Who named the Millennium Falcon?
Why is it called the Millennium Falcon?
How to show to the user a progression in a script? Homeomorphism is bijective Here is my question:  Is a homeomorphism, that is, a continuos function whose inverse function is also continuous, always bijective? NOTE: Signing up or Logging in via OpenID is no longer officially supported from any of our sites. Currently,  is still on the internet, but we are in the process of completely removing it.    Stack Overflow was an early and strong supporter of . We built our sign up/log in flow around it. We were idealistic and had , but these hopes weren't realized. Over the years people have wondered if  is . We've had to remove support as OpenID providers  or .  The time has come to part ways. The reality is OpenID support has created a ton of complexity in our codebase and the number of users actively using OpenID simply don't justify that cost. Users have spoken with their actions. You prefer Google, Facebook and Stack Exchange (aka email/PW) based account auth.     Of our 9,813,747 accounts around one-tenth of a percent are actively using OpenID (defined as having visited a site in the past 12 months). If you include all the inactive accounts it is still less than 2.9% of all accounts.  So, what is the plan?   DONE - Remove the ability to create a new account with most OpenID providers DONE - Add the ability to  DONE - Convert Launchpad (used on  ) to OAuth DONE - Remove creating new accounts with Steam from  (Steam removed support for OAuth. Only 454 accounts that use Steam have visited Arqade in the past 6 months.) DONE - We will email all users that rely on deprecated OpenID credentials and request that they  DONE - Run an announcement banner for any user who signs in with deprecated OpenID credentials urging them to  DONE - End support for logging in via OpenID  In progress - Remove  (note: only 0.2% of authorized sites in the last year have been non-Stack Exchange sites.)   Obviously, if your account relies on OpenID, then make life easy and go now and .   If you've forgotten what you credentials you use to login, then  to refresh your memory. If you don't see one of the three account types shown in the image below (Google, Facebook or Stack Exchange) then click the "add more logins" button to add one of them to your account.    Remember, SE OpenID accounts are really email/pw accounts handled through complicated OpenID code path. They will be migrated to a simpler email/pw only code path as a part of this change. No extra work required on your part. $n$ balls are thrown randomly into $k$ bins - how many are empty? A large number of variants of this question were already asked here, including these - , , which are close, but none seem to answer my question. Assume that $n$ balls are thrown randomly and independently into $k$ bins.  What is the probability of finding $x$ empty bins? What is the expectation of the number of empty bins? Got a bunch of .mp4 videos (info shows H.264 Video + AAC Audio) that are chapters of a larger video.  I'd like to merge them, but not sure which is the best way.  I don't have Quicktime Pro, but I'll get it if that's the best way to do this.  Edit: Just to clariify this is Mac OSX 10.6 I'm on. Changing Network Interfaces name Ubuntu 16.04 I tried to change network interfaces name on this new Ubuntu 16.04 LTS version but doesn't have the /etc/udev/rules.d/70-persistent-net.rules.  So, I tried to use /lib/udev/write_net_rules but it not exist.  Why do I need this modification? Because I'm using a tool to Simulate virtual platforms that use flex licensing and the authentication need to be in eth0 interface name.  Any suggestion?  The command ip link returns:  user@laptop:~$ ip link  1: lo: &lt;LOOPBACK,UP,LOWER_UP&gt; mtu 65536 qdisc noqueue state UNKNOWN mode DEFAULT group default qlen 1 link/loopback 00:00:00:00:00:00 brd 00:00:00:00:00:00 2: enp6s0: &lt;NO-CARRIER,BROADCAST,MULTICAST,UP&gt; mtu 1500 qdisc pfifo_fast state DOWN mode DEFAULT group default qlen 1000 link/ether b8:2a:xx:yy:xx:yy brd ff:ff:ff:ff:ff:ff 3: wlp7s0: &lt;BROADCAST,MULTICAST,UP,LOWER_UP&gt; mtu 1500 qdisc mq state UP mode DORMANT group default qlen 1000 link/ether 5c:e0:xx:yy:xx:yy brd ff:ff:ff:ff:ff:ff  user@laptop:~$ ifconfig  enp6s0    Link encap:Ethernet  HWaddr b8:2a:xx:yy:xx:yy   lo        Link encap:Local Loopback   wlp7s0    Link encap:Ethernet  HWaddr 5c:e0:xx:yy:xx:yy     (Some information were ignored and suppressed)  I already tried some links: In the season 8 episode finale of Game of Thrones the highest important lords met with Grey Worm to free      Tyrion and Jon. While it was pretty clear why they wanted to free Jon it was unclear to me why they also asked for Tyrion.   