Just happened to catch S1E3 yesterday (Parallax), and noticed that Paris refers to Voyager as the Voyager. Whenever we hear about other ships in Star Trek, they are always prefaced by "the" - "the Enterprise", "the Defiant", even "the Equinox" on Voyager as I recall. At some point in Voyager, the "the" was dropped - is there a reason why? It sounds awkward to me, but I usually rewatch the last few seasons of Voyager, which might bias me. Given that it seems the writers started out with using the definite article and later dropped it, I'm curious if the showrunners ever addressed this, as I'm sure people noticed the shift.
Some ships in Star Trek are referred to with definite articles. "The Enterprise", "The Defiant". Other ships are referred to without them. "Voyager", "Enterprise". Is there a pattern to or explanation for this behavior?
I saw a video where the person copied a material from one object to another (Cycles).  How can I accomplish this? At what age is the Force detectable in children? What are some practices for getting a name change so that people can find me more easily? For a similar thing, , where Jay Wacker managed to get people to call him by the name Jay even though he didn't need to get a legal name change. I'm not sure how to go about this though.  The main problem with me is that I have such a frustratingly common name that a lot of people cannot find me when they google me. So many of them simply don't notice the middle initial that I always use in between my first and last names, and this could actually become a major issue in academia, since people have little time and are prone to giving up quickly if they want to look for me (or for my papers) and can't find me at all (I know this having seen how several academics use the Internet and how they look for people's names). This, in turn, could easily ruin my citation count in the future (it's not just that - it's helpful to others when I have a less common name so that they can more easily locate my stuff). I already know at least several people who specifically told me that they tried to find my email but that they couldn't find it (and this isn't limited to just them - there are many, many more - including long-lost friends who have wanted to talk to me for a long time, but who couldn't find me due to said ultra-common name). Of course, people can go through the respective university directories, but how many people really do that? From my observations (when I've seen people look for someone else), very very few do it. Hell, there are even several people at my OWN university who share the same exact first+last name as me.  In Academia, this is even a bigger problem because the vast majority of your connections will be people who only vaguely recognize/know you, so they may know most of the search clues. Even a "full name" + university won't solve all the problems, because I may switch universities and people may only remember the old university that I was in. I'm also very very interdisciplinary, so I want to be searchable to people outside my field as well.  And even if I fix the issue for Google with a massive SEO operation or whatever (that may even be impossible for my ridiculously common name), it's still not going to fix the problem for all of the other ways that people use search.  I'm currently transitioning between undergrad and grad school, so now may be the perfect time for a name change? But I don't know what to do. Is it better for me to change my first name or my last name? The problem is that a citation like "Chen 2011" is going to produce so many entries that no one will ever find them, even though they frequently do google things like that (and I simply cannot prevent people from googling something like that). Chen is so frustratingly common that even a "Chen and Name2 2011" paper could come from some random medical paper rather than from something I wrote.  As an additional complication, the % of Chinese people using the English Internet is exponentially rising, and I can only expect the problem to get worse in the future because of that (and not just for my first name, but even variants of my first name too). Prove that $\frac{1}{\sin^2 z } = \sum\limits_{n= -\infty} ^ {+\infty} \frac{1}{(z-\pi n)^2} $ On the commutative property of multiplication (domain of integers, possibly reals) $ab = ba$  This is, inherently, true. Some texts drop it like an axiom without any justification. But I'm a bit curious where it stems from or basically why/how it works. If anyone could enlighten me a bit further, I'd be most grateful. Peano axioms? From set theory? Help.  What bugs me with this definition is it's real world "application":  Let's say I have a power output of 5 W ( $kg\cdot m^2 \cdot s^{-3}$) and I want 30% of that power output (to have some units and natural context).  