Why is the time in ubuntu different from that in windows?
Clock time is off on dual boot
I am currently developing a website which is going to take some time to finish. I need to move some 'stuff' to production frequently, while maintaining a under construction status. My front page will clearly mention that my website is under construction.  I know crawlers will access my URL before I am ready to go live. Should I tell them my site is under construction (no need to index)? If yes how?  P.S.: Eventually, I am ok to let them see my under construction front page. What happens when you target a "magical effect" with Dispel Magic? Generics and inheritance: need a complex Map instance There are many addons. Do you know how show it in two cloumns? How does the basic RS Flip-Flop (also known as latch) works? Why do some AC adapters and power supplies generate a whining noise, and what can I do about it? Change font size of labels in figures to any font size I want to change the font size of the label of a figure environment to any font size. It's possible to use the package caption with the argument font=Large for example. But I found no way to choose any font size, e.g. fontsize=28pt.  Here an example code:  \documentclass{standalone} \usepackage{anyfontsize}  \usepackage[font=Large]{caption} %\usepackage[fontsize=28pt]{caption} % not implemented   \begin{document} \parbox[t]{20cm}{  \fontsize{28}{30} \selectfont  \begin{figure}     %\includegraphics{figure.eps}     \caption{\label{fig_req_osnr} \fontsize{28}{30} \selectfont My caption.} \end{figure}  }  \end{document}     Do you have any ideas? Why does $\lim_{n\to\infty} z_n=A$ imply $\lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A$? I'm self-studying a bit of complex analysis, and I'm attempting to figure out the following.   Suppose $\lim_{n\to\infty}z_n=A$. How can I show that $$ \lim_{n\to\infty}\frac{1}{n}(z_1+\cdots+z_n)=A. $$  Is there a clever way to write the limit to make it more approachable? Thank you. natbib and hyperref create by default two links to the bibentry in the reference section|chapter of the document:   \begin{filecontents}{\jobname.bib} @book{author08, title = {{A Title}}, publisher = {Alpha}, year = {2008}, author = {Author, A}, address = {London} } @book{buthor90, title = {{B Title}}, publisher = {Bravo}, year = {1990}, editor = {Buthor, B}, address = {New York} } \end{filecontents}  \documentclass{article}  \usepackage{natbib} \usepackage{hyperref} \hypersetup{colorlinks,citecolor=red}  \begin{document}  \noindent citet: \citet{author08}, \citet[see][p. 20]{author08} \\ citep: \citep{author08}, \citep[see][p. 20]{author08}  \bibliographystyle{plainnat} \bibliography{\jobname}  \end{document}     In order to avoid two links to the bibliography (one for the [author] and one for the year [year], respectively), I took  to an older question and added it to my preamble.  \usepackage{etoolbox} \makeatletter  \pretocmd{\NAT@citex}{%   \let\NAT@hyper@\NAT@hyper@citex   \def\NAT@postnote{#2}%   \setcounter{NAT@total@cites}{0}%   \setcounter{NAT@count@cites}{0}%   \forcsvlist{\stepcounter{NAT@total@cites}\@gobble}{#3}}{}{} \newcounter{NAT@total@cites} \newcounter{NAT@count@cites} \def\NAT@postnote{}  % include postnote and \citet closing bracket in hyperlink \def\NAT@hyper@citex#1{%   \stepcounter{NAT@count@cites}%   \hyper@natlinkstart{\@citeb\@extra@b@citeb}#1%   \ifnumequal{\value{NAT@count@cites}}{\value{NAT@total@cites}}     {\ifNAT@swa\else\if*\NAT@postnote*\else%      \NAT@cmt\NAT@postnote\global\def\NAT@postnote{}\fi\fi}{}%   \ifNAT@swa\else\if\relax\NAT@date\relax   \else\NAT@@close\global\let\NAT@nm\@empty\fi\fi% avoid compact citations   \hyper@natlinkend} \renewcommand\hyper@natlinkbreak[2]{#1}  % avoid extraneous postnotes, closing brackets \patchcmd{\NAT@citex}   {\ifNAT@swa\else\if*#2*\else\NAT@cmt#2\fi    \if\relax\NAT@date\relax\else\NAT@@close\fi\fi}{}{}{} \patchcmd{\NAT@citex}   {\if\relax\NAT@date\relax\NAT@def@citea\else\NAT@def@citea@close\fi}   {\if\relax\NAT@date\relax\NAT@def@citea\else\NAT@def@citea@space\fi}{}{}  \makeatother     Which gets me almost where I want to be, but leaves me with two issues I would like to be solved:   while \citet includes pre- and postnotes in the link, \citep doesn't. How can \citep be changed to include postnotes? is there a way to color the brackets and commas that are typeset in black in the unmodified first example also in black in the modified second example without removing them from the link? For each $y \in \mathbb{R}$ either no $x$ with $f(x) = y$ or two such values of $x$. Show that $f$ is discontinuous. This is from Apostol's Mathematical Analysis.  Let $f$ be a function defined in $[0, 1]$ such that for each real number $y$ either there is no value of $x \in [0, 1]$ such that $f(x) = y$ or there are exactly two such values of $x$. Prove that a) $f$ must be discontinuous on $[0, 1]$. b) $f$ has an infinite number of discontinuities in $[0, 1]$.  Long back I had tried to solve the problem. I reasoned for a) as follows: Suppose that $f$ is continuous and let $m, M$ be its minimum and maximum values on $[0, 1]$. Clearly if $m = M$ then $f$ is constant and clearly for every $x \in [0, 1]$ we have $f(x) = m = M$ which contradicts given property of $f$. So we must have $m &lt; M$. Clearly there are now four distinct points $a, b, c, d$ in $[0, 1]$ such that $f(a) = f(b) = m, f(c) = f(d) = M$. By checking possible linear orderings of $a, b, c, d$ I could find a find a value of $y \in (m, M)$ for which there exist more than 2 values of $x$ such that $f(x) = y$. However this approach looks very clumsy and I suspect much simpler and easier proof is possible. Part b) seemed bit complex and my only guess was that probably $f$ was discontinuous on each subinterval of $[0, 1]$ but I could not prove this claim (may be the claim is wrong!). Perhaps there is a simpler solution to this part of the problem. Apostol's also asks to provide an example of such a function. Any hints on the overall problem will be appreciated.  