Today the commands in the terminal start with "public-docking-hg-2434:~ userName$"  What is this public-docking-hg? Why did it suddenly appear?
My computer name in System Preferences &gt; Sharing is set to &quot;archos&quot;, but is showing as &quot;iphone&quot; on Terminal. It just started doing this after I loaded Xcode for doing iPhone development: Last login: Mon Nov  7 14:46:55 on ttys001 iphone:~ travis$   Any ideas what could be causing this?
Previous posts                  As of today, the Moderator Flag Dashboard is officially graduated and the old version sent to the smelter's yard.  Based on stats from the past month, virtually all moderators were already on the new UI (thank you) so you hopefully won't notice any change.  That said, I would like to highlight the handful of improvements we made over the last month and a half based on your feedback:   A more responsive navigation bar at the top of moderator pages, improving the mobile experience. Similarly, , increasing room on the screen for the first question. . .  Not only do the headers look better, they now function better.  Now you scroll to and highlight the post/comment each time you click the link, not just the first time. .  Posts by the same user are grouped together and ordered newest first. .  We eliminated the VLQ (auto) side effect so now "Resolve all flags (auto)" doesn't do anything surprising.  If you still want to leave a comment or decline, you can do so with the buttons next to the individual flags.   Bugs and Feature Requests  Since this project is officially launched, all bugs and feature requests should be done in the form of questions tagged  on MSE, not answers on this post.  Feel free to post an answer here if you have specific concerns about the release process. Show that a positive operator on a complex Hilbert space is self-adjoint Let $(\mathcal{H}, (\cdot, \cdot))$ be a complex Hilbert space, and $A :  \mathcal{H} \to \mathcal{H}$ a positive, bounded operator ($A$ being positive means $(Ax,x) \ge 0$ for all $x \in \mathcal{H}$). Prove that $A$ is self-adjoint. That is, prove that $(Ax,y) = (x, Ay)$ for all $x,y \in \mathcal{H}$. Here's what I have so far. Because $A$ is positive we have $\mathbb{R} \ni (Ax,x) = \overline{(x,Ax)} = (x,Ax)$, all $x \in \mathcal{H}$. Next, I have seen some hints that tell me to apply the polarization identity: $$(x,y) = \frac{1}{4}((\lVert x+y \rVert^2 + \lVert x-y \rVert^2) - i(\lVert x + iy \rVert^2 - \lVert x - iy \rVert^2)),$$ where of course the norm is defined by $\lVert \cdot \rVert^2 = (\cdot, \cdot)$. So my guess is that I need to start with the expressions: $$(Ax,y) = \frac{1}{4}((\lVert Ax+y \rVert^2 + \lVert Ax-y \rVert^2) - i(\lVert Ax + iy \rVert^2 - \lVert Ax - iy \rVert^2)),$$ $$(x,Ay) = \frac{1}{4}((\lVert x+Ay \rVert^2 + \lVert x-Ay \rVert^2) - i(\lVert x + iAy \rVert^2 - \lVert x - iAy \rVert^2)),$$ and somehow show they are equal. But here is where I have gotten stuck. Hints or solutions are greatly appreciated. Horizontal enumeration in multiple columns Consider the following minimal example:  \documentclass{article}  \setlength{\parindent}{0mm}  \usepackage{paralist} \usepackage{tabto}  \begin{document}  \NumTabs{3} \begin{inparaenum} \item text \tab\item text \tab\item text \tab\item text \tab\item text \tab\item text    \end{inparaenum}  \end{document}   Which results in:    It does exactly what I want, but I was wondering whether it was possible to create an environment, say tabbedEnum so that we can eliminate the use of the \tab, in particular, that the first case doesn't need a tab but all others do. Something of the form:   \begin{tabbedEnum}[3] \item text \item text \item text \item text \item text \item text \end{tabbedEnum} Lost my old passport for Schengen visa application Image of concave mirror when object is farther than the focal point I have a set of concave/convex mirrors/lenses, and I have drawn out all of the ray tracing diagrams for each possible combination and type of image each optical element can produce. When I look at the ray tracing diagrams then reproduce the situation in the real world with myself as the &quot;object&quot;, all the scenarios make sense to me except one: when I as the object stand beyond the focal point of a concave mirror.  