Differences between Reference and Pointers
What are the differences between a pointer variable and a reference variable in C++?
What Happens to electrons after Alpha Decay and Nuclear Fission? I can't review posts anymore—was the minimum reputation changed? Is it just for me, or has the privilege to review new posts and late answers changed for everyone?  I recently started being more active in reviewing, but the rep requirement just shot up and I can not review anymore? The rep requirement has shot up to 500. Is this just for me, as a 'process", or has it been changed site wide? Linearization of PDE: $0$ is an eigenvalue since all translates of travelling waves are also travelling waves Consider the following PDE: $$ u_t=u_{xx}+f(u)-w,~~~~~w_t=\varepsilon (u-\gamma w),~~~~~~~~~(1) $$ where $f(u)=u(u-a)(1-u), 0&lt;a&lt;\frac{1}{2}, \varepsilon,\gamma &gt;0, \varepsilon\ll 1,\gamma\ll 1$.  A travelling wave for (1) is a solution that is a function of the single variable $\xi=x-ct$, i.e., $(u(\xi),w(\xi))$ satisfies $$ -cu'=u''+f(u)-w,~~~~~~-cw'=\varepsilon (u-\gamma w)~~~~~~~('=d/d\xi).~~~~~~(2) $$  If (1) is recast in a moving coordinate frame, i.e., in terms of variables $\xi=x-ct$ and $t$, it becomes $$ u_t=u_{\xi\xi}+cu_{\xi}+f(u)-w,~~~~~~w_t=cw_{\xi}+\varepsilon (u-\gamma w).~~~~~~~~(3) $$  One can see by (2), that a travelling wave is an equilibrium solution of (3). So one nearby question is to ask whether a travelling wave is stable.    For a fixed $\varepsilon$, let $U_{\varepsilon}(\xi)=(u_{\varepsilon}(\xi),w_{\varepsilon}(\xi))$ be a travelling wave for (1), i.e., an equilibrium solution for (3).          A standard technique for determining stability of $U_{\varepsilon}(\xi)$ is to linearize: If the right-hand side of (3) is linearized along $U_{\varepsilon}(\xi)$, the resulting operator is     $$ L\begin{pmatrix}p\\r\end{pmatrix}=\begin{pmatrix}p_{\xi\xi}+cp_{\xi}+f'(u_{\varepsilon})p-r\\cr_{\xi}+\varepsilon (p-\gamma r)\end{pmatrix}, $$     where $\begin{pmatrix}p\\r\end{pmatrix}(\xi)\in\text{BC}(\mathbb{R},\mathbb{R}^2):=\left\{u\colon\mathbb{R}\to\mathbb{R}^2 | u\text{ is bounded and uniformly continuous}\right\}$.      (The linearized criterion for stability of the travelling wave is that the spectrum of $L$ (except for $0$) lies in a left half-plane and $0$ is a simple eigenvalue.)  Thats the background.  Now, there is a statement that I could not understand yet.          Note, that $0$ must be in the spectrum because the translate of a travelling wave is another travelling wave.      I know that a translate of a travelling wave is again a travelling wave. But why does this imply that $0$ is in the spectrum of $L$?  Edit  After some looks at definitions, I have an idea:  Consider a solution $U(\xi,t)$ that is a translate of the travelling wave $U_{\varepsilon}(\xi)=(u_{\varepsilon}(\xi),w_{\varepsilon}(\xi))$, i.e.  $$ U(\xi,t)=U_{\varepsilon}(\xi-k)=(u_{\varepsilon}(\xi-k),w_{\varepsilon}(\xi -k)) $$ for some $k$. Then $U$ is an equilibrium solution of (3), too.  If I am not totally wrong, then we have  $$ p=u_{\varepsilon}(\xi-k)-u_{\varepsilon}(\xi),~~~~r=w_{\varepsilon}(\xi-k)-w_{\varepsilon}(\xi) $$ and  $$ p_t=0,~~~w_t=0, $$ because travelling waves are equilibria for (3) and a translate of a travelling wave is a travelling wave.  Hence $$ L\begin{pmatrix}p\\r\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}, $$ meaning that $$ \text{ker}(L-0\text{ id })\neq\left\{\begin{pmatrix}0\\0\end{pmatrix}\right\}, $$ because for each translate of the travelling wave, $p,r\neq 0$ and  $\begin{pmatrix}p\\r\end{pmatrix}\in\text{ker}(L-0\text{ id})$.  Hence $0$ is in the spectrum (point spectrum) of $L$.  