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Meaning of backtick in floating-point literal
Centos recently updated to 7.3 and there's been problems with sshd sftp group permissions. I have one user that is chrooted to it's home directory, and that user is in group sftponly. Then i have the /var/www directory, which has 775 permissions and owner is apache and owner group is sftponly. I have a bind link pointing from /home/user/files/web --> /var/www, so the user can access /var/www even though being chrooted to it's home directory. I can view files in /var/www with that user, but impossible to edit or add anything. This worked fine before the big Centos 7.3 update, and now it's stopped working. Any ideas? In what occasion could the word “precious” be taken disapprovingly, or sarcastically? I understand that the word “precious” in the following quote of Maureen Dowd’s article “Heart of Darkness” (NYT, March 20) literally means “2. valuable or important and not to be wasted” as defined in OALD:     Congressman Jones read an e-mail from a former boss of General   Allen’s, giving the congressman this unvarnished assessment:   “Attempting to find a true military and political answer to the   problems in Afghanistan would take decades. Would drain our nation of   precious resources, with the most precious being our sons and daughters. Simply put, the United States cannot solve the Afghan   problem, no matter how brave and determined our troops are.   But I didn’t know “precious” has another meaning as defined in OALD: 5. (disapproving, people and their behavior) very formal, exaggerated and not natural, and Readers English Dictionary: 2.(colloquial) terrible, almost worthless, until I was told by my respected forum mate a few days ago.   Being encouraged by his suggestion for me to “ask another question if I’m not sure of what he means by describing his input as “precious” is actually rather “precious” phrasing, I would like to ask:   What is an example of “precious” being used as “disapproving” or sarcastic way (as neither OALD nor CALD provides examples)? Is there an easy way to discern the instance “precious” being used as “disapproving” from the case being used in the sense of genuine “great value” as used in the above quote other than judging from the context? Let $a_k:=(-1)^k$ where $k\in\mathbb{N}$. $\mathbb{N}$ is the set of all non-negative integer.  And we define the partial sum $S_n:=\sum \limits_{k=0}^{n}a_k$. Notice that the sequence $\{S_k\}$ diverges which also implies that the infinite series $\sum \limits_{k=0}^{\infty}a_k$ cannot be defined.  If I consider the infinite sum $1-1+1-1+1-1+\cdots$, then this statement is equivalent that I just defined the infinite series $\sum \limits_{k=0}^{\infty}a_k$, which is contradiction.  But, suppose that the sum $1-1+1-1+1-1+\cdots$ exists and let the value of the sum be $S$. Then, we can easily observe that $S=1-S$, therefore $S=1/2$. The supposition of this proposition already proved as false, but if I ignore the definition of infinite series, it holds.  Which one is right? Or are these propositions depends on what we define? Gradient Boosting for Linear Regression - why does it not work? Is there a word to describe words that can be spelled different ways? Is there a word in English to describe words that can be spelled more than one way?  For example:   colour and color favourite and favorite   However, this shouldn't be limited to words spelled differently in different dialects, though I'm not sure what other case there would be. Get list of ALL offers from Amazon Product Advertising API File.Copy is giving exception Feature request - links underlined Biblatex citation order I'm facing a problem with biblatex citation order.  Using this configuration:  \documentclass {article} \usepackage[brazil]{babel} \usepackage[utf8]{inputenc} \hyphenation{es-ta-bi-li-da-de} \usepackage{setspace} \singlespacing \usepackage[natbib=true,style=numeric]{biblatex} \addbibresource{frameworks.bib}   Being frameworks.bib this:  @Article{JONHSON88, author = {Ralph Johnson, Brian Foote}, title = {Designin Reusable Classes}, journal = {Journal of Object-Oriented Programming SIGS}, year = {1988}, key = {Johnson:88}, volume = {1}, number = {5}, pages = {22-35}, month = jun }  @book{UML-F:00, author = {Fontoura M., Pree W., Rumpe B.}, editor = {Addison-Wesley}, title = {The UML Profile for Framework Architectures}, year = {2000}, }   The PDF is generated with a text like:  Something about something[4], and moo foo other children[1].   This citation [4] is the first in all text. How can I order this correctly to [1]...[2] ...? Restore open applications, documents, and window state When working on a project, I generally have a couple of Explorer windows open to relevant folders and a couple of applications (e.g. Eclipse and a form development tool) open to specific project-related documents.  There's a lot of value in that context, and I'd like to find a way to save a set of open apps under a project name, then restore that state at a later date.    Thus far all of the utilities I've found are virtual desktop products that don't save/restore state.  I'm specifically looking for something that will run the applications I had running, open the documents I had open, and position the windows as they were.  I'm amazed that such a seemingly simple concept as "Save my workspace state" doesn't seem to be available.  I'm on Windows 7.  Any suggestions? python multiprocessing on windows, if __name__ == "__main__" Trying to show that $\ln(x) = \lim_{n\to\infty} n(x^{1/n} -1)$ How do I show that $\ln(x) = \lim_{n\to\infty} n (x^{1/n} - 1)$?  