Have the bar at the top move down with you On some sites the bar at the top will move down with you when you move down the page. I think that something like this might look nice on Stack Exchange.  What I mean is for this bar:    To move down the page with you as you move down the page, and up when you move up the page so that you can always access it however far down the page you are. Is this something that SE might implement?
Should the top navigation be frozen (optional)? In Excel, the window can be frozen in place in such a way that you can always see the name of the column even though you go way down below the 'fold' of the screen.  On SE, when you scroll, the whole page goes up (or down). This means the top navigation bar is scrolled off screen. I'm not sure if it would be a good idea or not to fix the top navigation to the top of the screen and let the rest of the content scroll.  What do you think about this, in terms of pros and cons?  [Update]  With the new Top navigation bar, kudos to the design team, I think this question becomes relevant again. I'm aware of the busload of downvotes, but still.  The top navigation bar doesn't occupy much space and has two notification areas:   Inbox Achievements   Having these two notifications visible at all times would be nice, IMO.  [Update]  I've just noticed that I haven't used the word "optional" in this discussion... My bad. In Excel, the Freeze function is controlled by the user. You can optionally Freeze part of the interface, and you can eliminate the Freeze at will.
Problem:  Start page numbers from chapter one and also reflect this in the TOC.  Currently the code for TOC is:  \renewcommand{\contentsname}{Table of contents}  \setcounter{secnumdepth}{2} \setcounter{tocdepth}{2} \tableofcontents   I would appreciate if I can be pointed in the right direction. So far I have understood that you can use \chapter* in order not to include a chapter in the TOC.  Questions:  1) How would you force Latex to start numbering from chapter 1 and forward?  2) How can you start page numbering from the first chapter? I've seen many of the solutions on the topic, all posted in this site. But I haven't seen any of them mentioning Apple's adapter,   I have that, which was used to connect to my old MacBook Pro 2013.   Now, I'm assuming that all I need is to connect the Mini DisplayPort to an adapter, such as  (as mentioned in one of the posts) and the USB cable to a USB-C adaptor.   Is this correct? I noticed with the holiday patch that there are new trees and shrubs that spawn on my land.  One popped up today, being specifically an X-Mas tree - which costs 25,000 gold to remove.   Is there any benefit to removing this X-Mas tree?  Is there any benefit to leaving it there?    It is noticeably the most expensive obstacle to remove yet. Relationship between SVD and PCA. How to use SVD to perform PCA? Solving ODE with negative expansion power series I am solving a series of ODE, such that each DE is equal to some degree of term that I'm expanding to. For instance, one DE is this:  $\xi^r\partial_r g_{rr}+2g_{tt}\partial_t\xi^t=\mathcal{O}(r)$  $g_{ii}$ are given, from metric, but that's not important. I need to assume that the solution (since I'm looking for components of $\xi^\mu$, which is a vector with components $\xi^t,\xi^r,\xi^\phi$) is given with power series of the form:  $\xi^\mu=\sum\limits_{n}\xi^\mu_n(t,\phi)r^n$, and this is to be seen as expansion around $1/r$ (expansion around $r=\infty$).  Now when I plug this in the ODE I get this  $\frac{2}{l^2}\sum_n\xi^r_nr^{n+1}+2\sum_n\xi^t_{n,t} r^n+\frac{2}{l^2}\sum_n\xi^t_{n,t}r^{n+2}=\mathcal{O}(r)$, where   $\xi^\mu_{n,i}$   is the derivative with the respect to i-th component.   What troubles me is, how to expand this? Do I set n=0,-1,-2,... until my $\mathcal{O}(r)$ terms cancel each other out? Or?  I'm kinda stuck :\ Badge overflow on users with too many badges When a user has &quot;too much&quot; reputation and &quot;too many&quot; badges, the design of the user card breaks:    The badge counters overflowing to another line. Can anything be done about this? For example, expanding the rep and badges to their own lines:  NameRepBadges How do I install Wi-Fi drivers for Realtek RTL8723DE device in Ubuntu 16.04?  lspci -v | grep -i network 02:00.0 Network controller: Realtek Semiconductor Co., Ltd. Device d723 Finding Radii of Convergence for $\sum a_n z^{2n}$ and $\sum a_n^2 z^n$ Setting: Let $\sum a_n z^n$ have radius of convergence $R$.  We have that  $$ R = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} $$  via Hadamard's formula for the radius of convergence.  Question: What are the radii of convergence for (i) $\sum a_n z^{2n}$ and (ii) $\sum a_n^2 z^n$?  Attempt for $\sum a_n^2 z^n$:   For ease of notation, let $R_1$ and $R_2$ denote the radii of convergence for power series $\sum a_n z^{2n}$ and $\sum a_n^2 z^n$ respectively. From Hadamard's formula, we then have  $$ R_2 = \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n^2 \right|^{1/n}} = \left( \frac{1}{\underset{n \rightarrow \infty}{\limsup} \left|a_n \right|^{1/n}} \right)^2 = R^2 $$   But what about $R_1$? New passport and entrance Schengen I am from Brazil and travel a lot in Schengen countries. Under the rules of Schengen visa (3 months stay in a 6 months period). In my passport are the entrance and exit stamps, also to chec the days I stay in Schengen.  