I have a problem with SSD free space.  It says, that free space on drive C is around 5GB but  it says, that I used only 44GB. If i'm not bat at maths there is still around 75GB of free space but my PC says, that I have only 5GB.  and I'm looking forward for your answers.  Explanatory Notes :   "Všetko v priečinku C:" - everything in C: folder  "Využité miesto" - used space  "Voľné miesto" - free space
I have an external hard disk, and for some reason, it is reporting wrong free space.  In Windows Explorer, it is showing the used space is 378 GB.    I tried running chkdsk H: /f /r, it seems to report the same usage as Windows Explorer, and it says that there is no problem.    But when I tried selecting all the files inside this drive, it is showing much lesser usage.    I've already set the setting to show all hidden files and folders, also show the protected OS files, but it still cannot find about those missing files. And the difference is huge (282 GB). I've already cleaned the Recycle bin as well.    What I am missing here and how do I identify those missing spaces?  [UPDATE]: I used  to help identify and it seems to find some gigantic files, of which filename is quite weird. Anyone knows what this is and is it safe to delete?
How can I draw a dot under a symbol in math mode? The  claims that likelihood and probability are distinct concepts.     In non-technical parlance, "likelihood" is usually a synonym for "probability," but in statistical usage there is a clear distinction in perspective: the number that is the probability of some observed outcomes given a set of parameter values is regarded as the likelihood of the set of parameter values given the observed outcomes.    Can someone give a more down-to-earth description of what this means?  In addition, some examples of how "probability" and "likelihood" disagree would be nice. Javascript Array.sort implementation? Which algorithm does the JavaScript Array#sort() function use?  I understand that it can take all manner of arguments and functions to perform different kinds of sorts, I'm simply interested in which algorithm the vanilla sort uses. Grant the OP a binding close as duplicate vote Since the  to the close as duplicate mechanism it is a lot more visible to the OP that there is/might be a duplicate to their question. Empirically, I've noticed a lot more people agreeing, in the comments under their question, that their question is a dupe.   For instance, :        I'm aware that allowing  is  (back in the stone age) but now Stack Exchange is so much larger, and Stack Overflow is groaning under the weight of questions, I think this should be partially revisited.  Based on the principle that the simplest way to keep a site clean is to have as many people as possible nibbling away, I propose that the OP gets granted a binding close-as-duplicate vote.  It should simply be an additional line in the banner, which is . It might be something like this:     This question may already have an answer here:   &nbsp; "" 3 answers   &nbsp; Does this fully answer your question?  Yes / No   I'm not wedded to the Yes / No thing I just can't think of a better way at the moment.  Clicking on Yes would take the OP to the close as duplicate dialogue with that question highlighted. If there are multiple duplicate questions you'd need to ask the OP multiple times. A No could be used to track the relative success of the project, i.e.   Whether the question subsequently gets closed. Whether the question gets reopened again. How many times the OP actually clicks. How many times the OP disagrees with the analysis of the community.   Potentially, a No could be used to open a comment dialogue asking the OP to explain why they think their question is not a duplicate thereby stopping them simply putting "this is not a duplicate" in the question and formalising this process somewhat. Not sure about this idea...  I do not think this should be restricted by reputation, anyone should be able to close their own question as a duplicate. It gives people with lower reputation a more positive experience of the closing process and emphasises the importance of duplicate finding.  Most importantly, self-closes should not, initially, count towards the question ban. We can help new users learn how to use a lot more of the site than they might otherwise see and encourage them to search for duplicates more thoroughly. If the OP then goes on to do the same thing 10 times in a row then you can start using it toward the ban as they're obviously using the users as a personalised search service but at least the first 3-4 times this should not have any effect.  