False beliefs about Lebesgue measure on $\mathbb{R}$

Question: I'm trying to develop intuition about Lebesgue measure on $\mathbb{R}$ and I'd like to build a list of false beliefs about it, for example: every set is measurable, every set of measure zero is countable, the border of a set has measure zero, etc. Can you help me sharing your experience or with some reference list?
Comment: @David: An easier counter example to "a set of positive measure $A$ contains an interval" is the irrationals.  Fat Cantor sets are a better counter example to "a set of of positive measure $A$ is dense somewhere"
Comment: False belief: a set of positive measure $A$ contains an interval (but by Steinhaus theorem we know that it's the case for $A+A$); take fat Cantors.
Comment: False belief: if a function is continuous almost everwhere, then it is equal almost everywhere to a continuous functions, and vice versa.
Comment: Counterexamples in Analysis - B. Gelbaum, J. Olmsted (Dover, 2003)  would be a good reference here. e.g., they show:  there is a measurable non-Borel set; there is a set of measure 0 that is not a countable union of closed sets;
Comment: This is not just about $\mathbb{R}$, but still: all $\mathbb{R}^n$'s (with the Lebesgue measure) are isomorphic as measure spaces; there is no "invariance of dimension" (as one might falsely believe)
Comment: I second David Mitra's advice to get a hold of Gelbaum & Olmsted's _Counterexamples in Analysis_.
Comment: I often find myself in situations where the Borel measure already measures every subset of $\mathbb R$. To say what sort of beliefs, one also have to specify the sort of axioms he assumes in the background.
Comment: False belief: Lebesgue is spelled with a q.
Comment: 1.There might be something in Counter-Examples In Point-Set Topology. 2.Borel found an error in Lebesgue's famous monograph : The assertion that the projection of a 2-dimensonal real Borel set onto 1 co-ordinate is 1-dimensionally Lebesgue-measurable, which is not always true.This led to the discovery of a larger class : The projective sets.
Comment: @user254665 I believe you're thinking of [Suslin](https://en.wikipedia.org/wiki/Mikhail_Yakovlevich_Suslin) (not Borel) who caught a famous error of Lebesgue. He did not (of course) prove that the projection of a Borel set could be non-measurable, but he did show that it was not always a Borel set.
Comment: Right .I was going on memory. & it was Suslin  too.
Answer: False belief: the continuous image of a measurable set is measurable.
A counterexample is provided by the Devil's staircase. Since the image of the Cantor set has full measure, it will have subsets, still measurable, which have non-measurable image. The same function also serves as a counterexample to the following:
False belief: if a continuous function has derivative zero almost everywhere, then it is constant.
Comment: @Tim, the Devil's staircase is continuous on the unit interval and has zero derivative almost everywhere (it is locally constant on the complement of the Cantor set), but it is not constant.
Comment: Nice! How is it false that "if a continuous function has derivative zero almost everywhere, then it is constant"
Answer: False belief: a subset of an interval that is both open and dense has the measure of the interval.
A counterexample is obtained by enumerating the rationals on $[0,1]$ and putting an open interval of length $(1/3)^k$ around the $k$th one. The union of these intervals is clearly dense because it contains a dense set (the rationals) as a subset, and it is clearly open because it is a union of open intervals. But meanwhile, its Lebesgue measure is $\leq \sum_1^\infty (1/3)^k = 1/2$.
Comment: In fact, we can choose this open subset of measure $\varepsilon$ where $0<\varepsilon< m(I)$, using the continuity of the map $\varepsilon \mapsto m\left(\bigcup_{j\in\mathbb N}\left(r_j-\varepsilon 2^{-j},r_j+\varepsilon 2^{-j}\right)\right)$.
Answer: More Cantor madness:
True belief:
There is a measurable set $A$ in $[0,1]$ such that for any   interval $U$ in $[0,1]$, both  $A\cap U$ and $A^c\cap U $ have positive measure.
False belief:
The continuous image of a set of measure 0 has measure 0.
Comment: @seriouslydivergent: Enumerate the intervals with rational endpoints. Inside the first, find two disjoint nowhere dense compact sets $A_1$ and $B_1$ with positive measure. Inside the second interval, find $A_2$ and $B_2$ disjoint from each other and from the previous $A_i$ and $B_i$, and so on. Let $A$ be the union of the $A_i$ and $B$ of the $B_i$: both have positive measure on each interval, and they are disjoint.
Comment: Could you explain how  you can get such a set $A$ in "true belief" statment?
Answer: False Belief: A nowhere dense subset of $\mathbb{R}$ has measure $0$. (Let me recall that a subset $A$ of $\mathbb{R}$ is said to be nowhere dense if the interior of its closure is empty.)
I leave the explanation as to why this is indeed a false belief as an exercise!
Comment: Furthermore, examples of full measure meagre sets: http://mathoverflow.net/questions/43478/is-there-a-measure-zero-set-which-isnt-meagre
Answer: Consider the following (true) statement:

If $(I_n)$ is a sequence of subintervals of the unit interval and the sum of their lengths is strictly less than $1$, then the $I_n$ do not cover the unit interval.

False belief: This can be proven just by translating $I_1$ to begin at $0$, translating $I_2$ to start end the end of $I_1$ etc. If this worked, then the same would be true for the unit interval in $\mathbb Q$ where the statement is false.
I obvious can't claim this to be original; I got it from MO.
Answer: True and crazy: There exists a subset $E$ of $[0,1]^2$ which meets every line (horizontal, vertical, or slanted) along a measure zero subset of that line, in fact, along at most two points (i.e., no three points of $E$ are ever aligned), yet such that $E$ does not have measure zero, in fact, $E$ meets every closed set of positive measure in $[0,1]^2$.  Also, we can arrange so that $E$ is the graph of a function.
This is due to Sierpiński in 1920.  His proof (in French) can be found here ("Sur un problème concernant les ensembles mesurables superficiellement", Fund. Math. 1, 112–115).  It's a more or less straightforward transfinite induction with length $2^{\aleph_0}$.
So in particular, you shouldn't believe that $\int_{[0,1]^2} f = 0$ follows from $\int_{[0,1]} f(x,y)\,dx = 0$ for every $y$ and $\int_{[0,1]} f(x,y)\,dy = 0$ for every $x$ (take $f$ to be the characteristic function of this $E$).
Along similar lines, using the Continuum Hypothesis, one can find $E \subseteq [0,1]^2$ which has measure $0$ on each horizontal line and $1$ on each vertical line (in fact, take the graph of any well-ordering of $[0,1]$ with order type $\omega_1$).
Edit: It is worth noting that the pathologic behaviour is due to the resulting sets not being measurable. For measurable sets such things cannot happen due to Fubini's Theorem.
Comment: +1. Lovely! Is it consistent with ZF that this can't happen? (I'm sure it is but I don't see the proof just yet.)
Comment: @NoahSchweber: It is consistent with ZF that $\mathbb{R}$ is a countable union of countable sets, so I really don't know what Lebesgue measure means. But you probably implied ZF+DC, in which case I don't know. In related news, there's a 1980 theorem by Friedman that it's consistent with ZFC the last counterexample I mentioned can't occur.
Comment: Why doesn't Fubini-Tonelly imply  $\int_{[0,1]^2} f = 0$?
Comment: @LordOfNumbers In [Fubini-Tonelli](https://en.wikipedia.org/wiki/Fubini%27s_theorem#Fubini%E2%80%93Tonelli_theorem), $f$ must be measurable. Look at the edit at the end of this answer.
Answer: False belief: A simple arc in the plane (i.e., a subset of the plane which is homeomorphic to the interval $[0,1]$) has planar Lebesgue measure zero.
Comment: .Interesting. Can you give a reference?
Comment: @user254665 I'm repeating [this old answer](http://math.stackexchange.com/questions/820686/obvious-theorems-that-are-actually-false/820961#820961). Quoting my comment there:  "it follows from a more general theorem of R. L. Moore and J. R. Kline (On the most general plane closed point-set through which it is possible to pass a simple continuous arc, Ann. of Math. (2) 20 (1919), 218-223) that every (homeomorph of the standard) Cantor set in the plane is contained in an arc; apply this to a fat Cantor set in the plane."


Partially overlapped wave function of two particles

Question: For indistinguishable particles, the wave function of them should be symmetric or antisymmetric, but in the real world, this might require that all the same type of particles should be considered when solving a specific problem, so I am trying to figure out when could I treat two particles as distinguishable. If the wave function of two particles is not overlapped, it would make sense to treat them as distinguishable, but what if they are overlapped not totally, only partially, would they be distinguishable?
Answer: If the single-particle wavefunctions of two particles overlap they might be indistinguishable  or they might be distinguishable. For two indistinguishable fundamental particles (emergent particles might be more complicated: anyons or nonabelions) we should have Bose or Fermi statistics:
$$\psi(x_1,x_2) = \psi_1(x_1) \psi_2(x_2) \pm \psi_2(x_1) \psi_1(x_2).$$
But the problem is that just looking at the single-particle wavefunctions $\psi_1, \psi_2$ won't tell you this. Also, even if your two particles really did obey the above relation, I don't see any way to rule out the possibility that they were entangled distinguishable particles that just happened to be in a state with Bose or Fermi statistics.


Magento 2 use group by in collection

Question: I've got a collection from the factory method getCollection(). The problem is that when I use get_class_methods($class) to list all the methods that the collection possesses I can't seem to find one which groups the records in the collection.
I found this How to use group by in magento collection with join query
But I don't have the group() method in my collection class. How can I group by in that case ?
Answer: You cannot use group directly from the collection.
But each collection has an instance of Zend_Db_Select associated to it. This is the actual query that is going to be executed when the collection is loaded.
So you can use:
$collection = ....
$collection->getSelect()->group('group_field_here');

Comment: i have create a custom product attribute "sample_attribute" with drop down value i want to add group by in product collection but this gives unknown column how can i add group by in product collection

Magento 2 use group by in collection


Cognito Forms: Reveal section based on Boolean field response

Question: I have created a form section which I would like to be conditional upon a checkbox on a previous field being checked or not. While I have sailed through all aspects of form creation, it seems that this condition is not available. Am I missing something?
Answer: I am a developer with Cognito Forms.
Cognito Forms does support conditional visibility logic. If the click on the section you would like to conditionally hide/show, in the field settings section on the left there is an option called "Show This Field". Choose the "When" option. This will cause a dialog to open that will allow you to edit when the selected section should be visible to the user.
For more information check out our Help Topic regarding this, and other, field settings.


What manga is Mur Mur reading?

