The roots of this polynomial are $(-1)\zeta^n$. Where $\zeta$ is the 5th root of unity.  So the splitting field should just be $\mathbb{Q}(\zeta)$.  This has galois group $\mathbb{Z}/4$ since it's just a cyclotomic extension.  Is this correct?  Source:
I need help to find the Galois Group of $x^5 +1$. I know that it has a 5-cycle and a 4 cycle and is not a subgroup of $A_5$. Thanks!
Allow diamond moderators to reverse question migrations? It has happened on occasion, due to the high amount of 10K moderators on SO that questions are migrated to the wrong site, and needs to be reversed. Although currently it is possible to re-open the question on the original site and delete the one on the destination site, it still leave the migrated too banner behind.  It would be useful if diamond moderators could reverse a migration, removing it from the destination site completely and re-activating it on the original site and at the same time removing the banners.  Previously it has been discussed educating the users around when to move questions, however this doesn't stop the problem.  As per Shog9  I have to admit I do like Shog9's suggestion regarding the post being closed as not programming related rather then re-opened, however, reasons for closing it on other sites should also be kept in mind, as this also applies to stuff being sent back forth between the sites. Animated movie that had black robots that turned people into stone late 1960s to 1970s It was a weird sci-fi animated movie I saw only a little bit of it. This is what I recall:   It was old but had color, so I assume it was late 1960s to early 1970s. The animation and style reminded me of  but it wasn't as psychedelic. The setting was an alien planet with all sorts of bizarre creatures. The tech was advanced but not a dominant aspect of the planet's culture. The people lived simple and peaceful. The protagonist is a man with bluish skin like the others. The antagonists were humanoid shaped black robots that shot a beam that would turn people into stone. Uniform continuity of $\sin(x\sin x )$ for all $x\in (0,\infty)$ Why is $5^{n+1}+2\cdot 3^n+1$ a multiple of $8$ for every natural number $n$? I have to show by induction that this function is a multiple of 8. I have tried everything but I can only show that is multiple of 4, some hints? The function is $$5^{n+1}+2\cdot 3^n+1 \hspace{1cm}\forall n\ge 0$$, because it is a multiple of 8, you can say that$$5^{n+1}+2\cdot 3^n+1=8\cdot m \hspace{1cm}\forall m\in\mathbb{N}$$. By accident, I used rm on a file I didn't want to delete. Is there a way that I can get it back under Linux? Suppose that C is a closed convex subset of $\mathbb R^n$ and $x \in \mathbb R^n$. The projection of $\mathbf x$ onto C is the closest point $\mathbf y \in C : \mathbf z = \mathbf y$ minimizes ||$\mathbf z - \mathbf x$||$_2$ over all $\mathbf z \in C$.  (a) Show that the minimum exists.  (b) Show that there is only one minimizer: that is, show that if $\mathbf y_1$ and $\mathbf y_2$ both minimize ||$\mathbf x - \mathbf z$|| then $\mathbf y_1 = \mathbf y_2$   My ideas:  For (a), to show that the minimum exists I will need to show that it is closed and bounded. We already know it's closed because C is a closed convex subset so I just need to show it's bounded. Am I wanting to minimize ||$\mathbf x - \mathbf z$||$_2$? This is just the Euclidean norm which would be $\sqrt{(\mathbf z - \mathbf x)^2}$=$\sqrt{\mathbf z^2 -\mathbf z \mathbf x -\mathbf x \mathbf z + \mathbf x^2}$. I'm not sure where to go from here...  For (b) I want to use proof by contradiction- so I will assume that $\mathbf y_1$ and  $\mathbf y_2$ both minimize the function and then I'll somehow see that they are in fact equal. How do I set up this argument? As discussed here:  offering actual money as a bounty sets the wrong incentives. However I would suggest giving users the ability to pay with money for offering a bounty regardless of their reputation. The bounty would still count as reputation for the receiving user but would not be discounted from the paying user's reputation. Classes and lessons at Hogwarts are presumably scheduled at fixed times of day. The first class might start, for example at nine in the morning.  To be on time for one's lessons, it is therefore crucial to know the current time but we know that electric watches of course do not work at Hogwarts because of all the magical interference. Are there magic watches? Is there maybe a spell to tell the time?  How do people at Hogwarts usually tell the time? Remove particularly tricky items from context menu I've found various sources of info about removing these annoying things which I never use... but I haven't succeeded in getting rid of the following.    (Avira-related:) - Scan selected files with Avira  (I believe all this is M$ crapware... I presume it's all been lovingly crafted so they're all impossible to remove. You never know, though: some super-uber-uber-geek might just have found a way) - Restore previous versions  - Cut - Copy - Delete  - Rename Seeking detailed description of qgs file structure? Is there a detailed description on the .qgs file structure somewhere?   I checked at  and got     The requested URL /qgis.dtd was not found on this server.   and at  which wasn't very detailed. Where is the rest of Yoda's species? We never seem to see any beings of Yoda's species - except, of course, Yoda himself.  This leads us to the question of where the rest of the li'l critt'rs are. Yoda had to have parents, correct? So what happened to them? Was Yoda, put simply, just one of the last of a dying species? Can you end Beast Form early? Is there any way to end beast form early? I find myself using it to clear out a particularly hard portion of a dungeon at times, but then I have to sit around for 2-5 minutes so I can be my human character again (I have to feed to keep my health up, so I'm a werewolf longer than the 2.5 minutes).  Is there anyway to end it early? Let $A ∈ M_{m\times n}(\mathbb{Q})$ and $B ∈ \mathbb{Q}^m$. Suppose that the system of linear equations $AX = B$ has a solution in $\mathbb{R}^n$. Does it necessarily have a solution in $\mathbb{Q}^n$?  Where do I start?    I feel I can use this to :  $$x_i=\frac{1}{a'_{ii}}\left(b'_i - \sum_{j=i+1}^{k} a'_{ij} x_j \right).$$    Here is a thought that just crossed my mind, but I'm not sure if this is legal:  Suppose $Y\in \{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $Y'\in\mathbb{Q}^n$ are solutions to $AX=B$. Then $AY=AY'=B$, or $A(Y-Y')=A(Y+(-Y'))=0$, but  addition between $\{\mathbb{R}\setminus \mathbb{Q}\}^n$ and $\mathbb{Q}^n$ is . . . How to fix lsregister failing associations after reboot? I saw this keyword for the first time and I was wondering if someone could explain to me what it does.   What is the continue keyword? How does it work? When is it used? Implementing the decorator pattern in Python Set of branch points isn't discrete, but branch points are isolated? I refer to , which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.) Question 1: Do I understand correctly, and is this indeed true? What else I understand:  $A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.  This overflow question () seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).  2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a ): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.   This stackexchange question () seems to suggest this is true.  I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the , which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.  Kinda related:    This page () says    Now, the ramification and branch points must form a discrete set  I don't think the page assumes anything like $F$ is proper.  The  text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.  Kinda related (see Question 2 below):     says    If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.   This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.  Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.  (again) Kinda related: Two domains (when SSL) on same directory Some months ago I watched  video of Blenderguru about "How to create a realistic earth in blender". In this video blender render is used and  now that I understand texturing in cycles a little better I am trying to rebuild this "realistic earth" with cycles.    In the video - starting at about 11:30 min - it is shown how to map the "earth day" and "earth night" image textures. Depending on the lights shining on the earth the earth or night image is projected. I tried to rebuild that in cycles but I think I a making mistakes. In the circled box the "night" nodes are found. What Is my mistake ?   P.S When you see other mistakes in this node tree, please let me know.  OK ... I now have implemented the advice of Uncle Snail (except moving the black slider). Below an image of the earth and the updated node tree (I also rearranged nodes for better overview).    In the Color Ramp, I did not move the slider. Is that really necessary ?? How can I insert a newline in a framebox? I have several frameboxes in my document, using the following command:  \framebox(115,115){ R-Sq: \newline For example}    I want to force a new line, but neither \newline or \\ appear to work. Is there any other way of doing this?