Is this what was meant? Opening a file from terminal When we want to open an application or file from the terminal, we type, say,  okular file.dvi   This opens the application, but also shows the status of the application. We cannot close the terminal, because it kills the process. Unfortunately, if you're trying to, for example, create a LaTeX file, you will need one tab for the text editor, one for the dvi file, and so on. And if you're trying to open all windows from the terminal, you can forget it. I'm trying use the terminal as much as possible, and while I have Yakuake, it is still a bother having so many tabs and seeing which of those have an application I've killed and so on.  So, is there a way to open an application/file from the terminal so that the status doesn't show and immediately gives the prompt so that we can use it to open more applications? Order of automorphism group of cyclic group Let $G$ be a cyclic group of order $m$. What is the order of $\text{Aut}(G)$?  I want to know the proof as well (elementary if possible). I would still accept the proof if one answers with $m = p$, a prime. Or on top of that, I would accept the answer with extra assumption: $q \equiv 1$ mod $p$ with another prime $p$. What is this "answer your own question" jazz? I showed up here to write a perfectly normal rant about misaligned pixels, and there's a big fat checkbox cluttering up the "Ask" page:    WHAT THE EEEEK IS THIS NEW ABOMINATION? Ireland Transit Visa when flying from IND - Abu Dhabi - Dublin - US Can I apply multiple background colors with CSS3? Visualforce Date formatting What exactly do quotation marks around the table name do? I thought that the quotation mark (") was simply a type of grouping marker but I'm debugging some NHibernate code and notice that while  SELECT * FROM site WHERE site_id = 3;   Works fine  SELECT * FROM "site" WHERE site_id = 3;   fails with a table or view does not exist error.  What gives? How tipe are this series?  $$\sum_{n=2}^\infty \frac 1 {n^2\ln n}$$ But $$\sum_{n=2}^\infty \frac 1 {n\ln n\ln(\ln n)}$$ I used Condensation Test, but I stack after that. Thanks! Let $A= a(i,j)$ and $B= b(i,j)$ be ($n\times n$) matrices that are positive definite such that $a(i,j) &lt; b(i,j)$ .Let $C= c(i,j)= a(i,j) - b(i,j)$, then $C$ is also positive definite. Why or why not?  what i know is $x^T(C)x= x^T (A-B) x = x^T(A)x - x^T(B)x$. Now both terms on right-hand side are greater than $0$ as $A$ and $B$ are positive definite. Also  $a(i,j) &lt; b(i,j)$. But in such case the sign of $x^T(C)x$ being positive or negative will also depend on the value of elements of vector $x$. So, we can't assure that $x^T(C)x &lt; 0$ always or that $C$ is not positive definite.  Am i right about this? After upgrading to 12.10 quantal, the packaged version of fglrx no longer works. I discovered that this is because there is a separate 'legacy' fglrx driver for the HD 2k-4k series cards, but it is incompatible with the xorg server on 12.10.  This is the most current version of the driver for HD 2000 through HD 4000 series cards. You can't use the non-legacy fglrx driver, but you can use the open-source radeon driver if you prefer your WM compositing to be laggy and your YouTube videos to play like they would on a Pentium MMX series:    Usually this driver can be installed in the following way, necessary because apt-get install fglrx would pull in the non-legacy driver:  wget http://www2.ati.com/drivers/legacy/amd-driver-installer-12.6-legacy-x86.x86_64.zip unzip amd-driver-installer-* sudo sh ./amd-driver-installer-*.run --buildpkg Ubuntu/quantal sudo dpkg -i fglrx*.deb sudo aticonfig --initial -f   If you use a different version of fglrx (for example, a newer 12.9 that doesn't support those cards) then the final command will give you an error no supported hardware detected or something