Using Dispel magic on Cleric's Divine Domain: Shadow, Favor of Darkness ability
What happens when you target a "magical effect" with Dispel Magic?
A doubt from Milnor's "Topology from a Differentiable Viewpoint". This is a doubt from Milnor's "Topology from a Differentiable Viewpoint".     For a smooth $f:M\to N$, with $M$ compact, and a regular value $y\in N$, we define $n(f^{-1}(y))$ to be the number of points in $f^{-1}(y)$. It is easy to see that $n(f^{-1}(y))$ is finite. It can also be seen that $n(f^{-1}(y))$ is locally constant; i.e. there is a neighbourhood of $y$ such that for every point $y'$ in that neighbourhood, $n(f^{-1}(y'))=n(f^{-1}(y))$.   The proof for the statement in bold is the following:     Let $x_1,x_2,\dots,x_n$ be the points of $f^{-1}(y)$, and choose disjoint neighbourhoods $U_1,U_2,\dots,U_n$ of these points, which are mapped diffeomorphically into neighbourhoods $V_1,V_2,\dots,V_n$ of $y$. The required neighbourhood is $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$.   I don't understand how $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$ is open. I know that $(V_1\cap V_2\cap\dots V_n)$ is open. However, how can one prove that $f(M-U_1-U_2-\dots U_n)$ is closed? I can see this working if $f$ is surjective, but not otherwise. Will C++ exceptions safely propagate through C code? Who was flying the Y-Wing when the Death Star exploded? Near the end of Star Wars IV: A New Hope, we see four spacecraft flying from the Death Star just moments before it explodes.     As you can see, there is the Millennium Falcon crewed by Han and Chewbacca, a pair of X-Wings which has to be Luke and Wedge, and then in the bottom corner, we see a Y-Wing. Who is this pilot and did he have any screen time to justify him actually existing in this moment? For as far as I can tell during the first trench runs, they were conducted by the Y-Wings and all of them seemed to have been destroyed during their run. So as I asked, who is flying this ship and how did survive to this moment? I am attempting to search jobs by tag on StackOverflow Careers. On SO, this is done using square brackets, but that is simply doing a keyword search, and I still get results that don't have the tag I am looking for.  For example, when I search for the keyword "android", I get this:        which is the same as when I search for the keyword "[android]":        Is there a format I can use to search jobs by tag? If not, I believe this would be a nice feature (seems intuitive for the Stack Exchange invite-only job site). If $U$ is unitary operator then spectrum $\sigma(U)$ is contained inside the unit circle In a Hilbert space, let $U$ be a continuous operator which it unitary. Prove $\sigma (U)\subseteq \Bbb{S}^1$.   It is important for me to know how I am doing, and I didn't come by a clear explanation so it would be appreciated to have your correcting and guiding.  Attempt: $U^{*}U=UU^{*}=I$ and therefore $U^{-1}=U$. I have $$ \| U x \|^2 = \langle U x, U x \rangle = \langle U^* U x, x \rangle = \langle x, x \rangle = \|x\|^2 $$ and the same goes for $U^*$ and therefore $\| U \| = \| U^{-1} \| = 1$.   Looking at $U-\lambda$ I can take $$ -\lambda U(U^{-1}-{1\over \lambda}I) =U-\lambda I. $$ Clearly, for $\lambda \ne 0$, $-\lambda U$ is defined and is a linear operator. $(U^{-1}-{1\over \lambda}I)$ is invertible if $|\lambda|\ge 1$ and therefore $\sigma (A^{-1})=\sigma (A)\subseteq \{|z|\le1\}$. But from the RHS, $U-\lambda I$ is invertible if $|\lambda|\le 1$ and therefore $\sigma(A)\subseteq \Bbb{S}^1$. I have noticed my claims tend to have holes in them many times which I cannot spot. If it is substantially true, how can I make sure it is completely formal? I understand that a black hole bends the fabric of space time to a point that no object can escape.   I understand that light travels in a straight line along spacetime unless distorted by gravity. If spacetime is being curved by gravity then light should follow that bend in spacetime.  In Newton's Law of Universal Gravitation, the mass of both objects must be entered, but photon has no mass, why should a massless photon  be affected by gravity in by Newton's equations?  What am I missing? How to create a desktop shortcut in Unity? Is there an easy way to create a desktop shortcut for an application in Unity? The ability to right click or drag and drop icons to the desktop in Unity does not seem to be there. How can I do the following in Git?  