Is environmentally-driven purchasing of coal mines a good idea?

Question: Is environmentally-driven purchasing of coal mines a good idea? Certainly environmental non-profits won't have enough funds to do it on a large scale. Governments have, but they are corrupt. There's a huge risk that they'll just buy up those mines under the pretext of climate change and after the reserves deplete elsewhere and the prices soar, it'll just be one colossal fossil fuel cartel that would dwarf OPEC immensely
Comment: Related: [Shortcut carbon emission trading by buying and sinking coal](https://sustainability.stackexchange.com/q/6984/3379)
Answer: No this won't work, because there's so much coal reserves, unless you buy all of them.
With oil it might just work. Buy an oilfield and promise to not utilize it in any manner. However, the fossil fuel industry is probably capable of finding more oilfields than there is room for carbon dioxide in the atmosphere, and electric car transition is rapidly progressing, so I wouldn't be surprised if we naturally end up with unused oilfields. So the plan to buy oilfields could only work if there are more buyers for oilfields to be left unutilized than what oil would be left unutilized naturally.
With natural gas it might work. Natural gas will be used as transition adjustable power in electricity grids (because we don't have enough hydropower resources), to be replaced later with hydrogen. If you buy natural gas fields and leave them unused, it might accelerate the transition towards hydrogen. Yet, I wouldn't be surprised if the hydrogen transition accelerates so much that natural gas would be left unused naturally too. So not certain it will work, like with oil.
With coal no way is it going to work. The problem is, we have massive massive coal reserves, so much that using them all would warm the planet at least 10 degrees Celsius if not more. No way would all of them be utilized naturally. So your plan would work only if you purchased every single coal mine on this planet, and would continue buying them at a great rate as new mines are found.
A simpler way, much cheaper, would be to just buy carbon dioxide permits and promise to leave them unused. However, if enough many people start doing this, the policymakers deciding the number of carbon dioxide permits will increase the number of future permits, knowing that part of them will be left unused by intentional buyers, so the effect might be negated.
Another simpler way, much cheaper, is to buy forest and create biochar from the trees. No need to find all that money for a single oilfield, gasfield or coal mine.
Also a very important (if not the most important) way to prevent dangerous climate change is to vote in every election for a candidate favoring the environment. Also you should change all home appliances to the most energy efficient appliances possible, and switch to an electric car. All of them are cheaper than buying coal mines. Any unused money should be invested in companies that do good for the environment.
Comment: *So your plan would work only if you purchased every single coal mine on this planet, and would continue buying them at a great rate as new mines are found.* There's an old joke, [You don't have to outrun the bear; you just have to outrun the other people.](http://folklore.usc.edu/meaning-behind-the-proverb-i-dont-have-to-outrun-the-bear/) Similarly, you don't have to buy up *all the reserves in the world*. The more you buy, the more it costs for coal consumers to get at the remaining coal left over; you'd need to buy *enough to make coal uneconomical*, *relative to other energy sources.*
Comment: It would almost certainly be cheaper to buy politicians, or at least join the auction to drive the price up for the fossil companies. But it's worth remembering that an awful lot of fossil reserves are owned by governments (Saudi Arabia is an obvious example), and there is no way to use money to change their behaviour - not so much that they have more of it than you, but they are soveriegn countries so other people's money is less important to them. Limiting their ability to spend their money would be more useful, but that';s back into "vote. campaign" territory
Comment: That might still be impracticable, possible even absurd, given how much you'd have to buy up to seriously move the market. On the other hand, coal is already economically pretty marginal in many developed markets (hence the widespread shift in the US etc. to cheaper natural gas). I have no way of knowing what's possible here, but it seems like the thing you should be looking at is "Could somebody lock up enough coal to move the commodities markets?" not "Could somebody lock up all the coal in the world?"
Comment: @AlabamaScholiast The problem of coal is that we have 10x the amount of reserves we could realistically use without causing dangerous climate change. So if you buy 95% of the coal reserves, in that case you could have some effect. However, buying less than 90% of coal reserves is not enough, as we probably are going to use only 10% of reserves anyway.
Comment: *However, buying less than 90% of coal reserves is not enough...* Why not? Again, you do not need to lock up all of the coal reserves that people might use. What you need to do is affect the price of coal so that it is prohibitively expensive to burn it, relative to other things you might use for energy. If the price of coal increases significantly, then ceteris paribus for-profit companies will buy and burn less coal, even if there is still lots of coal left over for them to buy.


Can I play GTA 5 story mode with no wifi connection?

Question: I won't have a wifi connection for a while, is there any way I can pick up the game where I left it without connecting to xbox live and without starting it over? And if not, can I still save the progress if I do start it over? I have an xbox 360.
Answer: Yes. I play with no internet connection too. You dont need to be connected to save/load your progress in story mode.


A table contains forbidden words which you cannot use in a comment - is it a value object in DDD?

Question: My aggregate root is "Post" and sub aggregate is "Comment". In my database I have a table called "ForbiddenWords" with one column "Name" (it contains words for example: "sex", "porn"). If someone wants to add a comment, he can't use forbidden words. I need to create a class "ForbiddenWords" to validate a comment. Is this class a Value Object?
Answer: A value object is an object without identity. According to your description, a forbidden word is a value object.
For example, you would not be able to distinguish in your table two rows with the same word. Moreover if you would change a word, it would be different word with no relation to the previous one (the identity would not persist after across the changes).
Comment: Where to put validation logic is a broad topic. There are a couple of questions here (for example https://softwareengineering.stackexchange.com/q/210799/209774 or https://softwareengineering.stackexchange.com/q/341564/209774 ). If you don't find a suitable answer, I'd suggest that you raise a separate question (and explain what makes you hesitate about not having it in the domain model).
Comment: OK, thank you, so you resolved my doubts. :) One more question - do you think I can check if a comment doesn't contain forbidden words in a command handler CreateCommentCommandHandler, or it has to be in the domain model Comment?
Comment: Command handlers should not contain business logic. Checking comments for forbidden words is business logic.
Comment: thank you Christophe :)


How to convert multiple csv files to shp using Java

Question: I have some 100 CSV files. I want to convert them in shp format using Java. The csv file contains lat long information in WGS84. I can understand it is a batch processing and it is difficult to do in ArcGIS. is there any spatial plugin for eclipse that will do the task? I had a look at Geotools but found it bit complicated. Anything straight forward?
Comment: I am using ArcGIS for Desktop. The problem is when I am trying to do the batch processing through Make XY event it is not showing me the CSV files.
Comment: You'll need to create a table from your CSV file first through Table To Table
Answer: You mention Java in your question title but in your question body and tags it seems that ArcGIS for Desktop is available/preferred.
The ArcGIS for Desktop procedure that I would use to do this is to perform the following three steps on one CSV file to make sure the workflow works, and then use Copy As Python Snippet to copy the syntax for the three tools into a Python script that you add iteration through your 100 files to:

Table To Table
Make XY Event Layer
Copy Features
Comment: Thank you PolyGeo. That was helpful. I am not that verse with python. I was wondering how can I export all the layers to shape file in the Table of Content in one go?
Comment: @user26161 That sounds like a new question which is best researched/asked separately. I suspect it has already been answered elsewhere on this site but if not it I think it would make a useful Q&A in its own right.
Answer: If you want to use Java, you have a few libraries available including GeoTools and the Java bindings for GDAL/OGR API. Using these you can code a solution (check out the documentation).  However, far simpler would be to install GDAL which comes with pre-compiled utilities and use ogr2ogr.  You can wrap that in a batch process as required.


240v capable computer power supply is it neutral or phase to phase

Question: I'm trying to get some new wiring installed into my garage so I can add some high power servers (mining rigs). Anyway I bought some Evga 1000w G3 power supplies which are 240v capable, but my electrician wants me to find out if they are neutral or phase to phase. Why do I need to find this out, isn't there a single standard in the US? But if I need to does anyone know? The manufacturer site does not have this info and their chat tech help does not know either.
Comment: Perhaps the electician is asking if it should be connected to a normal 120 volt circuit, or to a 240 volt circuit (that would be phase-to-phase).  A 240 Volt outlet would be "unusual" in North America - usually only required for high-powered tools - he wouldn't expect a computer to need 240 volts..
Comment: The product is certified for use in North America by TUV Rheinland which is an accredited testing laboratory, so it shouldn't matter if it's phase-to-neutral or phase-to-phase - the safety marks should be good enough for the electrician in my estimation.
Comment: I believe he is asking if it is a straight 240 with ground and no neutral, or is it leg to leg with a ground and a neutral, it would make a difference on the wire he pulls.
Comment: The nameplate rating is 100-240VAC so I don't see what the issue is.
Comment: What he really needs to know is what NEMA pin configuration that you need for your equipment, answer that question for him and all other questions are answered.
Comment: Yes, he is asking if you need 2 wire with ground, or 3 wire with ground. Do you need just 240 volts straight (like a baseboard heater) or 240 volt with a neutral (like a range or dryer). I would say the power supplies are straight 2 wire no neutral needed. You should get one of the supplies for the electrician to examine.
Answer: Current flows in loops, so the power supply has two conductors coming into it.  (In Codeland we don't count safety ground as a conductor).
A world-voltage server power supply could contend with five styles of power. I'm mentioning voltage compared to ground, and neutral is near ground by design.

Euro 240V polarized: one 220-240V hot, and one neutral. (Illustration, upper left)
Euro 240V non-polarized: either conductor could be 220-240V, the other neutral. (Upper right) To deal with this, the unit must be insulated for 240V on either conductor.
North American 120V: one 120V hot, and one neutral, polarized. (Both lower left)
North American 240V: two 120V hots, out of phase so they stack to 240V, often no neutral. (Both lower right)
NYC or commercial 208V: two of 3 phases, both 120V from ground. (Not shown)

In 3 of 5 instances, the world-power device must be insulated for both conductors to be potentially hot, as high as 240V.  So the item would need to be tested/listed for hot on both of its conductors. The electrician is asking if that is so.. It's a routine question.
He should be using /2 cable and fitting NEMA 6-15 receptacles.  If the breaker is 20A he can fit 6-15 or 6-20 receptacles.

