How do you unlock new costumes?

Question: The PC release of The Amazing Spider-Man seems to have changed the way you unlock special costumes. The tooltip says you unlock them by completing story and side missions, but which story and side missions unlock which costumes?
Answer: I know how to unlock some suits:

Black Suit: progress through the half of the game.
Future foundation: complete all crimes and green side missions.
Negative Zone: complete all xtreme challenges.
Scarlet Spider:collect all 700 comics.
Answer: I unlocked the cross species suit after completing the game.
I unlocked the Big Time suit after completing the chapter 02 "Escape Impossible".
I unlocked the classic black suit after making some progress in the game.
I also unlocked the Future Foundation suit by completing the green icon missions in which you have to collect secret information and take some photographs for the article in Oscorp's secret lab. Those icons are green-coloured and you will find them on your cell phone but be careful over there because security in these labs are not a laughing thing.
There are also some sewers type of green icon which is poisoning the drinking water of the people. you have to fight with some of the infected enemies, Rhino and the Chameleon.
Answer: I unlocked:

new black suit - complete game 100%.
Big Time - complete every side boss fight.
Future Foundation - do every Oscorp secret lap.
Classic Black suit - 50% game completion.
Negetive Zone - done all the Xtreme Challenges.
Scarlet Spider(2012) - collect 700 piece of the comic book.
Classic Suit - download any DLC.
Comment: By classic suit, he means the Sam Raimi interpretation of Spider-Man.

How do you unlock new costumes?


How to convert subtitles from IDX/SUB to SRT?

Question: My WD HD Media Player supports only SRTs. How to convert subtitles from IDX/SUB to SRT?
Answer: here's a link at AfterDawn.com
link text
although as NiklasS says, you'll probably have more success getting an SRT file from one of the subtitle sites. Converting can cause all sorts of issues with syncing the sound correctly, so it will save you more time getting the correct SRT file forthe film you've gotten hold of.


Are there situations when regarding isomorphic objects as identical leads to mistakes?

