moving elements in an vector
Why doesn't this C++0x code call the move constructor?
In what situations is it more appropriate to use an HTML IMG tag to display an image, as opposed to a CSS background-image, and vice-versa?  Factors may include accessibility, browser support, dynamic content, or any kind of technical limits or usability principles. Learning special relativity and working with Minkowski diagrams for a while now i am still trying to get my head around some oddity.   From my understanding time can often be seen as the dependent variable in those diagrams, so why put it to the vertical axis? What confuses me even more is the apparent convention to denote a point in space time $(t,x)$ even though $x$ is on the horizontal and $t$ is on the vertical axis, so the usual notation would be to refer to that vector as $(x, t)$.  Are there mostly historical reasons or why is it beneficial to use this notation and to violate usual conventions? Background Video is shown in Rendered mode but not when szene is rendered I was hoping to find a more "mathematical" proof, instead of proving logically $\displaystyle \sum_{k = 0}^n {n \choose k}^2= {2n \choose n}$.  I already know the logical Proof:  $${n \choose k}^2 = {n \choose k}{ n \choose n-k}$$  Hence summation can be expressed as:  $$\binom{n}{0}\binom{n}{n} + \binom{n}{1}\binom{n}{n-1} + \cdots + \binom{n}{n}\binom{n}{0}$$  One can think of it as choosing $n$ people from a group of $2n$  (imagine dividing a group of $2n$ into $2$ groups of $n$ people each. I can get $k$ people from group $1$ and another $n-k$ people from group $2$. We do this from $k = 0$ to $n$. As the title suggests, I'd like to prove that the sum $$ \sum_{n=2}^\infty (\ln n)^{- \ln n} $$ is finite.  The root and ratio test both fail here, but  suggests that there is a comparison that can be used to show convergence.  The only thought I have is that it may help to write the terms as $e^{-\ln(n)\ln(\ln(n))}$, but this has not led me to any particular insight.  Any ideas are appreciated. The divisibility rule for $3$ is well-known: if you add up the digits of $n$ and the sum is divisible by $3$, then $n$ is divisible by three.  This is quite helpful for determining if really large numbers are multiples of three, because we can recursively apply this rule:  $$1212582439 \rightarrow 37 \rightarrow 10\rightarrow 1 \implies 3\not\mid 1212582439$$ $$124524 \rightarrow 18 \rightarrow 9 \implies 3\mid 124524$$  This works for as many numbers as I've tried.  However, I'm not sure how this may be proven.  Thus, my question is:     Given a positive integer $n$ and that $3\mid\text{(the sum of the digits of $n$)}$, how may we prove that $3\mid n$? What's the difference between "either" and "neither"? Invert "if" statement to reduce nesting When I ran  on my code, for example:      if (some condition)     {         Some code...                 }   ReSharper gave me the above warning (Invert "if" statement to reduce nesting), and suggested the following correction:     if (!some condition) return;    Some code...   I would like to understand why that's better. I always thought that using "return" in the middle of a method problematic, somewhat like "goto". Console flashes on launching, then closes (ms-windows) I installed blender on a windows 7 64 bit system.  I opened it one time then went to the tutorials.  in the meantime I had a power failure and when I tried to start blender again I get a dos window that closes itself then nothing.  I reinstalled Blender and get the same thing.  I am going to try to uninstall then reinstall but in the mean time, any other suggestions are welcome. Non-empty intersection. Let $A_1,A_2,...,A_{2019}$ be sets s.t. $|A_i|$=45 for each $i=\overline{1,2019}$ and $|A_i\cap A_j|=1$ for every $i\neq j$. Is it true that $$\Big|\bigcap _{i=1}^{2019} A_i\Big|=1?$$ What I thought of doing is proving by induction by i, the base is obvious. However I'm stuck on the next step. How do I show that a two-qubit state is an entangled state? The Bell state $|\Phi^{+}\rangle = \frac{1}{\sqrt{2}}(|00\rangle + |11\rangle )$ is an entangled state. But why is that the case? How do I mathematically prove that? I am trying to insert a figure with a caption into a multicol article. When the figure was inserted on its own, it was fine: includegraphics[width=0.5cm]{image}  However, when I tried to insert the caption [see below], the caption did not appear anywhere in the article and the figure disappeared.  \documentclass{article} \usepackage[hmarginratio=1:1,top=30 mm,columnsep=20pt,hmargin=2.5cm]{geometry} \usepackage[font=it]{caption} \usepackage{paralist} \usepackage{multicol} \usepackage{setspace} \usepackage{graphicx}  \begin{document} \begin{multicols}{2} text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text  \begin{figure}[h!]  \caption{Fig.1 xxxxx} \includegraphics[width=0.5cm]{image} \end{figure}  text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text text  \end{multicols} \end{document}   What might be the issue and how could I fix it? How do I install Ubuntu? I would like to see a full how-to guide on how to install Ubuntu. small limits in the pointed category of a category with all small limits In Hovey, Model Categories, he says that given a category $\mathcal{C}$ with a small limits and colimits, if we form the category of pointed objects ,i.e. the category with objects as morphisms $*\rightarrow A$ where * is the terminal object, then this category $\mathcal{C}_*$also has all small limits and colimits.  