Or if you are coming out to a PDU, he can fit something larger like 30A or 50A.  But if you do that, have him run neutral and use NEMA 14-30 or 14-50 receptacles.  This will give you much more versatility: the PDU could support 120V switches or fans... the receptacle could support welders, EV chargers, you name it... Or the cable could support a future subpanel.
Answer: US 240V has both supply conductors at 120V relative to ground. European 230V normally has one supply conductor (the "line") at 230V relative to ground and the other supply conductor (the "neutral") at roughly ground potential.
Some countries in Europe have polarised plugs, others don't. Modern portable equipment intended for sale in Europe is designed to harmonised European standards meaning it is safe to use with Live and Neutral reversed. As a side affect of this it will also be safe and functional on a US style 240V system.
A couple of notes

You may need to be more careful about vintage equipment or equipment designed for fixed installation.
I strongly recommend you come up with a scheme for identifying which power cords have 240V on them so that you don't accidentally plug them into a 120V-only device.
Answer: This is very simple - The answer is: it does not matter. The PS is designed to connect safely to 120 VAC or 240 VAC supply. For 120 VAC, one line will be HOT (Phase) and one will be NEUTRAL. For  208 VAC or 240 VAC both lines will hot (phase to hhase). Any qualified electrician will install proper replaceable and cord cap for appropriate voltage and current. (I am a master electrician as well as power supply designer)


what size of sub panel do I need for a small in law unit

Question: Im converting my garage into a small inlaw unit for my daughter I have a 125 amp service to my house. she will have a burner plate, fridge and an mini split ac unit for air and heat. it has a gas water tank. large tv and computer. the Main panel is out side our house about 30 feet from garage.
there is not going to be laundry, unless you count the convienent laundry services provided by my wife in the main house!
Comment: Do you know how many amps or watts the mini-split will draw?
Comment: You should be able to get by with a 60 amp possibly a 50 amp feed to your panel, the size of the split can be a loose cannon here depending on the size but I would bet it will be a 30 amp max breaker size or smaller 220v unit (the last small single head I installed could have run on a 120v 20 amp circuit)  but most have been 240v
Comment: its a 230vac 18kBTU
Comment: It say to use a 35 amp breaker
Comment: How many square feet is this unit, and will it have its own laundry facilities?
Answer: I would use a 30 space panel, because that is a cost sweet-spot between "Make sure you never run out of spaces" and "sanely priced".  Electrical equipment is so cheap that there's no real economy in scrimping to the point of inconvenience.  And believe me, you will want to upgrade things later, if not in the near term, in the far.
The panel does not need a shutoff switch if it's in the same building. Generally if a shutoff switch were required, people simply get a panel with a main breaker; it's cheaper, but nevermind that.
If the panel has a bus rating or main breaker, that does not matter (as long as it's equal or larger than the feed breaker). So don't be bashful about getting a 150A or 200A panel.    The important thing is to get plenty of spaces.
I suggest a 60A cable feed and 60A breaker from the main panel.  The cable to use is #6 copper or #4 aluminum. (Aluminum starts to make sense at these large sizes).


Automount drive on startup

Question: I have an internal HDD with all my Steam games loaded on it. Everytime I boot my pc I have to open the HDD in the File Manager. Otherwise Steam does not recognise it and I have to manually add the Steam library back again.
Is there a good fix for that?
Answer: Lacking details on your setup, I can only give general guidance.
All you have to do is edit (as root) <code>/etc/fstab</code> and add the partition you wish to mount.  For details, options, etc, check out https://help.ubuntu.com/community/AutomaticallyMountPartitions

Automount drive on startup


How to prove this integral inequality $ \int_0^{2\pi} p(x)[p(x)+p''(x)] dx \int_0^{2\pi}\frac{1}{p(x)+p''(x)} dx\geq 2\pi \int _0^{2\pi} p(x) dx $?

Question: Let $p\in C^2(\mathbb{R})$ be a $2\pi$-periodic function such that $p(x)>0$ and $p(x)+p''(x)>0$ for all $x\in \mathbb{R}$. Then it holds
$$
\int_0^{2\pi} p(x)[p(x)+p''(x)] dx \int_0^{2\pi}\frac{1}{p(x)+p''(x)} dx\geq 2\pi \int _0^{2\pi} p(x) dx
$$
The identity holds when $p+p''$ is a constant.
Comment: Where is the source?
Comment: Can you provide some context? Where does the inequality come from? Why do you think that it holds?
Comment: @Icesea Thank you. I have fixed it
Comment: Is there not some missing power there?
Answer: I can verify this for
$$p(x)=a_0 \cos(0x)+a_1 \cos(1x) + a_2\cos(2x).$$
By hypothesis, $p(x)+p''(x)$ is positive.  So that sum is positive for $x=0$ and $x=\pi/2$, which means that $a_0+3a_2$ and $a_0-3a_2$ are both positive.  Averaging these, $a_0$ is positive. Multiplying these, $a_0^2-9a_2^2$ is positive. Dividing by three and adding a square, $2a_0^2-3a_2^2$ is also positive.
Now when we evaluate all the integrals, the inequality becomes
$$\big[\pi(2a_0^2-3a_2^2)\big]\big[\frac{2\pi}{\sqrt{a_0^2-9a_2^2}}\big]\ge 2\pi \big[2\pi a_0\big].$$
Equivalently
$$\frac{2a_0^2-3a_2^2}{\sqrt{a_0^2-9a_2^2}}\ge 2a_0.$$
Since the numerator, denominator, and right-hand side are all positive, this is equivalent to
$$\frac{(2a_0^2-3a_2^2)^2}{a_0^2-9a_2^2}\ge 4a_0^2.$$
That in turn simplifies to
$$24a_0^2 a_2^2+9a_2^4 \ge 0,$$
which is true.  Can we generalize this to other periodic $p$, probably by using Holder's inequality?
Comment: Well, edit the solution with Hölder's inequality in!
Comment: @ploosu2, if I knew how to do that I would have done it.


Why do Muslims add "peace be upon him" after names of important people?

Question: In websites, books and TV programs, I frequently hear that people use this phrase.
Example:

Muhammad (peace be upon him) was born in 571.

What is the meaning and purpose of this?
Answer: Islam is very respectful. In general, the phrase is usually a du'a (prayer) for the person mentioned. The common groups you will see are:

Prophet Muhammad (peace be upon him)
Other prophets (Adam, Noah, Jesus, Moses, Abraham, and others)
Companions of the Prophet Muhammad
Scholars of Islam

Specifically for the prophet Muhammad, the Qur'an commands us to send peace and blessings on him specifically:

Indeed, Allah confers blessing upon the Prophet, and His angels [ask
  Him to do so]. O you who have believed, ask [ Allah to confer]
  blessing upon him and ask [ Allah to grant him] peace. [Qur'an: Surah Ahzab,
  verse 56]
Answer: The phrase "peace be upon him" is a prayer. Praying for others is encouraged in Islam. This particular prayer is one of the most common prayers in Quran. In using this prayer we follow the tradition set by God (e.g. verse 33:56, 19:15), his angles (verse 33:56), and his prophets (e.g. 19:33).
The phrase is translation of

عليه السلام

It is usually translated as "peace be upon him" but it does not completely preserve the meaning (as with any translation). The word "سلام" is often translated as "peace" but it is only one of the meanings. It is also one of God's names (verse 59:23).
Note that the Arabic root of "سلام" is "َسَلَم" which is also the root of Islam which can be translated to "submission [to God]" (c.f. verse 2:131).
Answer: Actually this is not mandatory or sunnah. In nabi Muhammad's time, his fellows called his name barely, or just called him "rasulullah" which means "messenger of Allah".
Peace be upon him means aleyhisselam in Arabic. Also, there is a longer version, this p.b.u.h is a short of that longer version. The longer version is sallallahu aleyhi vessellem. It means peace and salat of Allah be upon him. Yes, the salat here is the same salat which is used in English as prayer.
Some people claim that, when nabi Muhammad's name is heard, it is fard (order of Allah) to say sallallahu aleyhi wessellem. Their this claim relies on this ayat:

إِنَّ اللَّهَ وَمَلَائِكَتَهُ يُصَلُّونَ عَلَى النَّبِيِّ يَا أَيُّهَا الَّذِينَ آمَنُوا صَلُّوا عَلَيْهِ وَسَلِّمُوا تَسْلِيمًا
  (It is for sure that) Allah and his malaks (angels) do salat upon the nabi (the prophet). You, who believes, do salat upon him and greet ("sellimu"; be peaceful against) him with obedience.
  Ahzab 56 (33/56)

The misunderstanding comes from the meaning of salat. It has a very wide meaning, and used in a lot of places in Quran, and translated with various different words in English translations. The median of meaning of salat is support, or support the way via education as a term. When we do prayer we support "something" by reading Quran; this is a totally different and big discussion topic.
Since the meaning of this word is so wide, and Muslims felt themselves too lazy to support the prophet with actual affords. They found it easier to claim that this ayat only ordered us to repeat the phrase sallallahu aleyhi wessellem (salat and salam of Allah be upon him) whenever the nabi's name is mentioned. I know it is sad but, this is the real reason why we are repeating it so much.
Only salat-ing by month is nothing but implying "You and Allah go and do all the works for us. Don't request any support from us!`. We are some lazy bums who love this dunya life very much.".

قَالُواْ يَا مُوسَى إِنَّا لَن نَّدْخُلَهَا أَبَدًا مَّا دَامُواْ فِيهَا فَاذْهَبْ أَنتَ وَرَبُّكَ فَقَاتِلا إِنَّا هَاهُنَا قَاعِدُونَ
  They said, "O Moses, indeed we will not enter it, ever, as long as they are within it; so go, you and your Lord, and fight. Indeed, we are remaining right here."
  Maide 24 (5/24)
Comment: "Muslims felt themselves too lazy to support the prophet with actual affords."
Comment: Isn't that a matter of fact? Muslims are really lazy. They don't work in the path of Allah. That's way we are being tyrannized by other nations. - Do you think that you do your supporting job when you say "O Muhammad, Allah will support you."?
Comment: Whether they are or not, but you have accusing language against muslims, especially ulema, whether you are right or wrong at the end. I think we must warn each other but "suizan" is not good.
Comment: What do ulema have to do with my words? I didn't direct my criticization to any group of people. I don't think you are considering my answer objectively. It was a self-criticism, but you took at as offence.