Question: If there are not, then would it be easier to say that 2 objects are identical as ordered fields as opposed to being isomorphic as ordered fields? Or is the word isomorphism used to emphasise the fact that the objects are different as sets?
Comment: Yes, it is a Bad Idea to identify a finite-dimensional vector space with its dual.
Comment: There is a natural map from a Hilbert space to its dual, but it's antilinear.
Comment: Riesz's representation theorem is a great thing indeed, but it's the source of 90% of misconceptions physicists have about mathematical matters. :)
Comment: The whole idea of grupoids, stacks and $(\infty, 1)$-stacks is to consider higher automorphisms to get a universal object. So automorphisms are important...
Comment: Is it safe to identify a Hilbert space with its dual?
Answer: Inside of the complex numbers there are lots of examples of distinct fields which are isomorphic.  For instance, there are three subfields of the form ${\mathbf Q}(\alpha)$ where $\alpha^3 = 2$: take for $\alpha$ any of the three complex cube roots of 2 and you get a different subfield.  What are the consequences of treating them as literally equal? You can't make any sense of Galois theory if you do that!  Similarly, all $p$-Sylow subgroups of a finite group are isomorphic (since conjugate subgroups are isomorphic groups), but it would kind of destroy a lot of the content of the Sylow theorems by
trying to say the $p$-Sylow subgroups are identical.
More generally, anytime you have isomorphic but unequal objects inside a larger object, it can lead to confusion if not outright incomprehensibility if you try to regard them all as identical.  (There was a paper by Chevalley about unit groups in number fields where he made a genuine error by an abuse of the "square root" notation and I think one might be able to express the mistake in the form of an isomorphism being confused with an equality, but I'd have to look at the paper again to be sure about this.)
The word isomorphism does not emphasize that two objects are different; any group or vector space admits an isomorphism with itself using the identity map.  The word emphasizes that in a structural way the two objects look like each other even though they are not literally the same. Never say two objects are identical if they are not actually identical.  Having said that, I must admit that in mathematics one meets phrases like "since $X$ and $Y$ are isomorphic we can identify $X$ with $Y$" and then $X$ is replaced with $Y$. The usefulness of doing this depends on the application you have in mind.  Note, however, that replacing $X$ with $Y$ is not saying that $X$ and $Y$ are the same thing.
This question sounds like it is being asked by someone who hasn't had a lot of experience with isomorphisms and is trying to get a feel for what it means. In a year or two, after seeing more appearances of the concept and its uses, you'll get a better feel for it, but for now do not think the word isomorphism is a synonym for identical.
Comment: A very simple example for the danger of identifying isomorphic subobjects being "$ℤ = 2ℤ$, so $ℤ/2ℤ = ℤ/ℤ = 0$."
Answer: A slightly more general answer: if the objects in question have no non-trivial automorphisms (i.e. non-identity isomorphisms from itself to itself) then no danger will come from treating isomorphisms as the identity.  (E.g. the real numbers has no automorphisms as a field, and so
any two copies of the real numbers can be unambiguously identified as fields.)
But if $X$ and $Y$ are isomorphic but Aut($Y$) (or, equivalently, Aut($X$)) is non-trivial,
then the identification of $X$ and $Y$ is not uniquely determined (because you could always
postcompose with a non-trivial automorphism of $Y$, or precompose with a non-trivial automorphism of $X$, to obtain a different identification), and hence in this case there is
not an unambiguous identification.  (A typical example of this is the isomorphism between
a finite-dim'l vector space and its dual; there is not a uniquely determined such isomorphism,
and so one should not identify the two, although they are isomorphic.)
In some contexts, although there is not a uniquely determined isomorphism, there is a canonical choice, e.g. the identification of a finite dimensional space with its double dual.
Such isomorphisms are normally described in the language of natural isomorphisms between functors (e.g. the identify functor and the double duality functor are naturally isomorphic as functors from the category of finite-dimensional vector spaces --- over some given field --- to itself.)  Often it is safe to identify objects that are identified by a natural isomorphism in some category theoretic framework that is suitable for the problem at hand.
But even in these cases, sometimes one has to actually know an explicit description of the natural isomorphism (e.g. because you might need to make a computation involving both objects, which will of course require you to know how they have been identified).  In particular, when identifying two objects via some natural isomorphism, it is good form to explicitly describe the natural isomorphism at some point, unless the natural isomorphism
in question is utterly conventional (e.g. the double duality isomorphism for a finite-dimensional vector space; and even then, one might write something like "we identify $V$
and $V^{\vee\vee}$ via the usual double duality isomorphism").
Added: This answer is very similar to that of Charles Staats, which was posted while I was writing.
Comment: Even canonical, natural isomorphisms can lead to problems. For example, $A\times B\cong B\times A$ is not the identity for $A=B$.
Comment: It is possible for non-isomorphic rigid structures to become isomorphic by a unique isomorphism in a forcing extension. In this case, your view would have us say that non-identical objects become identical when enlarging the universe. I think this way of thinking will be very confusing.
Answer: If $p$ is a prime, then the multiplicative group modulo $p$ is isomorphic to the additive group modulo $p-1$, which is to say it's a cyclic group of order $p-1$. It's trivial to find a generator of the additive group, but a fair bit of work to find one for the multiplicative. Related to this, it's easy to solve $ax=c$ in the additive group, using the Euclidean algorithm, but the difficulty of solving the corresponding equation $a^x=c$ in the multiplicative group - the discrete logarithm problem - is the basis for some cryptography systems.
Answer: First of all, as other posters have said, you never claim that two objects are identical unless they actually are equal as sets. However, there are times when it is appropriate to "identify" two isomorphic objects. (This is really just an issue of language).
Exactly when such identification is appropriate is a matter of some subtlety. If there are many different isomorphisms between two objects, and no natural way of selecting one of them, then it is generally a bad idea to identify them (although sometimes this is done implicitly, as when we talk about the field with $q$ elements). This applies in particular to a vector space and its dual. Generally speaking, if you have a unique isomorphism, or even a "natural isomorphism" (this can be made precise using category theory if desired), you are safe identifying them; however, this might conceivably produce problems if they are distinct subsets of some larger set you are interested in.
To further complicate the matter, in many, or even most, cases, we don't actually care about the identity of the set in question, only its (external or internal) structure. Thus, for instance, any questions about the real numbers that cannot be answered using the fact that they form a complete ordered field (e.g., is $1 \in 2$?) are not really questions about the real numbers at all.  This produces interesting issues in philosophy of mathematics; see this website for a discussion. Things get even more interesting when you start defining objects (e.g., tensor products, limits, colimits) using universal properties; in this case, the objects are only defined up to natural isomorphism, so there is no single construction that "is" the desired object.
There are also times when either of two identifications may be appropriate, but not both at once.  For instance, $\mathbb{Z}$ is naturally isomorphic to a fixed subset of the p-adic numbers $\mathbb{Z}_p$, and to a fixed subset of the real numbers $\mathbb{R}$, but no one would ever write $\mathbb{Z} \subset \mathbb{Z}_p \cap \mathbb{R}$ (except to discuss when it is not appropriate to make identifications).
In short, it's a subtle question, and the answer is based more on experience than on any codified set of principles.
Answer: One paper I looked at needed to check every subgroup of $S_n$ for small $n$. There's a lot of symmetry there, which the author was trying to exploit. Although $S_4$ contains three subgroups isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$, the author only checked one of them; without loss of generality (``by symmetry'') that was sufficient. Problem is, one of those subgroups is normal and the other two are not, and that could have made a difference.
The punchline is that the additional case checked out fine---the logical error didn't snowball.
Comment: Sorry for the anal heckling ten years after the fact, but $S_4$ actually has got three nonnormal subgroups isomorphic to $C_2 \times C_2$ (one for each partition of $\{1,2,3,4\}$ in two pairs).
Comment: So I guess the author of this article did not explicitely use the relation of the subgroups to $A_n$ ?
Answer: Expanding on Andy Putman's comment, beginners in linear algebra often think of all finite-dimensional vector spaces (over $\mathbb{R}$, say) as the same, in particular as $\mathbb{R}^n$ with the standard basis.  This can lead to mistakes in any kind of computation, say in multivariable calculus, where one has to compute simultaneously with a row vector and a column vector (e.g. an element of a vector space together with an element of its dual).  The problem is that matrices act differently on row and column vectors, hence if you end up changing basis you need to deal with row and column vectors in opposite ways.
Comment: For some unfathomable reason, mathematicians just can't spell my name correctly .
Comment: They can't pass the Putman test!
Answer: The answers are great so far, especially Emerton's. I would like just to add another example which I have encountered. It is the gluing construction of the structure sheaf of an affine scheme. I think the one given by "locally constant elements in the stalks" is better, but anway:
If $A$ is a ring and $X=Spec(A)$ its spectrum, then it can be shown that the assignment $\mathcal{O}_X(D(f)) := A_f$ is a sheaf on the basic open subsets. Now every sheaf on a basis of a topological space (which is closed under intersections) extends uniquely to a sheaf on the whole topological space. But wait, if $D(f)=D(g)$, it is not necessarily true that $A_f = A_g$! But we can fix that by observing that there is a unique $A$-algebra homomorphism $A_f \to A_g$ if $D(g) \subseteq D(f)$, and that this is actually an isomorphism if $D(f) = D(g)$. Thus in practice it really does not matter which $g$ with $D(f)=D(g)$ is chosen.
There are some ways of formalizing this. For example you could define $Q$ to be a big ring which contains all the $A_f$ as subrings, but the identifications become identities. Define a partial order on $A$ by $f \leq g \Leftrightarrow D(g) = D(f)$. Then $A_f$ is a direct system on this partial order and we can take $Q$ as its colimit.
Answer: In Group Theory there is a notion stronger than isomorphism, i.e. similarity. For two permutation groups to be "similar" there must exist both an isomorphism between the group and a compatible (commuting in the categorical sense) isomorphism between the sets being permuted.
The couple of these two isomorphisms is called a similarity. Two permutation groups can be isomorphic but not similar.
Similarity is useful to compare group constructions such as the wreath product, group presentations and classifications of subgroups of the symmetric groups.
In these situations, using only isomorphism would be a mistake and would lose information.
Comment: The first observation is the statement that one has also to consider the corresponding element in the Burnside ring encoding distinct (formal) permutation representations of finite groups, see http://en.wikipedia.org/wiki/Burnside_ring. It generalizes to the fact that two linear groups can be abstractly isomorphic but can correspond to non-isomorphic representations of the underlying abstract group.
Comment: @Roland Bacher: thanks to mention the Burnside ring. As you do, I believe this gives the proper context (and marks tables a good pedagogical tool) for similarity and isomorphism of permutation groups.
Answer: Spivak, Differential Geometry, Vol. 1, Chapter 3, page 16.
Background: $M$ is a manifold, $i$ is an embedding pf $M$ into a Euclidean space, and $T(M,i)$ is the tangent bundle on $M$, constructed assuming the existence of the immersion $i$, which is the way Spivak does carries out the construction at first. Then he notes that for a different choice $j$ of the immersion, we have a bundle equivalence between $T(M,i)$ and $T(M,j)$.
Then goes on(quoted verbatim):

... In other words, the dependence of $T(M,i)$ on $i$ is almost illusory; we could abbreviate $T(M,i)$ to $TM$, if we agreed that $TM$ really denotes an equivalence class of bundles, rather than one bundle. That is the sort of thing an algebraist might do, and it is undoubtedly ugly.

