I have a player that's a born lycanthrope (weretiger), that was bitten by a known werebear.  one of my other players is planning on casting remove curse on them.  Would the caster know that the spell was unsuccessful, or would it be a "weird feeling", or would it be a "find out at the next full moon" thing?  The reason for the question is that the other players don't know that this character is a born weretiger.  This player is waiting for an exciting moment or something to unveil their weretigerness and would prefer me to not give anything away, however I don't want to possibly change how remove curse works.  If the caster knows it didn't work i can phrase it so they're just confused, until the unveiling of the weretiger, but if they didn't know if it worked then there's nothing to worry about.
Suppose a cleric casts Zone of Truth on a captured enemy. Unbeknownst to them, the enemy is a rakshasa. Zone of Truth says (PHB, p. 289):     You know whether each creature succeeds or fails on its saving throw.   But (MM, p. 257):     The rakshasa is immune to spells of 6th level or lower unless it wishes to be affected.   Assuming the rakshasa doesn't wish to be affected by the Zone of Truth spell, does the caster know this?  More generally, does a caster know when targets are unaffected by a spell they have cast, either because they have made saving throws, or because they are outright immune to the spell or its effects (e.g., casting Ray of Sickness on a shadow demon, which is immune to poison damage and the poisoned condition)?
I have tried without success to plot the curve of the chi-squared distribution. Is there a generous soul who can come to my rescue. Change of flavour, Weak interaction I couldn't find a straight and clear answer to this question on the internet: Why is the weak interaction (charged), the only interaction which can change the flavours of the quarks? Film where aliens looking like humans are waiting in an airport closed due to snowstorm I saw the movie 6 - 7 years back on cable tv, from the looks of it it seemed to be made in the late 1990's or early 2000's. Unfortunately I could not see the complete movie. The story goes like--   Some people are waiting at a small airport because snowstorm activity has cancelled all flights in and out. There are two children in the group as well (one boy and a girl). Somehow they discover there are aliens looking like humans among them. They decide to check passengers through baggage scanners (checking the skeletal structure via x-ray). They find and kill 2 aliens.   and the big twist:      5. They then realize the two children are also aliens.   any ideas what the name of the movie is ??? I need some help with the exercise I was assigned in math class. We were given a 3x3 square made of unit squares, and we wanted to paint all the squares. We had 3 sets of paint, red, blue, and green. Each paint was distinct from each other, but two unit squares with the same color were the same. Additionally, there's only enough paint of each color to color 3 unit squares, and all the paint has to be used. The goal of the exercise was to find the total amount of distinct possible ways such that no two unit squares that share an edge (that is to say they are next to each other vertically or horizontally, having same color across a diagonal is fine). I've started by trying to take a look at each possible arrangement of the bottom 3 unit squares, but I can't seem to find a pattern between it. Is there a better way than continuing by brute force? Any ideas on how I should proceed from here? How to detect when a variable changes value How can I easily detect when a variable changes value? I would like the execution of the program to break on the debugger whenever a specified variable changes value. Right now I'm using Eclipse's debugger. 2 separate lights on two separate switches on one line - issue I am trying to copy specific attribute table rows from one shapefile to a new shapefile that will contain only the specified rows in its attribute table, that is, if a field GDP_TOP10 = 1.   I am using this at the moment:  #Grab rows using Search Cursor with WHERE clause rows = arcpy.SearchCursor("C:/CGDP.shp", "GDP_TOP10 = 1") for row in rows:     arcpy.CopyFeatures_management("CGDP.shp", "New.shp") del row del rows   However, this just copies the entire shapefile, not the rows I want to select with the cursor. If I execute a print statement under my for loop instead, it works and prints only the relevant fields, but the copy to new shapefile command does not seem to copy specifically.   