Question: In the anime Mur Mur reads manga sometimes. What are these manga?
Answer: It's a fictional manga magazine featuring a few different chapters from different series.  The word シュシュ in japanese means an elastic band of some sort.
Comment: I don't have any context to base this on, but シュシュhere could be an irregular way to write しゅうしゅう (collection/variety), which might make more sense than naming it after scrunchies.
Comment: @LoganM yeah, I believe you might be right on that one.
Answer: To me the first one looks like a manga magazine. So it includes more than one manga. And the magazine is called "Shushu". I assume it is a fictional magazine since I can't seem to find information about it.

What manga is Mur Mur reading?


SQL for large scale datasets

Question: Some job descriptions include sth such as "demonstrated skill on handling large scale(massive) datasets using SQL"
I would like to know which kinds of SQL-related skill sets are required for meeting the requirements of these jobs?
Answer: Designing a performant schema and knowing when to denormalize (and when you've got problems you can solve other ways.)
Efficient query design.
The intimate details of index design, to the point where you can make changes and get the results you expected.
How to build, support, and effectively make use of test data.
How to read all the breadcrumbs your server leaves in its trail (logs and query plan analyses in particular.)
How to tell how hardware, dbms software, and configuration work together and be able to adjust parameters and modify hardware without fear, and get the results you expected.
Answer: Everything related to SQL would be, IMO. Everything from query writing to DBA changes with large scale datasets.

SQL for large scale datasets


if I have 1 M/Hs mined for me how does this translate into a bitcoin?

Question: I was offered 1 M/Hs per day for 37.00. What do I do with this in terms of turning it into a bitcoin. I am new to this and trying to make sense of all the mathematical gibberish
Comment: It doesn't. It's so little hashpower that you'll never earn a Bitcoin.
Answer: Sounds like your renting a server in the cloud,you can use this server to mine for Bitcoin. Mining for Bitcoin means the server will use its fast computing power to decrypt a bunch of numbers constantly. When it does that the server is helping validate transaction for other people. For your investment in the equipment and time you get rewarded with Bitcoin. Over time it gets harder and harder to obtain Bitcoin because the encryption gets harder overtime. You need faster servers to combat this issue. This generates scarcity which in turn generates value for Bitcoin. You can mine alone or mine with many other people, but you have to share. You have to get mining software to start mining. This server you are renting might already have this setup. I hope this helps you understand. Cheers


Are deleted WhatsApp (or similar) messages recoverable?

Question: When one deletes a Whatsapp (or other similar apps) message from a device (assume that no Google Drive, ICloud backup), can this message be recovered from the device? Does Whatsapp save messages into SQLite DB as normal text/unencrypted format?
Comment: "Does Whatsapp save messages into SQLite DB as normal text/unencrypted format?" is *easily* searched online. We've even talked about it on this site. I took that entire line and popped it into a Google query and got technical details.
Comment: As for "whatsapp deleted messages", that, too, is easily searched. As this is more of a question about this one product, and isn't even a forensic concern, *and* both questions can be answered by the top hits on Google, I'm closing as not a security question.
Answer: Yes, it is possible, but like with many security questions it depends on the circumstances.
Generally the messages can't be recovered because WhatsApp state that they do not store chat histories on their servers.
If the message has been backed up using either iCloud Backup (for iOS devices) or Google Drive (for Android devices) then the message can be recovered when the user uninstalls WhatsApp and restores the backup.
To restore using the Google Drive method the user would need to restore to a local backup from the local WhatsApp database. This can be found in the WhatsApp > Database folder on the device. The filename is msgstore.db.crypt12 and there shoudl be various msgstore-YYYY-MM-DD.1.db.crypt12 files.
If the device is Android then it may also be possible to recover the message using a data recovery program such as Android Photo Recovery which utilises USB debugging mode to scan the device from a PC or a Mac.
Answer: Additionally to Mike Pooles answer, be aware that you sent this message to at least one other device.
If you don't have control over the other end device, it will be always possible to recover your message by acquiring the device or its backups.
Hiding the fact that you communicated with this party is difficult or even impossible because the metadata is available on the servers of Facebook, on the other end device and possible other third parties.
Comment: Thanks. I've edited the question to make it clearer.
Comment: That is not necessarily accurate. Many secure messaging platforms have the ability to delete the message from the remote side too, when deleted locally.  Obviously this discludes screenshots being taken as a way to retain the message contents.
Comment: @JesseP. But there's no guarantee that this feature works on the target device at all. Additionally, this doesn't work on backups as long as they are not deleted or overwritten with the new state. When the backup is on a server, it's also very likely that there are backups of the backup and so on. In the end, it strongly depends on the threat actor but in general this kind of feature should not be relied upon for highly sensitive information.
Comment: @Noir No guarantee that it will succeed but some give a response of whether it did or not, so at least users know what the result was.  Still, it's possible and an option none-the-less.


What is the expected size of the largest strongly connected component of a graph?

Question: Given a directed graph with n vertices and the probability of any edge existing being p, what is the size of the largest strongly connected component in the graph? What if its undirected graph? Can we also estimate the number of components too in terms of n and p?
Answer: Much is known about the behavior of the (undirected) random graph $G(n, p)$ in the Erdős–Rényi model, including the following:
$\langle$Begin quote$\rangle$

If $np < 1$, then a graph in $G(n, p)$ will almost surely have no connected components of size larger than $O(\log(n))$.
If $np = 1$, then a graph in $G(n, p)$ will almost surely have a largest component whose size is of order $n^{2/3}$.
If $np → c > 1$, where $c$ is a constant, then a graph in $G(n, p)$ will almost surely have a unique giant component containing a positive fraction of the vertices. No other component will contain more than $O(\log(n))$ vertices.
If $p<\tfrac{(1-\epsilon)\ln n}{n}$, then a graph in $G(n, p)$ will almost surely contain isolated vertices, and thus be disconnected.
If $p>\tfrac{(1+\epsilon) \ln n}{n}$, then a graph in $G(n, p)$ will almost surely be connected.
Thus $\tfrac{\ln n}{n}$ is a sharp threshold for the connectedness of $G(n, p)$.

$\langle$End quote$\rangle$
The directed case is studied here.
Comment: Quote: `What if its undirected graph?` // About the directed case: do you decide the existence of edge $xy$ and the existence of edge $yx$ independently?
Comment: See the last line of my answer.
Comment: (Previous two comments answering now-deleted queries by the OP.)
Answer: The random directed graph on $n$ vertices, where each potential arc is present with probability $p$ independent of any other arc is typically denoted by $D(n,p)$. As Did's link mentions, $p=\ln n/n$ is a sharp threshold for strong connectedness. The following results deal with sparse random digraphs.
A $complex$ strong component is a strong component with more arcs than vertices. In other words, a complex strong component is a strong component that is neither a single vertex or a cycle.
Karp (1990, "The transitive closure of a random digraph") and Luczak (1990, "The phase transition in the evolution of random digraphs") proved that for any $\omega(n)$ tending to $\infty$, with probability tending to 1,
1) each strong component of $D(n,p=c/n), c<1$ has size less than $\omega$ and there are no complex strong components.
2) there is an unique complex strong component in $D(n,p=c/n),c>1$ which is of size on the order of $n$, and each other strong component has size less than $\omega$.
As opposed to undirected random graphs, this $\omega$ can tend to $\infty$ however slowly!
The behavior of $D(n,p=1/n)$ has been difficult to establish. As far as I know, the most up to date information, proved by Luczak and Seierstad (2009, "The Critical Behavior of Random Digraphs"), is that there is some $a,b \in (0,1)$ such that for $n$ large enough
$$P(\text{there exists a complex component in }D(n,p=1/n)) \in [a,b]$$
In the same paper, they show analogous results to 1) and 2) above for $c$ tending to 1, but $n^{1/3}|c-1| \to \infty$.


How much more dangerous is it to fly in Mexico than Europe?

Question: Recently, Mexico's air safety rating was downgraded. How much more dangerous (in terms of statistical chances of fatal accident) is it to travel with a Mexican airline, than a European one?
Comment: If you mean flying as a passenger on a commercial flight, your question really belongs on the Travel site.
Comment: @jamesqf Disagree, aviation safety and aviation regulations are both explicitly on topic here. Questions may be on topic on multiple sites and that's OK.
Comment: @jamesqf It specifically says that even aviation enthusiasts are allowed to ask questions. What would really change, though, if the question started off with "I am just about to start pilot training in Mexico next week..." and then continued exactly as written... Would it be on topic? Would the answer change at all? If the answers are "yes" and "no" in that order, then why discriminate based on who's hand was on the keyboard?
Comment: Why comparing with Europe? It is FAA who downgraded, not EASA.
Comment: @user3067860: If the OP was asking as a pilot, it would be relevant.  As a passenger, it's not.  If it's on topic, it shouldn't be.
Comment: @user3067860 jamesqf has a hobby... closing questions.
Answer: To answer your title question with some numbers, here are the accident rates (number of accidents per million departures) for the world, Europe and Mexico from 2008 to 2020:

(Data from ICAO - Accident Statistics)
It's not quite what you asked about since the ICAO data does not filter by airline origin country and the accidents might include non-fatal ones (but not incidents, see What is the difference between aviation "accident" and "incident"?).
Nonetheless, it's clear that Mexico is not significantly less safe than Europe. The average accident rate since 2008 was 3.27 in Europe, but only 2.06 in Mexico. The plot above should make it clear that these numbers are fluctuating quite a lot because of the small number of accidents in each year. For example, the large peak for Mexico in 2010 is the result of only 6 accidents. One should therefore be careful when deriving any conclusions from these numbers.
Comment: @mins Yes, that's what I meant by "the ICAO data does not filter by airline origin country". Unfortunately, I couldn't find such a dataset. I found some with number of accidents by airline, but without a suitable denominator to get a *rate*, such a dataset is useless.
Comment: If there's anything that following US passenger rail has taught me, it's that the sample period has to be appropriate for the number of departures.  Amtrak and the airlines used to trade safety crowns ... Amtrak was on top until Big Bayou Canot, then airlines were on top until 9/11.  All this is inaccurate, and a very silly game, because the sample sizes are way too small and are whipsaw'd by single events.
Comment: Not disputing anything included in this answer, but it is very important to understand that reading stats and interpreting graphs is a whole science of it's own. To draw **any** conclusions, one has to be familiar with the subject in question, as well as how the stats were derived from data. Only thing that can be safely stated based on the graph above is that apart from 2010, flying is generally about as safe in Mexico as in the rest of the world.
Comment: @Jpe61 Exactly, hence my last sentence. I would even go as far as saying in 2010 Mexico was just as safe as in the other years. It was just a statistical fluctuation.
Comment: It's a good idea to provide these stats, but be aware the State criteria doesn't count accidents by operator State but by accident location. E.g in [France stats](https://i.stack.imgur.com/Rgoji.png) Air France AF447 in 2009 (over east of Brazil) isn't counted while Germanwings 4U9525 in 2015 (over France) is counted.
Comment: ...there is solid truth in the saying "lies, damned lies, and statistics"...
Comment: @Jpe61: "*Only thing that can be safely stated based on the graph above is that apart from 2010, flying is generally about as safe in Mexico as in the rest of the world*" From what I see Mexico *has been* at least 2x or 3x better than the rest of the world in the last 10 years, in term of accidents *over* their territory, but maybe all accidents with Mexican airlines took places elsewhere, therefore decreasing the number of accidents over Mexico and increasing it for the rest of the world...
Comment: Exactly @mins really hard to say pretty much anything, as there are only a couple of accidents annually. Any occurence carries a lot of "weight". Considering the graph is Mexico vs. Europe vs rest of the world, you'd expect the rest of the world be a lot higher, as the rest of the world includes states with, well, sub par safety culture. Also, accidents do not necessarily relate to local safety culture at all, as they may be simple technical failures, or out-of-region-crew errors etc. etc...
Answer: Short answer
Such audits are done by FAA to evaluate whether safety is correctly managed by the Mexican regulator in order to inform US DoT.
FAA found there is a problem in how Mexico is organized to supervise Mexican airlines, and consequently US DoT will prevent new Mexican carriers to enter the US market, freeze new routes creation, and will prohibit US carriers to sell seats on Mexican carrier aircraft, until this is fixed.
Nothing can really be drawn from the result. There are different domains evaluated, not being compliant in one domain is sufficient to be disqualified. FAA doesn't disclose the details of the non-compliance.
While this audit is performed by US for their own benefit, the framework used is the international framework of ICAO. ICAO also conducts audits in a continuous way. ICAO audit results can be viewed online with some details. Mexico was audited by ICAO in 2012. A comparison of results for Mexico (2012), US (2007) and France (2020, there is no "Europe" rating, ratings are by countries):