Naturally, 30% is 30 1/100 (% is the unit). For simplicity, let's express that as:  $30$%$ = 30/100 = 3/10 = 3d$ where $d = 1/10$, a simpler unit (because %W would look... Wrong?).  So 30% of power is then $5 W \cdot 3d$, and we can force it to "make sense" if we associate d with W or basically scale the unit of power by d (which is intuitively understood as being 10 times smaller then the output of 1 W) $dW$. And we want to scale these 5 units of $dW$ three times.  $5 dW \cdot 3$  But the commutative property says it is the same as scaling 3 units of $dW$ 5 times or:  $3 dW \cdot 5$  And the result is the same. This is the bit that hurts my head, the fact it is the same. I try to interpret the same as, for example 50 J of work, it's either applying 50 N over 1 m of distance or 1 N over 50 m of distance.    Down here is additional stuff I think I have (you don't have to read it if you know how to answer immediately). I don't claim it's correct, if anyone parses this information, feel free to correct me.  $m\sum\limits_{i=1}^{n}{1} = n\sum\limits_{i=1}^{m}{1}$  This is how I've been trying to break it down, as it is repeated addition. It is repeated $n$ times and multiplied by $m$, and reverse on the right side. There is $x = m - n$, so when I multiply $n$ $m$ times, each repeated addition of $n$ lacks exactly $x$ to $m$.  $mn = nm$  $x = m-n$  $m(m-x) = (m-x)m$  $m^2 - mx = m^2 - mx$  $ true $  And yes, sadly, I realise that I am using the very property I'm trying to prove ($xm$ shifted as $mx$ on the right-hand side.)  And it even calls in the distributive property of multiplication over addition which I derive (for myself, informally) from the very nature of multiplication and the way we "process" numbers:  For example, number $55$ times $2$  $55 = 5 \cdot 10^1 + 5 \cdot 10^0 = 5 \cdot 10 + 5 \cdot 1$  $(5 \cdot 10 + 5 \cdot 1) \cdot 2$  This is the basis of our positional notation which appends digits of varying orders of magnitude or units together in a way it makes "sense". Each one is b times bigger than the one to the right. Basic stuff. It is why I expressed 10 and 1 explicitly, those are what I consider units in this case.  Now, from the definition of multiplication, which is at its heart just repeated addition, it is truly the same if you add together $2$ $50$ times and then again $5$ times or "all at once" (figuratively, we are always doing the former mentally) $2$ $55$ times.  That added together gives 110, which is true.  To me, the distributive property (in case of integer multiplication) is a repercussion of the very definition of multiplication which is repeated addition (evading strict concepts). I am not certain whether this is a good way to look at it. I have just downloaded "ubuntu-14.04.1-desktop-amd64.iso" and "ubuntu-14.04.1-server-amd64.iso" through "".   Please can someone confirm the maximum RAM that is supported.  According to many sites, Maximum RAM supported by the OS  ubuntu-14.04.1-desktop : " 2^64 bytes " -- Kindly confirm  ubuntu-14.04.1-server : " 2^64 bytes " -- Kindly confirm Trouble understanding the Bohr model of the atom Let $(X, Y)$ be distributed as a multivariate normal with parameters  $$ \mu = \begin{bmatrix} \mu_X \\ \mu_Y \end{bmatrix} \qquad \Sigma = \begin{bmatrix} \sigma_X^2 &amp; \sigma_{XY} \\ \sigma_{XY} &amp; \sigma_Y^2 \end{bmatrix}. $$   I would like to calculate $E(X | y_1 &lt; Y &lt; y_2)$, where $y_1$ and $y_2$ are constants.  From , I have managed to work out that $$ E(X | Y &gt; y_1) = \mu_X + \frac{\sigma_{XY}}{\sigma_Y} \left[\frac{\phi\left(\frac{y_1 - \mu_y}{\sigma_Y}\right)}{1 - \Phi\left(\frac{y_1 - \mu_y}{\sigma_Y}\right)}\right]\\ E(X | Y &lt; y_2) = \mu_X - \frac{\sigma_{XY}}{\sigma_Y} \left[\frac{\phi\left(\frac{y_2 - \mu_y}{\sigma_Y}\right)}{\Phi\left(\frac{y_2 - \mu_y}{\sigma_Y}\right)}\right] ,$$ where $\phi(\cdot)$ and $\Phi(\cdot)$ are, respectively, the p.d.f. and the c.d.f. of the Normal distribution. I could not figure out how to calculate $E(X | y_1 &lt; Y &lt; y_2)$, though.  I have searched for an answer in similar posts from the Stack Exchange network, but I couldn't get a clue from them that would solve this issue. For the record, here are some of them: Can I still command/control concentration spells while in the Border Ethereal? If I have cast something like Animated Objects or Bigby's Hand and then follow up the next round with casting Etherealness, can I still command/control those spells? Can a non-EU citizen travel freely within the Schengen area without passport? My current situation is that I have residence permit of Germany (as a stamp in passport) and I wish to apply for UK visa in next few days from Germany and won't have my passport anymore during the processing time.  