To prove this, for any Functor $F:J\rightarrow \mathcal{C}_* $ where $J$ is small we compose it with the forgetful functor $\mathcal{C}_* \rightarrow \mathcal{C}$ and take the limit here. My question is how will this give a limit in $\mathcal{C}_*$? we will get an object in the category, but how do we get the arrow from $*$ to that object? Improving "demonstrate a minimal understanding" close reason Note: the eventual outcome of this discussion involved dropping this particular off-topic reason, along with other changes. See:  &nbsp; &mdash;Shog     I was using the reason "demonstrate a minimal understanding" myself quite frequently, but the more I review, the more I have doubts about it and I think this reason can either be dropped entirely or at least rephrased, so I started to use "problems with code you've written must describe the specific problem" instead:     Questions asking for code must demonstrate a minimal understanding of the problem being solved.   So the poster has posted some description, maybe even a snippet of code and in his mind that already proves that he has that minimal understanding required. After all he described the problem and what he tried to solve. So telling him that he lacks that understanding is rather rude, and from the posters point of view doesn't even apply. What both version of the reason not emphasize is that a description of a problem is not various versions of     it doesn't work   but that we rather mean really specific details (see below). I can often see in the reopen review queue that the "improved" version just includes a more verbatim description, while what usually would be expected is some specific error message, a trace or a log, or whatever applies.  Then it continues with:     Include attempted solutions, why they didn't work, and the expected results. See also: Stack Overflow question checklist   comparing it to:     problems with code you've written must describe the specific problem — and include valid code to reproduce it — in the question itself. See SSCCE.org for guidance.   Which says essentially the same, only in more detail.   So the "minimum understanding" reason is IMO a more rude version of the reason telling the user that an SSCCE is missing.  The intent of giving this reason is to tell the user that he needs to give exact details to be able to help him, so IMO a checklist would be much better. Something like:     Your question was closed because it may not provide enough details. Things you can do to improve your question to get an answer include:         Format the code in a way that makes it readable and easily understandable.   Provide the smallest working sample of code demonstrating the problem (link to SSCCE)   Include your input and output.   If applicable provide logs (i.e. Android debugging log, stack trace, etc.).   If applicable provide the exact error messages (compiler, etc.).      And something like this should be best included already in the banner that appears when a question is closed instead of being hidden in some pages you often enough have to actively looking for.  It might be helpful to adjust this, based on the tags or close reason, as I realize that I'm more focused on the programming area, and the above list may not apply to other areas of SO.  Update  Another thing I forgot to mention (but wanted to) is this:     Questions asking for code   Is IMO essentially an invitation of the type of questions     I have a problem, can you provide the solution? (with an optional please)   while     Questions concerning problems with code you've written   puts at least some emphasis on the fact that the poster has to put some effort himself in trying to solve the problem by providing the code which he used to attempt the solution. Changing affiliation on publication I think this is a duplicate of an old question without a good answer:  My child was doing schoolwork on her school-issued Macbook Air. Suddenly lighter greys are totally white. This generally gives the images on the screen a vague feeling of being &quot;off&quot;, and pixellates all curved edges. The biggest problem is that text on some of her sites is unreadable. I've checked the following:  Color profiles - it looked like it was using the default, but a cycled through a few built-ins. I could see noticeable changes to color balance, but the light greys remained the same as white. Also deleted contents of ~/Library/ColorSync/. Accessibility - my child has accidentally inverted colors before, this isn't that. No other accessibility settings seemed related to color. Display settings - Modifying brightness was also fruitless. The screen overall got brighter or darker, but still couldn't distinguish between light grey and white Rebooting - colors look fine until shortly after logging in - seems connected to user profile somehow.  Here are some images to demonstrate the issue. This is a generic color test image:  On the Mac, it looks like this (simulated with photoshop):  Any solution would be appreciated, but if there is a keyboard shortcut that can cause this, it would be the most likely explanation (8 year old says they &quot;might have rested their arms on the keyboard&quot; during a class meeting) I don't know much more about SEO, but it's interesting to learn about it. When I tried to search my website with an associated keyword or you can say with the exact name of the domain, I got some suggestions by Google. Here's .  First of all, I don't want to spamming here, I just want prove my question and want to give a hint about what I am talking here. The above link display the Did You Mean, I know the specific reason why Google shows this, but I want to know the Google's guidelines and some guidance by you experts so that if someone search for the particular keyword then Google hide that Did You Mean and display the result. If the universe is expanding, what is it expanding into? Why pendulum does not follow SHM for larger angular displacement? Considering an ideal case(neglecting drag of air, damping etc.), a pendulum follows SHM if the angular displacement is small (upto 10 degrees). But, for large angular displacement(more than 10 degree), it does not follow SHM. Why? What force cause the pendulum not to follow SHM?