This comparison shows the Mexican authority is below the mean level regarding it's organization but also better than the US for licensing, legislation and operations.
How these categories are evaluated is described in the audit framework  ICAO Doc 9734 part A: The Establishment and Management of a State Safety Oversight System (respectively §3.3, §3.6, §3.1 and §3.2). Exactly like ISO 9000 audits, both ICAO and FAA audits are related to the management system ability to detect and correct problems in carrier operations, they are not related to how airline actually perform their operations.
Bottom line: We don't know how Mexican carriers perform, we know they are overseen by a non-compliant regulator according to FAA, due to one or more unknown weaknesses.
Details related to obligations of ICAO members and audits follow for those interested.

Details
Chicago Convention: Obligations of the States
Mexico is a member of ICAO and signatory of the Chicago Convention. As such Mexico has legal obligations, which fulfillment is delegated by Mexico government to AFAC, the Mexican regulatory agency, counterpart of FAA for Mexico. Contracting States have safety-related obligations, in particular with art. 37 and 12:

(37): Each contracting State undertakes to collaborate in securing the highest practical degree of uniformity in regulations, standards, procedures and organization in relation to aircraft, personnel, airways and auxiliary services in all matters in which such uniformity will facilitate and improve air navigation.

(12): Each contracting State undertakes to adopt measures to insure that every aircraft flying over or maneuvering within its territory and that every aircraft carrying its nationality mark, wherever such aircraft may be, shall comply with the rules and regulations relating to the flight and maneuver of aircraft there in force.

Each contracting State undertakes to keep its own regulations in these respects uniform, to the greatest possible extent, with those established from time to time under this Convention [...] Each contracting State undertakes to insure the prosecution of all persons violating the regulations applicable.

ICAO audit: USOAP-CMA
ICAO members agreed on a continuous audit approach which abbreviation is USOAP-CMA (USOAP website) to evaluate how States fulfill their obligations. ICAO and FAA audits are based on the same reference, ICAO Doc 9734 part A: The Establishment and Management of a State Safety Oversight System, which includes 8 domains, either for rule-making or for rule-enforcement:

Audited domains under USOAP-CMA and IASA, source ICAO Doc 9734
FAA and IASA
FAA conducted their IASA in May 2021. Objectives, from FAA:

IASA assessments determine compliance with these international Standards by focusing on the eight critical elements of an effective aviation safety oversight authority specified in ICAO Document 9734, Safety Oversight Manual.

Those eight critical elements include primary aviation legislation; specific operating regulations; State civil aviation system and safety oversight functions; technical personnel qualification and training; technical guidance, tools and the provision of safety critical information; licensing, certification, authorization, and approval obligations; surveillance obligations; and resolution of safety concerns.

The results are the one you're talking about in the question. The fact that Mexico was found to be not compliant to ICAO standards, means they were found to have one or more of these problems (source):

Lack of advisory documentation;
Shortage of experienced airworthiness staff;
Lack of control on important airworthiness related items such as issuance and enforcement of Airworthiness Directives, Minimum Equipment Lists, investigation of Service Difficulty Reports, etc.;
Lack of adequate technical data;
Absence of Air Operator Certification (AOC) systems,
Nonconformance to the requirements of the AOC System
Lack or shortage of adequately trained flight operations inspectors including a lack of type ratings;
Lack of updated company manuals for the use by airmen;
Inadequate proficiency check procedures;
Inadequately trained cabin attendants.

Consequences from a FAA standpoint
From FAA:

While the new rating allows Mexican air carriers to continue existing service to the United States, it prohibits any new service and routes. U.S. airlines will no longer be able to market and sell tickets with their names and designator codes on Mexican-operated flights. The FAA will increase its scrutiny of Mexican airline flights to the United States.

Re-occurence

It is the second time that Mexico is in Category 2. The first time happened in 2010, and it took the government four months to regain its prior status.

Source
Comment: It's not just that new Mexican carriers won't be allowed, it's also that existing ones can't open new routes to the U.S. and, perhaps more importantly, U.S. carriers can't sell codeshares onto the existing routes operated by Mexican carriers. This would be a pretty big deal for Delta, for example, which not only has Aeromexico as a close SkyTeam alliance partner, but actually owns a 49% stake in it. So, for example, they can't sell DL8XXX tickets on ATL-MEX flights operated by AM. Other Mexican carriers (e.g. Volaris, etc.) are also banned from opening new routes to the U.S.
Comment: The airworthiness part is the most troubling, and clearly shows Mexico well below France and slightly below the U.S.
Comment: @reirab: Good addition.


Isekai manga where mc receives two blessings/skills

Question: I was reading a manga where the MC was reincarnated into another world after receiving two XP related skills/blessings from two goddesses as they had a mix-up and both gave him a skill/blessing when he should only have one, I believe they were something like 20 times growth and 1/20 XP needed, or something along those lines.
Comment: Hi, welcome to the site. Is there anything else you can tell us about this manga? For example, were there any other significant characters? Were there monsters? Was it a full colour manga, or mostly black & white?
Answer: As per Manga where a goddess gives a character 20xp, then another goddess gives him 1/20 Level Up but he stays jobless, this is likely I've Became Able to Do Anything with My Growth Cheat, but I Can't Seem to Get out of Being Jobless (AKA Seichou Cheat De Nandemo Dekiru You Ni Natta Ga, Mushoku Dake Wa Yamerarenai You Desu).

I, Ichinosuke, failed to get a job for 100 consecutive times.
Furthermore, my record is still being updated as unemployed.
On my way to a job interview, I was involved in an accident with a
truck and tragically passed away.
And then, a Goddess bestowed upon me the ability to grow at a speed
400 times faster than normal, before sending me to a game-like
otherworld.
However, knowing that I would not become stronger even if I continued
growing while jobless, I aimed to get a job, but I found out that it
was my terrible fate to remain jobless.
As my Growth Cheat and Jobless Cheat existed side-by-side in this
otherworld, an adventure record, no doubt certified by the
job-placement agency as a harmful book, begins here.
Comment: As per the [tour], you can accept an answer by clicking on the checkmark by the voting buttons. In this case, because it has been asked before, it would be marked as a dupe.
Comment: FWIW, your mention of the 1-to-20 ratio sparked the memory in my head that we'd had this one before.


ACA Individual mandate fulfilled using employer insurance

Question: I am a bit confused with the Affordable care act and employer insurance. If the employer mandate starts in 2015 and the individual mandate starts in 2014, how would an employee know if their employer provided insurance was acceptable in terms of the individual mandate?
Comment: I'm not sure what you're asking. If you're covered by your employer, I don't think you have to worry about the individual mandate.
Comment: My question:  How do I know if health insurance from an employer sponsored plan is sufficient enough to cover the mandate? (We asked my spouse's benefits department if the plans offered are grandfathered and they literally had no idea.)
Comment: I think they automatically are, I have to check.
Answer: If your employer is providing you with insurance, the individual mandate does not apply. Your employer provided insurance exempts you from an individual mandate, and as such, minimum standards of care are not your concern. They wouldn't be your concern either way: you just have to have insurance. It's up to insurance companies to ensure that that insurance meets minimum standards of care. In fact, insurance companies will be cancelling such policies before the employer mandate comes into effect.
From Kaiser Health News:

The Congressional Budget Office estimates that in 2014 nearly three out of five Americans will have coverage through an employer-provided plan and 12 percent through Medicaid and the Children's Health Insurance Program, federal-state programs that provide insurance to lower income Americans. If you have insurance in either of those ways, you are not affected.

(Emphasis mine).
Because you receive insurance from your employer, you do not have to worry about the individual mandate.
Answer: It seems like its up to your benefits provider to know if your employer sponsored health plan is actually compliant or not.  The answer seems to be that cannot find out right now.
from healthcare.gov

How to find out if your plan is grandfathered?
Check your plan's materials: Health plans must disclose if they are grandfathered in all materials describing plan benefits. They must offer contact information.
Check with your employer or your health plan's benefits administrator.
Answer: I am in this exact situation, as our company is in the benefits selection phase for Medical and Dental plans for our employees to rollover in January 2014. As such, we have company-wide web meetings with benefits providers, who START the webcast by stating that our company plan is in compliance with the Affordable Care Act.
Having said this, we are not required to, BUT CAN, seek medical insurance through the ACA, if we believe we can achieve a better, more comfortable or affordable plan for our money, in the medical marketplace. They also make a point of saying that our company plan provides an employer match, making our plan more fiscally advantageous than anything else we could find, since the match does not carry over to plans found in the marketplace.
Comment: That's really interesting that every web cast starts the same way.  I wonder who is liable if they say it is compliant and it is not actually compliant.


How to reject a tenant?