Instead of waiting for the process, I wish to travel to Scandinavian countries immediately on holidays, later receive my passport from visa center after I return back to Germany.   I am wondering if photocopy of passport and residence permit suffice as identity proof to border control (if I face them)? What else are valid documents that I can show as valid identity proof to border control? What are the exact problems I will face if I face the border control, and they are not satisfied with my travel documents? show that if $a | c$ and $b | c$, then $ab | c$ when $a$ is coprime to $b$. Given two numbers $a$ and $b$, where $a$ is co-prime to $b$,     Show that for any number $c$, if $a|c$ and $b | c$ then $ab| c$. Is the reverse also true? In other words, if $ab |c$ and $a$ is co-prime to $b$, then do we have $a | c$ as well as $b|c$? Find expected value of $\frac{1}{1+\xi}$ Find expected value of $\frac{1}{1+\xi}$ if $\xi$ has Poisson distribution.  My try:  $$E\left[ \frac{1}{1+\xi}\right]=\sum_{k=0}^{\infty}{x}_{k}{p}_{k}=\sum_{k=0}^{\infty}\frac{1}{k+1}\frac{\lambda^ke^{-\lambda}}{k!}$$  Is it right? I am running a web site design service for local small business owner. I want avoid legal issues on the images I use. Usually, the images found on Flickr are forbidden for commercial use. Where would be a normal channel to get pictures for commercial use with minimal cost? How can I validate that the picture I used has no legal risk? Polynomial ring $R[X] $ has zero Jacobson radical, for $R$ a domain How to setup Windows from an external hard drive? Is it possible to install Windows from an external hard drive?  I have Windows XP installed on my laptop, and I want to remove it and install Windows 7.  My Windows 7 installation is in an .iso format on an external USB hard drive.  Is it possible to install it or will I need to put in a DVD before I start installing? Installing QGIS on Ubuntu 14.04? Following the QGIS installation guide I have tried both UbuntuGIS and QGIS stable, but both fail with the error message:  The following packages have unmet dependencies:  python-qgis : Depends: python-qgis-common (= 2.6.1.1+trusty1) but it is not going to be installed             Depends: libqgispython2.6.1 but it is not going to be installed            Depends: libqgis-analysis2.6.1 but it is not going to be installed            Depends: libqgis-core2.6.1 but it is not going to be installed            Depends: libqgis-gui2.6.1 but it is not going to be installed            Depends: libqgis-networkanalysis2.6.1 but it is not going to be installed   qgis : Depends: libgdal1h (>= 1.8.0) but it is not going to be installed      Depends: libqgis-analysis2.6.1 but it is not going to be installed     Depends: libqgis-core2.6.1 but it is not going to be installed     Depends: libqgis-gui2.6.1 but it is not going to be installed     Depends: libqgis-networkanalysis2.6.1 but it is not going to be installed     Depends: qgis-providers (= 2.6.1.1+trusty1) but it is not going to be installed     Recommends: qgis-plugin-grass but it is not going to be installed     Recommends: qgis-plugin-globe but it is not going to be installed   E: Unable to correct problems, you have held broken packages.    I had QGISv.2.4 running on my machine with Ubuntu 12.04. After upgrading to Ubuntu 14.04 I was getting error message that pyqgis could not be loaded.  So I uninstalled QGIS using the Ubuntu Software Center and also run  sudo apt-get purge qgis   After that I tried to install the latest version of QGIS Debian Stable and UbuntuGIS unstable but getting the error messages described above.  The weird thing is that neither Ubuntu Software Center nor apt-get list have QGIS listed as installed, but if I run a search for qgis on my file system qgis and all its dependencies are still there. That's why I getting told to have broken packages. The Medibuntu Project has come to an end - what do I do now? Assigning strings to arrays of characters I am a little surprised by the following.  Example 1:  char s[100] = "abcd"; // declare and initialize - WORKS   Example 2:  char s[100]; // declare s = "hello"; // initalize - DOESN'T WORK ('lvalue required' error)   I'm wondering why the second approach doesn't work. It seems natural that it should (it works with other data types)? Could someone explain me the logic behind this? What does unwrapping a model do and why is it important? Rename all the files within a folder with prefix “Unix_” Rename all the files within a folder with prefix “Unix_” i.e. suppose a folder has two files a.txt and b.pdf than they both should be renamed from a single command to Unix_a.txt and Unix_b.pdf