I have a formula field. I need to display a certain value depending on the local time of the day
Email Templates: Conditional Greeting Based on Time of Day
Probability of Obtaining the Roots in a Quadratic Equation by Throwing a Die Three Times I want to know if there is a way to switch to console mode from the boot menu.  I have an NVIDIA 1070 and every time there is an update, the GUI stops working. All I need to do is re-install the drivers; however, to do that I need to be in console mode. EDIT: I should have added more information to this question but I'll comment on all answers. In short: I've tried text instead of quiet splash I do see booting activity in text mode; however, I don't get a login screen and ctrl + Alt + F1 doesn't help because GDM keeps trying to restart. Previously I managed to kill GDM through multiple tries but is not working since the latest update. EDIT: After searching for runlevel, I've found  but I think it required me to have a running system to make changes -- please correct me if I'm wrong. Kronecker delta confusion I'm confused about the Kronecker delta. In the context of four-dimensional spacetime, multiplying the metric tensor by its inverse, I've seen (where the upstairs and downstairs indices are the same):  $$g^{\mu\nu}g_{\mu\nu}=\delta_{\nu}^{\nu}=\delta_{0}^{0}+\delta_{1}^{1}+\delta_{2}^{2}+\delta_{3}^{3}=1+1+1+1=4.$$  But I've also seen (where the upstairs and downstairs indices are different):  $$g^{\mu\nu}g_{\nu\lambda}=\delta_{\lambda}^{\mu}=\left(\begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 1 \end{array}\right).$$  How can there be two different answers to (what appears to me to be) the same operation, ie multiplying the metric tensor by its inverse? Apologies if I've got this completely wrong. Java OS X Lion Set application name doesn't work Running Yosemite 10.0 with a clean erase and install, when I reboot, all my launchpad apps are totally disorganised, folders and un-official apps disappears...  After reorganizing the layout twice, I don't want to keep sorting my launchpad every time I restart my Mac. How to add a loopback interface? I want to add two more loopback interfaces.  I used the following command:  ifconfig lo: 127.0.0.2 netmask 255.0.0.0 up   and type in ifconfig, i can see the loopback address was added.  But if i want to add one more interface... (for example 127.0.0.3) the previous interface (127.0.0.2) was overwritten.  Also when i look in /etc/network/interfaces i see no entry.  How can i add multiple loopback interfaces permanently? Which notation and why: $\text{P}()$, $\Pr()$, $\text{Prob}()$, or $\mathbb{P}()$ Calculation of $\int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx$, where $n\in \mathbb{N}$ Compute the definite integral      $$ \int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx $$      where $n\in \mathbb{N}$.   My Attempt:  Using $\cos (x) = \frac{e^{ix}+e^{-ix}}{2}$, we get  $$ \begin{align} \int^{\pi/2}_{0}\cos^{n}(x)\cos (nx)\,dx&amp;=\int_{0}^{\pi/2} \left(\frac{e^{ix}+e^{-ix}}{2}\right)^n\left(\frac{e^{inx}+e^{-inx}}{2}\right)\,dx\\ &amp;= \frac{1}{2^n}\mathrm{Re}\left\{\int_{0}^{\pi/2} \left(e^{ix}+e^{-ix}\right)^n\cdot e^{inx}\,dx\right\}\\ &amp;=\frac{1}{2^n}\mathrm{Re}\left\{\int_{0}^{\pi/2}\left(e^{2ix}+1\right)^n\,dx\right\} \end{align} $$  Letting $z=e^{4ix}$ gives us  $$ \begin{align} e^{4ix}dx &amp;= \frac{1}{4i}dz\\ dx &amp;= \frac{dz}{4iz} \end{align} $$  So the integral becomes  $$\frac{1}{4\cdot 2^n}\mathrm{Re}\left\{\int_{C}\left(\sqrt{z}+1\right)\cdot \frac{dz}{iz}\right\}$$  How can I complete the solution from here? What is the dimension of $R[[X]]$ where $R$ is a Noetherian ring? If $R$ is a Noetherian ring then we know that the Krull dimension of the polynomial ring $R[X]$ is $\rm dim(R)+1.$ Is there any formula for the Krull dimension of the power series ring $R[[X]] $ When $R$ is a Noetherian ring ? A childish question (literally) -    My 8 year old asked me this morning: "Dad, what does 'physical' mean?" - and I found myself at loss for an ordinary language answer.   Every answer I could come up with ended up sounding like a definition of empiricism: "physical things are those things that can be observed", "physical means something that can be measured", "whatever can be verified and tested", etc....   Claiming that things that are physical are those things which are material or things that are described by physics seems circular.   Google:     phys·i·cal   ˈfizik(ə)l/   adjective   1.   relating to the body as opposed to the mind.   "a whole range of physical and mental challenges"   synonyms:   bodily, corporeal, corporal, somatic; More   2.   relating to things perceived through the senses as opposed to the mind; tangible or concrete.   "pleasant physical environments"   synonyms:   material, concrete, tangible, palpable, solid, substantial, real, actual, visible   "the physical universe"   And various dictionaries (here Merriam-Webster):      Definition of physical   1   a :  of or relating to natural science   b (1) :  of or relating to physics (2) :  characterized or produced by the forces and operations of physics   2   a :  having material existence :  perceptible especially through the senses and subject to the laws of nature   everything physical is measurable by weight, motion, and resistance — Thomas De Quincey   b :  of or relating to material things     lead to similar definitions.   How can one define "physical" in ordinary language terms? Is there a way of defining "physical" that doesn't presuppose empiricism? What is the rationalist definition of 'physical'?   Is there a definition that doesn't commit to either dualism or empiricism or to any epistemology? I got a integrated GD, which is Intel, and another discrete GD, AMD Radeon 6490 HD I am using ubuntu 11.10  1st, when I boot into Windows, my GD is indeed AMD Radeon 6490 HD, but when I boot into Ubuntu, it shows that I have AMD Radeon 6470 instead.  2nd, I try most methods only to install driver for my AMD/ATI graphic card. First by jockey(fail, boot with unity 2D instead) and install manually (fail, boot into blank screen).   