Changing usage of certain words
"Oriented" vs. "orientated"
Padding is missing under the &quot;Communities&quot; border line, and the list is also showing &quot;Communities (0)&quot; (maybe the reason is due to caching as it shows 0 Communities):  Generally, padding is present: Currently the Flag feature, on answers, offers five choices. Three of them request moderator attention:   Not an answer Very low quality "Other"   In the spirit of reducing the burden on moderators, and distributing that work a little bit more evenly around the community, I'm proposing a couple of small tweaks to the way answers are flagged.   Eliminate  as an option. Very low quality posts that are, nevertheless, genuine answers do not need moderator attention - they should just be downvoted. (Further justification for this is  and ) Replace  with four specific options:   Should have been a comment - this is not an answer, it is a comment on the question or on another answer. Flagger would then be required to choose which post it should have been a comment on. Should have been an edit - this is not an answer, it is additional information about the question from the original asker. Only appears if the answer is from the same user as the question. Me-too! non-answer - this answer consists of nothing more than "Me too!" or "Did anyone find the answer?" Conversational and irrelevant - this is not an answer, just a random conversation or rant, or meaningless typing that should be deleted    This makes life much easier on the moderators, because now the moderator is not expected to clean up after the bad answer, they merely have to peek at what happened and click "Agree" or "Disagree". If they agree:   Should have been a comment converts the answer to a comment Should have been an edit converts the answer to an edit and it is appended to the question Me-too! non-answer deletes the answer and leaves a form-message for the poster explaining that we don't like "Me too", but they should upvote the question if they think it's important or valuable Conversational and irrelevant - deletes the answer and leaves a form-message for the poster explaining that we are a Q&amp;A site, etc. etc.   In fact, if enough people with high enough flag weight or reputation (FORMULA TBD) apply the same flag, the moderation happens automatically, without moderator involvement.  To prevent the moderation queue from filling up with problems that might cure themselves, these flags don't even go in the moderator queue until some period of time (TBD) has passed.  Thoughts? Nicomachus theorem proof - what did I do wrong? The exercise asked me to proof $1^3=1$ $2^3=3+5$ $3^3=7+9+11$ $...$ I formulate the equation as (1)$$a^3 = \sum_{i=0}^{a-1} (a-1)a+1+2i$$ Proof base case: $$1^3=\sum_{i=0}^{1-1} (1-1)(1)+1+2i=1$$ Proof (a+1) case: (2) $$(a+1)^3=a^3+3a^2+3a+1$$ (3) $$(a+1)^3=\sum_{i=0}^{(a+1)-1}((a+1)-1)(a+1)+1+2i=\sum_{i=0}^{a}(a)(a+1)+1+2i$$ factor out last summation term, so number of iterations match (1): $$\sum_{i=0}^{a-1}[(a)(a+1)+1+2i]+(a)(a+1)+1+2a=\sum_{i=0}^{a-1}[(a)(a+1)+1+2i]+a^2+3a+1$$ rearrange the summation term, so part of equation match (1): $$\sum_{i=0}^{a-1}[(a)(a-1)+1+2i]+\sum_{i=0}^{a-1}(2a)+a^2+3a+1$$ substitute summation with (1): (4) $$a^3+\sum_{i=0}^{a-1}(2a)+a^2+3a+1=a^3+(2a)(a-1)+a^2+3a+1=a^3+3a^2+a+1$$ From (2) I got $a^3+3a^2+3a+1$ but from (3) &amp; (4) I got $a^3+3a^2+a+1$ My result with induction is off by 2a. Can someone please help me to understand where I did wrong? Residue theorem with $\frac{1}{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) d\theta$. The following integral should be doable using the residue theorum:     $$\frac1{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) \,d\theta$$ Asymptotes in a plot I'm using TikZ to plotting some functions for a document. For example, I want to plot 1/(x+1). So I did something like this:  \begin{tikzpicture}[&gt;=stealth,x=3mm,y=3mm] \draw (0,5) node[above]{\fimg $y=f(x+1)=\dfrac{1}{x+1}$}; \clip (-5,-5) rectangle (5,5); \ejes[1]{-4}{4}{-4}{4} \draw[samples=100] plot(\x,{1/(\x+1)}); \end{tikzpicture}   where \ejes[style]{xmin}{xmax}{ymin}{ymax} is command, that I have defined before to draw the axes (join with \fimg). My expected output for the code below was just the function, but I get this:    So, TikZ automagically put the vertical asymptote at x=-1. That isn't a bad thing, but now I want to have some control over it. For example I want dotted lines for vertical asymptotes. How can I do that? As far as I see the "Blatantly off-topic" flag      Blatantly off-topic (this question has nothing to do with %site topic%)    is only available for low-rep users without the close vote privilege.  What is the reason to remove this option for high-rep users? I think it could be very usefull in some cases where now we must select      Other (add a comment explaining what is wrong)    option and waste time to type obvious message for a reason. It is well-known that empty text files have zero bytes:    However, each of them contains , which according to my research, is stored in , and do use space.   Given this, it seems logical to me that it is possible to fill a disk by purely creating empty text files. Is this correct? If so, how many empty text files would I need to fill in a disk of, say, 1GB?    To do some checks, I run df -i but this apparently shows the % of inodes being used(?) rather than how much they weigh.  