A simple misunderstanding of SSL/TLS

Question: I just came to learn about Deffie Hellman Key Exchange. And a little bit about RSA.
I came to DH from something about SSL. From what I understood before is that SSL uses DH and TLS uses RSA, but after some research, came to know DH and RSA are both secure/insecure, slow/fast in their own ways, and Both SSL and TLS can use both of them. Here I am confused. What uses what? And how can RSA work on DH? Coz the public key in DH is made by an exchange which is kinda symmetric unlike RSA which is Pure asymmetric Public/Private key crypto.
Can you "Simply" put me in the right track?
Thanks :-)
Comment: Could you post where you get the ideas? RSA mostly used for signature, or key exchange named RSA-KEM. Also, `Transport Layer Security (TLS) is the successor protocol to SSL. TLS is an improved version of SSL. ` DH is not insecure, prone to mitm-attack which can be prevented with certificates.
Comment: @C0DEV3IL: To RSA: Do you mean authenticated DH?
Comment: Your questions are answered by googling, reading the wikipedia articles about the terms you don't understand, reading one of the many good answers on crypto.SE and security.SE for "how does SSL/TLS work?".
Answer:
From what I understood before is that SSL uses DH and TLS uses RSA

Well, no.
First of all, there are a sequence of published versions of the SSL protocol:
SSLv2, SSLv3, TLS1.0, TLS1.1, TLS1.2, TLS1.3
(the name changed from SSL to TLS when the IETF started being in charge of it; Mozilla supposedly had an SSLv1 version internally, but they never published it).
Now, SSLv2 and SSLv3 use only RSA to perform key exchange; no other option.
When you get to TLS 1.0, they allowed you to select between RSA or Diffie Hellman (DH) to do key exchange; the general tendency over time has been to rely less on RSA and more on DH, until you get to TLS 1.3 (the most recent version) that entirely drops the use of RSA to do key exchange.
BTW: nowadays you shouldn't be using a version earlier than TLS 1.2, and using TLS 1.3 is generally accepted as best practice.

came to know DH and RSA are both secure/insecure

Well, both can certainly used insecurely; however done correctly, both are secure.

Coz the public key in DH is made by an exchange which is kinda symmetric unlike RSA which is Pure asymmetric Public/Private key crypto.

Well, no, DH is considered asymmetric crypto (well, both sides actually perform the same operation, however that's not what we mean by symmetric).  In fact, DH was the first published asymmetric crypto protocol; that is, it was published in the same paper that Diffie and Hellman first publicly introduced the idea of public key crypto.
What it is not is a public key encryption algorithm (however, with a bit of tweaking, it can become one); each side has a private key (which, if leaks, breaks the security of the system) and a public key (which is the key share that is publicly sent over to the other side, and which we assume the attacker can learn).
Comment: SSL2 was RSA-encrypt-premaster only, but SSL3 (like TLS1.0-1.2) allowed DHE signed by RSA or DSA (spelled DSS). Although one important implementation, Windows through XP at least and thus Internet Explorer, supported DHE_DSS but not DHE_RSA, during a period when public CAs only issued RSA certs, making DHE problematic for many uses.
Comment: Got it nicely. But another thing. If RSA is completely abolished, what encryption is used where we could give the final key as the (Public) key and the internal key(s) as Private Key? Coz there are 2 private keys right? So is an asymmetric cipher even used? Thanks
Comment: @C0DEV3IL: TLS 1.3 always uses DH (or the elliptic curve version of DH), or Preshared Keys to derive symmetric keys (and again DH/ECDH is asymmetric crypto, it's just not a cipher).  Also, RSA isn't completely abolished; it can (and is) still be used to generate signatures; it's just not used anymore as a public key encryption method.
Answer: DH in the classic form can be vulnerable to MITM, but the form used in TLS (SSL is deprecated, and hardly used any more, TLS is the more secure successor) is that the server sends its DH-contribution (and parameters used) to the client trying to connect, and these are signed by RSA and the server also presents a valid certifacate with its public key (backed by certificate authorities) so that the client can check the validity of that signature and then sends its contribution (now knowing the parameters and the reliability of the received data). Now both parties can compute a shared masterkey (which is then mutually verified (again to exclude MITM again) and used to derive the symmetric MAC and/or encryption keys).
So DH (often over elliptic curves in the modern suites) is used, but signed/verified by RSA signatures server side.
The most recent form of TLS does not use RSA for key transport anymore (because if the private RSA key is lost, all previous traffic can be decrypted retroactively) and so DH is preferred to create a fresh problem for each connection essentially. RSA is then only used in certificates and signatures for MITM protection and authentication.
Comment: TLS1.0-1.2 allows DHE signed by RSA or DSA (but DSA is rarely used) and 'optionally' ECDHE (or X25519) signed by RSA or ECDSA (or EdDSA) (all are used). TLS1.3 allows DHE (but only rfc7919 groups) or ECDHE/X25519 (only a few standardized curves) signed by RSA-PSS or ECDSA/EdDSA..
Comment: So you are saying the DH (elliptic Curve) is used completely, and RSA just for Certificates and Signatures. Now RSA isn't a HASH right? So are you saying that the RSA signature is actually encrypted? And the other question is, After the keys has been derived, What Cipher is used to actually encrypt normal data afterwords?
Comment: @C0DEV3IL Typically an authenticated cipher is used in TLS1.3 (ChaCha20-Poly1305 or AES-GCM or AES-CCM) and SHA256 or SHA512 as a hash (for key derivation). TLS1.0-1.2 allows AES-CBC-HMAC-SHA256 as well and some other classical combos for the normal data.

A simple misunderstanding of SSL/TLS


Why is the frequency of the response same as that of the forcing function in a linear circuit?

Question: Why is the frequency of the response same as that of the forcing function in a linear circuit?
What's the case when the circuit isn't linear?
Comment: Hi, Adrash. You might have observed that your question attracted some downvotes. This is mainly because people feel that this question is a homework question, which is okay to post it here if you show some effort.
Comment: If this is actually a homework question, and depending on the way the course is taught, then they might be looking for an answer something like "because the complex exponentials are eigenfunctions of the derivative operator". The trick is relating that mathematical explanation to the real world.
Comment: @Hazem I get your point, but it's not a HW question, you're right I should have shown some effort. But I couldn't get a starting point.
Answer: If a circuit (or any system) is linear, then the output is governed by a linear differential equation. This means that the input signal and all its derivatives are not cubed, squared or anything like that.
So if out input is \$x(t)\$ and the output is \$y(t)\$ we can write an equation of the form
$$
a_0 + a_1(t)y(t) + a_2(t)\frac{dy(t)}{dt}+a_3(t)\frac{d^2y(t)}{dt^2} + ... = b_0 + b_1(t)x(t) + b_2(t)\frac{dx(t)}{dt}+b_3(t)\frac{d^2x(t)}{dt^2} + ...
$$
So if x(t) is \$sin(wt)\$ the RHS can contain only terms of sin or cosine to the first power. Thus the LHS can only contain terms of sin or cosine to the first power. These all have the same frequency.
If there was a squared term, or some higher power, then there would be a sine squared term which oscillates at twice the original frequency, or another multiple.
Comment: The equation I wrote was supposed to just be a general linear differential equation. Best to ignore it for now if you are familiar with the fact that KVL will give you an integro-differential equation. You are mistaken, however, that it can be at most of order 2. It can be of order N if there are N reactive elements (inductors and capacitors). The order of a linear differential equation refers to the highest derivative. No matter how many times you differentiate sine or cosine you can only get a sine or cosine of the same frequency. There will be no product terms like you wrote.
Comment: The equation you provided kinda confused me, I would appreciate if you could explain the LHS of it. What I understand of the equation is that for a linear circuit the KVL will be an integro-differential equation which when solved results in a term(if any) of order 2 at max. so the frequency remains the same, but even in this case the frequency of the term $$sin\omega t*cos\omega t$$ is twice as the original  frequency?
Comment: Yes you're right, i just misunderstood what you said.
Answer: 'Forcing function' is the generic term for an input signal. If the forcing function is the unit impulse then it's response is the transfer function, since the Laplace transform of the unit impulse is unity. A transfer function must be a linear representation of a system for this to apply.
Answer: A transfer function is defined as the output-to-input ratio as a function of frequency. It makes really no sense to find such a ratio for a sinusoidal input and an output that is non-sinusoidal (because the circuit is non-linear and introduces non-linear distortions).
For this reason, transfer functions are given for linear circuits only, or better and more correct: ....for circuits which are assumed to be linear.
In fact, there are no linear circuits - however, for sufficiently small amplitudes we treat the circuits as linear (and accept a small error).
Comment: "(because the circuit is non-linear and introduces non-linear distortions)." So what?
Comment: @AdarshKumar, ... so you cannot use any of the linear analysis tools, such as transfer function, hence can't use s -> jw to calculate a frequency response. Also cannot use superposition, etc...
Comment: I get the point, but what's the relation between the transfer function and the frequency. Are you saying that we can't transform from time domain to the frequency domain, if yes why, is it related to the the transform function that you mentioned, how?


Dash in text in Lilypond

Question: How does one add a dash to the text in Lilypond? I would like what corresponds to --- in LaTex and — in HTML. With \addlyrics I have tried -, -- and ---, but these attempts are not interpreted as punctuation that should not be associated with a note.
Answer: Well, you can just add — literally if your editor and/or input system has some way to generate it (mine works with Compose - - - for an emdash or Compose - - . for an endash) or copy and paste from some text with the right Unicode in it.
You can also write
\include "text-replacements.ly"
#(include-special-characters)

and then literally use — in your lyrics.
Comment: Well, then don't separate it with a space from notes.  Or if you do, enclose the whole in spaces or `\markup { ... }`.  Or if you don't actually mean that you want an emdash, try "__" or "--" after the syllable in question separated with a space.  Or maybe show an image of what you actually want.
Comment: `—` generates a dash all right, but the dash is associated with a note. I do not want it to be associated with a note (my language in the question is convoluted).
Answer: If you can't type it directly on your keyboard, lookup the unicode value, and enter it as a special character:

in decimal: \markup{ \char #8212 }, or
in hexadecimal: \markup{ \char ##x2014 }

As seen in this page of the documentation:
3.4.4 Special Characters - Unicode

Edit:
After reading you comment to the other answer, and then reading your question again more carefully,  I see you are trying to get the symbol into lyrics.
Two dashes i.e. -- is supposed to be used for syllable separation in lyrics, but I guess it's not as long as you want it.
Update:
If you want to make theses inter-syllable hyphens to appear longer, still type them as --, but add the following override to the \addlyrics block:
\override LyricHyphen.length = #2.0

Dash in text in Lilypond


Why gaming with mouse feels so weird?

Question: I recently bought a macbook pro retina 2012 and later tried some fps game, e.g. Team Fortress 2. I plugged an usb mouse and noticed my it's movement is kinda weird or like unresponsive or something. It seems like my mouse movement has some kind of inertia, smoothing, acceleration or something, but it doesn't feel comfortable when gaming. Sometimes even a touchpad seems easier to play with. What am I experiencing and how can I make it like in windows?
Answer: OS X has a bit of mouse input lag and acceleration.
You can fix this by installing SmoothMouse. It works great for me in CS:GO. You can enable and disable it separately for the mouse and trackpad from the preference pane.
Comment: As for actual input lag in Source games on OS X, I'm in the process of writing a post regarding just that aspect and how to fix it. It's not complete yet but I'll add it to this answer when it is.

