Every wizard has a wand, but do any carry a backup in case the first one gets broken or disarmed? I could see that being handy for the element of surprise.  Auror: Expelliarmus! Hah! Now you are harmless without your wand. Looks like it'll be a one-way ticket to Azkaban for y– Death Eater: Avada Kedavra! Whew, that was close! Good thing there's no rule that says I can't have two wands!  I also imagine that a wizard might want multiple wands to use their specialities, like owning both an aspen wand for charms and an ebony wand for transfiguration, since each excel in those types of magic. Do any wizards carry around a backup wand like this? Let's assume that they aren't just temporarily holding on to their enemy's wands or using them for subterfuge, they own the wands and intend to use they both indefinitely. And the Elder Wand doesn't count either.
Is there any canon information in Harry Potter that proves one way or another whether a wizard always owns only one wand as a matter of routine?  What I am talking about is a wizard buying and owning one wand and also a second wand (for backup, or simultaneous use, or separate purpose, or ceremonial+work wand), akin to people buying more than one pair of boots, guns or hammers.  A proof would be either canon info showing someone owning 2 wands simultaneously; or a statement showing why that does not happen except in rare exceptions (see below).    NOTE: I am discounting special situations such as winning an extra wand in a duel ala Harry (e.g. the second wand's obtainment wasn't intended/planned), or someone winning a Death Stick from prior owner (in which case it was intended as a replacement wand and not an extra one); or taking another's wand for a time being, ala Voldemort and Lucius Malfoy (since it was more of a "temporary replacement because main wand was useless for the purpose due to paired cores" - far from a normal situation).  Also, I meant wands intended for use - e.g. if Malfoy would have a (imagined by me) collection of rare antique wands, that's as irrelevant as someone having flintlocks in a display case.
What does it really mean that telephoto lenses "flatten" scenes? In Thor: The Dark World, Odin and Loki are discussing Loki's actions during The Avengers.  Loki argues that he would have ruled Earth as &quot;a benevolent god&quot;, and Odin replies:  Odin: We are not gods! We are born, we live, we die, just as humans do.  Now that's pretty definitive.  And Loki's snarky reply (&quot;Give or take five thousand years...&quot;) would seem to confirm that Asgardians have a limited lifespan, and will eventually die of old age.1  And we know that Asgardians can die in battle, because we've seen it happen over and over with supporting characters and Asgardian soldiers. Now, I've always been more of a fan of Thor stories (like Jason Aaron's Thor: God of Thunder arc) in which Thor and the Asgardians are gods.  Straight-up, genuine, higher-level-of-existence gods, not von-Däniken-esque wannabes playing on the tired old notion of magic just being super-advanced science.  BUT, despite my preferences, this exchange would seem to confirm once and for all that in the MCU, the Asgardians are not gods, but rather were mistaken for gods because of their ram-horns full of awesome-sauce and their magnificent bulging pecks. But wait!  Watch that movie for about 30 seconds longer and we come across this exchange:  Thor: She is ill. Odin: She is mortal. Illness is their defining trait.  Wait, what?  I thought Asgardians weren't immortal?  They can die in battle, and they naturally die of old age.  In my book, that makes them long-lived, but entirely &quot;mortal&quot;.  Surely Odin isn't splitting a hair so fine as to say that resistance to disease alone makes one &quot;immortal&quot;? So, if the Asgardians of the MCU aren't actually gods (grumblegrumblegrumble...), and they can't live forever (or even particularly long in a cosmic sense), and they can be killed in battle, in what sense are they &quot;immortal&quot; at all?  1 Fun fact: adjusted for a roughly 5,000 year lifespan, 1,048-year-old Loki is currently in the tail-end of his rebellious teenage phase. What is the purpose of a Windows Workgroup? In Microsoft Windows, what is the purpose of a Workgroup? I understand that a Domain allows the administration of security (user names, passwords, file access, etc.) to be centralized. But what does a Workgroup give you? What is the name of the tactic that politicians use to bury people with torrent of words? Some people write bloated books and long essays with skilful use of hooks, e.g. Jared Diamond; some others speak in long-drawn sentences with torrents of words, e.g. Noam Chomsky. It reminds me of a tactic politicians use in public discourse to waste people's time. I came across this word before but I can't retrieve it from either memory or the internet. Theme Activate Hook Is there a reason why "gn" in "reigning" is pronounced /n/ while in "regnant" it is pronounced [gn]? Both reigning and regnant are related to the same Latin noun, regnum.  