Is there a different ArcPy command to only copy rows that match a query? What are the pros and cons of an articulated display? 1.  Why is the wave function complex? I've collected some layman explanations but they are incomplete and unsatisfactory. However in the book by Merzbacher in the initial few pages he provides an explanation that I need some help with: that the de Broglie wavelength and the wavelength of an elastic wave do not show similar properties under a Galilean transformation. He basically says that both are equivalent under a gauge transform and also, separately by Lorentz transforms. This, accompanied with the observation that $\psi$ is not observable, so there is no "reason for it being real". Can someone give me an intuitive prelude by what is a gauge transform and why does it give the same result as a Lorentz tranformation in a non-relativistic setting? And eventually how in this "grand scheme" the complex nature of the wave function becomes evident.. in a way that a dummy like me can understand.  2.   A wavefunction can be thought of as a scalar field (has a scalar value in every point ($r,t$) given by $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{C}$ and also as a ray in Hilbert space (a vector). How are these two perspectives the same (this is possibly something elementary that I am missing out, or getting confused by definitions and terminology, if that is the case I am desperate for help ;)   3.   One way I have thought about the above question is that the wave function can be equivalently written in $\psi:\mathbb{R^3}\times \mathbb{R}\rightarrow \mathbb{R}^2 $ i.e, Since a wave function is complex, the Schroedinger equation could in principle be written equivalently as coupled differential equations in two real functions which staisfy the Cauchy-Riemann conditions. ie, if $$\psi(x,t) = u(x,t) + i v(x,t)$$ and $u_x=v_t$ ; $u_t = -v_x$ and we get $$\hbar \partial_t u  = -\frac{\hbar^2}{2m} \partial_x^2v + V v$$ $$\hbar \partial_t v  = \frac{\hbar^2}{2m} \partial_x^2u - V u$$ (..in 1-D) If this is correct what are the interpretations of the $u,v$.. and why isn't it useful. (I am assuming that physical problems always have an analytic $\psi(r,t)$). Another meta question identified a bug that was marked as designed. To summarize the bug, the little number that appears next the Review link in the header shows an incorrect value most of the time.         I would like to propose that the feature be dropped. Honestly, I would prefer it be fixed, but the response to the question above gives me little faith that it will be. Half of the time, I click that little brown box and find out that I was lied to. There's nothing to review.  The little number no longer does the job it was created to do.        Oh sweet! Three things I can review!        Just kidding... How do you normalize a zero vector Prove $\mathbb{Z}_2 \times \mathbb{Z}_2$ is not isomorphic to $\mathbb{Z}_4$ In trying to understand why there are two sets of groups of order 4.  I know that there exists the Vierergruppe of $\mathbb{Z}_2 \times \mathbb{Z}_2$ and the group $\mathbb{Z}_4$ but I do not understand why they are not isomorphic How to prove that $\sqrt[3] 2 + \sqrt[3] 4$ is irrational? So while doing all sorts of proving and disproving statements regarding irrational numbers, I ran into this one and it quite stumped me:  Prove that $\sqrt[3]{2} + \sqrt[3]{4}$ is irrational.  I tried all the usual suspects like playing with $\sqrt[3]{2} + \sqrt[3]{4} = \frac{a}{b}$ for $a,b\in \mathbb{Z}$ , but got nowhere.  I also figured maybe I should play with it this way:  $2^\frac{1}{3} + 4^\frac{1}{3}=2^\frac{1}{3} + (2^2)^\frac{1}{3}=2^\frac{1}{3} + 2^\frac{2}{3}=2^\frac{1}{3} + 2^\frac{1}{3}\times 2^\frac{1}{3}=2^\frac{1}{3}(1+2^\frac{1}{3})$  But there I got stumped again, because while $1+2^\frac{1}{3}$ is irrational, nothing promises me that $2^\frac{1}{3} \times (1+2^\frac{1}{3})$ is irrational, and I feel like trying to go further down this road is moot.  