And then he constructs the tangent bundle in the way independent of the immersion, ie the more modern approach, and expounds its merits.
Answer: Consider $V \subset H$ both Hilbert spaces in the setting of evolution equations. It is useful to work with a Gelfand triple setting $V \subset H \equiv H^* \subset V^*$ where $H$ is identified with its dual. But then identifying $V$ with its dual is a mistake and leads to confusion.
Comment: Are you saying that if $V \subset H$, then there is a natural inclusion $H^* \subset V^*$? It seems to me that if $V \subset H$, there should be a natural surjection $H^* \to V^*$ given by restriction, $f \mapsto f|_V$ for linear functionals $f \in H^*$. It is surjective by Hahn-Banach... no?
Answer: Once I've been attending a functional analysis seminar where I've heard what seemed to be a classical pun in the field: Since all Hilbert spaces are isomorphic, all probability spaces of centered squared integrable random variables are the same, and thus probabilists should focus on, say, the uniform law on $[0,1]$ only.


IOS 8.0.2 Calendar on iPhone 5 not syncing with MS Exchange properly

Question: Since I upgraded my iPhone 5 to iOS 8.0.2 I have started having issues with the calendar syncing through Microsoft Exchange. The event description created on the iPhone do not appear on Outlook, only the subject and address are synchronised. The description of the event remains blank on Outlook.
Answer: Try Settings; Mail, Contacts, Calendars; Default Calendar.
Make certain that your Exchange server calendar is the default calendar.


Should I only store fixed-size data elements in a database?

Question: If the amount of data stored within a given field of a database is unknown, and could be very large, should I store it in an external file rather than within a field in the database?
Answer: Do you need transactional semantics (commit, rollback) for the data?  If so, then using the external file system greatly complicates life -- use the DBMS.  If you don't need transactional semantics, then the file system may make sense.
Answer: That's why BLOB or Memo field types were designed.  They are very good at storing variable length / large objects.
That being said, I would still store certain large objects using the file system instead of using a database.  If the objects were very large (multi-MB each) I would consider just placing them on the file system and storing a pointer (filename) in the database.
Answer: You should choose a database management system which has the capability to handle large data efficiently. The database system might store it within the database file or in an external file linked to from within the database. SQL Server 2008 can do both, transparently; not sure what other systems offer.
Answer: It depends on your task. There are pluses and minuses of storing large data on the FS instead of the DB. As for the size, you can limit it in the field definition of most databases.
Answer: I think it might depend on how you were going to use the data.  For instance, if I were storing images that would be used in a web site (Flickr or something like that), I would rather store the images as files and keep a reference in the database, rather than store image data in the database and have to reconstitute the image data each time via some HTTP handler.  On the other hand, collections of large documents that need to be searched may be more useful if the text were stored in the database.


Slow Lan connection ONLY between my PC and a MAC

Question: I'm trying to transfer some files over LAN from a Windows 10 PC to a Mac and it's incredibly slow around 2 MB/s.
I've tried the following with exactly the same files.

Transfer from my PC to another PC everything works great around 100 MB/s
Transfer from another PC to the MAC again around 100 MB/s
Unplug Ethernet cable from my pc and switch to Wifi still slow
Plug the Ethernet cable from the other PC into mine and still slow.
Disable Windows AV
Update Windows
Turn off Remote Differential Compression API Support
Use IP address instead of machine name

I'm completely out of ideas, I'm pretty sure it's my machine considering that the transfer speed from the others is OK.
It really looks like it has something to do with my machine, I just can't figure out what.
Extra info

Windows 10 Pro 1709 Build 16299.371
Alienware 17 R4 with a Killer E2500 Gigabit Ethernet Controller
It's not a direct connection to the PC we use a router
Files are copied with Windows Explorer
Answer: Looks like the entire problem was caused by the an app that came with the Ethernet Card called "Killer Control Center". In order to fix it you have to go to Settings and turn off Advanced Stream Detect


Difference of Gas Giant Planet and Ice Giant?

Question: What is the Difference of or the Structure of a GAS PLANET AND AN ICE GIANT Why they called ICE GIANTS like Neptune and Uranus. Do they have a similarity ?
Answer: Gas Giants consist primarily of the gases hydrogen and helium, with a rocky/iron core.
The interiors of Ice Giants consist primarily of what astronomers call 'Ices'. These are compounds like water, methane and ammonia, and are so-named presumably because they are solid at cold (-200 C) temperatures, unlike hydrogen and helium, which remain gaseous up until very close to Absolute Zero.
Hydrogen and helium are very light, and it only takes a small amount of heat energy to make them fast enough to escape the gravitational field of most planets. Jupiter and Saturn are the only planets heavy enough to retain a significant fraction of hydrogen and helium (though the atmospheres of ice giants Uranus and Neptune consist mainly of hydrogen and helium, it's mainly ices further down).
Comment: Still, ice giant is just a subclass of gas giants, isn't it? Shouldn't Jupiter and Saturn then get an own subclass name?
Comment: The concept that the gas giants have a rocky/iron core is suspect. The Juno spacecraft has shown that Jupiter does not have a differentiated rocky/iron core; it looks more like a diffuse core with lots of hydrogen and helium mixed with heavier elements. A recent study suggests the same is true for Saturn.


My new Leopard gecko turned grey and is cold to the touch

Question: so I just bought a leopard gecko (It's my first reptile) and last night I had turned off it's heat mat. I had read that leopard geckos could withstand temps of 70s (which is the coldest it usually gets in my house). But this morning when I went to go feed it, it was pale and cold to the touch. It's alive, and it caught two of the crickets I put in the tank, but I'm worried for it's health. It's a juvenile leopard gecko.
Comment: it is not clear to me what you are asking,https://www.thesprucepets.com/leopard-geckos-1236911 you can do a search for leopard gecko on top of this page and see if you find what you are looking for.
Answer: I do not know a lot about geckos, but my turtle is a poikilothermic animal too. They need the warmth to get going. If they have a low body temperature, they cannot move fast or with much power. So they need a place to warm up (for my turtle this is the sun-like light bulb on his island) to become fit, i.e. to hunt.
My turtle has his light shining at the daytime only, but the water has thermostat so it stays at the moderate temperature all the time.
A good way to let your animal choose its body temperature itself is to provide a temperature gradient over the animal's space. For example, provide the heat mat on one side of the terrarium and a "cold spot" on the opposite one. With such an arrangement, the gecko can choose the distance to the warmth, and thus the temperature, according to its needs.
Answer: If he/she is grey or white, that could be a sign of shedding. It's not worrying - but having lasted for so long, it's probably not good.
As for the heating, problem I am not sure about babies - but an adult, average gecko's temperature during the daytime should be 24 - 27 °C (75 - 80 °F). I don't know about night temperatures; I own a leopard gecko and she is fine with her cage being 26 °C (78 °F) at night. By the way, I count ferrinhight.
Best of luck to your gecko.


