If two Lagrangian (densities) $\mathcal{L}$ give the same equations of motion, are they equivalent?
So I was reading this:  and while the answers for the first question are good, nobody gave much attention to the second one. In fact, people only said that it can be proved without giving any proof or any.  So, if I have a Lagrangian and ADD an arbitrary function of $\dot{q}$, $q$ and $t$ in such a way that the equations of motion are the same, does this extra function MUST be a total time derivative?   EDIT Ok, I changed my question a little bit:  Question: If I have a function that obeys the Euler-Lagrange equation off-shell, this implies that my function is a time derivative? (This was used in Qmechanic's answer of this other question: , equation 14.)  Also, why people only talk about things that change the Lagrangian only by a total derivative? If this is not always the case that keeps the equation of motion the same, so why is it so important? And why in the two questions I posted about the same statement on Landau&amp;Lifshitz's mechanics book only consider this kind of change in the Lagrangian?
Can older cards that refer to a player/opponent ever be used on planeswalkers? Many older cards refer to players or opponents, e.g. "deal 3 damage to target player". Can any of these be used on planeswalkers, beyond the fact that you can simply attack them with creatures?  For example:   Would my opponent discard a card if  damaged a planeswalker they control? Would my opponent, the planeswalker, neither, or both amass poison counters from an unblocked ? Can  target a planeswalker? Does  affect planeswalkers? Can  prevent damage to a Planeswalker? Would  be obligated to take damage being dealt to a planeswalker? Can a planeswalker gain loyalty from ? Can  prevent damage to a planeswalker? What are the details of "Objective-C Literals" mentioned in the Xcode 4.4 release notes? Values of $\sum_{n=0}^\infty x^n$ and $\sum_{n=0}^N x^n$ Why does the following hold:  \begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ? \end{equation*}  Can we generalize the above to      $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?   Are there some values of $x$ for which the above formula is invalid?  What about if we take only a finite number of terms? Is there a simpler formula?     $\displaystyle \sum_{n=0}^{N} x^n$   Is there a name for such a sequence?    This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: . Approximate order statistics for normal random variables Point Renderings Slightly Off in Mathematica I have an unused copy of Mass Effect 2 Collector's Edition bought via Amazon.com in the US, including a special code for Inferno Armour (also unused).  Will it work fine if I install and start playing the game in the UK? Does geographical location matter?  Thanks! I saw this several times in different movies, but I don't understand what is this supposed to mean. The main idea is when the scene is about to change, for one second or so we still see the picture from the old scene, but the sound is already coming from from the upcoming scene. Why do they do this? What's the purpose of this technique? If $f_k \to f$ a.e. and the $L^p$ norms converge, then $f_k \to f$ in $L^p$ Let $1\leq p &lt; \infty$. Suppose that    $\{f_k\} \subset L^p$ (the domain here does not necessarily have to be finite),  $f_k \to f$ almost everywhere, and  $\|f_k\|_{L^p} \to \|f\|_{L^p}$.    Why is it the case that $$\|f_k - f\|_{L^p} \to 0?$$   A statement in the other direction (i.e. $\|f_k - f\|_{L^p} \to 0 \Rightarrow \|f_k\|_{L^p} \to \|f\|_{L^p}$ ) follows pretty easily and is the one that I've seen most of the time. I'm not how to show the result above though. What are complex numbers, actually? Brightness fn key shortcut doesn't work on ASUS laptop The brightness shortcuts via fn key don't work.  I tried each solution from . After editing /etc/default/grub file I rebooted.  GRUB_CMDLINE_LINUX="acpi_osi=Linux" GRUB_CMDLINE_LINUX="acpi_backlight=vendor" GRUB_CMDLINE_LINUX="acpi_osi=linux acpi_backlight=vendor" GRUB_CMDLINE_LINUX="acpi_osi_Linux"   What other workarounds/solutions are there?  Relevant output:  $ sudo lshw -C display   *-display UNCLAIMED             description: 3D controller        product: GK107M [GeForce GT 750M]        vendor: NVIDIA Corporation        physical id: 0        bus info: pci@0000:01:00.0        version: a1        width: 64 bits        clock: 33MHz        capabilities: pm msi pciexpress bus_master cap_list        configuration: latency=0        resources: memory:f6000000-f6ffffff memory:e0000000-efffffff memory:f0000000-f1ffffff ioport:e000(size=128) memory:f7000000-f707ffff   *-display        description: VGA compatible controller        product: 4th Gen Core Processor Integrated Graphics Controller        vendor: Intel Corporation        physical id: 2        bus info: pci@0000:00:02.0        version: 06        width: 64 bits        clock: 33MHz        capabilities: msi pm vga_controller bus_master cap_list rom        configuration: driver=i915 latency=0        resources: irq:50 memory:f7400000-f77fffff memory:d0000000-dfffffff ioport:f000(size=64) Can I use my device on a different carrier? Sometimes it's desirable to use a device on a carrier that does not officially sell it. Often this is due to regional differences and travel (for example, using a US phone in Europe) but also because some carriers simply do not offer phones that other carriers do.  How can I determine if it's possible to use my phone on a different carrier, and which carriers it is compatible with?    This community wiki is intended to be a canonical answer to a common question. If you feel that improvements could be made then please edit the post(s). I need a software, that emulates a printer, converts the printed documents to PDF/TIFF and send it via eMail to a specified address (for windows).  