TITLE: Prove $A=\{\,ra+na\mid r\in R, n\in\Bbb{Z}\,\}$ is an ideal.
QUESTION [3 upvotes]: Let $R$ be a commutative ring, and let $a\in R$. Prove that $A=\{\,ra+na\mid r\in R, n\in\Bbb{Z}\,\}$ is an ideal belonging to $R$.
Remember that we cannot assume that the ring is unital.
I got stumped on this question, and could not proceed.
My approach: it is easy to prove that for all $p,q\in A$, $p-q\in A$. However, what is difficult to prove without assuming $R$ is unital is the fact that $A$ is closed under multiplication with elements of $R$.
Ay help would be appreciated.

REPLY [2 votes]: Note that $s\cdot(ra+na) = (sr+ns)a+0a$.

REPLY [0 votes]: Let $b = (ra+na)\in A$ and $c\in R$. We have
$$ bc = a(cn+cr) = a(c(n+r) + 0c). $$
Since $c(n+r)\in R$ and $0\in \mathbb{Z}$ we conclude that $bc\in A$ and thus the closeness under multiplication holds.


TITLE: Please check: $f:X\longrightarrow Y$ perfect map. Let $X$ be second countable, prove $Y$ is second countable
QUESTION [0 upvotes]: Let $B=\{B_n\}_{n\in\mathbb{N}}$ be countable base for $\tau_x$ y $V=\{V_n\mid V_n=\bigcup_{i\in F} B_i\}$, $F\subset\mathbb{N}$ finite and $|V|\leq\aleph_0$. We define $W_n=T-f(X-V_n)$ y $W=\{W_n\}$ $|W|\leq\aleph_0$. Let,s see $W$ is a base for $\tau_Y$: $\forall U\in\tau_Y\;\forall y\in W \,\exists W_x\in U$ such $x=\bigcup_{n\in\mathbb{N}} B_n$. $y\in W\Longrightarrow f^{-1}(y)\in f^{-1}(W)$ and $f^{-1}(y)$ compact $\Longrightarrow$ $f^{-1}(y) \subset\bigcup_{n\in\mathbb{N}} B_n\Longrightarrow f^{-1}(y)\in V_k\in V$. Finally $f^{-1}(y)\cap (X-V_k)=\emptyset\Longrightarrow y\notin f(X-V_k)\Longrightarrow y\in Y-f(X-V_k)=W_k$.

REPLY [5 votes]: You have the right idea, but your proof is just a mess of symbols and arrows and has no clear structure. Tell the proof like a story, with more explanation.
The same proof (slightly different notation) presented in my style:
Let $\{B_n: n \in \Bbb N\}$ be a countable base for $X$.
Let $\mathcal{B}$ be the set of all finite unions of base sets:
$$\mathcal{B}=\{\bigcup_{n \in F} B_n \mid F \subseteq \Bbb N \text{ finite }\}$$
and note that $|\mathcal{B}|\le \aleph_0$ because $\mathbb{N}$ has countably many finite subsets.
Now define $$\mathcal{B}' = \{Y\setminus f[X\setminus B] \mid B \in \mathcal{B}\}$$
And note that this is an at most countable (as $\mathcal{B}$ is) set of open subsets of $Y$ (as $X\setminus B$ is closed and $f$ is a closed map).
We claim that $\mathcal{B}'$ is a base for $Y$ and if we show this, it follows that $Y$ is second countable too. In order to show this, let $y \in Y$ and $O$ open in $Y$ with $y \in O$.
Then $f^{-1}[O]$ is open in $X$ (continuity of $f$) and contains $f^{-1}[\{y\}]$, which is compact (we must assume $f$ is onto so that this set is non-empty too), and so we can pick $n(x) \in \Bbb N$ such that $$x \in B_{n(x)} \subseteq f^{-1}[O]$$ for all $x \in f^{-1}[\{y\}]$. Then the $B_{n(x)}$ cover the compact set $f^{-1}[\{y\}]$ and so there is a finite subset $F$ of $\Bbb N$ such that
$$f^{-1}[\{y\}] \subseteq \bigcup_{n \in F} B_n\subseteq f^{-1}[O]$$
And $B_y:=\bigcup_{n \in F} B_n \in \mathcal{B}$ so that $Y\setminus f[X\setminus B_y] \in \mathcal{B}'$. The proof is finished by noting that
$$y \in Y\setminus f[X\setminus B_y] \subseteq O$$
(suppose $y \notin Y\setminus f[X\setminus B_y]$ then $y \in f[X\setminus B_y]$, so $y=f(x')$ for some $x' \notin B_y$, but then also $x' \in f^{-1}[\{y\}]$ so $x' \in B_y$ contradiction. The inclusion is shown similarly).
QED.


TITLE: Homeomorphism from interior of unit disk and 0 as a fixed point
QUESTION [1 upvotes]: $f$ is analytic in $|z|<1$, $|f(z)|<1$ and $f(0)=0$, $f'(0)=\alpha \ne 0$
Find a sharp estimate of the radius $r$ of a disc such that for:
$z1$,$z2$ in the disc: $|z1| \lt r$ and $|z2| \lt r$
$$ f(z1)=f(z2) \Rightarrow z1=z2$$
Cauchy's integral theorem can help me to prove that f is injective:
\begin{align}
f(z1) - f(z2) &= \frac{1}{2\pi i} \int_{C_r} (\frac{1}{t-z1}-\frac{1}{t-z2}) f(t) dt \\
&= \frac{z1-z2}{2\pi i}  \int_{C_r} \frac{f(t)} {(t-z1)(t-z2)}dt
\end{align}
And then use the fact $f'(0)$ being non zero then $z1=z2$ in the neighborhood of $0$.
But I don't think this is rigorous and also it does not give me a method to find the estimate.

REPLY [3 votes]: Theorem (Landau-Dieudonne):
Let $f$ be analytic in $|z|<1,$ $|f(z)|<1$ and $f(0)=0,$ $f^\prime(0)=\alpha \ne 0.$
Then $f(z)$ is univalent in $|z|<r=\frac{|\alpha |}{1+\sqrt{1-|\alpha |^2}}$.
The estimate $r=\frac{|\alpha |}{1+\sqrt{1-|\alpha |^2}}$ is sharp.
Addendum:
Lemma 1. Suppose that $g(z)$ is analytic in $|z|<1$, $|g(z)|<1$. Then $$
\frac{|g(0)|-|z|}{1-|g(0)||z|}\le |g(z)|\quad (|z|<1).$$
This is frequently asked in this site, see   Schwarz lemma problem
for instance.
Lemma 2. Suppose that $f(z)$ is analytic in $|z|<1$, $|f(z)|<1$, $f(0)=0$. Then $$
\frac{\left(|f^\prime(0)|-|z|\right)|z|}{1-|f^\prime(0)||z|}\le |f(z)|\quad (|z|<1).$$
Proof. Apply Lemma 1 to $g(z)=\frac{f(z)}{z}.$ Note that $g(0)=f^\prime (0).$
Lemma 3.  Suppose that $f(z)$ is analytic in $|z|<1$, $|f(z)|<1$, $f(0)=0$. If there exist $z_1,z_2$ $(z_1\ne z_2)$ with $|z_1|=|z_2|=r$, $f(z_1)=f(z_2)=\omega$, then $|\omega|\le r^2$.
Proof. Let $$
F(z)=\frac{f(z)-\omega}{1-\bar{\omega}f(z)}\cdot \frac{1-\bar{z_1}z}{z-z_1}\cdot\frac{1-\bar{z_2}z}{z-z_2}.$$
Note that $F(z)$ is analytic in $|z|<1$ and $|F(z)|\le 1$. Especially we have $$|F(0)|=\left|\frac{\omega}{z_1z_2}\right|\le 1.$$ So we have $|\omega|\le |z_1z_2|=r^2.$
Now we prove Landau-Dieudonne's Theorem.
Suppose that $f(z)$ is not univalent in $|z|< \rho $. Then there are two points $z_1,z_2$ ( $|z_1|<\rho ,\,|z_2|<\rho )$ such that $f(z_1)=f(z_2)=\omega .$
Let $C_\rho =\{z:\, |z|=\rho \}$ and $\Gamma _\rho =f(C_\rho ),$ the image of $C_\rho $ under $f$.
Since $f(z)$ has (at least) two $\omega $-points $z_1$ and $ z_2,$ $$
\frac{1}{2\pi} \int_{C_\rho } d\arg (f(z)-\omega )=\frac{1}{2\pi i}\int_{C_\rho }\frac{f^\prime(z)}{f(z)-\omega }dz\ge 2.\tag{1}$$
Suppose that $\Gamma _\rho $ is a Jordan curve. Then
$$
\frac{1}{2\pi} \int_{C_\rho } d\arg (f(z)-\omega )=1,$$
since $\omega $ lies inside $\Gamma _\rho$. This is contrary to $(1)$. Therefore $\Gamma _\rho $ is not a Jordan curve, and hence there are two points $\zeta _1,\,\zeta _2$ on $ C_\rho $ such that $f(\zeta _1)=f(\zeta _2)$. By lemma 3, we have $$
|f(\zeta_1)|\le \rho ^2.$$
Lemma 2 says that
$$
\frac{\left(|\alpha |-|\zeta _1|\right)|\zeta _1|}{1-|\alpha ||\zeta _1|}\le |f(\zeta _1)|,$$
and hence we have $$
\frac{\left(|\alpha |-\rho \right)\rho }{1-|\alpha |\rho }\le \rho ^2,$$
which leads to $$
\rho \ge \frac{|\alpha |}{1+\sqrt{1-|\alpha |^2}}.$$
Thus we have proven that $f(z)$ is univalent in $|z|<r=\frac{|\alpha |}{1+\sqrt{1-|\alpha |^2}}.$
About sharpness:
Consider $f(z)=\frac{z(\alpha -z)}{1-\alpha z}$ $(0<\alpha <1)$ with $f(0)=0,\, f^\prime(0)=\alpha $. This function satisfies $f^\prime(z_0)=0$ at $z_0=\frac{\alpha }{1+\sqrt{1-\alpha ^2}}$. Therefore $f(z)$ is univalent only in $|z|<\frac{\alpha }{1+\sqrt{1-\alpha^2}}$, in other words, the estimate is sharp.


text:
\section{Conclusions.}
In this paper we proposed a method for rigorous verification of uniform hyperbolicity for maps. The method has been successfully applied to a Poincar\'e map of a non-autonomous ODE in four dimensions. We believe the method can be applied to other systems.

