I hold Lithuania citizenship and currently live in Denmark. Is it possible to travel to UK from Denmark and back using only the identity card?  According to      You can enter the UK with either a valid passport or a national identity card issued by a EEA country. It must be valid for the whole of your stay.   I am fully aware that this question was asked a number of times, for example . However, since "Brexit" situation might have changed. Moreover, recently while travelling from Copenhagen airport to Dublin I was asked to prove validity of my identity card using additional document (e.g. driving license).
On 23rd of June, 2016, the UK . This leads to the following questions:   Are UK citizens traveling to the EU affected? Are EU citizens traveling to the UK affected? Are UK/EU family members traveling both way affected? Are Commonwealth citizens traveling to the UK affected? Are other visa-free nationals (such as US/Canada citizens) affected? What other consequences might 'Brexit' have on travelers?   Also, see the  on Expats.SE.
Do extension fields always belong to a bigger field? Let $F$ be a field, $E_1$ and $E_2$ are two distinct extension fields of $F$. Is it the case that we can always somehow find a field $G$ that contains both $E_1$ and $E_2$? In other words, could extensions of fields have different 'direction's such that they are incompatible?  Edit: I began to think about this problem while reading a proof. $F$ is a field. $a$ and $b$ are algebraic over $F$. $p(x)$ and $q(x)$ are two polynomials in $F[x]$ of minimum degree that respectively make $a$ and $b$ a zero. The proof claims that there is an extension $K$ of $F$ such that all distinct zeros of $p(x)$ and $q(x)$ lie in $K$. For a single polynomial, I know this kind of field exists because of the existence of splitting field, why it is true for two polynomials? What is the name for moments, a parameter that is calculated Why is Cat5/Cat5e Ethernet cable rated 100MHz? As the questions says, is there any specific reason why Ethernet cable is rated 100MHz?  For a 100Mbps Ethernet the data rate is 100Mbps, and the baud rate is 125Mbaud, because of the specific line coding that it uses. What is the relationship between 125Mbaud rating of data transfer and 100MHz rating of cable?  Thank you. I have a vector space $V$ over a field of characteristic $0$. If $M_1,\dots,M_k$ are proper subspaces of $V$, and $N$ is a subspace of $V$ such that $N\subseteq M_1\cup\cdots\cup M_k$, how can you tell $N\subseteq M_i$ for some $i$?  I was first trying to show it just in the case $N\subseteq M_1\cup M_2$, and hoping to extend it to finitely many $M_i$. If either of the $M_i$ contains the other, the claim follows, so I suppose neither $M_i$ contains the other.   In hopes of a contradiction, I suppose $N\not\subseteq M_1$ and $N\not\subseteq M_2$, Picking $x_1\in N\setminus M_1$ and $x_2\in N\setminus M_2$, I'd have $x_1+x_2\in N\subseteq M_1\cup M_2$. The only thing I could conclude was that actually $x_1,x_2\notin M_1$ and $x_1,x_2\notin M_2$, which seems like a dead end. What's the right way to do this? How do I bridge network interfaces? I have a laptop running 11.10 with 2 network interfaces, wired and wireless.  Wireless interface connects to the router with DHCP and connects to the internet. Wired interface is connected to a switch. I want the computers on the switch to get their IP addresses from the router not my laptop.     How would I go about setting this up? Bridging the interfaces? Why does \cal{M}_n give “M>”? How to solve $y^2=3x^4+3x^2+1$ for integers. If $x,y \in \mathbb Z$ , then find all the solutions of   $$y^2=3x^4+3x^2+1$$       I was asked this question by my friend who said that he encountered this while solving another problem. I have tried several things but am unable to solve this question. Moreover, this has to be done using elementary methods only. So far, I have tried to factorize and use Pell's equation. At the end, I'm getting $$2y_{n} + (2x^2_{n}+1)\sqrt{3}=(2+\sqrt{3})^{n}$$   where $n \in \mathbb Z^{+}$    But I'm not able to figure out how to show a contradiction from here. Can anyone please help me out? Thanks. WCF server-side timeout How can I programmatically delete a line from a Word document using c#? How can I deduce/identify an app from a notification icon? Presumably after an update, google chrome has very serious problems with page rendering. I want to downgrade it. (Switching to vmlinuz.old and initrd.img.old did not help).  $ apt-cache policy google-chrome-stable google-chrome-stable:   Installed: 35.0.1916.153-1   Candidate: 35.0.1916.153-1   Version table:  *** 35.0.1916.153-1 0         500 http://dl.google.com/linux/chrome/deb/ stable/main i386 Packages         100 /var/lib/dpkg/status   on a computer where google chrome works ok:  $ apt-cache policy google-chrome-stable google-chrome-stable:   Installed: 34.0.1847.132-1   Candidate: 35.0.1916.153-1   Version table:      35.0.1916.153-1 0         500 http://dl.google.com/linux/chrome/deb/ stable/main amd64 Packages  *** 34.0.1847.132-1 0         100 /var/lib/dpkg/status   (If I had the .deb on the 2nd computer, everything would be easier. But there's no .deb. And no old .deb's on the google site.)  So I try:  $ sudo aptitude install google-chrome-stable=34.0.1847.132-1 Unable to find a version "34.0.1847.132-1" for the package "google-chrome-stable" Unable to find a version "34.0.1847.132-1" for the package "google-chrome-stable" No packages will be installed, upgraded, or removed. 0 packages upgraded, 0 newly installed, 0 to remove and 0 not upgraded. Need to get 0 B of archives. After unpacking 0 B will be used.   You see, it does not work. The Synaptic manager also shows no version that I could set.  So, how do I downgrade google-chrome-stable?  Probably someone knows an archive of old google-chrome .deb's? Neg Binomial and the Jeffreys' Prior I'm trying to obtain the Jeffreys' prior for a negative binomial distribution. I can't see where I go wrong, so if someone could help point that out that would be appreciated.  Okay, so the situation is this: I am to compare the prior distributions obtained using a binomial and a negative binomial, where (in both cases) there are $n$ trials and $m$ successes. I get the right answer for the binomial case, but not for the negative binomial.   Let's call the Jeffreys' prior $\pi_J(\theta)$. Then,  $$ \pi_J(\theta)\propto [I(\theta)]^{1/2}. $$  Under the regularity conditions (fulfilled as we are dealing with the exponential family),  $$ I(\theta)=-E\left(\frac{\partial^2 \log L(\theta|x)}{\partial \theta^2}\right) $$ where for the negative binomial $n$ is $x$ in the above expression (the total number of successes $m$ is fixed, $n$ is not). The distribution --I think-- is  $$ p(m|\theta)\propto\theta^m(1-\theta)^{n-m} $$ since $\theta$ is defined as the probability of success and $m$ is the number of successes. This is also the likelihood, since $m$ is a scalar and not a vector. Hence,  $$ L(\theta|n)\propto\theta^m(1-\theta)^{n-m}\\ \log L(\theta|n)=m\log\theta +(n-m)\log (1-\theta)\\ \frac{\partial\log L(\theta|n)}{\partial \theta}=\frac{m}{\theta}-\frac{n-m}{1-\theta}\\ \frac{\partial^2\log L(\theta|n)}{\partial \theta^2}=-\frac{m}{\theta^2}-\frac{n-m}{(1-\theta)^2} $$ so the Fisher information is  $$ I(\theta)=-E\left(\frac{\partial^2\log L(\theta|n)}{\partial \theta^2}\right)=\frac{m}{\theta^2}+\frac{E(n)-m}{(1-\theta)^2}=\frac{m}{\theta^2}+\frac{\frac{m\theta}{1-\theta}-m}{(1-\theta)^2}\\ =\frac{m(1-\theta)^2+\frac{m\theta^3}{(1-\theta)}-m\theta^2}{\theta^2(1-\theta)^2}=\frac{m(1-2\theta)+\frac{m\theta^3}{(1-\theta)}}{\theta^2(1-\theta)^2}\\ =\frac{m(1-2\theta)(1-\theta)+m\theta^3}{\theta^2(1-\theta)^3}=\frac{m(1-3\theta+2\theta^2+\theta^3)}{\theta^2(1-\theta)^3}\\ \propto\frac{1-3\theta+2\theta^2+\theta^3}{\theta^2(1-\theta)^3} $$  This, however, does not give me the correct answer. The correct answer is  $$ \pi_J(\theta)\propto \frac{1}{\theta(1-\theta)^{1/2}} $$ which means that the Information I get should be   $$ I(\theta)=\frac{1}{\theta^2(1-\theta)} $$ since the prior should be proportional to the square root of the information.   Can anyone find any mistakes? I wouldn't be surprised if I screwed something up with the set up of the distribution (successes vs failures with their respective probabilities, etc).   I used the expected value from  and I know the correct answer from . Ubuntu Server VM: copy paste? I've installed Ubuntu Server under Virtualbox just to get a nice little shell. I've also installed the VirtualBox "Guest Additions" and set "Shared Clipboard: Bidirectional".  I still can't seem to get any copy-paste working. Am I missing something here?    I've installed  which gives mouse support in server mode (no X), and allows copy-paste using the mouse, but still the VirtualBox integration doesn't work. How do I run specific sudo commands without a password? On one particular machine I often need to run sudo commands every now and then. I am fine with entering password on sudo in most of the cases.  