Why gaming with mouse feels so weird?


how to remove myself

Question: I asked one question and got three answers from 2 people over four days showing neither ever read the question or the updates.  One suggested his death.  Not helpful and want nothing more to do with this group.
Answer: Your account is scheduled for deletion. If you change your mind, you can comment here and I can stop it.
I think your reaction is a little strong. I took a look at the question, there is only one answer along with a series of comments (comments are not answers). I didn't see a recommendation for euthanasia, it was acknowledgement that symptoms like you mentioned are often associated with terminal illnesses. Maybe an ill-considered attempt at "softening" the potential, but it was not a recommendation.
In any event, your question is well nigh impossible to answer without some ability to physically examine the cat and there's really nobody qualified on the site. There are loads of possibilities, a laundry list of potential causes is not really helpful other than to cause unnecessary worry. Aside from all that, this is a pure volunteer site, we're not paid employees, and so we don't have an obligation to answer.
Comment: As the author of thge comment that is being complained about: yeah, perhaps Ill considered. To me, the wish was first "good luck" and then, if the worst happens, best wishes... But I can certainly see why a stressed pet owner could misread it. Something to keep in mind. (And of course if I had known how it was being interpreted I'd have apologised and removed it.)
Comment: @keshlam - I wouldn't stress too much. I thought you meant well, but you can't win them all when it comes to an impersonal mechanism.
Comment: I think SE's concept of expert answers (or at least answers reviewed by the collective expertise if the community) leads some newcomers to expect that experts are standing by and that only experts will respond. That isn't really what we've got. If you want pet experts on call, a better bet would be to contact your local emergency vet (whoever your vet suggests for after hours coverage) or perhaps your local SPCA if they're good. Admittedly I'm very lucky; MSPCA runs 24-hour walk-in veterinary clinics within reasonable driving distance of my house, plus Angell Memorial vet hospital.
Answer: Sorry to hear you've had such a tough time on our site. If you want to delete your account, you can use the Contact Page to ask for your account to be deleted.
One thing to note, though, is that content you've posted on the site is owned by you but licensed to Stack Exchange, and you therefore cannot request that it be deleted with your account.

how to remove myself


Canon 50mm F1.8 quality

Question: I recently bought 50mm f1.8 lens online, and quality is very poor as described in most of the reviews. But when i shake it gently I can feel some of the parts moving inside. And when ever I slide the AF/MF switch I get same feeling. Is it normal for a new lens ?
Comment: Can you give some details on "quality is very poor as described"?  I find the it can spend a long time hunting for focus at times, but the pictures can be very sharp.  I used it almost exclusively for a year for taking pictures of my new baby indoors at night.
Comment: @khedron I meant wrt to build quality, and yeah after using for 2 days I got used to MF.
Comment: Most of the time I was able to use AF anyway, except when taking pictures of black cats in dim light, etc.  Very good to have the option, though.
Answer: I've got a copy of this lens: I can confirm that by (gently) shaking it you can feel something moving (which is the moving part of the focusing mechanism). Apart from a click, no noises are heard when switching the AF/MF lever.
My lens has been perfectly functioning until now (I hope that this gentle shaking didn't affect it ;-) ).
Comment: Even I thought its the focusing mechanism as when I kep it in fully focused position it didnt rattle too much. Hope it stays longer with me. Thanks for confirming, I was kinda worried that it was broken in transit.
Comment: Good news, then. Enjoy your lens: I carry it with me (maybe as a second lens, together with a wider one) all the time, given its low weight and the nice features that it offers.
Answer: It's impossible to answer this question fully without examining the lens in person. All lenses have some degree of play in the system, and the 50 f/1.8 is well known to be a very cheaply built lens. I can't say whether the degree of movement you're detecting is out of line without knowing what your basis for comparison is.
Do the images meet your expectations, and do they seem in line with the reviews? If so then I wouldn't worry about it.
Comment: I owned one back in my Canon days - It was loud when you shake it, most lenses are. Build quality felt cheap, almost embarrassingly so - but it's only a $100 lens, what do you expect? The image quality was _really_ solid though, and that's what counts.
Comment: Agreed. If it's really worrisome I'd find a seasoned photographer and ask them about it as they'll have a basis for comparison.
Comment: Yes, agreed that its not possible to tell the condition without looking at it, but just wanted to know if a little bit of rattling was ok or not.

Canon 50mm F1.8 quality


Flex: How to add horizontal axes to a chart?

Question: I have a dynamically created chart and I need to add a horizontal and vertical axis in a script. I can't put the chart into MXML because I'm dynamically changing the chart type, deleting and creating a new chart. Therefore, the variable that I use to create the chart, an instance of chartBase, doesn't have either a horizontalAxis or verticalAxis. How do I assign to them if they don't exist? Should I be using a different type? The graphs I want to use are line, bar, and HLOC and/or candlestick.
Answer: I think you'll get better results by instantiating your chart as a proper type.
For example, in a component I have a skeleton lineChart setup in the mxml...
<code><mx:LineChart id="lineChart" dataProvider="{chartData}" dataTipFunction="dataTipFunction"
    width="100%"
    height="100%"
    showDataTips="true"
    />
</code>
... but you could create it as a variable and add it to the container. Then for the axis I determine the type I want and create and apply them, here's some code I use for the horizontal axis...
<code>// Setup new horizontal axis.
switch (hAxisFieldsComboBox.selectedItem.dataType)
{
    case "Date":
        var hDtAxis:DateTimeAxis = new DateTimeAxis();
        hDtAxis.title = hAxisFieldsComboBox.selectedItem.label;
        hDtAxis.dataUnits = "days";
        hDtAxis.dataInterval = 1;
        hDtAxis.parseFunction = dateParser;
        lineChart.horizontalAxis = hDtAxis;
        break;
    case "Time":
        var hTAxis:DateTimeAxis = new DateTimeAxis();
        hTAxis.title = hAxisFieldsComboBox.selectedItem.label;
        hTAxis.dataInterval = 1;
        hTAxis.parseFunction = timeParser;
        lineChart.horizontalAxis = hTAxis;
        break;
    case "Number":
        var hLAxis:LinearAxis = new LinearAxis();
        hLAxis.title = hAxisFieldsComboBox.selectedItem.label;
        hLAxis.interval = 1;
        hLAxis.minimum = 0;
        lineChart.horizontalAxis = hLAxis;
        break;
    default:
        var hCatAxis:CategoryAxis = new CategoryAxis();
        hCatAxis.title = hAxisFieldsComboBox.selectedItem.label;
        hCatAxis.dataProvider = schemaUtil.aggregateAndSortCasesAC(chartData, vAxisFieldsComboBox.selectedItem.data, [hAxisFieldsComboBox.selectedItem.data]);
        hCatAxis.categoryField = hAxisFieldsComboBox.selectedItem.data;
        lineChart.horizontalAxis = hCatAxis;
}
</code>
Hope that helps.


Relation of parameters of the logistic link function with linear predictor parameters

Question: I am trying to self study Generlized Linear Models using the book "Doing Bayesian Data Analysis" by Kruschke. In the chapter on link functions, the author explains how the linear predictor y = β0 + β1 can be transformed using the logistic function to better fit the central tendency represented by the data. He then goes on to say that the logistic regression is more conveniently represented by parameters θ and γ, instead of β0 and β1
What I don't really understand is how we got from equation 15.13 to equation 15.14.
It doesnt seem to be a simple rearranging of the terms of the parameters, so β0 doesnt equal γ, and β1 doesnt equal θ. Could somebody exaplain it to me? Please note that my background is in biology, not mathematics, so the questions might be trivial, but nethertheless the answer would be greatly appreciated.
Thank you in advance,
Jacek
Answer: You just let $\gamma=\beta_1$ and $\theta=-\beta_0 / \beta_1$. This is just a reparameterization, as long as $\beta_1\neq 0$ then there is a $\gamma$ and $\theta$ so that the model in 15.14 is identical to the model in 15.13 with a particular $\beta_1$ and $\beta_0$.


Laplacian in polar and spherical cordinates

Question: why does the radial dependence of Laplacian in spherical and polar coordinate vary? ie, in polar coordinates if there is no $\theta$ dependence the laplacian goes as $\frac{1}{r} \frac{\partial}{\partial r} (r \frac{\partial}{\partial r}) $ and in spherical coordinates it is $\frac{1}{\rho^2} \frac{\partial}{\partial \rho} (\rho^2 \frac{\partial}{\partial \rho})$. if my function only depends on $r$ shouldn't the laplacian be the same in both the coordinate system so that when $z\rightarrow 0$ in polar cordinate, $\theta\rightarrow \pi/2$ in spherical cordinate and $\phi_{polar}=\phi_{spherical}$ such that on $z=0 (\theta=\pi/2)$ plane as $\rho \rightarrow r$ the value of laplacian remains invariant since it is a scalar quantity?
Comment: Might [math.se] be better suited for this math question?
Answer: The meaning of the number $r$ is different in both cases. Imagine you want to locate a point $P$, whereas in spherical coordinates $r$ means the distance from the origin to the point $P$, in cylindrical coordinates $r$ is the distance from the origin to the projection of $P$ onto the $z=0$ plane

To avoid confusion, label your coordinates differently, i.e. $r$ in the spherical case, $R$ in the cylindrical/polar case
Answer: In polar coordinates, there are only two dimensions $x$ and $y$ that contribute to $r$, as opposed to three dimensions for the spherical coordinate version of $r$. You would have a different form yet if you work out the Laplacian for the radial coordinate in spherical coordinates in higher dimensions.
Comment: Is there any intuitive way of understanding this? @Chris


Statistical videos to learn regression analysis and multivariate analysis?

