Why semigroups could be important?

Question: There is known a lot about the use of groups -- they just really appear a lot, and appear naturally. Is there any known nice use of semigroups in Maths to sort of prove they are indeed important in Mathematics? I understand that it is a research question, but may be somebody can hint me the direction to look on so that I would see sensibility of semigroups, if you see what I mean (so some replies like look for wikipedia are not working as they are anti-answers).
Comment: Shouldn't this be titled "Why *are* semigroups important?" :)
Comment: @Martin: ok, in PDE's, in computer science (say, particularly, semi-Thue systems), the so-called Cuntz-algebras in C*-algebras etc. -- all this is where semigroups appear is like there appears something with associative operation and that's it, no further miracle of that we then go looking what is going on with those semigroups with further yielding nice results about our initial problem. And, C_0-semigroups aren't really semigroups :-) I'm not saying that semigroups are not important, I just wonder if somebody knows where semigroups can do some tricks (hence it's a good research question)
Comment: This is an extremely general question, given that whole books have been written about semigroups or monoids.
Comment: @Jim: Yes, perhaps this question asks for a direction where to begin ...
Comment: @Victor: The wikipedia article shows that semigroups are important in applied mathematics, PDE, theoretial computer science and probability theory ... perhaps you should explain why you think that this is not evidence enough for their importance. Check also related articles such as http://en.wikipedia.org/wiki/C0-semigroup
Comment: The renormalization "group" is really a semigroup.
Comment: Important to *whom*, and for *what*? (Less facetiously; not everything in life is reversible.)
Comment: When you say C_0-semigroups aren't really semigroups, perhaps you could explain what really *is* a semigroup?
Comment: @Victor: your comment to Martin B suggests that perhaps your underlying question is really the following: is the general theory of semigroups as important in wider mathematics as the general theory of groups is? If that is roughly what you were driving at, then perhaps the answer is a guarded "no", because there is so little one can say about an arbitrary semigroup. For particular classes of semigroup, though, there does seem to be fruitful general theory
Comment: A related question from MSE: http://math.stackexchange.com/questions/101487/why-are-groups-more-important-than-semigroups?rq=1
Comment: Natural number form a semigroup, but not a group. Most rings are semigroups, but not groups relative to multiplication operation. Matrix determinant is a morphism from a multiplicative semigroup of matrices to the multiplicative semigroup of $\mathbb{R}$ or $\mathbb{C}$ (or whatever field your matrices were define over). Perhaps I don't understand the question, for it seems to me that semigroups are used all the time...
Answer: Am slightly surprised no one has mentioned the Galvin-Glazer proof of Hindman's theorem via the existence of semigroup structure on $\beta{\mathbb N}$, the Stone-Cech compactification of the positive integers (see, for instance, part of this note by Hindman.
The relevance to the original question is that knowing that ``compact right topological semigroups have idempotents'' may sound recondite, but it is just what was needed to answer Galvin's original question about translation-invariant ultrafilters, which was itself motivated by a "concrete" question in additive combinatorics.
On a related note, while it is in general not possible to embed a locally compact group as a dense subgroup of something compact (the map from a group to its Bohr compactification need not be injective), you can always embed it densely into various semigroups equipped with topological structure that interacts with the semigroup action: there are various of these, perhaps the most common being the WAP-compactification and the LUC-compactification. Unfortunately this often says more about the complicated behaviour of compact semitopological semigroups (and their one-sided versions) than about anything true for all locally compact groups, but the compactifications are a useful resource in some problems in analysis, and the semigroup structure gives one some extra grip on how points in this compactification behave. (Disclaimer: this is rather off my own fields of core competence.)
Comment: +1:  I was thinking of remarking on this.  Furstenberg and Katznelson's paper on idempotents in compact semigroups (http://math.stanford.edu/~katznel/colorla.pdf) is amazing.
Comment: The link in the post is now dead. I suppose it is supposed to be link to the paper Neil Hindman: *Algebra in the Stone-Cech compactification and its applications to Ramsey theory*, Scientiae Mathematicae Japonicae 62 (2005), 321-329. Here is [Wayback Machine version](http://web.archive.org/web/20111222124744/http://mysite.verizon.net/nhindman/research/jams.pdf) and als a new link -  http://www.nhindman.us/research/jams.pdf
Comment: Thanks Martin for pointing out the linkrot. It also gives me an excuse to reread Hindman's entertaining and informative survey!
Answer: Semigroups provide a fundamental, algebraic tool in the analysis of regular languages and finite automata. This book chapter (pdf) by J-E Pin gives a brief overview of this area.
Comment: I overlooked this answer while writing my own; Pin's book is a nice reference, thanks for mentioning it.
Answer: Victor, I don't understand your claim that $C^0$-semigroups aren't really semigroups. You are not free to decide for all the mathematical community what is a semigroup (I guess that you are interested only on discrete semigroups, aren't you ?).
$C^0$-semigroups are fundamental in PDEs (in probability too as mentioned by Steinhurst). The reason is that a lot of evolution PDEs (basically all parabolic ones, like the heat equation, or Navier-Stokes) can be solved only forward but not backward. In linear PDEs, this is a consequence of the Uniform Boundedness Principle (= Banach-Steinhaus Theorem). There is a nice theory relating operators and semigroups, the former being the generator of the latter. In the linear case, a fundamental result is the Hille-Yosida Theorem. Subsequent tools are Duhamel's principle and Trotter's formula. A part of the theory extends to nonlinear semigroups.
Edit. John B. expresses a doubt on the fundamental aspect of semigroups, compared with the evolution equations from which they arise. Let me say that semi-groups say much more, for the following reason. Evolutionary PDEs have classical solutions only when the initial data $u_0$ is smooth enough, typically when $u_0$ belongs to the so-called domain of the generator. This result can never be used to pass from a linear context to a non-linear one via the Duhamel's principle. In other words, in order to have a well-posed Cauchy-problem in Hadamard's sense, we need to invent a notion of weaker solutions ; this is where the semi-group theory comes into play.
Comment: While you say that $C_0$ semigroups are fundamental, isn't it true that all can one can do with them can be done by using the evolution families that give rise to them? I would call these evolution families fundamental, but I have difficulties agreeing with the same for the consequent semigroups.
Answer: An important application of semigroups and monoids is algebraic theory of formal languages, like regular languages of finite and infinite words or trees (one could argue this is more theoretical computer science than mathematics, but essentialy TCS is mathematics).
For example, regular languages can be characterized using finite state automata, but can also be described by homomorphisms into finite monoids. The algebraic approach simplifies many proofs (like determinization of Buchi automata for infinite words or proving that FO = LTL) and gives deeper insight into the structure of languages.
Comment: I just gave you a reference...  Here, is this better: N. Bourbaki _Éléments de Mathématique I: Théorie des Ensembles_.
Comment: Interestingly enough, N. Bourbaki describes exactly this use of the theory in his set theory book, in the appendix to chapter 1.
Comment: That's interesting - could you provide a reference?
Answer: Circuit complexity. See
Straubing, Howard,
Finite automata, formal logic, and circuit complexity.
Progress in Theoretical Computer Science. Birkhäuser Boston, Inc., Boston, MA, 1994.
If you want a research problem relating circuit complexity with (finite) semigroups, there are many in the book and papers by Straubing and others. See also Eilenberg and Schutzenberger (that is in addition to Pin's book mentioned in another answer) - about connections between finite semigroups and regular languages and automata.
Answer: Semigroups of bounded $L^{2}$ operators are very important in probability. They in fact provide one of the main ways show the very close connection between a self-adjoint operator and a `nice' Markov process (nice can be taken to mean strong Markov, cadlag, and quasi-left continuous.) So how does one get this semigroup from a Markov process? If $X_t$ is your process let $\mu_{t}(x,A)$ be measure with mass $\le 1$ with value $P(X_t \in A | X_0=x)$. Then $\int f(y) \mu_t(x,dy) = T_tf(x)$ gives a semigroup of bounded $L^{2}$ operators.
Why are such constructions important and natural? If $f$ is your initial distribution of something (heat for example) and $X_t$ is Brownian motion. Then $T_t$ acts by letting the heat distribution $f$ diffuse the way heat should. This then gives a nice way to connect PDE and probability theory. I'd end by offering that semigroups are important, in part, because they do arise is so many places and can bridge between disciplines. There are other reasons as well.
Comment: Not necessarily, although that is the most common usage. But there is a huge literature on non-commutative dynamical systems, motivated by quantum mechanical problems.
Comment: When people say "semigroup" in relation to functional analysis they really only mean semigroups isomorphic to the non-negative reals, right?
Comment: Thank you very much -- this is the type of answers I'm looking for!!!!!!!!!!!!!!!!!!!!!! I'll look in that direction.
Answer: (Commutative) semigroups and their analysis shows up in the theory of misère combinatorial games.  The "misère quotient" semigroup construction gives a natural generalization of the normal-play Sprague-Grundy theory to misere play which allows for complete analysis of (many) such games.  (See http://miseregames.org/ for various papers and presentations.)
Comment: Also finitely generated commutative semigroups are equivalent to Petri nets.
Answer: Though you say that $C_0$ semigroups are not really semigroups, the structure of compact semitopological semigroups plays an important role in the investigation of their asymptotic behaviour. For example, Glicksberg-DeLeeuw type decompositions or Tauberian theorems are obtained such a way, see Engel-Nagel: One-Parameter semigroups for Linear evolution Equations, Springer, 2000, Chapter V.2.
Answer: Toric varieties in Algebraic Geometry!! Indeed, the category of normal toric varieties is equivalent with the dual of the category of finitely generated, integral semigroups.
Answer: Why nobody explicitly mentioned one of the most natural examples: symmetric semigroup of a set? This is the semigroup analogue of a permutation group (https://en.wikipedia.org/wiki/Transformation_semigroup).
Comment: Actually, it does not answer the question. It's just yet another example of semigroups which appear in mathematics. The question (as I understand it) was: are there examples where presence of semigroup structure provides a new insight, leads to new results, simplifies proofs, etc.
Answer: Unary (1-variable) functions mapping a set X to itself under composition is a semigroup.  Cayley's Theorem (one of them) says that every semigroup is isomorphic to one of this kind.
Gerhard "Ask Me About System Design" Paseman, 2011.02.18
Comment: That is definitely true, but here you see semigroups appear just like they appear and nothing else. Furthermore, this sort of says that every semigroup can be realised by something. What I'd like is that something would be resolved (may in particular) by means of semigroups use. It's hard to explain what I mean.
Answer: Given a group $G$, the Block Monoid $B(G)$ consists of sequences of elements in $G$ that sum to zero.  So for example, an element of $B(\mathbb{Z})$ is $(-2,-3,1,1,3)$.  The monoid operation is concatenation, and the empty block is the identity element.
Given a Dedekind domain, one can take its ideal class group, and consider the block monoid over that group.  Note that in the obvious way, elements of the block monoid can be irreducible or not.  One can study irreducible factorization in the Dedekind domain by studying irreducible factorization in the block monoid.