Combining two pdf

Question: I would appreciate some insights on this question I've had this past days.
I'll try to phrase the problem as best as I can. I'll try to make some sense of it by considering a sales context.
Consider some sales time horizon $t\in \left [ 0,\infty  \right ]$. The sales over time $X$ for some product given some condition $y$ ( day of the week for example ) follows a clearly defined pdf.
Let's imagine that the sales over time $X$ for the same product given a condition $z$ ( weather ) also follows a clearly defined pdf.
If we have $p_{X}(t|y)$ and $p_{X}(t|z)$ can we build a third pdf i.e. $p_{X}(t|y,z)$ using this two pdf. This combined pdf would give us the probability of purchase for some day of the week and some specific weather condition ?
For simplicity we can assume that $p_{X}(t|y)$ and $p_{X}(t|z)$ belong to the same family i.e. they are both normal or binomial distributions or whatever they may be but they have an analytical form.
I've tried to explore a solution with the Bayes Theorem but I don't seem to get anywhere.
Answer: There are different ways to construct the third pdf; it may not be unique.
Just need to make sure $\int{f_{X|Y,Z}(t|y,z)f_Z(z)dz}=f_{X|Y}(t|y)$ and $\int{f_{X|Y,Z}(t|y,z)f_Y(y)dy}=f_{X|Z}(t|z)$.
Comment: Thank you for your response. I could try to solve both of these integral equations in order to find the combined pdf.

Combining two pdf


"Yesterday I watched a film" "Yesterday, I saw a film.'

Question: Aren't they both considered correct? "Yesterday I watched a film" "Yesterday, I saw a film.'
Comment: Are you seeing movies in your free time. Is this sentence ryt as past tence
Answer: In this case, let me first paraphrase Swan's "Practical English Usage" (#368):

See is used when to talk about visual impressions coming to our eyes; it's not always deliberate, might be accidental; seeing could happen without you thinking about it, or even without you realising that you've done it
Look (at) suggests intention or concentration
Watch suggests intention or concentration while looking at something that is happening, changing, moving, developing.
While we say "watch TV", we usually use "see" with public performances of plays and films

That being said:
- you can watch TV, but you can't see TV (unless in the sense of, "I saw a nice TV in the store the other day" - in which case you can't really watch it).

you can both watch and see movies. In some cases, the watch version may imply more of a deliberate looking than the see version:

I saw 'Gone with the Wind' yesterday.
I watched 'Gone with the Wind' yesterday.

however, note that if you're talking about movies you have seen at some point in the past (ie., you have watched them, and now you know what they're about), then "I've seen 'Gone with the Wind'" would seem a bit more natural to me than "I've watched 'Gone with the Wind'".
finally, if you're talking about watching a film/movie N times, you can say both "I've seen 'Gone with the Wind' 5 times" and "I've watched 'Gone with the Wind' 5 times" (and the same goes for sentences of the type "I watched/saw X for the Nth time today"), and which version you choose here is up to you.
Comment: The corollary to your final point is that if you were to say *"I XXXX'ed 'Gone with the Wind' **for the third time** yesterday"*, you'd be more likely to use ***watched*** than ***saw*** (but both are perfectly valid, and it would be fatuous to suggest either is "more correct" than the other).
Comment: Yes, I was planning on adding that comment, but then forgot :) Thanks for adding it. Will update the answer accordingly.
Comment: Was the question _really_ about the comma (or lack thereof), or **see** vs. **watch**?
Comment: @PhilPerry Did you meant to post this comment on Jolene's answer? :)
Comment: No, though it was triggered by her answer. Your answer was excellent, but hers got me to wondering just what was being asked.
Comment: @PhilPerry Ahh, now I understand :) It's a fair question. I'm going to assume it was the see vs watch since the OP accepted my answer, but yes - it wasn't 100% clear what was being asked :)
Answer: Yes, they are both correct and in both cases the comma is optional. To watch something is to intentionally pay attention to it for some period of time. So certainly you "watch" a movie. To have seen something just means that the thing was within your visual field. "Seeing" something can be instantaneous and unintentional. You "see" the flash of a camera, but you don't "watch" it. In the case of "seeing" or "watching" a film, I don't think either of the words is preferred over the other.
Answer: I would say that it depends on context.  It comes back to the question of "intent" or "concentration".
So, I could say "I went to see a play".  Intent (and presumably contration) are implied.
"I watched a movie."  Could have been from the comfort of my own home, the use of the word watched implies intent and concentration.  Had I gone to the cinema, I could have easily said "I went to see a movie".


What is the reputation cap for a day?

Question: The 'Just jesting' cap says that you need the rep cap for a day. Is this value the same for all sites and what is the value? Is there a list of rep caps that one can access?
Answer: As explained in the faq:

You can earn a maximum of 200 reputation per day.

Any votes that would increase your reputation past 200 for a given day will have no effect; You are said to have capped out.
Comment: Just to be clear, the rep cap ONLY applies to rep from upvotes. Rep from acceptance of answers (both +2 and +15) and bounties does NOT get limited. Also, see "Mortarboard" badge
Comment: @bitmask It is relevant in that if you get +18 on your answers, +1 on a question, and 1 "accept" on an answer you posted - you hit +200 rep, but that doesn't hit the cap.
Comment: Thanks! I missed that.
Comment: @DVK: I didn't mention that be cause it is irrelevant for the question, as the OP is not concerned with the actual consequences of capping, but with the condition to be met (in order to get hatwarded).

What is the reputation cap for a day?


What does "barefoot notes" mean?

Question: Can someone please help me understand what "barefoot notes" means?
Comment: Please indicate where you've seen the term - preferably with a link - and what effort you've made to look it up. Please refer to [Where can I find answers to simple and basic questions?](http://english.stackexchange.com/help/on-topic)
Comment: The fact that there are a few Google hits for a string of two words doesn't mean that it is an idiom. It looks to me like a title someone's made up like 'Letters from a Small Island' or 'News from Lake Wobegon'. However, if it becomes as popular as Bryson's books, it may one day be recognised as an idiom or at least as a collocation.
Comment: No one called anyone an idiot, or dumb, or any such adjective. Calm down. :)
Comment: One reason I asked "where you've seen the term - preferably with a link" is because it is not a familiar term, and therefore the context may actually help us to understand it so that we can help explain it to you. That has nothing whatsoever to do with you being a non-native speaker - and I had no idea whether you were a native or non-native speaker. And *no-one called you "dumb", "idiot" OR "stupid"* - nor even implied it. I merely pointed you to some of the site guidance because, to quote your own words, you are "a very new user [and] not familiar with how to ask a question".
Comment: As you are a non-native speaker, you may also find our sister site useful: [ell.se].
Comment: Did you hear this or see it written? Could it be a mistake for *bare footnotes*?
Answer: This reminds me of the barefoot doctor in Mao's China.
In a similar vein barefoot notes probably refers to them being basic and sketchy.
Answer: I've never heard the term "barefoot notes".  The other comment was not intended to criticize you, just to show that he's never heard of it either.  It does show up in a web search, but there is no consistent meaning.  My best guess is that it is just another untranslatable English phrase, having as many meanings as there are people using it.
If I had to guess, I'd say that it described very informal notes or thoughts about things - going barefoot in this country is very informal.