Kuznetsov \cite{K2} proposed also a non-autonomous system on the plane which apparently possesses an attractor of Plykin type in the Poincar\'e map. The equations are quite complicated and given by
\begin{equation}\label{eq:plykin}
  \begin{array}{rcl}
 \dot x &=& -2\varepsilon y^2\Omega_1(x,y,t)
      \left(
          \cos\left(\frac{\pi}{4}\cos\frac{\pi}{2}t\right)
          -x\sin\left(\frac{\pi}{4}\cos\frac{\pi}{2}t\right)
      \right) + \\
      &&  Ky\Omega_2(x,y,t)\left(
          \cos\left(\frac{\pi}{4}\sin\frac{\pi}{2}t\right)
          -x\sin\left(\frac{\pi}{4}\sin\frac{\pi}{2}t\right)
      \right)\sin\frac{\pi}{2}t,\\\\
  \dot y &=& \varepsilon y\Omega_1(x,y,t)
      \left(
        2x\cos\left(\frac{\pi}{4}\cos\frac{\pi}{2}t\right)+(1-x^2+y^2)\sin\left(\frac{\pi}{4}\cos\frac{\pi}{2}t\right)
      \right) -\\
    && K\Omega_2(x,y,t)\left(
      x\cos\left(\frac{\pi}{4}\sin\frac{\pi}{2}t\right)+\frac{1}{2}(1-x^2+y^2)\sin\left(\frac{\pi}{4}\sin\frac{\pi}{2}t\right)
    \right)\sin\frac{\pi}{2}t,
  \end{array}
\end{equation}
where
\begin{eqnarray*}
 \Omega_1(x,y,t) &=& \frac{2x\cos\left(\frac{\pi}{4}\cos\frac{\pi}{2}t\right)+(1-x^2-y^2)\sin\left(\frac{\pi}{4}\cos\frac{\pi}{2}t\right)}{\left(1+x^2+y^2\right)^2},\\
 \Omega_2(x,y,t) &=& \frac{-2x\sin\left(\frac{\pi}{4}\sin\frac{\pi}{2}t\right)+(1-x^2-y^2)\cos\left(\frac{\pi}{4}\sin\frac{\pi}{2}t\right)}{\left(1+x^2+y^2\right)^2} + \frac{1}{\sqrt 2}.
\end{eqnarray*}
Kuznetsov \cite{K2} did a numerical study of this system and observed that for a wide range of parameter values around $K=1.9$ and $\varepsilon=0.72$ the Poincar\'e map has a hyperbolic attractor.
A rigorous enclosure of the attractor for $K=1.9$, $\varepsilon=0.72$ resulting from Algorithm~\ref{alg:enclosure} with the resolution $k=9$ is shown in Fig.~\ref{fig:plykin}. The resolution $k=9$ was not enough to perform the step of verification of uniform hyperbolicity of this map. We also enclosed this attractor with resolution $k=12$ but again this was not enough to verify the hyperbolicity. Further subdivision was not possible due to memory limitations in the computer that we used.

\begin{figure}[htbp]
 \centerline{
    \includegraphics[width=0.75\textwidth]{plykin.pdf}
  }
\caption{Enclosure of the attractor of the Poincar\'e map for the system (\ref{eq:plykin}) with parameter values $K=1.9$, $\varepsilon=0.72$ and with the resolution $k=9$. \label{fig:plykin}}
\end{figure}

Computations for the system (\ref{eq:plykin}) are much harder than for (\ref{eq:vpo}). First - the vector field is a quite large expression and even if the system is lower dimensional, it is more complicated for rigorous integration. Second, and most important; as easily seen in Fig.~\ref{fig:plykin}, the attractor is very thick and harder to cover than the almost one dimensional Smale solenoid. An approximate Hausdorff dimension reported in \cite{K2} is ca $1.7$. We reached the limit of available memory on our computer when enclosing the attractor with a thinner cover. We believe, however, that some further optimizations in data structures will help to prove hyperbolicity of this system. A
straightforward idea is to use non-uniform covers.

\subsection{General invariant sets.}

In this paper we verified hyperbolicity of attractors. But the method can be easily extended to invariant sets. The only part which requires some modification is the method for generating an enclosure of an invariant set. If the invariant set is an attractor, the strategy of ``inner'' enclosure is better - Algorithm~\ref{alg:enclosure}. For a general invariant set an ``outer'' approximation is necessary. We also implemented ``outer'' approximation for enclosing of general invariant sets but we do not report details here, since it is rather standard procedure. As a test case we proved that the well known H\'enon map \cite{H}
\begin{equation*}
\mathcal H_{a,b}(x,y) = (1 + y- ax^2,bx)
\end{equation*}
is uniformly hyperbolic on the invariant set for the parameter values $a=5.4$ and $b=-1$.

As mentioned above, we used an ``outer'' strategy to enclose the invariant set -- the result is shown in Fig.~\ref{fig:henon} and it consists of $8832$ boxes. Then we applied algorithms for finding cycles and periodic points, computing of coordinate systems and verification of the cone condition with the quadratic form $$Q=\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}.$$

The program executes within less than $1$ second on a laptop-type computer with the Intel Core 2 Duo 2GHz processor. A computer assisted proof for the same parameter values is presented in \cite{A,MT}. In \cite{MT} the authors report the time of computations less than $10$ seconds on a comparable CPU.

\begin{figure}[htbp]
 \centerline{
    \includegraphics[width=0.75\textwidth]{henon.pdf}
  }
\caption{An enclosure of the invariant set for $\mathcal H_{a,b}$ map with the parameter values $a=5.4$, $b=-1$. \label{fig:henon}}
\end{figure}

\subsection{Implementation.}
The C++ program for verifying the hyperbolicity has been implemented by the author and the source code is available at \cite{W}. The code is highly parallelized using the Open MP library supported by the compilers gcc-4.2 or newer. The code is written in a very generic way; it heavily uses template techniques. In fact, an input to the algorithms is an abstract map (template parameter) for which we assume that we know how to compute values and derivatives. In the case of Poincar\'e maps we used for this purpose the $C^0$ and $C^1$ solvers from the \cite{CAPD} library -- see also \cite{Z1} for an efficient algorithm for integration of first order variational equations.

All computations regarding Theorem~\ref{thm:main} and Theorem~\ref{thm:nontrivAttractor} were performed on a cluster with $7$ computers, each having $8$ Quad-Core AMD Opteron(tm)
8354, 2.2GHz processors with 64GB of RAM. During the computations the maximal usage of the memory was 32\%.

config: arxiv


text:
\begin{document}

\subsection{Mutual Information in Video Streams}
\label{sec:mi}
We consider the problem of transferring information from an input visual stream to the output space of a multi-layer convolutional network with $\ell$ layers, that processes each frame and yields $m$ \textit{pixel-wise} predictions. This corresponds to the maximization of the Mutual Information (MI) from the pixels of the input frames to the $m$-dimensional output space yielded by the $m$ units of the last layer, being $m$ the size of the filter bank in layer $\ell$. Hyperbolic tangent is used as activation function in each layer $j < \ell$, while the last layer is equipped with a softmax activation, generating $m$ probabilities \draftA{maybe $m+1$?}
$p(w,x,u)=(p_0(w,x,u),\ldots,p_{m}(w,x,u))$, being $x$ a pair of pixel coordinates and $u$ the processed frame.
This problem is studied in \cite{betti-nn} and related papers \cite{betti-ijcai,betti-tnnls}, where single-layer models (or stacks of sequentially trained single-layer models) are considered, while, in this paper, we exploit a deep network trained end-to-end. Previous approaches based on kernel machines can be found in \cite{gori2016semantic,10.1007/978-3-642-33783-3_62}.

In order to define the MI index, we consider a generic, non-time dependent weight configuration $\omega\in\R^n$.
We introduce the average output activation on the video portion between time instants $t_1$ and $t_2$,
\begin{equation}
P(\omega;\nu;t_1,t_2) = \int_{t_1}^{t_2} \int_{\Retina} p(\omega,x,u(\tau))\, d\nu(x,\tau)
\label{eq:quantities}
\end{equation}
where $\Retina$ is the set of points that constitute the retina and $\nu$ is a measure in $\Retina\times[t_1,t_2]\subset\R^2\times\R$.
The MI index over the video portion $[t_1,t_2]$, is defined as
\begin{eqnarray}
\nonumber \hskip-4mm \MI(X,Y;\omega;\nu;t_1,t_2) &\hskip-2mm=\hskip-2mm& - H(Y|X;\omega) + H(Y;\omega)   \\
&\hskip-60mm=\hskip-2mm&\hskip-30mm  -\sum_{j=0}^{m} \int_{t_1}^{t_2}  \hskip-2.5mm\int_{X}\hskip-1mm p_j(\omega,x,u(\tau))\log p_j(\omega,x,u(\tau))\, d\nu(x,\tau) +\sum_{j=0}^{m}P_j(\omega,t) \log P_j(\omega,t)
\label{eq:mi}
\end{eqnarray}
where $H(\,\cdot\,;\omega)$ is the entropy function, and $X$ and $Y$ are random variables ($Y$ is discrete) associated with the
input\footnote{Since we are dealing with convolutional feature a realization of the random variable $X$ is specified
by the coordinates of a point $x\in\Retina$, the value of the temporal instant $t$ and the value of the video $u(t)$.}
and output space, respectively\footnote{When selecting a $\log$ in base $m$, the MI is in $[0,1]$, that is what we will assume in the rest of the paper.}. When no further information is available, $\nu$ is commonly assumed to be uniform in time and space $d\nu(x,t)=C\, dxdt$ and it is normalized such that
$\int_{t_1}^{t_2}\int_\Retina 1\, d\nu= 1$.
\draftA{$\mu_t\otimes \delta_t$ where $\mu_t$ has density function $\mu(\cdot,t)$}
Online learning requires local in time lagrangian, thus
\[
U(\omega,u(t)) = -\MI(X,Y;\omega;\mu_t\otimes \delta_t,0,T),
\]
where $\mu_t\otimes \delta_t$ is the product measure of $\mu_t$ which we assume to have
density function $\mu(x,t)$ and the dirac measure at temporal point $t$.