Question: I have had some prospective tenants apply to lease my Philadelphia, PA condo. As I listed the property on Zillow, the official 'application' is on Zillow as well. Zillow only has an 'accept' button for completed applications, but no 'reject' button. How do I respectfully reject a tenant based on their application? Via email? Do I inform them why (low credit score, income, etc...)?
Comment: Some advice from a fellow landlord: You should come up with a system for approving/rejecting applications and follow it very consistently. Rejecting candidates for the wrong reasons can put you in legal jeopardy. Usually you are better off just telling them you accepted another candidate and not being too specific. Especially if you aren't going to change your mind. Giving them a reason invites them to negotiate.
Comment: For example, number the applications when they come in and all else being equal always pick the first person that applied. Also don't reject one candidate for a 650 credit score and then accept another with a 550. At least don't do that and state the credit score as the reason for the rejection. Even if you are not being discriminatory at all in your selection process you can be accused of it, and you need to have evidence that your selection process was fair to all applicants including those in protected classes.
Comment: @JohnFx that certainly makes sense. What do you mean by a 'system'? Here is my current process:
1. Tour;
2. Email screening questions (to avoid forcing people to pay Zillow application fee if I know I won't select them);
3. Zillow application;
4. Email acceptance/rejection;
Is this sufficient?
Comment: I don't do tours first, I screen first, saves everyone time unless you're doing open houses. Screen by stating your criteria clearly (No smoking, no pets, gross income requirement is 3x rent, no felony convictions.....) then screening questions to confirm they meet the clearly stated criteria can save you a lot of tours.
Comment: Generally I wouldn't recommend disclosing the reason to the potential tenant. There is no good reason for you to do so, and if you choose to do so; it could be used against you.
Answer: Email would be just fine. Even if Zillow had a built-in way to reject applicants, I would send an email. Someone might not have their notifications set up, or have used a burner email for their Zillow account (different then what they put on the application).
Whether you tell them why you rejected the application is up to you. If you have a justifiable reason then it may be helpful for the applicant to know why. They may have even forgot to give you some information (like a secondary income source).
Personally, we rented and then purchased a home that was originally found on Zillow and none of the communication with the landlord/realtor was actually done through the Zillow website other than the initial contact. It was all done through phone and email.

How to reject a tenant?


How do I find out the DOM node at cursor in a browser's editable content window using Javascript?

Question: I am looking for a solution that works cross browser i.e. IE, Firefox and Safari.
Answer: By "editable content window" I'm going to assume you mean an element with contenteditable turned on or a document with designMode turned on.
There are also two cases to consider: the case when the user has made a selection and the case where there is just a caret. The code below will work in both cases, and will give you the innermost element that completely contains the selection. If the selection is completely contained within a text node it's slightly complicated to get that text node in IE (trivial in other browsers), so I haven't provided that code here. If you need it, I can dig it out.
function getSelectionContainerElement() {
    var range, sel, container;
    if (document.selection && document.selection.createRange) {
        // IE case
        range = document.selection.createRange();
        return range.parentElement();
    } else if (window.getSelection) {
        sel = window.getSelection();
        if (sel.getRangeAt) {
            if (sel.rangeCount > 0) {
                range = sel.getRangeAt(0);
            }
        } else {
            // Old WebKit selection object has no getRangeAt, so
            // create a range from other selection properties
            range = document.createRange();
            range.setStart(sel.anchorNode, sel.anchorOffset);
            range.setEnd(sel.focusNode, sel.focusOffset);

            // Handle the case when the selection was selected backwards (from the end to the start in the document)
            if (range.collapsed !== sel.isCollapsed) {
                range.setStart(sel.focusNode, sel.focusOffset);
                range.setEnd(sel.anchorNode, sel.anchorOffset);
            }
        }

        if (range) {
           container = range.commonAncestorContainer;

           // Check if the container is a text node and return its parent if so
           return container.nodeType === 3 ? container.parentNode : container;
        }
    }
}

Comment: Still works for me in IE. Can you post an example showing your problem?
Comment: Tim, this is exactly what I was looking for. I would appreciate if you can also dig in the IE case where selection is contained with in a text node. Thanks.
Comment: Btw, the IE case where you return range.parentElement() always seems to return the body element even though the caret was in text that inside an anchor node eg. In this HTML xyz if the caret was in between say 'x' and 'y', shouldn't parentElement() return the 'a' node?
Comment: Yes, it should. I haven't had problems with `parentElement()` before - have you got a page I can look at showing the problem?
Comment: The IE code is quite involved. I use my own optimizations of the algorithms in IERange (http://code.google.com/p/ierange/).
Comment: @Tim, great code. But there is a problem with IE when you want to get the parent without selecting any text, just positioning a caret. How to fix that?
Answer: You can also use the Rangy Library:
elementAtCursor = rangy.getSelection().anchorNode.parentNode


How is market depth data useful

Question: I was wondering how market depth data is useful if the orders which change the price would not be available . If we consider the orders which change the price , these are the orders where the bid on one side and the ask on the other are executed. i this is the case , then market depth would show orders which cannot execute and dont change the stock price.
Comment: I think this answer could be helpful in explaining why market depth could be useful: https://quant.stackexchange.com/questions/54786/how-can-we-estimate-new-stock-price-after-a-large-purchase/54788#54788

Although not directly answering your question, it illustrates an example. Otherwise, market depth at the very least tells you the size of an order that one could execute **immediately** if one were to take out the full liquidity available in the market: i.e. it tells you the maximum available number of securities you can buy or sell at any point in time: which is very useful indeed.
Answer: The "price" of something is a historical value, i.e. it's the price at which the last trade happened, in the past. The information in the order book, together with the size and direction of incoming orders are what will determine the prices of the next trades. That's what you're really interested in.
Here's a toy example: Suppose the best bid is at 100, but only for 50 units and then the bids at 99 total 50 units, too and then the bids at 98 are lots of units. And suppose you need to send a market order to sell 100 units. So you will probably end up selling 50 of them at 100, and then 50 more of them at 99. Two things: 1) To know the prices at which your order was likely to execute, you needed to know more than the best bid. If you sent a larger market order to sell, you'd need to know more depth on the bid side to understand the likely execution of the order. And 2) Immediately after the order has executed (but, for the purpose of the example, before any other bids arrive), what is the best bid? Well it's been pushed down to 98 or lower. The more I know about the depth, the more I know where the price will go in the face of aggressive sell orders.
This is related to the idea of "microprice" (I think there's even "milliprice" nowadays but I don't know what it means). Some notions of microprice use info from deeper in the book to produce a weighted average of bid and ask that are different from the midprice. The weight is a decreasing function of the volume on the bid/ask sides because the thinking is that if there is more volume on the bid side, then the price is likely to go up, so we should weight the 'ask' value more heavily in our price estimate.

How is market depth data useful


How can I upgrade my units with the Ikanda

Question: I didn't get at all how to use the Ikanda by the in game pedia. In this page, it says
"[units] have access to the special upgrade path when built in a city with an Ikanda". Is it the only way to receive the Ikanda's bonuses, by creating military units within the city with the Ikanda? It also says, "if you manage to acquire all three promotions of the chain, you will get a very nice army!". How do I "manage to acquire" the promotions?
Thanks.
Comment: @DavidToh Ikandas are a building, they don't defeat enemies. (somehow this reminds me of a similar situation with gazebos)
Comment: the promotions are easy to get. merely use your ikandas to defeat other enemies, then as you slowly earn promotion chances, you can promote your unit till you unlock all the abilities in its particular ability chain.
Comment: It is important to note that ***only*** warriors, spearman, impi, horsemen, knights, and longswordsman receive the benefits from the Ikanda.  Archers and other units do not and the Ikanda acts as a standard barracks.
Answer: Units built in a city with an Ikanda get a special upgrade (Buffalo Horns) when they are produced.
Due to the way upgrades work in Civ V, having the Buffalo Horns upgrade is what lets you later take the Buffalo Chest and Buffalo Loins upgrades.
So to acquire all 3 upgrades, you simply need to train one of your units up to level 3.
Comment: So, after a unit built in a city with an Ikanda gets promoted by fighting, in addition to the regular bonuses he gets from the upgrade, he also gets the Buffalo Chest and later the Buffalo Loins? Did I understand you correctly? Thanks.
Comment: I would assume that the upgrades are options you can pick when you level, not free upgrades you get in addition.
Comment: @user52549 Michael is correct -- the Buffalo Chest / Loins are additional options (albeit pretty good options!), so you will have to choose them over, say, Shock I, rather than getting the latter 2 for free.