Update: After looking at Marc's answer, I think I have found a way to show that there are infinitely many discontinuities. Earlier the solution was a part of the question itself but it made the question look quite lengthy hence I moved it to a separate answer. Which of those functions are PRFs? How much does it cost to have a custom ASIC made? SSMS wont give up file associations What is the purpose of a Windows Workgroup? In Microsoft Windows, what is the purpose of a Workgroup? I understand that a Domain allows the administration of security (user names, passwords, file access, etc.) to be centralized. But what does a Workgroup give you? In Hitchcock's Vertigo, there is one part of the film that has never made sense to me and I believe that it was a cheap ploy to mislead the audience until the ultimate reveal (I won't spoil it here).    In this scene Jimmy Stewart watches Kim Novak enter a rooming house and afterwards sees her standing by the window on the 2nd floor.  When he enters the house and talks to the inn-keeper, she tells him that Kim Novaks's character positively never came in and even shows him her room key that is still behind the front desk.  They go up to the room and find it empty.   This scene seems to imply that the Kim Novak character may have had a ghost like quality. But given the reveal at the end she clearly does not and there was no way for her to enter the room without a key and without being seen by the inn-keeper.  If this was done by some other director I probably wouldn't have given this much thought but Hitchcock was a perfectionist (especially for this film) so I never understood the significance of this scene other than to mislead the audience (with a cheap ploy IMO).  Does anyone who is familiar with this film have a different explanation/perspective about this scene in an otherwise brilliant movie. If sup A $\lt$ sup B show that an element of $B$ is an upper bound of $A$ (a) If sup A &lt; sup B, show that there exists an element of $b \in B$ that is an upper bound for $A$.  I have argued that if sup A $\lt$ sup B, then choose an $\epsilon&gt;0$ such that sup A +$\epsilon \in B$. Since $a \le $ sup A for all $a \in A$, it follows $a \lt $ sup A + $\epsilon$ hence sup A + $\epsilon $ is an upper bound of A as well as an element of B.  (b) Give an example to show that this is not always the case if we only assume sup A $\le$ sup B.  I am having trouble extending my argument into an example which makes me think that it may not be correct.   Question: Is my argument for (a) correct? If not how would one show this fact? Can you give an example asked for in (b)? Thanks in advance.  I need to understand how I am assuming what I am supposed to prove in my approach. Can someone elaborate on that please? The other question doesn't really address this issue. Database synchronization between dev/staging and production How to understand dot product is the angle's cosine? How can one see that a dot product gives the angle's cosine between two vectors. (assuming they are normalized)  Thinking about how to prove this in the most intuitive way resulted in proving a trigonometric identity: $\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$.  But even after proving this successfully, the connection between and cosine and dot product does not immediately stick out and instead I rely on remembering that this is valid while taking comfort in the fact that I've seen the proof in the past.  My questions are:   How do you see this connection? How do you extend the notion of dot product vs. angle to higher dimensions - 4 and higher? Update: this is a lengthy post, you can     Note: The quick language list refers to the list in the screenshot below, accessible throught the shortcut Win + Space bar  I have two languages for windows spell check in my system, English (United States) and French (Switzerland), but despite both languages being properly configured with "Swiss French" input keyboard, US Keyboard input keeps appearing on this list:    Here's the settings for English (United States) language, you can see the only input is "Swiss French":    Same for Français (Suisse), the only input is "Swiss French":    In the Control Panel, both are also set as Swiss French:    All my settings are set to use "language list":     Here's a few things that I already tried:   I already managed to remove it by some language configuration gymnastics, by adding and removing languages, it always eventually comes back I have already done this:  There's no mention of US Keyboard in my registry settings either:     Please help, I'm desperate, this is my third computer with Windows 10 and they all do the same thing. How do I prevent directories mounted with 'bind' from appearing on 'Devices' on nautilus? I have these lines in the fstab   # binds /media/DataNtfs/Music           /home/can/Music                  none    rw,bind  /media/DataNtfs/Pictures        /home/can/Pictures               none    rw,bind  /media/DataNtfs/Downloads       /home/can/Downloads              none    rw,bind  /media/DataNtfs/Documents       /home/can/Documents              none    rw,bind  /media/DataNtfs/Backups         /home/can/Backups                none    rw,bind  /media/DataNtfs/Notes           /home/can/Notes                  none    rw,bind  /media/DataNtfs/Other           /home/can/Other                  none    rw,bind  /media/DataNtfs/Packages        /home/can/Packages               none    rw,bind  /media/DataNtfs/Photos          /home/can/Photos                 none    rw,bind  /media/DataNtfs/Videos          /home/can/Videos                 none    rw,bind  /media/DataNtfs/WorkSpace       /home/can/WorkSpace              none    rw,bind