TITLE: proving that if $x\le y$ and $y\le x$ then $x=y$
QUESTION [0 upvotes]: Here's what the authors of a textbook that I've been following argue:

Let $x,y\in \mathbb{R}$. Assume that $x \le y$ and $y ≤ x$. We claim
  that $x = y$. If false, then either, $x < y$ or $y < x$ by law of
  trichotomy. Assume that we have $x < y$. Since $y ≤ x$, either $x = y$ or $y < x$. Neither can be true, since we assumed $x \not= y$ and
  hence concluded $x < y$ from the first inequality $x ≤ y$. Hence we
  conclude that the second inequality cannot be true, a contradiction.
  Thus our assumption that $x \not= y$ is not tenable.

My argument is as follows: We assume that $x \le y$ and $y ≤ x$. Suppose $x\not=y$. Then $x \le y$ and $y ≤ x$ reduces to $x < y$ and $y < x$. But both inequalities  $x < y$ and $y < x$ cannot hold together at the same time due to trichotomy. Thus we achieved a contradiction to our hypothesis. Is this argument correct?
I need a clarification on proof done by the authors. They say "Assume that we have $x < y$" and on another line they say "hence concluded $x < y$ from the first inequality $x ≤ y$". Did they deduce it or assume it?

REPLY [1 votes]: Their argument:
Given $x \le y$ and $y \le x$.
Assume $x \ne y$.
$x\le y$ means $x=y$ or $x < y$.  We ruled out $x=y$ so that leaves $x < y$.  Stick a pin in that.
$y \le x$ means that $y=x$ or $y < x$. Remove the pin and compare with $x < y$.  Neither $y =x$ nor $y < x$ is compatible with $x < y$.
So we have a contradiction.
So $x =y$.
Your argument:
Given $x \le y$ and $y \le x$.
Assume $x \ne y$.
$x\ne y$ means $x=y$ or $x < y$.  We ruled out $x=y$ so that leaves $x < y$.
$y \le x$ means that $y=x$ or $y < x$. We ruled out $y=x$ so that leaves $y < x$.
$x < y$ and $y<x$ are mutually incompatible.
So we have a contradiction.
So $x =y$
====
Both your arguments are valid.
They took a "$x<y$" to a Contradiction fight and you took a "$x \ne y$" the the same contradiction fight.  They were both equally good weapons.
....
But I prefer your method.  It's more symmetric which aesthtically pleases me and I find more convincing.
The one thing they did that is both better and worse than what you did, is the really went to the basic axiomatic deductions:  They spelled out that $x \le y$ means $x=y$ or $x < y$ and if we had assumed that $x\ne y$ then we deduce $x < y$ whereas you took it as obvious.
It's better in that it is pure axiomatic deduction.  It's worse in that it is hard to read as it is easy to get lost in the tedium.
But I like that you used the same argument for $x \le y\implies x < y$ and $y \le x \implies y < x$ whereas I dislike that they used different $x \le y\implies x <y$ and $x<y \land y \le x$ are incompatible arguments.


TITLE: What is the difference between a straight line and a linear equation?
QUESTION [0 upvotes]: The books I am reading do not make clear distinction between the two. Also for understanding line I need to know basics of Geometry, similarly what prior knowledge is useful before one can understand linear equations?

REPLY [1 votes]: When there is a direct relation between $x,y$ we have $ y =m x. $ This is a straight line passing through the origin.This is direct proportional dependence.
In a relationship $ y=mx+c$  the graph does not pass through the origin. At start $x=0,$ there is already a constant value available.
Both are linear in their relationship between the two variables and their graphs are represented by a straight line.


TITLE: Prove that there is no cycle of at least length 3 in $G'$
QUESTION [0 upvotes]: Given the following facts:

Fact 1. For a natural number $n$, the integer logarithm of that number, denoted by $\sigma (n)$, is the sum, with repetitions, of the prime
  factors of $n$. For example: $\sigma (0) = \sigma (1) = 0,\sigma
 (63) = 13$,...
  Fact 2. $\forall n \in \mathbb{N} (n > 1 \Rightarrow n \ge \sigma (n))$

Let $\tau $ be the relation on $\mathbb{N}$ such that:

$\forall n_{1},n_{2}\in \mathbb{N} ({n_1} \tau {n_2} \Leftrightarrow {n_2} =
 \sigma ({n_1}))$

Let $G$ be the directed graph representing the relation $\tau$ on $\mathbb{N}$. Let $G'$ be the underlying undirected graph of $G$. Prove that there is no cycle of at least length 3 in $G'$.

REPLY [0 votes]: The greatest vertex in the cycle would have to be connected to two vertices less than itself, but given the given facts, each vertex $n$ is connected to exactly one vertex less than or equal to itself, namely to $\sigma(n)$.
(The restriction to $n\gt1$ in fact $2$ is superfluous, since $1\ge\sigma(1)=0$.)


TITLE: How can I solve prove that $8(1-a)(1-b)(1-c)\le abc$ with the conditions below?
QUESTION [1 upvotes]: There was a homework about inequalities (that why I ask a bunch of inequality problems). But I couldn't solve the following:

If $0<a,b,c<1$ and $a+b+c=2$, prove that $8(1-a)(1-b)(1-c)\le abc$

I tried many times, and finally I used Muirhead, but it failed!

$\begin{split}L.H.S.-R.H.S.&=8(1-a)(1-b)(1-c)-abc\\&=8-8(a+b+c) +8(ab+bc+ca)-9abc\\&=-(a^3+b^3+c^3)+(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)-3abc\\&=\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}a^3\Bigg)+\frac{1}{2}\Bigg(\sum_{sym}a^2b-\sum_{sym}abc\Bigg)\end{split}$
But as $(3,0,0)$ majorizes $(2,1,0)$ but $(2,1,0)$ majorizes $(1,1,1)$, so it fails.

Could someone help? Any help is appreciated!

REPLY [5 votes]: By AM-GM you have
$$(1 - a)^{1\over 2}(1 - b)^{1 \over 2} \leq {1 - a + 1 - b \over 2} = {c \over 2}$$
$$(1 - b)^{1\over 2}(1 - c)^{1 \over 2} \leq {1 - b + 1 - c \over 2} = {a \over 2}$$
$$(1 - c)^{1\over 2}(1 - a)^{1 \over 2} \leq {1 - c + 1 - a \over 2} = {b \over 2}$$
Multiplying these three inequalities together gives the desired inequality.


TITLE: Center of gravity $\frac{1}{\lambda_d(E)}\int_E xdx \space$
QUESTION [0 upvotes]: We are looking at $\mathbb{R}^d$ with the Lebesgue measure. Let $E \subset \mathbb{R}^d$ be bounded and measurable with $\lambda_d(E) > 0$. The center of gravity is component-wise defined as $$\frac{1}{\lambda_d(E)}\int_E xdx \space.$$
How do I calculate the center of gravity of the cube $[0,T]$, $T>0$?

REPLY [0 votes]: The cube $[0,T]^{3}\subset \mathbb{R}^{3}$ has centre of mass at some point $y=(y_{1},y_{2},y_{3})$. We can compute $y_{i}$ by the integral
$$ y_{i} = \frac{\displaystyle\int_{[0,T]^{3}}p_{i}(x) dx}{\displaystyle\int_{[0,T]^{3}}dx} = \frac{\displaystyle\int_{[0,T]^{3}}p_{i}(x) dx}{\lambda_{3}([0,T]^{3})}$$
where $p_{i}$ is the projection of $x$ into the $i$th (canonical) basis vector of $\mathbb{R}^{3}$ i.e. for $x=(x_{1},x_{2},x_{3})$, $p_{i}(x)=x_{i}$.
I assume you are familiar with triple integrals and thus this simply becomes an exercise in computing them.


TITLE: Finding Percentages of Solder Based on Specific Gravity
QUESTION [0 upvotes]: The Question:
Different types of solder are alloys of different proportions of copper, tin and lead. The specific gravity of tin copper and lead are 8, 6 and 14 respectively while the specific gravities of three different solders are 9.6, 9.7 and 9.1.
Find the percentage composition of each solder.
The Issue:
I do not understand how it is possible to use the specific densities of the metals to find the densities of the solders as there are more variables than there are equations that can be formed (as far as I can decern).
Additionally, it may be helpful to know that this is a question from a chapter focused on Gaussian elimination.
Any help would be appreciated.

REPLY [0 votes]: The unstated assumption is that the volume of the mix is the sum of the volumes of the components.  The densities are in g/cm^3.  Under this assumption, if you mix $x$ grams of copper, $y$ grams of tin, and $z$  grams of lead the volume of the product will be $\frac x8+\frac y6+\frac z{14}$, so the specific gravity of the product will be $\frac {x+y+z}{\frac x8+\frac y6+\frac z{14}}$.  For the first mix we have to solve $\frac {x+y+z}{\frac x8+\frac y6+\frac z{14}}=9.6$  This looks like one equation in three unknowns, but you can eliminate one  because you can specify the total amount you make.  That still leaves you with one extra unknown and there is not a unique solution.


TITLE: Is the speed of light constant in all directions?
QUESTION [1 upvotes]: Consider the shape of the waveform emitted by a moving source:

(with the horizontal axis "x" and the vertical axis "y").
According to the stationary observer, the wavefront of a short pulse emitted like this will reach a given distance along y before it will along x.
Does the speed of light depend on direction?
EDIT: After researching the answers I personally found it easiest to understand by looking at the emission of the first pulse in this video. The picture above is misleading because it makes you think the red dot is the original center of each wave.

REPLY [2 votes]: Actually the answer is "we don't know", because we have never measured speed of light in only one direction. All experiments measure "there and back again" speed.
Speed of light being the same in all directions is an axiom postulated by Einstein. That also means that you can't use SR/GR to prove/disprove it (see Kuhn Cycle) because c being constant is foundational for the current paradigm.
P.S. as mentioned by others, Doppler shit does not affect the speed of light.
Update: there is a video: https://www.youtube.com/watch?v=pTn6Ewhb27k


TITLE: Limit related to $ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $?
QUESTION [5 upvotes]: Let $n$ be a positive integer.
Let $b = 2 n - 1$.
Let $x$ be a positive integer.
Define $f(x)$ as :
$$ f(x) = \prod_{i = 1}^x \left( \sin\left( i \frac{\pi}{n}\right) + \frac{5}{4}\right) $$
Then it appears that
$$ f(b) = \frac{4}{5} + C(n)$$
And $C(n)$ is close to zero.
In fact
$$ \lim_{n \to \infty} C(n) = 0 $$
How do we prove this ??