Why semigroups could be important?


Paste timestamp hotkey?

Question: I would like to press a selected hotkey to paste a timestamp in the YYYYMMDDHHmmSS format 20180509145153 immediately, like AutoHotKey allows on Windows.
Comment: What is your operating environment? Linux, NetBSD, Solaris?  And GUI (which?) or text console?
Answer: You could do it at different levels depending on where you want it to happen. For instance, you could configure your shell or prefered text editor to  do that, but that would only affect them.
Or you could configure screen/tmux or your terminal emulator to do it, and it would affect commands running in that terminal emulator when the terminal emulator is the active window.
Under X11, you could use xbindkey to handle a specific key or key combination and have it simulate a series of keypress and keyrelease events (using xte or xdotool for instance). Which would work for any application connecting to the same X server and is active.
For instance, add to your ~/.xbindkeysrc:
"sleep 0.1; xdotool type --clearmodifiers $(date +%Y%m%d%H%M%S)"
  control+space

for that timestamp to be sent as a series of key press upon Ctrl+Space. (I find that if I don't add the sleep 0.1, I'm sometimes missing the start of the timestamp).
Another option would be to store the timestamp as the PRIMARY X11 selection, and simulate a middle mouse button click, which for many application causes the content of the PRIMARY selection to be inserted at the current mouse pointer position. In ~/.xbindkeysrc:
"date +%Y%m%d%H%M%S | tr -d '\n' | xclip; sleep 0.1; exec xdotool click --clearmodifiers 2"
  control+space

Answer: You could do it with different scripts in different OS.
AutoKey in Linux
The AutoKey Python script language is used for the following solution. Tested in my  KUbuntu version 20.04 (based on Ubuntu).
Set the Hotkey in the AutoKey GUI is self-explanatory (for e.g. Meta+n)
# output = system.exec_command("date") # Sa 20. Feb 09:38:39 CET 2021
output = system.exec_command("(date +%Y%m%d%H%M%S)") # 20210220093839
keyboard.send_keys(output)

AutoHotkey in Linux
You can also use AutoHotkey in Linux (some settings with Wine are necessary).
Here is an example Github solution: https://github.com/sl5net/Lintalist4Linux . Its using the tip's written in https://ubuntuhandbook.org/index.php/2020/01/install-wine-5-0-stable-ubuntu-18-04-19-10/

Paste timestamp hotkey?


How to use any module that has no library for arduino?

Question: I am 2 years old in this Arduino world.
I am currently working on a project where I am about to make use of a fingerprint sensor. Now my question is that could we be able to make use of a sensor that we usually find in our mobile phones with ATmega328p microcontroller chip. And what if that there were no libraries written for Arduino so that will make my life easier. And I wonder how these tech companies make use of any sensor they would like to use without any trouble while there is no existing textbook that taught these employees while pursuing a degree in CSE or ECE about how to write code for a particular sensor. Now I know that there are certain protocols like SPI, I2C, UART, etc. But how do they do it, and how do they know about microcontrollers and IC's that were never taught to them in any degree or course? I always scratch my head while looking at the consumer electronics, and the IC's and Chips that they have on their PCB board were never found on google, And by chance, if an IC or microcontroller is found then I barely understand the datasheet or the documentation which was written in Chinese or some other language. I hope someone could help me understand how all this happens.
Comment: I assume the datasheet of the IC is the main thing you need. If that's not available, it's like reverse-engineering, but for a non trivial IC this mostly is not worth trying.
Answer: About the fingerprint sensors from mobile phones: You might find the answers to this question useful. They explain in detail, why it is not really worth the effort to try interfacing such a fingerprint sensor with Arduino.

How to use any module that has no library for arduino?

You need to read the datasheet of the module. It should explain in detail, how the module is used. When you understood, how it works, (and understanding it is really really important) you can start implementing it in code, for example by writing your own library. A library is just a wrapper around the interfacing code for that module. If the module uses common used interfaces or is even a commonly used component, you can often stand on the shoulders of giants by making use of other libraries in your library (for example using the Wire library in a library for a specific I2C sensor). That makes it easier.

But how do they do it, and how do they know about microcontrollers and IC's that were never taught to them in any degree or course?

The engineers don't get taught to use specific controllers or ICs. They get taught the fundamental principles, including common interfaces like I2C, SPI, UART, I2S ... . In the end it all comes down to the datasheet of a component. It should explain all details, that are needed to use the component correctly, or at least reference to further documents. For example most microcontroller datasheets directly explain, how I2C is implemented in these microcontrollers. You can use that to implement I2C communication, even without a preexisting library. Yes, that can be a tedious and complex task, but it is totally possible. Keep in mind, that most Arduino libraries are not written by the manufacturer of the used component, but by third party people (sometimes the community, sometimes resellers like Adafruit, e.g.).
I would say "it is not rocket sience", but yes, it can be rather difficult to dig into the very detail of interfacing a component. Without considering myself an especially smart guy, I'm writing a library for multi-master I2C communication on Attiny's (without having any degree in that field). I could start with code from others, which made the entry easier. But it also involves much time, to dig into it, write the code and test it correctly. It is difficult, but it is possible.

And by chance, if an IC or microcontroller is found then I barely understand the datasheet or the documentation which was written in Chinese or some other language.

Good manufacturers and resellers provide a datasheet in english for their products. That cannot be said for every product, that you can purchase. Writing documentation takes time and thus costs money. Chips directly from China (where most electronics is produced these days) might have less good or no documentation at all, though these chips are often way cheaper. As a hobbyist you should stick to the good documentated components, if you don't want to really dig into it.

the IC's and Chips that they have on their PCB board were never found on google

PCB designs are only found in the internet, if they are published under some open source license (or if they were illegally leaked by someone). If a company doesn't want to publish it's product designs, that is totally ok.
About the components in consumer electronics: I think you refer to that the chips often don't have a googleable product number/marking to find a datasheet. That is because often these chips are custom chips, made maybe only for this company or even this specific product. Why would they want to publish a datasheet for it? They are not meant for public use for all. The design of components is often a good kept secret, that the companies are not willing to share, since it generates their revenue.
Hobbyists and small companies use publicly available, good and publicly documented parts to make their life and investment in the project easier. Big companies can easily hire people to design a custom chip, that will only contain the needed features. (With mass produced products you can save money this way, by not always needing to buy even the unused resources of a standard controller). Another reason to use a custom (maybe even a not programmable) chip is to keep the code in it secret.

All in all keep in mind, that the hobbyist world of Arduino and the world of consumer electronics like smart phones are rather different. Sure, they rely on the same principles, but the way, how things are done and created, is very different.
Comment: Electronics of IC internals are like electronics of discrete components, but much more complex. You first need to understand electronics in general, before you can even start with thinking about IC internals. I think this is out of your reach. Instead you can simply use publicly available parts to make your product and sell it. That will be way easier, especially if you want to start selling your product in this century. Take your passion and put it into normal electronics, not in designing ICs. When you have enough experience, you can still look into that topic.
Comment: Reading a datasheet is a bit of an art, and trying to start with something like an SPI sensor that lacks even demo code makes for a very steep learning curve.  If you want to learn about interfacing over one of these buses, try getting something like an LM75 temperature sensor, for which there are decent datasheets.  Then try to interface without too much googling - and without using a single line of code you don't understand
Comment: Thank you so much for such a piece of detailed information. So what should I consider to learn so that I could be knowledgeable enough to develop my own custom IC? Are there any books and references for this.
Comment: To be honest: No I cannot recommend things to learn for this. There are certainly many good books out there, but I don't know enough about this field to recommend anything. Designing your own IC is way over anything here. It is not something, that you can easily learn. And you are on the wrong site here to ask for it, since Arduino is not about custom IC design. Why do you want to design your own IC in the first place? From your question I assumed, that you are not doing this as a job
Comment: I just asked to give it a try . I am actually not ready to design any custom IC. And by the way I am very much passionate to create a product and sell it. I have an idea and I wanted my imagination to get real . Since I am new to electronics as I am a CSE student. I always stick to software engineering projects. I have never really thought that I will be so much passionate about electronics as I am today. I wanted to tinker a new product that is making rounds all day in my head. But due to my background in electronics I feel like I am lost.


How to deal with a client who redesigned their own site badly?