I ran into this identity on . I haven't been able to find any proof online and my efforts to get from $\ln(x) := \int_1^x \frac{\mathrm dt}t$ to that limit have been a failure. My Ubuntu 12.04 was normally updated through the Update Manager, something about gtk. Then I rebooted and now all that shows after the splash screen is a blue screen that only shows the cursor. How To Create A 'Two-Sided' Unique Index On Two Fields? How can I efficiently create a unique index on two fields in a table like this: create table t (a integer, b integer);  where any unique combination of two different numbers cannot appear more than once on the same row in the table.  In order words if a row exists such that a=1 and b=2, another row cannot exist where a=2 and b=1 or a=1 and b=2.  In other words two numbers cannot appear together more than once in any order.  I have no idea what such a constraint is called, hence the 'two-sided unique index' name in the title.  Update: If I have a composite key on columns (a,b), and a row (1,2) exists in the database, it is possible to insert another row (2,1) without an error.  What I'm looking for is a way to prevent the same pair of numbers from being used more than once in any order... reconfiguration issues with dpkg I ran into some problems installing a package with apt. The terminal hung up during an install and I saw no other choice but to close the terminal. The problem, however, is that I can't seem to solve the issues that this has caused.   As is typical, I get:  E: Could not get lock /var/lib/dpkg/lock - open (11: Resource temporarily unavailable) E: Unable to lock the administration directory (/var/lib/dpkg/), is another process using it?   This gets me nowhere and I'm locked out until I run sudo dpkg --configure -a. Makes sense. However, this process spits the following out and then hangs on the last line:  Setting up virtualbox-dkms (5.2.10-dfsg-6) ... Removing old virtualbox-5.2.10 DKMS files...  -------- Uninstall Beginning -------- Module:  virtualbox Version: 5.2.10 Kernel:  4.15.0-23-generic (x86_64) -------------------------------------  Status: This module version was INACTIVE for this kernel. depmod...  DKMS: uninstall completed.  ------------------------------ Deleting module version: 5.2.10 completely from the DKMS tree. ------------------------------ Done. Loading new virtualbox-5.2.10 DKMS files... Building for 4.15.0-23-generic Building initial module for 4.15.0-23-generic   A typical solution I've read is that you need to fix the broken packages. When I try to fix the broken packages (sudo apt install -f), the system also hangs on the same line, although the output is different:  Reading package lists... Done Building dependency tree        Reading state information... Done 0 upgraded, 0 newly installed, 0 to remove and 0 not upgraded. 3 not fully installed or removed. After this operation, 0 B of additional disk space will be used. Setting up virtualbox-dkms (5.2.10-dfsg-6) ... Removing old virtualbox-5.2.1  0 DKMS files...  ------------------------------ Deleting module version: 5.2.10 completely from the DKMS tree. ------------------------------ Done. Loading new virtualbox-5.2.10 DKMS files... Building for 4.15.0-23-generic Building initial module for 4.15.0-23-generic   I am unable to get out of it unless I kill the process. This I do by checking what is currently running related to apt (ps aux |grep apt), and then I kill the relevant processes with sudo kill -9 process_number_here.  If I try to run things after that, it tells me I still need to reconfigure dpkg.  I tried looking around for more solutions to dpkg and all I really see is the nuclear option:  sudo rm /var/lib/dpkg/lock sudo rm /var/cache/apt/archives/lock sudo rm /var/lib/apt/lists/lock   I also delete the files in /var/lib/dpkg/updates because when I did some digging one of the files inside had the exact same text as the above long output... but I fully admit to being out of my league here and I don't know what to do.   Taking a look at the error messages, I recognize that it hangs on building something related to the kernel. But I have no idea where to even start solving that. Any help is appreciated. My system is ubuntu 18.04 How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort? I will keep it short and take only an extract (most important part) of   the old task.      $$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$   What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /  If anyone wants see, here is my long solution which I'm not happy with:  $$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$  $$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$  $$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$  $$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$ Are Goodberries affected by the Life domain class features? STI vs STD vs Sexually Transmitted Virus Reference to undeclared entity exception while working with XML I am trying to set the innerxml of a xmldoc but get the exception: Reference to undeclared entity  XmlDocument xmldoc = new XmlDocument(); string text = "Hello, I am text &amp;alpha; &amp;nbsp; &amp;ndash; &amp;mdash;" xmldoc.InnerXml = "&lt;p&gt;" + text + "&lt;/p&gt;";   This throws the exception:      Reference to undeclared entity 'alpha'. Line 2, position 2..    How would I go about solving this problem? What happened to Startups SE? I really enjoyed that site. I've only posted on it once, but I enjoyed seeing other people's posts and appreciated reading about entrepreneurial dilemmas.