Looking at past events by observing sky?
If you view the Earth from far enough away can you observe its past?
I want to prove that if we have a commutative integral domain $D$ with field of fractions $F\neq D$ then $F$ is not finitely generated as a $D$-module. (In this question it may be the case that $1\not\in D$.)    My original plan was to localize at a maximal ideal and then use Nakayama's lemma but as we don't have a $1$ our localization may not be local (in fact we need not have maximal ideals).  So I'm a bit lost as to where to go?  Thanks for any help. Is it supposed to be a HTML or an HTML Mathematical notation around the world How to do the same crop on a batch of photos? I've got a batch of photos (an owl coughing up a pellet in case you're curious).  I want to make an animated GIF out of them, but the result would be improved if they were all given an identical crop first.  I'm currently using Picasa for my basic photo editing on a Windows 7 PC.  While Picasa does have batch editing, crop is not one of the items on the batch menu.  Does Lightroom have a batch crop?  If there's a way to get Picasa to show me the coordinates of the crop selection rectangle, I could also use that information to do the same crop repeatedly.  Are there any tools that let you crop an animated GIF?  I know I can create the animated GIF by uploading the component images to Google+ and then looking in the Auto Awesome area. I don't know about tools for working with animated GIFs. Consider the following example:  \documentclass{article} \newtheorem{Theorem}{Theorem} \begin{document} \begin{Theorem} On the other hand, this one has only one claim. \end{Theorem} \begin{Theorem} \begin{enumerate} \item First claim \item Second claim \end{enumerate} \end{Theorem} \end{document}   Typesetting it results in:    Notice there is extra horizontal space before the first item in the list of the second theorem.     Is there a sensible way to get rid of that?   Ideally, the solution would work with amsthm, thmtools, enumitem.  N.B.: this is related to  but not quite the same. Is there a way to substring a string in Python, to get a new string from the third character to the end of the string?  Maybe like myString[2:end]?  If leaving the second part means 'till the end', and if you leave the first part, does it start from the start? Data inconsistency prohibition if a table refers to another via two many-to-many relationships I have following database design (by =&gt; foreign key constraints are depicted):   Company[id] CompanyRealm[id, company_id=>Company.id]many2many  Project[id, company_id=>Company.id] ProjectRealm[id, project_id=>Project.id, company_realm_id=>CompanyRealm.id]many2many   Problem is that provided database design allows inconsistent data. For example:   Company1(id=1) Company2(id=2) CompanyRealm(id=11, company_id=1) Project(id=33, company_id=2) ProjectRealm(id=44, project_id=33, company_realm_id=11)   (Company can have many Realms, Project belong to Company and  may concern any number of company's Realms)  ProjectRealm refers to two different companies:   Company1 (via CompanyRealm); and Company2 (via Project).    Is something wrong with my DB design? If yes - which prescriptions are violated? If no - how to prevent incorrect data insertion (via constraint? trigger with exception?) Does iCloud Two-factor authentication send a code to the same device as a log in request? My Apple-ID (e.g., on  ) is set up to require entry of a 6-digit code sent to my registered Apple products.  This code is being sent to the same device I'm currently requesting to log in. I feel this absolutely defeats the purpose of this 2-Factor Authentication: If a person knows my password, this person can receive the code on that device, and access my apple account without any problems.  Is this a bug or can I set up / change how this security step works? Pass a PHP string to a JavaScript variable (and escape newlines) Does "the same number of people" behave as singular or plural? natbib and hyperref create by default two links to the bibentry in the reference section|chapter of the document:   \begin{filecontents}{\jobname.bib} @book{author08, title = {{A Title}}, publisher = {Alpha}, year = {2008}, author = {Author, A}, address = {London} } @book{buthor90, title = {{B Title}}, publisher = {Bravo}, year = {1990}, editor = {Buthor, B}, address = {New York} } \end{filecontents}  \documentclass{article}  \usepackage{natbib} \usepackage{hyperref} \hypersetup{colorlinks,citecolor=red}  \begin{document}  \noindent citet: \citet{author08}, \citet[see][p. 20]{author08} \\ citep: \citep{author08}, \citep[see][p. 20]{author08}  \bibliographystyle{plainnat} \bibliography{\jobname}  \end{document}     In order to avoid two links to the bibliography (one for the [author] and one for the year [year], respectively), I took  to an older question and added it to my preamble.  \usepackage{etoolbox} \makeatletter  \pretocmd{\NAT@citex}{%   \let\NAT@hyper@\NAT@hyper@citex   \def\NAT@postnote{#2}%   \setcounter{NAT@total@cites}{0}%   \setcounter{NAT@count@cites}{0}%   \forcsvlist{\stepcounter{NAT@total@cites}\@gobble}{#3}}{}{} \newcounter{NAT@total@cites} \newcounter{NAT@count@cites} \def\NAT@postnote{}  % include postnote and \citet closing bracket in hyperlink \def\NAT@hyper@citex#1{%   \stepcounter{NAT@count@cites}%   \hyper@natlinkstart{\@citeb\@extra@b@citeb}#1%   \ifnumequal{\value{NAT@count@cites}}{\value{NAT@total@cites}}     {\ifNAT@swa\else\if*\NAT@postnote*\else%      \NAT@cmt\NAT@postnote\global\def\NAT@postnote{}\fi\fi}{}%   \ifNAT@swa\else\if\relax\NAT@date\relax   \else\NAT@@close\global\let\NAT@nm\@empty\fi\fi% avoid compact citations   \hyper@natlinkend} \renewcommand\hyper@natlinkbreak[2]{#1}  % avoid extraneous postnotes, closing brackets \patchcmd{\NAT@citex}   {\ifNAT@swa\else\if*#2*\else\NAT@cmt#2\fi    \if\relax\NAT@date\relax\else\NAT@@close\fi\fi}{}{}{} \patchcmd{\NAT@citex}   {\if\relax\NAT@date\relax\NAT@def@citea\else\NAT@def@citea@close\fi}   {\if\relax\NAT@date\relax\NAT@def@citea\else\NAT@def@citea@space\fi}{}{}  \makeatother     Which gets me almost where I want to be, but leaves me with two issues I would like to be solved:   while \citet includes pre- and postnotes in the link, \citep doesn't. How can \citep be changed to include postnotes? is there a way to color the brackets and commas that are typeset in black in the unmodified first example also in black in the modified second example without removing them from the link? Is the external micro SD card path the same for all devices? I am using a Samsung Tab 2.  There is a path: /mnt/external_sdcard/myfile. Here, I can save, read, or write my file to the external micro SD card.  I'm just wondering if this path is the same for all devices, or if it may be different. How to zoom in/out an UIImage object when user pinches screen? Does the proof of Bolzano-Weierstrass theorem require axiom of choice? When selecting the terms of subsequence from each bisections, I thought axiom of choice might be required. But I'm not so sure whether or not, so please tell me. [edited] I'm sorry for the lack of explanation. I want to prove this statement: Let $a_1, a_2, \ldots \in \mathbf{R}$, and $(a_n)_{n\in\mathbf{N}}$ is bounded, then $(a_n)$ has some convergent subsequence.  The proof is as follows. Since $(a_n)$ is bounded, for all $n\in\mathbf{N}$,   $a_n \in I = [b, c]$.  Now, let $I_0 = I$ and if $I_n = [b_n, c_n]$,  we define $d_n = (b_n+c_n)/2$ and  if infinite terms of $(a_n)$ is included in $[b_n, d_n]$(resp. $[d_n, c_n]$), we will define  $I_{n+1} = [b_n, d_n]$(resp. $[d_n, c_n]$).If both intervals contain infinite terms, let $I_{n+1}$ be $[d_n, c_n]$. For all $n\in \mathbf{N}$, infinite numbers of $m \in \mathbf{N}$ exist such that $a_m \in I_n$ suffices. We take the sequence of natural numbers $(n(k))_{k\in\mathbf{N}}$which suffices $n(0) &lt; n(1) &lt; \cdots &lt; n(k) &lt; \cdots$ following this procedure: Now we have already selected $a_{n(1)}, \ldots, a_{n(k)}$, there are infinite numbers of $m\in \mathbf{N}$ which suffices $n(k)&lt;m, a_m \in I_{k+1}$, so let's take the minimum m out of it. Applying this process recursively, we obtain a infinite convergent subsequence(?). I think intuitively, by only repeating this process we can't obtain countable infinite terms of subsequence because we have to repeat infinite times. I'm reading a proof about Lagrangian => Hamiltonian and one part of it just doesn't make sense to me.  The Lagrangian is written $L(q, \dot q, t)$, and is convex in $\dot q$, and then the Hamiltonian is defined via the Legendre transform  $$H(p,q,t) = \max_{\dot q} [p \cdot \dot q - L(q, \dot q, t)]$$  Under the right conditions there exists a function $\dot Q (p,q,t)$ such that  $$H(p,q,t) = p \cdot \dot Q(p,q,t) - L(q, \dot Q(p,q,t), t)$$  i.e. when some $\dot Q(p,q,t)$ satisfies $p = \frac{\partial L}{\partial \dot q}\rvert_{(q, \dot Q(p,q,t), t)} = \frac{\partial L}{\partial \dot q}(q, \dot Q(p,q,t), t)$  (Finding this function is usually called "inverting p")  By taking partials in the $p$ variable and using the relationship, we can obtain the relationship  $$\dot Q = \frac{\partial H}{\partial p}$$  Because of the notation I chose, I get the strong urge to say $\dot q = \frac{\partial H}{\partial p}$  , and in fact this is what the textbook does. But have we proved this?    In other words, how can we deduce that  $$q'(t) = \frac{\partial H}{\partial p}(p(t), q'(t), t)$$ for any differentiable vector valued function $q$? (or maybe there are more conditions we need on $q$? Here  $$p(t) = \frac{\partial L}{\partial \dot q}(q(t), q'(t), t)$$ according to Lagrange's equations. How to get the directory of open blend file? (from Python) What's the best way to get the full directory of the current Blender file in Python?  It should be cross-platform. Distribution of range of Uniform $(0,1)$ distribution Let $X_1,...,X_n$ be iid as $U \sim [0,1]$ with pdf $f(x) = 1(0 &lt;x&lt; 1)$. Determine the range distribution.   I have managed to determine the distribution of the minimum and the maximum, however I cannot/do not know how to determine the distribution of range.   Minimum:  $f_N = n(1-x)^{n-1}$ and maximum $f_M= nx^{n-1}$   I am aware that the calculation of range distribution is based on convolution: $f_R=\int f_{N,M}(x,x-r)dx$, However I cannot get the correct result, which according to my lecture slides is : $f_R= n(n-1)r^{n-2}(1-r)$  Could someone break this down for me ? My math background is not great Open Finder window from current Terminal location? If I am in a specific path in a Terminal window, how can I open that same window in a new Finder window?   Note: This is the opposite of . Why were the people of Rhun and Harad allied with Sauron?  Did they join him out of fear, or did Sauron say he would give them something if they joined him?  I know the men of Dunland were angry at Rohan; did Rhun and Harad have something against Gondor? Put the book back on the table.   I'm having trouble. I think it is a preposition.