REPLY [5 votes]: Using
$$
\begin{align}
\sin\left(\frac{k\pi}n\right)+\frac54
&=\frac{e^{ik\pi/n}-e^{-ik\pi/n}}{2i}+\frac54\tag1\\
&=\frac{e^{-ik\pi/n}}{2i}\left(e^{ik\pi/n}+2i\right)\left(e^{ik\pi/n}+i/2\right)\tag2
\end{align}
$$
we get
$$
\begin{align}
\prod_{k=1}^{2n-1}\left(\sin\left(\frac{k\pi}n\right)+\frac54\right)
&=\frac45\prod_{k=0}^{2n-1}\color{#00F}{\frac{e^{-ik\pi/n}}{2i}}\color{#C00}{\left(e^{ik\pi/n}+2i\right)}\color{#090}{\left(e^{ik\pi/n}+i/2\right)}\tag3\\
&=\frac45\color{#00F}{\frac{(-1)^{n-1}}{2^{2n}}}\color{#C00}{\left[z^{2n}-1\right]_{z=2i}}\color{#090}{\left[z^{2n}-1\right]_{z=i/2}}\tag4\\[3pt]
&=\frac45\frac{(-1)^{n-1}}{2^{2n}}\left(2-(-1)^n\left(2^{2n}+2^{-2n}\right)\right)\tag5\\[6pt]
&=\frac45+\color{#90F}{\frac45\left((-1)^{n-1}2^{1-2n}+2^{-4n}\right)}\tag6\\[9pt]
&=\frac45+\color{#90F}{C(n)}\tag7
\end{align}
$$
Explanation:
$(3)$: $\frac45$ of the $k=0$ term is $1$; apply $(2)$
$(4)$: $\prod\limits_{k=0}^{2n-1}\!\left(z+e^{ik\pi/n}\right)=z^{2n}-1$
$(5)$: evaluate at $z=2i$ and $z=i/2$
$(6)$: simplify
$(7)$: the product is $\frac45+C(n)$
Thus,
$$
\begin{align}
C(n)
&=\frac45\left((-1)^{n-1}2^{1-2n}+2^{-4n}\right)\tag8\\[6pt]
&=O\!\left(4^{-n}\right)\tag9
\end{align}
$$


TITLE: Existence of reparametrization by arc length
QUESTION [1 upvotes]: I have got some issues in understanding the proof of the existence of reparametrization by arc length for any regular parameterized curve $\alpha :I\rightarrow R^n$.
I know the key is to construct a diffeomorphism $\phi :J \rightarrow I$ such that the speed of $\alpha \circ\phi$ is identially 1.
Why the following function's inverse is the diffeomorphism $\phi$ ?
$$\psi(t)=\int_{t_0}^t \left\lVert \alpha '(u)\right\rVert du $$
where $t_0$ is a point in I
By computation, $\psi '(t)=\left\lVert \alpha '(t)\right\rVert$ and it is always greater than 0 so there is no change of sign implies  $\left\lVert \alpha '(t)\right\rVert$  is smooth. Is it suffice to say that  $\left\lVert \alpha (t)\right\rVert$ is also smooth?
Is this fact also implies that the inverse of $\psi$ is smooth?

REPLY [2 votes]: You can calculate it:
Let $\tau:(0,L(\gamma))\mapsto (a,b) $ be the inverse of $\psi$ s.t. $\psi(\tau(s))=s$ for all $s$. By differenttiating we get
$$\frac{d}{ds}(\psi(\tau(s)) = \psi'(\tau(s))~\tau(s)=1. $$
Now we define $$\gamma:=\alpha(\tau)$$ then by the chain rule we get
$$\gamma(s)'=\alpha'(\tau(s))~\tau'(s).$$ And there for
$$|\gamma'(s)|=|\alpha'(\tau(s))|~\tau'(s)=\psi'(\tau(s))\tau'(s)=1$$
So the function $\tau$ is bijective and $\gamma$ is parametrises $\alpha(I)$ by arclength.


TITLE: Integer Congruence Explanation
QUESTION [2 upvotes]: Edit: I don't think I've expressed myself very well. This is another way of putting my question.
We can say two integers, a and b, are congruent mod m (where m is a natural number) if both numbers divided by m produce the same remainder. In other words, m must evenly divide their difference, a - b
Now, why are the two statements equivalent? (Why can we say "in other words"?)
I.e.
How to prove
$m\mid (a-b) \iff$ $a/m$ and $b/m$ have the same remainder
*** Original question below.
I understand that one definition of integer congruence is:
For a positive integer n, two numbers a and b are said to be congruent modulo n, if their difference a − b is an integer multiple of n
I also understand that the modulus operation (for mod n) can be seen as the remainder on division by n.
What I can't see is how these two concepts relate. Could someone please explain (perhaps using a number line) why the first definition holds?

REPLY [0 votes]: Okay so $$a-b=mx\implies a=mx+b$$ Now assume for a minute, $b>m$ and let $r<m$ be b's remainder on division by m, k it's integer quotient with m. We now have via substitution :$$a=mx+mk+r$$ regrouping the parts with m, we get $$a=m(x+k)+r$$ or by the first definition r is congruent to a mod m, because they are different by a multiple of m. Therefore, r is the remainder of both b and a, on division by m. They share the same remainder.


text:
\section{Discussion and Conclusion}\label{sec.conclusions}

We have studied the average-case complexity of teaching convex polytopes with halfspace queries, and showed that if the hyperplane arrangement is in $d'$-relaxed general position, then the average teaching complexity is $\bigTheta{d'}$. In contrast, the average-case sample complexity is $\bigTheta{d'\log n}$ for active learning and $\bigTheta{n}$ for passive learning. We showed that our insights could be applied to teaching $\phi$-separable dichotomies. Moreover, as discussed in details in the \appref{appendix: teaching ranking}, we further show that our insights in \secref{sec:avgtd} could be further generalized to the problem of teaching rankings over $n$ points $\{x_1, \dots, x_n\} \subseteq \mathbb{R}^d$ (encoded by their distances to an \emph{unknown} reference point $r \in \mathbb{R}^d$) via pairwise comparisons (e.g., ``is $x_i$ closer to $r$ than $x_j$''?).

One interesting line of future work is to understand whether our result could be extended to more general hyperplane arrangement settings. We believe our results provide useful geometrical insights for analyzing the average-case complexity for more complex hypothesis classes.

config: arxiv


text:
\begin{document}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}{Lemma}[section]
\newtheorem{observation}{Observation}[section]
\newtheorem{definition}{Definition}[section]
\newtheorem{corollary}{Corollary}[section]
\newtheorem{assumption}{Assumption}[section]
\newtheorem{property}{Property}[section]
\newtheorem{remark}{Remark}[section]
\newtheorem{example}{Example}[section]

\numberwithin{equation}{section}

\title{Numerical simulations of  X-rays Free Electron Lasers (XFEL)}
\thanks{Z. Huang was partially supported by the
NSFC Projects No.~11322113, ~91330203, the National Basic Research Program of
China under the grant 2011CB309705. P. Markowich
acknowledges support from the Royal Society through his Wolfson
Research Merit Award and the Award No. KUK-I1-007-43, funded by the
King Abdullah University of Science and Technology (KAUST)}

\author[P. Antonelli]{Paolo Antonelli}
\address[P. Antonelli]
{Centro di Ricerca Matematica Ennio De Giorgi, Scuola Normale Superiore, Piazza dei Cavalieri, 3, 56100 Pisa, Italy}
\email{paolo.antonelli@sns.it}

\author[A. Athanassoulis]{Agissilaos Athanassoulis}
\address[A. Athanassoulis]
{Department of
Applied Mathematics, University of Crete, Heraklion 71409, Greece}
\email{athanassoulis@tem.uoc.gr}

\author[Z.Y. Huang]{Zhongyi Huang$^*$}
\address[Z.Y. Huang]
{Dept. of Mathematical Sciences, Tsinghua University, Beijing
100084, China}
\email{zhuang@math.tsinghua.edu.cn}

\author[P. Markowich]{Peter A. Markowich}
\address[P. Markowich]
{King Abdullah University of Science and Technology (KAUST), MCSE
Division, Thuwal 23955-6900, Saudi Arabia}
\email{P.Markowich@kaust.edu.sa}

\thanks{$^*$ Corresponding author}
\begin{abstract}
We study a nonlinear Schr\"odinger equation which arises as an effective single particle
model in X-ray Free Electron Lasers (XFEL). This equation appears as a first-principles
model for the beam-matter interactions that would take place in an XFEL molecular imaging
experiment in \cite{frat1}. Since XFEL is more powerful by several orders of magnitude
than more conventional lasers, the systematic investigation of many of the standard
assumptions and approximations has attracted increased attention.

In this model the electrons move under a rapidly oscillating electromagnetic
field,  and the convergence of the problem to an effective time-averaged one is examined.
We use an operator splitting pseudo-spectral method to investigate numerically the
behaviour of the model versus its time-averaged version in complex situations, namely the
energy subcritical/mass supercritical case, and in the presence of a periodic lattice.

We find the time averaged model to be an effective approximation, even close to blowup,
for fast enough oscillations of the external field. This work extends previous analytical
results for simpler cases \cite{xfel1}.
\end{abstract}

\subjclass[2000]{65M70, 74Q10, 35B27, 81Q20} \keywords{X-ray free electron laser, nonlinear Schr\"odinger equation, time-splitting spectral method,
Bloch decomposition}
\maketitle

\tableofcontents

\newpage

\section{Introduction}

In this paper we study a first principles model for beam-matter interaction in X-ray free
electron lasers (XFEL) \cite{Ch,frat1}. Recent developments using XFEL include the
observation of the motion of atoms \cite{Br}, measuring the dynamics of atomic vibrations
\cite{Fri}, biomolecular imaging \cite{Neu} etc. The fundamental model for XFEL is the
following nonlinear Schr\"odinger equation
\beq \label{eq000} \left\{ \begin{aligned}
&i\hbar\d_t\psi=\frac{1}{2m}(i\hbar\nabla-e\Ab)^2\psi+Ze\frac{1}{|\xb|}\psi+e^2(|\cdot|^{-
1}\ast|\psi|^2)\psi -\lambda|\psi|^\sigma\psi,\\ &\psi(t,\xb)|_{t=0}=\psi_0(\xb),
\end{aligned}\right. \ee
on $\mathbb{R}^3$. Here $m_e$ is the electron mass, $e$ the
electron charge, $Z$ the atomic number and $\hbar$ is the scaled (by $2\pi$) Planck
constant.  The constant $\lambda>0$ measures the strenght of the local attractive
nonlinearity with power $\sigma>0$, in particular $\sigma=\frac{2}{3}$ and
$\lambda=3e^2\left(\frac{3}{8\pi}\right)^{1/3}$ for the local Hartree-Fock approximation.

A solution $\psi=\psi(t, \xb)$  of this Schr\"odinger equation can be considered as the
wave function of an electron beam, interacting self-consistently through the repulsive
Coulomb (Hartree) force, the attractive local Fock-type approximation with strength
$\lambda$ and exponent $\sigma$ and interacting repulsively with an atomic nucleus,
located at the origin. The vector field ${\Ab}={\Ab}(t)$ represents an external
electromagnetic field, which we shall assume to depend on time $t$ only (not on position
$\xb$).

In the experimental setup of interest \cite{Ch,frat1} the magnetic field is rapidly
oscillating. This effect of oscillating external magnetic field appears often in the
modelling of undulators; here the aim is to justify systematically the use of an
effective, time-averaged magnetic field, in the presence of beam-matter interaction.

 In \cite{xfel1} it was shown that, under appropriate assumptions, the  solution of
 \eqref{eq000} can be approximated, in the asymptotic regime $|\omega| \gg 1$, by the
 solution of an effective, time-averaged Schr\"odinger initial value problem. The present
 work is a continuation of that line of investigation in more general contexts. More
 specifically, with the help of numerical simulations, we are able to tackle more detailed
 questions with respect to the role of the fast magnetic field, the mode of convergence,
 and attack problems with blowup. Moreover, we investigate the effect of a periodic
 lattice on the model.

\subsection{The model}

We scale the equation (1.1) by choosing a characteristic time scale $t_c$ and a characteristic spatial scale $x_c$. Defining the semiclassical parameter $\eps=\frac{\hbar t_c}{2m_ex_c^2}$ we obtain after rescaling

\begin{equation*}
i\eps\d_t\psi=(i\eps\nabla-\Ab)^2\psi+\frac{c}{|x|}\psi+C_1(|\cdot|^{-1}\ast|\psi|^2)\psi-a|\psi|^\sigma\psi
\end{equation*}
where $c=eZ\frac{t_c^2}{2m_ex_c}$, $C_1=e^2\frac{t_c^6}{8m_e^3x_c^4}$,
$a=\lambda\left(\frac{t_c^2}{2m_ex_c^2}\right)^{\sigma+1}$.
Here also $\psi$ and $\Ab$ were scaled appropriately.

