that $\frac{p}{q} < \infty$ (a.e.).
        Thus the integrand is bounded above (a.e.).
        Following the same steps as in the previous part, we conclude that
        $\lim_{\alpha \uparrow 1} h_{\alpha}(p;q)=h(p;q)$.
        \qed

        We next show that the  \renyi \ cross-entropy is decreasing in $\alpha>0$, $\alpha \neq 1$.
\begin{theorem}

        The \renyi \ cross-entropy $h_{\alpha}(p; q)$ between $p$ and $q$ is monotonically decreasing in $\alpha$, for $\alpha > 0$, $\alpha \neq 1$.
\end{theorem}

{\bf Proof }
First note that
\begin{eqnarray*}
	\frac{\der}{\der \alpha} h_{\alpha}(p; q)

	&=& \frac{1}{(1 - \alpha)^2} \log \left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right) \\
	&& \qquad	+ \frac{1}{1 - \alpha} \frac{1}{\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu}
		\frac{\der}{\der \alpha} \left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right). \nonumber
\end{eqnarray*}
We next consider
\begin{equation*}
	\frac{\der}{\der \alpha} \left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right)
		= \lim_{\epsilon \downarrow 0} \int_{\mathcal{R}} p \left( \frac{q^{\alpha + \epsilon - 1} - q^{\alpha - 1}}{\epsilon} \right) \der \mu
	= \lim_{\epsilon \downarrow 0} \int_{\mathcal{R}} p q^{\alpha - 1} \left( \frac{q^{\epsilon} - 1}{\epsilon} \right) \der \mu.
\end{equation*}
In order to apply the monotone convergence theorem, we prove the integrand is decreasing in $\epsilon \in (0, 1)$ as $\epsilon \downarrow 0$ and is bounded above.
Note that
\begin{eqnarray*}
	\frac{\der}{\der \epsilon} \frac{q^{\epsilon} - 1}{\epsilon} &=& \frac{1 + q^{\epsilon} (\epsilon \log(q) - 1)}{\epsilon^2},
\end{eqnarray*}
hence, it is enough to show that

	$1 + q^{\epsilon} (\epsilon \log(q) - 1) \geq 0.$

Indeed, we have
\begin{equation}
	1 + q^{\epsilon} (\log(q^{\epsilon}) - 1) \geq 1 + q^{\epsilon} \left(1 - \frac{1}{q^{\epsilon}} - 1 \right)
	= 0, \label{eqn:mct_renyi_monotone}
\end{equation}
where~\eqref{eqn:mct_renyi_monotone} holds since $1 - \frac{1}{x} \leq \log(x)$, for all $x>0$.
We now show that the integrand is bounded above by letting $\epsilon \rightarrow 1$:
\begin{eqnarray*}
	\lim_{\epsilon \rightarrow 1} \frac{q^{\epsilon} - 1}{\epsilon} = q - 1 < \infty.
\end{eqnarray*}
Hence by the monotone convergence theorem we have
\begin{eqnarray*}
	\lim_{\epsilon \downarrow 0} \int_{\mathcal{R}} p q^{\alpha - 1} \left( \frac{q^{\epsilon} - 1}{\epsilon} \right) \der \mu
		= \int_{\mathcal{R}} \lim_{\epsilon \downarrow 0} p q^{\alpha - 1} \left( \frac{q^{\epsilon} - 1}{\epsilon} \right) \der \mu.
\end{eqnarray*}
By L'H\^{o}pital's rule, we obtain that
\begin{eqnarray*}
	\frac{\der}{\der \alpha} \left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right) = \int_{\mathcal{R}} p q^{\alpha - 1} \log(q) \der \mu.
\end{eqnarray*}
Therefore,
\begin{eqnarray*}
	\frac{\der}{\der \alpha} h_{\alpha}(p; q) &=&  \frac{1}{(1 - \alpha)^2} \log \left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right) \nonumber\\
       && \qquad         + \frac{1}{1 - \alpha} \frac{1}{\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu} \int_{\mathcal{R}} p q^{\alpha - 1} \log(q) \der \mu \\
	&=&  \frac{\int_{\mathcal{R}} p q^{\alpha - 1} (h_{\alpha}(p;q) + \log(q)) \der \mu}{(1 - \alpha)\left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right)}.
\end{eqnarray*}
We next consider two cases, $\alpha < 1$ and $\alpha > 1$.
For $\alpha < 1$, we have
\begin{eqnarray}
	h_{\alpha}(p;q) + \log(q) &=& \frac{1}{1 - \alpha} \log\left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right) + \frac{\log(q^{1 - \alpha})}{1 - \alpha} \nonumber \\
	&=& \frac{\log(q^{1 - \alpha} \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu)}{1 - \alpha} \nonumber \\
	&\leq& \frac{q^{1 - \alpha} \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu - 1}{1 - \alpha}, \label{eqn:log_lower_renyi_monotone}
\end{eqnarray}
where~\eqref{eqn:log_lower_renyi_monotone} holds since $\log(x) \leq x - 1$, for all $x > 0$.
We thus have
\begin{equation*}
	\frac{\der}{\der \alpha} h_{\alpha}(p; q)
		\leq \frac{\int_{\mathcal{R}} p q^{\alpha - 1}
		(q^{1 - \alpha} \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu - 1) \der \mu}{(1 - \alpha)^2 \left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right)} \\
	= \frac{\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu - \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu}
		{(1 - \alpha)^2 \left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right)}
	= 0.
\end{equation*}
Finally, for $\alpha > 1$, we consider
\begin{equation}
        h_{\alpha}(p;q) + \log(q) = \frac{\log(q^{1 - \alpha} \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu)}{1 - \alpha}
	\geq \frac{q^{1 - \alpha} \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu - 1}{1 - \alpha}, \label{eqn:log_lower_renyi_monotone_2}
\end{equation}
where~\eqref{eqn:log_lower_renyi_monotone_2} holds since $-\log(x) \geq -(x - 1)$, for all $x > 0$.
Thus, we have
\begin{equation*}
        \frac{\der}{\der \alpha} h_{\alpha}(p; q)
                \leq \frac{\int_{\mathcal{R}} p q^{\alpha - 1}
                (q^{1 - \alpha} \int_{\mathcal{R}} p q^{\alpha - 1} \der \mu - 1) \der \mu}{(1 - \alpha)^2 \left(\int_{\mathcal{R}} p q^{\alpha - 1} \der \mu \right)}
	= 0.
\end{equation*}

\qed

The above definition of differential
\renyi \ cross-entropy can be extended (assuming the integral exists)
by only requiring $q$ to be a non-negative function
(such as a non-normalized density function); in this case we call the resulting
measure as the (differential) {\em \renyi \ cross-entropy functional} and denote it by
$\mathcal{H}_{\alpha}(p;q)$. Similarly, we henceforth denote the Shannon cross-entropy functional by
$\mathcal{H}(p;q)$.

\begin{definition}\label{def: jensen-renyi-div}
The \textbf{Jensen-\renyi \ divergence of order} $\alpha$ between $p$ and $q$,
where $\alpha > 0$, $\alpha \neq 1$, is given by
  {\rm       \begin{eqnarray} \label{jensenRenyiDivergenceFormula}
	\hspace{-0.17in}	\JR_{\alpha}(p \Vert q) &:=& \frac{1}{2} \D_{\alpha}\left( p \bigg\Vert \frac{p + q}{2} \right )

+ \frac{1}{2} \D_{\alpha}\left( q \bigg \Vert \frac{p + q}{2} \right).
	\end{eqnarray}
	}
\end{definition}

By the non-negativity of the \renyi \ divergence, it follows by definition
that $\JR_{\alpha}(p \Vert q) \ge 0$ with equality iff $p = q$ (a.e.).
Finally since $\lim_{\alpha \rightarrow 1} \D_{\alpha}(p \Vert q) = \KL(p \Vert q)$, we have that
{\rm \begin{equation}
\lim_{\alpha \rightarrow 1} \JR_{\alpha}(p \Vert q) = \JSD(p \Vert q),
\end{equation}
}
where
{\rm\begin{equation}\JSD(p \Vert q) := \frac{1}{2} \KL \left( p \bigg\Vert \frac{p + q}{2} \right )
+\frac{1}{2} \KL \left( q \bigg\Vert \frac{p + q}{2} \right )\end{equation}
}
is the Jensen-Shannon divergence.

config: arxiv


TITLE: A relation between two elements
QUESTION [2 upvotes]: Let $G $ be a group and $ a, b\in G $ such that $a^{-1}b^2a=b^3$ and $b^{-1}a^2b=a^3$.

How can we show that $a=b$$\,$?

REPLY [3 votes]: $\bbox[5px,border:2px solid]{\begin{array}{c}b^2a=ab^3\\a^2b=ba^3\end{array}}\ $ these relations are equivalent to OP's since elements in a group are invertible.
If $G$ is commutative or if $a,b$ commute
We have $ab=ba$.
$b^3=a^{-1}b^2a=a^{-1}ab^2=b^2\Rightarrow b=1$
$a^3=b^{-1}a^2b=b^{-1}ba^2=a^2\Rightarrow a=1$
We get $a=b=1$.
G any group, but either $a$ or $b$ is of order $\le$ 4
For instance $a=1$
$b^3=a^{-1}b^2a=b^2\Rightarrow b=1$

For instance $a^2=1$
$a=a^3=b^{-1}a^2b=b^{-1}b=1$ and we conclude like previously

For instance $a^3=1$
$b=ba^3=a^2b\Rightarrow a^2=1$ and we conclude like previously

For instance $a^4=1$
$(ba)a^2=ba^3=a^2b=a^2ba^4=a^2(ba^3)a=a^2(a^2b)a=a^4(ba)=ba\Rightarrow a^2=1$ and we conclude like previously

Everything being symetrical in $a,b$, the same conclusion arises if we make the hypothesis on $b$.
So here also we get $a=b=1$
G any group, but either $a$ or $b$ order is finite
Let assume a^n=1

$a^2(ba)=(a^2b)a=(ba^3)a=(ba)a^3\Rightarrow \bbox[5px,border:2px solid]{\forall k\in\mathbb N,\ a^{2k}(ba)=(ba)a^{3k}}$
If n even then n=2p

$a^{2p}(ba)=a^n(ba)=(ba)=(ba)a^{3p}=(ba)a^na^{p}=(ba)a^{p}\Rightarrow a^{p}=1$ and since $p<n$ we can conclude by recurrence.
If n odd then n=2p+1

$a^{2p}(ba)=a^{n-1}(ba)=a^{-1}(ba)=(ba)a^{3p}=(ba)a^na^{p-1}=(ba)a^{p-1}\Rightarrow (ba)=(ab)a^p$
$b=(ab)a^{p-1}$

$(ab)b^2=ab^3=b^2a=(ab)a^{p-1}\,(ab)a^{p-1}\ a=(ab)a^{p-2}(a^2b)a^p=(ab)a^{p-2}(ba^3)a^p$
$b^2=a^{p-2}ba^{p+3}=a^{p-4}(a^2b)a^{p+3}=a^{p-4}(ba^3)a^{p+3}=a^{p-4}ba^{p+6}$
$\bbox[5px,border:2px solid]{\forall k\in\mathbb N,\ b^2=a^{p-2k}\,b\,a^{p+3k}}$
If p even

For $p=2k$ we have $b^2=ba^{5k}\Rightarrow b=a^{5k}$ then $a,b$ commute and we can conclude.
If p odd

For $p=2k+1$ we have $b^2=aba^{5k+1}\Rightarrow (ab)b^2=ab^3=b^2a=aba^{5k+2}=(ab)a^{5k+2}$
$b^2=a^{5k+2}\Rightarrow ab^3=b^2a=a^{5k+3}\Rightarrow b^3=a^{5k+2}=b^2\Rightarrow b=1$ and we can also conclude.

