Does the big bang theory state that space and time originated after the bang? So, was the singularity was converted into the vast space plus time?
Did spacetime start with the Big Bang?  I mean, was there any presence of this spacetime we are experiencing now before big bang? And could there be a presence/existence of any other space-time before the big bang?
A proof using Yoneda lemma If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$. If $x,y,z\in\mathbb R\setminus \{1\}$ and $xyz=1$, prove that $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 1$$  Without using calculus. There are a few ways I've tried solving this:  $1)$ We could try using the Cauchy-Schwarz inequality: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge \frac{(x+y+z)^2}{(x-1)^2+(y-1)^2+(z-1)^2}$$  But it's apparent that nothing's useful here.  $2)$ We could use AM-GM as well: $$\frac{x^2}{(x-1)^2}+\frac{y^2}{(y-1)^2}+\frac{z^2}{(z-1)^2}\ge 3\sqrt[3]{\frac{x^2y^2z^2}{(x-1)^2(y-1)^2(z-1)^2}}=\frac{3}{\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}}$$  So we only have to prove that: $$\sqrt[3]{(x-1)^2(y-1)^2(z-1)^2}\le 3$$  We could raise both sides to the power of 3: $$(x-1)^2(y-1)^2(z-1)^2\le 27$$  But this inequality doesn't hold.   $3)$ We could try cleaning the denominators by multiplying both sides by $(x-1)^2(y-1)^2(z-1)^2$. After a bunch of expanding and simplifying we get that: $$x^2y^2+y^2z^2+x^2z^2-6(xy+yz+xz)+2(x+y+z)+9\ge 0$$  I can't tell so easily whether the inequality is true or not. You could help me out on this one. Just don't forget that $x,y,z\in\mathbb R\setminus\{1\}$ and so we can't just simply use AM-GM, unless we're using it for squares that have to be non-negative. Photoshop To CSS: How do you translate Fill to CSS I have a content box in photoshop with the background color #bc0018 with a 65% opacity and 50% fill.  I am able to translate the backgroundcolor and opacity part to this, but without the 50% fill the backgroundcolor looks different.  .content-box {     background-color: rgba(188, 0, 24, 0.65); }   Any suggestions on also getting the proper "fill" in CSS?  EDIT  I did just divide the opacity by half and it appears to achieve the same effect as a 50% fill. Still, I would love to hear how other people do this? I definitely don't want to create an image just for this.  .content-box {     background-color: rgba(188, 0, 24, 0.325); }   Thanks! Custom blocks created as a plugin in a module not showing for Block::load() I have a couple of blocks that have been created using the ... extends BlockBase syntax in my module. They work perfectly fine, and I can add them on the blocks layout page, can add them into DS as a field, etc. They definitely exist fine.  However, when I try and render them using Block::load(), they won't show up, unlike other blocks that work fine.  For example, if I have the following docblock:  /**  * Provides a 'Cute Cats' Block.  *  * @Block(  *   id = "cute_cats",  *   admin_label = @Translation("Cute cats yo"),  * )  */  class CuteCats extends BlockBase { ... }   ... and then try and call it like this:  $block = \Drupal\block\Entity\Block::load('cute_cats'); return render(\Drupal::entityTypeManager()   -&gt;getViewBuilder('block')   -&gt;view($block));   Any idea why? Thanks in advance. Unique (non-repeating) random numbers in O(1)? This is something I'm supposed to be able to prove for an upcoming test, but I can't find anything to help me prove this in my notes or the chapter, which is on cosets and Lagrange's theorem. If all I start with is the group having no proper subsets, then that means any subset is the whole group. By Lagrange's theorem, that means the index is $1$, but that doesn't get me anywhere. Where do I start? Radial Schrodinger equation with inverse power law potential Recently I read a  about solving radial Schrodinger equation with inverse power law potential.  Consider the radial Schrodinger equation(simply set $\mu=\hbar=1$):  $$\left(-\frac{1}{2}\Delta+V(\mathbf{r})\right)\psi(\mathbf{r})=E\psi(\mathbf{r}).$$  A  gives a one-dimension equation:  $$-\frac{1}{2}D^2\phi(r)+\left(V(r)+\frac{1}{2}\frac{l(l+1)}{r^2}\right)\phi(r)=E\phi(r),$$  where $D=\dfrac{d}{dr}$, and $l$ is the azimuthal quantum number.  If we only consider the ground state, then $l=0$, so  $$-\frac{1}{2}D^2 \phi(r)+V(r)\phi(r)=E\phi(r).$$  We want to find eigenvalue $E$ such that $\phi(0)=\phi(+\infty)=0$.  The central potential discussed in the paper is of this form:  $$V(r)=\alpha r^{-\beta}.$$  It states (see page 4) that if $\beta&gt;2$ then the potential is repulsive (i.e. $\alpha&gt;0$).  My questions are:    Is this conclusion (i.e. If $\beta&gt;2$ then we must have $\alpha&gt;0$) generally valid in physics? What would happen if $\alpha&lt;0$ and $\beta&gt;2$? Is there 'ground state' in this condition? What about the condition $\alpha&gt;0$ and $\beta=1,2$? I have tried to solve the equation numerically with $\alpha=1,\beta=1,2$, and the ground state energy in this two conditions seem to be $0$, is my result correct?   P.S. When I try to find ground state when  $\alpha=-1,\beta=2$, the energy seem to be $-\infty$, which is qualitatively different from $\alpha=-1,\beta=1$. Given the following sentences:        I have even started using them in normal writing.   People can understand your writing better.      Are using and writing gerunds, verbs, or some other part of speech?  