However there are three sudo commands I want to run without entering password:   sudo reboot sudo shutdown -r now sudo shutdown -P now   How can I exclude these commands from password protection to sudo? Force newlines with cat wildcard printing I want to use cat with wildcards in bash in order to print multiple small files (every file is one sentence) to standard output. However, the separate file contents are not separated by a newline, which I'd like to for ease of reading.  How can I add a file delimiter of some sort to this command? For a while now I've noticed that if I take a cup of hot tea and pour it into two cups and leave it then both cups will cool faster than the single cup.  How is it that nature cools two cups of tea in parallel more quickly? It's still the same amount of tea but it gets cooled quicker if I have two cups. Similar to , what's the best way to stay up to date with the latest stable releases of Thunderbird? The output from uname:  root@debian:~ # uname -a Linux 5asnb 2.6.32-5-amd64 #1 SMP Mon Jun 13 05:49:32 UTC 2011 x86_64 GNU/Linux   However the /sbin/init executable shows up as 32-bit:  root@debian:~ # file /sbin/init /sbin/init: ELF 32-bit LSB executable, Intel 80386, version 1 (SYSV), dynamically linked (uses shared libs), for GNU/Linux 2.6.18, stripped   Other aspects of the system seem to contradict things as well:  root@debian:~ # echo $HOSTTYPE i486  root@debian:~ # getconf LONG_BIT 32 When I try to remove Java with sudo apt-get remove jdk1.8.0-40 I get this:    Reading package lists... Done Building dependency tree        Reading state information... Done The following packages will be REMOVED:   jdk1.8.0-40 0 upgraded, 0 newly installed, 1 to remove and 0 not upgraded. 1 not fully installed or removed. After this operation, 245 MB disk space will be freed. Do you want to continue? [Y/n] y (Reading database ... 231488 files and directories currently installed.) Removing jdk1.8.0-40 (1.8.040-1) ... find: `/usr/java/*': No such file or directory /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 586: /usr/sbin/alternatives: No such file or directory /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 598: /usr/sbin/alternatives: No such file or directory dpkg: error processing package jdk1.8.0-40 (--remove):  subprocess installed post-removal script returned error exit status 127 Errors were encountered while processing:  jdk1.8.0-40 E: Sub-process /usr/bin/dpkg returned an error code (1)   Output of  ls -1 /var/cache/apt/archives/*jdk*:  openjdk-7-jre_7u75-2.5.4-1~trusty1_amd64.deb openjdk-7-jre-headless_7u75-2.5.4-1~trusty1_amd64.deb   Output of sudo dpkg --purge --force-all jdk1.8.0-40:  (Reading database ... 231488 files and directories currently installed.) removing jdk1.8.0-40 (1.8.040-1) ... /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 586: /usr/sbin/alternatives: No such file or directory /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 598: /usr/sbin/alternatives: No such file or directory dpkg: error processing package jdk1.8.0-40 (--purge): subprocess installed post-removal script returned error exit status 127 Errors were encountered while processing:  jdk1.8.0-40   When trying the edit I get this:  (Reading database ... 231488 files and directories currently installed.) Removing jdk1.8.0-40 (1.8.040-1) ... /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 586: /usr/sbin/alternatives: Permission denied /var/lib/dpkg/info/jdk1.8.0-40.postrm: line 598: /usr/sbin/alternatives: Permission denied dpkg: error processing package jdk1.8.0-40 (--remove):  subprocess installed post-removal script returned error exit status 126 Errors were encountered while processing:  jdk1.8.0-40 E: Sub-process /usr/bin/dpkg returned an error code (1) Intersecting overlapping polylines using ArcGIS for Desktop? I have 4 multipart polyline shapefiles (download a sample dataset ).   The geometries are identical, except for the length of segments in each shapefile. They represent 4 variables describing an identical shoreline. From these 4 overlapping polylines, I would like to create a resulting polyline that has segments for each unique set of the 4 input lines.    My approach was to use an intersect. This, however, breaks down when I try to run it on 4 huge feature classes.  My question is whether there is a more efficient way to combine polyline attributes stored in separate layers. I thought a union would be good, but it only works on polygons. Spatial Join seems inefficient, as I would end up with many to many joins for a single segment. Turning the segment end points into a point layer, then splitting the line at those points?  I have access to arcpy, PostGIS, and shapely.  Has anyone had experience doing this before on large datasets? Is there a more efficient way to store these variables?