Movie or tv show with robots disguised as childen who try to kill progatonist
Movie with killer androids (glowing eyes, screaming)
How do I use the GUI to search for files in GNOME? On the command line I regularly use locate, or more often a locate x | grep y pattern to find a file I know exists but can't find!  I'd like something that did this quickly in the GUI. Can't find it if it's there.  Here's my confusion   In the applications/overview I can see Search and Indexing. This seems to be a configuration tool. But if this is a server, I can't find a client. In applications I find Nepomuk File Indexing -- I think this is a KDE thing? Ditto, no client? I've installed tracker. This is a strange mix of powerful and unreliable/limited. e.g. it doesn't find files like locate does (returns no results where locate returns 96 results), but is will unpack every tag I've added to my photos! And you can't limit by folder for a quick search within a folder. The old Gnome 2 Search for files... thing is not in the Accessories menu anymore.   I feel that this is basic/core-level functionality and I can't help feeling I'm missing something obvious. How do I get the local machine name? VARCHAR size limit in Excel spreadsheets Something's been bugging me about SO for a while now, that is the inaccuracy of "length of time" strings.  The best example I can give at the moment is .  If you look at the length of time he's been registered, it says 1 year.  He joined the site September 2008, which is only 3 months off two years.  It's the same with answers - you can give an answer 50 seconds before anyone else, but at certain times page views will show both as "5 minutes ago" or whatever.  That's not quite as bad, but with answers being sorted "randomly by votes" by default you could lose out on votes or being marked as the correct answer to the slower answer even if they're both the same.  There's probably a better method amongst these, IMO:   "Vulgar fraction" unicode characters     Member for 1½ years   answered 5¼ minutes ago asked 3⅓ hours ago  Different units   Member for 21 months (at least it's better than "1 year") Answered 306 seconds ago (ugly, I know) Asked 64 minutes ago (also ugly)  Combining units   Member for 1 year 9 months (fits easily on the profile screen) Answered 5 minutes 6 seconds ago (probably too long) Asked 3 hours 20 minutes ago (also long)  Rounding up   I personally prefer option 1, unicode fraction characters.  "Member for 21 months" is pretty good too because it's easy to quickly work out.  "Asked 452 seconds" ago is horrible though.  EDIT: I'm completely aware of the timestamp hover because I am forced into using it so much. It's a great feature, but it's not really in your face so not everyone knows about it. I still think there's a better "glancing" solution than having to hover over the time string. Why is "cupboard" pronounced with a silent "p"? According to Google at least, the word "cupboard" originated in late Middle English as denoting a board that held cups. Since then, the word has evolved to mean a kind of cabinet.  My question is, given its origin and spelling, why do we pronounce "cupboard" with a silent "p"? Has the pronunciation simply evolved because "cup-board" is too awkward to say, or is there a deeper pronunciation rule that I'm not aware of? Guinier regime for form factor Why is it such a good idea to plot the logarithm of the form factor vs $Q^2$ in . It seems arbitrary to me. How can I make a PATCH HTTP Callout from Apex? I need to make a PATCH http request.  I made the rather incredible discovery today that I can't set an HttpRequest to use the PATCH method, as when sending the request I get a callout exception:     System.CalloutException "Invalid HTTP method: PATCH"   The work around I've found on Stack Exchange and elsewhere is only applicable to a Salesforce Rest API request:    This is therefore not helpful to me as I'm calling out to a third party.  