Question: I'm having a tough time to understand concepts of Regression Analysis and Multivariate Analysis. I'm following the books of Johnston and Anderson, but some video lectures would help me learn it faster. Could anyone suggest me some videos of the same? Thanks.
Answer: You may want to take a look at the following;

MIT Opencourseware Linear Algebra course (especially lectures 15 and 16)
Coursera (especially Andrew Ng's machine learning course, there is also a specialised regression course on there too)
Answer: Coursera is an extremely good resource. At this point in time, there are at least three upcoming courses on regression-type topics. This one may be best suited to your needs.
One major advantage to Coursera is that you can ask any specific questions about topics not covered in the videos (like Box-Cox or other transformations) in the forums. There will be TAs or other participants who will be able to help you.
Comment: That particular Coursera [track](https://www.coursera.org/course/regmods) is horrible whenever the instructor is Peng. If you know regression analysis in other languages like Stata and want to see how it's done in R, then it's fine. If you're new, then it's nothing short of masochism.
Answer: Where are your problems?
I have heard praise for the machine learning course on coursera, which is free:
https://class.coursera.org/ml-003/lecture
Otherwise, I would recommend a book from Andy Field (learning statistics using...) which does a good job of teaching basics.
Comment: I need to learn about Transformation of variables, leverages, box-cox, box-tidwell etc..
Comment: Thanks for telling me about machine learning though.. I didn't know Andrew NG taught statistical concepts in that course!
Comment: Im afraid I do not know anything about box-cox etc. Sorry!
Where are these methods needed?
I want to take the course myself in the future - alas, I don't have time.
Answer: I agree that the mentioned Coursera courses are worth checking add. In addtion, for regression I'll recommend week 3  of this course: http://online.stanford.edu/course/statistical-learning-winter-2014 (taught by the ultimate masters of the field :) ). In week 10 you'll find some nice multivariate techniques, although the presentation is far from exhaustive.
Answer: http://www.statisticshell.com/  This is andy field's website. Particularly good video lectures for when you want a lecture from a different perspective that will keep you entertained.   You might find them a bit slow/lacking in detail in some areas depending on your level. Not sure transformation is covered.
Multiple regression is here: http://www.statisticshell.com/html/woodofsuicides.html and some other multivariate methods (e.g. Manova and SEM) can be found here: http://www.statisticshell.com/html/malbowges.html


How do I get to know if my question has been commented or answered?

Question: I'm new here and I don't know how things work here and don't have any knowledge about commenting or answering questions. How do I get to know if my question has been commented or answered?
Comment: You get a notification for comments on your posts and answers to your question. In the top bar, somewhat left, a the inbox symbol gets a red badge with a number (the number of notifications since you last checked).
Comment: First step, use *real* and *actual* words. Not internet abbreviation like "cuz".
Comment: Thanks a lot
Comment: For more info, visit http://math.stackexchange.com/tour
Answer: Check your inbox:

Like this:

The number of unread messages will show up in red:

(thanks to Daniel Fischer for providing the last image!)
Comment: Now you have ;)
Comment: Thanks! :D ${}{}{}$
Comment: Apart from inbox, there also is [responses tab](http://math.stackexchange.com/users/current?tab=responses) on the profile page. Inbox can be also accessed on the network profile, see [here](http://meta.stackexchange.com/questions/183963/does-global-inbox-have-a-separate-url-and-page/183970#183970). The difference between inbox and responses tab is that inbox shows notifications from the whole network (all sites you're using), responses tab only shows responses from one particular site.
Answer: Another way for this problem is to install the software of stackexchange for your device (phone or tablet with android operating system or IOS) and active notifications for this software to send you push notifications.


Reenable Google Play Services after disable

Question: I am developing an app that requires testing without google play services (I'm on a Samsung S3). I've turned google play services off in Application Manager and tested everything. Afterwards, I went back in and turned it back on, but I can't seem to get my app to recognize that google play services are re-enabled. What ways are there to check it Google Play Services is indeed enabled again on my phone?
Answer: After further debugging, google play services' connection was coming across as needing an update in my app's code. It turns out I had to clear Google Play Service's cache. Once I did that and turned on, it prompted me to install updates. I did that, and now it recognizes it as re-enabled as it should.
Answer: Under Application Manager, there should be two Google Play Services instances under "Running Apps".  These instances will also indicate which services are active, and that they were launched by an application.
Answer: The easiest way is reset app prefernces. Go to settings-app management-reset app prefernces. I tried all the above methods and nothing worked. I tried this by trial and error method and work done.
Answer: Go to Settings - Applications, then go to options tool and select "Reset preferences". It will start working again. I tried in Micromax.


How do I reimage OpenWRT?

Question: How do I reimage openwrt in such a way that all my settings will be lost. I've been having some issues, and I want to ensure that it's not a lingering setting, I want this to be a fresh install.
Comment: You might want to take a look into gargoyle, also an open-source router firmware for the WRT family. Personally, I've found it much easier to use.
Comment: @ken please give more info [here](http://unix.stackexchange.com/q/6277/29)
Answer: OpenWRT versions from Kamikaze onwards (which is basically Kamikaze and Backfire, but not White Russian) do not use NVRAM to store settings or configuration. It is all stored in the filesystem, either in the base squashfs image or the overlayed jffs image. This means you should be able to re-flash the image and get back to "factory defaults".
The way to flash an OpenWRT image is described at http://wiki.openwrt.org/doc/howto/installing . Once you have OpenWRT installed the first time, the easiest way to reflash is to use the "via the OpenWrt command line" method. Pay attention to the differences between .trx images and .bin images. .trx images are "raw" generic openwrt images used by the command line installation method. .bin images have vendor-specific headers and, so you need to have the appropriate image for your router.
There are some settings stored in NVRAM that are used by the bootloader but I don't think they should persist once the Linux image has booted. Possibly MAC addresses may persist, but can be overridden in the filesystem configuration anyway.
Whatever you do, do not indiscriminately wipe the NVRAM. You will almost brick the device, and may be bricked unless you can find on the net the appropriate settings to restore manually for your device.
Comment: what'st the appropriate way to reflash though? commands and such
Comment: @xenoterracide: see edit
Comment: man I swear what annoys me about openwrt is depending on which link I pull up, depends on which directions I get.
Comment: @xenoterracide: Keep at it. You'll find plenty more things to get annoyed about :-)
Comment: heh, well this whole thing came up because I'm once again playing with my firmware due to running into bugs in the latest openwrt, and I'm tired of working around them.
Answer: What is your issue?
If your issue is, that you are developing, and you want to be sure that your latest changes in your build process really are copied into the new firmware image, you don't have to "reimage OpenWRT" (whatever that should mean)
To analyze your freshly created openwrt bin image file you can use
binwalk -e imagefile.img

this will unpack the image so you can look inside the folder structure if everything is there in the correct place and has the correct permissions set (executable, readeable,...)

How do I reimage OpenWRT?


Ordered and random energy

Question: One of the books refer to potential energy being ordered whereas heat energy being random. Could someone explain what is ordered and random energy?
Comment: Ordered and random energy is merely an abuse of language.
Comment: Such a textbook should not be used which propagates wrong ideas.
Comment: Could you provide passages (quotes) from the textbook that explicitly refer to "ordered" and "random" energy?
Comment: A falling object moves in one direction, and only one : downwards. A pendulum may move in two directions : North-South and/or East-West or a combination of both. It has two degrees of freedom. Its movement is more random than a falling object. If now the pendulum hangs at a rubber band, it gets a third degree of freedom : up-downwards. Its movement is still more random. A hot body is made of particules that can move or vibrate like my elastic pendulum. It has three degrees of freedom. Its movement is more random.
Comment: Imagine a group of children at a summer camp trying to pick up a heavy wooden log/trunk a) orderly by command b) randomly. The present energy is the same, but in the latter case of less use or even useless. That is like the case of the thermal energy, than can be directly use as the energy source only by balancing temperature with a colder object. For other cases, you need other primary energy source to pump this energy in thermal pumps.
Answer: Consider the ability to predict something in a detailed manner as synonymous with "ordered", and the inability synonymous with "random". For instance, a perfect crystalline solid is more ordered than a gas because given the position of one particle in a solid and a few lattice constants we can derive to a good extent the positions of all particles in the solid. In a gas, knowing the position of one particle constrains the maximum distance of all others (given the temperature and pressure) and with some additional parameters provides perhaps a distribution function for the particles, but we can say little more about where the other particles actually are (we might say something detailed however about where they might be).
Consider an analogy useful to get physical intuition into why we might really refer to a system as "ordered" or "random". Consider a system of many balls colliding elastically. Up to some number of balls we can describe the motions of the balls with a high level of detail and maybe even predict future trajectories. We would consider this a relatively orderly process. But if the balls become too many, then we give up and instead try to describe the statistical properties of the collection of balls, and say the balls move randomly.
Usually one speaks of a system being ordered or a configuration being random, but not about energy itself. However, heat can be understood as energy being transferred through the random motion of particles, whereas work involves coherent motion of the same particles (in a gas some of the molecules happen to move coherently, that is, in what looks like an orderly parallel displacement. These are responsible for work). In that sense one might describe heat as involving a transfer of "random energy" and work as being a transfer of "ordered energy".
So what decides whether you call something random versus ordered (or heat or work) is then how much information you have to describe the system. The idea is that describing the detailed motion (positions and momenta) of a very large numbers of things, on the order of $\pu{6E23}$  particles (moles of stuff), and on top of that as a function of time, is difficult (actually it's impossible). So instead of even trying we speak of "heat" as the energy that is exchanged by all those particles when they collide with each other in ways we can't measure or predict in detail. We say that their motion is random, which is shorthand for saying we can't describe their motions except perhaps in a statistical sense, using measures such as entropy. Work (and change in potential energy) on the other hand is something we can measure in detail, because we use things such as volumes and distances of large objects to describe it. For instance when a gas expands we can measure how much the volume changed against an externally applied pressure (force per unit area). So we call this an orderly process, even though the molecules in the gas that do the work are moving about in individually undescribable ways.
Another example is the solar system versus a galaxy. The solar system does not appear random because we can describe the motions of the sun, planets and other large objects in great detail, and sometimes predict their motion far into the future. OTOH when we look at a galaxy the large number of objects make a detailed description more challenging, and it can be useful to use statistical measures to describe the collection of objects and just say that the distribution of stars looks pretty random. See for instance Ref 1 (the following from the introduction):

Except for interacting galaxies, it is reasonable to suppose  that  elliptical  galaxies  are,  at  least,  objects  in  thermodynamic quasi-equilibrium (the best evidence being the regularity in   the   luminosity   distribution).   Hence   we   will   proceed   by computing the entropy of a self-gravitating system using classical thermodynamics. Let us assume that the stars in a galaxy behave like   an   isolated,   self-gravitating   gas   of   equal-mass   particles. Consequently, we start by taking the differential of the entropy: ...