Why is the ‹gn› spelling pronounced [n] in the first word but [gn] in the second? I switch back and forth between two computers constantly:   OSX Snow Leopard  Ubuntu 10.10    I'd like to be able to make Ubuntu use the same keyboard shortcuts as OSX System Wide, for the following keyboard sequences only (ie I don't want to make Ctrl act like Super)   Super+C -> Copy Super+V -> Paste Super+T -> Open a new tab in whatever browser I'm in.  Super+W -> Close a tab in whatever browser I'm in.    So in short, is there anyway for me to map just these keyboard sequences to the following, system wide ?   Super+C -> Ctrl+C Super+V -> Ctrl+V Super+T -> Ctrl+T  Super+W -> Ctrl+W   I know there are ways for me to do this for vim, and Firefox, and I'm sure specifically for most applications... but I would prefer to have to do this just once and have it work that way system wide!  I'm using a standard PC keyboard, that is "Generic 105 key (intl) PC" on Ubuntu. I'm also using the same keyboard on my Mac mini. Study the convergence of the following series: $$\sum_{n=2}^\infty \frac{1}{n^\alpha \cdot\ln^\beta(n)} \text{ where }\alpha,\beta \geq 0 $$   Applying d'Alembert criterion I have that $$ \lim_{n\to\infty} \frac{n^\alpha  \ln^\beta(n)}{(n+1)^\alpha  \ln^\beta (n+1)} = \lim_{n\to\infty} \left(\frac{n}{n+1}\right)^\alpha\left(\frac{\ln(n)}{\ln (n+1)}\right)^\beta = 1$$ so the nature of the series is inconclusive.  If $\alpha = \beta = 0$, then the series diverges, since $\sum_{n=2}^\infty 1 = \infty$.  Should I study the rest of the cases (i.e. if $\alpha = 0, \beta &gt; 0$ the root test and the ratio test are also inconclusive). What is the best form to study the series?.  Thanks in advance New York is a great place to live. New York is a great place to live in.  I've seen the former usage a lot and I've started wondering what the grammar aspects of it are. The main question I'm asking involves the grammar when the preposition in the end of the sentence is deleted. Is it equally formal or correct to omit it? What are the grammar properties of this? Some more examples of similar usage:  This organisation is a great place to work. This is a great place to stay/sit/study/travel/go.  Just to show that 'live' isn't special in this case. Note: I'd posted the same question  before, but the answers were not satisfying enough (the answers there did not seem to provide the precise grammar aspects of this, those that could be in a serious grammar book), so I'm posting this here to get some more views on the question. I hope it is acceptable. One of my Fellows asked me whether total induction is applicable to real numbers, too ( or at least all real numbers ≥ 0) . We only used that for natural numbers so far.  Of course you have to change some things in the inductive step, when you want to use it on real numbers.  I guess that using induction on real numbers isn't really possible, since $[r,r+\epsilon]$ with $\epsilon > 0$, $r \in \mathbb R$ is never empty.  Can you either give a good reason, why it isn't possible or provide an example where it is used? Intuition behind the definition of linear transformation I have studied that given vector spaces $V_1$ and $V_2$, a function $T:V_1 \rightarrow V_2$ is called a linear transformation of $V_1$ into $V_2$, if following two properties are true for all $u, v \in V_1$ and scalar $c$:  $(1)$:  $T(u+v) = T(u) + T(v)$ and $(2):  T(cu) = c T(u) $.  My questions are   $1$: What is the geometrical interpretation of properties $1$ and $2$ which says that $T$ preserves additivity and scalar multiplication. I am not able to see this geometrically. What is the meaning of preserving additivity and scalar multiplication.  $2$: At some place I have studied that a linear transformation will be linear if it sends each line to line and planes to planes and so on. How can we interpret this based on these two properties.  I need help to understand this.   