To make things even more strange, they asked about his opinion of a potential solution which becomes the actual final one in the end      (choose king and nominate someone).   I thought they (besides Sansa and Bran) don't like him and wanted to judge him? Or am I missing something? How to compute azimuth between QgsPointV2s? The  class has an azimuth() method but the new  class does not. What's the best way to compute the azimuth between QgsPointV2s? I couldn't find a convenient way to convert V2 to the old version, so my fallback is to construct a QgsPoint from the V2's x and y:   self.traj = QgsLineStringV2(...)     p_i = self.traj.pointN(self.i) # this returns a QgsPointV2 p_j = self.traj.pointN(self.j) a = QgsPoint(p_i.x(),p_i.y()).azimuth(QgsPoint(p_j.x(),p_j.y()))   Is there a better way? Find the value of $\left[\frac{1}{\sqrt 2}+\frac{1}{ \sqrt 3}+......+\frac{1}{\sqrt {1000}}\right]$ Find the value of $$\left[\frac{1}{\sqrt 2}+\frac{1}{\sqrt 3}+.....+\frac{1}{\sqrt {1000}}\right]$$.Where [•] denote the greatest integer function.   I am very confused about this problem. I tried to find the upper and lower bound of the function. But I can't find any formulas to find the bounds. Somebody please help me. Let $D$ be an integral domain and let $a^m=b^m$ and $a^n=b^n$ where $m$ and $n$ are relatively prime integers, $a,b \in D$.   How do I show $a=b$? Form within a block I'm using drupal_get_form('event_signup_form') to embed my custom form inside a block. The form is being rendered as intended, but submitting the form triggers nothing.  The form function itself is called event_signup_form($form, &amp;$form_state) and the submit function is called event_signup_form_submit($form, &amp;$form_state) but it's not being called.  I am using  to show the form:  function event_block_view($delta = '') { $blocks = array(); if ($delta ...) { $blocks['content'] = drupal_render(drupal_get_form('event_signup_form')); } }   I managed to get this working - here is how (stub):  function hook_block_view($delta = '') { $blocks['content'] = drupal_render(drupal_get_form('event_signup_form')); }  function event_signup_form() { // Register form submit handler, same goes for validation handler $form['#submit'][] = 'event_signup_form_submit';  return $form; }  function event_signup_form_submit($form, &amp;$form_state) { // This is now triggered } How to disable Spotify indicator in the top panel? The Spotify indicator icon clashes with the native Unity theme, and is redundant, since it is already listed in the Sound indicator menu. Is there a way to remove or disable the Spotify indicator? Or at least to change the icon (couldn't grep it on my filesystem). I am using the Spotify "preview" with 12.04. Thanks! I want to calculate $$ \int_{0}^{\infty} \frac{\arctan{2t}}{e^{2\pi t}+1} \, {\rm d}t $$ and am wondering if there at all is an analytical expression for the result. I haven't tried much so far, but expressing the denominator as the geometric series and then substituting $z=2\pi t (n+1)$. Equivalently I get $$ \sum_{n=0}^{\infty} \frac{{\rm si}\left(\pi(n+1)\right)}{2\pi (n+1)} $$ where $$ {\rm si}(x) = -\int_{x}^{\infty} \frac{\sin(t)}{t} \, {\rm d}t $$ is the shifted ${\rm Si}(x)$ function. The value is approximately $0.03618247146$, in fact I know from comparison with something else that it is exactly $\frac{\ln \pi -1}{4}$, but there should be some (direct) derivation, or?  Thanks for your input if you have any.  Edit: I tried to follow your instructions, but am now stuck. So I wrote $$ {\cal L}^{-1} \left(\frac{\psi(t)-\psi(1)}{t-1}\right)(s) =\sum_{n=0}^{\infty} \frac{e^{-sn}}{n+1} = e^s \int_s^\infty \frac{{\rm d}x}{e^{x}-1} $$ and taking the Laplace-transform \begin{align} \psi(t) &amp;= \psi(1) + (t-1)\int_0^\infty {\rm d}s \, e^{-s(t-1)} \int_s^\infty \frac{{\rm d}x}{e^{x}-1} \\ &amp;= \psi(1) + (t-1) \int_0^\infty {\rm d}x \int_{0}^{x} {\rm d}s \, \frac{e^{-s(t-1)}}{e^{x}-1} \\ &amp;= \psi(1) + \int_{0}^{\infty} {\rm d}x \, \frac{1-e^{-x(t-1)}}{e^x-1} \tag{1} \\ &amp;= -\frac{1}{t} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-tx}}{e^x-1} \tag{2} \\ &amp;= \dots \\ &amp;=-\sum_{k=0}^{n} \frac{1}{t+k} + \psi(1) + \int_0^\infty {\rm d}x \, \frac{1-e^{-(t+n)x}}{e^x-1} \\ &amp;\stackrel{(\rm{2})}{=} -\frac{1}{2t} + \ln(t) - \int_0^\infty {\rm d}x \, \left\{ \frac{1}{2} - \frac{1}{x} + \frac{1}{e^x-1}\right\} e^{-tx} \tag{Binet 1} \end{align} where $\psi(1)=-\gamma$.  