I'm 100% sure I saw  about HDMI which may not have been asked more than 4 hours ago, but it was not in the result list when I entered HDMI in the search field.  Then I noticed: when I enter "HDMI" as search term, it searches for the tag  instead. Is this the expected behavior? And is it really my task to find out that I have to enter "HDMI" to get other results?  Can we at least have an indicator that some magic happened, e.g.     You entered HDMI. We're looking for questions tagged  instead. Search for "HDMI" instead.
Asume the following search queries: osx, cocoa, iphone, xcode, applescript, mac, carbon    Searching for osx, cocoa, iphone or xcode like:   http://stackoverflow.com/search?q=osx   Results into searching by tag:   http://stackoverflow.com/questions/tagged/osx   While searching for applescript, mac and carbon like:   http://stackoverflow.com/search?q=applescript   Results into searching by string.  Yes, I know how to search by tags using brackets: [applescript] But why are the queries above interpreted differently? All queries are valid Stack Overflow tags! Why not convert all queries into search by string or search by tag? What is the logic behind? Tag popularity? It seems rather inconsistent to me.
This is my last homework problem and I've been looking at it for a while. I cannot nail down what is wrong with this proof even though its obvious it is wrong based on its conclusion. Here it is:     Find the flaw in the following bogus proof by strong induction that   for all $n \in \Bbb N$, $7n = 0$.      Let $P(n)$ denote the statement that $7n = 0$.      Base case: Show $P(0)$ holds.      Since $7 \cdot 0 = 0$, $P(0)$ holds.      Inductive step: Assume $7·j = 0$ for all natural numbers $j$ where $0 \le j \le k$ (induction hypothesis). Show $P(k + 1)$: $7(k + 1) = 0$.      Write $k + 1 = i + j$, where $i$ and $j$ are natural numbers less than $k + 1$.  Then, using the induction hypothesis, we get $7(k + 1) = 7(i + j) = 7i + 7j = 0 + 0 = 0$. So $P(k + 1)$ holds.      Therefore by strong induction, $P(n)$ holds for all $n \in \Bbb N$.   So the base case is true and I would be surprised if that's where the issue is.  The inductive step is likely where the flaw is. I don't see anything wrong with the strong induction declaration and hypothesis though and the math adds up! I feel like its so obvious that I'm just jumping over it in my head. I met an interesting issue about C#. I have code like below.  List&lt;Func&lt;int&gt;&gt; actions = new List&lt;Func&lt;int&gt;&gt;();  int variable = 0; while (variable &lt; 5) {     actions.Add(() =&gt; variable * 2);     ++ variable; }  foreach (var act in actions) {     Console.WriteLine(act.Invoke()); }   I expect it to output 0, 2, 4, 6, 8. However, it actually outputs five 10s.  It seems that it is due to all actions referring to one captured variable. As a result, when they get invoked, they all have same output.  Is there a way to work round this limit to have each action instance have its own captured variable? Is $\sum\limits_{n=1}^{\infty}\frac{n!}{n^n}$ convergent? System program problem detected? underline omitting the descenders I'm very well aware that it is considered best practice not to use underlining other than on typewriters and in handwriting. But sometimes it is needed nonetheless.  In my opinion the typographically least offending way to underline in print is to leave out the descenders. This is also how I learned to do it in handwriting. Very few fonts offer a pre-made underlined variant, but some text-processors (e.g. OS X text-engine) do -- depending on the font used -- an acceptable job at emulating this effect:    How would I tweak any of the existing ways/packages for underlining to do the same?  As a bonus, I'd like to be able to tweak the behaviour (extent of the omitted part, apply to which letters) per font and text style, preferably working with XeLaTeX. Why is vim frozen? I own a lot of games on Steam, and I'm overall a big fan of the service.  However, one thing about it really bothers me, and I'm wondering if anything can be done about it.  Let's say I have two different online, multiplayer games, both of which I legally own.   If I bought these games normally, I could be playing one of them online, while my fiance could play the other one online on another computer.     However, with Steam, I can only be signed in on one computer at a time.  So, despite the fact that I have bought about 40 games, if I'm playing one of them, someone else in my house is not able to play any of the others.  Is there a way to get around this?  I don't think this is illegal or immoral as I own both of the games in question, and only one person is playing the game at a given time (same IP address and the same household, so it should be easy for Steam to verify that I am not sharing my games unfairly).    