3rd, I give up using the AMD/ATI graphic card. Then I got a new problem. Both my graphic cards are powered on on startup, which causes my laptop to over heat.   I view the content of /sys/kernel/debug/vgaswitcheroo/switch  0:DIS: :Pwr:0000:01:00.0 1:IGD:+:Pwr:0000:00:02.0   Then I switch off IGD using echo OFF &gt; /sys/kernel/debug/vgaswitcheroo/switch and I get  0:DIS: :Off:0000:01:00.0 1:IGD:+:Pwr:0000:00:02.0   But after I reboot, the discrete graphic card is powered on again. How to solve this? I want to simplify the following:  $\sum_{n=0}^x 1 - (1 - I[a \ne 0] * I[a \ge b]) * (1 - I[b \ne 0] * I[b \ge c]) * (1 - I[c \ne 0] * I[c \ge d]) * (1 - I[d = 0])$ If you search for something on "all sites" (using the searchbox at the top of , you'll reach the search results, e.g.  .  If you click directly on a tag on that page, e.g.        You get 404 Page Not Found error, e.g.     Steps to reproduce   Go to  and enter e.g. "keyforge" into the "Search all sites" field. Arrive at  Click on the "keyforge" tag below a search entry. Reach  (404 page not found)   Expected behaviour  Reach   System Info  Happens in both Firefox and Vivaldi browsers on Windows 10. Does decrease in temperature affect mass $E=mc^2$? My understanding of Quantum physics and String Theory is very basic and I don't yet have a grasp on the maths, but in my research I have come up with a question.  Does a decrease in temperature also signify/create a decrease in mass?  I am quite willing to believe the concepts I think I am beginning to understand are totally misconceptions on my part.  But, as I see it mass is a product/mark of disruption to the Higgs field as a particle moves.  If these particles are ultimately composed of vibrating Strings, their mass is given by the rate at which they vibrate and how it interacts with the Higgs field.  I don't have any idea if, as the temperature (as we know it) approaches 0k, the string/quark/sparticle would change its state in any way or if the changing of the state of the String would change what type of particle it represents.  It seems I'm more full of questions than answers, but that's why I love these topics!  I know what I'm learning is new if the best and brightest are still working on the 'basics' to just make their systems work :). Why does the command shuf file > file leave an empty file, but similar commands do not? I know this is sort of a duplicate of another question () but I wanted to expand on the question in response to the answers given.  The command  shuf example.txt &gt; example.txt  Returns a blank file, because the shell truncates the file before shuffling it, leaving only a blank file to shuffle. However,  cat example.txt | shuf &gt; example.txt  will produce a shuffled file as expected.  Why does the pipeline method work when the simple redirection doesn't? If the file is truncated before the commands are run, shouldn't the second method also leave an empty file? Disk space full but logical space available in database We have a fairly big MS SQL 2008R2 database that resides on a SSD drive. The drive itself only has ~110Gb of space, and the database files are the only files on the drive.  The database is in "Simple" recovery mode, and only has two files, .MDF and .LDF.   The disk is now nearly full: the MDF is currently 109Gb in size. However, SSMS tells me there's nearly 18Gb of "Space Available" (in the 'General' properties page), and if I go through the motions of Shrinking the file it also tells me there's 18Gb of space free. SSMS also tells me the database size is around 132Gb, which surprises me - that wouldn't fit on the drive!  From what I've read, shrink is a really bad idea. However, I'm starting to see replication errors (could not allocate space for object). We've previously tried shrinking the database, but within a few hours the file was back to it's original size.  How should we proceed - given there's apparently 18Gb of free space, should SQL automatically use that free space? Or is it as simple as: we really need more disk space? If the electric field is the gradient of the potential, then can we say that whenever potential is zero, the electric field is zero? For example, in a dipole, at the center of the two charges making up the dipole, the potential is zero but the electric field is non-zero. But if $E = -\operatorname{grad}V$, then why is $E$ not zero?  Another example: In the case of two positive charges, the electric field is zero at the mid-point of line joining them, but potential is non-zero. What am I misunderstanding here? I have heard a few times that one way of describing quantum computers is that they essentially use the computing power of their counterparts in alternate realities that they access through superposition. My first question is, of course,    is this actually an accurate description of how quantum computers work, or just a misrepresentation?   Also, if it were to be assumed true and taken literally, then presumably all the possible outcomes of any given computation done would be experienced by each of these alternate realities. I have a few questions regarding the implications of this:   Would it be correct to say that our universe essentially "spawns" these alternate realities to use for each computation, or that they all exist simultaneously in a higher dimension and only come into contact via the initiation of superposition by each computer at the same time? Would it be possible to create contact between universes via this connection? I'd expect that, if it was, it would be entirely random, and so not only would not cause any information transfer, it also would be completely undetectable, because the deviations would be within the expected probabilities of error rates - but is there any way to circumvent this? Can ordered list produce result that looks like 1.1, 1.2, 1.3 (instead of just 1, 2, 3, ...) with css? Series as an integral (sophomore's dream) I need help with this exercise.  I need to prove $$\int_{0}^{1}x^{-x}\ dx=\sum_{n=1}^{\infty}n^{-n}$$  I think I should use some convergence theorem, but I'm stuck.  Thanks a lot!