Everything being symetrical in $a,b$ the same conclusion arises if we make the hypothesis $b^n=1$.
Thus we have proved it in all cases either directly if $n$ odd, either by recurrence if $n$ is even.
G infinite non-commutative group, order of $a,b$ infinite
I have no idea... :-(


TITLE: Understanding a Basis
QUESTION [1 upvotes]: I'm having a very difficult time understanding a basis.  I understand that it's basically just coordinates that span a subspace, but can someone help me to understand how a set can span a subspace and be a linearly independent set?
If a set spans a subspace doesn't this mean that any vector can be represented by a linear combination of the vectors in the subspace?  If that is the case, isn't this a direct contradiction to the set being linearly independent?
What am I missing here?

REPLY [0 votes]: I am currently reading through Paul R. Halmos' "Finite dimensional linear algebra". The way a basis is defined in his book is as follows:
Given a finite dimensional vector space $V$, a (not "the") basis of $V$, say $S$, is a linearly independent set of vectors, such that every vector in $V$ is equal to some linear combination of the vectors in $S$.
Now, to answer your question, let us assume that a set of vectors, again say $S$, spans a vector space $V$. i.e, we assume that every vector in $V$ is equal to some linear combination of the vectors in $S$. Linear independence states that, given a linear combination of the vectors in $S$, if this combination is equal to the zero vector, then all the coefficients in the linear combination must be equal to zero.
Your statement is the following: $S$ is linearly independent $\implies$ that $S$ does not span $V$. Taking the contrapositive of this statement, we have that $S$ spans $V$ $\implies$ $S$ is linearly dependent. Now, we can simply provide a counter example. Say then that $V = \mathbb{R}^2$. Then, $V$ forms a vector space, and if we let $S = \{(1, 0), (0, 1)\}$, we have a linearly independent basis that spans $V$. $\square$
Note that some intricacies have been omitted (it's 1am!), such as proving that $S$ does actually span $V$, however I believe my answer to be sufficient as it stands. Do let me know if anything anything is missing. Clearly this idea extends to your question of a subspace if we set $V$ to be equal to the subspace you are speaking of (any will do).


TITLE: Solving $2^{2x+1} - 2^{x+4} = 2^3 - 2^x$
QUESTION [8 upvotes]: $$2^{2x+1} - 2^{x+4} = 2^3 - 2^x$$
How can I solve an exponential equation that has many terms as the one above. Include more than one method if available.

REPLY [0 votes]: Hint $\ ac = bd\,\Rightarrow\, b(a\!-\!b\!-\!c\!+\!d) = (a\!-\!b)(b\!+\!c).\ $ Let $\ a,b,c,d\, =\, 2^{2x+1},2^{x+4},2^3,2^x$


TITLE: Can every set be a group?
QUESTION [18 upvotes]: Lets consider a set A. Can we prove that  there exists at least one operation * such that under it the set forms a group. May I know how to approach this question?

REPLY [5 votes]: The answers are already given, but since you asked how you should approach this, I'll try to address that issue.
The first thing to understand is that if you have a bijection $f:A\to B$ between any two sets, then any group structure on $A$ can be transferred along $f$ to a group structure on $B$. More precisely, there is precisely one group structure on $B$ for which $f$ is a group homomorphism (necessarily isomorphism). If this is not clear to you, spend some time and prove it.
Now, the question becomes equivalent to, given any cardinality $\kappa$ (other than $\kappa = 0$), is there some set of that cardinality which supports a group structure. (Under the axiom of choice,) every set has a cardinality and so if for every cardinality there is some  group of that cardinality then, using the result above, every set supports a group structure (upon choosing a bijection, so the axiom of choice is used again).
So, it boils down to constructing groups of all possible cardinalities. Finite ones are dealt with easily by the finite cyclic groups. Infinite ones are constructed by taking products of sufficiently many finite groups, using basic rules of cardinal arithmetic (assuming the axiom of choice again).


TITLE: Bad proof that if $a + b + ab = 2020$ then $a+b=88$
QUESTION [8 upvotes]: Can you prove this:

Let $a,b \in \mathbb{N}$. If $a + b + ab = 2020$ then $a+b=88$.

This is the attempt given:
$\frac{2020-88}{a b}=1$
$a+b=88$
Substituting for $b$ using the $2$nd equation.
$2020-88 = a (88-a)$
That is a quadratic that is easily solved and gets $a = 42$ and $a = 46$.
So we have $a = 42$ and $b = 46$ or $a = 46$ and $b = 42$.
Was the question answered? A valid proof?

REPLY [1 votes]: You demonstrated the that both can co-exist and have a solution:

$\exists! (a, b) \in ℕ$ such as $a+b=88$ and $a+b+ab=2020$

Now if you can prove the uniqueness of the second part

$\exists! (a, b) \in ℕ$ such as $a+b+ab=2020$

That would be sufficient. (however, proving the uniqueness of the solution is quite likely to result in actually finding that solution, rendering the first part of the reasoning useless)


TITLE: How are Chi Square probabilities calculated?
QUESTION [0 upvotes]: What steps would one follow to calculate the values in a Chi Square probability table such as https://people.richland.edu/james/lecture/m170/tbl-chi.html?
Say you had 15 degrees of freedom and wanted p to be 0.105, could someone please write a step-by-step process for getting from those two numbers to the probability value?

REPLY [1 votes]: There is excellent open source software to which you could link
or which you might use as a guide. For large df there are good
approximations. Much of your approach will be governed by
whether you need only tail probabilities and how many places of
accuracy you need. Google 'approximate CDF of chi-squared'
and similar phrases to tap into rich literature on this topic.
IMSL maintains a library of compiled or compilable FORTRAN
code of well-vetted procedures for a wide variety of commonly
used distributions. I don't know what the licensing arrangements
are. And there are sources of excellent open source code.
All of this is to say that re-inventing this wheel, by now
very well refined and engineered by experts, seems a waste of time.
Nevertheless, here are a couple of additional observations.
Numerical integration works OK in many cases:
For example, if $Q \sim Chisq(df = 30)$, the R code
qchisq(.95, 30) returns 143.77297, and pchisq(43.77, 30)
returns 0.9499692.
If you mean P-value in the usual sense of statistical inference,
the P-value corresponding to an observed value 43.77 would
be 1 - pchisq(43.77, 30) or 0.05003083, just on the
borderline of significance at the 5% level. So it is essentially
the CDF you'd want to compute by numerical integration.
Representing the easily computable density function by dchisq,
we'd have the R code.
 grid = seq(0, 43.77, len = 10^5)
 height = dchisq(grid, 30)
 mean(height)*43.77
 ## 0.949962

Which gives the correct answer to reasonable accuracy. (Refinements
are possible and worthwhile, but this gives the central idea.)
Another approach, widely used in these days of cheap memory,
is to include a detailed chi-square table for reference.
Interpolation works well over short increments; not at all
between df's.
I am reasonably knowledgeable about what is behind the
code of the software I use regularly--in a general sense, but certainly not an expert on
the technical details. Going further would put me at serious risk of 'telling' you more.
than I know.


TITLE: How to predict direction of force of friction on car in banked roads?
QUESTION [0 upvotes]: Consider a car moving on a banked road   which is at an angle say x with the horizontal. Now if we consider friction t9 be zero then the ideal velocity with which car should move is $√(rgtanx)$ .
But what about if it moves with velocity greater or less than this velocity ? How do we predict the frictional force direction then ?
Some say that according to normal reaction being more or less than required as centripetal car has different tendancies to move but I dont understand why also if we consider friction of the road then how can we find minimum velocity with which car should move in a circle of radius R ?
http://www.batesville.k12.in.us/physics/phynet/mechanics/circular%20motion/banked_with_friction.htm

REPLY [2 votes]: You are setting a constraint on the car which is that it has to go around a banked corner at a certain speed $v$ and given radius $r$ without a frictional force being present.
Applying that constraint you get the condition $v = \sqrt {rg\tan x}$ where $x$ the angle of banking.
The force to provide the necessary centripetal acceleration $\frac{v^2}{r}$ was supplied by the horizontal component of the normal reaction of the slope on the car.
If now the speed of the car is increased then that horizontal component of the normal reaction is not sufficient to provide the necessary centripetal acceleration to keep the radius of the car's path the same.
With no frictional force the only thing the car can do is start moving along a path with a larger radius.
Put another way the force provided by the horizontal component of the normal reaction cannot change the direction of motion of the car sufficiently so the car starts moving up the slope.
To prevent this happen and keep the car moving at a faster speed along a path with the same radius as before there must be an extra inward force to increase the centripetal acceleration.
That inward force can be provided by the horizontal component of a frictional force acting down the slope.
With the car slowing down the horizontal component of the normal reaction changes the direction of motion of the car too much and so the car would start moving down the slope, so there must be a frictional force up the slope.

The easy way to remember which direction the frictional force should action is to think of the car not moving on a (frictionless) slope and the direction of a frictional force which will stop it sliding down the slope.


TITLE: Path connectedness of a subset of $\mathbb{R}^2$
QUESTION [0 upvotes]: I know about Connectedness and Path Connectedness in Topology. Recently, I got stuck into a problem. Let $A$ be the following subset of $\mathbb{R}^2:$
$$A=\{(x,y) \mid (x+1)^2 + y^2 \leq 1\}\cup \{(x,y) \mid y=x\sin(\frac{1} {x}),\ x>0\}$$ The connectedness is obvious from if $A$ is connected then $cl(A)$ is also connected. But I can't prove the path connectedness ( The answer is given that $A$ is both connected & Path connected). Please help. Thanks in advance.

REPLY [1 votes]: Let $B$ be the second set. It should be clear that $B$ is path connected.
Since $A$ is convex, it is path connected.
Let $p(x) = \begin{cases} (x,x \sin { 1\over x}) , & x \in (0,1]\\
(0,0), & x = 0 \end{cases}$.
Note that $p$ is continuous, $p(0) \in A$ and $p(1) \in B$.
Hence any two points of $A\cup B$ can be connected by a path.