Density of a functional space

Question: Let $D$ be a bounded domain of $\mathbb{R}^n$ with smooth boundary $\partial D$. Is the following subspace dense in space $L^2(D)\times L^2(\partial D)$:
$$\{(f,f\rvert_{\partial D}) : f\in C^\infty(\overline{D}), (\Delta f)\rvert_{\partial D}=0\}.$$
I tried to use density of $C_c^\infty(D)$ in $L^2(D)$, but I didn't get a final answer. This question was motivated by the density of the subspace
$$\{(f,f\rvert_{\partial D}) : f\in C^\infty(\overline{D})\}.$$
Thank you for any hint.
Comment: @Fan Zheng do you mean that the space is dense? Sorry, I don't get your hint.
Comment: Yes, and your attempt showed how to approximate when the boundary value is 0. Then you only need to approximate the boundary value in $L^2$, while keeping the Laplacian 0 at the boundary, which can be done using the hint.
Comment: Thank you, this is what I understood: let $(f_n)\subset C^\infty(\overline{D})$ approximating $(f,f\rvert_{\partial D})$. Consider the solution of: $\Delta h_n=0$ on $\overline{D}$ and $h_n\rvert_{\partial D}=f_n\rvert_{\partial D}$. Then $h_n \in C^\infty(\overline{D})$ and $h_n\rvert_{\partial D} \to f\rvert_{\partial D}$. But why $h_n \to f$?
Comment: Sorry, it's "approximating $(f,g)$" in my previous comment.
Comment: $(h_n)$ doesn't converge necessary to $f$, right?
Comment: @Fan Zheng I think it converges to the solution of $\Delta h=0$ in $\overline{D}$, $h\rvert_{\partial D}=f\rvert_{\partial D}$.
Comment: @Dolii The convergence needs to be in $L^2(D)$. This amounts to the boundedness of the the harmonic extension from $L^2(\partial D)$ to $L^2(D)$. I believe this to be true, at least assuming $\partial D$ is smooth enough, but I have yet to find a proof. There is a Springer book *The Dirichlet Problem with L2-Boundary Data for Elliptic Linear Equations* on that particular problem, which may be helpful to you.
Comment: @FanZheng I didn't find the desired result in the book. Maybe the missed result is the one in Corollary 1.5 page 6 in the following paper, with $\sigma=1/p=1/2$.
  : https://hal.archives-ouvertes.fr/hal-00629123/document
Comment: Hint: to approximate the boundary value in $L^2$, use a smooth function and then solve the Dirichlet problem with that as the boundary value.
Comment: @Fan Zheng, could you write a rigorous answer?
Comment: @Dolii What is needed is that $h_n$ converges to something in $L^2(D)$. But this is more difficult than what I have thought.
Answer: I believe that Daniele Tampieri's idea of using completeness Fichera's Theorem is the right one. However I think there is a simpler proof.
In view of the Hahn-Banach Theorem,  proving the density of the space
\begin{equation}
\left\{(f,f|_{\partial D})\ |\ f\in S \right\}, \quad \text{where }
S=\left\{ f\in C^{\infty}(\overline{D})\ |\ \Delta f|_{\partial D} =0 \right\},
\end{equation}
in $L^{2}(D)\times L^{2}(\partial D)$
is equivalent to show that, if $(F,G)\in L^{2}(D)\times
L^{2}(\partial D)$ is such that
\begin{equation}
\int_{D}F\, f\, dx + \int_{\partial D}G\, f\, d\sigma = 0
\end{equation}
for any $f\in S$, then $F=0$ a.e. in $D$
and $G=0$ a.e. on $\partial D$.
So suppose that these orthogonality conditions hold. In particular we have
$$
\int_{D}F\, f\, dx   = 0
$$
for any $f\in C^{\infty}_{c}(D)$. It is well known that this implies
$F=0$ a.e. in $D$.
Therefore
$$
\int_{\partial D}G\, f\, d\sigma = 0
$$
for any $f\in S$. Harmonic polynomials obviously belong to $S$ and then
$$
\int_{\partial D}G\, \omega\, d\sigma = 0
$$
for any harmonic polynomial $\omega$. Fichera's theorem implies
 $G=0$ a.e. on $\partial D$.
Comment: @Dolli You are right, in my proof I was considering domain $D$ such that $\mathbb{R}^n \setminus D$ is connected. If this condition is not satisfied the harmonic polynomials are not dense. However in Fichera's paper also mutiple connected domains are considered. In this case, in order to have the completeness of the system on the boundary, you have to add to harmonic polynomials  certain harmonic rational functions whose poles are in the "holes". So my proof works also in this case. You have just to take $\omega$ in this more complicated system.
Comment: @Daniele and Cialdea thank you for the clarification. I meant for my case I think I don't really need the density of harmonic polynomials, maybe a large class is sufficient to conclude without assuming that $\mathbb{R}^n\setminus \overline{D}$ is connected, and I think also that Fichera's theorem is more stronger here.
Comment: @Cialdea Thank you! I tried this method first but I missed Fichera's theorem. Do we need really this theorem? I ask this question because I don't want to add the assumption $\mathbb{R}^n\setminus \overline{D}$ is connected and I didn't see this assumption needed in Fan Zheng's method. Maybe this is not a necessary assumption in this context.
Comment: @Dolii you've been mislead by the references I've given in my answer. In the original paper of Fichera, nor $D$ nor $\Bbb R^n\setminus\overline{D}$ need to be connected. Fichera (loc. cit. p. 2) considers explicitly $D$ as a "*domain with $p\,(\in \Bbb N)$ holes*": however his paper is written in Italian and, to my knowledge, this material has not been translated in English. However, Alberto Cialdea could say something more on this: on my side, I'll add a note to my answer.
Answer: This is not an answer, but an elaboration of my comments above.
Let $(f,g)\in L^2(D)\times L^2(\partial D)$. Find $g_n\in C^\infty(\partial D)$ such that $g_n\to g$ in $L^2(\partial D)$. Let $h_n\in C^\infty(\bar D)$ be the harmonic extension of $g_n$ to $\bar D$, i.e., the solution of the Dirichlet problem $\Delta h_n=0$ on $D$ and $h_n=g_n$ on $\partial D$. By classical results such an $h_n$ exists and is unique. I will next assume that

(1) Harmonic extension is a bounded map from $L^2(\partial D)$ to
  $L^2(D)$.

I believe this is true, at least when $\partial D$ is smooth enough, and I think the book The Dirichlet Problem with L2-Boundary Data for Elliptic Linear Equations will be of help, but being far from an expert in elliptic equations with nonsmooth boundary values, I have yet to assemble a proof from a brief glimpse of the book so far. Maybe someone else can help?
Now let's just assume (1), and continue our proof. Using $g_n\to g$ in $L^2(\partial D)$ and (1), we know that $h_n$ converges to some $h\in L^2(D)$. Then $f-h\in L^2(D)$, and we can find $F_n \in C_c^\infty(D)$ such that $F_n\to f-h$ in $L^2(D)$. Now let $f_n=h_n+F_n$. Then $f_n\in C^\infty(\bar D)$, $f_n\to f$ in $L^2(D)$, $f_n|_{\partial D}=h_n|_{\partial D}=g_n\to g$ in $L^2(\partial D)$, and $\Delta f_n|_{\partial D}=\Delta F_n|_{\partial D}=0$.
P.S. It is likely that there is a simpler proof than consulting the Springer book.
Answer: The answer is affirmative and as a consequence of the following old result proved by Gaetano Fichera in [4] (see also the survey [2] pp. 54-59).
Theorem (Fichera [4]). Let $D$ be a bounded domain in $\Bbb R^n$, $n\ge 2$, with $C^2$ boundary $\partial D$ and such that $\Bbb R^n\setminus\overline{D}$ is connected. Denoting with $\{\omega_k\}_{k\in\Bbb N}$ the sequence of homogeneous harmonic polynomials, the following three properties hold

$\{\omega_k\}_{k\in\Bbb N}$ is complete in $L^2(\partial D)$
$\{\partial_\nu\omega_k\}_{k\in\Bbb N}$ is complete in the space
$$
\left\{v\in L^2(\partial D)\,\,\Bigg|\, \int\limits_{\partial D}v\mathrm{d}\sigma=0\right\}
$$
Given a partition $\Sigma_1, \Sigma_2$ of $\partial D$ (i.e. couple of  subsets of $\partial D$ such that $\Sigma_1\cap\Sigma_2=\emptyset$ and $\Sigma_1\cup\Sigma_2=\partial D$), the sequence $\big\{(\omega_k,\partial_\nu \omega_k)\big\}_{k\in\Bbb N}$ is complete in $L^2(\Sigma_1)\times L^2(\Sigma_2)$.

Note. The theorem is stated and proved for domains with $C^2$ boundaries, but Fichera's proof is easily adapted to domains with Lyapunov boundaries, i.e. $C^{1,\alpha}$ boundaries, and by using the methods described by Cialdea ([1], [2]), it can be finally extended to domains with $C^1$ boundaries.
Now, defining
$$
\begin{align}
\mathscr{C_H^\infty}(\overline{D})&=\big\{ f\in C^\infty(\overline{D}): (\Delta f)\rvert_{\partial D}=0\big\}\\
\operatorname{diag}\mathscr{C_H^\infty}(\overline{D})\times&\mathscr{C_H^\infty}(\overline{D})= \big\{(f,f|_{\partial{D}}): f\in\mathscr{C_H^\infty}(\overline{D})\big\}
\end{align}
$$
as the space of infinitely smooth function with harmonic trace on a domain $D$, by using Fichera's theorem result, for each $(h,g)\in L^2(D)\times L^2(\partial D)$ it is possible to construct a sequence
$$
\big\{(f_n,f_n|_{\partial D})\big\}_{n\in\Bbb N}\Subset \operatorname{diag}\mathscr{C_H^\infty}(\overline{D})\times\mathscr{C_H^\infty}(\overline{D}),$$
converging to it.
Step 1. Construct an open cover of the domain $D$ such that the open cover $\{U_n\}_{n\in\Bbb N}$ satisfies the following conditions
$$
\begin{cases}
D_n\Subset D&\\
\operatorname{dist}(D_n,D) \ge \dfrac{\epsilon}{2^n}&\text{for a properly chosen and fixed }\epsilon >0 \\
\end{cases}
$$
where $\operatorname{dist}(A,B)$ is the euclidean distance between the two sets $A, B\in\Bbb R^n$.
Step 2. Define $v_n=\sum_{k=0}^n a_k\omega_k$, where $a_k\in\Bbb R$, $k\in\Bbb N$ are the Fourier coefficients of the expansion of $g$ respect to the complete system of homogeneous harmonic polynomials $\{\omega_k\}_{k\in\Bbb N}$ and, as such,
$$
\lim_{n\to\infty}\Vert v_n -g\Vert_{L^2(\partial D)} = 0
$$
Now put
$$
f_n(x)=v_n(x)+\int\limits_{D_n}\psi_{\small\frac{\epsilon}{2^{n+1}}}\big(x-y\big)\big[h(y)-v_n(y)\big]\mathrm{d}y\quad\forall n\in\Bbb N
$$
where $\psi_\cdot$ is the standard mollifier in $\Bbb R^n$: the sequence $\{f_n\}_{n\in\Bbb N}$ defines the sequence $\big\{(f_n,f_n|_{\partial D})\big\}_{n\in\Bbb N}$ which clearly satisfies the following limit requirement
$$
\lim_{n\to\infty}\big\Vert(f_n,f_n|_{\partial D})-(h,g)\big\Vert_{L^2(D)\times L^2(\partial D)}=0\quad \forall (h,g)\in L^2(D)\times L^2(\partial D).
$$
Notes

The intuition of Fang Zheng was the spark that triggered my proof: while difficult to apply as it was stated, his correct suggestion made me remember of the completeness in the sense of Picone for polynomial solutions of PDEs, and the circle of ideas that was developed by his school, notably Gaetano Fichera and his pupil Alberto Cialdea. In particular, the papers [1], [2] and [3] give state of the art results as well as an historical survey and relevant references. Apart from this, this kind of procedure is common when using the variational approch for the solution of PDE problems, since it avoids the Prym-Hadamard phenomenon.
Edit: Fichera's theorem does not require $D$ nor $\Bbb R^n\setminus\overline{D}$ to be connected: Fichera himself ([4] p. 2) precises this by describing $D$ as a domain with $p$ holes, $p\in\Bbb N$.
Statement 1 and 2 of Fichera's theorem are easily extendible to the space $L^p(\partial D)$, $1\le p\le+\infty$. For statement 3, things are different: while holding true for any $1\le p\le 2$ as implied by the $p=2$ proof, its validity for $p>2$ is still an open problem. However, this implies that
$$
\operatorname{diag}\mathscr{C_H^\infty}(\overline{D})\times\mathscr{C_H^\infty}(\overline{D})\text{ is dense in }L^p(D)\times L^p(\partial D)
$$
Note that $\Delta f_n\neq 0$ for all $(h,g)\in L^2(D)\times L^2(\partial D)$ (or $L^p(D)\times L^p(\partial D)$, $1\le p\le\infty$) for all $x\in\overline{D}$ but only in a neighborhood of $\partial D$. Requiring $f_n$ to be harmonic on the whole $D$ implies the failure of the density property as simple counterexamples can show.