This question is, deliberately, only requesting 20% of the linked question (from 2009) and then only based on the new UI. The linked question asks for binding migration votes etc, which I do not want to see and I am not requesting a binding close-as-duplicate for the OP until someone else has already voted to close (though I think if the OP has 3k rep this might be allowable). This, I hope, addresses the concern's raised in the linked question and explains why this is not a duplicate of that question. Is it safe to put hot food in the fridge? I heard that putting really warm food in the fridge right after cooking it is not the right thing to do. It might cause bacterial growth? Is this true? How do I change permissions on a FAT32 formatted drive? I am using Microsoft Visual C++ on Windows and Qt Creator on Ubuntu. I can't change permissions on my flash drive (FAT32), even as root.  How can I change permissions on a FAT32 flash-drive? How to create a Customized Ubuntu Server ISO? Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$ I found the following formula   $$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$  and it is cited that Euler proved the formula above , but how ?  Do there exist other proofs ?  Can we have a general formula for the alternating form   $$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^q}$$ If $n\ne 4$ is composite, then $n$ divides $(n-1)!$. I have a proof and need some feedback. It seems really obvious that the statement is true but it is always the obvious ones that are a little trickier to prove. So I would appreciate any feedback. Thank you!  Here is what I am asked to prove:     If $n$ is composite then $(n-1)! \equiv 0 \pmod n$.   Proof:  $n$ is composite $\implies n=ab$ where $a,b \in \mathbb{Z}$ and $0&lt;a,b&lt;n$.  Case 1: If $a=b$ then $n=a^{2}$. Now $n \mid (n-1)! \implies a \mid (n-1)!$, so $$\begin{aligned} (n-1)! &amp;\equiv 1\times 2\times \dotsb \times a \times\dotsb\times (n-a)\times\dotsb\times (n-1) \\ &amp;\equiv 1\times 2\times \dotsb\times a \times\dotsb\times -a\times\dotsb\times -1 \\ &amp;\equiv 0 \pmod n \end{aligned}$$  Case 2: $0&lt;a&lt;b&lt;n$.  Then, since $a \mid n$, $b \mid n$ and $n \mid (n-1)!$ we have that $a \mid (n-1)!$ and $b \mid (n-1)!$.  So this implies $(n-1)! \equiv 1\times 2\times \dotsb\times a \times\dotsb\times b\times\dotsb\times (n-1) \equiv 0 \pmod n$, Q.E.D. How do I limit the number of rows returned by an Oracle query after ordering? Is there a way to make an Oracle query behave like it contains a MySQL limit clause? In MySQL, I can do this: select *  from sometable order by name limit 20,10  to get the 21st to the 30th rows (skip the first 20, give the next 10). The rows are selected after the order by, so it really starts on the 20th name alphabetically. In Oracle, the only thing people mention is the rownum pseudo-column, but it is evaluated before order by, which means this: select *  from sometable where rownum &lt;= 10 order by name  will return a random set of ten rows ordered by name, which is not usually what I want. It also doesn't allow for specifying an offset. Windows 7 task bar at autohide, make it totally disapear? My desktop look is empty so all I have is a nice wallpaper. My Windows taskbar has the "Taskbar Appearance" setting to "Auto-hide the taskbar". Although hidden, it leaves a thin top layer so you can still see it a bit, and where the pinned shortcuts are. Is it possible to totally have it hidden without a trace? This is a serious question. It really bugs m. Recovering an accidentally deleted  folder in OSX - no Time Machine or backup available I am a longtime Windows user, recently switched to Mac. I replaced folder with another folder (they had the same name) thinking that OS X will merge the contents of both folders. But OS X deleted the original folder and replaced it with the new one. Windows merges both the folders (I do this often in Windows). I have now lost hundreds of files. I haven't enabled Time Machine. I've tried using Finder's Undo, but that doesn't work.  Currently, I am trying to recover my data using . It shows estimated time as 7 hours, 31 minutes. Does anyone know any free data recovery software that's faster and more powerful? Is wildcard search possible in Gmail? diagram software for Ubuntu? What is a "static" function in C? The question was about plain  functions, not  static methods, as clarified in comments.  I understand what a static variable is, but what is a static function?  