Context:  I am prefixing the words in some phrases with abbreviations. Some of the words are giving me trouble in classification. (The message is supposed to be an implicit proof of why adding prefixes to table names in a database is terrible. But I'd like to get it right.)  Here's the full set of sentences for your amusement. However, please restrict your comments to the stated question, for the most part.        com-Don't ver-Listen prep-To adj-Those adj-Other nou-People.    pro-You aux-Should adv-Always ver-Use nou-Prefixes prep-With pro-Your adj-Table nou-Names.    pro-I aux-Have adv-Even ver-Started ver-Using pro-Them prep-In adj-Normal nou-Writing.   com-See adv-How adj-Effective pr-It ver-Is?    nou-People aux-Can ver-Understand pro-Your nou-Writing adv-Better!      (where com means command verb)  This question is part of 3 related questions:     This question   It originally came from Cardinality of cartesian product of an infinite set with N Prove: $\gcd(a,b) = \gcd(a, b + at)$. Connect iPhone to Mac Without Charging I am developing iOS apps and I hate, that always when I connect my iPhone to my Mac to install and test an app, the iPhone gets charged. Is there any possibility to connect my iPhone to my Mac without charging the phone?    My Environment: -  iPhone 4S (No Jailbreak) -  MacBook Air3,2 (Late 2010) Ok. So I read books as a kid, this may have been mid to late 1980s. Definitely before 1993. These details I remember, they might be from the same book or from several:   Slower than light space travel with severe time dilation. There were boardings of space ships, but the fighting odds could be very uneven because a foe could be from your future or your past, depending on when you started your space journey, because of relativistic time dilation. No faster than light travel, no time travel either. Just regular physics. To fight better in melee, soldiers would have (more or less willingly, I think they knew beforehand) false memories or notions implanted of the enemy as horrible human beings so soldiers could kill them more easily without remorse. Towards the end of the story, veterans of the war are described as not quite feeling at home after the war. One planet (home?) or country is described as inhabited entirely of clones, Adams and Eves, I think homosexual, or that sexuality could be switched deliberately. I think the veterans find at the end a refuge somewhere more like the home they are used to. The reason so much has changed is of course the time dilation. They spent maybe decades in the war, while on the planets centuries or millenia passed. I think one moral (there may be many more I missed as a child) of the story was the futility of war. If I recall correctly, the war was resolved in a way that made the efforts of the veterans seem for nothing. List extending strange behaviour Is there any advantage in marketing my product for £99.99 instead of £100? Some time ago, we announced that . This has gone well and we're extremely thankful to everyone for their support and feedback. Because it has been successful so far, we're going to be extending the program to the majority of our graduated sites. Well, we've pretty much already done that, but there are quite a few sites that have ads enabled, yet still display nothing. The thing is, we don't really have an abundance of clients offering products that a  would find relevant, but Amazon has stand mixers, a selection of super fine chinois, deliciously sharp cutlery and enough cast iron to fill the Gulf Of Mexico with . Beginning as early as this week (week of 5 February 2018), our ad team is going to be hand selecting affiliate ads to be displayed on select sites. The use of inventory will of course depend on the availability of relevant ads to display which means:  Some sites might see very few, if any ads. Some topics simply don't come remotely close to orbiting anything worth advertising. Furthermore, while graduated, some sites simply don't have enough activity for ads to make sense.  Since these are hand-selected, it could take a while for ads to appear initially.  These would of course yield to inventory from clients if clients had products that were relevant to any given topic.   Nothing changes when it comes to the . If you're used to occasionally seeing a community promotion ad, you'll just notice that you'll also occasionally see an ad for something that is likely to be of interest to someone interested in the topic that your site covers. In fact, to be clear, this is simply a continuation of the announcement we made in November of 2016. We just wanted folks to know that we're going to be continuing the program where there's good potential to show relevant advertisements, even if we don't have clients lined up with a demand for it. Where can I give feedback about a certain ad on [site]? Open a meta post on your per-site meta and let us know. Please remember, the bulk of advertisements are an effort to get something small back from folks that hit the sites through search engines and are extremely unlikely to ever become contributors. Their soufflé stopped falling, and we'll probably never see them again. Maybe we can get a few bucks once they realized that copper ions help egg whites stiffen properly and buy a bowl or a proper bloom whisk. However, you're the experts on these topics and we're very likely going to listen to you. What sites, specifically? Are any sites exempt? If you're a former Stack Exchange 1.0 site and / or subsequently joined the network under a special