What does "barefoot notes" mean?


IRS Payer/Receiver swap dv01

Question: I was just wondering whether it was possible for long payer swaps to have a negative dv01 and vice versa for a short payer to have a positive dv01? Intuitively by definition of payers and receivers I didnt think this was possible, but I was recently pricing some payers/receivers on Bloomberg and saw that this was in fact possible? Any help/insight would be greatly appreciated.
Comment: Just posted a screenshot!
Answer: The delta/DV01 is expressed with regards to a downward shift in rates. So in this case for a long payer, a 1bp downward shift would result in a loss of \$40k. For the short payer it'd be the opposite, i.e. a \$30k gain.
The DV01 itself is calculated as the average move in P&L for a 10bp shift up and down (i.e. numerically and not analytically).
Comment: Think it's specific to Bloomberg but can be explained by how risk is expressed for bonds. E.g. bond duration is usually positive which means if yields decrease by 1%, the bond price appreciates.
Comment: brilliant thank you so much! I just wanted to ask, is this a general convention for the quotation of swap deltas or just specific to bloomberg?
Comment: very much appreciated, thank you.

IRS Payer/Receiver swap dv01


better TCP performance over a "high delay network"

Question: I'm trying to improve my TCP throughput over a "high delay network" between Linux machines.
I set tcp_mem, tcp_wmem and tcp_rmem to "8192 7061504 7061504".
I set rmem_max, wmem_max, rmem_default and wmem_default to "7061504".
I set netdev_max_backlog and txqueuelen to 10000.
I set tcp_congestion_control to "scalable".
I'm using "nist" (cnistnet) to simulate a delay of 100ms, and the BW I reach is about 200mbps (without delay I reach about 790mbps).
I'm using iperf to perform the tests and TCPTrace to analyze the results, and here is what I got:
On the receiver side:
max win adv:         5294720 bytes
avg win adv:         5273959 bytes
sack pkts sent:            0
On the sender side:
actual data bytes: 3085179704
rexmt data bytes:    9018144
max owin:            5294577 bytes
avg owin:            3317125 bytes
RTT min:                19.2 ms
RTT max:               218.2 ms
RTT avg:                98.0 ms
Why do I reach only 200mbps? I suspect the "owin" has something to do with it, but I'm not sure (these results are of a test of 2 minute. A 1 minutes test had an "avg owin" of 1552900)…
Am I wrong to expect the throughput to be almost 790mbps even if the delay is 100ms?
(I tried using bigger numbers in the window configurations but it didn't seem to have an effect)
Comment: You got real hardware here. TCP take cpu, NIC have its own buffer, ACPI have its own limit, etc..
Answer: This is a common TCP issue called "Long Fat Pipe".  If you Google that phrase and TCP you'll find a lot of information on this problem and possible solutions.
This thread has a bunch of calculations and suggestions on tuning the Linux TCP stack for this sort of thing.
Answer: The site
http://www.psc.edu/networking/projects/tcptune/
mentions that as Linux nowadays autotunes TCP settings, messing with the values will likely not improve things.
That being said, maybe 100 ms together with a large bandwidth (at least 790 mbps) might lead to an enormous BDP, so maybe the autotuning decides that something is wrong and doesn't go far enough..
Comment: Depending on the kernel version, i've seen the auto-tuning to go well beyond 20MB.
Comment: Looks like this has moved to https://www.psc.edu/index.php/networking/641-tcp-tune
Answer: Try setting the iperf window size to really mach the bandwidth-delay-product of that link. So avg. RTT * 1Gbps should give you 10MB roughly. See if that improves things.
Answer: The only way you can really start to understand what is going on is to get more data -- else you are just guessing or asking other people to guess. I recommend getting a system level view (cpu, memory, interrupts etc) with sar from the iostat package. Also, you should get a packet dump with Wireshark or tcpdump. You can then use Wireshark to analyze it as it has a lot of tools for this. You can graph the window size over time, packet loss, etc.
Even a little packet loss on a high latency link tends to hurt bandwidth quite a bit. Alhough being simulated -- this is a bit strange. Lots of small packets might also cause high interrupts (even though those might be simulated as well?).
So in short, get TCPDump and Sar to see what is going on at the packet level and with your system resources.
Answer: How much memory does this machine have? The tcp_mem settings seems to be insane, it configured 28gb (7061504 * 4kb) for TCP data globally. (But this is not your perf problem since you most likely do not hit that limit on a few-socket test run. Just wanted to mention it since setting tcp_mem to tcp_xmem values shows a very common missconception).
The 7mb you have configured for default seems ok. The maximum however can go up much higher on large delay pipes. For testing I would use 64MB as the max number for tcp_wmem and tcp_rmem, then you can rule out that this is your limiting factor. (This does bloat your buffers, so it only works if you have limited concurrency and the connection has low jitter and drops).

better TCP performance over a “high delay network”


Diophantine $xy+yz+zx=4(x+y+z)$

Question: How do you solve the Diophantine equation $xy+yz+zx=4(x+y+z)$ for positive integers $x,y,z$? My approach was to consider $d=\gcd(x,y,z)$. I could just about show that the equation has no positive-integer solution for $d=2$, $d=3$ and $d>4$; for $d=4$ there is just one solution $(x,y,z)=(4,4,4)$. However I am stuck with the case $d=1$. How would you tackle this case? Any help appreciated. Thanks.
Comment: You can use this formula.   http://www.artofproblemsolving.com/community/c3046h1047580_  And to try all possible options.
Answer: Write $xy + yz+zx=4x+4y+4z$.
Thus they can't be all $>4$. The problem is symmetric.
Thus just set $x=0,1,2,3,4$.
Comment: I don't see all of the variables grater than 4... As I said, the problem is symmetric, u get this solution twice, setting x=4 and x=1 and permutating.
Answer: HINT
Here is how you could proceed.
$$z=\frac{-xy+4x+4y}{x+y-4}$$
If $x+y>4$, note that $-xy+4x+4y>0$. This implies that $(x-4)(y-4)<16$
Thus $x \ge 8$ and $y \ge 8$ cannot hold both at once. Assume without loss of generality that $x <8$. There are only a finite number of such $x$.
If $x+y \le 4$, there are only finite such $x,y$.