Reference

G. B. Lima Neto D. Gerbal and I. Marquez. The specific entropy of elliptical galaxies: an explanation for profile-shapedistance indicators?Mon. Not. R. Astron. Soc.309, 481±495 (1999)

Ordered and random energy


Installing magento in a sub-folder

Question: I currently have a live website (Say www.abc123.com)in my folder "public_html/abc123.com". It is currently on magento 18.104.22.168 and I have been told that to upgrade to 1.9.1 is sometimes complicated. So my question is am I able to install a fresh new magento 1.9.1 in a sub folder (public_html/abc123.com/magento), work on it, then once it is done and ready to go, I simply point my domain to the sub-folder. This way, I won't get any downtime as well. I also understand that my database and order history will start all over.
Answer: Yes. You have to make sure to use another database and base_urls to the subfolder
Comment: You shouldn't change the document root of the domain "abc123.com". Revert that and it should work.
Comment: I figured it out. I had to go change the base URL in my database to the original http://www.abc123.com/ and clear my VAR/CAC folder.
Comment: Hi I just did a test with one not live website. I installed the magento in the magento folder with a new database, went to my cpanel - add on domains and changed the document root to "public_html/abc123.com/magento". However when I go to my browser to type www.abc123.com. It redirects to me to www.abc123.com/magento/. Then I get Index of /magento, Parent Directory, Apache Server at www.abc123.com Port 80.
Comment: ok but if I revert it back, when my customer types www.abc123.com, it goes to the old website. Only if they type www.abc123.com/magento, then they will get to the new one.
Comment: correct. Isn't that what you wanted? When you are ready to launch, you change the baseurl of the new shop and either, move your code to the abc.com directory or change the webroot.

Installing magento in a sub-folder


Simulation hypothesis and solipsism

Question: Is Nick Bostrom's simulation hypothesis a solipsistic idea? This hypothesis says that only I have consciousness, and all other people are unconscious philosophical zombies? Which version of the simulation hypothesis is more plausible: 1.Solipsistic in which all but me are philosophical zombies. 2. Nonsoliptic where all people, like me, have consciousness, sensation, thoughts and emotions.
Comment: [Bostrom's simulation hypothesis](https://en.wikipedia.org/wiki/Simulation_hypothesis) has nothing to do with solipsism. His very argument for it presupposes other minds:"*the vast majority of minds like ours do not belong to the original race but rather to people simulated by the advanced descendants of an original race.*"
Comment: Thank you, you are one of the best professionals in your field!
Comment: @Ron, you can say that again!
Answer: It is, in the way you put it, undoubtedly solipsistic. Solipsism is very hard to argue about in a logical framework, as the solipsistic side always has the choice of claiming that the logic itself is a product of their own mind. So this is always outside the scope of logic.
To analyse the two cases you pointed out, we do not even need to consider the validity of simulation hypothesis. It is no different than the paradigm of Abrahamic religions: A creator -who is omnipotent- created the universe with a set of rules and parameters, can interfere with its processes anytime it desires.
The positive or logical approaches you can utilize do not seem to matter, as despite of the background, questions of the solipsistic person is the same:

Is my perception all there is? or
Do these people have the same experiences as I do, or are they not sentient unlike me?

As you can see, regardless of the ontological or epistemological paradigm (i.e. Abrahamic religion or simulation hypothesis), you can still ask the same questions. As long as you cannot get to experience the qualia of other people, you may still question their reality. That is why, I cannot see any reason why one might be more plausible than other, i.e. coming up with a statistical "likelihood" seems impossible.
Second one happens to be more solid in a logical framework though, as we can measure through EEG, fMRI etc. that similar experiences give rise to similar neural activity in different people. However then again, one can argue that these all are "the products of their own mind". There is no way of arguing against that under the assumptions of logic.
To summarize, both the simulation hypothesis and solipsism are outside the scope of science, or even philosophy, as these matters turn out to be beliefs at the end of the day.
Comment: The OP is about Bostrom's [simulation argument](https://www.simulation-argument.com/simulation.pdf), which is mainly about the possibility that you're part of a far-future "ancestor simulation", and that *all* the simulated ancestors are conscious, not just you--the first paragraph says "Suppose that these simulated people are conscious (as they would be if the simulations were sufficiently fine‐grained and if a certain quite widely accepted position in the philosophy of mind is correct)". On p. 12-13 he does briefly discuss simulations w/ only one conscious being, but seems mostly dismissive.
Answer: Well. Thomas Breuer had shown that for any observer the system which properly includes themselves has different physical properties (and is not predictable) due to self-reference. This is a mathematical result.
This means, that from the point of view of any observer, all others with whom he can interact are predictable (probabiblistically or deterministically). In other words, Turing-simulable. In other words, machines or zombies that just follow physical laws.
But if we consider the Many-worlds interpretation of quantum mechanics, we can have several such observers in the multiverse. But they will never meet in the same branch of wavefunction (or, rather, if they occasionally meet  for a moment, they instantly diverge, so they cannot observe each other's behavior).
Answer: I feel like your conclusion is a bit illogical in terms of just because you have an individual frame of reference doesn't mean no one you interact with is conscious. For example someone could measure me and therefore I would be Turing-simuable, however I am conscious. I'm under the assumption that to conscious observers can interact and influence each other. They just have individual viewpoints aka (frame of reference).

Simulation hypothesis and solipsism


Is there a way I can use a Galaxy Note on Virgin Mobile?

Question: I want to know if it is possible to use a Samsung Galaxy Note on Virgin Mobile's network.  If so how would I go about setting it up?  Virgin doesn't currently offer the Note, so I can't just buy it from them.  So would buying an unlocked version and calling Virgin's customer service to activate the phone on their network work?
Comment: What are you asking for? Can you re-edit your question to make it clearer :)
Comment: @t0mm13b Are the edits clearer?
Comment: Makes more sense :) Just realized by looking at Chahk's answer below, is your Galaxy Note CDMA radio based? I would have thought there'd be a CDMA based version of the Note? :o
Comment: @t0mm13b I don't have a Galaxy Note.  I was just thinking about getting one.
Answer: Not in the USA. Virgin Mobile USA's network is CDMA, while the Galaxy Note is a GSM phone.
Virgin Mobile UK is a GSM network, so Galaxy Note will work on it.  Although depending on whether the phone and the carrier use the same frequencies for 3G, you may only get 2G speeds.
Comment: Well that sucks.  The Galaxy Note looks awesome.
Comment: Rumors are that because original GNote sold so well, Samsung is readying a Galaxy Note 2 for this fall.  Hopefully they will release it on more carriers this time around.
Comment: Yes, I saw the tweet about the Gnote2 coming out later on this autumn... http://www.gsmarena.com/exclusive_galaxy_note_2_to_have_55_screen_remains_narrower_than_original_note-news-4455.php
Comment: Yeah, I heard about it as well.  Hopefully they make a CDMA version.


Why doesn't the exponential smoothing forecast package in R provide confidence intervals for the fitted values?

Question: The upper and lower prediction intervals for the forecast periods are provided by the forecast() function.
However, neither prediction or confidence intervals seem to be available for the fitted values within the range of the actual data.
Why is this?
Answer: The fitted values are one-step forecasts, so prediction intervals can be obtained by adding/subtracting a suitable multiple of the standard deviation of the residuals. E.g., assuming normal errors, an approximate 95% prediction interval is given by $\hat{y}_t\pm 1.96\hat{\sigma}$ where $\hat\sigma^2$ is the variance of the residuals.
Presumably you mean a confidence interval for the mean of the one-step forecast distributions. I'm not sure why you would want such a thing but you could obtain it by reformulating the model in state space form and using a Kalman filter.
You ask why the forecast() function does not provide these. Simply, because they are hardly ever useful.
Comment: Yes, it makes sense that you really only care about the precision of your forecast. I am accustomed to the intervals returned by predict.lm. Thanks for taking time to explain.


Induced character for non-injective homomorphisms

Question: Any group homomorphism $\phi\colon H\to G$ gives rise to an induction/restriction adjunction between $G$-representations and $H$-representations:
$$ \hom_G(\phi_! M, N) \cong \hom_H(M, \phi^* N) $$
However, it seems that most textbooks and web pages about representation theory inexplicably consider only the case when $\phi$ is injective, i.e. exhibits $H$ as a subgroup of $G$.  In this case, there are formulas for the character of $\phi_! M$ in terms of the character of $M$:
$$ \chi_{\phi_!(M)}(g) = \frac{1}{|H|} \sum_{k\in G \atop k^{-1} g k \in H} \chi_M(k^{-1} g k) = \sum_{\text{cosets } k H \atop k^{-1} g k \in H} \chi_M(k^{-1} g k) . $$
Can someone give a reference for versions of these formulas when $\phi$ is not injective?
Comment: Isn't it enough to replace $H$ bu $\phi (H)$ to reduce to the injective case ?
Answer: Exercise 7.1 in Serre's Linear Representations of Finite Groups gives a formula (without proof) in the case where $\phi$ is surjective. It is probably straightforward to compose this formula with your formula for the injective case to get the general formula.
Comment: This should yield the general formula by induction by steps.
Comment: Thanks!  I've put them together into the general formula at http://nlab.mathforge.org/nlab/show/induced+character .  I'd still be interested to hear any other references for the general (or just the surjective) case.


Add a link about jTab to the editing help page

Question: I know I can find it if I search for it, but wouldn't it be cool if it were linked from the editing-help page?
Comment: Presumably it's possible -- math.se has LaTeX info on their editing help page: http://math.stackexchange.com/editing-help#latex. Looks like we probably need SE staff to do it, though.
Comment: Now that ABCjs is available, this should also be linked here
Comment: MathJax is implemented on a bunch of sites, whereas jTab and ABCjs are implemented on just this site – so I don't foresee this happening for either one.
Answer: We still lack the documentation AND the mention

 prevents jTab to render.
I was kind of lost when I wrote this answer: https://music.stackexchange.com/posts/31438/revisions


What happens if I mix green and orange coolant in a 2008 Chevy cobalt?

Question: It's a 2008 Chevy cobalt ss. I recently bought the car from a friend, and it sat for awhile before I bought it. I'm now trying to update all its needs, and I wasn't sure what coolant was in it to begin with. I was told it had green coolant/antifreeze in it. But I had bought the orange kind. 50/50 (did not need to add water. Ready to use). I'm concerned that I might have messed but and I need some help please
Answer: Generally you shouldn't mix coolants, but really you've not killed anything. You've mixed two kinds of fluid which makes for ugly coolant. Green coolant is good for around 2 years, while the orange coolant (orange is usually a Dexcool variant, which is probably what should have been in your car in the first place) usually has a change interval of 5 years or 150k miles. They are two different types. When you mix them, you've basically reduced the orange coolant down to the green coolant longevity. By mixing the two you'll have not damaged the cooling system any, so don't worry too much about it at this point. When you get a chance and the means to do so, completely flush the system and put the correct coolant type in and you'll be golden.
NOTE: Some will argue you should never mix coolants. I personally have never found verifiable information showing where there's an issue doing so. Most is internet conjecture.
Comment: I have now started to have issues with my car starting. Is this because of the coolant being mixed?
Comment: @HarleyOthersen - Your coolant should have absolutely nothing to do with your car starting, unless the vehicle was allowed to overheat because of lack of coolant. If your car had zero coolant, you would be able to start your engine, though you wouldn't want to run it for too long. Mixing coolant would not cause you starting issues whatsoever.


