SciFi story from before 1976 that involves multiple dimensions
Children's book with "square triangles" in parallel universe?
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? Why is there a white space on the left side?  \documentclass{standalone} \usepackage{asymptote} \begin{document} \noindent \begin{asy} size(10cm,10cm); draw((0,0)--(100,100)); draw((0,100)--(100,0)); dot((50,50)); \end{asy} \end{document} My host is x64 bits Windows 8.1.  I downloaded the latest Virtual Box (4.3) and I'm trying to create a VM with a 64 bits Ubuntu OS (ubuntu-12.04.3-desktop-amd64).  When I go to New VM wizard, it doesn't give me option to select "Ubuntu (x64)" as I have seen in other people's screenshots, only just "Ubuntu". As a result, the ISO can't boot. I tried in another PC and Virtual Box gives the x64 variants to most listed OS...  Control Panel shows x64 OS, x64 processor. My host laptop is a Sony Vaio VPCZ22UGX/N, Intel® Core™ i7-2640M processor. CPUz shows Vx-t is available on my processor, of course.  Here is what I tried so far:   I enabled IO APIC as required in the docs. I have virtualization enabled in the BIOS. It works fine in VMware. Check that Hyper-V is not running or even installed on my Windows. Same for VMware. I also tried running the command:  VBoxManage modifyvm [vmname] --longmode on    for that VM, but no change.. I think the issue is really that I can't select x64 variant of the Ubuntu OS for that VM. Other people seem to indicate that's a requirement, but I don't get that option for some reason.  I spent a lot of time and can't find what's wrong... Anyone knows what could be missing here?  Thank you very much!!  Eduardo Cardinality of quotient ring $\mathbb{Z_6}[X]/(2X+4)$ I've tried latexdef which does not handle --- as the operand. And \texemdash is defined as a character. So that leaves me once again leaning on the collective wisdom of the group :) If $\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$,then prove that $\sin y=\frac{\sin x(3+\sin^2 x)}{(1+3\sin^2 x)}$ If $\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$,then prove that $\sin y=\frac{\sin x(3+\sin^2 x)}{(1+3\sin^2 x)}$    $\tan(\frac{\pi}{4}+\frac{y}{2})=\tan^3(\frac{\pi}{4}+\frac{x}{2})$ $\frac{1+\tan\frac{y}{2}}{1-\tan\frac{y}{2}}=(\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}})^3$ Applying componendo and dividendo rule on both sides, $\tan\frac{y}{2}=\frac{\tan\frac{x}{2}(3+\tan^2\frac{x}{2})}{(1+3\tan^2\frac{x}{2})}$ I am stuck here. I read  and thought that would easily be solved (not that it isn't solvable without) if one could write:   @Override public String toString() {     return super.super.toString(); }   I'm not sure if it is useful in many cases, but I wonder why it isn't and if something like this exists in other languages.  What do you guys think?  EDIT: To clarify: yes I know, that's impossible in Java and I don't really miss it. This is nothing I expected to work and was surprised getting a compiler error. I just had the idea and like to discuss it. Smoke Generator does not work When I try to run the smoke simulator in Blender 2.62 I get no smoke at all. When I attempt to render, the only output is a black screen. What am I doing wrong here that is making the smoke simulation not render?  Is it a mistake I made, or a problem with Blender version 2.62? Where can I find a repository for older unsupported releases? Erf as term for plot of land in a town or city. Is this word only used in southern Africa? The Free Dictionary website states and so does Wikipedia:     erf [ɜːf]   n pl erven [ˈɜːvən]   (Engineering / Civil Engineering) South African a plot of land, usually urban, marked off for building purposes   [Afrikaans]   The term erf is used in legal documentation and technical English throughout South Africa and Namibia, maybe even further afield in the region.  Is it used anywhere else, except maybe the Dutch/Flemish regions in Europe, where I assume the term originated?  An example:      The client purchased Erf 132, Windhoek for a large sum of money. He intends to build a house on this erf.   