Question: I am a freelance designer/developer. I am starting a registered business and will be offering other services soon and I am in the process of expanding the portfolio section of my existing website but have noticed that one of my previous clients has had a bit of trouble and has tried to mimic the design I created for him and do it himself on a Wordpress site.
I originally built him an actual website, on Windows hosting environment and the new version he has created (trying to mimic what I had created) looks terrible.
This is going to reflect badly on me since I am showcasing his website in my portfolio. Now when people go to check out his site, they will no longer see what I built, they'll be seeing a terrible clone of what I made.
How would you deal with this?
I have recently found out that my client had lost my number and spoke to my friend about contacting me but didn't. And based on what was said when I finished the website years ago, there's no reason to assume that he didn't want to contact me again.
Would you contact this client and see if they would like the website put back to its original state (and make any new updates that they wanted)?
Or would you just take the link down from your website's portfolio, and just include a screenshot of the work you did for them?
Or would you just show a list of logos on your portfolio for the people you've done work for?
Or would you include the link to the client's site, along with a screenshot and also some disclaimer somewhere stating that while the screenshots were correct at the time, they may change in the future?
I'd really like to contact this client to see if I can help them, because I know he likes his site to look perfect, and given what he said years ago, there's no reason for me to think that he likes what he's created. But I haven't contacted him yet because I don't want to be rude - because there's always a chance that he has changed his mind, or may be a little proud.
Answer: Joey, you asked several questions there, some answers are going to be personal preference, some others I have done myself and think they might apply to you as well. So, here it goes:
Deal with it
The web changes. A lot. Work done several years ago cannot and should not apply today. I have no clue on how many work references you have, but assuming you are in the business of creating websites and you do it well, your portfolio should either be filled with stuff or better, only showcasing a select few, aka. your absolute best work. Your work done a couple of years ago, does not interest me as a client. I want to see your current work, my expected output if I pay you to do work for me.
If I knew, I could
Contacting old clients is great service. Politely bringing your name back on the table is not considered rude. At least, I wouldn't feel that way. Actually, I would be flattered knowing that my designer is monitoring his old work and wanting to make sure that everything is in order. Even if it's years old.
Showcasing work, aka. what you write about your work
This is a tricky one. Basically, I am trying to only showcase work potential clients are interested in and are able to understand. Why? As a designer/programmer you tend to focus on your personal achievements during projects and not the actual outcome for the client. I always try to touch base with clients after work is done and ask questions like "how did your business improve after relaunch" and "what is the most awesome part you like with the new code/design". Sometimes this leads to follow-up projects, sometimes it leads to disappointing answers, but sometimes this leads to testimonials which are "golden". Conveying your unique selling proposition with the words of clients is big. In fact, my highest converting showcases portfolio work are the ones where the client does the talking.
Showcasing work, aka. what you show
I like to keep it simple. Every project gets a dedicated page, with more than one screenshot. I only focus on parts that I actually did. There is no point in screenshoting the homepage, if you didn't do most of it. I always link to the client's website. No matter, if the work is actually present or even made for the web. It's something I call pride. And customers do see the backlinks in their Analytics or their SEO tools. I am also trying to only showcase my top references, knowing that old ones will be replaced. This is unfortunate for the reference, since I am really proud about some older projects of mine, but do show them if I am asked to present more references to potential clients.
What's next
No matter if you are large or small, you should always touch base with older clients from time to time. Don't be shy to call them up, write them an email or try to sell them more. This is business and should be treated as such. There is no time in being an introverted business man. This also applies, if not even more, with clients it didn't work out so well. In fact, they might be the best ones to actually understand what didn't work and how to improve your business.
Comment: Thank you so much @DKOATED, I now see this in a whole new light. I am going to contact the client
Comment: Good answer. Relationships to former clients is a very good thing for every small business / freelance worker, not only for webmasters and designers.
Answer: It seems to me that you have nothing to lose in contacting your old client to see if they require work to be done to their website. You will achieve nothing by not asking and it could be profitable if you do. You don't have to be so blunt by saying "your site looks terrible", simply regaining contact and enquiring about work could be sufficient. You cannot force them to make changes as they might be (naively) happy with what they have, or simply not have the cash to do more.

Or would you just take the link down from your website's portfolio, and just include a screenshot of the work you did for them?

I don't think you should link to the website if it is not your own work, or you are unhappy with the result. This can only count against you. By all means include screenshots of your original design, date it and place a footnote stating that the live site has since been updated by a 3rd party - or something to that effect.

Or would you just show a list of logos on your portfolio for the people you've done work for?

Logos might be OK for large established organisations and perhaps where you have done development work, not necessarily associated with the design. But if you are wanting to build a design portfolio then a logo isn't particularly helpful; unless you have designed the logo.
Comment: Thank you @w3d, based on what you and DKOATED have said I am going to contact them.
Answer: I'm in a similar situation.  I built a custom magento site for a client in 2009.  I quit the agency I was working for and moved to the West Coast.  He hired a very inexperienced webdev that didn't know magento but he knew Wordpress so he recreated the rest of the site in wordpress.  The design is great but there are errors all over the place.  I assumed he was happy with the current guy who was helping him but little did I know he was trying desperately to track me down! :(
One day out of the blue (after 3 years) he found me!  Now I'm stuck with a two wordpress installs (several duplicate databases that cross reference one another), an oscommerce install, and one very out of date magento install.  He never deleted any files, just kept adding on!
With all that said, the client is very happy to finally have it back on track and I'm happy to add him back to my client list.
Final answer: Contact him - keep it as friendly as possible. Perhaps just let him know you were checking up on previous clients to see if they needed any help or additional maintenance.


What counts towards 100% in Red Dead Redemption?

Question: I played the whole game and I'm at 79%. But I can't find anything else to do, what counts to 100% completion? I found all the places, there aren't any side missions that I'm aware. I'm stuck and I'm not sure what else to do. Well, I haven't completed all the challenges in the game, but, I don't think that's the only thing I'm missing.
Comment: Do you have saves before the ending?
Comment: @Brok3n No, I don´t but for what i read, there is no missable missions in the game. Just "I know you?" or something like that, and I alredy finish that one. Why do you mention it?
Comment: @Zarathustrax "I know you" is not required for 100% as it is missable.
Answer: Rockstar used to have a web checklist for this, but it appears to be taken down. Anyway, 100% requires all:

Story missions (57)
Outfits (9)
Rare weapons purchased (5)
Houses owned (the two you can 'rent' do not count, 13)
Gang hideouts (7, the exclusive PS3 hideout does not count)
Challenges done (10 for sharpshooter, survival, hunting each, but 9 for treasure hunter)
Named bounties (8 in Mexico, 8 in New Austin, 4 in West Elizabeth)
Stranger tasks (19, note that these CAN be completed post-game)
"Win" at every minigame (generally, break even or better)
Map locations (you can buy maps at general stores)

That should cover everything for 100%. For more in-depth information, I recommend the following GameFAQs guide:


Possible quantum states and eternity

Question: According to a TED talk by Sean Carroll (about 12' 40" in) if our universe is going to expand forever according the observed expansion rate we have calculated and the possible quantum states for our universe will occur in time span of $10^{10^{120}}$ years. The question is what would happen after that time span would pass?
Comment: Could you include a reference for where you got that number?
Comment: Yes ,from a lecture by Dr sean carroll at Ted talks
Comment: Your question can be made a good one, but only by you. First, the logic of your question is not clear, you missed multiple internal steps. Second, cite some references, "lecture by X" in a comment is not enough. Third, try to fix your grammar. Your question is unclear in its current state, what is a close reason.
Comment: First I would like to thank all the people that would trouble themselves to read and answer my questions.second I apologize for my English Grammer .regarding the references I have another reference but as the first one is in a lecture by Dr alexander velinkin he also gave the same number for the possible quantum states in our universe.
Answer: This number looks to be the Poincare recurrence time. In a world with $10^{90}$ particles the Poincare recurrence time is about $T~\simeq~10^{10^{90}}$ time units. The time units are comparable to the Planck time. This is similar to the number quoted here. This is also comparable to the stability of the de Sitter vacuum. The question is then; What happens next? That is not certain, but it may be the cosmos quantum fluctuates into a new cosmology.
There is also another recurrence time. This is the time it takes for the quantum states to return to the phase they had at some prior time. This is exponentially larger, $T~\simeq~10^{10^{10^{90}}}$ seconds. All of this may have some connection to multiverse, and the physics of duplication or near duplication of quantum states. If you were able to travel $10^{10^{90}}$ Planck units of distance out you might encounter a copy of the world here. These worlds though might be many world quantum alternative worlds to our world. The recurrence of the quantum phase might have something to do with the number of possible world branches in the yggdrasillian set of future possibilities.
Of course with this speculation there may never be any way to verify any of it. The many worlds interpretation of quantum mechanics is like all other interpretations likely not really testable. By the same there is no way we can observe anything that far in the distance, even if space is an infinite $\mathbb R^3$ space.
Comment: Thanks a lot. I thought that after the universe would pass in all quantum states it will return to the first (quantum state) thats a new big bang again.
Answer: Sean Carroll's talk is based on the idea that dark energy is, or behaves like, a cosmological constant and therefore in the far future the universe will be well described by the de Sitter geometry. Exactly what this means is somewhat involved, but for Carroll's talk the significance is that in a de Sitter universe the size of the observable universe is constant. For an observer in a de Sitter universe the bit of the universe they can see is surrounded by a horizon that is similar to a black hole event horizon, and this horizon is at a fixed distance that doesn't change with time.
The second key idea, which Sean Carroll doesn't mention in the talk, is that there is an idea called the holographic principle that states the maximum number of possible configurations of a system is proportional to its surface area. Specifically, the number of configurations is equal to the area in Planck units. This idea arose from studies on black hole entropy, though we should emphasise that the idea is poorly understood and highly speculative.
Anyhow, if you accept both the ideas discussed above then the maximum possible number of states of a de Sitter universe is the area of its horizon. Based on what we know about our universe we expect its de Sitter horizon to have a radius of about $10^{60}$ Plank distances, so the area will be about $\pi$ times this radius squared or about $10^{120}$ Planck areas (this is such a vague calculation we can simply ignore the factor of $\pi$).
And finally we can see where Carroll got his number $10^{10^{120}}$. If we have an observable universe with a fixed size, so it has a fixed maximum possible number of states $10^{120}$, then the amount of time taken for the universe to explore all these configurations is about ten to the power of the number of states or:
$$ T = 10^{10^{120}} $$
But don't take all this seriously because it depends on all sorts of ideas and assumptions that are exceedingly speculative to say the least. The point of my answer is purely to explain how Carroll got that magic number - it doesn't mean I believe a word of it!
Comment: But Sir let's assume that what Sean Carroll is saying is the true state of our universe,what dose that mean for a person like me who is not a specialist in physics (rather an architect)that means that the universe have to go through the whole possible quantum state again (that means we would live again without remembering any thing about our current lives) or I  have misunderstood Dr scrolls notion?