similar. However, everything works at this point and you will get a reasonable xorg.conf:  ... other stuff  Section "Device"         Identifier  "aticonfig-Device[0]-0"         Driver      "fglrx"         BusID       "PCI:1:5:0" EndSection  ... other stuff   At this point you're supposed to reboot and everything will be working with the fglrx driver. However, upon rebooting, you'll be treated to the following errors in Xorg.0.log when fglrx attempts to load:  (EE) Failed to load /usr/lib/xorg/modules/drivers/fglrx_drv.so: /usr/lib/xorg/modules/drivers/fglrx_drv.so: undefined symbol: noXFree86DRIExtension   Some searching around will show that this is a problem with the legacy ATI drivers not supporting xserver 1.13 or newer. () ATI has released a fixed driver for its most recent (HD 5000 series or later) cards, but not for the 'legacy' cards yet. The non-legacy ATI drivers can't be used with the old cards.  What should an Ubuntu user, using one of these HD 2000-4000 series cards, do?    Wait for an updated 'legacy' ATI driver that properly works with xserver 1.13?   Downgrade back to 12.04 Precise, which uses xserver 1.11? Try to downgrade xserver on 12.10 Quantal to 1.12, which could possibly break Unity and GNOME? Forced upgrade to HD 5000 series or later card? (Not possible with integrated graphics...) Some other 1337 action that fixes this problem painlessly? Proof for $e^z = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$ Q1: Could someone provide a proof for this equation (please focus on this question):   $$e^z = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{z}{x} \right)^x$$  Q2: Is there any corelationn between above equation and the equation below (this would benefit me a lot because i at least know how to proove this one):  $$e = \lim \limits_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right)^x$$ Do I need an airport transit visa through Switzerland (Schengen) if I have no Schengen visa? Do I need an airport transit visa through Switzerland if I have no Schengen visa? Am traveling from Cairo to Moscow via Swiss Air What is the translation of what's written below?  I think it may be the poem of the One Ring, but I'm not sure. How to define binary operation on arbitrary set in order to create a group structure. Why are old initrd files of uninstalled kernels filling up /boot partition? When looking at most sites (including SO), most of them use:  &lt;input type="button" /&gt;   instead of:  &lt;button&gt;&lt;/button&gt;    What are the main differences between the two, if any? Are there valid reasons to use one instead of the other? Are there valid reasons to use combine them? Does using &lt;button&gt; come with compatibility issues, seeing it is not very widely used? listings package in XeLaTeX: every tab in lstlisting compiles to "^^I" in the pdf In beamer class, when i compile using LaTeX, the output pdf results perfectly the way i expect it to. But when i change it to XeLaTeX, in the lstlisting environment, every tab in the code results in a "^^I"-Placeholder (See output-pdf below). How do I fix this?   if( Bedingung ) {  ^^I//Code, der nur ausgefuehrt wird, wenn die Bedingung wahr ist   }  ^^I^^I   LaTeX-Code (Reduced to the minimum, one frame as an example):  \begin{document}     \begin{frame}[fragile]{if-Statements}         \begin{block}{Syntax}                \begin{lstlisting} if( Bedingung ) {     //Code, der nur ausgefuehrt wird, wenn die Bedingung wahr ist }                    \end{lstlisting}         \end{block}         \begin{block}{Logik}                 \begin{itemize}                  \item Alles innerhalb eines if-Blocks wird nur ausgeführt, wenn die Bedingung (in Klammern) wahr ist.             \end{itemize}         \end{block}     \end{frame} \end{document}