Is IDENTITY continuous in new versions of PostgreSQL?
Challenges with Key with Gaps?
How to use testfor with a mob with a two-word name? How can I test for mob who has a name with two words? For example villager named Grumpy Bill. For example I want to kill this villager.  This command   /kill @e[name=Bill]   would work just fine but when I try  /kill @e[name=Grumpy Bill]   it says "The entity UUID provided is in an invalid format"  What is the proper command for this action? How do I make a floor plan with specific dimensions? How can I set the maximum draw type for multiple objects? I'm trying to delete some old flow which are inactive via workbench. DestructiveChanges.xml &lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt; &lt;Package xmlns=&quot;http://soap.sforce.com/2006/04/metadata&quot;&gt;     &lt;types&gt;         &lt;members&gt;Advanced_Round_Robin&lt;/members&gt;         &lt;name&gt;Flow&lt;/name&gt;     &lt;/types&gt;Ï &lt;/Package&gt;  package.xml &lt;?xml version=&quot;1.0&quot; encoding=&quot;UTF-8&quot;?&gt; &lt;Package xmlns=&quot;http://soap.sforce.com/2006/04/metadata&quot;&gt;       &lt;version&gt;48.0&lt;/version&gt; &lt;/Package&gt;  I've added both file in zip and in workbench when I deploy using this zip file I get below error.  insufficient access rights on cross-reference id   What permission I need to have to perform this action? Also is there way I can delete few version of same flow only? Like keep only last 3-4 version and delete all previous version of given flow? Thank you. Gradient steepest direction and normal to surface? From this , I now understand the gradient to be the directional derivative that returns the steepest slope at a point. However, reading my textbooks, they all say that the gradient is normal to the tangent plane. In the above SE question, the gradient sounds like it should be parallel to a direction in the $xy$ plane, not actually up on the surface.   My question is therefore do I have a misunderstanding of the tangent plane, gradient or directional derivative?   In addition, I question how the gradient always passes through the origin or the $xy$ plane? If for instance in the graphic, the gradient had been another direction (say horizontal), then it wouldn't pass through the origin! In addition, the gradient of other points in the graphic would satisfy this example. Many thanks! i am finding nowhere a solution I am running UBUNTU 12.04 and using GNOME desktop, which I installed additionally. Logging into classic GNOME I see the following:     WHAT is this "earth globe" icon coming from?Seems to be named "starter". It is configured to open a website, from my bookmarks. WHY is it not showing the TIME?   Using dconf-editor I had no luck in answering the above questions.  Any idea? Is it true that one should not use NSLog() on production code? I was told this a few times in this very site, but I wanted to make sure this is really the case.  I was expecting to be able to sprinkle NSLog function calls throughout my code, and that Xcode/gcc would automatically strip those calls out when building my Release/Distribution builds.  Should I avoid using this? If so, what alternatives are most common between experienced Objective-C programmers? Are ores replenished over time? Trying to remember a book that had people (I think they turned white?) segregated into a compound Capitalizing "the" when mid-sentence in a company name? Consider a company called:     The Association of Examples Ltd.   When you refer to this company, should you have to capitalize "the" when it's mid-sentence and would have been used even if it wasn't in the name?   It looks odd to me:     In other news, The Association of Examples announced that...      Sarah is the founder of The Association of Examples and has...   But I imagine it's the correct way.  For the grammar-conscious, why or why not?    To give you context, I am asking also when branding ones own company like that. The Economist did it (as did ) and , but I can imagine it can cause confusion. I am trying to figure out how I create a downloadable PDF template of a wedding invitation design that my customers can edit themselves. They will be approx 2 fonts used on each template and the customer will need to be able to edit all text themselves.  e.g. change their names, wedding date, location etc. etc.  Basically I want to be able to sell an editable downloadable template.  So I will need to lock the artwork and "open" the text so they can edit the text only.  If you think of something like this image. That has artwork that I want to be uneditable and then has text that I want the customer to be able to edit and print out themselves at home etc. How do I go about proving this?  If $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$. I'm very confused with gcd proofs. Surjectivity implies injectivity and conversely Curious badge bug I saw that it has been already reported that also they have earned courious badge but the badge has not been given. but their answers say that they must wait a day and the badge will be awarded.  In my case its 8/5 and it was 5/5 long time ago and I have never received a message telling that I have earned it. its just telling me that the next badge to be awarded is Curious badge.  