Diophantine $xy+yz+zx=4(x+y+z)$


Gmail: I send myself a file to keep in a label forever, but then I delete my sents and it deletes from my label. Help?

Question: I have to constantly delete my sents because I send about 10 video clips (6 MEG each) per day, thus I am always filling up my free storage that Google allows. That's fine with me, I don't need to keep those sent files. What I do need to keep are files with attachments that I send myself (from other computers) so that I can keep them in labels in my iPhone and iPad Gmails for future use forever. (This is all under my one Gmail account.). But if I select all my sents, my labeled emails also get deleted. How can I stop this or find a work around?  It's painstaking (takes about 45 mins each time) to go thru each of my sents and check the ones I need to keep and the ones I can get rid of, so I'm wondering if there's another workaround rather than that obvious one.  (Yep, I send hundreds of emails per day.)
Answer: To delete a subset of email from your sent mail while avoiding those emails you sent to yourself do the following.

You will need to execute a search, this can be entered in the search bar:
<code>is:sent to:(-email@example.com) has:attachment
</code>
Note: the minus sign is crucial.
You should now see all the messages that met the criteria, you sent it, but not to you, and there is an attachment. It will show you 20 messages at a time. I don't know of a way to expand this beyond 20 at a time
Scan the list and if it looks OK delete them.
If you have enabled Quick Links from gmail labs you can add the search string as a quick link so you don't have to remember the syntax of the search string.
Comment: Edited the query to make it in line with the OP's request (delete emails with attachments sent to self).
Comment: I had to edit it back: the key phrase is "need to keep are files with attachments that I send myself". Therefore  the search had to be for messages that I didn't send to me, then delete those messages.
Comment: Hmm. The wording is a bit weird. I thought the OP wanted to delete the 2nd copy of the email.


Could a waterborne virus contaminate the entire ocean?

Question: Is it plausible that a virus could contaminate the entire ocean to where humans that enter it are highly likely to contract the virus? It doesn't necessarily have to be the whole ocean what with the pressure from the deep depths and the temperatures of the N&S poles, but anywhere that a person could swim in.
Comment: Relevant article: https://www.whoi.edu/oceanus/feature/a-drop-in-the-ocean-is-teeming-with-life/. Loads of bacteria, viruses, and other micororganisms are already in the ocean environment. Happily humans have strong, layered, experienced defenses to cope with the onslaught, and most survive their ocean dip. Humans are currently most at risk from known human-adapted waterborne pathogens emitted from our own communities - like *E. Coli* in sewage. Crossover viruses that can thrive in both the ocean and in us (very different environments!) are rare.
Answer: Virus and bacteria, like any other life form, need a suitable environment to stay active.
Virus in particular need a host to reproduce, while if they are in the outer environment they usually degrade and become inactive.
The time a virus can remain active outside of a host is variable from virus to virus, but in principle it is possible for one to stay dormant for a long enough time.
The main problem I see is that, being humans not exactly a water species, it is unlikely to have a dedicated virus attacking specifically humans in water, since the average time to find a host would be extremely long. It has to be an opportunistic virus attacking water animals/mammals.


Are On-Screen Keyboards Really More Secure?

Question: A quick question: I use on-screen keyboard when entering my bank passwords, credit card numbers, etc., in an effort to guard against malware. Is this a good idea or am I wasting my time? Some internet banking website have an on-screen keyboard for entering in a password are these any more secure?
Answer: The built-in on-screen keyboard that comes with many operating systems is designed to help people who are unable to use a physical keyboard because of disabilities. Due to this, an on-screen keyboard behaves as much like a real keyboard as possible and it's activity will most likely be logged by a keylogger.
On-screen keyboards specifically designed for security (on a bank's website, for example) are a different story and are likely more secure against keyloggers.
http://www.viruslist.com/en/analysis?pubid=204791931
Comment: Thank you, I had an inkling but I wasn't sure but "More important than the quest for certainty is the quest for clarity"
Answer: On-screen keyboards prevent a keyboard dongle from recording your keystrokes. There are other methods for recording keystrokes an on-screen keyboard will not prevent, though.
Comment: The bank ones that I've seen tend to move the keyboard button position around a bit on each login so the relative mouse position changes, making it hard to know what buttons a user clicked on.
Answer: Using an on-screen keyboard would make your password readable by people with no connection to your computer.  :D
http://en.wikipedia.org/wiki/Van_Eck_phreaking
http://en.wikipedia.org/wiki/TEMPEST#Public_research
Answer: Note that an OSK is not secured against things that also track mouse movements, though figuring out what you clicked is much harder. I like to keep parts of my password in autohotkey hotstrings. (because "password" is so much easier to work out than "p66j#wokk#d")
Comment: If your OSK changes its layout every few characters--which is annoying, mouse movements become harder to track--now images of the screen have to be saved too.
Answer: Spyware can detect mouse movements, allowing to calculate which buttons were clicked and even take screenshots on mouse release. Not sure if standard keyboard that goes preinstalled with Windows protects against this.
However, there is a number of so called anti-keylogger on screen keyboards that claim to provide protection against such attacks. I am not providing names here but you can google them. Some people even test and compare them, laying out results on the web.
The question is how trustworthy such analysis is. The best way would be to test them yourself, for that you would have to install such a keyboard on a testing machine along with a keylogger and see how it goes.
PS: An interesting comparison is here: https://www.raymond.cc/blog/how-to-beat-keyloggers-to-protect-your-identity/
Answer: The best solution to beating keyloggers I can think of is this: use a cloud computing virtual machine and within that virtual machine use a password manager to autofill passwords. since the password is never displayed, screen monitoring from the host computer won't do anything and the keylogger won't be on the virtual machine itself to log traffic there and since you're using a password manager on the virtual machine, no keystrokes or mouse patterns from activating the password manager will reveal anything useful to the key logger on your computer. furthermore, all data going to and from the cloud virtual machine is encrypted with a VPN, and any data concerning the password won't be sent to your computer other than the black dots obscuring the password. provided you never download anything other than what is absolutely necessary for the virtual machine's security and what important/sensitive things you need to use it for, you shouldn't ever get malware of any kind on the virtual machine. I don't think even a hardware based keylogger on the host computer can log that. you'd basically be immune to key logging.
Comment: As it's currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-answer).


Argument for Fermions and Bosons as eigenfunctions of particle exchange operator $P$

Question: Some time earlier, my prof took me through an argument leading to emergence of fermions and bosons by the application of particle exchange operator on the multiparticle wavefunction. I will try to present the argument below:
We have a multiparticle wavefunction written as $\psi(x_1..x_i..x_j..x_n)$ representing a system of $N$ identical particles. And there is a parity operator $P$ which exchanges two particles in our system on $N$ particles. $P$ acts as follows,
$$P_{\boldsymbol{i,j}}\psi(x_1...\boldsymbol{x_i}...\boldsymbol{x_j}..x_n) = \psi(x_1...\boldsymbol{x_j}..\boldsymbol{x_i}..x_n)$$
Now, if we operate $P$ on $\psi$ twice, we are supposed to get back the same system we started with before exchanging particles $i$ and $j$, leading to $P^2 = \mathbb{I}$. This tells us that $P$ can have only two possible eigenvalues $\pm1$ and thus we call the eigenfunctions  with eigenvalue +1 as bosons and -1 as fermions.

