Unwanted glue with boxes in Picture environment
How to troubleshoot unwanted white spaces smartly?
Nature of gravity: gravitons, curvature of space-time or both? General relativity tells us that what we perceive as gravity is curvature of space-time.   On the other hand (as I understand it) gravity can be understood as a force between objects which are exchanging (hypothetical) virtual particles called gravitons, similar to the way electromagnetic forces are due to objects exchanging virtual photons?  At least at first glance, the two concepts seem mutually exclusive. Is there a description of gravity which includes both, or is this contradiction one of the problems in combining GR with quantum mechanics? Encouraging people to explain downvotes Note: If you're looking for a simple explanation as to why comments aren't required on downvotes, see .   I used to get "upset" (though that is too strong a term) when I got downvoted without comment. If my answer isn't good enough then I'd like to know why. Not only does it improve the answer for the OP but it improves my knowledge too.  Where the down-vote has been explained I've found it useful &amp; it has improved my answer, or forced me to delete the answer if it was totally wrong.  So is there any way we can encourage people to leave a comment? Perhaps they don't lose rep if they explain their down-vote?  I must admit though that I haven't always explained my down-vote either so you could call me a hypocrite. I've also grown a thicker skin over the months of using SO (it seems to have come with the higher rep score ;-)), so I'm less bothered about this now. Glossaries: Make a glossary entry recognizable in printing How do you make it visible to the reader that the printed word/term is an entry in the glossary? In my PDF on the computer, I just see it because I use the hyperref package and there's a red box around it. But when printing it, there's just nothing that makes it visible to the user.  Is there a typographically good and common way of doing this? What are the convergent sequences in the cofinite topology I consider the co-finite topology on $\mathbb{R}$ that is $$\{G\subset \mathbb{R}, card(\mathbb{R}\setminus G)&lt;+\infty\}\cup \{\emptyset\}$$  What are the convergent sequences in this space .  thank you Is it possible to reconstruct the Hamiltonian from knowledge of its ground state wave function? Is it possible to "construct" the Hamiltonian of a system if its ground state wave function (or functional) is known? I understand one should not expect this to be generically true since the Hamiltonian contains more information (the full spectrum) than a single state vector. But are there any special cases where it's possible to obtain the Hamiltonian? Some examples would be really helpful. Problem:  Let $A_{n.n}$ be square complex matrix. Prove the following: $$\left \| A \right \|=\left \| A \right \|_{HS}\Leftrightarrow  rank(A)\leqslant 1$$. Where $\left \| . \right \|_{HS} $ is the Hilbert Schmidt Norm.  Please read my solution and tell me whether it is correct. If not, let me know where the mistake is.  Proof of the implication $\Leftarrow $:   If $rank(A)=0$, then in this case $A=0$. It follows that $\left \| A \right \|=\left \| A \right \|_{HS}=0$ If $rank(A)=1$, than: $ A=\begin{pmatrix} A_{1}\\  \alpha _{2}A_{1}\\  ...\\  \alpha _{n}A_{1} \end{pmatrix}$ where $A_{1}$ is the first row of $A$   and in this case: $\left \| A \right \|=\left \| A \right \|_{HS}=\left \| A_{1} \right \|\sqrt{1+\alpha _{1}^{2}+...+\alpha _{n}^{2}}$  Proof of the implication $\Rightarrow $   We know that $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|$  On the other hand: $ A=\begin{pmatrix} A_{1}\\  A_{2}\\  ... \\ A_{n}  \end{pmatrix}$. So, $$\left \| Ax \right \|=\left \| \begin{pmatrix} \left \langle A_{1},x \right \rangle\\  \left \langle A_{2},x \right \rangle\\  ...\\   \left \langle A_{n},x \right \rangle\end{pmatrix} \right \|=\sqrt{\left \langle A_{1},x \right \rangle^{2}+\left \langle A_{2},x \right \rangle^{2}+...+\left \langle A_{n},x \right \rangle^{2}} $$  Where $ \left \langle A_{i},x \right \rangle$ is the inner product of $A_{i}$ and $x$  Using the Cauchy-Schwarz inequality: $\left \langle A_{i},x \right \rangle^{2}\leq \left \| A_{i} \right \|^{2}\left \| x \right \|^{2}$, we get: $\left \| A \right \|=max_{\left \| x \right \|=1}\left \| Ax \right \|\leq max_{\left \| x \right \|=1}\left \| x \right \|\left \| A \right \|_{HS}=\left \| A \right \|_{HS}$. In order to have the equality: $\left \| A \right \|=\left \| A \right \|_{HS}$, we should have: $\left \langle A_{i},x \right \rangle^{2}=\left \| A_{i} \right \|^{2}\left \| x \right \|^{2} $ which occurs only if $A_{i}=\lambda _{i} x$. So, the rows of A are dependent, which implies that $rank(A)=1$. Note that the case $rank(A)=0$ happens when $A=0$  Please let me know if my solution makes sense. How does java do modulus calculations with negative numbers? How to Disable new terms (free tagging) with Taxonomies? I have a "term reference" field, in a form that uses an auto-complete widget. I want to limit user input to terms in the vocabulary only, and output a message when a user tries to enter a non existing term.  