Performing maximum-MI-based online learning of $w$ using the CALs in the time horizon $[0,T]$ is not straightforward. Once we restore the dependency of $w$ on time, $\hat{w}=w(t)$, we cannot simply plug (minus) the MI index as a potential loss $U$ in the Lagrangian due to the lack of temporal locality. As a matter of fact, in order to implement online learning dynamics, $U$ is temporally local, i.e., it depends on $u(t)$ only. The authors of \cite{betti-nn} relax the MI definition, assuming that the potential loss in the Lagrangian is limited to $u(t)$, that is $U(w(t),u(t)) = -\MI(X,Y|w(t),t,t)$ \draftA{questa quantit\`a \`e 0}, and the integration over time is restored by the outer integration in the functional of Eq.~(\ref{eq:fun3}) (or, equivalently, in the one of  Eq.~(\ref{eq:fun2})).
A drawback of this formulation is that, due to this temporal assumption, it could lead to a loose approximation of the original term $H(Y|\cdot,\cdot,\cdot)$ of Eq.~(\ref{eq:mi}), for which the inner integration on time (Eq.~(\ref{eq:quantities})) is lost, and replaced by the outer integration of the functional. In order to better cope with the optimization dynamics, the two entropy terms are commonly weighted by positive scalars $\lambda_c$, $\lambda_e$. \draftA{Io farei cos\`i: per ottenere $U(w(t),u(t))$ sceglerei $d\mu(x,s)=dG(x,t)d\delta_t(s)$, poi per
il fuoco dell'attenzione baster\`a porre $G(x,t)=g(x-a(t))$ con $t\mapsto a(t)$ data.}

In addiction to the vanilla case we just described (referred to as \textsc{PLA}), we explore two other alternative criteria to mitigate the impact of time locality, that we will evaluate in Section~\ref{sec:exp}. The first one (\textsc{VAR}) consists in introducing an additional auxiliary variable $s(t)$, that is used to replace $P(w(t),t)$, while its variation, $\dot{s}(t)$, is constrained to be almost equivalent to $P(w(t),t)$. The Lagrangian is augmented with $\lambda_s \left|\dot{s}(t) - P(w(t),t) \right|^2$, a soft-constraint that enforces $s(t)$ to approximate the case in which $P(w(t),t)$ is not limited to the last frame ($\lambda_s>0$).\footnote{Probabilistic normalization must be enforced after every update of $s(t)$.}
This idea is presented in \cite{betti-nn} but not followed-up in any experimentation. As a second criterion (\textsc{AVG}), we propose to replace $P(w(t),t)$ with the outcome of an averaging operation that keeps track of the past activation of the output units, i.e., $\zeta_s P(w(t'),t') + (1-\zeta_s) P(w(t),t)$, for two consecutive time instants $t>t'$.

\section{Focus of Attention}
\label{sec:foa}

The way video data is commonly processed by machines usually lacks a key property of the human visual perception, that is the capability of exploiting eye movements to perform shifts in selective visual attention. High visual acuity is restricted to a small area in the center of the retina (fovea), and the purpose of the Focus Of Attention (FOA) is to selectively orient the gaze toward relevant areas with high information, filtering out irrelevant information from cluttered visual scenes \cite{McMains2009, kowler2010attention,zanca2019gravitational}.

In the context of Section~\ref{sec:mi}, we consider a visual stream and a neural architecture with $m$ output dimensions (per pixel), and we aim at developing the network weights $w$ such that the MI index is maximized as strongly as possible with respect to the model capacity.
Of course, restricting the attention to a subset of the spatio-temporal coordinates of the video, due to a FOA mechanism, seems to inherently carry less information than when considering the whole video. However, in the latter case, the processed data will be characterized by a larger variability, mixing up noisy/background information with what could be more useful to understand the video. Such mixture of data could be harder to disentangle by a learning model than well-selected information coming from a human-like FOA trajectory, leading to a worse MI estimate. Curiously, the learning process restricted to the FOA trajectory could end-up in facilitating the development of the weights, so that the MI computed on the whole frame area could be larger than when learning without restrictions.

Following the notation of Eq.~(\ref{eq:mi}), the MI maximization, for each $t$, is based on the spatial distribution ${\mu}(x,t)$ . Such distribution models the relevance of each coordinate $x$ when learning from frame $u(t)$. In \cite{betti-tnnls,betti-ijcai}, ${\mu}(x,t)$ is assumed to be uniform over the frame area, while in \cite{betti-nn} it is also described the idea of considering ${\mu}$ ($f$ in \cite{betti-nn}) as the most natural candidate for implementing a FOA-based mechanism. Let us assume that $a(t)$ are the spatial coordinates of the FOA at time $t$, then we define
\begin{equation}
\mu(x,t) = g(x - a(t)),
\label{eq:foa}
\end{equation}
being $g$ a function that is peaked on $a(t)$.

Following this parametrization of $\mu$, we borrow a state-of-the art model for scanpath $a(t)$ prediction defined in \cite{zanca2019gravitational}, that shares a physics-inspired formulation as CALs.
Such FOA model has been proven to be strongly human-like in free-viewing conditions \cite{DBLP:journals/corr/abs-2002-04407}. It is based on the intuition that the attention emerges as a gravitational process, in which  both low-level (gradient, contours, motion) or high-level features (objects, context) may act as gravitational masses. In particular, given the gravitational field $E(t,a(t))$ generated by the selected features, the law that drives the attention is given by
\begin{equation}
\ddot{a}(t) + \rho \dot{a}(t) -E(t,a(t)) = 0,
\label{eq:geymol}
\end{equation}
that is indeed another 2nd order model as the one we proposed in Section~\ref{sec:2nd} (see \cite{zanca2019gravitational} for more details). The dissipation is controlled by $\rho>0$, and the importance of each mass can also be tuned. Interestingly, Eq.~(\ref{eq:geymol}) describes the dynamics of the FOA, and it is not based on pre-computed or given saliency maps.
In this paper, following \cite{zanca2019gravitational},  we consider two basic (low-level) perceptive features as masses, the spatial gradient of the brightness and the strength of the motion field. The trajectories simulated by the model show the same patterns of movement characteristic of human eyes: \textit{fixations}, when the gaze remains still in a location of interest; \textit{saccades}, rapid movement to reallocate attention on a new target; \textit{smooth pursuit}, slow movements performed in the presence of a visual feedback with the purpose of tracking a stimulus.

Different choices on $g$ are possible. In Section~\ref{sec:exp} we will consider the extreme case in which $g(x-a(t))$ is a Dirac delta on the coordinates $a(t)$ (we will refer it to as \textsc{FOA}), so that $\mu(x,t)$ is essentially a mono-dimensional signal. A less extreme setting is the one in which $g(x-a(t))$ is, for example, a rectangular widow centered in $a(t)$ that covers a small fraction of the frame (\textsc{FOAW}), while the most-relaxed setting is when $g$ is simply uniform on the whole frame (\textsc{UNI}), i.e., $a(t)$ is not used.

\begin{table*}
\centering
\footnotesize
\caption{Main result. Mutual Information (MI) index in three video streams, considering three neural architectures (\textsc{S}, \textsc{D}, \textsc{DL}). Each column (starting from the third one) is about the results of network trained using an input probability densities taken from \{\textsc{UNI}, \textsc{FOA}, \textsc{FOAW}\}, and tested measuring the MI index in all the three density cases (labeled in column
``Test'').}
\resizebox{0.98\textwidth}{!}{
\begin{tabular}{cc|ccc|ccc|ccc}

\toprule

& &  \multicolumn{3}{c}{S} & \multicolumn{3}{c}{D} & \multicolumn{3}{c}{DL}  \\
\cmidrule(l){3-5}    \cmidrule(l){6-8}  \cmidrule(l){9-11}
Stream & Test & 	UNI & FOA & FOAW  & 	UNI & FOA & FOAW  & 	UNI & FOA & FOAW \\
\toprule
 &    UNI   & 0.017 &  \textbf{0.112} &  0.078 &  0.004 &  \textbf{0.144} &  0.020 &  0.012 &  \textbf{0.132} &  0.026 \\
SparseMNIST & FOA & 0.239 &  \textbf{0.486} &  0.391 &  0.103 &  \textbf{0.431} &  0.229 &  0.146 &  \textbf{0.350} &  0.194\\
& FOAW & 0.154 &  \textbf{0.209} &  0.197 &  0.144 &  \textbf{0.255} &  0.157 &  0.117 &  \textbf{0.215} &  0.131 \\
\midrule

& UNI & \textbf{0.776} &  0.601 &  0.695 &  0.653 &  0.556 & \textbf{0.745} &  0.445 &  0.292 &  \textbf{0.496}  \\
Carpark   &   FOA & \textbf{0.742} &  0.675 &  0.694 &  0.678 &  0.639 &  \textbf{0.768} &  0.477 &  0.315 &  \textbf{0.529}  \\
& FOAW & \textbf{0.719} &  0.629 &  0.671 &  0.653 &  0.601 &  \textbf{0.721} &  0.501 &  0.357 &  \textbf{0.532} \\

\midrule
& UNI &  \textbf{0.329} &  0.314 &  0.315 &  0.339 &  \textbf{0.556} &  0.350 &  0.208 &  \textbf{0.304} &  0.218  \\
Call &   FOA   & \textbf{0.405} &  \textbf{0.405} & 0.371 &  0.430 &  \textbf{0.582} &  0.492 &  0.246 &  \textbf{0.365} &  0.270  \\  & FOAW & \textbf{0.429} & 0.420 &  0.413 &  0.442 &  \textbf{0.566} &  0.457 &  0.304 &  \textbf{0.374} &  0.310  \\
\bottomrule
\end{tabular}}
\label{tab:exp_mi_main}
\end{table*}

\end{document}


TITLE: Triangle with circumcircle- calculate angle measure
QUESTION [0 upvotes]: Moderator Note: At the time that this question was posted, it was from an ongoing contest. The relevant deadline has now passed.