Possible quantum states and eternity


converting multiple png to mp4, all frames are the first image

Question: I am trying to convert multiple images (.png) into a video mp4.
I am using avconv, I already made it work before but I can't remember how ...
I am using ubuntu 14.04, I had the avconv 9.something, I just updated it to the version 11.6 by downloading it on the libav website.
I have around 70 png that I want to convert into a mp4.
I am typing the following command:
avconv -r 10  -i implant%4d.png -b:v 1000k test.mp4

or
avconv -r 10 -start_number 1 -stats -i "implant%4d.png" -b:v 1000k test.mp4

the video in output as the good number of frame, but all of them are the first image.
the output on the terminal is the following:
avconv version 11.6, Copyright (c) 2000-2014 the Libav developers
  built on Apr 28 2016 08:46:48 with gcc 4.8 (Ubuntu 4.8.4-2ubuntu1~14.04.1)
Input #0, image2, from 'implant%4d.png':
  Duration: 00:00:02.88, start: 0.000000, bitrate: N/A
    Stream #0.0: Video: png, rgb24, 1350x982, 25 fps, 25 tbn
File 'test.mp4' already exists. Overwrite ? [y/N] y
Output #0, mp4, to 'test.mp4':
  Metadata:
    encoder         : Lavf56.1.0
    Stream #0.0: Video: mpeg4, yuv420p, 1350x982, q=2-31, 1000 kb/s, 10  fps, 10 tbn, 10 tbc
   Metadata:
      encoder         : Lavc56.1.0 mpeg4
Stream mapping:
  Stream #0:0 -> #0:0 (png (native) -> mpeg4 (native))
Press ctrl-c to stop encoding
frame=   72 fps= 26 q=2.0 Lsize=     362kB time=7.10 bitrate= 417.4kbits/s

Answer: You need to change the wildcard.
From implant%4d.png to implant%04d.png
Works for me.
Comment: `implant%4d.png` means `implant1.png` to `implant9999.png`, while `implant%04d.png` means `implant0001.png` to `implant9999.png` (note the noughts)


How do enemies scale in Borderlands 2?

Question: I am considering buying Borderlands 2 today, but I've read in a few places that enemies don't scale properly. Some say that this only occurs on normal while others say otherwise. I just want to make sure that if I do get it - that the enemies will scale properly as I go along. So is it true or false that enemies don't scale properly?
Comment: Somewhat I think. I was not sure about DLC because I know that some DLC (I forget which) adds 10 or 11 levels or something along those lines so I figure that it has to modify the scaling of enemies somehow.
Comment: Enemies scale based on what area they're in, not based on your level. If there's a problem with the scaling, it has more to do with the order you do areas in than with the enemies themselves.
Comment: @Oblivious Sage so is it kind of like Pokemon (yes I know different games, but bare with me)? where different levels of Pokemon are in certain routes/areas?
Comment: See the question Yi Jiang's Evil Clone linked, but yes, until you meet certain conditions it's like Pokemon.
Answer: In Borderlands, enemies do scale properly.
Wherever enemies scale, it does not introduce problems or imbalances.
Player's health and gun damage (values easily inspected) increase by 13% per level. Other variables, like enemy damage and enemy health, scale accordingly - while I can't readily prove that it is 13% per level, it is definitely adequate.
Some areas of the game have enemies of fixed level, while other spawn enemies of levels (player level + some constant).

How do enemies scale in Borderlands 2?


Origin of the "white feather as a symbol of cowardice"

Question: I'm researching the origins of the white feather as a symbol of  cowardice. The earliest reference I can find is in Francis Grose's 1785 Dictionary of the Vulgar Tongue. Does anyone know of any earlier occurrences?
Comment: Is this question about using white feathers themselves as symbols or about the metaphorical use of the **phrase** *white feather*? The former would be outside the domain of this site.
Comment: Related: https://history.stackexchange.com/questions/42520/when-was-the-white-feather-first-used-as-a-symbol-of-cowardice
Answer: The OED uses Grose's dictionary as its first source and adds

A white feather as a symbol of cowardice. [...]With allusion to the fact that a white feather in a game bird's tail was considered a mark of inferior breeding.

1785   F. Grose Classical Dict. Vulgar Tongue   White feather, he has a white feather, he is a coward, an allusion to a game cock, where having a white feather, is a proof he is not of the true game breed.

Grose's example, game cock/breed is a use of "game" in the hunting sense and OED game, adj. 1. Of a person or animal: full of pluck, spirit, or fight; spirited, plucky; intrepid. Also applied to actions, attributes, etc.


Prove that white noise + normality = independence

Question: If the time series process is linear, then the ARIMA model is specified.
The residuals from this model are $(1.)$ no autocorrelation $(2.)$ mean equals zero $(3.)$ constant variance. We say that this process is a white noise process. But under white Gaussian noise, how to proof no autocorrelation = independent? And how to proof the white noise from Gaussian distribution?
Comment: Gaussian distribution does not imply white noise, so you cannot prove that.
Answer: No autocorrelation is not independent. To statistically check if the given time series is autocorrelated you can use Ljung-Box-Test, there is also many other test that have the same target. You can also plot the histogram, the Q-Q-Normal, and the Residuals vs Predicted graph, with those you can visually check if the residuals follow a normal distribution or not.


What effect does Brexit have on Britons' Right to be Forgotten?

Question: As I understand it, the Right to be Forgotten was promulgated solely as an EU regulation1.
Assuming that the UK follows through and leaves the EU, would search engines no longer have to remove results related to British subjects?
Answer: The 'right to be forgotten', as currently being applied throughout the European Union (EU), does not come from the General Data Protection Regulation (GDPR), which will come into force 25 May 2018. Rather, the current basis exists in the Data Protection Directive, 95/46/EC, article 12(b):

Right of access
Member States shall guarantee every data subject the right to obtain
  from the controller:
[...]
(b) as appropriate the rectification, erasure or blocking of data the
  processing of which does not comply with the provisions of this
  Directive, in particular because of the incomplete or inaccurate
  nature of the data;

This was one of the main points upon which the Court of Justice of the European Union (CJEU) decided Google Spain v Agencia Española de Protección de Datos (AEPD) and Mario Costeja González, the judgement which allowed for individuals to ask for search engines to remove results containing personal data. The CJEU explained that the 'right to be forgotten' was not absolute and should take into account economic considerations as well as other rights (paragraphs 85 - 88).
The GDPR, article 17, for purposes of contrast, has a more robust and explicit right of erasure than currently in force:

Article 17
Right to erasure ('right to be forgotten')

The data subject shall have the right to obtain from the controller the erasure of personal data concerning him or her without undue delay and the controller shall have the obligation to erase
  personal data without undue delay where one of the following grounds
  applies:

[...]
  Shortened for clarity

GDPR, article 17 fleshes out much of the lack of specifics that the Directive's article 12(b).
It should be noted that the UK's Data Protection Act 1998 is a transposition of the Data Protection Directive. This means that the full effect of the rights and obligations in the Directive are contained within the Data Protection Act 1998. Furthermore, the interpretation of the rights and obligations of the Data Protection Directive by the CJEU are highly influential on interpretation of the Data Protection Act 1998. This is due to the European Communities Act 1972, s 3(2) which states:

Judicial notice shall be taken of the Treaties, of the Official
  Journal of the Communities and of any decision of, or expression of
  opinion by, the European Court on any such question as aforesaid; and
  the Official Journal shall be admissible as evidence of any instrument
  or other act thereby communicated of any of the Communities or of any
  Community institution

Even without the above provision, it would be extremely likely that British courts would take into account Google Spain, given that the foundation of the Data Protection Act 1998 is the Data Protection Directive. The most likely outcome is for British courts to uphold the interpretation in Google Spain and allow for data subjects to request removal from search engines.
Of course, it is not entirely possible to know without a test case, but it is doubtful any major search engine would attempt to deviate from the current situation as there is currently legal certainty. The Data Protection Act 1998 will still be in force after the future Brexit agreement and is certainly not contingent on continued EU membership.
Update
The current DPA will be repealed when Brexit comes in to play and will be replaced by a new one. Read more about the new data protection bill on the ICOs web site, which is based on the GDPR with certain derogations and additions.
Answer: Something like "right to be forgotten" is based on British law, not on the UK's EU membership. Sure, the EU asked the UK (and all its other members) politely to create such a law, and the British government followed that polite suggestion by the EU, but the law is British law. It stays in place until it is changed or revoked by the British government. It doesn't just disappear if or when Britain leaves the EU.
Answer: Britain's exit from the EU is a minimum of 2 and a maximum of 7 years away. During this time there will need to be a whole swathe of domestic legislation that will deal with, among other things, this.
So, right now Britons are still covered.
Comment: It could happen sooner than in two years, if all the EU countries agree. The EU countries seem eager to effect the separation as soon as possible, so that agreement may not be all that hard to reach. (Also, which *Rhine* were you speaking of?)


What are some code coverage tools for Perl?