I found the  module. However this does not work with forms that contain a field collection or a field group.  There is also a contributed patch . But I was wondering whether there was a simpler/cleaner solution like a module or maybe I missed an already existing way of doing it with latest drupal and taxonomy versions. It seems common that PBX and other telephone hardware use a positive-ground power supply, where the "hot" line is at -48v.  What's the reason for that? Drawing on an image with TikZ I saw  hosted by Stephen Hawkins.  And if I didn't get it wrong, it says that there was no time before the big bang, time was created there.  So how can anything happen when there is no time (eg: the creation of time and the universe)? Is this what he calls a 'singularity'? Which I could think like this: when there is no time nothing can happen, except during this singularity.  Of course I don't know of a law that says 'when there is no time nothing can happen..', I just came up with it, as something intuitive. Prove without using graphing calculators that $f: \mathbb R\to \mathbb R,\,f(x)=x+\sin x$ is both one-to-one, onto (bijective) function. Prove that the function $f:\mathbb R\to \mathbb R$ defined by $f(x)=x+\sin x$ for $x\in \mathbb R$ is a bijective function.    The codomain of the $f(x)=x+\sin x$ is $\mathbb R$ and the range is also $\mathbb R$. So this function is an onto function. But I am confused in proving this function is one-to-one. I know about its graph and I know that if a function passes the  horizontal line test (i.e horizontal lines should not cut the function at more than one point), then it is a one-to-one function. The graph of this function looks like the graph of $y=x$ with sinusoids going along the $y=x$ line. If I use  a graphing calculator at hand, then I can tell that it is a one-to-one function and $f(x)=\frac{x}{2}+\sin x$ or $\frac{x}{3}+\sin x$ functions are not, but in the examination I need to prove this function is one-to-one theoritically, without graphing calculators. I tried the method which we generally use to prove a function is one-to-one but no success. Let $f(x_1)=f(x_2)$ and we have to prove that $x_1=x_2$ in order fot the function  to be one-to-one. Let $x_1+\sin x_1=x_2+\sin x_2$ But I am stuck here and could not proceed further. How do I use/change Animation Extrapolation? (Can't find F-Curves / Animation Modifiers in 2.8(1)) There are several reasons to be looking for a ROM:   Upgrading or downgrading your stock ROM Looking for firmware from a different region/language Flashing a custom ROM for different features, removing bloat, etc.   But where can one find ROMs in the first place?  What sites are likely to have the ROM I need?  ; please add to the wiki answer instead of posting separate answers.  See also: Can you make a non-polar electrolytic capacitor out of two regular electrolytic capacitors? There was some discussion on this question        which I don't see as being conclusively resolved:   "turns out that what might LOOK like two ordinary electrolytics are not, in fact, two ordinary electrolytics." "No, do not do this. It will act as a capacitor also, but once you pass a few volts it will blow out the insulator." 'Kind of like "you can't make a BJT from two diodes"' "it is a process that a tinkerer cannot do"   So is a non-polar (NP) electrolytic cap electrically identical to two electrolytic caps in reverse series, or not?  Does it not survive the same voltages?  What happens to the reverse-biased cap when a large voltage is placed across the combination?  Are there practical limitations other than physical size?  Does it matter which polarity is on the outside?  I don't see what the difference is, but a lot of people seem to think there is one.  Summary:   As posted in one of the comments, there's a sort of electrochemical diode going on:     The film is permeable to free electrons but substantially impermeable to ions, provided the temperature of the cell is not high. When the metal underlying the film is at a negative potential, free electrons are available in this electrode and the current flows through the film of the cell. With the polarity reversed, the electrolyte is subjected to the negative potential, but as there are only ions and no free electrons in the electrolyte the current is blocked. —    Normally a capacitor cannot be reverse-biased for long, or large currents will flow and "destroy the center layer of dielectric material via electrochemical reduction":     An electrolytic can withstand a reverse bias for a short period, but will conduct significant current and not act as a very good capacitor. —    However, when you have two back-to-back, the forward-biased capacitor prevents a prolonged DC current from flowing.  :     For circuit positions when reverse voltage excursions are unavoidable,   two similar capacitors in series connected “back to back” ... will create a   non-polar capacitor function ... This works because almost all the circuit voltage   is dropped across the forward biased capacitor, so that the reverse   biased device sees only a negligible voltage.   :     The oxide dielectric construction that is used in tantalum capacitors   has a basic rectified property which blocks current flow in one   direction and at the same time offers a low resistance path in the   opposite direction. Prove that $A = B = C = D$ $A, B, C, D$ are sets, where $A \not = \emptyset$ and $B \not = \emptyset$. Equation $(A \times B) \cup (B \times A) = C \times D$ is true.  Prove that $A = B = C = D$.  I tried to find $A, B, C, D$ separately, but that didn't get me anywhere. Any ideas how to solve it? How to store date/time and timestamps in UTC time zone with JPA and Hibernate Can I speed up iCloud photo sync between iPhone and Macbook? I have an iPhone 5 running iOS 9.1 and a Macbook Pro running Yosemite.  Both devices are set up to sync photos to and from iCloud.  When I take a photo on the iPhone (for further processing on the Macbook), it takes an inordinate amount of time until the photo shows up in the Macbook's photos library - sometimes up to 10-20 minutes. Sometimes it seems to work faster if I quit the Photos app on both devices, and start it again.  I often end up sending myself photos through E-Mail because it's quicker. That seems ridiculous!  Is there a way to force a sync on the iPhone, or to speed up the process some other way? Making crafted items usable in adventure mode In a map I'm making the player has to survive while locked in a maze. To stop them from breaking out, they will be in adventure mode. However, I still want them to be able to craft items and use them with specified blocks. I have tried several methods already but they all have a flaw.  One method is to have a scoreboard objective called item and giving certain item entities a specific score using something like:  /scoreboard players set @e[type=Item] item 1 {Item:{id:"minecraft:sapling"}}   Then it is possible to select these items with:  /entitydata @e[type=Item,score_item=1,score_item_min=1] {Item:{tag:{CanPlaceOn:["dirt"],HideFlags:24}}}   This works but it requires the item to be dropped, which isn't a problem for saplings, but for tools and other craftable items, it becomes a hassle.  To fix this, I'm currently clearing tools without damage and giving the player a slightly damaged tool in return:  /clear @a minecraft:stone_axe 0 1   /give @a minecraft:stone_axe 1 1 {CanDestroy:["log","log2","leaves","leaves2"],HideFlags:24}   This doesn't work for items without damage values (e.g. torches), but I don't mind these having to be dropped as there aren't too many of these. However, there is another problem. In the nether, the player will have access to an enchantment table and an anvil. This means they can repair enchanted tools, which will cause them to be cleared and replaced with unenchanted ones as soon as they go back to the overworld.  I also thought about using a stat.craftItem scoreboard objective and giving them the usable item if they craft it. However, this would result in the player having a duplicate of every item they crafted, as I think it is impossible to select an item without NBT-data in the /clear command.  Is there any way to make this work? And if not, would it be better to have the player drop all crafted items or have duplicates of them all? Test the series whose general term is: $$\sqrt{n^2 + 1} - n$$   This is from Higher Algebra by Henry Sinclair. I'm struggling to understand how this series is divergent. If you do the following expansion: $$\sqrt{n^2 + 1} - n$$ $$= n\biggl(1 + \frac{1}{n^2}\biggr)^\frac{1}{2} - n$$ $$= n + \frac{1}{2n} - \frac{1}{8n^3} + ... - n$$ $$= \frac{1}{2n} - \frac{1}{8n^3} + ... $$  The remaining series appears to converge to zero as $n$ approaches infinity.  Their explanation is similar to mine, but at the end they say that the above series approaches $\frac{1}{2n}$. From there they directly say that the series is divergent. Earlier in the chapter, it's proved that the expansion of $(1 + x)^n$ is convergent for $x &lt; 1$. In this case, $\frac{1}{n^2} &lt; 1$ as $n$ approaches infinity, so I would assume that the series converges.  The solution to the problem is given  (number 17)  Any help is appreciated. Thanks.