Let $ABC$ be a triangle with circumcicle $\omega$. Then the bisector of $\angle ABC$ meets $AC$ at $D$ and circle $\omega$ at $M \ne B$. The circumcircle of $\triangle BDC$ meets $AB$ at $E\ne B$, and $CE$ meets $\omega$ at $P\ne C$. The bisector of $\angle PMC$ meets $AC$ at $Q \ne C$. If $PQ = MC$, what is $\angle ABC$?

REPLY [1 votes]: This sounds like homework. Here's a hint: use the inscribed angle theorem, which states two things of interest to you:

The angle $\angle ABC$ inscribed within a circle by points $A$, $B$, and $C$ located on the edge of the circle is always exactly half of the central angle of the intercepted arc between B and C.
Opposing vertices of any quadrangle that can be inscribed by a circumcircle (a cyclic quadrangle) have supplementary angles.

These allow you to form the following lemmas:

The arcs $AM$ and $MC$ each have the same central angle measure as $\angle ABC$.
The angles $\angle EDC$ and $\angle EBC$ are supplementary, and the angles $\angle BED$ and $\angle BCD$ are supplementary, because by definition they form a quadrangle within the circumcircle of $\triangle BCD$.
By the same token, angles $\angle PBC$ and $\angle PMC$ are supplementary, and $\angle BPM$ and $\angle BCM$ are supplementary.

Here's something that looks right to me, but my geometry is insufficient to prove it; I think if you can, it'll go a long way:

Quadrangles $BCDE$ and $PMCB$ are similar; $BCDE$ is upside-down relative to $PMCB$. Prove this, and you get $\angle ABC \cong \angle PMC$ and $\triangle PMC \cong \triangle ABC$, therefore $MC = BC$.


TITLE: Why are these SUVAT equations true?
QUESTION [0 upvotes]: \begin{align}
v&=u+at\\
s&=ut+\frac{1}{2}at^2\\
v^2&=u^2+2as\\
s&=\frac{(v+u)t}{2}
\end{align}
I have just started with learning acceleration in school and I don't really understand why these equations are the way they are. I know how to apply them and use them but not why they exist.

REPLY [4 votes]: For constant acceleration, $a$ is defined as the change in velocity per unit time,
$$a = \frac{v-u}{t}.$$
Rearrange to get your first equation.
Distance travelled is average velocity times time.
$$ s = \frac{v + u}2 t.$$
That is your last equation. Multiply these two equations together,
$$a s = \frac{v-u}{t} \frac{v + u}2 t = \frac{v^2-u^2}2,$$
and rearrange to get your third equation.
Substitute your first equation into your last equation,
$$ s= \frac{ ((u+at) + u)t}2 = \frac{ 2ut+at^2} 2 ,$$
and simplify to get your second equation.
(Note: There are really only two independent equations here. The definition of average acceleration, and the definition average velocity, applied to the case of constant acceleration. The rest is just algebraic manipulation to produce four (or sometimes five) equations in a form useful to apply to problems).


TITLE: Vector space structure on the tangent bundle of a scheme and relation to the tangent sheaf
QUESTION [4 upvotes]: First a word of warning: I am not a trained algebraic geometer, so it is possible (likely) that these questions are inappropriate for MO, if so: my apologies.
Said this: As far as I understand the tangent bundle $TX$ of a scheme $X$ is defined as the spectrum of the symmetric algebra of its sheaf of differentials, and so (I guess) pointwise nothing else than the associated scheme to the Zariski tangent space.
The "projection map" $\pi: TX \rightarrow X$, which is a morphism of schemes, gives a local description of the tangent bundle, by $TU \cong \pi^{-1}(U)$, where $U$ is an affine open set in $X$.
My (first) question is the following: (1) is there a "global" description of the vector space structure (which I suppose exists?!) on the tangent bundle of a scheme. I mean addition should somehow be defined by a scheme morphism $TX \times_X TX \rightarrow TX$, where $TX \times_X TX$ is the scheme fibre product, and scalar multiplication by a scheme morphism $\mathbb{C} \times TX \rightarrow TX$. Does there exist such a description? Or what other basic properties does the tangent bundle of scheme have? (A reference would also be appreciated)
As a second question: (2) Is the total space associated to the tangent sheaf (defined as the $\mathcal{O}_X$ dual of the cotangent sheaf, so the total space associated to a quasicoherent sheaf) the same as the tangent bundle? I guess there could be some issues with "taking the dual of sheaves"?!

REPLY [6 votes]: The tangent bundle of a scheme is not in general a vector bundle.  You need your sheaf of differentials to be locally free.  This is automatic if the scheme is smooth over a field.
However, the additive group structure arises canonically from the symmetric algebra construction of the tangent scheme with no smoothness hypotheses, because symmetric algebras are Hopf algebras, and the addition operation comes from applying the relative Spec functor to the comultiplication map.  Generalities on relative Spec and symmetric algebras are in EGA 2 and 1, respectively (maybe there's a more accessible reference?).
The process of taking duals is not involutive in general.  For example, if your scheme is a pro-finite dimensional affine space (i.e., the spectrum of a polynomial ring in infinitely many generators), you can get an extra big total space.  The constructions are identical if your scheme is smooth (and I imagine you can weaken the smoothness condition a bit, but I don't know how much).


TITLE: Many uncountable dense linear orders
QUESTION [3 upvotes]: A theorem of Shelah (which has been quoted several times on this website) states that if $T$ is a countable unstable theory and $\kappa$ is an uncountable cardinal, then $T$ has $2^\kappa$ non-isomorphic models of cardinality $\kappa$.
I was wondering if there is an elementary way ($\mathit{i.e.}$ requiring very little model and set theory) to illustrate this theorem in the special case of dense linear orders. In particular, I would like to construct explicitly $2^{\mathfrak{c}}$ pairwise non-isomorphic dense linear orders of cardinality $\mathfrak{c}$. My starting point was to consider the real line (“horizontally”) and to attach to every real number on the line a “vertical” fiber, which would be either $\mathbb{R}$ or $\mathbb{Q}$, and then order the whole subset of $\mathbb{R}^2$ obtained this way by the lexicographic order. Given any $A\subseteq\mathbb{R}$, I would then consider the dense linear order $\mathcal{M}_A$ which has $\mathbb{R}$ as a vertical fiber over every $x\in A$ and $\mathbb{Q}$ as a vertical fiber over every $x\notin A$. I have produced $2^{\mathfrak{c}}$ models of cardinality $\mathfrak{c}$, which unfortunately are not pairwise non-isomorphic (for example, whenever $A$ and $B$ are singletons, $\mathcal{M}_A$ and $\mathcal{M}_B$ are isomorphic). So I'm not sure this is the way to go. Any thoughts?

REPLY [6 votes]: Start with $\kappa$ and replace each ordinal with either $(\omega_1^*+\omega_1)\times \mathbb{Q}$ or $(1+\omega_1^*+\omega_1)\times \mathbb{Q}$. This gives $2^{\kappa}$ many nonisomorphic dense orders.


TITLE: Hamiltonian vector fields for standard Poisson structure
QUESTION [1 upvotes]: Given a Poisson manifold $(M,\Pi)$, a vector field $X \in \mathfrak{X}(M)$ is called $\textit{Hamiltonian}$ if $\iota_X \Pi = df$ for some smooth $f$. A vector field $X$ that satisfies $\mathcal{L}_X\Pi=0$ is called $\textit{Poisson}$.
In general, Poisson vector fields need not be Hamiltonian. However, on $(\mathbb{R}^{2n}, \sum \frac{\partial}{\partial x_i} \wedge \frac{\partial}{\partial y_i})$, every Poisson vector field is Hamiltonian. I wanted to prove this, however I can't seem to find how to begin. My first guess would be to use some analogous "Cartan magic formula", but I don't think this makes sense in the case of multi-vector fields (at least not immediately).
Any help is apprciated.

