Im new with LaTeX, I have MikTek and TeXnic Center. The first time I tried to write a document I noticed I couldn't cancel terms in mathematical equations, so iI download the cancel package and installed it by hand.   Now the new package is working (or at least that is what I think), but for example the following simple document generates a Missing \begin{document} error and a first blank page:  \documentclass{article}  \usepackage[spanish,activeacute]{babel} \usepackage{amsmath} \usepackage{cancel}  \title{...} \author{...}  \begin{document}  \maketitle  \begin{equation} \cancel{a} \end{equation}  \end{document}   I cant copy all the build output from the TexnicCenter output window, but here you can see some part of it    Why is this?
Unicode applies the convention of using a  as signature at the beginning of a text stream, identifying the encoding used within it. The following three bytes at the beginning of a file: EF BB BF, identifies this file as a UTF8 file. Vi and most text editors gracefully ignore this signature. Open office does that as well, but it adds this signature at the beginning.  Now, if I open a TeX file with OpenOffic.org, and I do that quite a lot (for mixed directionality editing), the signature is added, and baffles LaTeX, which produces an error message such as:   ! LaTeX Error: Missing \begin{document}.   See the LaTeX manual or LaTeX Companion for explanation.  Type  H &lt;return&gt;  for immediate help.  ...                                                l.1   ��\documentclass{article}   whereas a simple dump of the file does not show the problem. Is there a way to eliminate this problem, while staying in the realm of latex?
How prove that $\lim\limits_{x\to+\infty}f(x)=\lim\limits_{x\to+\infty}f'(x)=0$ if $\lim\limits_{x\to+\infty}([f'(x)]^2+f^3(x))=0$? Question:     Let $f$ be differentiable on $[0,+\infty)$, such as$$\lim_{x\to+\infty}\left([f'(x)]^2+f^3(x)\right)=0$$show that $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$$   I think this problem is similar to  and (the  background)  For my problem I guess that if  $$\lim_{x\to+\infty}\left([f'(x)]^{2m}+f^{2n+1}(x)\right)=0,m,n\in N^{+},n,m&gt;1$$ then we have $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$. Random Sample vs. As Large a Sample as Possible? I'm struggling to differentiate scenarios where you would use random sampling vs. getting as large a sample as possible. For instance, one thing I am trying to test is the notion that being in a "home" arena boosts player performance.  I have collected individual football game data for every game over the past 10 years. I have over 34000 observations of players performance during this time span. Would it have been more prudent for me to take a random sample of say, 1000 players over the past 10 years?  I just don't understand why, in the above scenario, I'd take a random sample of 1000 when I can just take all of the players performances from the past 10 years. Change all file type associations from one program to another past simple or present perfect Please have a look at this sentence:     We haven't gotten a lot of Finnish groups that we liked here at Bananas but I'm pretty happy with this four tracker from Helsinki, apparently recorded in the same place where the Ramones played in the 70's.     The first verb is in present perfect (up to now) and the second is in simple past tense. I don't understand the simple past for the second verb.  Why "like" is not in its present perfect form as the first verb is? How is the skillset of an evolution being chosen? You may already know that if you catch two Pidgeys, they may have a different skillset.  I've also seen that evolving them from different skillsets will also create Pidgeottos with different skillsets.  Are those skills chosen randomly or do they follow a specific pattern? (Like Pidgey pattern 1 will always evolve into Pidgeotto pattern 1.)  If they are specific, is there a website to check for them? New messages flagged blue bubble does not go away When someone flags a message in chat all 10k+ users in chat get a little blue bubble to vote on. It shows up in the upper left hand corner of your gravatar (which is in the bottom left of the page).  When someone flags a message and it shows up, if I wait for a second to click on it, and then click on it, and it has already been handled, it shows the message  "There are no flags to display."  