By what manner of magic can I send a PATCH request from Salesforce?  The only work around I can think of is to not do it from Apex - this is not an option. Understanding a particular evaluation of $\prod\limits_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)$ I'm having a hard time understanding the following evaluation of the infinite product  $$ \prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right).$$ In particular, I don't understand how you go from line 2 to line 3. Here $\omega = -\frac{1}{2}+ i \frac{\sqrt{3}}{2}$, which is a primitive third root of unity.  $$\begin{align}\prod_{n=2}^{\infty}\left(1-\frac{1}{n^3}\right)  &amp;=\prod_{n=2}^{\infty}\frac{(n-1)(n^2+n+1)}{n^3} \\  &amp;=\lim_{m\to\infty}\frac{1}{m}\prod_{n=2}^m\frac{(n-\omega)(n-\omega^2)}{n^2} \\  &amp;=\lim_{m\to\infty}\frac{\Gamma(m+1-\omega)\Gamma(m+1-\omega^2)}{m(m!)^2\Gamma(-\omega)\Gamma(-\omega^2)(1-\omega)(1-\omega^2)(-\omega)(-\omega^2)} \\  &amp;=\frac{1}{3\Gamma(-\omega)\Gamma(-\omega^2)} \\ &amp;=\frac{\sin{\pi(-\omega)}}{3\pi} \\  &amp;=\frac{\cosh (\frac{\sqrt{3}\pi}{2})}{3\pi } \end{align}$$ EDIT: I think I'm starting to make sense out of this by writing out the terms. $$ \begin{align} \prod_{n=2}^{m} \frac{(n- \omega)(n-\omega^{2})}{n^{2}} &amp;= \frac{(2-\omega)(3-\omega) \cdots (m- \omega)(2-\omega^{2})(3-\omega^{2})\cdots (m- \omega^{2})}{2^{2} \cdot 3^{2} \cdot \cdots  \cdot m^{2}}  \\ &amp;= \frac{1}{(m!)^{2}} \frac{\Gamma(m+1-\omega)}{(1-\omega)(-\omega)\Gamma(-\omega)} \frac{\Gamma(m+1-\omega^{2})}{(1-\omega^{2})(-\omega^{2})\Gamma(-\omega^{2})} \end{align}$$ Now I have to figure out how to take the limit as $m \to \infty$. EntityFieldQuery display pager How do I display pager using EntityFieldQuery for fetch results? I can't find complete example.  I am doing simmilar to:   But I miss part to output simple pagination links (1,2,3... for every next page). What is the name of this castle part? There are lots (an infinitude) of smooth functions that coincide with $f(n)=n!$ on the integers. Is there a simple reason why Euler's Gamma function $\Gamma (z) = \int_0^\infty t^{z-1} e^{-t} dt$ is the "best"?  In particular, I'm looking for reasons that I can explain to first-year calculus students. Can I uninstall and reinstall Ubuntu Software Center? Since migrating to a true install and upgrading to 12.04 I have had several problems that seem to relate to the Ubuntu Software Center. Hanging at applying changes and blank screen on screen lock only when it is running, for instance.  My question is: Can I safely uninstall and reinstall Ubuntu Software Center and could this possibly fix those problems or are they just bugs I have to wait for updates for? If $u$ is harmonic and bounded in $0 < |z| < \rho$, show that the origin is a removable singularity This is a reworking of a previous question  which was marked as a duplicate. Some nice folks have referred me to solutions to similar problems. I still have a couple of questions, since one of the solutions used the Poisson kernel, which is not the desired solution in the text, because the Poisson kernel has not been introduced yet. The other solutions were for problems that were similar, but whose solutions were only partially applicable, as I understand them.  I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis:     If $u$ is harmonic and bounded in $0 &lt; |z| &lt; \rho$, show that the origin is a removable singularity in the sense that $u$ becomes harmonic in $|z|&lt;\rho$ when $u(0)$ is properly defined.   My Solution So Far:   Here's what I've got for a non-Poisson kernel solution: We want to find a harmonic conjugate for $u$ in the punctured disk, and then we can apply big Picard to show that $0$ is a removable singularity. The problem is that the punctured disc is not simply connected.   