TITLE: The cardinality of the set all symmetric relations on the set of natural numbers is $\mathfrak{c} = | \mathbb{R} | = 2^{\aleph_0}$
QUESTION [1 upvotes]: Prove that a set of all symmetric relations on the set of natural numbers has cardinality $\mathfrak{c} = | \mathbb{R} | = 2^{\aleph_0}$.

Here I think that the $(a,b)b$ - will be every number there is in $\mathbb N$ and hence we will never get pass ordered pair that contain $a$ as first number. If it's true can you please tell me how to write it?

REPLY [3 votes]: Hint: symmetric relations on $\Bbb N$ are in bijection with subsets of $\{(a,b)\in\Bbb N^2:a\le b\}$.
Alt route: any subset of $\{(c,c)\in\Bbb N^2:c\in\Bbb N\}$ is a symmetric relation.


TITLE: Has vol. 3A of Cullis's "Matrices and Determinoids" been scanned and vol. 3B been archived?
QUESTION [7 upvotes]: This is a borderline question, but I'm going to risk posing it.
Cuthbert Edmund Cullis (1875?-1955?) was a somewhat obscure British mathematician whose opus magnum was a multi-volume treatise called Matrices and Determinoids. While his notation is somewhat idiosyncratic, it was a fairly systematic exposition of most(?) of the linear algebra known by the 1920s (heavy on determinants and minors, but also including invariant factors and canonical forms, or at least things looking like them), along with lots of new results. Some of his innovations, I believe, would be of interest even to us modern people: in particular, in the first 200 pages of Volume 3A, he seems to algorithmically construct multi-resultants using multisymmetric polynomials (aka MacMahon symmetric polynomials). I have never seen this done anywhere else.
Three volumes of his treatise were published: volume 1, volume 2 and volume 3A. According to his necrologue, he left behind a (plan of?) volume 3B. Thus, I'm wondering:

Question 1. Has Volume 3A ever been digitized?
Question 2. Is the manuscript of Volume 3B available anywhere?

The reason for Question 2 is mostly curiosity, but I have a rather practical reason for Question 1: I'm interested in understanding multi-resultants elementarily and also in applying multisymmetric polynomials. The twist is that I have access to Volume 3A in hardcopy for the rest of this week, but before I leave this library I'd like to know if it's worth scanning the doorstopper (700 pages!) or someone has already done that work.

REPLY [6 votes]: Q1: Volume 3 part 1 (a.k.a. volume 3A) has been digitized and reissued as a paperback by Cambridge UP, see Amazon. The digital version is online in the HathiTrust digital library, but with limited "search only" access because of copyright restrictions.
With some further search I actually found a library in India that has placed the digital version online (a 54 MB download). Quite possibly not copyright compliant, but in the spirit of the National Emergency Library I guess in these difficult times a download can be justified.


TITLE: Understanding percentile computation
QUESTION [0 upvotes]: I understand percentile in the context of test scores with many examples (eg. you SAT score falls in the 99th percentile), but I am not sure I understand percentile in the following context and what is going on. Imagine a model outputs probabilities (on some days we have a lot of new data and outputted probabilities, and some days we don't). Imagine I want to compute the 99th percentile of outputted probabilities. Here are the probabilities for today:
a = np.array([0,0.2,0.4,0.7,1])
p = np.percentile(a,99)
print(p)

0.988

I don't understand how the 99th percentile is computed in this situation where there are only 5 outputted probabilities. How was the output computed? Thanks!

REPLY [1 votes]: The correct result would be the number at position $5$: $a_5 =1$.
A $p$-th percentile $P_p$ is characterized by the following two properties:

At most $p\%$ of the data is less than $P_p$
At most $(100-p)\%$ of the data is greater than $P_p$

Let $n$ be the number of data items. There are two cases:

If $n\cdot\frac{p}{100}$ is not an integer, then $P_p$ is uniquely determined. Then, the value of the data item at position $\left\lceil n\cdot\frac{p}{100} \right\rceil$ (rounding up) is the $p$-th percentile. In your case
$$5\cdot\frac{99}{100}=4.95 \stackrel{}{\longrightarrow}\lceil n\cdot\frac{p}{100}\rceil = 5$$
If $n\cdot\frac{p}{100}$ is an integer, then any value starting from the data item at position $n\cdot\frac{p}{100}$ till the item at position $n\cdot\frac{p}{100}+1$ satisfies the above given characterizations. This is the only case, where interpolation might be applied.

Summary:
The percentile function in "numpy" (np) is mathematically not correct.


TITLE: Explain why any coset in $R$ can be written as $g+\langle f \rangle$ where $\deg(g) \leq 2$.
QUESTION [0 upvotes]: Let $f=x^3+x^2+x+2 \in \mathbb{Z}_3[x]$. Let $R=\mathbb{Z}_3[x]\langle f \rangle$.

Explain why any coset in $R$ can be written as $g+\langle f \rangle$ where $\deg(g) \leq 2$.
Then explain why $g+\langle f \rangle \neq h+\langle f \rangle$ if $g\ne h$ are polynomials of degree less than or equal to $2$.
Lastly use the results of the above question to determine number of elements in $R$.

I am confused on the first question. My intuition for the second question is to show that $g-h \notin \langle f \rangle$. IS this correct?

REPLY [2 votes]: It is enough to show $x^3$ is linearly dependent on $1, x, x^2$. But clearly

$$x^3 = -x^2-x-2= 2x^2+2x+1$$

So now if you have any $x^n, n>2$ write it as $x^3\cdot x^{n-3} = 2x^{n-1}+2x^{n-2} +x^{n-3}$, so by induction and the well-ordering principle, we can reduce any degree polynomial to an equivalent one of degree at most $2$.


TITLE: Is the absolute value function a linear function?
QUESTION [8 upvotes]: I'm inclined to say yes, as it doesn't involve exponentiation, roots, logarithmic or trigonometric functions, but I watched a video where the teacher said that the absolute value function is "clearly non-linear". Why would he say that? Is he wrong?
Wikipedia's graph for abs:

REPLY [0 votes]: Think of the definition of absolute value.  It is a piecewise-defined function.
$|x| = x$, if $x\ge 0$, and $-x$ if $x<0$.  In other words, the graph of $y=|x|$ is formed by two pieces of two lines.
For the part of the domain where $x$-values are less than zero, the graph corresponds to the graph of $y=-x$.  For parts of the domain where $x$-values are greater than or equal to zero, the graph corresponds to the graph of $y=x$.
While the absolute value function does not satisfy the above definitions for linear functions, it is actually "parts" of two linear functions.


TITLE: Milnor's exact sequence and a certain proof
QUESTION [3 upvotes]: Please forgive me if this is not the right forum for this question.
Let $$ X = \cdots \rightarrow X_n  \rightarrow X_{n-1} \rightarrow \cdots \rightarrow  X_0 = \ast$$ be a tower of fibrations of simplicial sets. I am trying to understand a proof of Milnor's exact sequence found in Goerss-Jardine, but I am having some trouble and have been thinking about it for quite some time.
In the proof one considers the following model for the homotopy inverse limit of X, namely as the pullback of two morphisms where the first is $$j: \Pi_n Hom(\Delta^1,X_n) \rightarrow \Pi_n X_n \times X_n$$ given by the inclusion $$\partial \Delta^1 \rightarrow \Delta^1$$ and the second is given by $$(1,q): \Pi_n X_n \rightarrow \Pi_n X_n \times X_n .$$ Here , $(1,q)$ is the product of the maps $$(1,p_n): \Pi_n X_n \rightarrow X_n \times  X_{n+1} \rightarrow X_n \times X_n , $$ where the first map is projection followed by the identity times $p_n$. $p_n$ is the map $X_{n+1} \rightarrow X_n$ coming from the tower of fibrations. Let us call this pullback for $T(X).$ Then it is easy to see that the map $$T(X) \rightarrow \Pi_n X_n $$ is a fibration with fiber $$\Pi_n \Omega X_n .$$ Then, in the long exact sequeence in homotopy induced by this fibration, it is claimed that the boundary map $$\partial_i: \pi_{i+1}(\Pi_n X_n) \rightarrow \pi_i (\Pi_n \Omega X_n) $$ can be identified, under the iso $\pi_i(\Omega X) \cong \pi_{i+1}(X)$' with  the map $$f: \pi_{i+1} (\Pi_n X_n) \rightarrow \pi_{i+1}(\Pi_n X_n)$$ where $$f((a_i))= (a_i -p_n(a_{i+1})) ,$$  i.e a map whose kernel is the limit of the tower pf abelian groups $\pi_{i+1}(X). $
My question is, how can the boundary map be identified with the map $f$ as above? I tried to use naturality but to no real luck.
I would be very grateful for help.

REPLY [3 votes]: First take a space $X$, and try the fibration $Hom(\Delta_1,X)\to X\times X$.  The fiber over $(x,x')\in X\times X$ is the space of paths in $X$ which go from $x$ to $x'$ (which can sometimes be empty).  Over a point $(x,x)$, the fiber is $\Omega X$ (based loops at $x$).
Now consider the sequence
$$
\pi_n \Omega X \to \pi_n Hom(\Delta_1,X) \to \pi_n X\times \pi_n X \to \pi_{n-1}\Omega X,
$$
using $(x,x)$ as a basepoint in $X$, and the constant path at $x$ in the other spaces.  The middle map is really the diagonal inclusion $\pi_n X\to \pi_n X\times \pi_n X$, so we get a short exact sequence
$$
0 \to \pi_n X\to \pi_nX\times \pi_nX \xrightarrow{d} \pi_{n-1}\Omega X\approx \pi_n X \to 0.
$$
The only reasonable choice(s) for $d$ is $d(a,b)=ab^{-1}$ (up to reordering $a$ and $b$ and placing the $-1$) which we write as $a-b$ if the groups are abelian.  Showing that this is actually true requires an actual argument, which will probably have to know exactly how the map $d$ is defined.  (Or you could use naturality and the fact that $\pi_n S^n=Z$, and the fact that these are really the only possible choices in that case.)
In the argument you describe, you are dealing with an infinite product of maps of  this form as one side of your pullback square, so they work the same way.
In the homotopy inverse limit argument you describe, you actually need to use as a basepoint a path $\gamma$ from $x$ to $x'$ in $X$, which requires a little more care to read off what these groups are.  ($\pi_n (X,x)$ and $\pi_n (X,x')$ are not the same group, but we are given a canonical isomorphism between them by the choice of $\gamma$.)


text:
\begin{document}

\begin{abstract}
We completely describe in terms of Hausdorff measures the size of the set of points of the circle that are covered infinitely often by a sequence of random arcs with given lengths. We also show that this set is a set with large intersection.
\end{abstract}

\maketitle

\section{Introduction and statement of the results}

Let us consider a nonincreasing sequence $\ell=(\ell_{n})_{n\geq 1}$ of positive reals converging to zero. In 1956, A.~Dvoretzky~\cite{Dvoretzky:1956bs} raised the question to find a necessary and sufficient condition on the sequence $\ell$ to ensure that the whole circle $\T=\R/\Z$ is covered almost surely by a sequence of random arcs with lengths $\ell_{1}$, $\ell_{2}$, etc. To be specific, let $A(x,l)$ denote the open arc with center $x\in\T$ and length $l>0$, that is, the set of all $y\in\T$ such that $\dist(y,x)<l/2$, where $\dist$ denotes the usual quotient distance on $\T$. Then, let $(X_{n})_{n\geq 1}$ be a sequence of random points independently and uniformly distributed on $\T$ and let
\[
E_{\ell}=\limsup_{n\to\infty} A(X_{n},\ell_{n}).
\]
Dvoretzky's problem amounts to finding a necessary and sufficient condition to ensure that
\begin{equation}\label{ascover}
\as \qquad E_{\ell}=\T.
\end{equation}
This longstanding problem, along with several of its extensions, has raised the interest of, notably, P.~Billard, P.~Erd\H{o}s, J.-P.~Kahane, P.~L\'evy and B.~Mandelbrot, and has found various applications, such as the study of multiplicative processes and that of some random series of functions, see~\cite{Kahane:1985gc,Kahane:2000ao}.