Thank you very much for your time.. In theory, is it possible that the loss on the train set will increase together with the train set accuracy? (in the normal case the train set loss should go down) And vice versa, can the train loss decrease along with the accuracy?   Note: This is obviously different than the classic underfit / overfit cases which are easier to grasp. I am starting to learn using blender, and at some point I managed to do something to the 3D view camera.  Usually the camera orbits around an object, but now the rotation seems to happen (atleast very close to) around the "camera's origin". Neither the input setting in "User Preferences", the pivot point selection nor Ctrl-Alt-Shift-C (as suggested on some forum) do anything to the issue. How to create a table with merged columns and rows, and wrapped text, like in MS Word? Kernel did not upgrade between 14.04 LTS to 16.04 LTS When I'm using  find . -type f -name "*.htm*" -o -name "*.js*" -o -name "*.txt"  it finds all the types of file. But when I add -exec at the end:  find . -type f -name "*.htm*" -o -name "*.js*" -o -name "*.txt" -exec sh -c 'echo "$0"' {} \;  it seems it only prints .txt files. What am I doing wrong?  Note: using MINGW (Git Bash) Sum of indepedent random variables and a constant Let $X_1$ and $X_2$ be independent Normal random variables with mean $\mu_1$ and $\mu_2$, and variances $\sigma_1$ and $\sigma_2$. Let $Y = X_2-X_1 + c$, where $c$ is a constant. For notational simplicity, let $Y = X' + c$, where $X' = X_2-X_1$ The moment generating function of a Normal random variable $X_n\sim N(μ_n,σ_n^2)$ is $$M_{X_n } (t)=\text{exp}\Big(μ_n t+\frac{σ_n^2 t^2}{2}\Big)$$ Then, $$M_{X'} (t)=\text{exp}\Big[t(\mu_1+\mu_2)+\frac{t^2}{2}(\sigma^2_1+\sigma^2_2)\Big]$$ Thus, by the uniqueness property of moment generating function, $X'$ is said to follow a Normal distribution with mean $(\mu_1+\mu_2)$ and variance $(\sigma^2_1+\sigma^2_2)$. How do I proceed from here to derive the following:  First, I want to establish that $Y$ is also a Normal random variable using the moment generating function. The constant $c$ bugs me, but I cannot ignore it. As per my understanding, $Y\sim N\big(c+\mu_1+\mu_2,\sigma^2_1+\sigma^2_2\big)$ (and I could be completely wrong). But I don't understand how to apply the moment generating function to prove that.  Secondly, I want to derive $P(Y \leq a)$, where $a$ is a constant. If I am not mistaken, $P(Y \leq a)$ can be written as $$P(Y \leq a)=\Phi\Bigg(\frac{a-X_2-X_1 + c}{\sqrt{2(\sigma^2_1+\sigma^2_2)}}\Bigg).$$ But how do I arrive at that.   Any help is high solicited and appreciated. Is linear algebra more “fully understood” than other maths disciplines? In , it was discussed how LA is a foundation to other branches of mathematics, be they pure or applied. One answer argued that linear problems are fully understood, and hence a natural target to reduce pretty much anything to.   Now, it's evident enough that such a linearisation, if possible, tends to make hard things easier. Find a Hilbert space hidden in your domain, obtain an orthonormal basis, and bam, any point/state can be described as a mere sequence of numbers, any mapping boils down to a matrix; we have some good theorems for existence of inverses / eigenvectors/exponential objects / etc..  So, LA sure is convenient.  OTOH, it seems unlikely that any nontrivial mathematical system could ever be said to be thoroughly understood. Can't we always find new questions within any such framework that haven't been answered yet? I'm not firm enough with G&ouml;del's incompleteness theorems to judge whether they are relevant here. The first incompleteness theorem says that discrete disciplines like number theory can't be both complete and consistent. Surely this is all the more true for e.g. topology.  Is LA for some reason exempt from such arguments, or does it for some other reason deserve to be called the best understood branch of mathematics? Why isn't Arthur woken up by the kick? Prevent suggested edits under 10 reputation on protected questions Recently, . The specific  is protected:  protected by  ♦ Apr 27 '11 at 1:10 This question is protected to prevent &quot;thanks!&quot;, &quot;me too!&quot;, or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.  Great, so 10 reputation is required to answer the question. This doesn't stop anonymous and new users from spamming edit suggestions. Could the same restriction be applied to suggested edits to the protected question and its answers? As this would only apply to protected questions, which evidently attract not-so-good attention, it would not hinder new and anonymous users' ability to edit for the most part.