Question: Are there any good (and preferably free) code coverage tools out there for Perl?
Answer: As usual, CPAN is your friend: Have a look at Devel::Cover
Comment: Please don't link to a specific version of a module.  Try http://search.cpan.org/perldoc?Devel::Cover instead.
Comment: I still think the perldoc link is better in this case, since it shows you the documentation.  But the place to debate the merits of these link methods is http://stackoverflow.com/questions/43758/how-to-link-to-cpan-modules-in-answers
Comment: In this particular case, I think linking to the distribution is more appropriate, because there are a huge number of modules within it, as well as a tutorial.
Answer: Yes, Devel::Cover is the way to go.
If you develop a module, and use Module::Build to manage the installation, you even have a testcover target:
 perl Build.PL
 ./Build testcover

That runs the whole test suite, and makes a combined coverage report in nice HTML, where you can browse through your modules and watch their coverage.
Comment: thanks, I am just a bit puzzled , 3 replies to the same question in one day, two of them almost identical
Answer: As noted, Devel::Cover is your friend, but you'll want to google for it, too.  It's documentation is a bit sparse and if you change your environment radically, you'll need to reinstall it because it builds Devel::Cover::Inc with a bunch of information pulled from your environment at the time you install it.  This has caused plenty of problems for us at work as we have a shared CPAN environment and if one developer installs Devel::Cover and a different developer tries to run it, strange (and incorrect) results are common.
If you use this module, also check out Devel::CoverX::Covered. This module will capture much of the information which Devel::Cover throws away.  It's very handy.
Answer: Moritz discusses how modules built with Module::Build can use Devel::Cover easily.
For modules using ExtUtils::MakeMaker, an extension module exists to invoke the same functionality. Adding the following code before the call to WriteMakefile():
eval "use ExtUtils::MakeMaker::Coverage";
if( !$@ ) {
    print "Adding testcover target\n";
}

... will allow one to run the command 'make testcover' and have Devel::Cover perform its magic.


Is it possible to reset S.M.A.R.T. Spin_Retry_Count?

Question: I have a 20-bay nas that uses Norco 5 in 2 hard drive cages.
It's been at about half capacity most of it's life, but I recently added 5 new drives too it and ever since I've just been having sporadic problems but nothing that S.M.A.R.T would report on.  I've been swapping the drives around in different ports seeing if that made a difference (motherboard sata vs expansion on some ports), updating firmware, bios, drivers, etc. with little comfort it ever actually fixed the problem.
Today after another fix attempt, I noticed my top cage would no longer light up at all and I figured the cage itself might just be bad.  Then it hit me that it might be a power issue.  Looked at how I had them wired up and realized I had 15 drives all powered on the same lead from the power supply.  I re-configured to split it up over multiple lines and ever since it's been very stable.
In all of this chaos though, I'm now getting one of the drives I was moving around reporting "Spin_Retry_Count" errors.  The drive seems fine and I have a hot spare and proper backup strategy in place so I'm ok if it does keel over, but I'm fairly confident that this issue is related to bad power / swapping rather than the drive itself being a problem.
As such, I'd like to know if there is a way to reset this field so it won't bug me anymore on my smart reports.
Thanks for your help.  A linux command line option would be ideal.
Comment: This is a lifetime parameter and cannot be reset.
Answer: Per Michael Hampton's comment, it looks like Spin_Retry_Count is a lifetime parameter and cannot be reset.


How to display date/ time of current user base on time zone?

Question: Is it possible to display date/time of current user in his/her profile page. It should be base on his/her time zone settings. How can I achieve this goal?
Answer: You can modify the user pages (view/edit/etc) with the user templates, like the user profile template In the template's top documentation there's some good info like:
 * Use render($user_profile) to print all profile items, or print a subset
 * such as render($user_profile['user_picture']). Always call
 * render($user_profile) at the end in order to print all remaining items.

So you can first dpm the $user_profile array and find out the user's timezone, then add php code after the print render($user_profile); to print the date/time according to the timezone value you retrieved from the user profile array.


In Alliance War, is weapon damage boosted for sub-level 50 characters?

Question: I understand that player stats are boosted, but if you go in with a level 10 weapon does it still do only 20 damage, or does it get boosted as well?
Answer: All your stats are boosted, no matter whether they're based on your character (like health or stamina) or your actual items.
Just keep in mind that due to this items will scale similar to how your character is scaled. For example, if you're indeed level 10 and your weapon is level 5, then don't expect to have the equivalent of a level 50 weapon. It will be more likely around level 40 or even lower.
Answer: Stats do scale in Cyrodil, but it's important to remember that the system is far from perfect and doesn't scale a lot of things it probably should. For instance, I went in at lv12 on my Templar and it had me around 720 armor and resist, I'm now level 29 on my Templar and my armor/resist are both well over 900. This won't be a huge sticking point for doing the PvE content, but your contribution to PvP skirmishes will need to be either via healing, CC, or other support. Do not expect to rack up kills without outside help unless the opponent is also at or below your level. (Which will no longer be common)


How to expose API to developers securely?

Question: Hi I want to expose API of my web application to the developers so that they can build application on top of my API.
I want to develop it same as twitter does that means I want to build Consumer key,Consumer secret,Request token URL,Authorize URL,Access token URL,Callback URL,Access token,Access token secret.
I know some basics of Oauth that is first we have to generate request token. the request token then gets autherized on successful login by user. Then we can exchange autherized request token with access token.
I want to implement all this in my rails application. But not getting how to start with. I don't have any previous exprience on building and exposing API so any guidelines will be greatly helpful.Thanks in advance
Answer: Something like this should help:
http://stakeventures.com/articles/2007/11/26/how-to-turn-your-rails-site-into-an-oauth-provider


How do people get the proton number for each element from experiment?

Question: How did people determine the proton number for each element from experiment in each decade of 20th century?
Answer: Moseley, the physicist who 'fixed' the Periodic Table at the start of the 20th Century, did it by measuring X-ray spectra. The energy of the $K_\alpha$ X-ray emission line is proportional to $(Z^2 - 1)$, where $Z$ is the atomic number.
The results of Moseley's experiment fitted his formula so perfectly that he was able to predict the existence of several as-yet-undiscovered elements by looking at the gaps in his graphs. He also re-ordered the controversial placement of nickel and cobalt. Sadly he was killed in World War One before he was able to become the great scientific figure he surely would have been.
Comment: This is certainly true, but doesn't Rutherford scattering at large angles give you the nuclear charge too, and in a more direct manner?
Comment: Yes, and I'm sure there are other ways I haven't thought of (I know, for example, the X-ray scattering form factor depends on Z). I only gave one example in my answer because it seems superfluous to give several, and historically this was an important one. Not sure what you mean by 'more direct' though.
Comment: More direct in that it doesn't involve an uncontrolled inner-shell screening approximation, replacing Z by Z-1. The Rutherford scattering is clean at large angles, because it's off the nucleus itself. The x-ray scattering might be confounded by electronic scattering.


What is the first appearance of Elves in fantasy?

Question: Elves are a common appearance in many worlds in Fantasy. From Tokien, to Harry Potter, to Dungeons and Dragons, and Eragon. Many works feature Elves in some way, all with many common characteristics. But, I'm kind of curious, what's the earliest book that featured Elves?
Comment: fairies (and folklorish elves) appear in Shakespeare's A Midsummer Night's Dream - making them older than steam...
Comment: Do you mean elves as in the LoTR kind? immortal, wise etc.? I don't think that traditionally elves were supposed to be like that.
Comment: @apoor: read Terry Pratchett's "Lords and Ladies" for the exact opposite from Tolkien's elves.
Answer: The earliest novel-length published work with Elves in a recognizable "modern fantasy mode" is most likely Le Morte d'Arthur, by Sir Thomas Mallory, finished around 1470, and first published in 1485.
While Le Morte is literally nothing new, even when written, it's one of the earliest written works with recognizable, embodied fae in human shape, and armed and armored as men.
Note that Le Morte draws heavily on many tales and stories later recorded elsewhere; it comes together, however, as the definitive Arthurian tale, and the Fair Folk within are credited as inspiration for many later versions of Aelfdom...
Answer: The name elf has been borrowed from mythology, so the term itself has been used for perhaps several thousand years. However the term refers to different kind of beings depending on time and place.
The wiki page on elves gives an introduction to the concept of elves in different mythologies. In mythologies they are either treated as

Semi-divine and powerful magical beings; or
Small, mischievous beings (i.e. a subset of fairies)

Note that in the first case, they are either divided into light and dark elves, or represent light with dwarves representing the dark.
In fantasy, they have been popularised by Tolein into a very standard image, but people still use some alternate representations for them. This is discussed in detail at this wiki page - Elves in fantasy fiction and games. A summary of the article states that

The fantasy genre in the 20th century
  grows out of 19th century Romanticism.
  19th century scholars such as Andrew
  Lang and the Grimm brothers collected
  "fairy-stories" from popular folklore
  and in some cases retold them freely.
  A pioneering work of the genre that
  would come to be known as "fantasy"
  was The King of Elfland's Daughter, a
  1924 novel by Lord Dunsany. The Hobbit
  by J. R. R. Tolkien (1937) is seminal,
  predating the lecture On Fairy-Stories
  by the same author by a few years. In
  the 1939 lecture, Tolkien introduced
  the term "Fantasy" in a sense of
  "higher form of Art, indeed the most
  nearly pure form, and so (when
  achieved) the most potent". Elves
  played a central role in Tolkien's
  legendarium, notably The Silmarillion.
  Tolkien's writing has such popularity
  that in the 1960s and afterwards,
  Elves speaking an Elvish language
  similar to those in Tolkien's novels
  (like Quenya, and Sindarin) became
  staple non-human characters in high
  fantasy works and in fantasy
  role-playing games.
Post-Tolkien fantasy elves
  (popularized by the Dungeons & Dragons
  role-playing game) tend to be more
  beautiful and wiser than humans, with
  sharper senses and perceptions. They
  are said to be gifted in magic and
  mentally sharp and are characterized
  as lovers of nature, art, and song. A
  hallmark of fantasy elves is their
  long and pointed ears (a convention
  begun with a note of Tolkien's that
  the ears of elves were
  "leaf-shaped").[22]

Also note that in popular culture, Christmas elves are very different from Tolkienian elves, and are closer to fairies.
Comment: Also worth noting that in many fantasy series and games, elves follow the same sort of archetype as Tolkien used, but are also fey creatures. I don't know a whole lot about Middle Earth, but I don't think Tolkien's elves were associate with fey/fairies if such a concept even existed in that mythology. So while I'd agree that the 'tall, long-lived, nature-loving pretty boys' idea of elves originates (or at least is popularized by) Tolkien, it certainly feels like there is at least some influence from the original mythology as well.
Comment: HAHAHA nature loving pretty boys.  +1 for that @Asmor :)
Answer: I don't think there's going to be a definitive answer to this... Elves have been a part of folklore since long before any 'modern' books, and quite possibly before books even existed.
Answer: Brewer (17th Ed):

Elf [...] Originally a dwarfish being of Germanic mythology, possessed of magical powers which it used for the good or ill of mankind.