[1] Cialdea, Alberto, Completeness theorems for elliptic equations of higher order with constant coefficients, Georgian Mathematical Journal 14, No. 1, 81-97 (2007). MR2323374, Zbl 1135.42333.
[2] Cialdea, Alberto, Completeness theorems in the uniform norm connected to elliptic equations of higher order with constant coefficients, Analysis and Applications (Singapore) 10, No. 1, 1-20 (2012). MR2876933, Zbl 1243.42043.
[3] Cialdea, Alberto, Completeness Theorems: an example of the legacy of Gaetano Fichera, in: C. Sbordone (Ed.) Equazioni a derivate parziali nell'opera di Gaetano Fichera, Quaderno n. 60 Accademia Pontaniana, Giannini Editore Napoli, 49-68 (2014).
[4] Fichera, Gaetano, Teoremi di completezza sulla frontiera di un dominio per taluni sistemi di funzioni, Annali di Matematica Pura ed Applicata, IV. Serie 27, 1-28 (1948). MR0029014, Zbl 0035.34801.
Comment: Thank you so much for the complete and valuable answer. I need more time to understand some details. I think it's somehow equivalent to what is proposed by Fan Zheng and maybe the missed part in his answer is as in my last comment. What do you think.
Comment: Thank you very much. The comments, if not the answer, of @FanZheng have inspired mine, mostly because I've tried (and perhaps succeeded) to provide a simple explicit construction of the approximating functions on $\partial D$. Notably, the possibility of approximating any $g\in L^p(\partial D)$ by harmonic polynomials shows that the $(\Delta f)|_{\partial D}=0$ condition can be satisfied in a *classical sense* on a neighborhood ranging from a distance of $\epsilon/2^{n+1}$ from the boundary inside $D$ to the whole $\Bbb R^n\setminus D$: I think this is my main contribution.

Density of a functional space


Who can answer questions?

Question: Sunnis and Shias are considered to be Muslims, but then you have Qadiani who claim to be Sunnis and Alawis who claim to be Shia, but all within their respective schools have declared them kafir. How do we make sure that only real Muslims are answering or do you have to be Muslim to answer?
Comment: Remember votes (both up and down) on Meta have an entirely different meaning than on the main site. Especially for *questions* on meta. Don't take it pesonally.  You brought up an important issue, that had to be addressed sooner or later.
Comment: By the number of down votes, I take it this didn't go down too well with some people :) this was not an attack on anyones belief etc it was to provoke thought and dialogue
Comment: None taken personally at all :) I respect peoples different opinions even if contrary to mine.
Answer: What is a real Muslim?
All Muslims think of themselves as real Muslims.
Another point, why should we not accept answers from non-Muslims if they are correct?
Filtering answers according to the person's beliefs (or lack of them) is highly discriminatory and bordering on ad hominem attacks.
Answers should be judged on their own individual merits, rather than on who the answerer is. The SE format should take care of that.
Comment: As a Muslim are willing to take your rulings from a non-muslim, would you send your children to a church to learn about Islam?
Comment: @abushahin - If a non-Muslim (or even a Muslim for that matter) gave me advice that was bad I would not take it regardless of his identity. And if that hypothetical church had a good, well known, well reviewed teaching program that was better than anything else around then sure, I would. Asking questions here is only half the battle. Reviewing the different answers and doing your own thinking and research on them before accepting any is the other half. Lazy fatwa hunters need not apply.
Comment: I understand your point, I just want to dig deep and find out if Islam allows for muslims to take answers about Islam from non-muslims, as Islam claims to have answers for everything then surely it has a say whether you are allowed or not. To derive your own opinions and Hukms (rulings) in Islam requires a lot of work and an understanding of the arabic language (Mujtahid) to do Ijtihad! Not to be taken lightly.
Comment: @abushahin - True. That is what works so well in the SE format. Answers that show greater understanding and diligence will naturally float to the top, regardless of who wrote it. After all, how would we really know he *is* a Muslim? Because he wrote so in his profile? So it is far better to simply ignore that issue and concentrate on what truly matters; the quality of the answer itself.
Answer: We had this very discussion on Christianity.SE when it first started beta, and we came to a simple conclusion (You can see the relevant meta discussion here) (adapted for Islam.SE):

Islam Stack Exchange is not a game show. We aren't here to vote people off the island. For the purpose of this site, please assume that the answer to "is X Muslim?" is always the same as "does X self-identify as Muslim?" It's the only way a group with such diverse cultures and beliefs can get along and do something productive.

I think the only way to run a successful site is to adopt the same policy here.
Answer: There's no reason to make sure that only "real Muslims" are answering; what matters is the answer, not the person it comes from.  If anybody posts an answer that is wrong or outright harmful, it will be voted down (and possibly even deleted) by the community.
We don't want to divide the ummah, and I see no reason why those who follow Ahmad shouldn't be allowed to use the site, even if I (personally) disagree with a lot of their beliefs.  Regardless of what others tell me, I won't treat anyone who claims to be Muslim as a kafir without clear evidences, nor do I recommend we as a community do so either.

Who can answer questions?


Best free UML authoring tool

Question: What free UML authoring tools do you use and why is it better than others?
Comment: Look at the question rating. It needs to be opened. I have some useful information to add.
Answer: ArgoUML - I use it for its simplicity.
Comment: Pity it doesn't have features for UML 2.2
Comment: @Jonn, what would be those features ?
Answer: Although it's not exclusive to UML, I use Dia. It has the symbols used in most (if not all) of the UML diagrams, but it also supports flowcharts, network diagrams, and a few other things that I've occasionally used as a software engineer.
Comment: I used to use Argo, then I found Dia and found it to be more enjoyable to use.
Comment: Dia is great, and multiplatform. Works great on Windows and Linux.
Answer: I tried most of the aforementioned tools so let me state my opinion on it here:

Dia - an old veteran; builds reliable charts (not just UML) but is rather cumbersome to use (especially if your diagrams get bigger :-( ) almost no restrictions on what to connect to each other, laying out diagrams nicely needs lots of manual adjustment (a serious time killer!), the dialogue boxes are hard to use (e.g. obsolete shortcuts such as alt+O for 'OK' to close it), navigating in a diagram wrecks your nerves with an incomplete endless sheet metaphor (scrollbars only work if one of your objects is out of the viewport; not all the time [like in Inkscape]) etc. etc.;
To sum it up: robust and reliable, but aged (esp. in terms of usability); I used it a lot (and wasted lots of time -rearranging my diagrams).
StarUML and argoUML - just used them shortly as they only support UML 1.x; someone even wrote their thesis (in German) on StarUML's shortcomings!
Visual Paradigm - new, intelligent, but the community edition is very limited: you'll get an ugly watermark if you create more than one diagram type per project; you can, however, easily circumvent this by putting all your diagrams into one and cut it up with graphics app later,
This is my clear recommendation; you just save so much time when creating diagrams compared to Dia!
POPP/POI (Plain Old PowerPoint/Impress) - use your favourite office's graphics app! Dumb to the bones when it comes to what's allowed, but at least the connections flow nicely and aligning objects works like a charm!
Edit: 1/7/2013: The drawing component in Google Docs supports snapping and drawing. Still no "real UML tool", but works good enough and is easily shareable.
Online tools such as gliffy.com - mostly nice, but no good for any serious work ;-)
yEd - I just gave it a short try, but it seems as well suited as Visual Paradigm. Give it a try and see for yourself!
Red Koda - Was recommended on StackExchange in an article asking for UML learning resources; also interesting in a broader sense!

Nota bene: You will find shortcomings (unsupported features, wrong layout etc.) in almost any UML tool you'll use. Thus, IMHO the drawing apps supporting UML shapes or snapping are still the most useful.
Answer: There's yuml which is pretty cool as it allows you to create UML diagrams online, with no tools and so easily embeddable in blogs, wikis, emails, etc.
Answer: I use ArgoUML because it's free. and in polytechnic they introduce, teach and make us use IBM Rational Software Modeler (RSM) school license.
In the case of simplicity, IMHO I think all UML modeling software are much complicated due to the requirements of UML standards and its various diagrams.
Answer: I use Umlet. What I like with this software is that it's a drawing tool only. It doesn't force you to create and maintain a model, and it doesn't try to generate/parse code. Unlike a lot of UML tool I tried, I've always been able to draw the diagram I had in mind (the drawing primitives are quite complete, and they are extensible by code).
It works well with my other tools : the text based format is ok for working with my VCS, and the png/svg can be generated using a command line (I use it to automate the build of my doc).
Answer: Jude Community is my first choice. Even though they're Astah now, you can still find jude community 5.2 over the web. If you never used I'd give it a try.
PS: I personally didn't like StartUML.
EDIT:

Although they share some very common features and even though Jude is not developed anymore, I prefer Jude mostly because of its usability. I've used Jude some years ago for studying and have to work with starUML + VS UML Tool today (company requirement) so I've been an avid user of both tools. I find Jude way more friendly to use. That's why I said: I personally didn't like StartUML. Features compared I don't know how Jude would go, since we use starUML for documentation purposes only and Jude is discontinued. Regards.

Regards.
Comment: Your post would be more useful if you stated your reasons to prefer Jude over StartUML.
Comment: @Vitor Granted. Although they share some very common features and even though Jude is not developed anymore, I prefer Jude mostly because of its usability. I've used Jude some years ago for studying and have to work with starUML + VS UML Tool today (company requirement) so I've been an avid user of both tools. I find Jude way more friendly to use. That's why I said: _I **personally** didn't like StartUML_. Features compared I don't know how Jude would go, since we use starUML for documentation purposes only and Jude is discontinued. Regards.
Comment: Now it's informative. +1 :)
Comment: +1. [Jude Community](http://jude.change-vision.com/jude-web/product/community.html) is also known as [Astah](http://astah.change-vision.com/en/) nowadays. Another great program, multi-platform as well.
Answer: I am surprised that BOUML is not yet mentioned. I recommend BOUML.
It is full featured, open source and regularly maintained.
I also request the readers to visit BOUML Project status - https://stackoverflow.com/questions/3721008/bouml-project-status.
Comment: BOUML is maintained, but feature development has stopped due to the author's dispute with wikipedia
Answer: Personally, I like StarUML.
Very full featured and open-source!
From the website:

StarUML is an open source project to
  develop fast, flexible, extensible,
  featureful, and freely-available
  UML/MDA platform running on Win32
  platform. The goal of the StarUML
  project is to build a software
  modeling tool and also platform that
  is a compelling replacement of
  commercial UML tools such as Rational
  Rose, Together and so on.
Answer: I use yEd when the idea/app is in its infancy and migrate up to ArgoUML when it needs more bells and whistles. Liked Visio, to some degree, but not enough to buy
Answer: I've used many of the really expensive ones and hated them all. I even resorted to using drawing tools in a number of cases, but that was very limiting and leaves you without many of the benefits of having a UML tool to begin with. Anyways, at my latest company they were using one I never tried, Visual Paradigm. I have to say that it is by far the best I've used. It still has its issues but it is about the only one that I actually like to use. Visual Paradigm does pretty much everything the really expensive tools do but at a miniscule fraction of the cost.
They have a free community edition, that I know is available for non-commercial use. I'm not sure if they limited functionality in any other way. If I recall correctly, you can buy a commercial edition for around $100 bucks. So if you need it for commercial purposes and your employer won't spring for that then I'd really be looking for a new job really quickly.
The only thing we haven't looked into yet is multi-developer support, which all the expensive tools are able to do. But from the web-site it seems like they support it.
Answer: I use a licensed version of Visual Paradigm at work.  There is a free community edition that should be able to handle most of your basic UML needs.
Answer: Visual Paradigm Plus for UML is what I use on my mac. It's free and pretty easy to use.