My question is why can't we extend the above argument further to higher even powers of $P$. For example, if we apply the $P$ operator on our system 4 times, we will get an equation of the form $P^4 = \mathbb{I}$. Won't this lead us to two more eigenvalues for $P$, namely $\pm i$. what will we call the eigenfunctions of $P$ in these cases.

I'm not sure if I make much sense here. This has been in my mind since that class and I wish to get it clarified as to where am I going wrong with my argument.
Thanks in advance.
Answer: There are two problems here - one with your argument about eigenvalues, and another with the very common but deeply flawed argument for fermions and bosons.
The problem with your eigenvalue argument can be understood in the context of basic algebra.  The mistake is in assuming that if I start with some equation and then raise both sides to some power, the solution sets of the two equations remain the same, but that's not true. For example, $z=1$ has one (fairly obvious) solution, but when I square both sides I obtain $z^2=1$, which now has two.  A solution to the former is guaranteed to be a solution to the latter, but the converse is not true.
In the context of operator theory, this idea takes the following form.  If an operator $\hat A$ has an eigenvalue $\lambda$, then $\hat A^2$ is guaranteed to have $\lambda^2$ as an eigenvalue.  However, if $\hat A^2$ has some $\mu$ as an eigenvalue, then we cannot conclude that $\pm \sqrt{\mu}$ are both eigenvalues of $\hat A$.  A simple example of this is the identity operator $\mathbb I$, whose only eigenvalue is $1$.  $\mathbb I^2$ also has 1 as an eigenvalue (obviously), but we cannot from this conclude that the eigenvalues of $\mathbb I$ are $\pm \sqrt{1} = \pm 1$.

The flaw in the argument for fermions and bosons is more subtle.  The problem is intimately related to the rarely-discussed distinction between states and wavefunctions.
Let's restrict our attention to 2 particles for simplicity. A wavefunction for this system is loosely a function $\psi : \mathbb R^n \times \mathbb R^n \rightarrow \mathbb C$ such that $\int \mathrm d^3 x \mathrm d^3 y \ |\psi(\mathbf x,\mathbf y)|^2 < \infty$.  If an observable $\hat A$ has an eigenvalue $\lambda$ corresponding to an eigenvector $\phi_\lambda$, then the probability of measuring $\hat A$ to have that value is given by
$$\mathrm{Prob}_\psi(\hat A,\lambda):= \frac{|\langle \psi,\phi_\lambda\rangle|^2}{\langle \psi,\psi\rangle}$$
In that sense, $\psi$ allows us to compute the probability of any measurement outcome we can imagine.  This collection of probabilities is what we call a state.
The crucial point is that a given state can be represented by an infinity of different wavefunctions.  Its an easy exercise to see that if you look at the probability expression above and then scale $\psi$ by any non-zero complex number, the probability you obtain would be exactly the same. Therefore, a state should not be identified with an individual wavefunction but rather with an entire family of wavefunctions which are non-negative multiples of one another. In this spirit, we define the projective Hilbert space $\mathscr P(\mathcal H)$ to be the set of equivalence classes of elements of $\mathcal H$, where two elements of $\mathcal H$ belong to the same class if each is a non-zero multiple of the other.
Now, it turns out that working directly with some $\Psi\in\mathscr P(\mathcal H)$ is enormously inconvenient. Therefore, in practice we always simply choose a concrete representative $\psi\in \mathcal H$ and work with that instead.  Ordinarily this causes no particular issues, and so one can be forgiven for forgetting that $\psi$ is merely a representative of a state, but sometimes the distinction is crucial, and this is one of those times.
Note that any linear or antilinear map $M:\mathcal H \rightarrow \mathcal H$ induces a well-defined map $\widetilde M:\mathscr P(\mathcal H) \rightarrow \mathscr P(\mathcal H)$ on the space of states (note that $\widetilde M$ is not linear, because $\mathscr P(\mathcal H)$ is not a vector space!).  Because working at the level of $\mathcal H$ is vastly more convenient than working at the level of $\mathscr P(\mathcal H)$, this is normally what we would do to define a transformation on a state.  However, there is great ambiguity here!  For instance, if two operators $\hat A$ and $\hat B$ are related by $\hat A = c \hat B$ where $c$ is a non-zero complex number, then these two distinct operators on $\mathcal H$ induce precisely the same transformation of $\mathscr P(\mathcal H)$.
This leads us to the problem with your instructor's argument.  We want to define the action of the permutation group $S_2$ on the space of states, and so it is natural to define its action on $\mathcal H$ first.  But the family of pseudo-permutation operators
$$\hat P_\theta \psi(\mathbf x,\mathbf y) = e^{i\theta} \psi(\mathbf y,\mathbf x)$$
all induce precisely the same permutation map on $\mathscr P(\mathcal H)$. You have arbitrarily chosen $\hat P_0$ as the correct permutation operator but there is no physical justification for doing this.  Since $\hat P_\theta^2 = e^{2i\theta}\mathbb I \neq \mathbb I$ in general, the rest of the argument falls apart.
In summary, if we want to define the particle permutation map on the space of physical states $\mathscr P(\mathcal H)$, we can choose any of the family of pseudo-permutation operators $P_\theta$ at the level of the Hilbert space $\mathcal H$.  In the absence of any further constraints, all of these operators do the same job.  Since $\hat P_\theta^2 \neq \mathbb I$ in general, the requirement that the state be invariant under particle interchange implies only that the wavefunction be an eigenfunction of $\hat P_\theta$, without constraining those eigenvalues to be $\pm 1$.
Answer: Mathematically, doing a transposition twice in a row gives the identity, i.e. $P_{ij}^2=\hat 1$.  You cannot escape this mathematical fact, and it's baked in the permutation group.  If you don't want to use the permutation group, then you're just doing something different than permutation symmetry.
Note that not all permutations satisfy $P^2=\hat 1$, and in that way the permutations of two objects is quite boring and not in any way representative of the structure of the permutation group for $n$ objects.
For instance, in 3-particle systems, the 3-cycle $P_{123}$ which takes $(a,b,c)\to (c,a,b)$ so that $P_{123}P_{123}(a,b,c)=(b,a,c)$  clearly does not satisfy $P_{123}^2\ne \hat 1$.  However
transpositions, permutations which act on only two entries in a list, you will certainly have $P^2=\hat 1$ irrespective of the number of particles.
Of course part of the strategy in using the permutation group is that you have all the known machinery for that group available from the math literature, and it makes sense physically to "permute" coordinates using this group, at least in the simplest cases (you could do braiding instead of permutations for fancier systems like anyons).
Comment: I think i did not state my question much clearly, I was talking about transpositions only but applied n times, n being even integer. I was looking for an answer given by @By_Symmetry. Thanks for answering anyway :)
Answer: All eigenfunctions of $P$ must satisfy
$$
P^n \psi = \lambda^n \psi
$$
where $\lambda$ is the relevant eigenvalue, for all values of $n$. In particular they must satisfy the equation for the equation for $n = 2$, giving the eigenvalues $\pm1$. If you plug these values into the higher order equations you will find they are also satisfied, for example
$$
P^4 \psi = \hat{1}^2 \psi = (\pm1)^4 \psi
$$
On the other hand you cannot go in the other direction. Your "extra" solutions $\pm i$ satisfy the equation for $n = 4k$, $k\in\mathbb{Z}$, but they do not satisfy the equation for $n = 2$
Comment: Ahh yes yes. Thanks again for answering :)