The image ends up as &quot;real&quot; and appears to be in front of the mirror. Now I understand that in the situation above, I as the object would stand at 6m and this would result in the image appearing at 12m. So if I placed a wall or screen at the 12m mark, the image would form on that surface (just like a movie theater screen).... at least I think I interpreted that correctly. But I still see myself when I look in the mirror?! If the image is supposed to form and come into focus behind me, what in the world am I seeing when looking at the surface of the mirror? I still see myself, just inverted. The surface of the concave mirror still has a reflection on it. It doesn't look quite like a normal flat mirror, in that the image doesn't really look like it's &quot;inside&quot; the mirror, more like it's right on the surface of it. It's really frustrating because every other possible arrangement of optical elements makes total sense to me, except this one. Any suggestions on how to think about it? Thank you for your time! I am trying to write a bash script for testing that takes a parameter and sends it through curl to web site.  I need to url encode the value to make sure that special characters are processed properly.  What is the best way to do this?    Here is my basic script so far:  #!/bin/bash host=${1:?'bad host'} value=$2 shift shift curl -v -d "param=${value}" http://${host}/somepath $@ "Pandora's Box" metaphors The majority of definitions give the same meaning - "Pandora's box" is a synonym for "a source of extensive but unforeseen troubles or problems."  Are there any other metaphors with the same meaning? Handle application exit What's the @ in front of a string in C#? Does anyone have an intuitive explanation of why this is the case? How to stop Gedit, Gvim, Vim, Nano from adding End-of-File newline char? Proof of $\lim_{n\to \infty} \sqrt[n]{n}=1$ Thomson et al. provide a proof that $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$ in . It has to do with using an inequality that relies on the binomial theorem:  I have an alternative proof that I know (from elsewhere) as follows.  Proof. \begin{align} \lim_{n\rightarrow \infty} \frac{ \log n}{n} = 0 \end{align} Then using this, I can instead prove: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} &amp;= \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \newline &amp; = \exp{0} \newline &amp; = 1 \end{align}  On the one hand, it seems like a valid proof to me. On the other hand, I know I should be careful with infinite sequences. The step I'm most unsure of is: \begin{align} \lim_{n\rightarrow \infty} \sqrt[n]{n} = \lim_{n\rightarrow \infty} \exp{\frac{ \log n}{n}} \end{align} I know such an identity would hold for bounded $n$ but I'm not sure I can use this identity when $n\rightarrow \infty$. Question: If I am correct, then would there be any cases where I would be wrong? Specifically, given any sequence $x_n$, can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n = \lim_{n\rightarrow \infty} \exp(\log x_n) \end{align} Or are there sequences that invalidate that identity?  (Edited to expand the last question) given any sequence $x_n$, can I always assume: \begin{align} \lim_{n\rightarrow \infty} x_n &amp;=  \exp(\log \lim_{n\rightarrow \infty} x_n) \newline &amp;=  \exp(\lim_{n\rightarrow \infty} \log x_n) \newline &amp;=  \lim_{n\rightarrow \infty} \exp( \log x_n) \end{align} Or are there sequences that invalidate any of the above identities? (Edited to repurpose this question). Please also feel free to add different proofs of $\lim_{n\rightarrow \infty} \sqrt[n]{n}=1$. What is the mechanism behind the slowdown of light/photons in a transparent medium? What's the difference between MathOverflow & Mathematics.SE? If I have a maths question, on which site should I ask:  or ? Can you attack with different weapons using Extra Attack? The fighter's  feature (PHB, p. 72) says:     Beginning at 5th level, you can attack twice, instead of once, whenever you take the Attack action on your turn.   Assuming I am holding two different weapons, and don't want to spend my bonus action on Two-Weapon Fighting, can I attack once with each weapon? How to create an empty file from command line Is there any reference where Captain America's shield is made from stolen Wakanda vibranium? Comma — a punctuation error? Consider:     This book provides a solid psychological understanding of these experiments, and adds a few expansions and conclusions of its own. It also provides …   My question regards the comma. I have been told that having the comma there is an error.  According to  this seems to be the case according to “some style guides”. The reason is that the stuff after the comma is not independent; it’s dependent on what comes before.  What do you think? It would seem to me the comma helps a lot, because those two “and”s without any comma make the sentence rather cluttered. That’s why I put the comma there. So I’m not sure. Prevent people from downloading / copying / making a copy of data in Google doc scope resolution operator without a scope