A standard transform is used to simplify the original equation, and bring it to a form more amenable to analysis as well as computation. By setting
\beq
\label{eqlenrd}
u(t,\xb) = \psi\(t, \xb + b(t)\) \exp\(\frac{i}{\eps}\dpm\int_0^t{|\Ab(s)|^2ds}\),
\ee
where $b(t)=2\int_0^t\Ab(s)ds$,
one readily checks that $u$ satisfies the initial value problem
\begin{equation}
\label{eq:fast-1}
\left\{\begin{aligned}
&i\eps\partial_tu=-\eps^2\Delta u+\frac{c}{|\xb-b(t)|} u+C_1(|\cdot|^{-1}\ast|u|^2)u
-a|u|^\sigma u\\
&u^\omega(t,\xb)\big|_{t=0}=u_0(\xb),
\end{aligned}\right.
\end{equation}
where $u_0:=\psi_0$.
Now we choose the scales $x_c, t_c$ such that $\eps=1$ (thus relating $x_c$ and $t_c$ by $2m_ex_c^2=\hbar t_c$), since we are mostly interested in the fully quantum mechanical case, and we fix
\begin{equation*}
b(t)=e(t)\sin(2\pi\omega t).
\end{equation*}
where $e:\R\to\R^3$ is a smooth vector field, slowly varying in time.
We consider the large frequency case
\begin{equation*}
|\omega|\gg\frac{{\rm sec}}{t_c}=\frac{\hbar}{2m_e}\left(\frac{{\rm meter}}{x_c}\right)^2\approx 5\times10^{-3}\left(\frac{{\rm meter}}{x_c}\right)^2.
\end{equation*}
and study the asymptotic regime $|\omega|\to\infty$. In particular we shall investigate the convergence of solutions of
\begin{equation}\label{eq:fast}
\left\{\begin{aligned}
&i\partial_tu^\omega=-\Delta u^\omega+V^\omega u^\omega+C_1(|\cdot|^{-1}\ast|u^\omega|^2)u^\omega-a|u^\omega|^\sigma u^\omega\\
&u^\omega(t,\xb)\big|_{t=0}=u_0(\xb),
\end{aligned}\right.
\end{equation}
where the fast oscillatory potential is given by
\begin{equation}\label{eq:fast-p}
V^\omega(t, x):=\frac{c}{|\xb-e(t)\sin(2\pi\omega t)|},
\end{equation}
to solutions of the approximate problem (see \cite{xfel1} and Theorem 1 there)
\begin{equation}\label{eq:avr}
\left\{\begin{aligned}
&i\partial_tu=-\Delta u+\langle V\rangle u+C_1(|\cdot|^{-1}\ast|u|^2)u-a|u|^\sigma u\\
&u(t,\xb)\big|_{t=0}=u_0(\xb),
\end{aligned}\right.
\end{equation}
with $\langle V\rangle$ is the time \emph{averaged} potential given by
\begin{equation}\label{eq:avr-p}
\langle V\rangle(t, \xb):=\int_0^1\frac{c}{|\xb-e(t)\sin(2\pi s)|}ds.
\end{equation}
We remark that \eqref{eq:avr-p} is well defined as the weak limit of $V^\omega$ defined in \eqref{eq:fast-p}
in the space $L^\infty(\R;L^{3-}(\R^3)+L^{3+}(\R^3))$, as $|\omega|\to\infty$ (see \cite{xfel1} for more details).
Passing back and forth between $u$ and $\psi$ (models (\ref{eq:fast}) and (\ref{eq000})) is completely straightforward, by virtue of the transformation (\ref{eqlenrd}).
\begin{remark}
In molecular imaging (spatial characteristic scale of the order of magnitude of thousand nanometers, temporal characteristic scale of some femtoseconds), we compute
$\omega\gg10^{15}$. Therefore it makes sense to investigate the asymptotic regime. We shall demonstrate later on that this is reached already for much smaller values of $\omega$.
\end{remark}

\subsection{Main objectives}

The main result of \cite{xfel1} was the following

\begin{theorem}
Let $0<\sigma<4/3$, $u_0\in L^2(\R^3)$, $u^\omega, u\in C(\R; L^2(\R^3))$ be the unique global solutions of
\eqref{eq:fast}, \eqref{eq:avr}, respectively. Then
for each finite time $0<T<\infty$ and for each admissible Strichartz index pair $(q, r)$, we have
\begin{equation*}
\|u^\omega-u\|_{L^q([0, T]; L^r(\R^3))}\to0\qquad\textrm{as}\;|\omega|\to\infty.
\end{equation*}
\end{theorem}

We use this averaging result as a benchmark (example \ref{ex1}); the energy for the fast problem is observed to converge weakly to the energy for the
averaged problem (and not strongly $L^\infty$ in time). We proceed to investigate mass supercritical problems in example \ref{ex2}, i.e. the exponent $\sigma$ is taken to be larger than $\frac{4}3$. The point is to investigate whether the time-averaged model still gives a good description near blowup. We proceed to an example that highlights the interaction with the (rapidly oscillating) external electromagnetic potential in example \ref{ex:shot}, and with a two-time-scale dependent electromagnetic potential in example \ref{ex:td}. Finally we investigate the interaction with a periodic lattice in example \ref{ex:pl}.

\begin{figure}[htb!]
\centering
\includegraphics[width=0.35\textwidth]{omega640.pdf}
\includegraphics[width=0.35\textwidth]{wf_err_640.pdf}
\caption{
Convergence of the energy for problem (\ref{eq:fast}) to that of (\ref{eq:avr}) is a good way to keep track of the time-averaging,
as well as a diagnostic tool for the computation. Here we show the differences of the energies (left) and the wave functions (right)
between the fast potential model and time-averaged model. We take the time local average of $E^\og(t)$ over $[t,t+0.005]$.

}

\label{Fig222}
\end{figure}

\section{Numerical methods for nonlinear Schr\"odinger equations \eqref{eq:fast} and \eqref{eq:avr}}

As we have shown in \cite{BJM, HJMS, HJMS-2,HJMSZ} the time-splitting pseudo-spectral/Bloch-decomposition based
pseudo-spectral methods works well for (non)linear Schr\"odinger equation (with periodic potential), thus
we shall focus on these two kind of methods in this paper. We have shown that these two methods have spectral convergency in spatial discretization
\cite{BJM, HJMS, HJMS-2,HJMSZ}.
\subsection{Time-splitting spectral algorithm}\label{sec:2.1}
We solve \eqref{eq:fast} or \eqref{eq:avr}
by a classical time-splitting spectral scheme as follows \cite{BJM, HJMSZ}:

\textbf{Step 1.} We solve the equation
\begin{equation}
\label{ts1}
\begin{aligned}
&\, \I \partial _t u_1 = -\, \btu u_1, \\
\end{aligned}
\end{equation}
on a fixed time interval $\Delta t$, relying on the
\emph{pseudo-spectral} method.
Thus we can get the value of $u_1$ at new time
$t^{n+1}$ by
\[u^{n+1}_1=\(e^{-\I\btu t |\xi|^2}\hat{u}_1^{n}(\xi)\)^\vee,\]
here `$\wedge$' and `$\vee$' denote the Fourier and inverse Fourier
transform respectively.

Let $\rho(t^{n+1},\xb)=|u_1(t^{n+1},\xb)|^2$, then we compute the Hartree potential $V_{Hartree}$ by
\[V_{Hartree}(t^{n+1},\xb)=\(\frac{1}{|\xi|^2}\hat{\rho}(t^{n+1}, \xi)\)^\vee.\]

\textbf{Step 2.} We solve the ordinary differential equation
\begin{equation}
\label{ts2}
\begin{aligned}
& \,\I \partial _tu_2  = \left(C_1V_{Hartree}+V(t,\xb)-a|u_2|^\sg\right)u_2, \quad u_2(t^{n},\xb)=u_1(t^{n+1},\xb),
\end{aligned}
\end{equation}
on the same time-interval, here $V(t,\xb)$ is given by \eqref{eq:fast-p} or \eqref{eq:avr-p}.
It is easy to check that in this step the density $\rho(t,\xb)\equiv|u_2|^2(t,\xb)$ and consequently the Hartree potential do not change in time, hence
\[ u_2(t+\btu t,\xb)=u_2(t,\xb)\,\exp\({\dpm-\I \int_{t}^{t+\btu t}
\left(C_1V_{Hartree}(\xb)+V(s,\xb)-a\rho^\frac{\sg}{2}(\xb)\right) ds}\)
\] with $\rho(\xb)=|u_1(t^{n+1},\xb)|^2$, $V_{Hartree}(x)=\(\frac{1}{|\xi|^2}\hat{\rho}(\xi)\)^\vee$.

The key is to calculate the oscillatory integral,
\beq\int_{t}^{t+\btu t} \frac{1}{\left|\xb-b(s)\right|}ds\equiv \int_{t}^{t+\btu t} \frac{1}{\left|\xb-e(s)\sin(2\pi\og s)\right|}ds,\label{2.3}\ee
where $|\og|\gg1$, particularly due to the occurrence of the singularity in the kernel.
Therefore, we use a regularization technique. For example, we could take
\beq\int_{t}^{t+\btu t} \frac{1}{\left|\xb-b(s)\right|}ds\approx \int_{t}^{t+\btu t}\Phi^\eta \(\frac{1}{\left|\xb-b(s)\right|}\)ds,\label{2.4}\ee
where
\[\Phi^\eta \(\frac{1}{\left|\xb-b(s)\right|}\)=\(\frac{2}{\eta}\)^{\frac{3}{2}}
\int_{\R^3}e^{-2\pi\frac{|\xb-\xb'|^2}{\eta}}\frac{1}{\left|\xb'-b(s)\right|}d\xb'.\]

In Section \ref{anal:sect} we will discuss the stability with respect to perturbations of the potential.

The energies for two models are
\beq\label{eq:energy1}
E^\og(t)\df\int_{\R^3}\(|\btd u^\omega|^2+\frac{C_1}{2}|\btd V_{Hartree}^\omega|^2
+cV^\og|u^\omega|^2-a\frac{2}{\sg+2}|u^\omega|^{\sg+2}\)d\xb,
\ee
and
\beq\label{eq:energy2}
E(t)\df\int_{\R^3}\(|\btd u|^2+\frac{C_1}{2}|\btd V_{Hartree}|^2+{c}\langle V\rangle|u|^2-a\frac{2}{\sg+2}|u|^{\sg+2}\)d\xb,
\ee
respectively. The total mass
\beq\label{eq:mass}
M(t)\df\int_{\R^3}|u(t,\xb)|^2d\xb,
\ee
is conserved in both models.

\subsection{Bloch-decomposition based time-splitting spectral algorithm}\label{sec:2.2}
If we consider an external potential $V_{ext}\(\xb\)$ in the equation \eqref{eq000} \cite{Fri,MT},
we have
\[
i\eps\partial_t\psi=(i\eps\nabla-A)^2\psi+V_{ext}(\xb)\psi+c\frac{1}{|\xb|}\psi+C_1(|\cdot|^{-1}\ast|\psi|^2)\psi
-a|\psi|^\sigma\psi,
\]
by the same transformation \eqref{eqlenrd} and by fixing $\eps=1$, we obtain the following equation
\begin{equation}\label{ext-fast}
i\partial_tu^\omega=-\Delta u^\omega+V_{ext}\(\xb-b(t)\)u^\omega+V^\omega u
+C_1(|\cdot|^{-1}\ast|u^\omega|^2)u^\omega-a|u^\omega|^\sigma u^\omega,
\end{equation}
and its time-averaged problem is
\begin{equation}\label{ext-avr}
i\partial_tu=-\Delta u+\langle V_{ext}\rangle u+\langle V\rangle u+C_1(|\cdot|^{-1}\ast|u|^2)u-a|u|^\sigma u,
\end{equation}
with
\[\langle V_{ext}\rangle(t,\xb) = \dpm\int\limits_{-\frac{1}2}^{{\frac{1}2}} V_{ext}\left(\xb-e(t) \sin(2\pi  s)\right) ds.\]

In particular, if we consider a case with an periodic potential $V_\Gm\(\xb\)$ \cite{Fri,MT},
\ie $V_\Gm(\yb)$ is periodic w.r.t to a regular lattice $\Gm$,
\[V_\Gm(\yb+\gm)=V_\Gm(\yb), \quad \fa \gm\in\Gm, \yb\in \R^3,\]
the above two equations \eqref{ext-fast}--\eqref{ext-avr} become
\begin{equation}\label{bloch-fast}
i\partial_tu^\omega=-\Delta u^\omega+V_\Gm\(\xb-b(t)\)u^\omega+V^\omega u
+C_1(|\cdot|^{-1}\ast|u^\omega|^2)u^\omega-a|u^\omega|^\sigma u^\omega,
\end{equation}
and its time-averaged problem is
\begin{equation}\label{bloch-avr}
i\partial_tu=-\Delta u+\langle V_\Gm\rangle\(t,\xb\) u+\langle V\rangle u+C_1(|\cdot|^{-1}\ast|u|^2)u-a|u|^\sigma u,
\end{equation}
with
\[\langle V_{\Gm}\rangle(t,\xb) = \dpm\int\limits_{-\frac{1}2}^{{\frac{1}2}} V_{\Gm}\left(\xb-e(t) \sin(2\pi  s)\right) ds.\]

We solve \eqref{bloch-fast} or \eqref{bloch-avr} by using the \emph{Bloch-decomposition based time-splitting method}, see
\cite{HJMS, HJMS-2}.