charter, nothing changes - don't panic :) For the rest, we're just extending what we're currently doing with ads on graduated sites to also include those where there's an opportunity to make sensible use of affiliate ads. If a site has less than 15k (average) daily visits, or fewer than roughly 50k combined total posts - it might not yet be time to invest in selecting ads to be shown there. But, part of this exercise is also seeing what works best. I don't really have feedback, just some suggestions Ads aren't just an opportunity to sell stuff, they're an opportunity to really show people that we understand a given topic. That, of course, makes suggestions from people that do that stuff every day quite valuable. Post on your per-site meta and our community / ad critters will take a look. Questions? Comments? Kvetches? Let us know in an answer or comment below. If anything about the plan changes, we'll update this post to let folks know. Proof of centripetal acceleration formula ($a_c = v^2/r$) for non-uniform circular motion The formula for centripetal (radial) acceleration is well known, and there exist many proofs for it: $$||a_c|| = \frac{||v||^2}{r}$$  However, all the proofs I've seen rely on the fact that it is uniform circular motion and the magnitude of the tangential velocity vector does not change. For instance, take the classic proof using similar triangles — the similarity can only be established if the final tangential velocity vector and the initial one are of the same length.  In addition, take  calculus based proof from Khan Academy, outlined as follows:     For this proof to work, $(d\theta / dt)$ must be considered a constant, $\omega$, that does not depend on time. In the case of non-uniform circular motion, however, this is not always true as since there exists a tangential acceleration along with a radial one, $\omega$ must depend on time and is not necessarily a constant value.  Intuitively I understand that the centripetal/radial acceleration depends only on the difference in orientation between two tangential velocity vectors, and that their magnitudes do not matter -- hence the formula intuitively holds true in the non-uniform case. However, how would you go about modifying either of the proofs presented so that they are still valid in this case? Or alternatively, is there another proof that holds valid even when there exists tangential acceleration?    As per the suggestion of one of the answers, I let $\omega$ vary with time and took its derivative as $\alpha(t)$. This is my work so far. Unfortunately, I am stuck after the last step.  \begin{align} \overset{\rightharpoonup }{p}(t) &amp;= r \cos (\theta (t))\cdot\hat{i}+r \sin (\theta (t))\cdot\hat{j} \\ \overset{\rightharpoonup }{v}(t) &amp;= -r \sin (\theta (t))\cdot\omega(t)\cdot\hat{i}+r \cos (\theta (t))\cdot\omega(t)\cdot\hat{j} \\ \overset{\rightharpoonup }{a}(t) &amp;= (-r \cos(\theta(t))\cdot\omega(t)^2 - r \sin(\theta(t))\cdot\alpha(t))\hat{i} \\ &amp;+ (-r \sin(\theta(t))\cdot\omega(t)^2 + r \cos(\theta(t))\cdot\alpha(t))\hat{j} \end{align}     From here on, $\theta(t)$ is represented as just $\theta$ for brevity   and clarity   $$\overset{\rightharpoonup }{a}(t) = -\omega(t)^2(r \cos\theta\cdot\hat{i} + r \sin\theta\cdot\hat{j}) -\alpha(t)(r \sin\theta\cdot\hat{i} - r \cos\theta\cdot\hat{j})$$ Probability question with interarrival times (Why did the chicken cross the road?) A chicken wants to cross a single-lane road where the cars arrive according to a Poisson process with rate $\lambda$. She needs at least $k$ minutes to cross the road safely, so she will have to wait until she sees a gap of at least $k$ between the oncoming cars. If the gap between the car that just arrived and the next one is at least $k$ then she starts crossing the road immediately. Let $T$ denote the random time she needs to wait by the road.  Find the expected time needed to cross the road. How To Make A Solid Object Drop To The Ground Like It Was Liquid So I have a simple bouncing ball animation, but I just wanted to something "weird" for the ending. Is there a way to change the ball's material on the fly, where after the last bounce it changes materials and physical composition from a rigid body(baked animation now) to a fluid sim?  (within 30-40 frames) Work done changes between reference frames? (This is not homework; a friend shared with me this puzzler and neither of us can figure it out.) Suppose you are in a plane traveling at velocity $v_1$ relative to the ground. The flight attendent pushes a cart of mass, say, $m$, accelerating it from rest to $v_2$, relative to the plane (so relative to the ground, its velocity goes from $v_1$ to $v_1+v_2$). From your perspective, the work done is then $$W = \Delta E = \dfrac{1}{2}mv_2^2$$ but from the perspective of the ground, the work done is $$W = \Delta E = \dfrac{1}{2}m(v_1+v_2)^2 - \dfrac{1}{2}mv_1^2 = \dfrac{1}{2}mv_2^2 + mv_1v_2.$$ Why are these quantities unequal?  I've seen some similar problems, like a body colliding with a wall in different reference frames, but I don't have the physics experience to tell if that question is analogous to this one. So, my apologies if this is a duplicate. :) There is no "helpful flags" counter (with a link to ) on the mobile version of profile page. And I see no other way to reach flagging summary page and check your flagging history from mobile rather than switching to the full site view.  Desktop version   &nbsp; Mobile version