this is my profile: Block any query in BIND I've been trying to block ANY queries in my BIND server with no success.  I followed this post  which says that I can use minimal-responses but it doesn't work.  None of the answers fit my current configuration.  view world {     match-clients   { world-clients; };     allow-recursion { none; };     recursion       no;     allow-transfer  { key XXXXXX; };      minimal-responses yes;      forwarders      { };      include         "/etc/bind/world.conf"; }; Let $f \in L^1(\mathbb{R})$, for $a\in \mathbb{R}$ let $f_a(x)=f(x-a)$, prove that: $$\lim_{a\rightarrow 0}\|f_a -f \|_1=0$$  I know that there exists $g\in C(\mathbb{R})$ s.t $\|f-g\|_1 \leq \epsilon$, this is also true for $f_a$ and $g_a$.  Now I have the next estimation:  \begin{align*} \|f_a-f\|_1 &amp;= \int_{\mathbb{R}} |f(x)-f(x-a)| dx \\ &amp;= \int |f(x)-g(x)+g(x)-g(x-a)+g(x-a)-f(x-a)| \\ &amp;\leq \|f-g\|_1 + \|f_a-g_a\|_1 + \int |g(x)-g(x-a)| \end{align*} my question is: I can argue that $\lim_{a\to0} \int |g(x)-g(x-a)| = \lim_{a\to0} \lim_{T\to\infty} |g(x_0)-g(x_0-a)| 2T$, now I think that I can change the order of the limits here, but I am not sure why?  P.s $x_0$ is some point in $[-T,T]$, and the above is valid from the intermediate integral theorem for continuous functions, right?  Thanks. What's wrong with "I'll open you the door"? When I call the buzzer outside my girlfriend's flat, she sometimes says *"I'll open you the door". I correct this to "I'll open the door for you".  I've never heard a native speaker say it the first way, which is why I think it's wrong. But I can't explain why.  There is a pattern in lots of English phrases that would suggest both are correct. A few common examples:   Give the keys to me before you go — Give me the keys before you go I'll buy a coffee for you at the cafe — I'll buy you a coffee at the cafe We sent a text to Martin on his birthday — We sent Martin a text on his birthday   The left-hand side is more formal, and the right-hand side is more common in everyday speech.  What do you call this pattern?  What makes the 'open-the-door' sentence an exception? Reason of existense of the extra space in align environment and removal of it between two align environments This is a follow-up question regarding the extra vertical space in amsmath's align environment. Why when in use align uses so much vertical space in contrast with other environments like for example tabular or itemize?  \documentclass[12pt]{article} \usepackage[top=0.3in, bottom=1.2in, left=0.8in, right=0.8in]{geometry}   \usepackage[fleqn]{amsmath} \usepackage{unicode-math}  \setlength{\mathindent}{0cm} \newcommand{\3}{\vspace{0.3cm}}  \title{} \author{} \date{}  \begin{document}  70  \begin{align*} &amp;e^{jz}=\cos z+j\sin z\\ &amp;\cos z=(1/2)(2\cos z)=\\ &amp;=(1/2)(2\cos z+j\sin z-j\sin z)=\\ &amp;=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\ &amp;=(1/2)(e^{jz}+e^{-jz}) \end{align*}  80  \begin{itemize} \item Item 1  \item Item 2  \end{itemize}  90  \begin{tabular}{c|c}  Entry 1&amp; Entry 2  \\  Entry 3&amp; Entry 4 \end{tabular}  \end{document}     Also another image of the huge space between two align environments. Even without the blank lines the space is a lot.    \begin{document}  \begin{multicols*}{2}   \begin{align*} &amp;\text{70}\\ &amp;e^{jz}=\cos z+j\sin z\\ &amp;\cos z=(1/2)(2\cos z)=\\ &amp;=(1/2)(2\cos z+j\sin z-j\sin z)=\\ &amp;=(1/2)(\cos z+j\sin z+\cos z-j\sin z)=\\ &amp;=(1/2)(e^{jz}+e^{-jz}) \end{align*}    \begin{align*} &amp;\text{207}\\ &amp;x(t)=A_{c}\cos \theta(t)\\ &amp;\theta(t)=2\pi f_{c}t+\phi(t)\\ &amp;\phi(t)=K_{p}m(t)\\ &amp;x(t)=A_{c}\cos[2\pi f_{c}t+K_{p}m(t)] \end{align*}   Edit I:  As I mention in a comment I have tried the following code and there is no difference between two align environments, only between text and an align environment. Why doesn't work in the case of the two environments and how can I fix it so to work? Also what is the reason that this space exists?  \setlength{\abovedisplayskip}{0pt} \setlength{\belowdisplayskip}{0pt} \setlength{\abovedisplayshortskip}{0pt} \setlength{\belowdisplayshortskip}{0pt} My processor is using a big part of my RAM memory as cache and I want to clean it up because of that; will it prejudice something? The equivalence of open-set definition of continuous mappings to the $\epsilon-\delta$ definition A function $f: M\rightarrow N$ is defined to be continuous iff $\forall$ open set $U\subseteq N$, the preimage of $U$ is also open.  I was trying to prove that such a definition would be equivalent to the $\epsilon -\delta$ definition in a metric space, and I could only do so if I assume $f$ is surjective, or otherwise a counterexample would be $g: \mathbb{R}\rightarrow\mathbb{R}$ where $g(x)=x^2$. As if I take $U=(-1,1)$, then its preimage is $[0,1)$, which is neither open or closed. In fact, any $U$ containing elements that cannot be achieved by $g$ would face the same problem.  So I was wondering that have I remembered the definition of preimage wrong, or do I need to make sure that $U$ must be chosen from the image set of $f$?