\textbf{Step 1.} We solve the equation
\begin{equation}
\label{bd1}
\begin{aligned}
&\, \I \partial _t u_1 = -\, \btu u_1 +V_\Gm\(t,\xb\) u_1, \\
\end{aligned}
\end{equation}
on a fixed time interval $\Delta t$, relying on the
\emph{Bloch-decomposition based pseudo-spectral} method.

Let $\rho(t^{n+1},\xb)=|u_1(t^{n+1},\xb)|^2$, then we obtain the Hartree potential $V_{Hartree}$ by
\[V_{Hartree}\(t^{n+1},\xb\)=\(\frac{1}{|\xi|^2}\hat{\rho}(t^{n+1}, \xi)\)^\vee.\]

\textbf{Step 2.} Then, we solve the ordinary differential equation
\begin{equation}
\label{bd2}
\begin{aligned}
& \,\I \partial _tu_2  = \left(C_1V_{Hartree}+V(t,\xb)-a|u_2|^\sg\right)u_2, \quad u_2\(t^{n},\xb\)=u_1\(t^{n+1},\xb\),
\end{aligned}
\end{equation}
on the same time-interval, here $V(t,\xb)$ is given by \eqref{eq:fast-p} or \eqref{eq:avr-p}.
It is easy to check that  in this step the density $\rho(t,\xb)\equiv|u_2|^2(t,\xb)$ and consequently the Hartree potential do not change in time, hence
\[ u_2(t+\btu t,\xb)=u_2(t,\xb)\,\exp\({\dpm-\I \int_{t}^{t+\btu t}
\(C_1V_{Hartree}(\xb)+V(s,\xb)-a\rho^\frac{\sg}{2}(\xb)\) ds}\)
\] with $\rho(\xb)=\left|u_1(t^{n+1},\xb)\right|^2$, $V_{Hartree}(x)=\(\frac{1}{|\xi|^2}\hat{\rho}(\xi)\)^\vee$.

\begin{remark} Certainly, if there is no lattice potential $V_\Gamma$, the above two algorithms are coincided with each other.
But we must include the lattice potential $V_\Gamma$ when the interaction with laser or other lattice potenital are not eligible.
We will see that the Bloch-decomposition based method is much more efficient in the case with lattice potential $V_\Gamma$
than the traditional pseudo-spectral method (\cf Fig. \ref{Fig012}-\ref{Fig013}), \ie we could use much larger mesh size
than the traditional pseudo-spectral method to get the same accuracy.
\end{remark}
\section{Numerical results}

\subsection{Numerical experiments}

\subsubsection{Convergence to the time-averaged model}
\begin{example}\label{ex1} Here we set the semiclassical parameter $\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=\frac{2}{3}$, $a=50$, $C_1=20$, $c=1$.
We compare the time-averaged model with the fast-potential model for different values of $\og$.
We choose appropriate constants $C_1$ and $c$ to make the two potential components in the energy comparable.
\end{example}
\begin{table}[h t]
\begin{center}
\caption{Example \ref{ex1}: errors between two models vs frequency $\og$.}\label{tab:ex1}\vspace{1mm}

\begin{tabular}{c|cccc}\hline
 $\og$  & 5  & 10  & 20 & 40     \\ \hline
$\|u-u^\og\|_{L^2}$ & 1.1E-1 & 5.4E-2 & 2.9E-2 & 1.8E-2  \\ \hline
convergence rate &  &  1.0 & 0.9 & 0.7  \\ \hline
$\|E-E^\og\|_{L^1}$ & 7.1E-1  & 4.5E-1 & 2.9E-1 & 1.9E-1 \\ \hline
convergence rate &  & 0.6 & 0.6 & 0.6  \\ \hline
\end{tabular}
\end{center}
\end{table}

\begin{figure}[!ht]
\centering
\includegraphics[width=40mm]{energy_og_5.pdf}
\includegraphics[width=40mm]{energy_og_10.pdf}
\includegraphics[width=40mm]{energy_og_20.pdf}
\includegraphics[width=40mm]{energy_og_40.pdf}

\hspace{0.3cm}$\og=5$ \hspace{2.95cm} $\og=10$ \hspace{2.9cm} $\og=20$\hspace{2.9cm} $\og=40$
\caption{
Example \ref{ex1}: Comparison of the energies of two models with different $\omega$. Here we take the average of $E^\og(t)$ over $[t,t+0.1]$.}
\label{Fig001}
\end{figure}

\begin{figure}[!ht]
\centering
\includegraphics[width=40mm]{wf_err_5.pdf}
\includegraphics[width=40mm]{wf_err_10.pdf}
\includegraphics[width=40mm]{wf_err_20.pdf}
\includegraphics[width=40mm]{wf_err_40.pdf}

\hspace{0.3cm}$\og=5$ \hspace{2.95cm} $\og=10$ \hspace{2.9cm} $\og=20$\hspace{2.9cm} $\og=40$
\caption{
Example \ref{ex1}:
The relative distance between the wavefunctions of two models with different $\omega$: $\frac{\|u^\og-u\|_{L^2}}{\|u\|_{L^2}}$.}
\label{Fig001-1}
\end{figure}

\begin{figure}[!ht]
\centering
\includegraphics[width=50mm]{ground_s.pdf}
\includegraphics[width=50mm]{V_a.pdf}\vspace{-2mm}

Initial datum $u_0(x,y,0)$ \hspace{2cm} $\overline{V}(x,y,0)$ \hspace{1cm}\vspace{2mm}

\caption{Example \ref{ex1}: Graphs of the time-averaged model, the parameters are:
$\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=\frac{2}{3}$, $a=50$, $C_1=1$, $c=0.001$.
Our numerical results show the conservation of the energy $E(t)$ (given by \eqref{eq:energy2})
and mass $M(t)$ (given by \eqref{eq:mass}) in both time-averaged model and fast-potential model.
}
\label{Fig002}
\end{figure}

\begin{figure}[!ht]
\centering
\includegraphics[width=40mm]{omega080.pdf}
\includegraphics[width=40mm]{omega160.pdf}
\includegraphics[width=40mm]{omega320.pdf}
\includegraphics[width=40mm]{omega640.pdf}

\hspace{0.1cm} $\og=80$ \hspace{2.6cm} $\og=160$ \hspace{2.6cm} $\og=320$ \hspace{2.6cm} $\og=640$
\caption{
Example \ref{ex1}:
The distance between the energies of two models with larger $\omega$ vs. time.
Here we take the average of $E^\og(t)$ over $[t,t+0.005]$.}
\label{Fig001-add}
\end{figure}

\begin{remark} From Figures \ref{Fig001}--\ref{Fig002}, we can see that as $|\og|\to\infty$, the average of the energy $E^\og(t)$ converges to
the energy $E(t)$ of time-averaged model. The corresponding wave-function $u^\og(t,\xb)$ also converges to $u(t,\xb)$ at least in $L^2$-norm.
Figure \ref{Fig002} also shows the conservation of $E(t)$ and $M(t)$.
\end{remark}

\subsubsection{Blow-up tests}
\begin{example} \label{ex2} Here we consider an example with $\eps=1$, $e(t)=(0,0,1)^T$, $a=50$, $C_1=100$, $c=0.1$.
We start the time-averaged model from a ground state for $\sg=2/3$ (\cf plot (a) in Fig. \ref{Fig003}).
Then we consider \eqref{eq:fast} with a mass supercritical power-type nonlinearity, i.e. $\sg>4/3$, we analyse the blow-up of the norm
\beq\label{eq:H1}\|\btd u\|_{L^2}=\left(\int_{\R^3}|\btd u(t,\xb)|^2d\xb\right)^{1/2}.\ee
\end{example}

In Fig. \ref{Fig003}--\ref{Fig004}, we can see that the blow-up time decreases as $\sg$ increases.

\begin{figure}[!ht]
\centering
\includegraphics[width=50mm]{initial_40_100_64.pdf}
\includegraphics[width=50mm]{bl_2_end_psi_40_100_64.pdf}\vspace{-2mm}

Initial datum $u_0(x,y,0)$ \hspace{2cm} $|u(t,x,y,0)|$ \hspace{1cm}\vspace{2mm}

\includegraphics[width=50mm]{bl_H1_2_40_100_64.pdf}
\includegraphics[width=50mm]{gs_mass.pdf}

$\|\btd u\|_{L^2}$ (given by \eqref{eq:H1}) \hspace{1.5cm} $M(t)$ (given by \eqref{eq:mass})
\caption{Example \ref{ex2}: Graphs of the time-averaged model, the parameters are:
$\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=2$, $a=50$, $C_1=100$, $c=0.1$.
Blow-up case.}
\label{Fig003}
\end{figure}

\begin{figure}[h!t]
\centering
\includegraphics[width=50mm]{initial_40_100_64.pdf}
\includegraphics[width=50mm]{bl_15_end_psi_40_100_64.pdf}\vspace{-2mm}

Initial datum $u_0(x,y,0)$ \hspace{2cm} $|u(t,x,y,0)|$ \hspace{1cm}\vspace{2mm}

\includegraphics[width=50mm]{bl_H1_15_40_100_64.pdf}
\includegraphics[width=50mm]{gs_mass.pdf}

$\|\btd u\|_{L^2}$ (given by \eqref{eq:H1}) \hspace{1.5cm} $M(t)$ (given by \eqref{eq:mass})
\caption{Example \ref{ex2}: Graphs of the time-averaged model, the parameters are:
$\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=1.5$, $a=50$, $C_1=100$, $c=0.1$.
Blow-up case.}
\label{Fig004}
\end{figure}

\begin{figure}[!ht]
\centering
\includegraphics[width=50mm]{psi_end_fast_2.pdf}
\includegraphics[width=50mm]{H1_fast_2.pdf}

\hspace{1cm}$u(t,x,y,0)$ \hspace{2.5cm} $\|\btd u\|_{L^2}$ (given by \eqref{eq:H1})
\caption{Example \ref{ex2}: Graphs of the fast-potential model, the parameters are:
$\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=2$, $a=50$, $C_1=100$, $c=0.1$, $\og=10^4$.
Blow-up case.}
\label{Fig005}
\end{figure}

\begin{figure}[h!t]
\centering
\includegraphics[width=50mm]{psi_end_fast_15.pdf}
\includegraphics[width=50mm]{H1_fast_15.pdf}

\hspace{1cm}$u(t,x,y,0)$ \hspace{2.5cm} $\|\btd u\|_{L^2}$ (given by \eqref{eq:H1})
\caption{Example \ref{ex2}: Graphs of the fast-potential model, the parameters are:
$\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=1.5$, $a=50$, $C_1=100$, $c=0.1$, $\og=10^4$.}
\label{Fig006}
\end{figure}