I need a new passport and will get this in Brazil. So in this new passport are NOT the stamps. Does this mean I start new with the Schengen visa rule? I seems like. The new passport is empty. Or do I miss something? It would be great because I am almost on my 90 days stay in the period of 180 days. Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$  I have tried and it gets confusing. Coconut Water from White vs Brown Coconuts What are "schlieren" in English? The German word for these patterns is "Schliere, pl. Schlieren", and apparently this word has entered the English language as a loanword (cf. article  in Wikipedia or entry  in OED). But the German loanword is probably not something most native speakers of English understand without the help of a dictionary, so what would you call, or how would you describe, the following patterns?  Concrete:    Air:    Oil on water:    Coffee and milk:    Paint:    Smoke:    Rock:    Marbled paper:    And so on.    "Waves" is not applicable, because waves are vertical undulations of the surface, while these are a pattern on, or inside, the surface level, with no amplitude.  "Striæ" does not really fit either, because those refer more to the parallel layers, than to the curly forms. Take the following two lines of code:  for (int i = 0; i &lt; some_vector.size(); i++) {     //do stuff }   And this:  for (some_iterator = some_vector.begin(); some_iterator != some_vector.end();     some_iterator++) {     //do stuff }   I'm told that the second way is preferred. Why exactly is this? Review counts in top-bar and /review don't match So far, I thought that the number shown in the orange box in the top-bar was the number of items available for review. However, when I acually visit /review, the numbers don't match:    Note that top-bar shows 15 items (?) but the queues only add up to 8.  Since I can't think of a reason for this being by design (which, as experience shows, is not a good indicator for "this is not an SE feature) I'm tentatively labelling this a (UI) bug. Prove that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 10^3$. Prove that there are no positive integers $x$ and $y$ such that $x^3 + y^3 = 10^3$.   This is a homework question, and I understand that its part of Fermat's Last Theorem, but when I looked that up to try to figure out the homework, I realized that it is way further than what we've learned in class so far. We have to prove this, but the only way I can think to do it is by exhaustion, which would be extremely lengthy. Any little shove in the right direction would be appreciated. Thanks! zero matrix to the power of 0 Why $0^0=I$?   I'd tried prove that considering $N^0$ where N is a Nilpotent matrix and then using the Cayley -Hamilton theorem  Thanks in advance. Intuition for Rayleigh PDF We have ground-truth data $\mathbf{x}^* = (0,0)^T \in \mathbb{R}^{2}$. Furthermore, we have $N$ measurements $\mathbf{x}_i \in \mathbb{R}^{2}, i\in \{1,\ldots,N\}$. We calculate the $N$ error vectors $\mathbf{e}_i = \mathbf{x}_i - \mathbf{x}^*$. We assume that the two components of $\mathbf{e}_i = (e_{x,i}, e_{y,i})^T $ are independent of each other and normally distributed with equal mean $\mathbf{\mu} = (0,0)^T$ and variance $\sigma^2$. The absolute errors $e_i = \sqrt {e_{x,i}^2 + {e}_{y,i}^2} $ follow a Rayleigh distribution, is that correct?  The probability density function (PDF) of the Rayleigh function is zero at a value of zero. Why is this the case (intuitively)? I would assume that our errors should most likely be 0 (i.e. PDF should be highest at 0)... Downgrade SQL Server Enterprise Edition to SQL Server Standard Edition Schengen visa application rejected because my US visa stamp expires in December. What are my options? I went to the Italian consulate in Washington, DC today and they would not accept my visa application because they said my US visa stamp cannot expire less than 3 months after the end of my trip to Italy. Here's my relevant visa and trip info:   US visa stamp expires on 16 Dec 2015 I-797 valid until 3 Mar 2017 Trip to Italy is from 15 Oct to 25 Oct 2015   Is it true that my visa stamp needs to be valid for 3 months after my trip, even though I can legally stay in the US until 2017? Does anyone have any suggestions for getting a Schengen visa in my situation? Killing form on $\mathfrak{sp}(2n)$ I have the same question as  from a long time ago. Is there an easy way to see that the Killing form on $\mathfrak{sp}(2n)$ is $\kappa(x,y) = (4n+2) \mathrm{tr}(xy)$?  For example, the Killing form for $\mathfrak{sl}(n)$ can be found from the easier to calculate case of $\mathfrak{gl}(n)$ using the decomposition $$\mathfrak{gl}(n) = \mathfrak{sl}(n) \oplus \mathbb{C}.$$ I don't know a similar argument for $\mathfrak{sp}(2n)$ though.