In 1972, L.~Shepp completely solved Dvoretzky's problem by showing that~(\ref{ascover}) holds if, and only if,
\begin{equation}\label{seriesshepp}
\sum_{n=1}^\infty \frac{1}{n^2}\exp(\ell_{1}+\ldots+\ell_{n})=\infty.
\end{equation}
Still, several related questions remained open. For example, when the series in~(\ref{seriesshepp}) converges, it is natural to ask which proportion of the circle is actually covered by the random arcs. In other words, what is the size of the set $E_{\ell}$? A first answer may be given by computing the value of its Lebesgue measure $\leb(E_{\ell})$. This is in fact trivial, since Fubini's theorem and the Borel-Cantelli lemma directly imply that
\begin{equation}\label{valuelebesgue}
\as \qquad \leb(E_{\ell})=\begin{cases}
0 & \mbox{if } \sum_{n}\ell_{n}<\infty \\[2mm]
1 & \mbox{if } \sum_{n}\ell_{n}=\infty.
\end{cases}
\end{equation}
A typical way of refining the description of the size of the set $E_{\ell}$ is then to compute the value of its Hausdorff dimension. This has recently been done by A.-H.~Fan and J.~Wu~\cite{Fan:2004it} in the particular case where $\ell_{n}=a/n^\alpha$, for some $a>0$ and $\alpha>1$. Their result states that
\[
\forall a>0 \quad \forall\alpha>1 \quad \as \qquad \dim E_{(a/n^\alpha)}=\frac{1}{\alpha}.
\]
Corollary~\ref{cordim} below ensures that actually, for {\em any} nonincreasing sequence $\ell=(\ell_{n})_{n\geq 1}$ converging to zero, the Hausdorff dimension of the set $E_{\ell}$ is almost surely equal to
\begin{equation}\label{defsell}
s_{\ell}=\sup\bigl\{ s\in (0,1) \:\bigl|\: \sum\nolimits_{n} {\ell_{n}}^s=\infty \bigr\}=\inf\bigl\{ s\in (0,1) \:\bigl|\: \sum\nolimits_{n} {\ell_{n}}^s<\infty \bigr\},
\end{equation}
with the convention that $\sup\emptyset=0$ and $\inf\emptyset=1$. Corollary~\ref{cordim} follows from Theorem~\ref{thmsize} below, which in fact gives the value of the Hausdorff $g$-measure of the set $E_{\ell}$ for {\em any} gauge function $g$ and not only the monomial functions used to define the Hausdorff dimension, see Subsection~\ref{sizeprop}. Therefore, this theorem provides a complete description of the size of the set $E_{\ell}$ in terms of Hausdorff measures, for any sequence $\ell$.

We also show in this note that the set $E_{\ell}$ enjoys a remarkable property originally introduced by K.~Falconer~\cite{Falconer:1994hx}, namely, it is a set with large intersection. Roughly speaking, this means that $E_{\ell}$ is ``large and omnipresent'' in the circle in some strong measure theoretic sense, see Subsection~\ref{intprop}. Sets with large intersection have been shown to arise on other occasions in probability theory -- more precisely in multifractal analysis of stochastic processes, see~\cite{Durand:2007kx,Durand:2007fk} -- as well as in other fields of mathematics, namely, number theory and dynamical systems, see~\cite{Durand:2008jk,Durand:2007uq,Durand:2006uq} and the references therein.

\subsection{Size properties of the set $E_{\ell}$}\label{sizeprop}

A typical way of completely describing the size of a subset of the circle is to determine the value of its Hausdorff $g$-measure for any gauge function $g$, see~\cite{Falconer:2003oj,Rogers:1970wb}. We call a gauge function any function $g$ defined on $[0,\infty)$ that is nondecreasing near zero, enjoys $\lim_{0^+} g=g(0)=0$ and is such that $r\mapsto g(r)/r$ is nonincreasing and positive near zero. For any gauge function $g$, the Hausdorff $g$-measure of a set $F\subseteq\T$ is then defined by
\[
\hau^g(F)=\lim_{\delta\downarrow 0}\uparrow\hau^g_{\delta}(F) \qquad\mbox{with}\qquad \hau^g_{\delta}(F)=\inf_{F\subseteq\bigcup_{p} U_{p}\atop\diam{U_{p}}<\delta} \sum_{p=1}^\infty g(\diam{U_{p}}).
\]
The infimum is taken over all sequences $(U_{p})_{p\geq 1}$ of sets with $F\subseteq\bigcup_{p} U_{p}$ and $\diam{U_{p}}<\delta$ for all $p\geq 1$, where $\diam{\cdot}$ denotes diameter. Note that if $g(r)/r$ goes to infinity at zero, every nonempty open subset of $\T$ has infinite Hausdorff $g$-measure and that, otherwise, $\hau^g$ coincides up to a multiplicative constant with the Lebesgue measure on the Borel subsets of $\T$.

The size properties of $E_{\ell}$ are then completely described by the following result.

\begin{thm}\label{thmsize}
Let $\ell=(\ell_{n})_{n\geq 1}$ be a nonincreasing sequence of positive reals converging to zero and let $g$ be a gauge function. Then, with probability one, for any open subset $V$ of $\T$,
\[
\hau^g(E_{\ell}\cap V)=\begin{cases}
\hau^g(V) &\text{if } \sum_{n}g(\ell_{n})=\infty \\[1mm]
0 &\text{if } \sum_{n}g(\ell_{n})<\infty.
\end{cases}
\]
\end{thm}

Recall that the Hausdorff dimension of a nonempty set $F\subseteq\T$ is defined with the help of the monomial functions $\Id^s$ by
\[
\dim F=\sup\{s\in (0,1) \:|\: \hau^{\Id^s}(F)=\infty\}=\inf\{s\in (0,1) \:|\: \hau^{\Id^s}(F)=0\},
\]
with the same convention regarding the infimum and the supremum of the empty set as in~(\ref{defsell}). Also, it is customary to let $\dim\emptyset=-\infty$. Using Theorem~\ref{thmsize}, it is then possible to determine the value of the Hausdorff dimension of the set $E_{\ell}$, thereby generalizing the result of~\cite{Fan:2004it}.

\begin{cor}\label{cordim}
For every nonincreasing sequence $\ell=(\ell_{n})_{n\geq 1}$ of positive reals converging to zero, with probability one, $\dim E_{\ell}=s_{\ell}$, where $s_{\ell}$ is defined by~(\ref{defsell}).
\end{cor}

Theorem~\ref{thmsize} and Corollary~\ref{cordim} are proven in Sections~\ref{proofthm} and~\ref{proofcor}, respectively.

\subsection{Large intersection properties of the set $E_{\ell}$}\label{intprop}

Rigorously, the fact that $E_{\ell}$ is a set with large intersection means that it belongs to some classes $\grint^g(V)$ of subsets of the circle that we defined in~\cite[Section~5]{Durand:2007fk}. We refer to that paper, and also to~\cite{Durand:2007uq}, for a precise definition of those classes and we content ourselves with stressing on the fact that, for any gauge function $g$ and any nonempty open subset $V$ of the circle, one may define a class $\grint^g(V)$ of {\em sets with large intersection in $V$ with respect to $g$} which, among other properties, enjoys the following.

\begin{prp}\label{grintstable}
For any gauge function $g$ and any nonempty open $V\subseteq\T$,
\renewcommand{\theenumi}{\alph{enumi}}
\begin{enumerate}
\item\label{intersect} the class $\grint^g(V)$ is closed under countable intersections;
\item\label{relsizelargeint} every set $F\in\grint^g(V)$ enjoys $\hau^{\overline{g}}(F)=\infty$ for any gauge $\overline{g}$ with $\overline{g}\prec g$;
\item $\grint^g(V)=\bigcap_{\overline{g}}\grint^{\overline{g}}(V)$ where $\overline{g}$ is a gauge function enjoying $\overline{g}\prec g$;
\item\label{openincl} $\grint^g(V)=\bigcap_{U}\grint^g(U)$ where $U$ is a nonempty open subset of $V$;
\item every $G_{\delta}$-set with full Lebesgue measure in $V$ belongs to the class $\grint^g(V)$.
\end{enumerate}
\end{prp}

The notation $\overline{g}\prec g$ means that $\overline{g}/g$ monotonically tends to infinity at zero. In view of Proposition~\ref{grintstable}, every set in the class $\grint^g(V)$ has infinite Hausdorff $\overline{g}$-measure in every nonempty open subset of $V$ for any gauge function $\overline{g}\prec g$, and any countable intersection of such sets enjoys the same property. Therefore, the classes $\grint^g(V)$ provide a rigorous way of stating that a set is large and omnipresent in $V$ in a strong measure theoretic sense.

In order to describe the large intersection properties of the set $E_{\ell}$, we shall make use of the following result, which gives a simple sufficient condition for a limsup of arcs to be a set with large intersection in the circle. It may be seen as the analog for the periodic setting of the {\em ubiquity} result established in~\cite{Durand:2007uq}.

\begin{prp}\label{ubiquity}
Let $(y_{n})_{n\geq 1}$ be a sequence in $\T$ and let $(r_{n})_{n\geq 1}$ be a sequence of positive reals converging to zero. Then, for any gauge function $g$,
\[
\leb\left(\limsup_{n\to\infty} A(y_{n},2g(r_{n}))\right)=1 \qquad\Longrightarrow\qquad \limsup_{n\to\infty} A(y_{n},2r_{n})\in\grint^g(\T).
\]
\end{prp}

Proposition~\ref{ubiquity} may be interpreted as follows. Given that any gauge function $g$ is bounded below by the identity function (up to a multiplicative constant), the limsup of the arcs $A(y_{n},2r_{n})$ may be seen as a ``reduced'' version of the limsup of the arcs $A(y_{n},2g(r_{n}))$. If the latter limsup is large and omnipresent enough to contain Lebesgue-almost every point of the circle, then its reduced version is also large and omnipresent, in the weaker sense that it belongs to the class $\grint^g(\T)$. We refer to Section~\ref{proofubiquity} for a proof of Proposition~\ref{ubiquity}.