When excess electrons are in an insulator they don't move. Why is this?

Question: Let's say you have an insulator that is electrically neutral(has no net charge). Let's say you are able to add additional electrons into the same insulator resulting in the insulator having a net negative charge. These electrons that were added to the insulator will stay where they are and not move. How is this possible when same charges repel each other? Shouldn't those excess electrons added to the insulator repel away from each other?
Comment: Does this answer your question? [When you introduce excess charge into an insulator the charge stays still. Why is this?](https://physics.stackexchange.com/questions/583341/when-you-introduce-excess-charge-into-an-insulator-the-charge-stays-still-why-i)
Comment: @DKNguyen What is the force that is causing the electrons not able to move in the insulator?
Comment: Does this answer your question? [Excess charge on an insulator and conductor](https://physics.stackexchange.com/questions/38488/excess-charge-on-an-insulator-and-conductor)
Answer: Consider the following diagram of the quantum mechanical densities of electron states in different types of material:

The Fermi energy $E_F$ is the energy level new electrons are added at. Loosely said, the states below this energy are filled, the states above are empty, with a smooth transition around $E_F$. The conducting electrons are those in this smooth transition.
On the far left is a metal. There are plenty of states available around $E_F$, so adding an electron simply puts it among the already conducting electrons.
On the far right is an insulator. The Fermi energy is in a gap between the bands with no available electron states. The lower band is filled and cannot conduct, the upper band is empty and cannot conduct. If you add an electron to this material, you can only add it in the upper band, since the lower band is already full. When you do this, you effectively lift $E_F$ up to the upper band, and you no longer have a perfect insulator. The electron you added can "move" just fine, and conduct electricity. This is similar to the n-type situation in the diagram.
See also the Wikipedia article on the band gap.
Answer: It is possible to apply a force without having any motion (that should be obvious just from your day to day experiences, right?). So there is a repulsive force on the electrons but they don't move anywhere, at least nowhere fast because the force is not enough to push them through the insulator...at least until the insulator breaksdown.
Like sticking a bunch of people who hate each other into the same room full of stuff. They might want to get as far from each other as possible, but they can't go anywhere.
Comment: what force is making the electrons not move?
Comment: @FaheemAzeemi That I don't know. You'll have to wait for someone to understands quantum mechanics. Because what is responsible is whatever keeps electrons in the band they are in. It will probably be referred to as an energy rather than a force.
Comment: What do you mean by 'band'. The electrons that are added to the insulator and are not bound to any atom. How can they have a 'band'. (Sorry, if this is a silly question)
Comment: @FaheemAzeemi Well, when you inject electrons into a conductor they also fall into a band do they not? The conduction/valence band (since they are the overlap in conductors). Maybe do some light reading on energy bands?
Answer: If one charges an insulator, then yes, sometimes the excess charge stays on
the insulator.   A charged particle is attracted to any polarizable medium
(static cling, in your socks, is this effect, writ large).
Mobility of charge in such a situation is dependent on the LOCAL electric field,
not the large-scale "it has multiple positive charges" global situation.   The
polarized medium puts a local attractive charge next to the excess charge,
after all, at the expense of slightly altering (polarizing) the bound-to-molecules
charges in their molecular orbits.
The applied charges do NOT "stay where they are" however, they migrate to
the most attractive local sites (and will slowly drain away into air or
elsewhere, as thermal motion jostles them about).   This effect, leakage,
is slow, but important.   Detectors of ionizing radiation are basically looking at
the excess leakage that happens due to photons or other energetic particles
complicating the insulator's inherent insulation property.


Why the name "category"?

Question: I'm reading about category theory. A question I don't see answered anywhere is about the choice of the word "category" in this context. Why this concept was named "category"?
Comment: They're just supposed to be different categories of mathematical objects in the colloquial sense, e.g. groups, rings, topological spaces. I don't think it's particularly deep.
Comment: @QiaochuYuan The answer from Geoff suggests that the etymology is in fact pretty interesting, somewhat deep.
Comment: @QiaochuYuan Maybe you're right. Groups are another example. The colloquial concept of "group" anticipates nothing about its meaning in mathematics, I think.
Answer: The term ``category'' came from the Kantian perspective on a category being the most general form of thinking, although I may be mistaken on the finer points of the actual philosophy there. This is discussed in this philosophy.stackexchange post here  (see the accepted answer in particular) which speaks far more eloquently on the subject than I could hope to mimic.
Comment: Thanks, that link is a very interesting discussion!
Comment: I find this quote from the linked OP particularly illuminating:

*... my guess is that categories in mathematics don't usually mention individual elements of a set explicitly. They only define properties of elements via how various sets relate to each other. This in turn defines the elements.*

Why the name "category"?


Logic gate diagram for JK latch? (Not flip-flop)

Question: What would the logic gate diagram look like for a JK Latch? I know what a JK flip-flop looks like, but how would you build a latch with only J and K inputs, without a clock input?
Answer: This is pretty easy to derive by hand. A JK latch is just like an SR latch except with a 11 input, an SR latch does nothing, while a JK latch toggles. So, basically, you can write out the truth table and solve it with a Karnaugh map.
The lack of a clock makes toggling pretty useless here; I have never seen a JK latch in the wild but they are theoretically quite possible.
Comment: I believe you would also have to make sure the transition was defined regardless of which input went high first and that it only toggled once until both inputs had gone low, and this would also cause momentary blips in the signal if it was high and received Reset first or if it was low and received set first.
Answer: From Know all about Latches and Flip Flops:

JK latch is similar to RS latch. This latch consists of 2 inputs J and K as shown in the below figure. The ambiguous state has been eliminated here: when the inputs of Jk latch are high, then output toggles. The output feedback to inputs is the only difference we see here, which is not there in the RS latch.

Compared to the JK flip-flop:

So, you just remove the AND input for the clock.
Comment: This doesn't seem to work on logisim. There's an error around the NOR gates.


Protected ISO File

Question: I face a strange problem. Few days before I have downloaded matlab 2014a and it was downloaded in the form of ISO image file. But when I try to copy the ISO image file to my pen drive from the system it says "not enough space" even I was using 16GB empty pen drive. I want to know how to copy that ISO file in my pen drive.File size is 8GB.Can anybody help me. Thanks in advance...:)
Comment: The filesystem is likely to be fat32 for a pen drive. The FAT32 filesystem does not support files larger than 4 GB, so your ISO image does not fit. What does ls -l /path/to/matlab2014a.iso (or whatever the iso is called) give you? Add it to your request.
Comment: What's the filesystem of your pen drive?
Answer: Problem is solved.Its very simple.Normally when we format the pen drive then file system we choose is FAT32. but this time we have to choose the file system NTFS. and then format it. By this I have transferred the ISO file of 4 GB in 8 GB pen drive. Problem occured because FAT32 file system does not allow to transfer the file more than FAT32.

Protected ISO File


IS Network automation kills carries of network Engineers?

Question: Iam working as network engineer since long time and i enjoying working same . But now i am afraid after listening of network automation .Is network automation kills jobs of network engineer across globe .
Comment: Voting to close as opinion-based
Comment: Until we have AI and robots, I think my job is safe. Much of the automation in my life is stuff I've created so I'm not typing the same commands over and over. As Zac answered, we still need humans to design and build networks. Automation is helpful on a day-to-day basis, but people still have to install hardware, run cables, plan address spaces, and think about the future.
Answer: To some extent. However, networks tend to grow, so much of the automation gain is eaten up by growth.
Also, the more skilled you are the less automation can take over your job. After all, someone has to engineer and maintain the automation - and of course, create the architecture, do the troubleshooting, ...


Calculating the voltage dropped in a series/parallel circuit

Question:
I've been trying to figure out a way to solve this problem to no avail. I think it's got to do with a voltage divider or voltage dropped. I've seen a number of questions on this topic but never one like this, where the value of particular resistor wasn't given and has to be found, which also makes it difficult working out the total resistance in the circuit.
Comment: To get you started: if you know the voltage across R4, then you know the current through R4.  This is the same as the current through R3, which means you can find the voltage across R3.
Comment: If there's 5V across R4, how much current flows through R4? Now how much current flows through R3? So then what's the voltage across R3? So now you know the voltage across R3 & R4, so what's the voltage across R1? So if you then know the voltage across R1, what current flow through it? So this current "enters" the node at the top between R1, R2 & R3, and you've already calculated how much of that current goes down the R3 + R4 leg, how much goes down the R2 leg? So now you know the current through R2, and you already calculated the voltage across it earlier (it's in parallel with R3+R4), so ...
Answer: The others have said it, but I'll list it out here in order of how I would solve it:

Perform nodal voltage analysis - There will be two unknowns, one of them being R2, the other being V(R2).
The voltage they gave you is important, use V = I*R to solve for I4. Because R3 and R4 are in series, the current is the same.
You now know I4 = I3, therefore you also now know V(R3). Why is that helpful? Well, the voltage across V(R2) is in parallel with V(R3) + V(R4), otherwise, V(R2) = V(R3) + V(R4)
Going back to the original voltage nodal equation you did in step 1, now, there is only 1 unknown, the resistance value of R2.

Try and go through the process and do the numbers, and post what you get; otherwise, comment if you get stuck.
Cheers.
Comment: Thanks pal your analysis is similar to that of the others, it makes so much sense now
Comment: No problem! glad it helped
Answer: The question is asking you to apply both Kirchhoff's Voltage Law and Current Law, as well as Ohm's Law to solve.
We want to find R2. To do that we'll need to find its current, IR2 and voltage, VR2.
Besides the rest of the circuit values except for R2, they give you a key piece of information: voltage across R4 (VR4). Let's start with that.

Use Ohm's Law to find current IR4 = VR4/R4
Use Kirchhoff's Current Law to find current through R3 (IR3 = IR4, that is, recognize that they're the same)
Use Ohm's Law to find voltage VR3 = IR4*R3
Use Kirchhoff's Voltage Law to find VR2 = (VR3 + VR4).

With VR2 known, we move on to finding IR2:

Use Ohm's Law to calculate IR1 = (VR1/R1) = [(18V - VR2)/R1]
Use Kirchhoff's Current Law to calculate the net current through the R2 branch, as IR2 = (IR1 - IR4).

And finally:

Use Ohm's Law to calculate R2 = (IR2 / VR2).

Quick review of Kirchhoff's Laws:

Voltage Law: Directed sum of all voltages around a closed loop is zero

(Battery [18V] is a positive value, while all the IR drops in the resistors are negative. All add up to zero.)