If we take $\sigma = (u \circ \exp)(z): L \to \mathbb{C}$, where $L$ is the left half-plane, now we have $\sigma$ harmonic, and defined in a simply connected set $L$, so it has a harmonic conjugate $\tau$. Now we wish to take $v = (\tau \circ \log):\{0 &lt; |z| &lt; \rho\}\to \mathbb{C}$, but in order to do so we need $\tau$ to be $2\pi i$-periodic. This is where my proof breaks down.  One of the similar problems which readers referred me to used the following strategy: note that since $\sigma$ is $2\pi i$-periodic, and $\sigma, \tau$ satisfy the Cauchy-Riemann equations, we have that $\nabla\tau(z) = \nabla\tau(z + 2 \pi i)$ for all $z\in L$, or in other words, $\tau(z) - \tau(z+2\pi i)$ is a constant function, say, equal to $\alpha \in \mathbb{R}$. Then, the solution suggests, we can set $$g(z) = u + iv - \frac{c}{2\pi}z,$$  in order to make $g$ $2\pi i$-periodic, and then we can use $(g\circ \log)$ on $\{0 &lt; |z| &lt; \rho\}$. But the reason why this won't work for my purposes is: now we have changed $u$! On the other hand, we can't simply subtract $\frac{c}{2\pi}\Im{(z)}$, and try to leave $u$ untouched, because now $g$ is not analytic. How can we get around this?  Another strategy entirely might be to try to use the mean-value property. Ahlfors discusses that a lot in the preceding section, and his main example is the punctured disc. Since $u$ is harmonic, we know that the average over circles centered at zero is a constant... Infinite distinct factorizations into irreducibles for an element Consider the factorization into irreducibles of $6$ in $\mathbb{Z}[\sqrt{-5}]$. We have $6=2 \times 3$ and  $6=(1+\sqrt{-5}) \times (1-\sqrt{-5})$, i.e. $2$ distinct factorizations. And,  $$6^2=3 \times 3\times2\times2$$ $$=(1+\sqrt{-5}) \times (1-\sqrt{-5}) \times (1+\sqrt{-5}) \times (1-\sqrt{-5})$$ $$=(1+\sqrt{-5}) \times (1-\sqrt{-5})\times3\times2.$$ More generally, $6^n$ will have $n+1$ distinct factorizations into irreducibles in $\mathbb{Z}[\sqrt{-5}]$ by a simple combinatorial argument. But, can we construct a ring in which there exists an element that has an infinite number of distinct factorizations into irreducibles? To make life harder, can we construct an extension of $\mathbb{Z}$ in which this happens? I have been thinking about this for a while and have managed to find no foothold.. Any help is appreciated. How does the momentum operator act on state kets? We are developing a C# application for a web-service client. This will run on Windows XP PC's.  One of the fields returned by the web service is a DateTime field. The server returns a field in GMT format i.e. with a "Z" at the end.  However, we found that .NET seems to do some kind of implicit conversion and the time was always 12 hours out.  The following code sample resolves this to some extent in that the 12 hour difference has gone but it makes no allowance for NZ daylight saving.  CultureInfo ci = new CultureInfo("en-NZ"); string date = "Web service date".ToString("R", ci); DateTime convertedDate = DateTime.Parse(date);               As per :     UTC/GMT Offset       Standard time zone: UTC/GMT +12 hours   Daylight saving time: +1 hour   Current time zone offset: UTC/GMT +13 hours     How do we adjust for the extra hour? Can this be done programmatically or is this some kind of setting on the PC's? Where Android apps store data? Could you list all possible directories where Android apps may store data, providing description what kind of data are stored in each directory? Evaluating the integral $\int_0^\infty \frac{x \sin rx }{a^2+x^2} dx$ using only real analysis Calculate the integral$$ \int_0^\infty \frac{x \sin rx }{a^2+x^2} dx=\frac{1}{2}\int_{-\infty}^\infty \frac{x \sin rx }{a^2+x^2} dx,\quad a,r \in \mathbb{R}. $$  Edit: I was able to solve the integral using complex analysis, and now I want to try and solve it using only real analysis techniques. Android backup/restore: how to backup an internal database? "How are" or "How is" the wife and kid?