The large intersection properties of the set $E_{\ell}$ are then completely described by the following result.

\begin{thm}\label{thmint}
Let $\ell=(\ell_{n})_{n\geq 1}$ be a nonincreasing sequence of positive reals converging to zero and let $g$ be a gauge function. Then, almost surely, for any nonempty open subset $V$ of $\T$,
\[
E_{\ell}\in\grint^g(V) \qquad\Longleftrightarrow\qquad \sum\nolimits_{n} g(\ell_{n})=\infty.
\]
\end{thm}

The remainder of this note is organized as follows: Theorems~\ref{thmsize} and~\ref{thmint} are established in Section~\ref{proofthm}, Corollary~\ref{cordim} is proven in Section~\ref{proofcor}, and the proof of Proposition~\ref{ubiquity} is given in Section~\ref{proofubiquity}. Before detailing the proofs, let us mention that we shall basically only make use of the main properties of the classes $\grint^g(V)$ given by Proposition~\ref{grintstable}, the ubiquity result given by Proposition~\ref{ubiquity} and the value~(\ref{valuelebesgue}) of the Lebesgue measure of the set $E_{\ell}$. In particular, unlike the authors of~\cite{Fan:2004it}, we do not need to call upon any specific result on the spacings between the random centers $X_{n}$ of the arcs. This also means that our method can straightforwardly be extended to the case of random balls on the $d$-dimensional torus for any $d\geq 2$.

\section{Proofs of Theorems~\ref{thmsize} and~\ref{thmint}}\label{proofthm}

Theorems~\ref{thmsize} and~\ref{thmint} follow from four lemmas which we now state and prove. Throughout the section, $\ell=(\ell_{n})_{n\geq 1}$ is a nonincreasing sequence of positive reals converging to zero.

\begin{lem}\label{convhau}
For any gauge function $g$,
\[
\sum\nolimits_{n} g(\ell_{n})<\infty \qquad\Longrightarrow\qquad \forall V\mbox{ open} \quad \hau^g(E_{\ell}\cap V)=0.
\]
\end{lem}

\begin{proof}
For any $\delta>0$, there is an integer $n_{0}\geq 1$ such that $0<\ell_{n}<\delta$ for any $n\geq n_{0}$. Moreover, the set $E_{\ell}$ is covered by the arcs $A(X_{n},\ell_{n})$ for $n\geq n_{0}$, so that $\hau^g_{\delta}(E_{\ell})\leq\sum_{n=n_{0}}^\infty g(\ell_{n})$. If the series $\sum_{n} g(\ell_{n})$ converges, then letting $n_{0}$ tend to infinity and $\delta$ go to zero yields $\hau^g(E_{\ell})=0$.
\end{proof}

\begin{lem}
For any gauge function $g$,
\[
\sum\nolimits_{n} g(\ell_{n})<\infty \qquad\Longrightarrow\qquad \forall V\neq\emptyset\mbox{ open} \quad E_{\ell}\not\in\grint^g(V).
\]
\end{lem}

\begin{proof}
Let us assume that the series $\sum_{n} g(\ell_{n})$ converges. Then, one may build a gauge function $\overline{g}$ such that $\overline{g}\prec g$ and the series $\sum_{n} \overline{g}(\ell_{n})$ converges too, for example by adapting a construction given in the proof of~\cite[Theorem~3.5]{Durand:2007lr}. By Lemma~\ref{convhau}, the set $E_{\ell}$ has Hausdorff $\overline{g}$-measure zero in $V$ and thus cannot belong to the class $\grint^g(V)$, due to Proposition~\ref{grintstable}(\ref{relsizelargeint}).
\end{proof}

\begin{lem}\label{divgrint}
For any gauge function $g$,
\[
\sum\nolimits_{n} g(\ell_{n})=\infty \qquad\Longrightarrow\qquad \as\quad\forall V\neq\emptyset\mbox{ open} \quad E_{\ell}\in\grint^g(V).
\]
\end{lem}

\begin{proof}
If the series $\sum_{n} g(\ell_{n})$ diverges, then $\sum_{n} g(\ell_{n}/2)$ diverges as well (because $r\mapsto g(r)/r$ is nonincreasing near zero). Hence, thanks to~(\ref{valuelebesgue}), the limsup of the arcs $A(X_{n},2g(\ell_{n}/2))$ has Lebesgue measure one with probability one. We conclude using Proposition~\ref{ubiquity} and Proposition~\ref{grintstable}(\ref{openincl}).
\end{proof}

\begin{lem}
For any gauge function $g$,
\[
\sum\nolimits_{n} g(\ell_{n})=\infty \qquad\Longrightarrow\qquad \as\quad\forall V\mbox{ open} \quad \hau^g(E_{\ell}\cap V)=\hau^g(V).
\]
\end{lem}

\begin{proof}
We may obviously assume that $V$ is nonempty. Let us suppose that the series $\sum_{n} g(\ell_{n})$ diverges. Then, again by following a construction given in the proof of~\cite[Theorem~3.5]{Durand:2007lr}, it is possible to build a gauge function $\underline{g}$ such that $g\prec\underline{g}$ and the series $\sum_{n} \underline{g}(\ell_{n})$ diverges too, provided that $g\prec\Id$. Therefore, thanks to Lemma~\ref{divgrint}, the set $E_{\ell}$ belongs to the class $\grint^{\underline{g}}(V)$. Hence, $\hau^g(E_{\ell}\cap V)=\infty=\hau^g(V)$, owing to Proposition~\ref{grintstable}. In the case where $g\not\prec\Id$, the Hausdorff $g$-measure coincides, up to a multiplicative constant, with the Lebesgue measure on the Borel subsets of the circle and the result follows from~(\ref{valuelebesgue}).
\end{proof}

\section{Proof of Corollary~\ref{cordim}}\label{proofcor}

In order to prove Corollary~\ref{cordim}, let us consider a nonincreasing sequence $\ell=(\ell_{n})_{n\geq 1}$ of positive reals converging to zero. Theorem~\ref{thmsize}, along with the definition~(\ref{defsell}) of the real $s_{\ell}$, implies that for any real $s\in (0,1)$, with probability one,
\[
\hau^{\Id^s}(E_{\ell})=\begin{cases}
\infty &\text{if } s<s_{\ell} \\[1mm]
0 &\text{if } s>s_{\ell}.
\end{cases}
\]
Let us assume that $s_{\ell}\in (0,1]$. Then, for all $m$ large enough, with probability one, the set $E_{\ell}$ has infinite Hausdorff $\Id^{s_{\ell}-1/m}$-measure, so that its Hausdorff dimension is at least $s_{\ell}-1/m$. Therefore, the dimension of $E_{\ell}$ is almost surely at least $s_{\ell}$. Likewise, if $s_{\ell}\in [0,1)$, then $E_{\ell}$ has $\Id^{s_{\ell}+1/m}$-measure zero with probability one for all $m$ large enough, so that its Hausdorff dimension is almost surely at most $s_{\ell}$. As a result, with probability one, $\dim E_{\ell}=s_{\ell}$ if $s_{\ell}\in (0,1]$ and $\dim E_{\ell}\leq 0$ if $s_{\ell}=0$.

It remains to establish that $E_{\ell}$ is almost surely nonempty when $s_{\ell}=0$. Note that the set $E_{(1/n)}$, obtained by picking $\ell_{n}=1/n$, has Lebesgue measure one with probability one, by virtue of~(\ref{valuelebesgue}). Let us assume that this property holds. Furthermore, note that $\ell_{n}={\rm o}(1/n)$ as $n$ goes to infinity, thanks to Olivier's theorem~\cite{Olivier:1827eu}. In particular, $\ell_{n}\leq 1/n$ for any integer $n$ greater than or equal to some $n_{1}\geq 1$. Let $I_{1}=A(X_{n_{1}},\ell_{n_{1}}/2)$. The union over $n>\max\{n_{1},8/\ell_{n_{1}}\}$ of the arcs $A(X_{n},1/n)$ has full Lebesgue measure in the circle, so its intersection with the arc $A(X_{n_{1}},\ell_{n_{1}}/4)$ is nonempty. Therefore, there is an integer $n_{2}>\max\{n_{1},8/\ell_{n_{1}}\}$ such that $A(X_{n_{2}},1/n_{2})\subseteq I_{1}$. Then, let $I_{2}=A(X_{n_{2}},\ell_{n_{2}}/2)$. Repeating this procedure, one may obtain a nested sequence of open arcs $I_{n}$ and the intersection of their closures yields a point that belongs to the set $E_{\ell}$.

\section{Proof of Proposition~\ref{ubiquity}}\label{proofubiquity}

Let $g$ be a gauge function such that the limsup of the arcs $A(y_{n},2g(r_{n}))$ has Lebesgue measure one. Thus, following the terminology of~\cite{Durand:2007uq}, the family $(k+\dot y_{n},g(r_{n}))_{(k,n)\in\Z\times\N}$ is a homogeneous ubiquitous system in $\R$. Here, each $\dot y_{n}$ is the only real in $[0,1)$ such that $\phi(\dot y_{n})=y_{n}$, where $\phi$ denotes the canonical surjection from $\R$ onto $\T$. Thanks to~\cite[Theorem~2]{Durand:2007uq}, the set of all reals $x$ such that $|x-k-\dot y_{n}|<r_{n}$ for infinitely many $(k,n)\in\Z\times\N$ belongs to the class $\grint^g(\R)$ of sets with large intersection in $\R$ with respect to the gauge function $g$, which is defined in~\cite{Durand:2007uq}. Equivalently, the inverse image under $\phi$ of the limsup of the arcs $A(y_{n},2r_{n})$ belongs to $\grint^g(\R)$, which ensures that this limsup belongs to the class $\grint^g(\T)$, see~\cite[Section~5]{Durand:2007fk}.


TITLE: Showing this function is increasing or decreasing
QUESTION [1 upvotes]: We have a function
$$f(x) = e^x - x\int_{0}^{1} z^n e^{-xz} dz$$ that is defined and continuous for all $x \in [-a, 0]$. We also know that $n\in\mathbb{N}$.
I'm trying to determine how $f(x)$ changes as $x$ varies in $[-a, 0]$ but I'm having a hard time with it.
I tried to show that $f(x_1) < f(x_2)$ for $x_1 < x_2$ but it wasn't successful. I also tried to use some results in Karlin, S. 1968. Total Positivity. Vol. 1. Stanford University Press, but it was very complicated for me.
I need some guidance and any comments and hints are very appreciated.