Current Law: Sum of all currents flowing into and out of one node is zero

(Current through R1 into R2/R3/R4 node branches into currents out via R2 and R3+R4.)
Comment: Wonderful analysis and explaination, this has been bothering me for hours but you just made it look too easy Thanks pal


SOAP Async Continuous object

Question: I generated two class from apex2wsdl function, but I'm a little confused with async class, method get to parameter System.Continuous object. It is big different between callout method with future annotation and with System.Continuous in param?
Answer: The difference is that the Continuation object is meant to be used when you need the response back in the UI, but you don't want to lock up the UI while you're waiting for a response. If you don't need the response back in the UI, you can use a future method instead. As an example of each, the former can be used to show prices for products retrieved from a third-party server, while the latter might be used just to determine the city and state of a record, which would then be updated asynchronously.
Answer: @sfdcfox because I cannot comment, does this mean continuation class does not need to be called asynchronously?
Comment: The Continuation class will be called asynchronously, so there is no need to call it in a future method or other asynchronous code.

SOAP Async Continuous object


Well-definedness of first obstruction (to extending a map or homotopy of maps over the skeleta of a CW complex) lying in a nonzero group

Question: Let $X,Y$ be CW complexes and suppose for simplicity that $Y$ is simply connected.
There is an obstruction $O(f_n)$ to extending a map of CW complexes $f_n:X^n\to Y$ defined only on the $n$-skeleton $X^n$ of $X$, to a map $f_{n+1}:X^{n+1}\to Y$, and this obstruction lives in $H^{n+1}(X;\pi_n(Y))$. The precise statement is that there exists an extension of $f_n|_{X^{n-1}}$ to $f_{n+1}$ if and only if $O(f_n)=0$.
Similarly, given two maps $f,g:X\to Y$, there is an obstruction $O(f,g,H_n)$ to extending a homotopy $H_n:X^n\times I\to Y$ between $f|_{X^n}$ and $g|_{X^n}$ to a homotopy $H_{n+1}:X^{n+1}\times I\to Y$ between $f_{X^{n+1}}$ and $g|_{X^{n+1}}$, and this obstruction lives in $H^{n+1}(X; \pi_{n+1}(Y))$. This second fact is actually a special case of (the relative version of) the first fact.
My question is to prove (or disprove!) the following claim, left explicitly as an exercise, in the book "Rational Homotopy Theory and Differential Forms" by Griffiths and Morgan:
Fix $f,g:X\to Y$ and suppose $H^i(X;\pi_i(Y))=0$ for $i\leq n$, so that by the above we can inductively construct a (not necessarily unique!) homotopy $H_n:X_n\times I\to Y$. Then $O(f,g,H_n)\in H^{n+1}(X;\pi_{n+1}(Y))$ is independent of the choice of homotopy $H_n$. This is often phrased as the statement "the first obstruction lying in a non-zero group is well defined".
Additional Comments:
I am aware of a proof of this result (in fact of a version of this result for the more general case of extending maps, from which this result follows) but under the stronger hypothesis that the actual coefficient groups vanish: $\pi_i(Y)=0$ for $i\leq n$. However, this proof relies strongly on this extra hypothesis and provides little insight into how to proceed without it.
As well, the analogous claim for the more general case of extending maps is clearly untrue without some additional hypothesis, because we could always choose the map $f_n:X^n\to Y$ to be a map to a point, which would of course have vanishing obstruction $O(f_n)$, and if the obstruction were independent of this choice then this would imply the obstruction always vanished!
Answer: A version of this result is proved by Theorem 6.4 of Griffiths and Morgan: "Rational Homotopy Theory and Differential Forms".  For the definition of the obstruction class for a homotopy between $f$ and $g$ on $X^{(n)}$, one needs an additional assumption on the vanishing of $H^{i-1}(X;\pi_{i}(Y))$. It is not clear to me whether the original statement is true.


Isometry group of SO(n) and SE(n)

Question: We know that E(n) is the isometry group of euclidean space. But considering E(n) or SE(n) or SO(n) itself as a Lie group with a left/right invariant Riemannian metric, what is the isometry group of E(n),SE(n) and SO(n)?
Answer: The isometry group of $SO(n)$ (with the metric induced by the spectrum, e.g., the operator norm) is well-known.  It is given by all $X\mapsto O'OX^{\pm 1}O^{-1}$, $O,O'\in SO(n)$ unless $n=4$, in which case we need to extend by $C_2$, swapping (1,4) and (2,3), and (4,1) and (3,2) entries in $\mathfrak{so}(4)$.
Comment: I am not familiar with the concept of operator norm but is that same as the induced left or right invariant Riemannian metric on SO(n) as well? Can you mention some resources for operator norms and proving what you just answered? And how about SE(n)?   What is the meaning of $C_2$

Isometry group of SO(n) and SE(n)


How to add images in semantic media wiki

Question: I need some images to display on my main page in (semantic media wiki)
Answer: Semantic MediaWiki is just an extension that enables you to tag articles with properties and values understandable by computers, so you can query them later. It doesn't affect the other functionality of your MediaWiki, so you can upload files and insert them into your articles normally.
Please check this article to learn more about images and MediaWiki: http://www.mediawiki.org/wiki/Help:Images


log4net with .NET 4.0

Question: I've thrown together some code to tinker with the new .Net 4.0/VS 2010 pieces, but I can't seem to find a build of my logging framework of choice (log4net) for 4.0, and I'm getting reference errors with the 2.0 version.  Is there a 4.0 version available somewhere?  I'm not asking for new features, just a version that's already been rebuilt against the new assemblies.  Anyone know where I can find a build of 1.2.10 built for the 4.0 framework?
Comment: I think I remember seeing in-proc side-by-side for 2.0/4.0-compat, but that doesn't appear to allow my 4.0 code to reference the 2.0 log4net library.
Comment: Have you tried taking a log4net source project and running it through VS2010's upgrade process?
Comment: I have pulled the source for 1.2.10 from the SVN server and run it through the upgrade process.  It looks like so long as I define the _NET_2_0 symbol and add attribute to get the Level1 security-enforcement rules, things work.  Obviously, there's more work involved for the complete 4.0 update, so I'm hoping someone has already started on that work.
Answer: log4net has a known issue of referencing a System.Web component which is not part of the .NET Framework 4 Client Profile and the VS2010 by default sets project target to be the lightweighted Client Profile.
The solution is to change the target to .NET Framework 4:
Right click project -> properties -> Application.
Change:
<code> .NET Framework 4 Client Profile
</code>
to be:
<code> .NET Framework 4
</code>
... and that should work, i.e. you can use log4net again...
Comment: Just downloaded latest binaries (1.2.11) and it contains a version compatible with 4.0 Client Profile in \bin\net-cp\4.0\release. Just tested it and works like a charm !
Comment: Downloaded today latest release, it doesn't support ClientProfile with .NET 4.0
Comment: I should say the latest releases already added support for Client profile, so no need to target the full one.

log4net with .NET 4.0


How to tell if handset is on WiMAX

Question: I use TelephonyManager to get the network type as in getNetworkType(). Currently I see only CDMA and GSM networks only. How do I detect WiMAX as the networkType. Sprint's HTC EVO 4G phone supports WiMAX and when I use getNetworkType(), I get NETWORK_TYPE_EVDO_A even though the handset is on WiMAX network.
Answer: Check out this document here
Section 2.1 gives a sample and method calls to determine if the WiMax connection is enabled. This may only work for the Evo and Epic - it doesn't say if it works for the Evo Shift but my guess is that it will work fine for any Android Sprint 4G device.
Something like this:
ConnectivityManager cm = (ConnectivityManager) this.getSystemService(Context.CONNECTIVITY_SERVICE);
int networkType = cm.getActiveNetworkInfo().getType();

Edit:
I just noticed the document states

While 4G is the active data connection you can only detect this using the ConnectivityManager, under this condition the TelephonyManager will still reflect the status of the CDMA network which is providing voice services.  You will not see any 4G/WiMax information show up in the TelephonyManager.
Comment: Daniel, thanks for you quick reply. I tested this on Sprint's EVO 4G. I did see WIMAX as the returned value for cm.getActiveNetworkInfo().getTypeName(). Do you think if I try this on an LTE handset such as the one recently released by MetroPCS (Samsung) would show LTE as the TYPE?
Comment: No this is only for Sprint 4G Devices.  Since Android (I don't think) supports 4G technology natively it must be implemented by the carrier; which is why this method only works for Sprint.  I tried to look for a MetroPCS API but couldn't find anything (yet) - if I do I'll add it to my answer for you later.  I'm going to look for a Verizon one too.
Answer: Honeycomb API level has introduced updates to Telephony Manager including a number of new networks - be sure to filter by API 11:
int     NETWORK_TYPE_LTE    Current network is LTE
int     NETWORK_TYPE_UMTS   Current network is UMTS
Note that it will probably take at least 6 months until we start seeing this in the "wild", as new devices (Such as HTC Thunderbolt) are still shipping with Froyo. It will be interesting to see what workaround Verizon implemented in Froyo for LTE.

How to tell if handset is on WiMAX


How to get tile off of cement board in modular home

Question: We have a modular home. We are attempting to remodel one of our bathrooms. The trouble is, we can not get the tile off the cement board. My husband states if we take the cement board off the wall we will mess up the spray foam insulation. I'm not sure the tiles were put on with just thinset. How do we get the current tiles on wo damaging the cement board so it can be reused?
Comment: You will find that even if you could remove the tile, and the cement board remained intact, the surface would be so uneven from adhesive that stayed in place and that came off with the tile it would take a lot of work to get it flat enough to retile over. This is also being very optimistic.
Answer: I hate to be the bearer of bad news, but I don't think it can be done. Any repairs that I have made involving just one tile leaves a hole in the cement board. When comes to renovating, the whole wall is stripped down to bare studs.
If the foam is sprayed in, it should be sprayed in from the inside. I don't think it is ever sprayed from the outside against the interior wall, but I guess it can be done that way.... If the foam is an extruded rigid board, fitted between the studs, there will be no issue there either.
At the most in either case, creating hand holds to pull the cement board off will perhaps be the only place where damage to the foam will occur. If that is the case, touch it up with canned spray foam. Careful, doing this is really messy.
Comment: That's a no. *Finish the demo.* If you gotta fix the foam then you fix the foam. "touch it up with canned spray foam" or don't; it'll be fine. Even if it was wooden subfloor and therefor *possible* I'd still quote you on replacing the sub. Same thing every time you do a roof, site unseen; quote includes 'plywood as necessary'. Do-it-yourself becomes professional level when you quit dicking around like this.
Comment: @Mazura I take it that is a comment for the OP?
Comment: Yes, sorry. Any animosity there is from my own experience learning how to renovate and make money by the end of it. Too much time spent working around what shouldn't still be there by the time it's done, and half the time w/e it is ends going away anyway.