Comparing the $L^2$ norms of the gradient of the wave-function for the time averaged model and the fast model in the blow up test we can see that they are very similar in the super-critical case
(\cf Figures \ref{Fig005}--\ref{Fig006} and Figures \ref{Fig003}--\ref{Fig004}).
They exhibit more differences in the subcritical case (see Fig. \ref{Fig007}--\ref{Fig008}).

\begin{figure}[h!t]
\centering
\includegraphics[width=50mm]{fast_psi1.pdf}
\includegraphics[width=50mm]{fast_psi_pk1.pdf}
\includegraphics[width=50mm]{fast_sub1.pdf}

\hspace{0.6cm}$|u(t,x,y,0)|_{t=0.15}|$ \hspace{2cm} $|u(t,x,y,0)|_{t=0.11}|$ \hspace{1.8cm} $\|\btd u\|_{L^2}$ (given by \eqref{eq:H1})\vspace{2mm}

\includegraphics[width=50mm]{average_psi1.pdf}
\includegraphics[width=50mm]{average_psi_pk1.pdf}
\includegraphics[width=50mm]{average_sub1.pdf}

\hspace{0.6cm}$|u(t,x,y,0)|_{t=0.15}|$ \hspace{2cm} $|u(t,x,y,0)|_{t=0.11}|$ \hspace{1.8cm} $\|\btd u\|_{L^2}$ (given by \eqref{eq:H1})

\caption{Example \ref{ex2}: Graphs of the fast-potential model (first line) and time-averaged model (second line), the parameters are:
$\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=1$, $a=50$, $C_1=100$, $c=0.1$, $\omega=10^4$.}
\label{Fig007}
\end{figure}

\begin{figure}[h!t]
\centering
\includegraphics[width=50mm]{h1_fast_075.pdf}
\includegraphics[width=50mm]{h1_av_075.pdf}

$\|\btd u^\og\|_{L^2}$ (given by \eqref{eq:H1}) \hspace{1.1cm} $\|\btd u\|_{L^2}$ (given by \eqref{eq:H1})

\caption{Example \ref{ex2}: Graphs of the fast-potential model (left) and time-averaged model (right), the parameters are:
$\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=0.75$, $a=50$, $C_1=100$, $c=0.1$, $\omega=10^4$.}
\label{Fig008}
\end{figure}

Furthermore, from Figures \ref{Fig009}--\ref{Fig009-1}, we can see that the Coulomb potential significantly impacts the wave packet at the early stage. This becomes particularly clear with slower blow-up and smaller frequency $\og$.

\begin{figure}[h!t]
\centering
\includegraphics[width=50mm]{eng-super-bl-flat.pdf}
\includegraphics[width=50mm]{H1-super-bl-flat.pdf}
\includegraphics[width=50mm]{ini-super-bl-flat.pdf}

\hspace{0.6cm}$E(t)$ (given by \eqref{eq:energy1}) \hspace{1.6cm} $\|\btd u\|_{L^2}$ (given by \eqref{eq:H1})\hspace{2.2cm} $|u(t,x,y,0)|_{t=0}|$ \vspace{4mm}

\includegraphics[width=50mm]{psi1-super-bl-flat.pdf}
\includegraphics[width=50mm]{psi2-super-bl-flat.pdf}
\includegraphics[width=50mm]{psi3-super-bl-flat.pdf}

\hspace{0.6cm}$|u(t,x,y,0)|_{t=0.006}|$ \hspace{1.8cm} $|u(t,x,y,0)|_{t=0.0105}|$ \hspace{1.8cm} $|u(t,x,y,0)|_{t=0.0425}|$ \hspace{2mm}

\caption{Example \ref{ex2}: Graphs of the fast-potential model, the parameters are:
$\eps=1$, $e(t)=(1,1,1)^T$, $\sigma=1.35$, $a=40$, $C_1=200$, $c=10$, $\omega=10^2$.}
\label{Fig009}
\end{figure}

\begin{figure}[h!t]
\centering
\includegraphics[width=40mm]{ini-super-con.pdf}
\includegraphics[width=40mm]{H1-super-con.pdf}
\includegraphics[width=40mm]{psi1-super-con.pdf}
\includegraphics[width=40mm]{ct1-super-con.pdf}

$|u(0,x,y,0)|$ \hspace{1.2cm} $\|\btd u\|_{L^2}$ (given by \eqref{eq:H1})\hspace{0.5cm}  $|u(t,x,y,0)|_{t=0.002275}|$\hspace{1.2cm} contour plot\vspace{5mm}

\includegraphics[width=40mm]{psi3-super-con.pdf}
\includegraphics[width=40mm]{ct3-super-con.pdf}
\includegraphics[width=40mm]{psi6-super-con.pdf}
\includegraphics[width=40mm]{ct6-super-con.pdf}

$|u(t,x,y,0)|_{t=0.00819}|$\hspace{1.4cm} contour plot\hspace{1.4cm} $|u(t,x,y,0)|_{t=0.020475}|$\hspace{1.45cm} contour plot\hspace{0.5cm}\vspace{5mm}

\includegraphics[width=40mm]{psi8-super-con.pdf}
\includegraphics[width=40mm]{ct8-super-con.pdf}
\includegraphics[width=40mm]{psi10-super-con.pdf}
\includegraphics[width=40mm]{ct10-super-con.pdf}

$|u(t,x,y,0)|_{t=0.03276}|$\hspace{1.45cm} contour plot\hspace{1.5cm} $|u(t,x,y,0)|_{t=0.045}|$\hspace{1.65cm} contour plot\hspace{0.6cm}

\caption{Example \ref{ex2}: Graphs of the fast-potential model, the parameters are:
$\eps=1$, $e(t)=(-1,-1,-1)^T$, $\sigma=1.35$, $a=40$, $C_1=200$, $c=10$, $\omega=5$.}
\label{Fig009-1}
\end{figure}

\subsubsection{Interaction of the magnetic field with the particle}
\begin{example}\label{ex:shot} Here we are interested in the patterns of interaction of the magnetic field with the particle, especially when a particle is shot towards the field, and scattered by it. We set $e=(0,0,1)^T$ in eq. (\ref{eq:fast-p}), and choose an initial datum of the form
\[
u_0(\xb)=e^{-4|\xb-(1,1,0)^T|^2+ i (x_1-x_2)}, \quad \mbox{ for } \xb=(x_1,x_2,x_3)^T.
\]
In this example, we add a harmonic trap potential
\[V(\xb)=50|\xb|^2,\]
\ie the equation in \eqref{eq:fast} becomes
\begin{equation}
i\d_tu^\omega=-\Delta u^\omega+V\(\xb-b(t)\)u^\omega+V^\omega u^\omega
+C_1(|\cdot|\ast|u^\omega|^2)u^\omega-a|u^\omega|^\sigma u^\omega
\end{equation}
\end{example}

The results are shown in Figure \ref{Fig010}. We can see that the wave packet moves under the interaction with the trap $V(\xb)$.

\begin{figure}[h!t]
\centering
\includegraphics[width=50mm]{exam3_0.pdf}
\includegraphics[width=50mm]{exam3_1.pdf}
\includegraphics[width=50mm]{exam3_2.pdf}\vspace{1mm}

\includegraphics[width=50mm]{exam3_3.pdf}
\includegraphics[width=50mm]{exam3_4.pdf}
\includegraphics[width=50mm]{exam3_5.pdf}\vspace{1mm}

\includegraphics[width=50mm]{exam3_6.pdf}
\includegraphics[width=50mm]{exam3_7.pdf}
\includegraphics[width=50mm]{exam3_8.pdf}\vspace{1mm}

\includegraphics[width=50mm]{exam3_9.pdf}
\includegraphics[width=50mm]{exam3_10.pdf}
\includegraphics[width=50mm]{exam3_e.pdf}
\caption{Example \ref{ex:shot}: (fast-potential model), the parameters are:
$\eps=1$, $e(t)=(0,0,1)^T$, $\sigma=\frac{2}{3}$, $a=50$, $C_1=20$, $c=1$, $\omega=10^2$.}
\label{Fig010}
\end{figure}

\subsubsection{Time-dependent vectors $e(t)$}
\begin{example}\label{ex:td} We consider a time-dependent vector $e(t)=(-1,-1,-1)\sin 2\pi t$ in eq. (\ref{eq:fast-p}), and an initial datum of the form
\[
u_0(x)=e^{-4|\xb|^2}.
\]
\end{example}

The results are shown in Figure \ref{Fig011}. As $e(t)$ is slowly varying in time, its effects are not very pronounced except when $e(t)$ is close to 0.

\begin{figure}[h!t]
\centering
\includegraphics[width=40mm]{H1-fast.pdf}
\includegraphics[width=40mm]{psi-3-fast.pdf}
\includegraphics[width=40mm]{psi-7-fast.pdf}
\includegraphics[width=40mm]{psi-10-fast.pdf} \vspace{5mm}

\includegraphics[width=40mm]{H1-average.pdf}
\includegraphics[width=40mm]{psi-3-average.pdf}
\includegraphics[width=40mm]{psi-7-average.pdf}
\includegraphics[width=40mm]{psi-10-average.pdf}
\caption{Example \ref{ex:td}: Graphs of the example \ref{ex:td} for fast-potential model (first line) and time-averaged model (second line), the parameters are:
$\eps=1$, $e(t)=(-1,-1,-1)^T\sin 2\pi t$, $\sigma=\frac{2}{3}$, $a=40$, $C_1=100$, $c=2$, $\omega=10^2$.}
\label{Fig011}
\end{figure}

\subsubsection{Interaction with periodic lattice}
\begin{example}\label{ex:pl} Finally, we consider a case with periodic lattice,
\ie the equation in \eqref{eq:fast} becomes
\begin{equation}
i\d_tu^\omega=-\Delta u^\omega+V^\omega u^\omega+V_\Gm\(\xb-b(t)\)u^\omega
+C_1(|\cdot|^{-1}\ast|u^\omega|^2)u^\omega-a|u^\omega|^\sigma u^\omega,
\end{equation}
and its time-averaged problem is
\begin{equation}
i\d_tu=-\Delta u+\langle V\rangle u+\langle V_\Gamma\rangle u+C_1(|\cdot|^{-1}\ast|u|^2)u-a|u|^\sigma u.
\end{equation}
We also take the initial datum of the form
\[
u_0(x)=e^{-4|\xb|^2}.
\]
Here $V_\Gm(\yb)$ is periodic w.r.t to a regular lattice $\Gm$,
\[V_\Gm(\yb+\gm)=V_\Gm(\yb), \quad \fa \gm\in\Gm,\ \yb\in \R^3.\]
For sake of simplicity, we choose \[V_\Gm(\yb)=\sum_{l=1}^3 \sin^2(\og_l y_l), \quad \mbox{ for }\yb=(y_1,y_2,y_3)^T\in\R^3, \]
with $\og_1=\og_2=\og_3=2\pi$.
\end{example}

The results are shown in Figures \ref{Fig012}-\ref{Fig013}. We can see the interaction between the lattice potential and the wave packet.
Certainly, the wave function of the fast model is more peaked than the solution of the time-averaged model. From these figures, we can also
find that the Bloch-decomposition based algorithm is more efficient than traditional pseudo-spectral method in this case.
\begin{figure}[h!t]
\centering
\includegraphics[width=50mm]{fast80.pdf}
\includegraphics[width=50mm]{fast160.pdf}
\includegraphics[width=50mm]{avr.pdf}
\caption{Graphs of $|u^\og(t,x,y,0)|^2$ and $|u(t,x,y,0)|^2$ of the example \ref{ex:pl} for fast-potential model (left $\og=80$, middle $\og=160$)
and time-averaged model (right) by our Bloch-decomposition based algorithm given in section \ref{sec:2.2}, the parameters are:
$\eps=\frac{1}{8}$, $e(t)=(0,0,1)^T$, $\sigma=\frac{2}{3}$, $a=1$, $C_1=1$, $c=2$, mesh size $h=\frac{1}{128}$.}
\label{Fig012}
\end{figure}

\begin{figure}[h!t]
\centering
\includegraphics[width=50mm]{fast_sp.pdf}
\includegraphics[width=50mm]{fast_sp_refine.pdf}
\caption{Graphs of $|u(t,x,y,0)|^2$ of the example \ref{ex:pl} for fast-potential model by the typical pseudo-spectral method given in section \ref{sec:2.1} with $\og=160$,
left: mesh size $h=\frac{1}{128}$, right: $h=\frac{1}{1024}$.}
\label{Fig013}
\end{figure}

\section{Stability estimate}\label{anal:sect}
In this Section we present the analytical results needed in order to rigorously justify the approximation used in Section 2. Because of the singularities of the Coulomb potential, we use a mollified version in the numerical simulation. Here we discuss the additional errors introduced by the smoothing.