The capitalisation of "Erf 132" is commonplace, since it is a definition of a location. This is similar to capitalisation of towns or cities. I find it slightly awkward, but that is how it is done. It's not consistent to leave an edit pending for approval (thus not making available for everyone else and locking its state) after a user who has those privileges have given his endorsement.   You could say why don't I improve the edit, but that makes no sense for something that is correct and it would not follow the "keep it simple" policy.  Update:  This is the current time line:   Someone who doesn't know about formatting makes a question. It's not legible A user without edit privileges amends the formatting. The correction is not visible, still not legible. I see the question and the pending revision and then cast my approval. It doesn't take effect yet, still not legible. Even for me if I didn't take care of copying the question source because it disappears from my own revision page!   I might agree that one's approval is indicative of both revision's quality and editor's competence. But we should retain that the main objective is to ask and answer questions. So, I propose this model:   Someone who doesn't know about formatting makes a question. It's not legible A user without edit privileges amends the formatting. The correction is not visible, still not legible. A user with edit privileges cast his approval. Revision takes effect. Although it keeps the pending approval (the Edit (1) button). If a second user with edit privileges cast a rejection, it also takes effect as a rollback. Set root domain record to be a CNAME I need to create an NS record for a domain that is a CNAME, for the purpose of having two domains pointed at one IP, and not having to maintain the current IP address in two different places.  The DNS provider for this domain is DynDNS, but they block this operation:     CNAME cannot be created with label   that is equal to zone name   I can do this with another domain whose DNS is served by 1and1:  root@srv-ubuntu:~# dig myseconddomain.co.uk  ; &lt;&lt;&gt;&gt; DiG 9.4.2-P1 &lt;&lt;&gt;&gt; myseconddomain.co.uk ;; global options:  printcmd ;; Got answer: ;; -&gt;&gt;HEADER&lt;&lt;- opcode: QUERY, status: NOERROR, id: 61795 ;; flags: qr rd ra; QUERY: 1, ANSWER: 2, AUTHORITY: 0, ADDITIONAL: 0  ;; QUESTION SECTION: ;myseconddomain.co.uk.             IN      A  ;; ANSWER SECTION: myseconddomain.co.uk.      71605   IN      CNAME   myfirstdomain.co.uk. myfirstdomain.co.uk.     59      IN      A       www.xxx.yyy.zzz  ;; Query time: 298 msec ;; SERVER: 10.0.0.10#53(10.0.0.10) ;; WHEN: Tue Aug 18 14:17:26 2009 ;; MSG SIZE  rcvd: 78   Is this a breach of the RFCs or does DynDNS have a legitimate reason for blocking this action?  Followup Thanks to the two answers already posted I now know that 1and1 IS breaching RFCs to do this. However it does work and they seem to support it. For a company that hosts so many domains it seems very odd that they get away with doing this on such a massive scale without objection.  More followup  The output of "dig myseconddomain.co.uk ns" as requested.  root@srv-ubuntu:~# dig myseconddomain.co.uk ns  ; &lt;&lt;&gt;&gt; DiG 9.4.2-P1 &lt;&lt;&gt;&gt; myseconddomain.co.uk ns ;; global options:  printcmd ;; Got answer: ;; -&gt;&gt;HEADER&lt;&lt;- opcode: QUERY, status: NOERROR, id: 18085 ;; flags: qr rd ra; QUERY: 1, ANSWER: 2, AUTHORITY: 0, ADDITIONAL: 2  ;; QUESTION SECTION: ; myseconddomain.co.uk.             IN      NS  ;; ANSWER SECTION: myseconddomain.co.uk.      4798    IN      NS      ns67.1and1.co.uk. myseconddomain.co.uk.      4798    IN      NS      ns68.1and1.co.uk.  ;; ADDITIONAL SECTION: ns67.1and1.co.uk.       78798   IN      A       195.20.224.201 ns68.1and1.co.uk.       86400   IN      A       212.227.123.89  ;; Query time: 59 msec ;; SERVER: 10.0.0.10#53(10.0.0.10) ;; WHEN: Wed Aug 19 12:54:58 2009 ;; MSG SIZE  rcvd: 111 I went through a , but the answer was informing how to hide the monitor, but not how to disable it. I do not want any new windows to creep into the laptop display when the external display is connected. I could not find any in the Display Settings (under System Preferences). I don't want to close the lid because I think it impacts the air flow of the MacBook Pro. I don't want to close the lid because the sound quality is reduced. Galois group of the product of two fields being the product of Galois groups? I'm just wondering if the following theorem is reasonable and whether the proof is makes sense or not? Also I have an application of it which I was trying to do earlier but I made a lot of mistakes so I would be very grateful if anyone would give some comments on this in case I have some very strong misunderstandings about Galois theory.    Theorem Let $E_1$ and $E_2$ be Galois field extensions of $F$ with trivial intersection ($E_1 \cap E_2 = F$), then $E_1 E_2$ has Galois group $\text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.  Proof   $E_1$ and $E_2$ are splitting fields (over $F$) of some polynomials $f_1$ and $f_2$ so $E_1 E_2$ is the splitting field of $f_1 f_2$ over $F$ and therefore it is Galois and we are justified using the "Gal" notation. Let $\sigma_1 \in \text{Gal}(E_1/F)$ and $\sigma_2 \in \text{Gal}(E_2/F)$ then by the fundamental theorem of Galois theory we can lift these both into elements of $\text{Gal}(E_1 E_2/F)$. The claim is that $\sigma_1 \sigma_2 = \sigma_2 \sigma_1$, to see this consider an element $\alpha \in E_1$, we have both $\sigma_2 \alpha = \alpha$ and $\sigma_2 \sigma_1 \alpha = \sigma_1 \alpha$ because it is invariant with respect to $\sigma_2$, but we can combine these equalities to get $\sigma_1 \sigma_2 \alpha = \sigma_2 \sigma_1 \alpha$. Both groups $\text{Gal}(E_i/F)$ ($i = 1,2$) are included in $\text{Gal}(E_1 E_2/F)$ and the degrees say there are no more elements, furthermore since they commute we can conclude $\text{Gal}(E_1 E_2/F) = \text{Gal}(E_1/F) \times \text{Gal}(E_2/F)$.     So hopefully that theorem is, if not correct, fixable.. and can be used to prove that $\mathbb Q (\sqrt{a_1},\sqrt{a_2},\cdots,\sqrt{a_n})$ has Galois group $C_2^m$ ($m \le n$) over $\mathbb Q$? Uniform convergence of derivatives, Tao 14.2.7. Minimal Sufficient Statistic for location family In Casella &amp; Berger Statistical Inference,in exercise 6.8, we're asked to prove that, for any distribution belonging to the location family, the order statistics are minimal sufficient.  How does one prove this? I have no idea on how to tackle this exercise.  Any help would be appreciated. Is there any way to block websites without Root access I am searching for a way to block websites without having root access. Is there any way to do it? Any help will be appreciated. Why is quadratic integer ring defined in that way? Quadratic integer ring $\mathcal{O}$ is defined by \begin{equation}    \mathcal{O}=\begin{cases}       \mathbb{Z}[\sqrt{D}] &amp; \text{if}\ D=2,3\ \pmod 4\\       \mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]\  &amp; \text{if}\ D=1\pmod 4    \end{cases} \end{equation} where $D$ is square-free.   I understand $\mathbb{Z}\left[\frac{1+\sqrt{D}}{2}\right]$ is not closed under multiplication if $D=2,3\pmod 4$. But still, isn't it more natural to define $\mathcal{O}=\mathbb{Z}[\sqrt{D}]$ for all square-free $D$? (in that case, it really seems like 'quadratic integer') I wonder what is the motivation of this definition. Separate Models vs Flags in the same model I have customer data from 2 brands. The data structure are the same, but I expected the customer behaviour to be different in different brand.  So I could train 2 models, 1 for each brand, or I could train one model on both brands, but flag which brand the customers belongs to within the data.  So my question is, would training on both brands with flags be too much more work for the Machine Learning (I'm using neural network), and therefore be harder to achieve good accuracy than to have dedicated models for each brand? Can we get x comment votes per y minutes, rather than one vote per 5 seconds?