REPLY [1 votes]: Since
$$\int_{0}^{1} z^n e^{-xz}\, dz=x^{-(n+1)} (\Gamma (n+1)-\Gamma (n+1,x))$$
$$f(x)=e^x-x^{-n} (\Gamma (n+1)-\Gamma (n+1,x))$$
Expanding as series
$$\Gamma(n+1,x)=\Gamma(n+1)+x^n \sum_{k=1}^\infty (-1)^k\frac{ x^k}{(k-1)!\,\, (k+n)}$$
$$f(x)=e^x+\sum_{k=1}^\infty (-1)^k\frac{ x^k}{(k-1)!\,\, (k+n)}=\frac 1{e^{-x}}+\sum_{k=1}^\infty \frac{ (-x)^k}{(k-1)!\,\, (k+n)}$$ and you consider the case where $x <0$. Letting $x=-y$, we have
$$g(y)=\frac 1{e^{y}}+\sum_{k=1}^\infty \frac{ y^k}{(k-1)!\,\, (k+n)}$$ where all terms are now positive for $0< y <a$.
I hope that this could help you.


TITLE: Probability, Mass Function
QUESTION [0 upvotes]: For what values of constant C does the function $p(k)=C3^{−k}+4^{−k−1}$ defined on the non-negative integers k=0, 1, 2, … constitutes a (probability) mass function?
The analogous probability "density" for continuous random variables can be interpreted in the same way. If $f\left(x\right)$ is your probability density, then
$$
\int_a^b f\left(x\right) dx
$$
can be interpreted as a "mass".
But I don't understand this when i tried to compute some values, i found that the possible values are 3/4 and 1/3, but i´m not sure at all, please if you can help me.

REPLY [1 votes]: For $\{p_k\}$ to be a probability mass function, we must have
$$\sum_{k=0}^\infty p_k=\sum_{k=0}^\infty (C3^{-k} + 4^{-k-1}) = \frac32 C + \frac13=1, $$
which implies that $C = \frac49$.


text:
\section{Regular Space is Semiregular Space}
Tags: Semiregular Spaces, Regular Spaces

\begin{theorem}
Let $\struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
Then $\struct {S, \tau}$ is also a [[Definition:Semiregular Space|semiregular space]].
\end{theorem}

\begin{proof}
Let $T = \struct {S, \tau}$ be a [[Definition:Regular Space|regular space]].
From the [[Definition:Regular Space|definition]]:
:$\struct {S, \tau}$ is a [[Definition:T3 Space|$T_3$ space]]
:$\struct {S, \tau}$ is a [[Definition:Kolmogorov Space|$T_0$ (Kolmogorov) space]].
We also have that a [[T3 Space is Semiregular|$T_3$ Space is Semiregular]].
Hence the result, by definition of [[Definition:Semiregular Space|semiregular space]].
{{qed}}
\end{proof}

config: wiki


TITLE: Constrained maximization problem with linear function
QUESTION [0 upvotes]: I have the following problem. Given this function
$E[\pi] = (1-r)[\alpha b- (1-p)C-K]+T $
I would like to find the maximum w.r.t. $r$ given this constraint:
$U = (1-r)b-T \geq 0$.
It is an economic problem that I am formalizing, but this is not relevant to its solution, I am only interested in the mathematical resolution of the problem. I provide some background. Our variable $r$ is a number between $0$ and $1$, $p$ is some probability, $\alpha,b,C,K$ and $T$ are all positive constants. If it is useful for the resolution of the problem, we can also assume that $\alpha$ is between $0$ and $1$. An important assumption (namely, assumption $\&$) is that $(1-p)C+K>\alpha b$. Readers used to economics may recognize that the objective function is a sort of expected profit and the constraint is a sort of expected utility.
I tried with Kuhn-Tucker, but with miserable results since deriving w.r.t. $r$ does not yield any expression with $r$.
Now I'm following a more intuitive approach. I start by assuming that $U=0$ is the constraint, then I can find an expression for $r$ from the constraint and I substitute it in the target function. After easy steps, I get this
$T \frac{\alpha b - (1-p)C-K}{b} + T$.
At this point, I can use assumption $\&$ to conclude that the first $T$ above will be negative, but I'm not able to conclude whether it will be lower, equal or higher to/than the second $T$ because it is multiplied by some constant.
At this point I'm stuck. I do not know if my approach can work. Could you please suggest a nicer way to proceed? I am on the right track or it's a dead end?
For the admin: This is a new post since the old one was blocked because it lacked many information. I hope that in its current form it is acceptable to start a discussion. Let me know if I need to add anything else. Thank you.

REPLY [1 votes]: Hint.
Calling $\phi = (1-p)C+K-\alpha b\gt 0$ and considering
$$
U = (1-r)b-T \geq 0\Rightarrow r \le 1-\frac Tb
$$
we can formulate the problem as
$$
\min_r (r-1)\phi+T,\ \ \text{s. t.}\ \ \cases{r\ge 0\\
r\le1-\frac Tb}
$$
now as $(r-1)\phi+T$ is linear, the solution is one of the set
$$
\left\{T-\phi,T-\phi\frac Tb\right\}
$$


TITLE: Why doesn't integrating the area of the square give the volume of the cube?
QUESTION [21 upvotes]: I had a calculus course this semester in which I was taught that the integration of the area gives the size (volume):
$$V = \int\limits_a^b {A(x)dx}$$
But this doesn't seem to work with the square. Since the size of the area of the square is $x^2$ then $A(x) = {x^2}$, then:
$$V = \int\limits_{ - r}^r {{x^2}dx}  = \left[ {\frac{{{x^3}}}{3}} \right]_{ - r}^r = \frac{{{r^3}}}{3} - \frac{{ - {r^3}}}{3} = \frac{2}{3}{r^3}$$
It's clear that this is not the volume of the cube. Why is this the case? Am I misunderstanding something?

REPLY [1 votes]: In general if a solid has cross-sectional area $A$ which is constant along its normal length say $L$ then the volume of such solid is $$\color{blue}{V=\int_{0}^L Adx=A\int_{0}^L dx=A[L-0]=AL}$$
In fact, the area of the square is not varying with the distance $x$ in a cube. It is $a^2$ which is constant along entire length $x=a$ hence, volume $V$ of cube =\int_{0}^a a^2dx=a^2\int_{0}^a dx=a^2[a-0]=a^3$$


TITLE: A doubt regarding circular permutations
QUESTION [1 upvotes]: In my textbook it was given that the number of circular permutations of n different things taken r at a time (regarding anticlockwise and clockwise arrangements as different) is (nPr)/r. Any help regarding its proof will be appreciated.

REPLY [1 votes]: Suppose we're dealing with persons being assigned to chairs around a circular table. Suppose there are $n$ people and the chairs are numbered $1\ldots r$.
The first task is to pick an (ordered) list of $r$ people out of $n$ people to sit in the chairs— someone to sit in chair #1, #2, and so on. There are $nPr$ different seating assignments when considered this way.
But because the chairs are arranged in a circle, some of the different seating assignments are actually equivalent except for a rotation: a seating assignment like ABCDE should be considered the same as BCDEA and CDEAB, and DEABC, and EABCD.
In fact, for each circular permutation there are $r$ equivalent permutations where a different person sits in "Seat #1" while keeping the same seating order around the table. We divide by $r$ to avoid overcounting these variant permutations.


TITLE: How to show if $\vdash \alpha \leftrightarrow \beta$ then $\alpha \equiv \beta$
QUESTION [0 upvotes]: We know that $\alpha \equiv \beta$ if $\beta \vDash \alpha$ and $\alpha \vDash \beta$.
In first order logic:
$(M, v) \vDash \alpha$ iff $(M, v) \vDash \beta$ for every interpretation $(M, v)$.
But how to show starting from $\vdash \alpha \leftrightarrow \beta$?

REPLY [1 votes]: Suppose that $\alpha\leftrightarrow\beta$ is a valid sentence. Assume that $(M,v)\models\alpha$, since $\alpha\leftrightarrow\beta$ is a valid sentence, it must be the case that $(M,v)\models\alpha\leftrightarrow\beta$...
Continue from there.


TITLE: Elliptic functions and Weierstrass $\wp$-function
QUESTION [6 upvotes]: Question that seems pretty easy, but I can't formalize it:
Let  $L \subset C$ be a lattice, and $f(z)$ be an elliptic function for $L$, that is a meromorphic function so that $f(z+w) = f(z)$ for all $\omega \in L$. Assume that $f$ is analytic except for double poles at each point of the lattice $L$. Show that $f = a\wp + b$ for some constants $a,b$.
What I tried:
$\displaystyle f(z) = \prod_{\omega \in L} {\frac{g(z)}{(z-\omega)^2}}$ , $g(z)$ is analytic and therefore constant in the fundamental domain. Now what is left to do, is to take the product apart to partial fractions, and then I get almost what needed, except it's not one constant $a$ and $b = \sum_{\omega \in L} -\frac1{\omega^2}$.
Am I right? How do I proceed?
Thanks in advance.

REPLY [5 votes]: Here is a hint to help orient you: Suppose $f$ is doubly periodic on the lattice $L_{\tau} = \mathbb{Z} \oplus \tau \mathbb{Z}$, where $\tau \in \mathbb{H}$, and $f$ is assumed analytic everywhere save for double poles on the lattice points of $L_{\tau}$. What can you say about the function $f/\wp$ on $L_{\tau}$?
Compare the position of the poles of both $f$ and $\wp$. If they occur at the same place, then $f/\wp$ has no poles (provided that the order of the poles is the same), and so is doubly periodic as well as analytic, hence constant of $\Lambda_{\tau}$.
Now generalize $f$ to an arbitrary lattice $\omega_1 \mathbb{Z} \oplus \omega_2 \mathbb{Z}$, as the one you post, and consider the function $g = (f - a)/\wp = f/\wp - a/\wp$, where $a$ is an arbitrary constant. What properties will $g$ have on the lattice?


TITLE: If an arrow category is like a path what's like a cylinder?
QUESTION [4 upvotes]: The arrow category is isomorphic to a functor category out of the interval category.
$$\mathrm{Arr}( C) = [\mathrm I, C]$$
It is like a very simple/degenerate case of a path object.
A "loop" speaking loosely would just be an arrow (possibly identity?) with equal source and target.
You would expect it to have an adjoint something like
$$? = C \times \mathrm{I}$$
And a sort of "suspension" category would be some sort of quoitent or pushout that I don't quite understand equating (x, 0) and (y, 1).
But I can't really figure out how to conceptualize these as concrete things. What does it mean to be an object of a "cylinder" category?
Maybe you could think of it as mapping between source/targets?

REPLY [2 votes]: The way I like to think about $\mathcal C \times I$ is that it classifies natural transformations.
Just like $X \times I$ classifies homotopies for topological spaces $X$.
Natural transformations $\phi:\mathcal F \rightarrow \mathcal G$ between two functors $\mathcal C \rightarrow \mathcal D$ are in bijection with functors $H$ that fit into the diagram.