In the past, after showing this message, the blue (1) would disappear. However, lately it has not been going away and requires a page refresh in order to remove. Can this be fixed?  This issue is addressed here:  from September 12th 2012. But since it is still occurring 10 months later, I believe the cause is different. Can I have my iPhone play sound for MacBook? Say I am playing a video on my MacBook Pro, and I don't want to distract others with the sound. The problem is, I don't have a 3.5 mm Headphone or AirPods at my hand, but I have an iPhone 8 Plus and its EarPods with Lightning Connector. Can I play the sound from the video playing on macOS with the iPhone, preferably without connecting the iPhone to the Mac so that I can plug in the EarPods to the iPhone's socket?  This is basically the reverse of Can someone give me an example of a function $f: I\to \mathbb R$ which is uniform continuous but not Hölder continuous (for any $\alpha$)? Here $I$ is an interval. If $I$ is closed and bounded then we are essentially assuming that $f$ is merely continuous.   By saying that $f$ is not Hölder continuous for any $\alpha$, I mean for all $\alpha &gt;0$,   $$\sup_{x,y\in I, x\neq y} \frac{|f(x) - f(y)|}{|x-y|^\alpha} = \infty.$$  That is, I need to find a function $f$ so that for all $\alpha$ and $M&gt;0$, there are $x, y\in I$ so that   $$ \frac{|f(x) - f(y)|}{|x-y|^\alpha} \ge M.$$  I know that $f(x) = x^\alpha$ are $\alpha$-Hölder continuous. Thus, in some sense, I am looking for $f$ which are worst than the functions $x^\alpha$ for all $\alpha &gt;0$. How to plot a surface from a set of data? Disclaimer: I already read questions like  and  but these did not solve my problem.  I have this set of data:  x       y       z 2.17    0.001   0.82044815 2.17    0.002   0.82345825 2.17    0.004   0.82679255 2.17    0.008   0.83334715 2.17    0.016   0.84395915 2.17    0.032   0.8584953 2.21    0.001   0.77582165 2.21    0.003   0.78520505 2.21    0.009   0.80205985 2.21    0.027   0.83085105 2.24    0.001   0.7227885 2.24    0.002   0.73391615 2.24    0.005   0.7543979 2.24    0.015   0.78798745 2.24    0.003   0.74176635 2.24    0.009   0.77064805 2.24    0.027   0.81042375 2.26    0.001   0.66545585 2.26    0.003   0.7012046 2.26    0.005   0.721067 2.26    0.009   0.7447984 2.26    0.015   0.76715245 2.26    0.027   0.794177 2.27    0.001   0.62916195 2.27    0.003   0.6774642 2.27    0.009   0.72961785 2.27    0.027   0.7861086 2.28    0.001   0.5750828 2.28    0.003   0.65059675 2.28    0.005   0.6802631 2.28    0.009   0.7145367 2.28    0.015   0.74447695 2.28    0.027   0.7774403 2.29    0.001   0.51357255 2.29    0.002   0.581053 2.29    0.003   0.6173075 2.29    0.009   0.6972096 2.29    0.027   0.76793225 2.31    0.001   0.36997965 2.31    0.002   0.474415 2.31    0.003   0.53649295 2.31    0.009   0.6587164 2.31    0.016   0.70870255 2.31    0.027   0.7482423 2.31    0.05    0.7912395 2.34    0.001   0.2204104 2.34    0.002   0.316308 2.34    0.003   0.39256745 2.34    0.004   0.45240835 2.34    0.009   0.5883453 2.34    0.016   0.6590771 2.34    0.027   0.71444205 2.34    0.05    0.7690014 2.38    0.001   0.13286995 2.38    0.002   0.1828288 2.38    0.004   0.2980268 2.38    0.008   0.4507145 2.38    0.016   0.58417075 2.38    0.032   0.6833616   And if I plot it in Mathematica (with ListPlot3D[data]) it looks, without further tweaking and fiddling around, like:    But if I try this pgfplots code  \begin{tikzpicture}   \begin{axis}     \addplot3 [surf] table {data.txt};   \end{axis} \end{tikzpicture}   I only get this:    What do I have to do to get a surface like in Mathematica (color, smoothness, style, etc. does not matter, just a closed surface)? Let $G$ be a group. Show that a nonempty subset $H$ is a subgroup of $G$ if any only if $ab^{-1} \in H $ for any $a,b \in H$.  The forward direction is quite easy. Suppose $H$ is a subgroup. Then by closure, $ab \in H$ for any $a,b \in H$. Every element has an inverse. Hence, if $b \in H $, then $b^{-1} \in H$. Hence, by closure again, $ab^{-1} \in H$.   Backward direction, suppose $ab^{-1} \in H $ for any $a,b \in H$. Let $a=b$. Then we have $bb^{-1}=1_H\in H$. Let $a=1_H$ , we have $1_Hb^{-1}=b^{-1} \in H$. I don't know how to show closure.  Reference: Fraleigh p. 58  Question 5.45 in A First Course in Abstract Algebra I am planning to visit London for 6 days from the US. I have given my biometrics and when I was preparing to mail my documents to the UK visa processing center in the US, I used this link from the UK government website to cross-check the documents to be sent to the processing center.     (Guide to supporting documents).  Something that really surprised me was in Section 4 (documents that should NOT be sent along with the package) was hotel bookings and flights bookings.     Section 4: documents you should not send unless specifically requested      This page provides guidance on the types of documents that are not required to consider your application.         hotel bookings   flight bookings      Typically I have always been asked to provide my travel plans, accommodation plans and proof of them. This is the first time that travel plans and accommodation plans are not asked to be sent. What might be the reason behind it? This really confuses me. Any help in clearing my confusion would be greatly welcome. Encrypting a folder I'm familiar with bitlocker, having been using it for a while, but I am looking at just encrypting a folder instead. But I just don't understand the options it's giving me.  When I right-click a folder > properties > advanced > [x] encrypt contents to secure data > OK &lt; apply > ...  I get:  () apply changes to this folder () apply changes to this folder, sub-folder and files.   This doesn't make sense to me at all, if I encrypt a folder then the whole folder, everything within it is encrypted, then why should I be given these options, how can I have a folder encrypted and files within that folder unencrypted, that seems stupid to me, it doesn't make sense at all.  Can anybody explain why these options exist and how can I simply encrypt a folder? Is it possible to produce such design hints (on elements padding, spacing, width, height) automatically in photoshop? Is there any plugin that can potentially do this? Why do PBX systems use -48 V? It seems common that PBX and other telephone hardware use a positive-ground power supply, where the "hot" line is at -48v.  What's the reason for that? How strong of magnetic field would noticibly attract a person? There is iron in our blood, which is magnetic. Roughly how strong would a magnet have to be to induce a noticeable attraction? It would be nice to know this for several distances. Also, do electromagnets that strong exist? How to access the shared folder between two computers running ubuntu I have to computers running Ubuntu and I want to share files between those computer. these computers are already connected through a cross cable and sharing internet connection using firestarter. Both computers have samba installed.   In Desktop computer I have a shared folder How do I access it on the other one. How do I enable Cortana after the Anniversary Update (1607)? "It's cold today" -- what term do linguists use to call "it" when it's used as the subject of a sentence, but has no real antecedent? What are some examples of induction where the base case is difficult but the inductive step is trivial? According to :     ...proofs by mathematical induction   have two parts: the "base case" that   shows that the theorem is true for a   particular initial value such as n = 0   or n = 1 and then an inductive step   that shows that if the theorem is true   for a certain value of n, it is also   true for the value n + 1. The base   case is often trivial and is   identified as such, although there are   cases where the base case is difficult   but the inductive step is trivial.   What are some examples of proofs by induction where the base case is difficult but the inductive step is trivial? Does every Abelian group admit a ring structure? Given some Abelian group $(G, +)$, does there always exist a binary operation $*$ such that $(G, +, *)$ is a ring? That is, $*$ is associative and distributive:  \begin{align*} &amp;a * (b * c) = (a*b) * c \\ &amp;a * (b + c) = a * b + a * c \\ &amp;(a + b) * c = a * c + b * c \\ \end{align*}  We also might have multiplicative identity $1 \in G$, with $a * 1 = 1 * a = a$ for any $a \in G$. Multiplication may or may not be commutative.  Depending on the definition, the answer could be no in the case of the group with one element: then $1 = 0$. But the trivial ring is not a very interesting case. For cyclic groups the statement is certainly true, since $(\mathbb{Z}_n, +, \cdot)$ and $(\mathbb{Z}, +, \cdot)$ are both rings. What about in general? Is there some procedure to give arbitrary abelian groups ring structure?