I've recently seen latest mac os is having invisible scroll bars which will be visible only once your mouse comes onto it. I think it is a bit tough for the user to know if there is a page overflow or not. Is it not a bad User experience.
The scroll bar is hidden on MacBook if the user doesn't move the mouse for some time.  This pattern is also used in other devices too, for example Android devices.  The scroll bar does not take much place but it informs a user of where on page he is.  Is it a good approach hiding the scroll bar when the user is presumed viewing the content?
Antisymmetric Relations Let $a,b,c\in \mathbb{Z}$. If $ac&gt;bc$ and $c&gt;0$, then $a&gt;b$   I think I would use contrapositive to prove this. But I am not sure proper contrapositive statement.   Is it If $a&lt;b$, then $ac&lt;bc$, and $c&lt;0$.? If so, contrapositive is not a good choice, isn't it? Is this correct?     The person with whom I'm doing the project should be here soon.   If it is, is with always a dative preposition (like mit in German)? xslt processing on select attribute on apply-templates What is the difference between conditioning on regressors vs. treating them as fixed? What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$  Here is my attempt:    Letting    $$f(z)=\frac{1}{az^2+bz+c}$$  The poles of $f(z)$ are located at   $$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$  and  $$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$  Then  $$ b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)= \lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}= \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b} $$  Using L'Hopital's rule. Continuing, we have the limit is  $$ \lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}= \frac{\pi\cot (\pi z_0)}{2az_0+b} $$  For $z_0 \ne 0$  Similarly, we find  $$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$  Then   $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\ -\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)=  -\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)= \frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}= \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}} $$  Then we have  $$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$    Is this correct?  I feel like I made a mistake somewhere.  Could someone correct me?  Is there an easier way to evaluate this sum? Hey all I downloaded this file () but can't seem to be able to uninstall it. I tried looking online as well as the README file but there is only instruction for installing, not uninstalling the package.   Attempted solutions:  sudo apt-get remove laptop_mode   returned   E: Unable to locate package laptop_mode   but when I type in laptop_mode I do get   Unhandled kernel version: 3.13 ('uname -r' = '3.13.0-32-generic')   Is there any way I can uninstall laptop-mode-tools? How align the qedsymbol after of a big formula in display mode? Description of the problem My problem is this. For the proof of a math theorem i use the proof environment of the package amsthm. It adds a \qedsymbol in the last line of the proof, but if the proof ends with a displayed formula the qedsymbol appears in the next line leaving an unwanted blank space. The documentation of  teaches to fix this problem placing the command \qedhere after of the displayed formula but inside of the appropriate environment. This works well if the formula is of one line like \[ax^2+bx+c=0.\qedhere \]  but if the proof ends in a matrix, or cases with a &quot;great height&quot; the qedsymbol is fixed in the baseline at right and it this looks higher than expected (for my taste). The result is showed in this MWE \documentclass{article} \usepackage[width=7cm]{geometry} % Only for reducing space in this MWE  \usepackage{amsmath} % For `bmatrix` \usepackage{amsthm} % For `proof` environment \begin{document} \begin{proof} And this proof ends with \[ \begin{bmatrix}     1 &amp; 0 \\     0 &amp; 1 \\     0 &amp; 0 \end{bmatrix}.\qedhere \] \end{proof} \noindent Text \end{document}  that produces  What I want I'd like something like  with the qedsymbol aligned with the bottom of the matrix. Is this possible? How to influence the position of float environments like figure and table in LaTeX? How to deal with very long lines that do not contain spaces? What do the different colors mean in ls? Suppose $f$ is continuous on $[a,b]$ and differentiable on (a,b). Does it follow that $f'$ is continuous on $(a,b)$? What is the volatile keyword useful for? At work today, I came across the volatile keyword in Java. Not being very familiar with it, I found . Given the detail in which that article explains the keyword in question, do you ever use it or could you ever see a case in which you could use this keyword in the correct manner? Debugging with command-line parameters in Visual Studio I'm developing a C++ command-line application in Visual Studio and need to debug it with command-line arguments. At the moment I just run the generated EXE file with the arguments I need (like this program.exe -file.txt) , but this way I can't debug. Is there somewhere I can specify the arguments for debugging? How to construct an example of a probability distribution for which $\mathbb{E}\left(\frac{1}{X}\right)=\frac{1}{\mathbb{E}(X)}$ holds, assuming $\mathbb{P}(X\ne0)=1$?    The inequality which follows from Jensen's inequality for a positive-valued RV $X$ is like $\mathbb{E}\left(\frac{1}{X}\right)\ge\frac{1}{\mathbb{E}(X)}$ (the reverse inequality if $X&lt;0$). This is because the mapping $x\mapsto\frac{1}{x}$ is convex for $x&gt;0$ and concave for $x&lt;0$. Following the equality condition in Jensen's inequality, I guess the distribution has to be degenerate for the required equality to hold. A trivial case where the equality holds is of course if $X=1$ a.e. Here is an example I found in a problem book: Consider a discrete random variable $X$ such that $\mathbb{P}(X=-1)=\frac{1}{9}, \mathbb{P}(X=\frac{1}{2})=\mathbb{P}(X=2)=\frac{4}{9}$. It is then easily verified that $\mathbb{E}\left(\frac{1}{X}\right)=\frac{1}{\mathbb{E}(X)}=1$.   This example shows that $X$ need not be positive (or negative) a.e. for the equality in the title to hold. The distribution here is not degenerate either.  How do I construct an example, possibly like the one I found in the book? Is there any motivation? “...than I am” vs. “...than me” I read an English grammar article in which the author talked about the correct use of pronouns.  He writes:     If the extended sentence is “She’s more likely to ask him than I am,” in which the comparison is between the subject and the writer, I is correct. However, if the intent is to convey that the man referred to as him is more likely to be asked something by the subject than the writer is, the correct pronoun form is me, but that distinction should be clarified with a revision like “She’s more likely to ask him than ask me.”   Before reading this piece, I thought that the two sentences had the same meaning. While the second example is quite clear for me (“She’s more likely to ask him than ask me”), I don't think that I have correctly understood the first one. Could anyone explain it to me, please? Can wizards create their own spells? In the Harry Potter series, are wizards capable of creating their own spells, or are they limited to the spells that they're taught/learn?  I seem to remember something about Snape creating sectumsempra, but I'm not sure if that really is the case. I'm looking for a term for audio in form of the word visual.  Visual is defined as      of or relating to the sense of sight   What could you call the sense of hearing? Also, what do you call this form of a word? (i.e verb, noun...)  It should fit the x in the following sentence:     Is it possible to fall in love with someone without visual or x contact?   I'm not a native English speaker, so I couldn't find the terms for searching this. Number Theory: Prove that $m | \phi(a^m - 1)$ How was Thor able to dent Captain America's Shield? I was reading about Adamantium on the      Proto-Adamantium   This is the original Adamantium created by Dr. MacLain. Proto-Adamantium is the official name for the metal that makes up Captain America's Shield. It has never been recreated and is the absolute most invulnerable item in all of existence.    This lead me to read about      The true shield was once slightly dented by an enraged Odinforce empowered Thor, using Mjolnir, but he later reconciled and pounded out the small dent.    Was the dent caused by Odinforce or Thor's Hammer? Is Mjolnir or Captain America's Shield "the absolute most invulnerable item in all of existence"?