Now we integrate over $t$ from $1$ to $y$ yielding $$ \ln \Gamma(y) \stackrel{({\rm 1})}{=} \psi(1)\,(y-1) + \int_0^\infty {\rm d}x \, \frac{y-1+\frac{e^{-x(y-1)}-1}{x}}{e^x-1} $$ but what now? Substitute? What is the difference between "I've been to China" and "I've gone to China"? Both are in the present perfect, but one uses the verb 'to go' and the other 'to be'. Is there a rule for this?  Is there any chance that the differences between "I have been" vs. "I have gone", are differences in English dialect (I don't know if this is the right word).  Edit Could I say that North American English speakers use "I have gone", and maybe U.K speakers say "I've been"? Or vice versa. When working with an image in Photoshop, I'd like to have a simple keyboard shortcut, like in , that automatically does:   Export > Export As or Export > Save For Web  Resize to biggest dimension = 600px (either width or height) Add suffix _600.jpg to filename   How to do that?  I'm not sure how to edit a recorded macro to do that (add the suffix and choose 600px for the biggest dimension).  Example: you open myphoto.png, you press F6 and then you automatically get myphoto_600.jpg. If I don't want to set any special behavior, is it OK if I don't bother to have a robots.txt file? Is errno thread-safe? How to count derangements? I was going through .Suppose that there are n persons who are numbered 1, 2, ..., n. Let there be n hats also numbered 1, 2, ..., n. We have to find the number of ways in which no one gets the hat having same number as his/her number. Let us assume that the first person takes hat i. There are n − 1 ways for the first person to make such a choice. There are now two possibilities, depending on whether or not person i takes hat 1 in return:  A) Person i does not take the hat 1. This case is equivalent to solving the problem with n − 1 persons and n − 1 hats: each of the remaining n − 1 people has precisely 1 forbidden choice from among the remaining n − 1 hats (i's forbidden choice is hat 1).  B) Person i takes the hat 1. Now the problem reduces to n − 2 persons and n − 2 hats.  From this, the following relation is derived: !n = (n - 1) (!(n-1) + !(n-2))  I have a hard time understanding point B and the equation derived. I was trying to understand with an example of 4 people named A, B, C and D and the corresponding hats A, B, C and D. Everyone of them are supposed to wear any hat except their own matching hat i.e. A can wear B, C or D.  If suppose A wears B hat so we will be left with only 3 people and 3 hats. So I can understand "(n-1)!(n-1)" part but what i didn't understand is part B(marked above). Can anyone help? What canonical momenta are the "right" ones? I'm doing some classical field theory exercises with the Lagrangian $$\mathscr{L} = -\frac{1}{4}F_{\mu \nu}F^{\mu \nu}$$ where $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. To find the conjugate momenta $\pi^\mu_{\ \ \ \nu} = \partial \mathscr{L} / \partial(\partial_\mu A^\nu)$, I can use two methods.  First method: directly apply this to $\mathscr{L}$. We get a a factor of $2$ since there are two $F$'s, and another factor of $2$ since each $F$ contains two $\partial_\mu A_\nu$ terms, giving $$\pi^\mu_{\ \ \ \nu} = -F^\mu_{\ \ \ \nu}.$$  Second method: get $\mathscr{L}$ in terms of $A$ by expanding and integrating by parts, yielding $$\mathscr{L} = \frac{1}{2}(\partial_\mu A^\mu)^2 - \frac{1}{2}(\partial_\mu A^\nu)^2.$$ Differentiating this gets factors of $2$ and gives $$\pi^\mu_{\ \ \ \nu} = \partial_\rho A^\rho \delta^\mu_\nu - \partial^\mu A_\nu.$$  These two answers are different! (They do give the same equations of motion, at least.) I guess that means doing the integration by parts changed the canonical momenta.  Is this something I should be worried about? In particular, I have another exercise that wants me to show that one of the canonical momenta vanishes -- this isn't true for the ones I get from the second method! Plus, my stress-energy tensor is changed too. When a problem asks for "the" canonical momenta, am I forbidden from integrating by parts? Similes and Metaphors - are similes a subset of metaphors?