We simply set $H((c,0) \rightarrow (c,1)) = \phi_c$ to establish this bijection.


text:
\section{Empirical Evaluation: Real Data}\label{section:real_empirical}

We have carried out an exploratory analysis of publicly available data for COVID-19 from \cite{Data_COVID} for the five countries mentioned above under the proposed framework. We have used testing data for a period of \rev{eleven months from 1st March 2020 to 31st January 2021}. The testing data is comprised of the number of tests carried out ($z(k)$), the number of confirmed positive cases ($c(k)$), and the number of recoveries ($d(k)$). Note that the death counts are included in the recovery data. Some dates have missing data on the number of tests being carried out. In such cases, linear interpolation of the past and next available data points are used \cite{ciss}. We have filtered the original data by taking a $7-$day moving average to reduce the effect of outliers. \rev{The plots in the left panel of Fig. \ref{fig:real_scalar_inf} show the test positivity rate ($c(k)/z(k)$) and active cases for two countries, Germany (DE) and Italy (IT), respectively.}

\rev{We emphasize that the results in this section are obtained by the proposed methodology which is based on the SIR epidemic dynamic and other assumptions stated in the paper. In the absence of large-scale serology tests and knowledge of testing policies, it is difficult to quantify the accuracy of the inference results in general, and the same caveat applies to the results presented in this paper as well.}

\begin{figure*}
\centering
    \includegraphics[scale=0.6]{cfr_de_it.pdf}
    ~\includegraphics[scale=0.6]{scalar_de_inf_nosero.pdf}
    ~\includegraphics[scale=0.6]{scalar_it_inf_nosero.pdf}
    \\
    \includegraphics[scale=0.6]{at_de_it.pdf}
    ~\includegraphics[scale=0.6]{scalar_de_suc_nosero.pdf}
    ~\includegraphics[scale=0.6]{scalar_it_suc_nosero.pdf}
    \caption{Real testing data and inferred/predicted states for Germany (DE) and Italy (IT) from 1st March 2020 to 31st January 2021 with values of $\alpha$ as stated in Table \ref{tab:inf_alpha_real} for the respective countries and values of $T_2$.}
    \label{fig:real_scalar_inf}
\end{figure*}

\subsection{Inferring epidemic states}

From the above data, we infer the underlying epidemic states, (i.e., the proportions of susceptible, infected, and recovered subpopulations in all five countries) \rev{and the SIR epidemic parameters following the methodology proposed in Section \ref{subsection_testing_risk} and Section \ref{section:least-squareb}.} \rev{We first analyze the testing data for each of the five countries separately with value of $\tau$ set to $0$. For each country, we consider the available testing data from $T_1 = 1$ to different values of $T_2 \in \{90,150,210,270,330\}$ to examine how the inferred states change as more and more testing data becomes available. For each value of $T_2$, we solve the least squares problem in \eqref{eq:ls_opt_new} assuming that $\beta$ and $\gamma$ parameters change every $30$ days. We then learn the initial inferred states and these parameters for integer values of $\alpha$ ranging from $1$ to $100$. The values of $\alpha$ that achieve the smallest least squares cost are reported in Table \ref{tab:inf_alpha_real}.}

\begin{table}[htb]
\centering
\caption{\small \rev{Values of $\alpha$ that achieve the smallest least squares cost for real testing data from $T_1 = 1$ and different values of $T_2$.}}
\begin{tabular}{lclclclclc|c|}
\toprule
$T_2$ & DE  & FR & AT & IT & CH
\\ \midrule
$90$ & $6$  & $12$  & $6$ & 12 & $10$     \\
150   & 6      & 14    & 6    & 14 & 10      \\
210   & 8      & 19     & 10  & 15   & 12     \\
270   & 12    & 26  & 24  & 26 & 36      \\
330   & 24    & 30    & 24  & 31 & 40
\\ \bottomrule
\end{tabular}
\label{tab:inf_alpha_real}
\end{table}

\rev{The values in Table \ref{tab:inf_alpha_real} show that as more testing data becomes available, the value of $\alpha$ that best explains the testing data increases. This is evident from the inferred and predicted quantities shown in the middle and right panel of Fig. \ref{fig:real_scalar_inf} for Germany and Italy, respectively. The solid (red) curve shows the predicted infected and susceptible proportions obtained via solving \eqref{eq:for} with the learned values of initial states and parameters when testing data from $T_1 = 1$ to $T_2 = 90$ is used with the value of $\alpha$ shown in the first row of Table~\ref{tab:inf_alpha_real}.} \rev{Fig. \ref{fig:real_scalar_inf} shows that a significant proportion of individuals became infected during the first wave of COVID-19. In fact, the peak infected proportion was more than $0.4$ for Italy.}

\begin{figure*}[tb]
	\centering
	\includegraphics[scale=0.6]{net_gamma.pdf}~
	\includegraphics[scale=0.6]{net_all_inf.pdf}~
	\includegraphics[scale=0.6]{net_all_suc.pdf}~
	\\
	\includegraphics[scale=0.6]{net_it_inf.pdf}~
	\includegraphics[scale=0.6]{de_lock_vs_noloc.pdf}~
	\includegraphics[scale=0.6]{it_lock_vs_noloc.pdf}~
	\caption{\footnotesize{The top left panel shows the learned values of recovery/curing rates for different sub-intervals. The middle and right panels in the top row show the predicted proportions of infected and susceptible individuals. The bottom left panel compares the inferred and predicted infected proportions for Italy. The middle and right panels in the bottom row show the predicted and projected infections without NPIs for Germany and Italy; $\widehat{x}^2_{\mathtt{LS}}$ is the forecasted trajectory if the parameters were kept unchanged after the second sub-interval.}}
    \label{fig:IT_ls}
\end{figure*}

\rev{However, data from the second wave of infections indicates that the fraction infected during the first wave was in fact significantly smaller. As a result, the previous value of $\alpha$ (learned using testing data till $T_2 = 90$) results in an infeasible trajectory under testing data from $T_1 = 1$ to $T_2 = 330$. The dashed (black) line in Fig. \ref{fig:real_scalar_inf} shows the inferred/predicted quantities with value of $\alpha$ shown in the bottom row of Table~\ref{tab:inf_alpha_real}. The figure also indicates that the learned initial proportion of infected nodes was around $0.1$ for Italy at the beginning of March while it was negligible for Germany.}

\rev{Thus, our analysis provides valuable insights into the effect of the tuning parameter $\alpha$, the significance of learning the initial infected proportion, and how to tune the inference technique as more and more testing data becomes available. Nevertheless, we clarify that while the values of $\alpha$ reported in Table \ref{tab:inf_alpha_real} should be viewed as quantities that best explain the data assuming that $\alpha$ remains unchanged over the entire time period. In practice, this parameter, which captures the ratio of the likelihood of an infected individual undergoing testing and a healthy individual undergoing testing, should be chosen/learned in collaboration with public health officials. This parameter is also likely time-varying as testing practices evolve over the course of the pandemic. The inference can be further tuned using outcomes of large-scale serological surveys by imposing suitable lower bounds on the proportion of susceptible individuals at different points of time. Further investigations along these lines are beyond the scope of this paper and will be explored in a follow up work.}

\subsection{Inference and estimation in the networked model}

\rev{We now learn the parameters of the networked SIR epidemic dynamic, with the network structure as shown in Fig. \ref{fig:baseline}, by solving the least squares problem in \eqref{eq:ls_opt_new} for the first wave data of $90$ days from 1st March 2020 to 29th May 2020. We divide the time-span into four sub-intervals from day $1$ to $15$, $16$ to $30$, $31$ to $60$ and $61$ to $90$ to better capture the changing lockdown measures being put in place during those periods. Specifically, we learn four different sets of $\beta_{ij}$ and $\gamma_i$ parameters where each corresponds to different sub-intervals. We choose $\alpha = 40$ for our inference as it would be consistent with both first and second wave data as discussed earlier. We also impose the constraint that the initial infected proportion to not exceed $0.05$ at any of the nodes.}

\rev{The learned values of the curing or recovery rates $\gamma$ for different countries and sub-intervals are shown in the top left panel of Fig. \ref{fig:IT_ls}. The learned parameters show that during the initial period, the recovery rates were small which contributed to a rapid increase in the number of infections. However, the recovery rates showed a gradual increase over time except in France and Italy where the variation was not significant. The predicted infected and susceptible proportions (by solving \eqref{eq:for} starting from the initial inferred states and the learned $\beta$ and $\gamma$ parameters) are shown in the middle and right panels in the top row of Fig. \ref{fig:IT_ls}.} The inferred states with delay $\tau=8$, are found to be very similar to a shifted version of the results obtained with $\tau=0$. These results are shown in the supplementary appendix.

\rev{In the bottom left panel of Fig. \ref{fig:IT_ls}, we compare the inferred infected proportion defined via the learned initial proportions and the testing data following \eqref{eq:inf_x_final}, and the predicted infected proportion determined following \eqref{eq:for} starting from the learned initial proportions and using the learned $\widehat{\beta}$ and $\widehat{\gamma}$ parameters for the different sub-intervals for Italy. Both quantities are largely in agreement showing that the learned $\beta$ and $\gamma$ parameters well explain the inferred states. However, note that this plot does not indicate the accuracy of the inferred states since the ground truth is not known. The learned $\beta$ parameters are shown in the supplementary appendix.}

\subsection{Learned vs. optimal NPIs}

\begin{figure*}[tb]
	\centering
	\includegraphics[scale=0.6]{opt_lambda.pdf}~
	\includegraphics[scale=0.6]{opt_de.pdf}~
	\includegraphics[scale=0.6]{opt_fr.pdf}
	\caption{Predicted and optimized trajectories of the growth rate and the infected proportions at nodes DE and FR under optimal recovery rates computed by solving \eqref{eq:GP_SIR} with the original cost (C) and twice the original cost (2C) as budgets. Time/day $0$ on the left panel corresponds to time/day $17$ in the middle and right panels.}
    \label{fig:NPI_Real1}
\end{figure*}

We now illustrate the impact of learned NPIs by computing the trajectory under the parameters learned for the first and the second sub-intervals. Specifically, let $\widehat{x}^2_{\mathtt{LS}}$ be the trajectory of the proportion of infected individuals if the infection and recovery rates were kept equal to the parameters learned from data corresponding to the second sub-interval (days $16$ to $30$). \rev{The plots in the middle and right panels in the bottom row of Fig. \ref{fig:IT_ls}} show that NPIs such as imposing lock-down measures and augmenting healthcare resources during the second half of March significantly reduced the infected proportion compared to the initial growth rate, \rev{which if unchecked, would have led to a significantly higher peak.}

We further explore the optimality of the recovery rate parameters learned from the observed data. \rev{We determine the total cost (C) of the learned recovery rates under the cost function $g_i$ from \eqref{eq:NPI_cost_fn} for different sub-intervals to be $1.388, 1.194, 2.076$ and $2.567$, respectively. In the cost function, we set the upper and lower bounds to be the largest and smallest among the learned parameters for each country over all sub-intervals.\footnote{\rev{Note that the cost function is inversely proportional to $\bar{\gamma}$ and the bounds play the role of a normalization constant. If we allow the upper and lower bounds to be more spread out than the learned parameters, then the denominator in $g_i$ will increase and as a result, performance of the optimized trajectory would be ``better" due to more relaxed constraints and a smaller cost value. Thus, our choice of bounds allows for a fair comparison with the learned parameters.}}}

We now compute the optimal recovery rates to minimize the growth rate of the epidemic subject to the budget constraints on the cost of the recovery rates. Specifically, we solve \eqref{eq:GP_SIR} for the optimal $\gamma$ once for each sub-interval starting from day $16$. We do not treat the infection rates as decision variables, and use their learned values for the respective sub-intervals. We choose the budgets to be the total cost of the learned NPIs stated above (C) (that is, a reallocation of the existing resources), and twice the cost (2C). The resulting growth rates and the trajectory of the infected proportions for DE and FR are shown in Fig. \ref{fig:NPI_Real1}. The figure shows that with the same total cost of NPIs, the optimal recovery rate allocation would have significantly reduced the infected proportion for FR with a relatively minor increase in the infected proportion of DE. When the optimal NPIs are imposed under twice the budget, the performance is significantly better for all the nodes for all sub-intervals. Thus, our results in this section highlight the importance of a closed-loop approach, i.e., using the inferred states as feedback to compute optimal NPIs in order to mitigate and control networked epidemics such as COVID-19.