And then goes on to quote Shakespeare. So in answer to:

what's the earliest book that featured Elves?

A Midsummer Night's Dream is 1595 (strictly not a book on that date) so likely earlier.
Answer: I think faeries and elves came out of Irish mythology.  The Tuatha De Danann who are their gods were powerful beings skilled in magic.  Then the Christians (there's never any peace with them around) converted the Irish.  But the people still believed in the gods, only as faeries.
Answer: The idea of Elves may be very, very old. Hesiod (a contemporary of Homer) mentions the waves of the Golden race, Silver race, Bronze race. Tolkien's  chronology associates elves with the Silver race (as does the linguistic similarity: silver--silvan).
But the idea my also be autochthonous. There's an Inuit legend about a village of petite people who had the power of making their bodies lighter or heavier at will.
Comment: Silver in greek is άργυρος - arguros - and in latin, argentum. Sylvan is an adjective from the latin Silva, "forest"...


Adobe Flash Player 11.2 Security?

Question: The most recent Flash Player for Windows is 11.4.x, for Linux it is 11.2.x. Are there already any security issues? Are security patches still released for the Linux versions?
Answer: Adobe has changed their support for Flash on Linux.

NOTE: Adobe Flash Player 11.2 will be the last version to target Linux
  as a supported platform. Adobe will continue to provide security
  backports to Flash Player 11.2 for Linux.

Adobe is saying that they will provide security updates:

Mozilla, Firefox, SeaMonkey (Flash Player 11.2 is the last supported
  Flash Player version for Linux. Adobe will continue to provide
  security updates.)

link
---- content merged from Tom Brossman's answer ----
Instead of a standalone plugin, the latest version is available only in Google's Chrome browser for Linux.
Reading their press release from the marketing team, it seems that they are  'partnering' with Google.
Answer: Flash 11.2 regularly receives security updates, backported from the latest version by Adobe. These are installed through Update Manager.
Flash on Linux is not as "abandoned" as some suggest.
Comment: I'm pretty sure I recall seeing an update come across last week.

Adobe Flash Player 11.2 Security?


Am I doing anything wrong by getting a greeting card for an employee who is leaving?

Question: An employee gave her two weeks notice. This employee had worked for the company for 4 years then left the company. Then came back to the company after 6 months into a new position. Now one year later, the employee is leaving again. Since I have worked for this company, about 7 months. I have seen employees sign a going away card for the leaving employee and give a gift. I have gotten close to the employee who is leaving. So I asked the manager if we would be doing anything or getting anything for the leaving employee. His response was no because this employee left the company once before. I feel that was a little cold and I feel the leaving employee should at least get a card. I plan to buy a goodbye greeting card for the leaving employee and pass it around for employees to sign. I plan to not give it to the manager. Am I doing anything wrong by getting the card? The card would be paid for from my own pocket. If I had known the Manager would react like that. I never would have asked him.
Comment: Your title says 'greeting card', I thought you were being clever by suggesting she would be back with another new job again..
Comment: "*So I asked the manager if we would be doing anything or getting anything for the leaving employee*" Is that what you literally asked? Because if that's what you literally said I'd say no too. How am I supposed to know you're not talking dinner, party or gift?
Comment: I ended up getting the card and just walking it around getting people to sign it. I will leave it on his desk on the day he is leaving. That's all.
Comment: "I have seen employees sign a going away card for the leaving employee" So why not just sign this card? Or did you mean you've seen this for other leaving employees?
Comment: Do you know if there was any bad feeling or dispute between this employee and your manager? If there was, and it's related to the employee leaving, your manager might be touchy about this because they don't want you to shine a light on the situation.
Comment: I guess I didn't think of it like that. All I was thinking was maybe a card and maybe a $50 Starbucks gift card. Now I can see that maybe I was thinking of something more elaborate.
Comment: @DiligentWorker25 Yeah. It's like if I want to go to lunch with a bunch of coworkers. We don't ask. We just go. But if I ask my manager if a bunch of coworkers could go to lunch together he is going to interpret that as asking the company to pay, because why else would I ask? Personally, I think the company paying for a $50 gift card would be too much since this is the employee's second time leaving, and a card is so cheap that there's no reason to ask the company to pay (which, incidentally, is a large part of why it would not even occur to me if an employee asked me that)
Comment: I can see that now. I decided to get a card only and just walk it around the office getting different people to sign it. The employee leaving is popular and many people thanked me for including them. I did not give the card to the Manager to sign. He made his stance, so I didn't think he would be interested.
Comment: @DiligentWorker25 It only makes sense to ask people who care anyways.
Comment: If I had never asked the manager and just got the card and passed it around to sign, this would not be an issue. In one of the posts, someone made the point that maybe my manager thought I wanted to take the employee for lunch or some big gesture. That was not my thinking at all. So now I can see why the manager said no. I don't think a greeting card only will be harmful. And it's paid for from my pocket.
Answer: You are making a mistake by doing this behind the manager's back. The manager might feel disrespected after you asked and ignored the answer, or even that you are challenging the manager's authority by doing something the manager decided wasn't earned.
If you want to get her the card out of your own pocket, that's fine, but at least tell the manager your intentions if you want to pass it around, allow the manager to dissuade you, and offer the manager to sign it. If the manager pushes back again, just present the card with only your remarks.
Comment: I'd agree with this answer considering the limited information we have. But actually we have no idea if the manager's opinion is "Naahhh.... we gave her one already", or "It's strict company policy: One leaving card per person", or the manager's personality at all, or the mood / culture in the company. There are a lot of nuances involved in a question like that probably can't be conveyed on an anonmymous public forum.
Comment: It may be that the manager just doesn't like this employee and was annoyed that she was rehired, but would be in full support of a card for other employees leaving a 2nd time. Who knows... Fully support the OP giving a goodbye card signed only by him/herself.
Comment: I'd argue this is also a "don't know the circumstances" - if coworker has been asked to leave or is being pushed out the door, you might not celebrate in any way
Answer: Nope not doing anything wrong. You've done it of your own volition and out of your own pocket.
I understand the manager's reason(s) - if they've left previously and come back - it's a little petty, but I get it.
You are fine though.
Comment: @Ertai87 I would hardly call getting a card with some well meaning messages a "gift" worth farming. We're talking about a few cents of whatever is the local currency.
Comment: Each person gets 1 leaving gift.  You can't "farm" gifts from the company by joining and leaving in this way (of course there are other considerations as to why someone may not want to/be able to do so, but anyway the manager says they can't, absent those other concerns)
Comment: @user3067860 So they shouldn't be considered as persons worthy of a few cents spent on them and a token of gratitude for the months working alongside them? This is so petty I can't even begin to describe it. You also make it sound like someone selling their services (employees) are getting something they're not worthy of.
Comment: Can you explain? I don't get the Manager's reasoning.
Comment: I read this as though your coworker had been gone for 6 months, and back for 6 months at first, which as a manager would have been super annoying. But, yeah, you can't really get frustrated at people who leave after a year.
Comment: "it's a little petty" - more than petty, it sounds to me like the manager had some dispute with this person (probably part of the reason they're leaving relatively soon), there's a bad cloud in the air, and the manager doesn't want anyone shining a light on the situation.
Comment: In addition to @Ertai87 answer - there's a length of time aspect - when I left my last company, I had been there for 6-7 years. I got a nice company meal and a plane ticket to fly down to our capitol city to see one of my favourite extreme metal bands - about $300 worth of present. If I'd come back and only worked for a year, it would be extremely presumptious to expect the same treatment.
Comment: @DiligentWorker25 Usually people who left and came back were given a _generous_ raise or bonus to get them to come back. If they're now leaving again quickly, especially if it's after exactly one year (a lot of bonuses have conditions like you must stay one year) it's not a good look.
Comment: @Jonast92 It's not really the card, but I'd rather have the bonus go to someone who stayed and worked hard and didn't have a foot out the door the whole time. It's like parents who spend all the time and attention on the "bad kid" and just ignore the "good kid" who never causes trouble. It reinforces the whole environment of you have to threaten to quit if you ever want to get a good raise, because you're never getting one otherwise.
Comment: @user3067860 I think in fairness the whole "you have to threaten to quit to get a raise" is more the fault of the employer for not seeing that long term employees should be valued and rewarded - I think that's where I'd get worked up, rather than at this
Comment: @terdon OP didn't ask the manager about a card though. If OP's words are accurate OP asked about "doing anything or getting anything". If I heard this, a card would not be the first thing that would come to my mind.
Comment: @user56reinstatemonica8 See my above comment.
Comment: @Jonast92 See my above comment.
Comment: @DKNguyen the OP's words are: *I feel that was a little cold and I feel the leaving employee should at least get a card. I plan to buy a goodbye greeting card for the leaving employee and pass it around for employees to sign.*
Comment: @terdon The manager isn't a mind reader. What you quoted is not what the OP actually said to the manager. What the OP actually asked was: "*if we would be doing anything or getting anything for the leaving employee*". If I was manager and an employee asked me that, a card would not be what I had in mind. I would interpret that as a request for me to organize a dinner or gift, for an employee that is re-leaving the company, no less. I'd say no to that too and it would be completely reasonable.
Answer: I understand how you feel, but ultimately the only reason why I would do this would be if I was personally close to the departing employee.  I would not give a gift to a random coworker, particularly against the wishes of my manager and the company at-large, and I would certainly not involve other coworkers in doing so.
If you are personally close to this coworker, by all means, buy this person a gift or a card and give it to them, in person, and in private.
Comment: In that case, give them a gift as a friend if you wish, but make it clear that it is not from the company, but from you personally as a friend.  There's nothing wrong with a friend giving a gift to another friend.
Comment: Make the gift personal and inexpensive. I once gave a jar of homemade jam in the same situation. He almost got tears in his eyes.
Comment: @DiligentWorker25 your comment would be an _excellent_ addition to your original question. Please edit that info in so everyone sees it when they read your question.
Comment: I am close to the person. I made the mistake of going to the manager first. His reaction was not what I was expecting.
Answer: I understand why you want to do this, but you shouldn't. Two different relationship dynamics are in conflict here:
(A) You and this individual as friends.
(B) You and this individual as employees of the company under a common manager.
As a professional you need to clearly separate those out, and act appropriately.