In order to avoid technicalities we only state a Proposition which shows that equation \eqref{eq:fast} is stable under perturbations of the potential. The proof is indeed quite standard, as similar results are well known in the literature, see below for the references.
The proof of stability of equation \eqref{eq:fast} with respect to perturbations of the potential is done by adapting in a straightforward way the local well-posedness result. For furhter details we refer the reader to \cite{Kato87}, \cite{Caz} Section 4.
\begin{theorem}
Let $V, \tilde V$ be two potentials such that $V, \tilde V\in L^\infty(\R; L^{\frac{3}{1+3\delta}}(\R^3)+L^{\frac{3}{1-3\delta}}(\R^3))$,
$\nabla V, \nabla\tilde V\in L^\infty(\R; L^{\frac{3}{2+3\delta}}(\R^3)+L^{\frac{3}{2-3\delta}}(\R^3))$, for some $\delta>0$ small.
Let $u, \tilde u$ be the solutions of the Cauchy problems
\begin{equation*}
\left\{\begin{aligned}
&i\partial_tu=-\Delta u+V u+C_1(|\cdot|^{-1}\ast|u|^2)u-a|u|^\sigma u\\
&u(t,\xb)\big|_{t=0}=u_0(\xb),
\end{aligned}\right.
\end{equation*}
  with the potential $V$ and respectively with $V$ replaced by $\tilde V$, and initial data $u_0$ and, respectively $\tilde u_0\in H^1(\R^3)$, respectively. Let $0<T<\min\{T_{max}, \tilde T_{max}\}$, where $T_{max}, \tilde T_{max}$ are the maximal existence times for $u, \tilde u$, respectively. Then, for any admissible pair $(q, r)$, we have
\begin{equation}\label{eq:stab1}
\|u-\tilde u\|_{L^q((0, T); W^{1, r}(\R^3))}\leq C\pare{
\|u_0-\tilde u_0\|_{H^1}+\|V-\tilde V\|_{L^\infty_t(L^{\frac{3}{1+3\delta}}_x+L^{\frac{3}{1-3\delta}}_x)}
+\|\nabla(V-\tilde V)\|_{L^\infty_t(L^{\frac{3}{2+3\delta}}_x+L^{\frac{3}{2-3\delta}}_x)}}.
\end{equation}

\end{theorem}

\section{Conclusion}
In this paper, we propose the time-splitting (Bloch-decomposition based) pseudo-spectral
methods to simulate the XFEL Schr\"odinger equation (with and without periodic potential). Our simulation results go far beyond known analytical results \cite{xfel1}. In particular we demonstrate numerically that the time-averaging procedure works in mass supercritical/energy subcritical cases, even up to blow-up time. Moreover, we show numerically that the energy of the oscillatory problem converges weakly in time to the energy of the time averaged problem (which is constant when the vector $e(t)$ is not dependent on time). Also, we demonstrate the impact of external trapping potentials and periodic lattice potentials on the electron beam wave function and numerically verify the time average procedure in those cases.


TITLE: If $f\in \mathcal{C}(\mathbb{S}^n,\mathbb{S}^n)$ and $f$ is not surjective, then $f$ has a fixed point
QUESTION [3 upvotes]: If $f\in \mathcal{C}(\mathbb{S}^n,\mathbb{S}^n)$ and $f(\mathbb{S}^n)\neq \mathbb{S}^n$, where $\mathbb{S}^n=\left \{x\in\mathbb{R}^{n+1}:\|x\|=1\right \}$. Prove that $f$ has a fixed point.

This is an exercise on the chapter about Brouwer's fixed-point theorem in a Differential Topology book (Chinese version). I think we may need to extend $f$ to a map from $D^{n+1}=\left \{x\in\mathbb{R}^{n+1}:\|x\|\leq 1\right \}$ to itself, so that we can apply Brouwer's fixed-point theorem to get some useful conclusions. However, I don't know how to achieve the goal.
Another thought: If $y_0\in \mathbb{S}^n\setminus f(\mathbb{S}^n)$, using stereographic projection, there is a homeomorphism $\varphi :\mathbb{S}^n\setminus \{y_0\}\to \mathbb{R}^n$. Hence $\varphi \circ f:\mathbb{S}^n\to \mathbb{R}^n$ is a continuous function. However, I don't know how to proceed from this.
I'm not good at geometric stuffs. If my question is silly, forgive me please. Any help would be appreciated!

REPLY [4 votes]: Since $f(\mathbb S^n)$ is a closed subset of $\mathbb S^n$ and is not the whole $\mathbb S^n$, there is a subset $D$ of $\mathbb S^n$ homeomorphic to the closed unit disc $\mathbb D^n$ so that $f(\mathbb S^n) \subset D$. Thus $f|_D : D\to D$ is a continuous map and by Brouwer's fixed point theorem it has a fixed point $x_0$. This fixed point is of course also a fixed point of $f$.
Remark on how to choose $D$: Assume that $y_0\in \mathbb S^n$ is not in the image of $f$. Let $g : \mathbb S^n \to \mathbb R$ be the function $g(x) = x\cdot y_0$ (usual dot product). Then $g\circ f: \mathbb S^n \to \mathbb R$ is a continuous function so that $g\circ f (y) <1$ for all $y\in \mathbb S^n$. Since $\mathbb S^n$ is compact, there is $\delta \in (0,1)$ so that $g\circ f (y) \le 1-\delta$ for all $y\in \mathbb S^n$. Now choose $D = \{ y\in \mathbb S^n : g(y) \le 1-\delta\}$.


TITLE: Subset of Lebesgue measurable sets is not empty
QUESTION [1 upvotes]: Let $l$ be the Lebesgue measure on the $\sigma$-algebra $L$ of Lebesgue measurable subsets of $\mathbb{R}$ and $A\in L$ with $l(A)>0$.
Let $U$ be an open interval such that for $0<\epsilon<1$ we have $$l(A\cap U)>\epsilon\cdot l(U).$$
Let $u\in(-1/2\cdot l(U),1/2\cdot l(U))$ and $A\cap U_{+u}=\{a+u:a\in A\cap U\}$

How do I show that $A\cap U$ and $A\cap U_{+u}$ have non-empty intersection?

I thought that we could prove this by contradiction.
Translation doesn't change the measure, so $l(A\cap U_{+u})=l(A\cap U)$. If the intersection is empty, then
$$l[(A\cap U)\cup(A\cap U_{+u})]=l(A\cap U)+l(A\cap U_{+u})=2l(A\cap U)>2\epsilon\cdot l(U).$$
So now I have to show that $l[(A\cap U)\cup(A\cap U_{+u})]\leq 2\epsilon\cdot l(U)$. How would I do that?

REPLY [0 votes]: The quantifier on your $\epsilon $ is unclear. If the only requirement is existance, then the theorem fails, since $A=(0,\frac{1}{3}) $, $U=(0,2) $, and $u=\frac{1}{2} $ demonstrates.
Assuming that it is for all epsilon, take $\epsilon=\frac{3}{4} $. Assuming without loss of generality that $u\geq 0$, it then follows that $[A\cap U]\cup [A\cap U_{+u}]\subseteq U\cup (U+u) $. Thus for $U=(a,b) $, $[A\cap U]\cup [A\cap U_{+u}]\subseteq (a,b+u) $. Hence $l ([A\cap U]\cup [A\cap U_{+u}])\leq l ((a,b+u))=u+(b-a)\leq \frac {1}{2}l (U)+l (U)$, giving the desired contradiction.


TITLE: Just one vertex not in a cycle means there is a vertex cut?
QUESTION [2 upvotes]: I saw this assertion:

Any connected graph $G$ with $|V(G)| \geq 3$ and at least $2$ vertices that are not in a cycle must have a vertex cut

and realise that we just need one vertex not in any cycle. I don't know if I made a mistake or my claim is ok:

First, if $G$ does not have a cycle we are done because every vertex that is not a leaf is a vertex cut
So let $C$ be a cycle and $v_1$ a vertex such that $v_1u_1 \in E(G)$ for some $u_1 \in C$ and $v_1$ is not in any cycle. These $C$,$v_1$ exists because if we have any vertex $v$ out of a cycle, we take a path $P$ between $v$ and some cycle $C$, minimal in this sense (so $P$ does not go through any other cycle). Then $P \cap C =\{u_1\}$ with $u_1 \in C$ and $u_1v_1 \in E\,$ for some $v_1\in P$, that is not in any cycle.
For any $u_2 \in C$ there is no $(v_1,u_2)-$path that do not use $u_1$. Otherwise $u_1v_1Pu_2u_1$ would be a cycle that contains $v_1$. Then $u_1$ is a vertex cut (it separates $v_1$ from $C$)

is this ok? or am I missing something?

REPLY [3 votes]: You are correct, and to make it even easier:

For the graph $G$ with $|V(G)|\geq 3$ find a vertex $a$ not in any cycle.
If $a$ has degree $1$ (a leaf), its sole adjacent vertex is a cut vertex, separating $a$ from the rest of $G$
If $\text{deg}(a)>1$, $a$ is a cut vertex, since there is no other path between any two of its adjacent vertices, otherwise $a$ would be in a cycle.


TITLE: Express this in terms of conditional probabilities and express this in terms of conditional probabilities
QUESTION [0 upvotes]: Let $N$ equal the number of time a certain device may be used before it breaks. The probability is $p$ that it will break on any one try given that it did not break on any of the previous tries.
(a) Express this in terms of conditional probabilities.
I'm confused I have two interpretations, but What is the correct?
Let $s$ the  number of future tries and let $t$ the previous tries (the device  did not break)
$Pr( t < X < t+s | X > t) = p$
or
$Pr( X  > t+s | X > t) = p$
(b) Express it in terms of a density function, and find the density funtion
Can Someone  help me there??  Thanks for your help :)

REPLY [1 votes]: I would say that for part (a), you're looking for
$$
P(N=t+1\mid N>t)=p.
$$
That is, the probability that it breaks on the $(t+1)$-st attempt to use it, given that it did not break in the first $t$, is $p$.
I don't want to make any comment on a density function here, however; this is, as stated, a discrete process -- so, you could talk about its mass function, but it shouldn't have a density.  Unless there are further assumptions here that aren't stated?


TITLE: Generating normal random variables with mean and variance
QUESTION [0 upvotes]: I wish to generate normals $X$, $Y$, and $Z$ with the correlation matrix $R$ but with means $0$, $1$, and $2$, as well as variances $4$, $16$, and $25$, respectively. How would you do this?

REPLY [0 votes]: The methodology for such a process is as follows:
1) Generate i.i.d standard normals (X,Y,Z). You can think of this as a vector $z$
2) Do a Cholesky Decomposition on your specified correlation matrix $R$.
3) Multiple your vector $z$ by your newly decomposed vector. This will generate 3 correlated standard normals.
4) These correlated standard normals can then be scaled without an effect on the correlation.
$U = \sigma_1X + \mu_1$,
$V= \sigma_2Y + \mu_2$,
$W = \sigma_3Z + \mu_3$
The values of $\sigma$ and $\mu$ are those specified in your question.