Best free UML authoring tool


Increasing height by back/neck straightening exercises?

Question: Is it possible that an adult could add height (or rather appear taller) by posture correcting exercises? Besides Wall Angel exercise and stretching exercises, what would help? How much could this affect the height?
Answer: From what I understand you cannot naturally increase your height as an adult. However you might be able to appear taller with correct posture; doing exercises back and core exercises will help with this.
Some recommendations: Deadlift, Squat, Russian twist, Reverse flyes.
Remember to do those exercises with proper technique otherwise it maybe detrimental in the long run.


will allah fgorgive me and make my mother`s heart?

Question: Assalam o Alaikum
I`m here to tell my sins and ...... the problems I'm facing. I was in a haram relationship and my mom caught me last night talking to him this was my first time ... I was never like this I was a sweet, innocent girl. I want to repent of my sins. So my question is will my mother and Allah ever forgive me
Answer: As salaamu alaikum never tell your sins publicly.
On the authority of Anas (may Allah be pleased with him), who said:
I heard prophet muhammadﷺ say: Allah the Almighty said:
"O son of Adam, as long as you call upon Me and put your hope in Me, I have forgiven you for what you have done and I do not mind. O son of Adam, if your sins were to reach the clouds of the sky and then you seek My forgiveness, I would forgive you. O son of Adam, if you were to come to Me with sins that are close to filling the earth and then you would meet Me without ascribing any partners with Me, I would certainly bring you forgiveness close to filling it." [Al-Tirmidhi] (Grade: Hasan)
Say, 0 Prophet, that Allah says, 1 "O My servants who have exceeded the limits against their souls! Do not lose hope in Allah's mercy, for Allah certainly forgives all sins. He is indeed the All-Forgiving, Most Merciful.
Az-Zumar (39) : 53
Show kindness and seek forgiveness from your mother, who you may have disappointed. Request her forgiveness gently and reassure her that you have also asked for forgiveness from Allah. Remember, you only have one mother, so treat her with respect. Listen to her without arguing if she yells, and continue to be kind even if she does not forgive you immediately. Pray for her forgiveness and strive to always be good to her.


Set Battery Saver Always On

Question: On Android 6, the battery saver automatically turns on when the battery is low and off after charging. I would like it remain on all constantly without having to manually turn it back on. How can I do this?
Answer: If you're using an unrooted Android, you can consider using BatterySaver Lollipop Shortcut with Tasker (userguide), AutoShortcut and Nova Launcher.
Fire up Tasker and follow these instructions:

Profile: Event → System → Intent Received:

Action: android.os.action.POWER_SAVE_MODE_CHANGING
Priority: Highest

Task: (Actions):

Task → If → Condition: %mode eq false
Task → Wait → Seconds: 1
Optional: Alert → Flash → Enabling Battery Saver
Plugin → AutoShortcut:

Configuration: Activities (next to Nova launcher's icon) → tap Battery Saver entry and select the entry listing the activity com.fd.batterysaver.ToggleActivity

Profile: State → Power → Power:

Source: Any
tick Invert

Task: (Actions):

Optional: Alert → Flash → Enabling Battery Saver
Plugin → AutoShortcut →

Configuration: Activities (next to Nova launcher's icon) → tap Battery Saver entry and select the entry listing the activity com.fd.batterysaver.ToggleActivity

You can use QuickShortcutMaker in place of Nova Launcher. However, I noticed that when Nova Launcher was not set as default launcher in my stock Android 6.0.1, although Settings app reported Battery Saver mode being enabled, I didn't get to see red shade in status bar and navigation bar. You may have to do some experiments here.
If you've root access, instead of using a shortcut to that battery saver app, you can setup the action Code → Run Shell and enter the following in the command field:
settings put global low_power 1

Make sure to tick Use Root in that action. (I came to know about this trick from this answer of Paweł Zmarzły.)
Whether the Android is rooted or not, the pro version of Battery Saver Lollipop claims to provide support for automation. It might be helpful considering that the free version causes few glitches during toggle. I cannot confirm though.

Information about said Tasker profiles
Alright, so the first profile would make sure that as long as the device is not plugged in to any power source, whenever you would disable Batter Saver mode, whether accidentally or deliberately, the mode would automatically be turned on!
The second profile would make sure that Battery Saver mode gets turned on whenever the device is unplugged from  a power source.
Comment: The tasks are flexible. Use them as you wish.

Set Battery Saver Always On


Are SQL Server statistics stored in database or buffer pool?

Question: Just wondering are statistics kept in the database but not in the memory? If I backup/restore the database from a prod server to a development server, would it keep the same statistics so that execution plans won't be much different if I do it on the development server?
Answer: The buffer pool is a cache of the database. There is never an 'or', things that are in the buffer pool are also in the database, always. And anything read from the database must be, even temporarily, present in the buffer pool.
As for the question: statistics are in the database so a backup/restore will preserve the statistics.
Note though that preserving statistics is not a guarantee to plan reproducibility. Other factors influence plan generation, like number of CPUs and amount of RAM.
Comment: ok, make sense. thanks for the clarification!
Answer: Addition to what Remus has mentioned, I would suggest you read --

SQL Server Statistics Questions We Were Too Shy to Ask
Aarons answer to - Where are Statistics physically stored in SQL Server?
UNDERSTANDING SQL SERVER STATISTICS
Answer: As Remus said, table statistics are stored in the database similar to other objects like tables and indexes. They play a big role in selecting the execution plan, but there are other factors.
That being said, SQL Server knows another type of statistics, statistics that give us information about recent behavior.
For example the DMVs sys.dm_db_index_usage_stats and sys.dm_db_index_operational_stats return statistics about how the index is used.
Those statistics are stored in memory only. The get lost on server restart and do not travel with the backup.
However, the optimizer does not use these to generate a plan either.
Comment: Yeah, this is why I had the question/confusion in the first place. I was thinking statistics are similar to DMVs. They could be gone when server restarts or db restored to another server.  Thanks for making the point.


2004 Dodge Ram 1500 5.7 Liter Hemi 4x4 Automatic cranks, but won't start

Question: I was driving 2004 Dodge Ram  1500 5.7 Liter Hemi 4x4 Automatic and it just died on me as I was going down the road,  before I could get it parked,  so I put it into neutral and I tried to start it back up and it backfired...I got it towed, now when I  go to try and start it it will crank over but it will not fire up and still continues to backfire. Here is a list of things that I have done to it so far...The fuel pump works, it's got fuel pressure to the engine, all of the fuses and the relays are good, it is getting spark/fire... Here is what I found that was working properly... It has low oil pressure and hardly any compression at all whatsoever...I'm completely stumped,I just don't know what to do.... If you could help me in any way whatsoever I would really appreciate it...Could it be clogged up catalytic converters, or does it need a  new oil pump? Or could it be the timing, and if so is it worth fixing,if so how much is it going to cost? Would I be better off if I was to just replace the whole engine? Please help me.....
Comment: Welcome to Motor Vehicle Maintenance & Repair! First question I'd ask you is, have you read the codes? I'm wondering if a cam or crank sensor has gone out? This would usually pop a code and you could experience symptoms like you're talking about (except the low/no compression). As far as the compression is concerned, how did you test this?
Comment: When you say "low compression" what exactly do you mean?
Answer: I don't think we have enough information to isolate this problem
As Paulster2 mentioned, any codes might help. Can you get us the codes? Loss of a sensor or the Engine Control Unit, ECU, the engine computer could cause it to die and just pop. It could be a crank sensor, it could be the ECU. It does not explain the low oil pressure. However, some engines will not start with low oil pressure, but usually it does this by disabling the spark and you say it has good spark and/or fuel. Most of these sensor or ECU problems will show in the OBD2 codes. A low battery will also cause ECU or injector problems. Is the battery good?
An engine needs three things to at least pop.

Compression
Ignition
Fuel

You already said you had hardly any compression. How do you know this, is it the sound of the starter or did you check compression? Is the compression low across all the cylinders or just some? Could you give us the compression values?
It could be a timing chain. This often causes drop in compression because of the altered cam timing. There should be a timing mark on the crankshaft. The number 1 cylinder should have both the valves completely closed on one of two revolutions. If they aren't you know the timing chain has slipped. If the valves aren't moving, then you have probably lost a timing chain, cam gear or similar.
A engine with clogged catalytic converter(s) should still have compression.
You don't say how many miles are on the engine. Eventually metal fatigue catches up with engines. It could be a catastrophic failure of the crankshaft, camshaft, flywheel, snapped oil pump shaft, etc. Check for any unusual holes in the engine, dents in the block or covers, or metallic noises when cranking the engine. Look for any unexpected oil leaks or pools of oil.


What is the smell after quartz is rubbed together?