REPLY [2 votes]: A side note: you inverted the vector field $X$ and the 1-form $df$ in the definition of Hamiltonian vector field. $X$ is Hamiltonian if it is a Poisson gradient, and some equivalent notations are $X_{f} = \{f, \cdot\} = \pi(df, \cdot) = (df)^{\sharp} = \iota_{df}\pi $.
On a similar line, since you cannot feed vector fields to a bivector as you do with a 2-form, you cannot have a strict analogue of Cartan's magic formula for the Lie derivative of a bivector.
Back to your question
Page 122 and around there of

Paulette Libermann and Charles-Michel Marle. Symplectic Geometry and
  Analytical Mechanics

can be of help without invoking Poisson cohomology, that requires building Poisson calculus first (that is, defining brackets of 1-forms from usual Poisson brackets of smooth functions).
As @studiosus said, your Poisson manifold is special in two independent ways:
A. in $\mathbb{R}^n$ differential forms are closed iff exact (all de Rham cohomology groups are trivial); and
B. the Poisson bivector is non-degenerate everywhere (making the manifold symplectic).
These implications show how these two facts make any Poisson vector field a Hamiltonian vector field:

The arrows marked with (A) or (B) hold only if the respective condition holds; let's start climbing up this chain from the bottom.
A vector field $X$ is called Poisson if $\mathcal{L}_X \pi = 0$.
The first difficulty we face is that a degenerate bivector $\pi$ defines a non-bijective homomorphism $\sharp$ mapping a 1-form into a vector field, so for a generic vector field there may not exist a 1-form $\alpha$ such that $\alpha^{\sharp} = X$.
If a Poisson vector field belongs to the image of $\sharp$, so $\mathcal{L}_X \pi = 0$ and $X = \alpha^{\sharp}$ for some 1-form $\alpha$, we say that $X$ is locally Hamiltonian.
Proposition 10.5 at page 122 of Libermann's book shows that a vector field is locally Hamiltonian iff the related 1-form is c-closed, namely $d\alpha$ vanishes on hamiltonian vector fields.
Now, if B. holds, the Poisson bivector is non-degenerate, and it defines an isomorphism between vector fields and 1-forms: in particular every vector field is associated to precisely one 1-form [* footnote].
This means that, if B. holds, every Poisson vector field is locally hamiltonian, and it can be shown that every c-closed 1-form is closed (every vector can be written as the image of a Hamiltonian vector field at one point).
To summarize, if the Poisson bivector is non-degenerate, there is a 1-to-1 correspondence between Poisson vector fields and closed 1-forms.
Finally, as said in A., in $\mathbb{R}^n$ a closed 1-form is exact: $\alpha = df$ for some function $f$; so the Poisson vector field we started with, $X = \alpha^{\sharp}$, is indeed Hamiltonian: $X = (df)^{\sharp}$.

[* footnote] If B. holds the Poisson manifold is symplectic. The symplectic form $\omega$ can be defined from the bivector since $\sharp^{-1}$ now makes sense; the non-degeneracy of $\pi$ grants the non-degeneracy of $\omega$; and interestingly the Jacobi property of the Poisson structure, or equivalently the vanishing Schouten brackets $[\pi,\pi]_S = 0$, assures that $d \omega = 0$.


text:
\begin{document}

\begin{abstract}
Let $E$ be an elliptic curve over $\bbq$, with L-function
$L_E(s)$. For any primitive Dirichlet character $\chi$, let
$L_E(s, \chi)$ be the L-function of $E$ twisted by $\chi$. In this
paper, we use random matrix theory to study vanishing of the
twisted L-functions $L_E(s, \chi)$ at the central value $s=1$. In
particular, random matrix theory predicts that there are
infinitely many characters of order 3 and 5 such that $L_E(1,
\chi)=0$, but that for any fixed prime $k \geq 7$, there are only
finitely many character of order $k$ such that $L_E(1, \chi)$
vanishes. With the Birch and Swinnerton-Dyer Conjecture, those
conjectures can be restated to predict the number of cyclic
extensions $K/ \bbq$ of prime degree such that $E$ acquires new
rank over $K$.
\end{abstract}

\maketitle

\section{Introduction}

Let $E$ be an elliptic curve defined over $\bbq$ with conductor
$N_E$. For any number field $K/\bbq$, let $E(K)$ be the group of
points of $E$ defined over $K$. By the Mordell-Weil Theorem,
$E(K)$ is a finitely generated abelian group.
Let $L_E(s, K)$ be the L-function of $E$ over the field $K$.

\begin{conjecture}[Birch and Swinnerton-Dyer conjecture over number fields]
$L_E(s, K)$ has analytic continuation to the whole complex plane, and
\begin{eqnarray*}
\displaystyle \mbox{ord}_{s=1} L_E(s, K) &=& r_K(E)
\end{eqnarray*}
where $r_K(E)$ is the rank of $E(K)$.
\end{conjecture}

In this paper, we fix $E$ an elliptic curve over $\bbq$, and we
study how  the rank varies over abelian fields $K / \bbq$ of fixed prime
degree. For example, are there infinitely many such number fields
where $E$ acquires new rank over $K$ (i.e. $r_K(E) > r_\bbq(E)$)? With the Birch and
Swinnerton-Dyer conjecture, one can rephrase the question in terms
of vanishing of the L-function $L_E(s,K)$ at $s=1$. Let $K$ be an
abelian extension of $\bbq$ with Galois group $G$ and conductor
$m$. Let $\hat{G}$ be the group of characters of $G$ which can be identified with a set
of Dirichlet characters
$$
\chi: \left( \bbz / m \bbz \right)^* \rightarrow \bbc^* .
$$
Let
$$L_E(s) = L_E(s, \bbq) = \sum_{n \geq 1} \frac{a_n}{n^s}$$
be the L-function of $E$ over $\bbq$. For each primitive Dirichlet character
$\chi$, let
$$L_E(s, \chi) = \sum_{n \geq 1} \frac{\chi(n) \, a_n}{n^s}$$
be the L-function of $E$ over $\bbq$ twisted by the character
$\chi$. By the work of \cite{wiles, taylorwiles, BCDT}, $L_E(s)$ and
$L_E(s, \chi)$ have analytic continuation to the whole complex plane.
It also follows from properties of number fields that
\begin{eqnarray}
\label{numberfields}
 L_E(s, K) = \prod_{\chi \in \hat{G}} L_E(s, \chi) ,
\end{eqnarray}
and the vanishing of the twisted L-functions $L_E(s, \chi)$ at
$s=1$ is
equivalent, via the Birch and Swinnerton-Dyer conjecture, to the
existence of rational points of infinite order on $E(K)$.

In this paper, we use random matrix theory to
study the vanishing of the twisted L-functions
$L_E(s, \chi)$ at $s=1$. It has been known
since the work of Montgomery \cite{montgomery} that certain
statistics on probability spaces of random matrices (as pair
correlation between the eigenangles of the matrices) are similar
to the same statistics on the zeroes of the Riemann zeta function.
This intuition is supported by the extensive computations of
Odlyzko \cite{odlyzko} on the critical zeroes of the Riemann zeta
function.

This was explored further in the work of Katz and
Sarnak \cite{KaSa1, KaSa2}, who extend the analogy between other
probability spaces of matrices, and families of L-functions. Katz
and Sarnak also studied the case of function fields, where they
can actually prove some of those mysterious connections between
random matrices and families of L-functions.

In order to study different statistics of number theoretic
objects, Keating and Snaith \cite{KS1, KS2} introduced a new
random variable on spaces of random matrices, the characteristic
polynomial of the matrix evaluated at a given point. They computed
the probability distribution of this new variable, which led to
striking conjectures for the asymptotic behavior of the moments of
the Riemann zeta function on the critical line.

The ideas of Keating
and Snaith have been applied to study vanishing of L-functions in
families \cite{CKRS, DFK, watkins}. Families of quadratic twists
are studied in \cite{CKRS}, where a conjectural asymptotic for
the number of quadratic twists with even non-zero rank is
presented. This refines a conjecture of Goldfeld \cite{goldfeld}
which predicts that quadratic twists with rank greater than one
have density zero. In \cite{DFK}, the authors used the ideas of
Keating and Snaith to obtain a conjectural asymptotic for the
number of cubic Dirichlet characters $\chi$ such that $L_E(1,
\chi)$ vanishes. We present in this paper the case of
characters of order $k$, for $k$ any odd prime. More precisely,
the conjectures that we obtain from the random matrix model are
(the case $k=3$ of \cite{DFK} is included for completeness)

\begin{conjecture} \label{mainconjecture}
Let $k$ be an odd prime, let $E$ be an elliptic curve defined over
$\bbq$, and let
\begin{eqnarray*}
N_{E,k}(X)  &=& \# \left\{ \chi \;\mbox{of order $k$} \;:\;
\mbox{cond}(\chi) \leq X \;\mbox{and}\; L_E(1 ,\chi) = 0
\right\} .

\end{eqnarray*}

If $k = 3$, then
\begin{eqnarray*}
\log{N_{E,k}(X)} &\sim&  \frac{1}{2} \log{X} \;\;\;\;
\mbox{as $X \rightarrow \infty$.}
\end{eqnarray*}

If $k = 5$, then
$N_{E,k}(X)$ is unbounded, but
$N_{E,k}(X) \ll X^\epsilon$ for any $\epsilon > 0$
as $X \rightarrow \infty$.

If $k \geq 7$, then
$N_{E,k}(X)$ is bounded.

\end{conjecture}

In the light of (\ref{numberfields}), and
under the Birch and Swinnerton-Dyer conjecture,
one can rewrite $N_{E,k}(X)$ as
\begin{eqnarray*}
\label{twistsandfields}
N_{E,k}(X) &=& (k-1) \;  \# \left\{ K/\bbq \; \mbox{cyclic of degree $k$} \;:\; \right.\\
&& \;\;\;\; \left. \mbox{cond}(K) \leq X  \;\mbox{and}\; r_K(E)
> r_\bbq(E) \right\}.
\end{eqnarray*}

The structure of the paper is as follows. In the second section,
we use modular symbols to rewrite the special values $L_E(1,
\chi)$ as a product of terms depending only on $E$, and some
algebraic integer $n_E(\chi)$ depending on the character. In the
third section, we use the embedding of number fields as lattices
in $\bbc$ to give a discretisation of the algebraic integers
$n_E(\chi)$. In the fourth section, we use this discretisation and
the work of Keating and Snaith to obtain conjectures on the
asymptotic behavior of $N_{E,k}(X)$. Finally, the last section
presents some experimental results.