Filesystem             Inodes  IUsed    IFree IUse% Mounted on udev                   947470    556   946914    1% /dev tmpfs                  952593    805   951788    1% /run /dev/sda2            28786688 667980 28118708    3% / tmpfs                  952593     25   952568    1% /dev/shm tmpfs                  952593      5   952588    1% /run/lock tmpfs                  952593     16   952577    1% /sys/fs/cgroup /dev/sda1                   0      0        0     - /boot/efi tmpfs                  952593     25   952568    1% /run/user/1000 /home/lucho/.Private 28786688 667980 28118708    3% /home/lucho Can we see emission lines in stars? I read that stars show both absorption and emission lines. While how we see absorption lines is clear to me, I don't understand how (and when) we see the emission lines. Based on what I read in several of your threads such as  we should not be seeing emission lines at all, since all the emissions would come from the hotter interior but those can't reach the outside (unless I misunderstood this point), and the only emission we see is blackbody radiation. But then how do we see the few emission lines such as in the spectrum shown ? Can a willing lycanthrope be True Polymorphed? Can a lycanthrope or other shapeshifter be willingly affected by the spell True Polymorph? I know that it can be affected by regular Polymorph, but the text for the enhanced version seems to differ. I have trouble comparing 2 double in Excel VBA  suppose that I have the following code  Dim a as double Dim b as double a = 0.15 b = 0.01   After a few manipulations on b, b is now equal to 0.6  however the imprecision related to the double data type gives me headache because  if a = b then  //this will never trigger end if   Do you know how I can remove the trailing imprecision on the double type? Must present perfect tense be used if the action takes place more than once? I was told that if an action is completed once in the past, the simple past is used. Ex:  I saw that movie.  If the action is completed more than once, the present perfect tense must be used. Ex:  I have seen that movie twelve times.    So, "I have seen that movie twelve times on Tuesday." is correct and "I saw that movie twelve times on Tuesday." is incorrect? Reputation and inbox notification shows up again after refresh How to disprove this fallacy that derivatives of $x^2$ and $x+x+x+\dots\quad(x\text{ times})$ are not same. Possible Duplicate:          \begin{align*} x^2 &amp;= \underbrace{x + x + x + \dots + x}_{x \text{ times}}, \\ \therefore \frac{\mathrm{d}}{\mathrm{d}x} (x^2) &amp;= \frac{\mathrm{d}}{\mathrm{d}x} (\underbrace{x + x + x + \dots + x}_{x \text{ times}}) \\ &amp;= \underbrace{1 + 1 + 1 + \dots + 1}_{x \text{ times}} \\ &amp;= x. \end{align*}  But we know that $$ \frac{\mathrm{d}}{\mathrm{d}x} (x^2) = 2x. $$  So what is the problem?  My take is that we cannot differentiate both sides because $\underbrace{{x+x+x+\cdots+x}}_{x \text{  times}}$ is not fixed and thus $1$ is not equal to $2$. Bluetooth earphones work fine until sleep. After resuming from sleep however, they appear to connect for a brief moment before disconnecting. On blueman, the error given is Resource temporarily unavailable. This issue arose only after updating to 18.04 LTS.   Here's the terminal output for lsusb:  Bus 001 Device 002: ID 8087:8001 Intel Corp.  Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 003 Device 001: ID 1d6b:0003 Linux Foundation 3.0 root hub Bus 002 Device 004: ID 1bcf:0002 Sunplus Innovation Technology Inc.  Bus 002 Device 003: ID 04f2:b477 Chicony Electronics Co., Ltd  Bus 002 Device 002: ID 0a5c:21f1 Broadcom Corp. HP Portable Bumble Bee Bus 002 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub If $|f(z)|\le |g(z)|$, can we say $f$ is constant? Let $f$ and $g$ be entire functions and $g(z)\neq 0$ for all $z\in \mathbb{C}$. If $|f(z)|\le |g(z)|$, can we say $f$ is constant?  Liouville theorem says that an entire bounded function is constant, but g is not given bounded here. So I think it should be false. What else can we say about $f$? Prove the following using epsilon-delta definition:  Let $a \in \mathbb{R}$ and $f,g$ be defined on $\mathbb{R}$.  If  $$\lim_{x\to a} f(x) = 647 \tag 1$$  and $$\lim_{x\to a} g(x) = \infty \tag 2$$ then   $$\lim_{x\to a} f(x) + g(x) = \infty.$$  $$(1) \implies\forall \epsilon &gt; 0, \exists \delta &gt; 0, s.t. |x-a|&lt;\delta \implies |f(x)-647| &lt; \epsilon$$  $$(2) \implies \forall M &gt; 0, \exists \delta_2 &gt; 0, s.t. |x-a|&lt;\delta_2 \implies |g(x)| &gt; M$$  Not sure about this... I've set up the definitions. I believe we will have to take a minimum between the deltas for first two limits then use it in the third.  Anyone have any clues? I have a .txt file with numbers ordered like this (on the same row):  106849_01373    106849_01967    106850_00082    23025.7_01059      I would like to convert them like that:  106849_01373 106849_01967 106850_00082 23025.7_01059   I have no idea which command to use. Can somebody help me with this? The identification of an electron as a particle and the positron as an antiparticle is a matter of convention. We see lots of electrons around us so they become the normal particle and the rare and unusual positrons become the .  My question is, when you have made the choice of the electron and positron as particle and anti-particle does this automatically identify every other particle (every other fermion?) as normal or anti?  For example the proton is a particle, or rather the quarks inside are. By considering the interactions of an electron with a quark inside a proton can we find something, e.g. a conserved quantity, that naturally identifies that quark as a particle rather than an antiparticle? Or do we also just have to extend our convention so say that a proton is a particle rather than an antiparticle? To complete the family I guess the same question would apply to the neutrinos. I have two tables A &amp; B. Table A id | name | num  Table B id | date | roll  id in table B is a foreign key which relates to the id in table A. I want to get the contents of table B where id is given and also along with that data, I need the name whose id is given from table A. NOTE: cannot use multiple queries separated by commas; need to do both the SELECTs in one query. Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$