I plan to buy a goodbye greeting card... The card is paid for from my own pocket.

This is completely fine. You're doing a nice thing for a friend. Relationship A dominates. Relationship B has little relevance. For example you doing this outside working hours and even premises would not affect the gesture - in fact you may well consider doing that!

(I plan to) pass it around for employees to sign. I plan to not give it to the manager.

With this, though, relationship B dominates. You've involved other employees of the company as a collective, and by definition will likely have to do so in working hours and/or premises. Worse, you've indirectly involved the manager too by defying their wishes.
It'll still be a nice gesture from their point of view, of course, but relationship B just makes it overall inappropriate, and a negative for you, given the circumstances.
Your 7 month tenure at the company is a factor here too - you've likely not yet built the social capital at the company necessary to counterbalance any negative effect on you, in a way a more tenured employee may be able to.
Really, I think it's unwise to do this. I'd advise only doing the first part of buying a personal card/gift yourself and leave the gesture at that.
Answer:
So I asked the manager if we would be doing anything or getting anything for the leaving employee.

If this is literally what you asked then the manager probably interpreted that as requesting the company to organize a dinner or gift. What other reason could there possible be to bring it up with a manager? For an employee leaving for the second time, it would be completely reasonable to say no to that.
A card is so inexpensive that there's little reason to get the manager or company involved which would be why it might not even occur to the manager that was what you had in mind.


Sql Server 2005 how to change dbo login name

Question: I have a database with user 'dbo' that has a login name "domain\xzy".  How do I change it from "domain\xzy" to "domain\abc".
Answer: I figured it out. Within SQL Management Studio you have to right-click on the database -> Properties -> Files -> Owner field. Change this field to the login name/account that you want associated with the "dbo" username for that database. Please keep in mind that the login name/account you choose must already be setup in the sql server under Security -> Logins
Comment: Or make the same change via code:  use someDB;go;exec sp_changedbowner 'newOwner'
Answer: If you are trying to remap a login to a db user you can use sp_change_user_login
exec sp_change_user_login 'Update_One', 'user', 'login'
Comment: The sproc appears to be 'sp_change_users_login' (i.e. _users_). FYI the BOL also states that it's deprecated and ALTER USER should be used instead in the future.
Comment: Although usually this would work, the stored proc quite specifically **rejects** an attempt to alter the **dbo** user in this manner.
Answer: PhantomTypist gives a good answer using the GUI.  For achieving the same result with TSQL, you can use this code:
<code>USE [My_Database_Name]
GO
EXEC dbo.sp_changedbowner @loginame = N'domain\abc', @map = false
GO
</code>
Answer: This is a Windows login, not a SQL Server login, so you cannot 'change' the login name since it is linked to the user account in Active Directory.
Create a new Server Login (Windows) mapped to the new windows user (and remove the old one if necessary). Then in login's Security > User Mapping, permission that login to the appropriate database as user 'dbo' (or assign to the db_owner role)
Comment: I suspect you're not properly understanding the SQL Server security model. The *whole server instance* has **logins** associated with the server. These may be SQL or Windows logins. Each database on the server has a number of **users** with permissions to access that database. The database **user** is associated with a SQL **login** which are usually (but not required to be) the same. **dbo** is a *special user* on each database. OP is not trying to and a new arbitrary user with db_owner rights to the database - he's trying to associate a different server login with the standard **dbo** user.


What is Tension in a string? How is it produced at a molecular level?

Question: I can't understand the direction of tension. Why is the direction of tension at the ends of a string away from the object or block of mass? Can someone tell me what happens internally in a string? ps: string is massless
Comment: Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
Comment: Duplicate/related [Conceptual doubt in Tension force](https://physics.stackexchange.com/questions/267987/conceptual-doubt-in-tension-force/268002#268002)
Answer:
I can't understand the direction of tension. Why is the direction of tension at the ends of a string away from the object or block of mass?

I too had many misconceptions about tension and used to struggle with it. Now, this is how I treat it:
Tension is no 'extra force' that you need to learn about.
Given that it is massless and inextensible, it is simply the means of transfer of forces between 2 objects tied to the 2 ends.
Imagine 2 objects A and B tied to each other using a massless, inextensible string. Start pulling A. As you do so, the string becomes taut, A pulls B with a certain force that is equal to the force with which B pulls A, by Newton's third law. This would answer your question about the direction of tension in the string. It pulls A towards B because B is pulling A and pulls B towards A because A is pulling B ( the pair of 3rd law forces between A and B).
Now, in the same case of pulling, if there had  been a massless rod, the effect would have been the same.( only in case of pulling. Note that it doesn't work in the case where you push A towards B because a rod would stay stiff between them and a string would become slack)
Thus, an inextensible, massless string is just like a 'connector' between 2 objects.
Observe that I stress on the string being massless and inextensible.
If it is not, the string will be an extra object (say C) between the 2 masses and there will be 2 more pairs of 3rd  law forces: That between A and C and another (and a different force) between B and C.
Your second question has already been answered by others. In summary, a string is an assembly of many particles in a line, which tend to remain at a constant distance from its neighbours (if it is an inextensible string) due to electromagnetic forces, and hence simply transmit the force by object A till the other end at B.( apply Newton's third law (edit : and also second law) on every pair of particles in the string for better understanding)
Comment: I used only 3ed law on each and every particle on a string just as you told and it works. But here I will be linking a video from youtube that uses 2nd law as well as the 3ed law... KInda different approach.... Is this approach also correct? https://youtu.be/VXu2gatnMWE
Comment: Yes,  the second law should also be used, and you would have done so implicitly if the string is not moving, because net force on the fragments on the string will be zero (and that's when it becomes easier and you simply move on with the third law ). Otherwise, you will  have to use the 2nd law as done in the video.  Also, it is not a 'different' approach. Rather, my writing was less precise by not including the 2nd law. Remember that both the second and the third law are always applicable where ever there is force involved.
Comment: To understand why the application of second law is necessary, consider a string with some mass ( let it be inextensible to make it simple) and divide it into several fragments $m_1,m_2, etc.$. Object A pulls object B, both connected by the string, and the whole system accelerates. When  A pulls $m_1$, $m_1$ pulls A with the same force. However, the force exerted by $m_1$ on $m_2$ is a little lesser. Why?   $m_1$ is itself accelerating and hence net force on it is now non-zero. So, as you move towards end B, the tension is lesser than the force A exerts on that end of the string attached to it.
Comment: Now, construct and solve a problem to understand better.  A scenario similar to the one mentioned previously. Consider the string with mass m. Apply a force F1 on A (Take your favorite numbers for forces/masses). If the string is inextensible, the whole system moves with the same acceleration a (implying that A, the string and B have the same 'a'=$F_1/(m_A+m_B+m)$). Use the second (you know the acceleration and mass of each object) and the third laws to calculate force of string on B and that by B on string. (tension in the string at the pint of attachment of string to B). Similarly at A.
Answer: Tension force lies in electromagnetic category of the four fundamental forces.
When you try to stretch a string more than its length, you are actually trying to increase the intermolecular spaces between the atoms of the string. Atoms contain charged particles (initially in equilibrium) and when you disturb them with external force, they show reluctance (opposing force). This opposing force which is electromagnetic in nature, is the fundamental cause of tension in the string.
For direction, it is totally justified that tension (opposing force) will be applied in opposite direction of stretch tendency.
Answer: At the ends, the molecules of the object tries to stretch the string by pulling it's end molecule, so the string wants to regain it's shape and pulls back the molecule away from the object.
Microscopic interactions can be approximated as spring force at very small scale.
Answer: Tie one end of a string to a wall and pull on it.  Pull so it's taut and pull some more.  The string stretches just as a spring would.  The string is then in tension.  It is stretched, and just like a spring, exerts a force to bring it back to an unstretched position.  At each end, it exerts a force towards its center.
Unlike a spring, a string has no strength in compression.  Push on the ends of a string and it just folds, giving no resistance.  Push on a spring or even a solid bar, however, will put into compression, causing it to push outward from its center.  Compression is negative tension.


How do I breed an Evergreen Dragon?

Question: I see a new dragon, the Evergreen dragon has been created. How do I breed this dragon? How do I know when I'm going to get one? Are there any special requirements?
Answer: The Evergreen Dragon can be bred by breeding a Plant Dragon and a Cold Dragon. If you get a breeding time of 10 hours, that means you'll get an Evergreen dragon. If you get a time of 6 hours, you got a Lichen dragon instead as this is the other possible result.
You'll need to be Level 9 to get this dragon, since that's the earliest you can get a Cold Dragon of any kind.
Can't find an Egg picture yet, but the Evergreen dragon looks like this:
Answer: I get it with lichen and plant both level 16 in EBI
Comment: Welcome to Arqade! Please avoid posting one line answer. Your answer lack any reference to back it up.