\section{Special values and modular symbols}

The notation of this section follows the introduction of
\cite{MTT}. Let $E$ be an elliptic curve over $\bbq$, and let
$f(z) = \sum_{n \geq 1} a_n \; e^{2 \pi i n z}$ be the Fourier
expansion of $E$. Then, the L-function $L_E(s)$ is the Mellin
transform
\begin{equation}
\label{mellintransform} L_E(s) = \frac{(2 \pi)^s}{\Gamma(s)}
\int_{0}^{\infty} f (it) t^{s-1} \; dt .
\end{equation}
By the work of \cite{wiles, taylorwiles, BCDT}, $f$ is modular and
$L_E(s)$ has analytic continuation to the whole complex plane, and
satisfies the functional equation
\begin{equation}
\label{functionalequation} \Lambda_E(s) = \left(
\frac{\sqrt{N_E}}{2 \pi} \right)^s \Gamma(s) L_E(s) = w_E
\Lambda_E(2-s)
\end{equation}
where $w_E = \pm 1$ is called the root number.
From (\ref{mellintransform}), we have that $L_E(1) = 2 \pi
\int_{0}^{\infty} f(it) \; dt.$ For $a,m \in \bbq, m> 0$, one
defines the modular symbols
\begin{eqnarray}
\label{defmodularsymbol} \lambda(a,m;E) &=& \lambda(a,m;f) = 2 \pi
\int_{0}^{\infty} f \left( it - \frac{a}{m} \right) dt.
\end{eqnarray}
Let $\chi$ be a primitive character of modulus $m$ with Gauss sum
$$\tau(\chi) = \sum_{a \;\rm{mod}\;  m} \chi(a) e^{2 \pi i a/m}. $$
The twisted L-function $L_E(s,\chi)$ satisfies the functional
equation
\begin{eqnarray*}
\label{generalFE} \Lambda_E(s, \chi) &=& \left( \frac{m
\sqrt{N_E}}{2\pi} \right)^s \Gamma(s) L_E(s, \chi) \\
&=&\frac{w_E \chi(N_E) \tau(\chi)^2}{m} \Lambda_E(2-s,
\overline{\chi}).
\end{eqnarray*}
From the identity
$$
\chi(n) = \frac{1}{\tau(\overline{\chi})}\sum_{a \;\rm{mod}\;  m}
\overline{\chi}(a) e^{2 \pi i a n /m},
$$
we have
\begin{eqnarray*}
f_\chi(z) &=& \sum_{n \geq 1} \chi(n) a_n e^{2 \pi i n z}\\
&=& \frac{1}{\tau(\overline{\chi})}\sum_{a \;\rm{mod}\;  m}
\overline{\chi}(a) f \left( z + a/m \right)
\end{eqnarray*}
by rearranging the sums (Birch's lemma).

It then follows that
\begin{eqnarray}
\label{specialvalue1} L_E(1, \chi) &=&
\frac{1}{\tau(\overline{\chi})} \sum_{a \;\rm{mod}\;  m}
\overline{\chi}(a) \lambda(a,m;E).
\end{eqnarray}

We define
\begin{eqnarray*}
\lambda^+(a,m;E) &=& \lambda(a,m;E) + \lambda(-a,m;E) \\
&=& {2 \pi} \int_{0}^{\infty} \sum_{n \geq 1} a_n \;e^{-2 \pi n t}
\;\left( e^{2 \pi i a n/m} + e^{-2 \pi i a n / m} \right) dt \\
&=& 4 \pi \sum_{n \geq 1} a_n \left( \mbox{Re}\left\{ e^{2 \pi i a
n /
m} \right\} \right)\int_{0}^{\infty} e^{-2 \pi n t} dt \\
&=& 2 \sum_{n \geq 1} \frac{a_n}{n} \;\mbox{Re}\left\{ e^{2 \pi i
a n / m} \right\}
\end{eqnarray*}
which is a real number as $E$ is defined over $\bbq$. Then, all
$\lambda^+(a,m;E)$ can be written as a rational multiple of the
real period, which we denote by $\Omega_E$. We define
$$
\Lambda(a,m;E)  = \frac{\lambda^+(a,m;E)}{\Omega_E} \in \bbz.
$$
In this paper, $\chi$ has prime order $k \geq 3$. Then, $\chi(-1)
= 1$, and we can rewrite (\ref{specialvalue1}) as
\begin{eqnarray}
\label{specialvalue2} L_E(1, \chi) &=& \frac{\Omega_E}{2
\tau(\overline \chi)} \sum_{a \;\rm{mod}\;  m} \overline{\chi}(a)
\Lambda(a,m;E)
\end{eqnarray}
where the integers $\Lambda(a,m;E)$ do not depend on the character $\chi$,
but only on the conductor $m$.
Then, we see from (\ref{specialvalue2}) that
$$
\frac{2 \tau(\overline \chi) L_E(1, \chi)}{\Omega_E}
$$
is an algebraic integer in the field obtained by adding a $k$th root of unity.

In fact, one can prove the stronger result which is critical to
the  discretisation of the next section. The particular case $k=3$
was proven in \cite{DFK}. For any odd prime $k$,
let $\bbq(\xi_k)$ be the cyclotomic
field obtained by adding a primitive $k$th root of unity $\xi_k$,
and let $\bbq(\xi_k)^+$ be the maximal real extension $\bbq
\subseteq \bbq(\xi_k)^+ \subseteq \bbq(\xi_k)$ of degree
$(k-1)/2$ over $\bbq$. The ring of integers of $\bbq(\xi_k)^+$ will be
denoted by $\bbz[\xi_k]^+$.

\begin{theorem}
\label{maximalreal} Let $k$ be an odd prime, and let $\chi$ be a primitive character of order
$k$. Then,
\begin{eqnarray*}
\frac{2 \tau(\overline \chi) L_E(1, \chi)}{\Omega_E} = \left\{
\begin{array}{ll} \chi(N_E)^{(k+1)/2} \;n_E(\chi) &
 \mbox{when $w_E = 1$} \\
 \\
\displaystyle \left( \xi_k^{-1} - \xi_k \right)^{-1}
 \chi(N_E)^{(k+1)/2} \;n_E(\chi) &
 \mbox{when $w_E = -1$} \end{array} \right.
\end{eqnarray*}
where
$n_E(\chi) \in \bbz[\xi_k] \cap \bbr = \bbz[\xi_k]^+$.
\end{theorem}

\noindent{\bf Proof:} From the functional equation, we have
\begin{eqnarray*}
L_E^{alg}(1, \chi) &=& \frac{2 \tau(\overline \chi) L_E(1, \chi)}{\Omega_E} =
\frac{2 \tau(\overline \chi) w_E \chi(N_E) \tau(\chi)^2}{m \Omega_E} L_E(1, \overline \chi) \\
&=& w_E \chi(N_E) \overline{\frac{2 \tau(\overline \chi) L_E(1, \chi)}{\Omega_E}}
= w_E \chi(N_E) \overline{L_E^{alg}(1,\chi)}
\end{eqnarray*}
Let $z \in \bbc^*$ satisfying $z = w_E \chi(N_E) \overline{z}$.
Then, $L_E^{alg}(1,\chi) = \alpha \overline{z}^{-1}$ with $\alpha$ real.
If $w_E  = 1$, we take $z = \chi(N_E)^{(k+1)/2}$, and $L_E^{alg}(1,\chi) =
\alpha \overline{z}^{-1}$ with $\alpha \in \bbr \cap \bbz[\xi_k] = \bbz[\xi_k]^+$,
which gives the result. \\
If $w_E = -1$, we take $z = \left( \xi_k - \xi_k^{-1} \right)
\chi(N_E)^{(k+1)/2}$, and
$L_E^{alg}(1, \chi) = \alpha \overline{z}^{-1}$ with $\alpha \in \bbr \cap \bbz[\xi_k] = \bbz[\xi_k]^+$,
which gives the result.
\qed

\section{Discretisation}
\label{discretisation}

By Theorem \ref{maximalreal},
$n_E(\chi)$ is an algebraic integer in $\bbz[\xi_k]^+$, and there is then a natural
discretisation on the algebraic integer $n_E(\chi)$ given by the geometry of numbers.
Let $\phi$ be the map
\begin{eqnarray*}
\phi :   \bbz[\xi_k]^+ &\rightarrow& \bbr^{(k-1)/2} \\
\alpha &\mapsto& (\sigma_1(\alpha), \sigma_2(\alpha), \dots, \sigma_{(k-1)/2}(\alpha) ) \\
\end{eqnarray*}
where $\mbox{Gal}(\bbq(\xi_k)^+/\bbq) = \left\{ \sigma_1 = 1, \sigma_2, \dots, \sigma_{(k-1)/2} \right\}$.
Let $\alpha_1, \dots, \alpha_{(k-1)/2}$ be an integral basis for $ \bbz[\xi_k]^+$.
The image of $ \bbz[\xi_k]^+$ in $\bbr^{(k-1)/2}$ is the lattice generated by
the linearly independent vectors
$$\omega_1 = \phi(\alpha_1), \;\dots, \;\omega_{(k-1)/2} = \phi(\alpha_{(k-1)/2}).$$
Let $R \subseteq \bbr^{(k-1)/2}$ be the region
\begin{eqnarray*}
R &=& \left\{\; a_1 \omega_1 + a_2 \omega_2 + \dots + a_{(k-1)/2}
\omega_{(k-1)/2}: \right. \\
&& \;\;\; \left. -1 < a_i < 1 \;\mbox{for $1 \leq i \leq (k-1)/2$}
\;\right\}.
\end{eqnarray*}
The discretisation given by the embedding of
$\bbz[\xi_k]^+$ in $\bbr^{(k-1)/2}$ is then
\begin{eqnarray}
\label{geometryofnumbers} n_E(\chi) = 0 \iff \phi(n_E(\chi)) \in R.
\end{eqnarray}
Let $\chi$ be any character of conductor $m$ and order $k$. For
any automorphism $\sigma \in \mbox{Gal}(\bbq(\xi_k) / \bbq)$, let
$\chi^{\sigma}$ be the character
\begin{eqnarray*}
\chi^{\sigma} : \left( \bbz / m \bbz \right)^* &\rightarrow&
\left< \xi_k \right>
\subseteq \bbc^* \\
a &\mapsto& \sigma \left( \chi(a) \right)
\end{eqnarray*}
Then, $\chi^\sigma$ is also a character of conductor $m$ and order
$k$.