TITLE: Solving for possible orientations of 3 objects on a 3x3 grid
QUESTION [1 upvotes]: Say you have a 3x3 grid, and 3 objects to work with. Each occupies one space. How would I go about solving the amount of ways they can lay on the grid.
Example: (pardon my bad ASCII art)
[][][]
[][][]
[x][x][x]
Any help would be much appreciated, thanks!

REPLY [1 votes]: To reiterate what was said in the comments, imagine temporarily that your grid has its positions labeled.
$\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix}$
A specific arrangement of the objects will then correspond to a selection of three numbers from $1$ to $9$.
$\begin{bmatrix}\circ&\times&\circ\\\circ&\circ&\times\\\circ&\circ&\times\end{bmatrix}$ would correspond to the set $\{2,6,9\}$
Assuming you consider $\begin{bmatrix}\circ&\times&\circ\\\circ&\circ&\times\\\circ&\circ&\times\end{bmatrix}$ to be a different arrangement than $\begin{bmatrix}\circ&\times&\times\\\times&\circ&\circ\\\circ&\circ&\circ\end{bmatrix}$, one can move freely between the set of desired arrangements and the set of subsets of size three of $\{1,2,\dots,9\}$ via a natural bijection.
The fact that there is a bijection between the sets in question implies that if we know how to count how many arrangements one of these questions has, we know how to count both.
We have thus boiled the question down to asking the question, "How many size three subsets exist of the set $\{1,2,\dots,9\}$?"
This is one of the many problems that binomial coefficients were designed to answer.  The answer being $\binom{9}{3}=\frac{9!}{3!6!}=\frac{9\cdot 8\cdot 7}{3\cdot 2\cdot 1} = 84$ possible arrangements.

If you consider rotations and/or reflections of your grids to be considered "the same", for example the two arrangements pictured above, the question becomes harder since we will have over counted.  Approach instead by Burnside's lemma.


TITLE: Trying to find $\lim_{x\to 0^+} \frac{x^2\sin(1/x)}{\sin x}$, I get $\frac{\infty}{0}$, what is that?
QUESTION [2 upvotes]: I'm trying to find $\lim\limits_{x\to 0^+} \dfrac{x^2\sin(1/x)}{\sin x}$, I get $\dfrac{\infty}{0}$.
If $ \frac{\infty}{0} $is not an indeterminate form (like $ \infty \times 0, 1^\infty, \infty-\infty, \infty^0, 0^0  $) then what is it?
I know $ \frac{0}{\infty}$, for example, is zero...so what is $\frac{\infty}{0}$?
I'm asking because I'm trying to do $ \lim_{ x\to 0^+} [ (x^2)(\frac{\sin \frac{1}{x}}{\sin x }] $.
When I plug in zero I get $ 0\times \frac{\infty}{0} $...so I'm not sure what to do after this.
I tried separating the function like $\lim _{x\to 0^+}\frac{x^2}{\sin x}\times
 \lim_{x\to 0^+}\frac{\sin \frac{1}{x}}{\sin x} $....Then I got $\frac{0}{0}\times \frac{\infty}{0}$. For the $\frac{0}{0}$ one I applied L'hopitals and got 0. But then I still have that infinity/0. What do I do with this?
By the way, when I'm doing these "l'hopital's" problems...when do I know to stop or to keep going/rewriting. For example, if I get $ \frac{\infty}{0}$ do I just stop and assume that the limit does not exist...or do I try to rewrite and try to see if something works. Because I know that for those indeterminate forms I mentioned earlier, whenever I come across one of those, I know that I should just rewrite until I get  $\frac{0}{0}$  or $\frac{\infty}{\infty}$ to apply L'hopitals. But any time I get something other then these forms, I'm unsure of what to do.

REPLY [1 votes]: While the question of how to handle your limit has been answered, I will attempt to answer your question of "What is $\frac{\infty}{0}$?".
Firstly, I'll clarify what is a "form" in this calculus context. A form like $\infty - \infty$ means: $\lim_{x} f(x)-g(x)$ where $\lim_{x} f(x)=\infty$ and similarly for $g$. In general, we talk about these things because we want to know to what extent the limits of the parts determine the limits of the whole.
Unambiguous forms: Things like $5-7$ (where one part has a limit of 5 and the other has a limit of 7) or $10/3$ are unambiguous forms (this is not a technical term), because of the limits of the parts exist and have those values, and the arithmetic makes sense and doesn't involve $\infty$, then theorems (often called "limit laws") guarantee that the limit of the whole is just what you get when you do the operation on the parts. There are also other unambiguous forms like $7*\infty$ (the limit of the whole is always $\infty$) or $0^\infty$ (the limit of the whole is always $0$) or $\frac{7}{\infty}$ (the limit of the whole is always $0$).
Indeterminate forms: The indeterminate forms you encounter are so called because they tell you essentially nothing about the limit of the whole. If you have a limit with the "form" $\infty-\infty$, then the limit of the whole might be $\infty$ or $-\infty$ or $17$ or "not exist due to oscillation" or whatever.
But there are other forms, which I will call "slightly ambiguous forms": like $1/0$, $\infty/0$, $0^{-\infty}$, etc. These forms are not indeterminate because if you take the limit of the absolute value of the whole function you always get $\infty$. For example, $\lim_{x\to\infty}\dfrac{1}{\frac{\sin x}{x}}$ has the form $1/0$, and the limit does not exist (due to "oscillation"), but $\lim_{x\to\infty}\left|\dfrac{1}{\frac{\sin x}{x}}\right|=\infty$. $\lim_{x\to\infty}\dfrac{x}{\frac{1}{x}}$ has the form $\frac{\infty}{0}$ but the limit simply equals $\infty$.


TITLE: Number of matrices satisfying given conditions
QUESTION [0 upvotes]: Let A be the set of all $3*3$ symmetric matrices all of whose entries
  are either $0$ or $1$.Five of these entries are $1$ and four of them are
  $0$.

$1)$ The number of matrices in $A$ is ?
$2)$ The number of matrices A in A for which the system of linear equations $A\begin{bmatrix} x \\ y \\ z \end{bmatrix}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}$ has unique solution is ?
$3)$ The number of matrices $A$ in $A$ for which the above system of linear equations is inconsistent is?
Can someone give me some hints please atleast for the first two parts?

REPLY [0 votes]: Hint: If you didn't have the symmetric condition, this would be the same as choosing $4$ positions for the $0$'s from the $9$ positions in the matrix.  This would give $\binom{9}{4}=126$ possible matrices.
In your case, however, the matrices are required to be symmetric.  So, we can divide the matrix into the diagonal and the portion above the diagonal, let $n_d$ be the number of $0$'s on the diagonal and $n_u$ the number of $0$'s above the diagonal.  Since the matrix is symmetric, the number of $0$'s below the diagonal is the same as the number above the diagonal.  Hence, $4=2n_u+n_d$.

If $n_u=0$, then $n_d=4$, but there are only $3$ positions on the diagonal.
If $n_u=1$, then there are $2$ $0$'s on the diagonal and $1$ above the diagonal.  There are $3$ positions on the diagonal and $3$ positions above the diagonal.  This gives $9$ matrices.
If $n_u=2$, then there are $0$ $0$'s on the diagonal and $2$ $0$'s above the diagonal.  Since there are $3$ positions above the diagonal, this gives $3$ matrices.


TITLE: Can the covering transformation group of a 6-sheeted covering map have 12 elements?
QUESTION [3 upvotes]: Can the covering transformation group of a 6-sheeted covering map $p : X \to Y$ have exactly 12 elements?
I suspect that the answer is negative, but I cannot see an invariant that shows this. We don't have any limitations on the nature of spaces involved (i.e. they are not necessarily CW-complexes, etc.).

REPLY [5 votes]: Edit: I almost completely rewritten my original answer.
Let $p: X\to Y$ be a covering map of topological spaces. The group of deck-transformations $Deck(p)$ is the subgroup of $Homeo(X)$ consisting of all homeomorphisms $g$ such that $p\circ g=p$. In particular, such a group has to act freely on $X$, provided that $X$ is connected. Now, pick a point $x\in X$. Then $p$ is constant on the $G$-orbit $G\cdot x$, which projects to a point $y\in Y$. This orbit has cardinality equal $|G|$. Therefore,
$$
d=|p^{-1}(y)|\ge |G|$$
where $|Z|$ denotes the cardinality of a set $Z$. Thus, for connected $X$, one cannot have $|G|=12, d=6$.
For disconnected $X$ the picture is much messier. Here is an example with $|G|=12, d=6$:
Let $p_1: X_1\to Y_1$ be a degree 6 cyclic regular cover (say, circle covering itself). The deck-group of $p_1$ is $G_1=Z_6$. Let $p_2: X_2\to Y_2$ be a degree 6 covering map with the deck-group $G_2=Z_2$. To construct such, note that if $H$ is the dihedral group of order $12$ and $H'<H$ is the order 2 reflection subgroup, then the
normalizer of $H'$ in $H$ is $Z_2\times Z_2$ and, hence, $H'$ is index 2 subgroup in its normalizer.
Furthermore, $6=|H:H'|$. Now, consider a connected complex $Y_2$ with the fundamental group $H$ and let $p_2: X_2\to Y_2$ denote the 6-fold covering map associated with the subgroup $H'<H$ as above. Then its deck-transformation group is $Z_2$.
Lastly, let $X=X_1\sqcup X_2$, $Y=Y_1\sqcup Y_2$, $p: X\to Y$ be the map whose restriction to $X_i$ is $p_i$. Then $p$ is a 6-fold covering; its deck-transformation group is $G_1\times G_2$, it has order $12=6\times 2$.


TITLE: Show $\sum_{k=1}^\infty|a_k|^q$ converges
QUESTION [0 upvotes]: Let $1 \leq p < q < \infty$. Show that if $x=(a_k)∈ℓ^p$
i.e. the condition that the series $$\sum_{k=1}^\infty|a_k|^p$$
converges holds, then $x∈ℓ^q$
i.e. $$\sum_{k=1}^\infty|a_k|^q$$
 converges.
I did try to use the ratio test but I don't think it will work because of the power on the terms in the series.

REPLY [1 votes]: Because $\sum_{n=1}^{\infty} |a_n|^p < \infty,$ $|a_n|^p \to 0.$ Thus there exists $N$ such that $|a_n|^p< 1$ for $n\ge N.$ For such $n$ we can say that, since $q/p > 1,$
$$|a_n|^q = (|a_n|^p)^{q/p}\le |a_n|^p.$$
Thus $\sum_{n=1}^{\infty} |a_n|^q$ converges by the comparison test.


TITLE: What spaces are homeomorphic to $\mathbb{Q}^\omega$ = $\mathbb{Q}^\mathbb{N}$ = $\mathbb{Q}^\infty$?
QUESTION [3 upvotes]: What spaces are homeomorphic to $\mathbb{Q}^\omega$ =  $\mathbb{Q}^\mathbb{N}$ =  $\mathbb{Q}^\infty$? (The space of all rational sequences, considered with the standard product topology).
I have found interesting characterization in a paper by Engelen "Characterizations of the countable infinite product of rationals and some related problems" that says:

Let $X = \{ (x_i)_{i \in \mathbb{N}} \in \mathbb{N}^\omega: lim_{i
\mapsto \infty} x_i = \infty \}$. Then $X \simeq \mathbb{Q}^\omega$.

I find this interesting and my questions are

What are some more "practical" examples of spaces with the condition above?
What are other spaces or conditions for spaces being homeomorphic to $\mathbb{Q}^\omega$?

Thank you for any advice.

REPLY [1 votes]: Let $f(z)=\exp(z)-1$. The "zero-dimensional remainder" of $\mathbb Q ^\omega$ is homeomorphic to the set of points in the complex plane $z\in \mathbb C$ such that $f^n(z)$ goes to neither $0$ nor $\infty$ https://arxiv.org/pdf/2010.13876.pdf. By zero-dimensional remainder, I mean the complement of a dense copy of $\mathbb Q ^\omega$ in a complete zero-dimensional space. For example, $\mathbb P ^\omega\setminus (\mathbb Q+\pi)^\omega$. I am unaware of a copy of $\mathbb Q ^\omega$ that is generated very simply in this manner, in complex dynamics.