The virtual printer / PDF conversion is possible with a lot of PDF-converters. But I need a software, which is also able to directly mail me these PDFs. It's always the same mailaddress, and I have an SMTP server.  Any suggestions? Passing a dictionary to a function as keyword parameters I'd like to call a function in python using a dictionary.  Here is some code:  d = dict(param='test')  def f(param):     print(param)  f(d)   This prints {'param': 'test'} but I'd like it to just print test.  I'd like it to work similarly for more parameters:  d = dict(p1=1, p2=2) def f2(p1, p2):     print(p1, p2) f2(d)   Is this possible? I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.     Proof $\liminf E_k \subset \limsup E_k $   If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define \begin{align*} \limsup E_k &amp; = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\ &amp;= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\ &amp;= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots. \end{align*}  Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$. \begin{align*} \liminf E_k &amp; = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\ &amp;= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\ &amp;= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots. \end{align*} Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.  Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$. Would raising the review limit help with the Close Votes queue? Bootstrap CSS Active Navigation How can I get the size of a memory block allocated using malloc()? Lightning component : Get the id of a record In a lightning component I display a list of articles (Article__c custom object) so I ca see, for example, the name of the article, the price, and a button that I'd like redirect to another component which display the details of the article selected.   For now I'm just trying to get the id of the article, in a basic alert in javascript, but I doesn't work : it says the id is null.   Here is a part of the component :  &lt;aura:handler name="init" value="{!this}" action="{!c.doInit}" /&gt; &lt;aura:attribute name="wrappers" type="ListeArticlesWrapper" /&gt; &lt;aura:attribute name="articles" type="Article__c" /&gt;  &lt;aura:attribute name="Detail" type="String"/&gt; &lt;aura:registerEvent name="navigate" type="c:NavigateToDetail"/&gt;  &lt;div class="page"&gt;     &lt;h1&gt;Liste des articles&lt;/h1&gt;     &lt;div class="slds"&gt;         &lt;table class="slds-table slds-table--bordered slds-table--cell-buffer"&gt;             &lt;thead&gt;                 &lt;tr class="slds-text-heading--label"&gt;                     &lt;th scope="col" class="head"&gt;&lt;/th&gt;                     &lt;th scope="col" class="head"&gt;Nom&lt;/th&gt;                     &lt;th scope="col" class="head"&gt;Type&lt;/th&gt;                     &lt;th scope="col" class="head"&gt;Prix&lt;/th&gt;                     &lt;th scope="col" class="head"&gt;&lt;/th&gt;                 &lt;/tr&gt;             &lt;/thead&gt;             &lt;tbody&gt;                 &lt;aura:iteration items="{!v.wrappers}" var="wrap"&gt;                     &lt;tr&gt;                         &lt;td class="cell"&gt;                             &lt;ui:inputCheckbox value="{!wrap.selected}" /&gt;                         &lt;/td&gt;                         &lt;td class="cell"&gt;                            &lt;ui:outputText value="{!wrap.art.Name}" /&gt;                         &lt;/td&gt;                         &lt;td class="cell"&gt;                             &lt;ui:outputText value="{!wrap.art.Type__c}" /&gt;                         &lt;/td&gt;                         &lt;td class="cell"&gt;                             &lt;ui:outputText value="{!wrap.art.Prix__c}" /&gt;                         &lt;/td&gt;                         &lt;td class="cell"&gt;                             &lt;button class="slds-button slds-button--neutral slds-not-selected"                                      onclick="{!c.getDetail}" data="{!wrap.art.Id}"&gt;Details&lt;/button&gt;                         &lt;/td&gt;                     &lt;/tr&gt;                 &lt;/aura:iteration&gt;             &lt;/tbody&gt;         &lt;/table&gt;   The associated controller js (with the function getDetail)  ({ doInit : function(component, event, helper) {     helper.init(component, event); },  getArticles : function(component, event, helper) {     helper.getArticles(component, event); },  getDetail : function(component, event, helper){     var id = event.target.getAttribute("data-data") || event.target.parentNode.getAttribute("data-data")     alert('detail article : id = ' + id); //it gives me null } })   Thanks !  EDIT  And the wrapper class in case you wonder what is the "art" :  public with sharing class ListeArticlesWrapper { //wrapper pour récupérer les articles selectionnés @AuraEnabled public Article__c art {get; set;}  @AuraEnabled public Boolean selected {get; set;} } Understanding ROC curve Making a (simple) directory tree I'm trying to make a simple directory tree representation in LaTeX. Something like:  spam  |  +-- ham  +-- eggs      |      +-- more spam      +-- dead parrots   The tree won't be that long (say, 10--15 entries or so), with three or so levels, and is a one-off thing. So, quick-and-dirty solutions are welcome. When recommending tikz, please note that I have almost no experience with it, so I'd appreciate a clean simple solution, even if it doesn't look quite perfect.   On Tex.SE tikz is almost like the jquery recurring joke on StackOverflow..    Edit: As per Thorsten's suggestion,  is a nice and easy-to-use solution. The example from above could be written as:  \usepackage{dirtree} ... \dirtree{% .1 spam. .2 ham. .2 eggs. .3 more spam. .3 dead parrots. }