\begin{lemma}
\label{discrete}
Let $k$ be an odd prime, and $\chi$ a character of
order $k$ and conductor $m$. For any $\sigma$ in
Gal$(\bbq(\xi_k)/\bbq)$, we have
\begin{eqnarray*}
\vert L_E(1, \chi^\sigma) \vert &=& \frac{c_{E,k}}{m^{1/2}} \vert
n_E(\chi)^\sigma \vert
\end{eqnarray*}
where $c_{E,k}$ is an explicit constant depending only on $E$ and
$k$.
\end{lemma}
\noindent{\bf Proof:}  Using (\ref{specialvalue2}), we have
\begin{eqnarray*}
L_E^{alg}(1,\chi)^\sigma &=& \left( \frac{2 \tau(\overline \chi) L_E(1,
\chi)}{\Omega_E} \right)^\sigma \\
&=& \sum_{a \;\rm{mod}\; m} \overline{\chi}^\sigma(a)
\Lambda(a,m;E) = L_E^{alg}(1,\chi^\sigma).
\end{eqnarray*}
Suppose first that $\omega_E=1$. Then,
\begin{eqnarray*}
n_E(\chi)^\sigma &=& L_E^{alg}(1,\chi)^\sigma \left( \chi(N_E)^{(k+1)/2}
\right)^\sigma \\
&=& \frac{ 2 \tau(\overline{\chi}^\sigma) L_E(1,
\chi^\sigma)}{\Omega_E} \left( \chi(N_E)^{(k+1)/2} \right)^\sigma,
\end{eqnarray*}
and taking absolute values we get the result with $c_{E,k} =
\Omega_E / 2$. The proof for $\omega_E = -1$ is similar, with a
different explicit constant $c_{E,k}$. \qed

We first consider the case $k=5$. We have that
$\bbz[\xi_5]^+ = \bbz\left[ \alpha \right]$ with $\alpha = (1 +
\sqrt{5})/2$, and $G_5 = \left< 1, \tau \right>$, where the
non-trivial automorphism $\tau$ sends $\sqrt{5}$ to $-
\sqrt{5}$. Then, the lattice of $\bbz\left[ \alpha \right]$ in
$\bbr^2$ is generated by $\omega_1 = (\alpha, \alpha^\tau)$ and
$\omega_2 = (\alpha^\tau, \alpha)$.
Let $R$ be the region
$$
R = \left\{ a \omega_1 + b \omega_2 : -1 < a < 1, -1 < b < 1 \right\}.
$$
By (\ref{geometryofnumbers}), $n_E(\chi) = 0$ if and only if
$\phi(n_E(\chi)) = (n_E(\chi), n_E(\chi)^\tau) \in R.$ As the region
$R$ is not symmetric with respect to the absolute value, and we
have a probability model for $\left \vert L_E(1, \chi) \right
\vert$, we also consider the two regions of $\bbr^2$
\begin{eqnarray*}
R_1 &=& \left\{ (x,y) \; : \;  - 1 < x,y <   1 \right\} \\
R_2 &=& \left\{ (x,y) \; : \;  -\sqrt{5} < x,y <  \sqrt{5}  \right\}
\end{eqnarray*}
with the property that $R_1 \subseteq R \subseteq R_2$, and
\begin{eqnarray}
\label{absolutevalue}
 (n_E(\chi), n_E(\chi)^\tau) \in R_i &\iff&
\left( \vert n_E(\chi) \vert, \vert n_E(\chi)^\tau \vert \right)
\in \vert R_i \vert
\end{eqnarray}
where
$$
\vert R_i \vert = \left\{ (x,y) \in R_i \; : \; x,y \geq 0 \right\} .
$$
The following lemma is now immediate from  (\ref{absolutevalue})
and Lemma \ref{discrete}
\begin{lemma}
\label{fork5} Let $\sigma \in \mbox{Gal}(\bbq(\xi_5)/\bbq)$ be an
automorphism which restricts to the non-trivial automorphism of
$\bbq(\sqrt{5})$. For $i=1,2$, we have
\begin{eqnarray*}
\left( n_E(\chi), n_E(\chi)^\sigma \right) \in R_i &\iff&   \vert L_E(1, \chi)
\vert, \vert L_E(1, \chi^\sigma) \vert \leq \frac{c_i}{\sqrt{m}}
\end{eqnarray*}
where $c_1, c_2$ are explicit constants depending only on $E$.
\end{lemma}

We now suppose that $k \geq 7$. Let $B$ be the non-zero constant
$$
B = \max_{1 \leq i \leq (k-1)/2} \sum_{j=1}^{(k-1)/2} \vert \sigma_i ( \alpha_j)
\vert ,
$$
and let $R  \subseteq R' \subseteq \bbr^{(k-1)/2}$ be the region
$$
R' = \left\{ (x_1, \dots, x_{(k-1)/2} ) \; : \; -B \leq x_i \leq B
\;\mbox{for $1 \leq i \leq (k-1)/2.$} \right\} $$ Then, $n_E(\chi)
= 0  \Rightarrow \phi(n_E(\chi)) \in R'$ by
(\ref{geometryofnumbers}). The following lemma is now immediate
from  Lemma \ref{discrete}
\begin{lemma}
\label{fork7} Let $\sigma_1, \dots, \sigma_{(k-1)/2} \in
\mbox{Gal}(\bbq(\xi_k)/\bbq)$ be a set of representatives for
$G_k = \mbox{Gal}(\bbq(\xi_k)^+/\bbq)$. Then,
$\phi(n_E(\chi)) \in R'$ if and only if
\begin{eqnarray*}
\vert L_E(1, \chi^{\sigma_i}) \vert \leq \frac{c_k}{m^{1/2}} \;\;\;
\mbox{for $1 \leq i \leq (k-1)/2$}
\end{eqnarray*}
where $c_k$ is an explicit constant depending only $E$ and $k$.
\end{lemma}

\section{Unitary Random Matrices}

Let $U(N)$ be the set of unitary $N \times N$ matrices with complex
coefficients which forms a probability space with respect to the
Haar measure. For each $A \in U(N)$, let
\begin{eqnarray*}
P_A(\lambda) = \mbox{det} (A - \lambda I)
\end{eqnarray*}
be the characteristic polynomial of $A$. For any $s \in \bbc$, let
$$
M_N(s) = \int_{U(N)} |P_A(1)|^s \;d{\mbox {Haar}} $$ be the
moments for the distribution of $|P_A(1)|$ in $U(N)$ with respect
to the Haar measure. In \cite{KS1}, Keating and Snaith proved that
\begin{eqnarray}
\label{MUSN} M_U(s,N)  &=& \prod_{j=1}^{N} \frac{\Gamma(j)
\Gamma(j+s)}{\Gamma^2(j+s/2)} ,
\end{eqnarray}
and then $M_U(s,N)$ is analytic for $\mbox{Re}(s) > -1$, and has
meromorphic continuation to the whole complex plane. By Fourier inversion, the
probability density of $|P_A(1)|$ is
$$
p(x) = \frac{1}{2\pi i} \int_{(c)} M_N(s) x^{-s-1} ds
$$
for some $c > -1$. Then, for any $I  \subseteq \bbr$,
$$
\mbox{Prob} \left( |P_A(1)| \in I \right) = \int_{I} p(x) \; dx .
$$
In our application to the vanishing of twisted L-functions, we will
be interested only in small values of $x$ where  the value of
$p(x)$ is determined by the first pole of $M_U(s,N)$ at $s=-1$.
More precisely, for
$$x \leq N^{-1/2},$$
one can show that
$$
p(x) \sim G^{2}(1/2) N^{1/4} \;\;\;\;\; \mbox{as $N \rightarrow
\infty$},
$$
where $G(z)$ is the Barnes G-function, with special value
$$G(1/2) = \exp{\left( \frac{3}{2} \zeta'(-1) - \frac{1}{4} \log{\pi} + \frac{1}{24} \log{2} \right)}$$
(see \cite[p. 81]{KS1} or \cite[p. 58]{hughesthesis} for more details).

We now consider the moments for the special values of L-functions
in families of twists. Fix $k \geq 3$, and let
\begin{eqnarray*}
S_k(X) &=& \left\{ \chi \; \mbox{of order $k$ and conductor
$\leq X$} \right\} \\
N_k(X) &=& \# S_k(X) \sim b_k X
\end{eqnarray*}
with an explicit constant $b_k$ (see for example \cite{cohenetal}). We then define
for any $s \in \bbc$
\begin{eqnarray}
\label{allmoments} M_E(s,X) &=& \frac{1}{N_k(X)} \sum_{\chi \in
S_k(X)} |L_E(1, \chi)|^s
\end{eqnarray}
The family of twists of order $k$ has unitary symmetry, as the
values  $|\zeta(1/2 + it)|$ on the critical line. Then

\begin{conjecture} [Keating and Snaith Conjecture for twists of order $k$]
\label{conjectureKS}
\begin{eqnarray*}
M_E(s,X) &\sim& a_E(s/2) M_U(s,N)  \;\;\;\;\; \mbox{as $N = 2 \log{X} \rightarrow \infty$},
\end{eqnarray*}
where $a_E(s/2)$ is an arithmetic factor
depending only on the curve $E$.
\end{conjecture}
In the conjecture, the relation between $N$ and $X$ is obtained by
equating the mean density of eigenangles of matrices in the
unitary group, and the mean density of non-trivial zeroes of the
twisted L-functions $L_E(s, \chi)$ at a fixed height (see
\cite{DFK}). The arithmetic factor $a_E(s)$ can not be obtained
from the random matrix theory, and has to be determined separately
for each family from its arithmetic. This was done for the family
of cubic twists in \cite{DFK}, and could be done for the family of
twists of order $k$ for each $k$. The arithmetic factor $a_E(s)$
would then be a meromorphic function for all $s \in \bbc$. As it
will be seen below, the only influence of the arithmetic factor
$a_E(s)$ in our application is that the special value $a_E(-1/2)$
will be part of the constant of the conjectural asymptotic of
$N_{E,k}(X)$. This would not provide any further information to
the cases $k \geq 5$ considered in this paper considered in this
paper in view of Conjecture \ref{mainconjecture}.