If I had bought simple boxed copies, this wouldn't be an issue, but since I became such a fan of Steam, I can't play any of those games at the same time as my fiance.    Is there anything that can be done? Is every complex (smooth) manifold a scheme? The question in the title doesn't quite make sense. I was always wondering if the scheme is the generalization of manifold. The precise statement should be like following:  If $X$ is a complex (smooth)manifold, is there a scheme $Y$ over $\mathbb{C}$ such that the corresponding analytic space $Y^{an}$ isomorphic to $X$?  I guess if $X$ is projective, then this is Chow's theorem. And I doubt the question has an affirmative answer if $X$ is non-compact (I guess that $X: y-e^x=0$ in $\mathbb{C}^2$ might be an example, but I do not know how to show this).  Edit 1:  There are also example (as @hunter pointed out) of a complex torus but not an abelian variety shows that a compact manifold might not be a variety, however, is it possible to be a scheme ?(according to Hartshorne, the category of varieties is a subcategory of quasi-projective, integral, separated scheme. But could there be some non-quasiprojective scheme($Y$ has to be integral, separated in our case according to GAGA, I guess) )?  Edit 2: Another question I think of is: is (Artin) Stack a generalization of manifold? I really want to know if manifold could be think of some subcategory of some category of with "algebraic objects". How to download all OpenStreetMap highways in a large area? I need to download all highways as line features from OpenStreetMap in a large area (about 10.000km²). Therefore I wanted to use overpass-turbo with the following statement in the wizard:   highway=*   in my chosen bounding box.  Unfortunately it looks like my bounding box is too big, because I get the error:   runtime error: Query run out of memory using about 2048 MB of RAM.   Do I have to change my overpass turbo query or is it not possible to download large datasets in overpass turbo? Which alternative method could I use to download all highways as line features in a larger region. I need to have all attributes, like the type of the highway, "foot=yes/no/...", "access="yes/no/....", etc. and if I use something like geofabrik, these attributes do not appear. Example 4.21 in Baby Rudin: Why is the inverse of this function not continuous at this point? Was the "Eye of Sauron" Sauron's actual physical form? During the time frame of the Lord of the Rings, was the Eye of Sauron Sauron's actual physical form? If not, what form did he have? How do I go about porting Ubuntu Touch to different devices? I would like to port Ubuntu Touch to a device that is currently not supported. What are the requirements for doing this and where do I begin? Prove that $\lim_{t\to \infty} t\mu(\{x:f(x)\geq t\})=0$ Problem  Suppose $f$ is a non-negative integrable function on a measure space $(X,\mathcal{A},\mu).$ Prove that $$\lim_{t\to \infty} t\mu(\{x:f(x)\geq t\})=0$$  Attempt  Let $E_t=\{x:f(x)\geq t\}$ Note that $f\chi_{E_t}\leq f$. Since $f$ is integrable, the Lebesgue dominated convergence theorem tells us that $$\lim_{t\to\infty} \int f\chi_{E_t}\rightarrow \int \lim_{t\to\infty} f\chi_{E_t}=0,$$ since $\lim_{t\to\infty} f\chi_{E_t}=0$ a.e. $\ ^{(1)}$ The result follows from the fact that $$t\mu(E_t)\leq\int f\chi_{E_t}.$$  Question  I am having trouble proving $(1)$. That is, how can I show that $f\chi_{E_t}\rightarrow 0$ a.e? I have this rule in my iptables:  iptables -A INPUT -s 192.168.11.0/24 -j LOG   My question is:  Where is the iptables log file, and how can I change that? Animating Dynamic Paint with noise texture What are the Minecraft Easter eggs? I was looking at  and I didn't realize that Minecraft had some Easter eggs in it. Besides the Christmas chest Easter egg, and the sheep Easter egg, what else is there? I would like to know how to write the Exec command of a .desktop file to open a new terminal and execute a shell script in it. The shell script is working and accessible by all users. When launching the script from the terminal everything works, but it doesn't when trying to launch the script from a .desktop file.  Here are some combinations I have already tried:  Exec=gnome-terminal -x sh -c 'echo hello' Exec=sh -c 'gnome-terminal echo hello' Exec=sh -c 'echo hello' Exec=echo hello   The .desktop terminal option is set to true. What are good initial weights in a neural network? I have an entry-level Nikon DSLR, and I cant decide if I should upgrade my camera body or upgrade from a  AF-S DX Nikkor 18-200mm to whatever else might be recommended? Age limit is too strict