TITLE: Kronecker product on a matrix with structured blocks
QUESTION [0 upvotes]: I'm currently attempting to write a symmetric matrix with structured blocks into Kronecker-factorized form, but I'm not sure if the task is possible at all. My matrix takes the following form:
$$
M=
\begin{bmatrix}
a I_1 & \mathbf{0} & gI_1 & \mathbf{0} & iI_1 & \mathbf{0} \\
\mathbf{0} & bI_2 & \mathbf{0} & hI_2 & \mathbf{0} & jI_2 \\
g I_1 & \mathbf{0} & cI_1 & \mathbf{0} & kI_1 & \mathbf{0} \\
\mathbf{0} & hI_2 & \mathbf{0} & dI_2 & \mathbf{0} & lI_2 \\
i I_1 & \mathbf{0} & kI_1 & \mathbf{0} & eI_1 & \mathbf{0} \\
\mathbf{0} & j I_2 & \mathbf{0} & lI_2 & \mathbf{0} & fI_2
\end{bmatrix}
$$
where $I_1, I_2$ are identity matrices in $\mathbb{R}^{d_1},\mathbb{R}^{d_2}$ respectively, and $a,b,\cdots,l$ are arbitrary constants. Obviously, it'd be desirable to have written
$$
M = W\otimes\begin{bmatrix}
I_1 & \mathbf{0} \\
\mathbf{0} & I_2
\end{bmatrix}
$$
but I suspect it'd be feasible at all for arbitrary constants. Any feedback/suggestion would be of great help. Thanks!

REPLY [0 votes]: To begin, note that there exists a permutation matrix $P$ such that
$$
P^TMP = \pmatrix{
aI_1 & gI_1 & iI_1\\
g I_1 & c I_1 & kI_1\\
iI_1 & k I_1 & eI_1\\
&&& bI_2 & h I_2 & j I_2\\
&&& h I_2 & dI_2 & l I_2\\
&&& j I_2 & l I_2 & fI_2
}
$$
where the blank spaces denote zeros. We can write each of these blocks as a Kronecker product.  That is, we have
$$
P^TMP = \pmatrix{\pmatrix{a&g&i\\g&c&k\\i&k&e} \otimes I_1
\\ & \pmatrix{b&h&j\\h&d&l\\j&l&f} \otimes I_2}
$$


TITLE: Probability of this Champions League Quarter-Final's Draw Falling along "relative seeding strength"?
QUESTION [0 upvotes]: This morning's CL draw fell along lines that are likely to generate some suspicious of it being rigged - i.e. the strongest teams all avoided each other. I was trying to figure out the approximate probability of this draw. I imagine it's a relatively simple question and I want to check my thoughts.
Basically I would divide the 8 teams into 2 pots: Pot 1-Stronger and Pot 2-Weaker (Dortmund and Juventus are probably the only debatable choices in my lists)
Pot 1:
Barcelona
Real Madrid
Bayern Munich
Dortmund
Pot 2:
Juventus
PSG
Gala
Malaga
My question is: what is the probability of all 4 matches containing exactly one Pot 1 team and one Pot 2 team?
For any given draw the probability is 4/7 that you get a team from the opposite pool but what happens then?
How do I sum the probabilities for the remaining draws (presumably the drawn teams are taken out of the pool for the next round)?
Thanks!

REPLY [0 votes]: I was wondering the exact same question as user1141785 posted, what would be the probability that no teams from the same country would meet each other in the quarter finals? And what about the europa league, in which there were 3 english clubs and the remaining clubs are from 5 different countries. What are the chances of no match between clubs from the same country in both CL and EL combined, as happened this week?


TITLE: Conserved positive charge for a PDE
QUESTION [7 upvotes]: Let $(x,t) \in \mathbb{R}^2$ and consider the following partial differential equation for the real-valued function $U(x,t)$:
\begin{equation}
\frac{\partial^2 U}{\partial t^2} = - \frac{\hbar^2}{4m^2} \frac{\partial^4 U}{\partial x^4},
\end{equation}
where $m$ and $\hbar$ are positive constants. In the following we shall be quite sloppy, and we shall assume that given (smooth enough) initial conditions $U(x,0)$ and $\frac{\partial U}{\partial t}(x,0)$ (lying in some space) there exists a unique (smooth enough) solution $U$ (lying in some space) to this partial differential equation. Let us call the set of solutions $\mathcal{E}$.
Let us define for ease of notation $D_{x}^k F=\left( F, \frac{\partial F}{\partial x}, \dots, \frac{\partial^k F}{\partial x^k} \right)$  for every non-negative integer $k$ and for every smooth enough function $F(x,t)$. I ask whether there exist some non non-negative integer $k$ and some (smooth enough) real-valued functions $p \geq 0$ and $j$, with $p$ non-constant, such that, by setting
\begin{equation}
P(x,t)=p \left((D_{x}^k U)(x,t), \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right), \\
J(x,t)=j \left((D_{x}^k U)(x,t),  \left(D_{x}^{k} \frac{\partial U}{\partial t}\right)(x,t) \right),
\end{equation}
the following properties hold:
(i) for every $U \in \mathcal{E}$ the following continuity equation holds
\begin{equation}
\frac{\partial P}{\partial t} + \frac{\partial J}{\partial x}=0;
\end{equation}
(ii) for the special solution $U(x,t)=\cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right)$, we have that $P(x,t)$ is independent of $\omega > 0$.
The answer should be negative, but I have no idea of a possible proof. Obviously, since we have not formulated the problem in a rigorous way, we do not expect to get a rigorous proof, but only some heuristic, convincing argument in this direction.
NOTE (1) Let us explicitly note a trivial consequence of property (ii). Since for the special solution
\begin{equation}
U(x,t)=\cos\left(\sqrt{\frac{2m \omega}{\hbar}}x-\omega t \right),
\end{equation}
$P(x,t)$ is independent of $\omega > 0$ for every $(x,t) \in \mathbb{R}^2$, we have that $(x,t) \mapsto P(x,t)$ is a constant function. Indeed, for this special solution, we have that $P(x,t)=F(x-vt)$ for some function $F$, where $v=\sqrt{\frac{\hbar \omega}{2m}}$. If we had $F'(\xi)\neq 0$ for some $\xi \in \mathbb{R}$, we would get for $x-vt=\xi$:
\begin{equation}
\left. \frac{\partial P}{\partial t}(x,t) \middle/ \frac{\partial P}{\partial x}(x,t) \right.  = -v = \sqrt{\frac{\hbar \omega}{2m}},
\end{equation}
a contradiction.
NOTE (2) This problem, as the notation shows, has a physical background, and the mathematical formulation of the problem that I give here is my personal interpretation of a physical exposition given by the great XXth century physicist David Bohm in his wonderful treatise Quantum Theory published in 1951. Bohm explicitly states that the problem has a negative answer, without giving any proof or heuristic argument. For all the physical details about this problem see my post Nonexistence of a Probability for Real Wave Functions.

REPLY [1 votes]: This is not an answer, but only a remark too long to be posted as a comment.
We can make the formulation of the problem more explicit in the following way. Let $y=(y_0,y_1,\dots,y_k) \in \mathbb{R}^{k+1}$ and $z=(z_0,z_1,\dots,z_k) \in \mathbb{R}^{k+1}$. Then to say that the functions $(y,z) \mapsto p(y,z)$ and $(y,z) \mapsto j(y,z)$ satisfy property (i) means that for every $U \in \mathcal{E}$
\begin{equation}
\nabla_{y} p \cdot D_{x}^{k} \frac{\partial U}{\partial t} + \nabla_{z} p \cdot D_{x}^{k} \frac{\partial^2 U}{\partial t^2} + \nabla_{y} j \cdot D_{x}^{k} \frac{\partial U}{\partial x} + \nabla_{z} j \cdot D_{x}^{k} \frac{\partial^2 U}{\partial x \partial t}=0,
\end{equation}
which is equivalent to say that
\begin{equation}
\nabla_{y} p \cdot D_{x}^{k} \frac{\partial U}{\partial t} - \frac{\hbar^2}{4 m^2} \nabla_{z} p \cdot D_{x}^{k} \frac{\partial^4 U}{\partial x^4} + \nabla_{y} j \cdot D_{x}^{k} \frac{\partial U}{\partial x} + \nabla_{z} j \cdot D_{x}^{k} \frac{\partial^2 U}{\partial x \partial t}=0.
\end{equation}
Now Borel Theorem tells us that for every real sequence $(a_{n})_{n=0}^{\infty}$ and every fixed $x_0 \in \mathbb{R}$ there exists a smooth function $f:\mathbb{R} \rightarrow \mathbb{R}$ such that
\begin{equation}
\frac{d^m f}{d x^m}(x_0)=a_{n} \quad (n=0,1,2,\dots),
\end{equation}
so that from the arbitrariness of the initial data $x \mapsto U(x,0)$ and $x \mapsto \frac{\partial U}{\partial t}(x,0)$, by replacing $\frac{\partial^m U}{\partial x^m}$ by $y_m \in \mathbb{R}$ and $\frac{\partial^{m+1} U}{\partial^m x \partial t}$ by $z_m \in \mathbb{R}$ for every non-negative integer $m$, we conclude that $p$ and $j$ satisfy the following equation:
\begin{equation}
\sum_{m=0}^{k}\frac{\partial p}{\partial y_m} z_{m}  - \frac{\hbar^2}{4 m^2}\sum_{m=0}^{k} \frac{\partial p}{\partial z_m} y_{m+4} +
\sum_{m=0}^{k} \frac{\partial j}{\partial y_m} y_{m+1} + \sum_{m=0}^{k} \frac{\partial j}{\partial z_m} z_{m+1} =0 \qquad (I).
\end{equation}
From this equation we immediately deduce that $p$ does not depend on $z_{k-2}, z_{k-1}, z_k$ and that $j$ does not depend on $z_k$. Moreoever, we also get that
 \begin{equation}
 - \frac{\hbar^2}{4m^2} \frac{\partial p}{\partial z_{k-3}}+ \frac{\partial j}{\partial y_k} = 0,
\end{equation}
and since $p$ and $j$ do not depend on $z_k$ it also holds
\begin{equation}
 \frac{\partial p}{\partial y_k} +  \frac{\partial j}{\partial z_{k-1}}=0.
\end{equation}
There are no other simple consequences of Equation (I). Now, the issue in the post amounts to see whether there exist solutions $p \geq 0$ and $j$ of Equation (I), with $p$ non-constant and satisfying property (ii). This latter request makes all the matter quite involved.
Let us explicitly note there are simple non-constant functions $p(y,z) \geq 0$ which satisfy (ii). Consider e.g. $p(y,z)=(y_1 z_1 - z_0 y_2)^2$. It is easy to see that in this case there is no $k$ and no function $j$ such that $p$ and $j$ satisfy equation (I). Indeed, equation (I) would imply in our case that $j$ depends only on $y_0,\dots, y_4, z_0,z_1$. Moreover it would also imply that
\begin{equation}
\frac{\partial j}{\partial y_4}=  \frac{\hbar^2}{2 m^2}(y_1 z_1 - z_0 y_2) y_1, \\
\frac{\partial j}{\partial z_1}= 2(y_1 z_1 - z_0 y_2) z_0.
\end{equation}
From the last equation we would get for some function $g$:
\begin{equation}
j=2(y_1 z_1 - z_0 y_2) z_0 z_1 + g(y_0,\dots,y_4,z_0),
\end{equation}
so that we should have
\begin{equation}
\frac{\partial g}{\partial y_4} = \frac{\partial j}{\partial y_4} = \frac{\hbar^2}{2 m^2}(y_1 z_1 - z_0 y_2) y_1,
\end{equation}
which is a contradiction since $g$ does not depend on $z_1$.
The difficulty of the problem is to show in general that conditions (i) and (ii) are not compatible, and I have no idea for now of how to tackle the problem.