Question: When I rub quartz together, it glows due to triboluminescence but it also creates a burnt smell. What causes that smell?
Comment: It smells like burnt hair to me.
Answer: A good question that has long been asked, as shown by a 1900 Nature letter  inquiring about The Smell emitted by Quartz when Rubbed.
Rubbed quartz can produce more than one smell, so this is an extension of the original answer in light of comments from others and new information from the original poster that the smell is not specific to triboluminescent quartz.
As noted in the answer to Milky quartz stones give off odour after being struck - persistent tribo-odorosity(?), an odour from rubbed quartz could be due to the release of hydrogen sulfide (H2S) trapped in tiny inclusions in the quartz.  When you rub together two pieces of quartz, you break open some of the microscopic inclusions and release the hydrogen sulphide.
As @akhmeteli's answer citing an 1858 Scientific American article shows, the basics of this have long been known, but the details have been more recently confirmed. The presence of hydrogen sulfide in inclusions has been noted by studies such as this 1984 article on "Gas chromatographic analysis of volatiles in fluid and gas inclusions" which reported that hydrogen sulfide was the dominant gas released when a sample of "fetid quartz" was crushed. Another study the same year reported on Characterization of H2S bearing fluid inclusions in quartz, fluorite, and calcite.
As correctly noted in @user346760's answer and @kryan's comment, the question describes the smell as "burnt", not the usual "rotten egg" description of low concentrations of hydrogen sulfide.
The 1858 cited Scientific American article mentions a rotten egg smell, but Thomas Wedgwood described it simply as "strong" in a 1791 Royal Society report,  Delius referred to it as "sulphurous" in 1748, and other common descriptions include "metallic".
If the smell is "burnt" and not "rotten eggs", then also as noted by @user346760, the triboluminscence could be igniting contents of the inclusions.  Hydrogen sulfide is a volatile flammable gas which burns to sulphur dioxide and water.  Sulphur dioxide is a pungent gas which causes the smell of burnt matches. It is also possible that any elemental sulphur inclusions could be ignited producing sulphur dioxide, but such inclusions seem to be less frequently mentioned than H$_2$S inclusions in the scientific literature.
As noted by @KRyan, it is also possible other compounds could contribute to the smell. The most common other volatile components of quartz inclusions are odorless, i.e. water, methane, and carbon dioxide, but the surface of a quartz pebble could absorb organic materials from its environment.
Although I believe all the above discussion is accurate, it may be moot in light of new helpful information from the original poster that the same smell is produced when two random rocks are rubbed together. Both the original poster and myself have confirmed that the same smell is produced by rubbing together a variety of rocks and even from simply vigorously rubbing two glass jars together, but not when metal or plastics are rubbed together. Since the smell can be produced by rubbing glass together, this is clearly not a sulphur related smell due to inclusions.  Personally, the smell makes me think of electric motors, e.g. ozone. Others with more sensitive and experienced noses may be able to identify it better.
That ozone may be significant component of the smell is certainly plausible.  Ozone is produced when granite blocks are crushed or when grinding granite, basalt, schist, rhyolite, or gneiss. If ozone is being produced, then as noted in the answer to Where does the smell of electrostatic charge come from?, we can also expect nitrogen oxides. This is similar (but on a much smaller scale) to gas production by of lightning.
The smell of ozone has been variously described as "chlorine, metal, burnt wire"
Nitrogen oxide is a "sharp sweet-smelling gas" and nitrogen dioxide has "a strong, harsh odor".
It seems likely that the general smell of rubbing rocks (or glass) together is mostly due to some combination of these.
Comment: It would be interesting for this answer to investigate if compounds other than H₂S might be in play here, since the described "burnt" smell is rather different from the distinctive smell of hydrogen sulfide.
Comment: I don't know the smell of rotten eggs so I wasn't sure. I also took 2 random rocks I found on the ground and rubbed it together it also created the same smell. it doesn't smell like something rotten but a smell that would come from a burning thing.
Comment: Thanks! Your second observation is very interesting and I have reproduced your observations myself. Rubbing quartz together produces light and and a "burnt" smell that made me think ozone more than H2S or SO4.  Following your lead, I also rubbed random pebbles together and notice no light but the same odour. Even rubbing two glass jars together creates the same smell, which probably rules out inclusions. There is no smell when I rub metals or plastics together. My nose is not infallible, so others may want to try their own experiments. I'll revise my answer again after some more research.
Comment: I also did what you did and got the same results. I tested a lot of rocks like gneiss, processed and unprocessed marble, flint, white granite, bricks and concrete. All of them created the smell. glass also created it but metal or plastic didn't. its probably not because of the contents of the rock.
Comment: I upvoted already. In the case of "fetid fluorite", the smell of the crushed mineral is due to elemental fluorine, as per a link in this nice answer by @uhoh: https://chemistry.stackexchange.com/a/104258/79678.
Answer: For what it's worth:

When certain varieties of quartz and limestone are strongly rubbed,
they give off the odor of rotton eggs. This peculiar smell is
occasioned by the evolution of sulphureted hydrogen

I guess "sulphureted hydrogen" is the same as hydrogen sulfide.
Comment: @RussellMcMahon : But it is a quote.
Comment: Minor nit-pick: "Evolution" suggests it is produced at te time and did not exist before then. Something like "release" would be better.
Answer: It is the smell of sulfur dioxide gas ($\text{SO}_2$) being released, as the elemental sulfur inclusions are getting oxidized in the air due to the heat caused by friction.
It is NOT, as all the other replies assume, hydrogen sulfide. A quick check of the properties of those two gases confirms that it could not have been $\text{H}_2\text{S}_{}$, as it has the pungent and foul odor of rotten eggs. Meanwhile, sulfur dioxide has the smell of just-struck match, or burnt fireworks.
Comment: The question outright mentions "burnt smell". That is completely different than the smell of $H_{2}S$, which is that of rotten eggs. Those smells are not even in the same chapter, not in the same book, they aren't even the same literature genre. It would be nice if people actually read and acknowledged such important details placed in questions, instead of quickly jumping to write vague guess-answers based on loose assumptions, in order to game the system and get dozens of exposure votes.
Comment: On the other hand, this answer has no citation related to quartz, only citations to support that the compounds smell a certain way.
Comment: @DavidBailey Hey thanks a lot for your response and acknowledgement, much respect for updating your answer!
Comment: You make a valid point and I will edit my answer. I did wonder about the description of the smell as "burnt" instead of "rotten eggs", but based on the long history in the literature describing the smell of rubbed quartz, I assumed it was a casual and perhaps not precise description. It would certainly be interesting for the original poster to clarify if "burnt" was a precise description inconsistent with "rotten eggs". Your suggestion that that triboluminescence could be igniting whatever is in the inclusions is also a good one.


How do I install Ubuntu on one hard drive, and use a second hard drive for storage?

Question: I have a pc with 2 500gb HDDs. I am currently running Windows 8.1 but am migrating to Ubuntu. I want to know how to setup the partitions during Installation, so that I would have Ubuntu installed on the first Hard Drive, with the whole system being there, while still being able to store random files in the second Hard Drive?
Thanks.
Answer: You don't need to do anything special.  Just start Ubuntu installation and select the option to erase everything and install Ubuntu.  Your first hard disk will be selected automatically.  The other one won't be used at all for the installation.  It will keep all the partitions and the data it holds.
Answer: Like Stormlord said, when you do the installation it won't touch the second hard drive unless you explicitly allow it to. So make sure you don't accidentally approve writing over it! One thing of note, though (would have just commented on his answer but I don't have a high enough level yet), is that Linux and Windows tend towards different file systems. You should have an option during the installation (or immediately afterwards - Ubuntu has a nice program for partitions and file systems which you can use) to format the second hard drive into something more usable. Don't forget to back things up first!
Answer:
If you want just using Ubuntu and get rid of Windows, first go to the windows and then shutdown it. it may cause mount problem if you are using Windows 8 or higher without this.
Ubuntu dosn't change anything on your 2nd Hard disk until you want it. if it's your first time, Just for safety Remove your 2nd Hard disk and then install it. after install you can set it up.


Long cat5e cable

Question: Cabling contractors put wires according to plan however I think the distance is longer than 100 meters as I cannot see devices connected on another end. When the cat5e cable is longer than 100 meters, will I still see lights on the port that is connected to ? IS there any way of test 'signal' strength on the end of the cable ?
Comment: every contractor I worked with handed over test protocols of a network cable tester when they were done. You should require yours to do the same.
Comment: I can tell you, from experience, cables over 100 m _can_ work. They may not work _well_ though. Any decent cable tester should be able to verify the cable, measure length, test attenuation, cross-talk, etc.
Answer: You can use (good) Cat5e cables with length > 100 meters. They may not provide with full quality and speed, but may work.
You can test the cable and measure the length with some (pro) cable testers like Fluke Networks CableIQ. Some switches can do this, too. The feature is called TDR (Time-Domain-Reflectometry).
Lights should be on as soon as the cable makes the right contact between to active devices. Lights on do not implicitly mean there is a data connection.
Regards
Comment: Is the issue only about signal strength? Originally there was a requirement that packets had to be long enough to reach from one end of the cable to the other. Does that still apply? Or was that only for half-duplex links?
Answer: Lights usually mean you have an electrical connection, they do not necessarily mean that the signal is getting through perfectly. You can have lights but still see such high packet-loss that communication is impossible. Imagine a phone call where you can reach the person you call, but their voice is scrambled with static.
The 100 meter length is not a hard limit - It is a recommendation based on expected signal loss over long runs. I have seen 120 meter cables work ok (with some packet loss), and I have also seen 80m cables fail completely. It depends on a lot of factors.
The issue is signal degradation leading to packet loss, not latency. Signal degredation can have many causes, such as EMI (electro-magnetic interference), wire resistance, cable quality, and shielding just to name a few. If you need a longer run, use high-quality shielded cable, and high-quality ends. Make sure that your cable runs do not go near any obvious sources of EMI, such as fluorescent lights, transformers, electric motors, or high-voltage wiring. Don't run them in the same conduit as power cables. Avoid crimps in the middle of the wire.
As other posters have suggested, you need a decent cable tester to verify the quality of the connection. If you cannot get it to work, you can either shorten the cable, or insert a switch or signal booster in the middle of the run.
Answer: You would probably need a professional meter that would measure all sorts parameters, as the strength of the signal is just one of them, and by all means not the most important. Noise and cross talk are examples of parameters I would personally look at. You can rent a professional Fluke meter for a reasonable price.

Long cat5e cable


Apart from visually inspecting the graph, is there a way of calculating when a trend line will cross another series?

Question: Is there a way to find the point of intersection of a trend line and another series on a chart in Excel? The other series may be another trend line or just the interpolated line drawn by Excel.
Comment: I do not understand what "trendline will cross another series" means.
Comment: A Trendline can be added to a series - basically a prediction of the data extended beyond the known data. I want to know exactly when the trendline will intersect another series or even another trendline.
Answer: Get equations for both trendlines. Put them in the form y = f(x). Set them equal to each other. Solve for x.
Comment: Is the equation available as an object then? - Excel calculates the trendline.
Comment: For my current use - its just easier to look at the chart! -
Answer: You would have to compute the slope of the line vs. every data point that you are looking at.  Any data point that has a higher X axis value than the corresponding computed point on the slope would exceed it.  You could optimize the search for this point by first computing your slope, then sorting your data points from the highest X axis value, in decending order, then interating through the list of data points and computing the corresponding X axis value of the trendline slope vs. the examined X axis value of the data point.