How do I breed an Evergreen Dragon?


BLOCKS ARE NOT BRAKING

Question: I finished my house in minecraft and went back into Gamemode s. I tried to break my bed and it did not work. I tried to break other blocks such as wood and it would not break! I hope some one can help.
Comment: Please provide a video or more detailed description. There's a lot of things that could've gone wrong, from bug to lag or insufficient permissions.
Comment: "Game mode s", do you mean survival?  Are you on a server or single player?  Does it show the block breaking animation?  Does the block temporarily disappear and then reappear?
Answer: These are the reason I can think of that can cause this behavior

In case it's a simple bug you should try restarting your Minecraft.
You also might not be in the correct gamemode.  Confusion happens when you mix up using numbers and letters such as /gamemode 2 and /gamemode a.  Just to be safe you should try typing it all the way out like /gamemode survival
It could also be that your Minecraft is running slow so it is
causing the blocks to lag.  You can reduce this by going to your
Video Setting and turning your Render Distance down, as well as
changing your Graphics from Fancy to Fast, turning Clouds:
OFF, and turning Particles to Minimal
Make sure you are holding down the attack button.  If you have
played in creative a lot you may be used to instant breaking or maybe you are right clicking instead.  Just to make sure you
should check under controls to see which button your
Attack/Destroy is set to.

BLOCKS ARE NOT BRAKING


What is a ground loop in an electrical service?

Question: I can't figure out what a ground loop is in an electrical service (residential, commercial, industrial). All I can figure out is that it has something to do with different potentials in the system, but what is the exact issue and flow of electricity? Thank you.
Comment: We need a bit more info are you talking about a loop of wire buried in the ground to creat your grounding electrode? OR a parallel path in the grounding system like when a sub panel has the neutral and grounds connected not isolated BAD.
Answer: "Ground loop" is a phrase from audio design. Ask an audio engineer.
It is practically not a thing in mains electrical, and let's think about why.  Safety ground has two jobs:

Return natural current (ESD, lightning) to source (being earth)
Return human-generated fault current to source (being neutral).

The subject of a "ground loop" is all about managing and coordinating currents flowing on safety ground -- that is the logic, yes?  Which currents are those?  No currents. System design calls for no current whatsoever on safety ground, except during fault conditions (which should only be long enough to trip a breaker or GFCI).
From a mains system design perspective, safety ground is the one case where we want a "web" of connections.  It's perfectly OK for grounds to criss-cross among different circuits and even services.  We run a ground wire to an outbuilding which has a ground rod too (partly: we don't want lightning traversing the wire to the main building).  The more's the merrier, the more's the safer.
And that's why we give audio engineers isolated grounds :)
Now, "ground loops" are concern in audio and computer networking; but understand in the electronics world, "GND" is a completely different animal: GND is the "common" or "zero-voltage reference" or what we in mains call "neutral". And this is often used as "signal ground", or the "zero point" in a referential signal (like RS-232, which has 1 common and many signal wires).   As distinct from a "differential" signal like RS-422 (which has 2 signal wires per signal, and only the difference between them matters).
Particularly in networking/distance signaling, if this "common" is bridged to AC mains safety ground at two places, it becomes vulnerable to differences in voltage on that safety ground network - which shouldn't exist.
But it's possible to have an ongoing ground fault that is too little to trip a breaker and GFCIs are not installed... and that can introduce millivolts of gradient on the safety ground network, pulsing at 50/60Hz of course. Or a signal processing device can leak noise back through its power supply onto the AC mains, which can then feed it onto safety ground via capacitive coupling.
The lesson is that audio and network designers need to be careful about treating AC mains safety ground as some sort of signal zero reference.  Also, they need to be careful about binding wires or shields in their cable to AC mains safety ground - lest that wire suddenly be asked to carry dozens or hundreds of amps during a bolted ground fault.
Comment: It's not only audio (though you may be right that that's where the term originated) -- it's anything where your signal of interest is in the microvolt range, and you have to amplify it many orders of magnitude to use it.  Another common example is that most hospital rooms are wired with isolated grounds for their EEG-type equipment.  Any time you're working with a signal that small, any RF picked up by a ground loop functioning as an antenna can capacitively couple to your signal wires and appear in your output.
Comment: @NateS. Yeah. Safety ground shares this with GFCI - it improves life safety and is compatible with *almost* everything; and so it was clomped onto *absolutely everything* in a forceful and rough-shod manner, because that was the only way to overcome people's resistance to pay for it lol. So just as we have fridges we don't want on GFCI, we have hospital and signal gear we don't want on grounds.   For hospital/signal application my first thought is "separately derived service" but can't they just do that inside their own power supplies? 2-prong plug and ground is irrelevant.
Comment: They do, but they also need ground for another purpose -- shielding.  In many cases, when first detecting a real-world signal, it has to be single ended, so you can't use differential techniques to stop the sensor and wire from picking up RF as well, so the only option is to shield it to ground.  The ideal arrangement for these kinds of things would be a connection to a ground rod and *not* to the building's main ground system, but in practice a dedicated ground wire to the main panel bus works okay and is compatible with things that need a ground for safety.
Answer: The term "ground loop" usually refers to a circuit in which various "end items" have a ground wire attached, and the ground wire from one item goes to the ground attachement on the next, and so on, with an earth ground connection somewhere in this continuous loop of wiring.   The "end items" could be components on a circuit board or they could be the various consumer electrical items like outlets and stoves and lights in a house.
The problem with a loop like this is that any conducting loop is happy to act as an antenna, picking up pretty much any electromagnetic waves in the area (to within certain limits on the frequencies).  This then dumps noise into your electrical system, because the ground wiring is everywhere.
The preferred diagram is called "star-grounding," in which each end item's ground wire goes directly back to a common earth ground point.  In this way, the antenna gain is much less, and further because each ground wire dumps directly to earth, rather than passing by the other nodes, far less noise is transmitted into the devices of interest.


Eigenvalues that are complex numbers

Question: Have a square matrix problem that involves complex numbers and am at a loss.  $M$ is a square matrix with real entries.  $\lambda = a + ib$ is a complex eigenvalue of $M$, show that the complex conjugate $\bar{\lambda} = a - ib$ of $\lambda$ is also an eigenvalue of $M$.  Does solving this relate to a matrix having a characteristic polynomial of $(t^2-4)$? Could the community please explain?
Comment: Where id $t^2-4$ come from?
Answer: $M$ having real entries, its characteristic polynomial will have real coefficients. Say this polynomial is $P(X)=a_nX^n+a_{n-1}X^{n-1}+\cdots+a_1X+a_0$  One has by definition of eigenvalues that $P(\lambda)=0$ taking the complex conjugate of that identity and considering that all $a_i$ are real one gets $P(\bar\lambda)=0$ and this means $\bar\lambda$ is as well an eigenvalue.
Answer: If $Mv=\lambda v$ where $v=(v_1,...,v_n)$ is a vector of complex numbers, then $$M\bar v=\overline{Mv}=\overline{\lambda v}=\bar\lambda \bar v$$
Here by writing a line over a matrix (or vector) I mean to take the complex conjugate of every entry; the first equality holds because $M$ is real, so $\bar M=M$, and because as you can check directly $\bar M\bar v=\overline{Mv}$ for any complex matrix $M$ and vector $v$. So $\bar v$ is a $\bar\lambda$-eigenvector of $M$.
This does relate to the characteristic polynomial $\chi M$: every eigenvalue is a root of $\chi M$, and if $M$ is real then $\chi M$ has real coefficients, and if a real polynomial $p$ has a complex root $\lambda$ then $\bar \lambda$ is a root as well.
Answer: let the eigenvector corresponding to the eigenvalue $a+ib$ be $u+iv$ with both $u$ and $v$ real and $|u|^2 + |v|^2 \neq 0$. then we have $M(u+iv) = (a+ib)(u+iv)$ can be separated to give $Mu = au - bv, Mv = av + bu.$
now, $M(u -iv) = Mu - iMv = (au-bv) -i(av+bu) = a(u-iv) -ib(u-iv) = (a-ib)(u-iv)$ this shows that $a-ib$ is an eigenvalue of $M.$
Comment: *And* you get the eigenvector for $a - bi$ to boot!  Well done, ***Plus One!!!***
Answer: If you consider complex eigenvalues of a matrix $M$, then you are implicitly interpreting it as the matrix of a complex-linear operator $\phi$ acting on a complex vector space$~V$, even if $M$ has only real entries. The fact that $M$ has real entries then is basis dependent (it will probably not hold for the matrix of$~\phi$ with respect to any basis of$~V$); if $V_\Re$ denotes the real span of the chosen basis and $V_\Im$ the real span of the basis vectors multiplied by $\def\ii{\mathbf i}\ii$, then the fact that $M$ has real entries means that $\phi$ commutes, not only with multiplication by$~\ii$ as any complex-linear operator must, but also with "complex conjugation": the $\Bbb R$-linear operator $J:V\to V$ that acts by a scalar$~1$ on$~V_\Re$ and by a scalar$~{-}1$ on$~V_\Im$. Note that $J$ anti-commutes with multiplication by$~\ii$, so it is not complex-linear, but conjugate-linear: (it is $\Bbb R$-linear and) $J(\lambda v)=\overline\lambda J(v)$ for all $\lambda\in\Bbb C$ and $v\in V$.
Now if $v\in V$ is an eigenvector of$~\phi$ for an eigenvalue$~\lambda\in\Bbb C$, this means that $\phi(v)=\lambda v$. Now from the above $\phi(J(v))=J(\phi(v))=J(\lambda v)=\overline\lambda J(v)$, so $J(v)$ is an eigenvector for $\overline\lambda$.

Eigenvalues that are complex numbers