My current branch is branch1 and I have made some local changes. However, I now realize that I actually meant to be applying these changes to branch2. Is there a way to apply/merge these changes so that they become local changes on branch2 without committing them on branch1? I've had a very strange Finer error caused by recent macOS High Sierra betas that affected Macs running macOS 10.13 as well as macOS 10.12 Sierra. This error causes an entire Finder window to show visual artifacts when opening certain folders on either an APFS or HFS+ volume.  After moving the cursor or the affected Finder window, the window may look something like this;    Attempting to use Disk Utility to repair the affected volume does not resolve this issue, nor does restarting the affected Mac (even in Safe Mode).  Without formatting this disk, how can I prevent Finder from exhibiting this graphical error when opening particular folders? Concecutive last zeroes in expansion of $100!$ I saw  hosted by Stephen Hawkins.  And if I didn't get it wrong, it says that there was no time before the big bang, time was created there.  So how can anything happen when there is no time (eg: the creation of time and the universe)? Is this what he calls a 'singularity'? Which I could think like this: when there is no time nothing can happen, except during this singularity.  Of course I don't know of a law that says 'when there is no time nothing can happen..', I just came up with it, as something intuitive. , where a user had taken the time to look at the link on a post and realized that the post wasn't just off-topic, it was actually half-decently-disguised spam. So he flagged it as spam, and an overworked mod declined it, giving the questioner the benefit of the doubt. And we probably want mods to be giving the questioner the benefit of the doubt.  We also want to get rid of spam ASAP, and make the lives of mods easier. I'd like to suggest a feature the mod in question mused might be useful: An optional textbox where you can add a note saying why you think it's spam, so the mod can see it easily. In this case, the helpful user who went the extra mile could have written "Site linked doesn't relate at all to the question, instead solicits BitCoins" or similar.  One can leave a comment on the question, but:   People sometimes leave...unhelpful...comments on spam questions, and we don't want mods to have to take the time to read through them hoping for something useful It allows the user flagging the spam to do so and say why without calling attention to themselves publicly     Update: We may well want this for all flags, but in particular I had reason to want it for an NAA flag, so I've posted . Confirmation that someone is listening to another person's speech When someone is telling you a very long and detailed story he usually wants to hear some "confirmations" (or response) that you are listening to his story. In Russian we often use something like "tak" (which has a meaning of "ok" and "well, proceed further"), "a-ha" or "uh-huh".  What word serves the same purpose in English and American English? Geometric interpretation of the multiplication of complex numbers? I've always been taught that one way to look at complex numbers is as a Cartesian space, where the real part is the $x$ component and the imaginary part is the $y$ component.  In this sense, these complex numbers are like vectors, and they can be added geometrically like normal vectors can.  However, is there a geometric interpretation for the multiplication of two complex numbers?  I tried out two test ones, $3+i$ and $-2+3i$, which multiply to $-9+7i$.  But no geometrical significance seems to be found.  Is there a geometric significance for the multiplication of complex numbers? 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? In Ubuntu ISO MBR, why does a GPT partition overlap the boot partition, and make MBR uneditable? Replace Builtin commands with Custom Commands for CMD.exe So I have a torus, and I need it to "telescope" open -- kind of like Pacman's death animation, but with a torus instead of a sphere.  I'm very new to 3D modeling/animating so I really have no clue how to tackle this problem.  Any advice to help me get pointed in the right direction would be most appreciated :) warning message at command line `sudo` Just noticed the "meta user" link under the picture and reputation on the Stack Exchange user's page. It doesn't seem to work for all users. Error is "Page Not Found".   Not working:  /  Working:  /    My guess is this is because they've never logged into the meta site.