>From Conjecture \ref{conjectureKS}, the probability density $p_E(x)$ for
the distribution of the special values $|L_E(1, \chi)|$ for
characters of order $k$ is
\begin{eqnarray}
\nonumber
p_E(x) &=& \frac{1}{2\pi i} \int_{(c)} M_E(s,X) \,x^{-s-1} \,ds \\
\label{computep} &\sim& \frac{1}{2\pi i} \int_{(c)} a_E(s/2)
\,M_U(s,N) \,x^{-s-1} \,ds
\end{eqnarray}
as $N = 2 \log{X} \rightarrow \infty$.
As above, when $x \leq N^{-1/2}$, the value of $p_E(x)$ is determined
by the residue of $M_U(s,N)$ at $s=-1$, and it follows from (\ref{computep})
that
\begin{eqnarray}
\label{smallx}
p_E(x) \sim C_E \log^{1/4}{X}
\end{eqnarray}
for $x \leq (2 \log{X})^{-1/2}$, $X \rightarrow \infty$, and
$C_E = 2^{1/4} a_E(-1/2) G^2(1/2)$.

Let $\chi$ be a character of order $k \geq 3$ and conductor $m$.
We apply the above model to find the probability that $|L_E(1, \chi)| < c m^{-1/2}$,
for some constant $c > 0$.
For $x < c m^{-1/2} < (2 \log{m})^{-1/2}$ (for $m$ large enough), we have
$p_E(x) \sim C_E \log^{1/4}{m}$, and then
\begin{eqnarray}
\nonumber
\mbox{Prob} \left( |L_E(1, \chi)| < c m^{-1/2} \right) &\sim&
\int_0^{c m^{-1/2}} C_E \log^{1/4}{m} \; dx \\
\label{probability} &=& c \,C_E \frac{\log^{1/4}{m}}{m^{1/2}} .
\end{eqnarray}

We now use the probability density of the random matrix model
with the discretisation of Section \ref{discretisation} to obtain
conjectures for the vanishing of the L-values $L_E(1, \chi)$. We
first suppose that $k=5$, and as in the previous section, let
$\sigma \in \mbox{Gal}(\bbq(\xi_5)/\bbq)$ which restricts to the
non-trivial automorphism of $\bbq(\sqrt{5})$. We saw in the
previous two sections that
\begin{eqnarray*}
L_E(1, \chi) = 0
&\iff& \phi(n_E(\chi)) = (n_E(\chi), n_E(\chi)^\sigma) \in R.
\end{eqnarray*}
As $R_1 \subseteq R \subseteq R_2$, and using Lemma \ref{fork5},
the probability that
$L_E(1, \chi)$ is zero is bounded below by
\begin{eqnarray*}

\mbox{Prob} \left( \vert L_E(1, \chi) \vert <
\frac{c_1}{\sqrt{m}} \right) \mbox{Prob} \left( \vert L_E(1,
\chi^{\sigma}) \vert < \frac{c_1}{\sqrt{m}} \right)
\end{eqnarray*}
and bounded above by
\begin{eqnarray*}

\mbox{Prob} \left( \vert L_E(1, \chi) \vert <
\frac{c_2}{\sqrt{m}} \right) \mbox{Prob} \left( \vert L_E(1,
\chi^{\sigma}) \vert < \frac{c_2}{\sqrt{m}} \right) .
\end{eqnarray*}
Assuming that $|L_E(1, \chi)$ and $|L_E(1, \chi^\sigma)|$ are independent identically distributed
random variables, and using (\ref{probability}), we get that the probability that
$L_E(1, \chi)$ is zero is about
$$
\frac{\log^{1/2}{m}}{m} ,
$$
neglecting all constants which are not significant here. The sum
of the probabilities is
\begin{eqnarray}
\sum_{\chi \in S_3(X)} \frac{\log^{1/2}{m}}{m} &\sim& \frac{2
b_3}{3} \log^{3/2}{X} .
\end{eqnarray}
As discussed in \cite{DFK}, the exact power of $\log{X}$ that is
obtained with the random matrix approach depends subtly on the
discretisation, and is difficult to predict. For examples,
rational torsion of order three on the elliptic curve seemed to
cause extra vanishing of the twisted L-values $L_E(1, \chi)$ for
cubic characters, and changed the power of logarithm in the
conjectural asymptotic for $N_{E,k}(X)$ of \cite{DFK}. For $k =
5$, the sum of the probabilities is just on the border between
convergence and divergence, and the random matrix model seems to
indicate that the number of quintic twists such that $L_E(1,
\chi)$ vanishes is infinite, but that $N_{E,k}(X) \ll X^\epsilon$
for any $\epsilon > 0$. This agrees with the empirical evidence of
Section \ref{jack}.

We now suppose that $k \geq 7$. Let $\sigma_1 = 1, \dots, \sigma_{(k-1)/2}$
be elements of the Galois group of $\bbq(\xi_k) / \bbq$ which form a set
of representatives for the Galois group of $\bbq(\xi_k)^+ / \bbq$.
As we saw in the two previous sections,
\begin{eqnarray*}
L_E(1, \chi) = 0
&\iff& \phi(n_E(\chi)) = \left( n_E(\chi)^{\sigma_1}, \dots, n_E(\chi)^{\sigma_{(k-1)/2}} \right) \in R.
\end{eqnarray*}
As $R \subseteq R'$, and using Lemma \ref{fork7},
the probability that
$L_E(1, \chi)$ is zero is bounded by
\begin{eqnarray*}

\mbox{Prob} \left( \vert L_E(1, \chi^{\sigma_1}) \vert <
\frac{c_k}{\sqrt{m}} \right), \dots, \mbox{Prob} \left( \vert L_E(1,
\chi^{\sigma_{(k-1)/2}}) \vert < \frac{c_k}{\sqrt{m}} \right) .
\end{eqnarray*}
Assuming that $|L_E(1, \chi^{\sigma_i})$  are  independent identically distributed
random variables for $1 \leq i \leq (k-1)/2$, and using (\ref{probability}), we get that the probability that
$L_E(1, \chi)$ is zero is
$$
\frac{\log^{(k-1)/8}{m}}{m^{(k-1)/4}}.
$$
neglecting all constants which are not significant here. Summing
the probabilities, this gives for $k \geq 7$
\begin{eqnarray}
\sum_{\chi \in S_3(X)}  \frac{\log^{(k-1)/8}{m}}{m^{(k-1)/4}} &=& O(1).
\end{eqnarray}
From the random matrix model, we then conjecture that the number of twists of order $k \geq 7$
such that $L_E(1, \chi)$ vanishes is bounded. This also agrees with the empirical evidence of Section \ref{jack}.

\section{Numerical Evidence}
\label{jack}

The following table shows the observed number of vanishing twists
$L_E(1, \chi)$ for characters of orders three, five and seven, and
for the first three elliptic curves in the Cremona catalogue
\cite{cremona}. For each elliptic curve $E$, the characters with
conductor prime to $N_E$ and less than two million were
considered. Any two characters of conductor $m$ and order $k$
generating the same cyclic subgroup of the character group are
conjugate, and hence the special values $L_E(1, \chi)$ vanish
simultaneously by Lemma \ref{discrete}. The number in the table
records one of each class of conjugate characters for which the
special value vanishes, which is $1/(k-1)$ of the number of
characters with vanishing special value. The twists of order
eleven in the same range for the curve $E14$ were also computed,
and no vanishing were found.

\bigskip

\centerline{
\begin{tabular}{|c|c|c|c|}
\hline Curve & Cubic & Quintic & Septic  \\
 & vanishing & vanishing & vanishing  \\
\hline
\hline E11   & 1152   &  15   & 2      \\
\hline E14   & 4347   & 10  & 0       \\
\hline E15   &  2050      & 11    & 0    \\
\hline
\end{tabular}}

\bigskip

The results for cubic twists have been analyzed in \cite{DFK} and support Conjecture \ref{mainconjecture}.
The results for quintic and septic twists are too sparse to either support or refute Conjecture \ref{mainconjecture},
but they nevertheless illustrate the extreme scarcity of vanishing in higher order twists which is
predicted by the conjecture.

\bigskip

\footnotesize{ {\bf Acknowledgments} This paper was first presented
at the ``Ranks of Elliptic Curves and Random Matrix Theory'' workshop held at the Isaac Newton
Institute in February 2004. The first and second authors would like to thank the
organizers of the workshop and the Isaac Newton Institute for their hospitality and
financial support. The first author would also
like to thank B. Birch, B. Conrey and C. Hughes for helpful discussions.}


TITLE: Surface integrals-Important parametrization of a surface
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Well, my question is, how shall I parametrize the surface $z=5-(x^2/2)-y^2$ in terms of the radius($r$) and $\theta$?

REPLY [0 votes]: We can rewrite the surface as $z=5-r^2$ and put $x=\sqrt{2}rcos{\theta}$ and $y=rsin{\theta}$. We obtain ${\beta(r,\theta)=(\sqrt{2}rcos\theta,rsin\theta,5-r^2)}$


TITLE: A sequence 〈a_n〉 is defined recursively ,and find infinite series
QUESTION [1 upvotes]: A sequence $〈a_n〉$ is defined recursively by the equations
$a_0=a_1=1$
$n(n-1)a_n=(n-1)(n-2)a_{n-1}-(n-3)a_{n-2}$
find infinite series $\sum_{n=0}^{\infty} a_n=$ ?

REPLY [2 votes]: Hint: Find the first few terms of the series.
Hint: Prove the pattern.

 Show that $a_n = \frac{1}{n!} $.

Once you know what you're supposed to show, it's straightforward induction to prove the sequence.
Hint: What is the Maclaurin series for $e^x$?

 Hence conclude that $ \sum a_i = e$.