And why is it that if I declare a function, let's say void print_matrix, in let's say a.c (WITHOUT a.h) and include "a.c" -  I get "print_matrix@@....) already defined in a.obj", BUT if I declare it as static void print_matrix then it compiles?  UPDATE Just to clear things up - I know that including .c is bad, as many of you pointed out. I just do it to temporarily clear space in main.c until I have a better idea of how to group all those functions into proper .h and .c files. Just a temporary, quick solution. Find all functions $f\colon \mathbb R\rightarrow \mathbb R$ such that for all reals $x$ and $y$:   $$f(f(x)f(y))+f(x+y)=f(xy).$$   It was six hours ago in IMO 2017 (problem 2).  I tried  the standard way: $x=0$, $x=y$, $x=1$,etc. but without any success.   Let $f(0)=a$. Hence, for all $x\in\mathbb R$ we have $f(a(f(x))+f(x)=a$. What is the rest?   Also we have $(x-1)(y-1)-1+x+y-1=xy-1$ and  $1-(1-x)(1-y)+1-x-y=1-xy$. Apologies in advance for the long question.  My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to $x^\mu(\lambda_b)$, then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length; $$   t_{ba} = \int_{\lambda_a}^{\lambda_b}d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\dot x^\mu(\lambda)\dot x^\nu(\lambda)} $$ will have elapsed where $g_{\mu\nu}$ is a metric on spacetime with signature $(-,+,+,+)$.    Why is this so?     Here is how I would attempt to justify this fact in special relativity with $g_{\mu\nu} = \eta_{\mu\nu}$.  Consider an inertial observer $O$ in $\mathbb R^{3,1}$, and suppose that this observer sees a clock, which I'll call observer $O'$ moving around on a curve $x^\mu(\lambda)$.  If $O'$ were also an inertial observer, then given any event with coordinates $x^\mu$ as measured by $O$, observer $O'$ would measure the coordinates of the event to be $x'^\mu = \Lambda^\mu_{\phantom\mu\nu} x^\nu + x_0^\mu$ for some Lorentz transformation $\Lambda$.  If $O'$ is not inertial, then this is no longer true, and there is some more complicated family of transformations, say $T_\lambda$ between events as seen by both observers.  I would argue, however, that if we were to partition the interval $[\lambda_a, \lambda_b]$ into a large number $N$ of intervals $I_1=[\lambda_a, \lambda_i], \dots, I_N=[\lambda_{N-1}, \lambda_b]$ with $\lambda_n = \lambda_a+n\epsilon_N$ and $\epsilon_N=(\lambda_b-\lambda_a)/N$, then on each interval $I_n$, $O'$ is approximately an inertial observer in the sense that $$   T_{\lambda_n} = P_n + \mathcal O(\epsilon_N), \qquad (\star) $$ for some Poincare transformation $P_n$.  Then we would note that since $O'$ is stationary in his own reference frame, he measures his worldline to have the property $\dot x'^\mu(\lambda) = (\dot t(\lambda), \mathbf 0)$ so that $$   I_{ba}=\int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x'^\mu\dot x'^\mu} = \int_{\lambda_a}^{\lambda_b} d\lambda \, \sqrt{\dot t^2} = t(\lambda_b) - t(\lambda_a) = t_{ba} $$ On the other hand the integral on the left can be written as a Riemann sum using the partition above, and we can invoke ($\star$) above to get \begin{align}   I_{ba}  &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x'^\mu(\lambda_n)\dot x'^\nu(\lambda_n)}\right] \notag\\ &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x^\mu(\lambda_n)\dot x^\nu(\lambda_n)} + \mathcal{O}(\epsilon_N^2)\right] \notag\\ &amp;= \int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\mu} \end{align} Combining these two computations gives the desired result.  How do others feel about this argument?  I'm not completely comfortable with it because of the assumption $(\star)$ I made on $T_\lambda$.  I imagine that in GR a similar argument could be made by invoking local flatness of the metric. Mathematica not plotting whole logarithmic function Configuring dpkg stops processing when calling an HTTP request to Adobe Flash Player Do I need swap space if I have more than enough amount of RAM? From what I understand, the purpose of a swap partition in Linux is to free up some "not as frequently accessed" information from RAM and move it to a specific partition on your harddrive (at the cost of making it slower to read from or write to), essentially allowing active applications more of the "high speed memory".  This is great for when you are on a machine with a small amount of RAM and don't want to run into problems if you run out. However, if your system has 16 GB or 32 GB of RAM, and assuming you aren't running a MySQL database for StackExchange or editing a 1080p full length movie in Linux, should a swap partition be used?