So what am I missing (other than sleep and food)? What route should I take to  prove this? Thanks in advance! Proof of centripetal acceleration formula ($a_c = v^2/r$) for non-uniform circular motion The formula for centripetal (radial) acceleration is well known, and there exist many proofs for it: $$||a_c|| = \frac{||v||^2}{r}$$  However, all the proofs I've seen rely on the fact that it is uniform circular motion and the magnitude of the tangential velocity vector does not change. For instance, take the classic proof using similar triangles — the similarity can only be established if the final tangential velocity vector and the initial one are of the same length.  In addition, take  calculus based proof from Khan Academy, outlined as follows:     For this proof to work, $(d\theta / dt)$ must be considered a constant, $\omega$, that does not depend on time. In the case of non-uniform circular motion, however, this is not always true as since there exists a tangential acceleration along with a radial one, $\omega$ must depend on time and is not necessarily a constant value.  Intuitively I understand that the centripetal/radial acceleration depends only on the difference in orientation between two tangential velocity vectors, and that their magnitudes do not matter -- hence the formula intuitively holds true in the non-uniform case. However, how would you go about modifying either of the proofs presented so that they are still valid in this case? Or alternatively, is there another proof that holds valid even when there exists tangential acceleration?    As per the suggestion of one of the answers, I let $\omega$ vary with time and took its derivative as $\alpha(t)$. This is my work so far. Unfortunately, I am stuck after the last step.  \begin{align} \overset{\rightharpoonup }{p}(t) &amp;= r \cos (\theta (t))\cdot\hat{i}+r \sin (\theta (t))\cdot\hat{j} \\ \overset{\rightharpoonup }{v}(t) &amp;= -r \sin (\theta (t))\cdot\omega(t)\cdot\hat{i}+r \cos (\theta (t))\cdot\omega(t)\cdot\hat{j} \\ \overset{\rightharpoonup }{a}(t) &amp;= (-r \cos(\theta(t))\cdot\omega(t)^2 - r \sin(\theta(t))\cdot\alpha(t))\hat{i} \\ &amp;+ (-r \sin(\theta(t))\cdot\omega(t)^2 + r \cos(\theta(t))\cdot\alpha(t))\hat{j} \end{align}     From here on, $\theta(t)$ is represented as just $\theta$ for brevity   and clarity   $$\overset{\rightharpoonup }{a}(t) = -\omega(t)^2(r \cos\theta\cdot\hat{i} + r \sin\theta\cdot\hat{j}) -\alpha(t)(r \sin\theta\cdot\hat{i} - r \cos\theta\cdot\hat{j})$$ LINK : fatal error LNK1104: cannot open file 'D:\...\MyProj.exe' No notification (+15) for accepted answer A list of inconveniences between quantum mechanics and (general) relativity? Can US citizens go to Cuba freely and vice versa after the recent diplomatic relation restoration? A recent full diplomatic relation restoration between the US and Cuba took place, does that mean that US citizens can go to Cuba freely and vice versa? I observe the following strange characters ^@ in syslog at the time when my machine freezes.  Then i have to hardboot my machine.  What could be wrong any ideas?  Oct 12 12:20:52 nanak-P570WM anacron[1074]: Normal exit (1 job run) ^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@^@  Oct 12 12:24:28 nanak-P570WM kernel: imklog 5.8.11, log source = /proc/kmsg started. Oct 12 12:24:28 nanak-P570WM rsyslogd: [origin software="rsyslogd" swVersion="5.8.11" x-pid="543" x-info="http://www.rsyslog.com"] start I've come across some strange interaction between overlay specification and \pause. Basically I have an itemize and after some text(or block). I want to step through the itemize and after that show the text/block.  I've tried to put a simple \pause between the itemize and the text, but in this way two slides are created, both with the full itemize and no text.  If I don't put the \pause the text is visible from the start(which I don't want).  I've already found a simple work-around: instead of using itemize[&lt;+-&gt;] I can specify the slides on the items(e.g item&lt;1-&gt;, ...item&lt;2-&gt;,... \pause), but I don't like this solution for some reasons:   I have to type more than before I have to hard-code the slide numbers This is solution is not robust. Modification on the item order, removal or addition of new items would break it. I think there has to be some better solution.   So, how should I do?  By the way, a minimal example:  \documentclass{beamer}  \usepackage[utf8]{inputenc} \usepackage{default}  \begin{document}  \begin{frame}     \begin{itemize}[&lt;+-&gt;]         \item A         \item B         \item C     \end{itemize}      \pause  % double pause here      Some text. \end{frame}  \end{document}