TITLE: Prove $\int_{-\pi}^{\pi} \frac{dk}{2 \pi} \frac{\cos(nk)-1}{\cos(k)-1} = n$
QUESTION [0 upvotes]: The following integral comes up in the theory of Brownian motion in 1D:
$$\int_{-\pi}^{\pi} \frac{dk}{2 \pi} \frac{\cos(nk)-1}{\cos(k)-1} = n$$
where $n \in \mathbf{N}$. Does anybody know a short proof of this formula?

REPLY [2 votes]: Let $z=e^{ik}$ and then
$$ \cos k=\frac12(z+\frac1z), \cos nk=\frac12(z^n+\frac1{z^n}). $$
So
\begin{eqnarray*}
\int_{-\pi}^{\pi} \frac{dk}{2 \pi} \frac{\cos(nk)-1}{\cos(k)-1}&=&\frac{1}{2\pi}\int_{|z|=1}\frac{\frac12(z^n+\frac1{z^n})-1}{\frac12(z+\frac1z)-1}\frac{dz}{iz}\\
&=&\frac{1}{2\pi i}\int_{|z|=1}\frac{(1-z^n)^2}{z^n(1-z)^2}dz\\
&=&\frac{1}{2\pi i}\int_{|z|=1}\frac{1}{z^n}(\sum_{k=0}^{n-1}z^{k})^2\\
&=&\frac{1}{2\pi i}\int_{|z|=1}\frac{1}{z^n}(\cdots+nz^{n-1}+\cdots)dz\\
&=&\frac{1}{2\pi i}2\pi i n=n.
\end{eqnarray*}


TITLE: Continuity of a function from the reals to the reals that fixes rational numbers
QUESTION [0 upvotes]: Define $f:\mathbb{R} \to\mathbb{R}$ by $x\mapsto x$ if $x$ is rational and $x\mapsto 0$ if $x$ is irratoinal.
Prove that $f$ is continuous at a point $a\in\mathbb{R}$ if and only if $a=0$.
I'm struggling to grasp how to approach this, could someone point me in the right direction?

REPLY [0 votes]: Continuity means if we force $x$ and $y$ close together we force $f(x)$ and $f(y)$ close together.
If $x \ne 0$ then know matter how close we force $x$ and $y$ together there will be both irrationals and rationals close $x$ and but the differences between $f(rational)$ and $f(irrational)$ will about $x$.  So we aren't forcing them closer than that.
Or more formally:
Let $x \ne 0$
Let's let $x > \epsilon > 0$.
Case 1: $x$ is rational. $f(x) = x > \epsilon$.  For any $\delta > 0$ we can always find an irrational $y$ such that $|x - y| < \delta$.  But $|f(x) - f(y)| = |x - 0| = x > \epsilon$.  We can not find a $\delta > 0$ where $|x - y| < \delta$ forces $|f(x) - f(y)| < \epsilon$.  So $f$ is not contionuous at x.
Case 2: $x$ is irrational:  For any $\delta > 0$ we can always find a rational $r$ such that $x > r > \max(x - \delta, \epsilon)$.  So $|x -r| < \delta$ and $|f(x) - f(r)| = |0 - r| = r > \epsilon$.  We can not find a $\delta > 0$ where $|x - y| < \delta$ forces $|f(x) - f(y)| < \epsilon$.  So $f$ is not contionuous at x.


text:
/-
Copyright (c) 2021 Yakov Pechersky. All rights reserved.
Released under Apache 2.0 license as described in the file LICENSE.
Authors: Yakov Pechersky
-/

import data.fintype.basic
import group_theory.perm.sign
import logic.equiv.basic

/-! # Equivalence between fintypes

This file contains some basic results on equivalences where one or both
sides of the equivalence are `fintype`s.

# Main definitions

 - `function.embedding.to_equiv_range`: computably turn an embedding of a
   fintype into an `equiv` of the domain to its range
 - `equiv.perm.via_fintype_embedding : perm α → (α ↪ β) → perm β` extends the domain of
   a permutation, fixing everything outside the range of the embedding

# Implementation details

 - `function.embedding.to_equiv_range` uses a computable inverse, but one that has poor
   computational performance, since it operates by exhaustive search over the input `fintype`s.
-/

variables {α β : Type*} [fintype α] [decidable_eq β] (e : equiv.perm α) (f : α ↪ β)

/--
Computably turn an embedding `f : α ↪ β` into an equiv `α ≃ set.range f`,
if `α` is a `fintype`. Has poor computational performance, due to exhaustive searching in
constructed inverse. When a better inverse is known, use `equiv.of_left_inverse'` or
`equiv.of_left_inverse` instead. This is the computable version of `equiv.of_injective`.
-/
def function.embedding.to_equiv_range : α ≃ set.range f :=
⟨λ a, ⟨f a, set.mem_range_self a⟩, f.inv_of_mem_range, λ _, by simp, λ _, by simp⟩

@[simp] lemma function.embedding.to_equiv_range_apply (a : α) :
  f.to_equiv_range a = ⟨f a, set.mem_range_self a⟩ := rfl

@[simp] lemma function.embedding.to_equiv_range_symm_apply_self (a : α) :
  f.to_equiv_range.symm ⟨f a, set.mem_range_self a⟩ = a :=
by simp [equiv.symm_apply_eq]

lemma function.embedding.to_equiv_range_eq_of_injective :
  f.to_equiv_range = equiv.of_injective f f.injective :=
by { ext, simp }

/--
Extend the domain of `e : equiv.perm α`, mapping it through `f : α ↪ β`.
Everything outside of `set.range f` is kept fixed. Has poor computational performance,
due to exhaustive searching in constructed inverse due to using `function.embedding.to_equiv_range`.
When a better `α ≃ set.range f` is known, use `equiv.perm.via_set_range`.
When `[fintype α]` is not available, a noncomputable version is available as
`equiv.perm.via_embedding`.
-/
def equiv.perm.via_fintype_embedding : equiv.perm β :=
e.extend_domain f.to_equiv_range

@[simp] lemma equiv.perm.via_fintype_embedding_apply_image (a : α) :
  e.via_fintype_embedding f (f a) = f (e a) :=
begin
  rw equiv.perm.via_fintype_embedding,
  convert equiv.perm.extend_domain_apply_image e _ _
end

lemma equiv.perm.via_fintype_embedding_apply_mem_range {b : β} (h : b ∈ set.range f) :
  e.via_fintype_embedding f b = f (e (f.inv_of_mem_range ⟨b, h⟩)) :=
by simpa [equiv.perm.via_fintype_embedding, equiv.perm.extend_domain_apply_subtype, h]

lemma equiv.perm.via_fintype_embedding_apply_not_mem_range {b : β} (h : b ∉ set.range f) :
  e.via_fintype_embedding f b = b :=
by rwa [equiv.perm.via_fintype_embedding, equiv.perm.extend_domain_apply_not_subtype]

@[simp] lemma equiv.perm.via_fintype_embedding_sign [decidable_eq α] [fintype β] :
  equiv.perm.sign (e.via_fintype_embedding f) = equiv.perm.sign e :=
by simp [equiv.perm.via_fintype_embedding]

namespace equiv
variables {p q : α → Prop} [decidable_pred p] [decidable_pred q]

/-- If `e` is an equivalence between two subtypes of a fintype `α`, `e.to_compl`
is an equivalence between the complement of those subtypes.

See also `equiv.compl`, for a computable version when a term of type
`{e' : α ≃ α // ∀ x : {x // p x}, e' x = e x}` is known. -/
noncomputable def to_compl (e : {x // p x} ≃ {x // q x}) : {x // ¬ p x} ≃ {x // ¬ q x} :=
classical.choice (fintype.card_eq.mp (fintype.card_compl_eq_card_compl _ _ (fintype.card_congr e)))

/-- If `e` is an equivalence between two subtypes of a fintype `α`, `e.extend_subtype`
is a permutation of `α` acting like `e` on the subtypes and doing something arbitrary outside.

Note that when `p = q`, `equiv.perm.subtype_congr e (equiv.refl _)` can be used instead. -/
noncomputable abbreviation extend_subtype (e : {x // p x} ≃ {x // q x}) : perm α :=
subtype_congr e e.to_compl

lemma extend_subtype_apply_of_mem (e : {x // p x} ≃ {x // q x}) (x) (hx : p x) :
  e.extend_subtype x = e ⟨x, hx⟩ :=
by { dunfold extend_subtype,
     simp only [subtype_congr, equiv.trans_apply, equiv.sum_congr_apply],
     rw [sum_compl_apply_symm_of_pos _ _ hx, sum.map_inl, sum_compl_apply_inl] }

lemma extend_subtype_mem (e : {x // p x} ≃ {x // q x}) (x) (hx : p x) :
  q (e.extend_subtype x) :=
by { convert (e ⟨x, hx⟩).2,
     rw [e.extend_subtype_apply_of_mem _ hx, subtype.val_eq_coe] }

lemma extend_subtype_apply_of_not_mem (e : {x // p x} ≃ {x // q x}) (x) (hx : ¬ p x) :
  e.extend_subtype x = e.to_compl ⟨x, hx⟩ :=
by { dunfold extend_subtype,
     simp only [subtype_congr, equiv.trans_apply, equiv.sum_congr_apply],
     rw [sum_compl_apply_symm_of_neg _ _ hx, sum.map_inr, sum_compl_apply_inr] }

lemma extend_subtype_not_mem (e : {x // p x} ≃ {x // q x}) (x) (hx : ¬ p x) :
  ¬ q (e.extend_subtype x) :=
by { convert (e.to_compl ⟨x, hx⟩).2,
     rw [e.extend_subtype_apply_of_not_mem _ hx, subtype.val_eq_coe] }

end equiv

subset_name: curated