How long does it typically take for an edit to be peer reviewed?

Question: I'm an English teacher first (30 years' experience - 14 of them with scientists and engineers) and developer second, so I often find it relatively easy to grasp what a non-native speaker is trying to ask without always feeling confident enough to answer the technical points posed in the question. In these cases, my most useful contribution is to improve the way the question is phrased in the hope that some whiz kid out there can provide the technical answer. What is involved in the review process, and how long might it typically take?
Comment: while such edits are certainly welcome, please bear in mind that questions that do not demonstrate sufficient *effort* on the behalf of the asker will probably be closed, even if you graciously translate them from l33t speek to English. Try to contribute edits on questions that are fundamentally salvageable per the guidelines at http://stackoverflow.com/questions/how-to-ask
Comment: I all fairness, I would have made almost exactly the same edit as @michael made. I don't think this necessarily was a comment on your question or part of the discussion. What is more, I would have also rejected [your accepted edit](http://stackoverflow.com/suggested-edits/290561). There is no need to mention " [Now corrected by a professional English teacher]" within the question. Keep them clean. Don't muddy the question with content that should not be there.
Comment: your edit should be a true edit to the post not a suggestion like [this one](http://stackoverflow.com/suggested-edits/285142). Fix everything that seems adequate but don't mention that you edited in the edit :)
Comment: Good point Jeff.  I'm more qualified to fix the English than to judge whether it's a good technical question but I found myself sympathising with someone struggling with both the technical problem and the language - I imagine trying to do it in Spanish! I've found a few questions that seemed reasonable to me - just hard to read.
Comment: As a side note, it looks like it generally takes [just over an hour](http://data.stackexchange.com/stackoverflow/query/73162) for a suggested edit to be reviewed (either approved or rejected) on Stack Overflow.
Comment: I edited your question text to improve it. "whizz kid" is ridiculous. And you're the one who chose to bring the edit into the discussion, though in a way that prevented me from seeing your comment ([use "@" if you wish a user to be notified of your comments](http://meta.stackexchange.com/questions/43019/how-do-comment-replies-work)).
Comment: Bart, since my question is about editing and this is a place for meta discussion, it seems worth bringing this in.  For instance you should have written "In all fairness", but it's obviously just a typo and not worth considering.  I would like to contribute by helping bring a mangled non-native speaker question up to a reasonable level of comprehensibility for experts out there to answer and I'm exploring how best to do that.
Comment: Give with out be anxious, and with our expecting anything.
Comment: @MichaelPetrotta Thanks Michael.  I might eventually get round to putting my real name when I get comfortable with this very public discussion.  I've spent my working life helping non-native speakers with English so I might give someone an "excellent" for a text you would probably feel it's just a rough first draft.  You help the native speakers if you feel they need it. I'll stick to what I know - with the help of all these useful comments I've seen today! As per Webster, I'm putting back the whizz kid!
Comment: (on a sorta related note, when I initially saw phrases like "do the needful", I wanted to "fix" that as well. Then I looked it up - it's a [standard Indian English phrase](http://en.wikipedia.org/wiki/Indian_English#Vocabulary_and_colloquialisms), and was actually common in UK and US English up until about a century ago.)
Comment: Start by reading [How do suggested edits work?](http://meta.stackexchange.com/questions/76251/how-do-suggested-edits-work)
Comment: Michael Petrotta, I presume you're trying to make a point with your editing.  I'm trying to help non-native speakers with their English, not those with already good English. e.g The Oxford English Dictionary suggests "whizz-kid" or "whiz-kid". So, we're both wrong: you with your spelling and I with my hyphenation.  Writing "scientist" instead of "scientists" is a typo that I would let go. The way you re-phrased the "about answering..." is a matter of taste. I prefer my original. How did that edit contribute to this discussion?
Comment: Bart and Aristos.  Thank you. That's what I was trying to get clear.  This collaborative editing is a new and awkward experience. Now I understand that presenting credentials (professional English teacher) does not re-assure anyone. You feel it muddies the question.
Comment: @MichaelPetrotta Sorry Michael, I didn't know about the "@" convention.  I wasn't deliberately trying to prevent you seeing anything. Do we have the usual transatlantic problem? I don't think so.  Websters suggests "whiz kid" and "whizz kid" i.e. the same as the OUP but not hyphenated. What do you mean by ridiculous?
Comment: Well, if you're on the other side of the pond, then maybe. "Whizz kid" sounds like a child with bladder control problems ("wiz" is short for "wizard", while "whiz" and "whizz" are childish slang terms for "urinate". Which reminds me of the transatlantic difference in the usage of "pissed").
Comment: @MichaelPetrotta I knew the American slang term for urine but the combination with "kid" seems to shift the meaning - at least according to your American dictionary. How on earth can we both avoid every questionable term like that? Remember, I'm qualified to help get a non-native speaker's English up to speed. I'm not the editor of the Herald Tribune.
Comment: @örs Thanks örs.  That helps a lot.  I'm finding it hard to get used to where to attempt an edit, where to attempt a comment and where to have a go at answering a question.
Comment: I'm not suggesting that we do, user341180 (may I call you user341180?). If you don't like that edit, feel free to revert it. If you review my history, especially on Stack Overflow, you'll see that I edit for English a lot. I am sorry if you felt singled out - that wasn't my intent.
Comment: Whoops! Another typo: Should have been "you would probably feel *is* just a rough first draft"
Comment: @Bart I've paid close attention to eveyone's comments.  With them in mind, here is the kind of edit I am hoping to provide at the level most appropriate for a TEFL teacher like myself: http://stackoverflow.com/questions/11064447/editing-html-in-joomla.  Bart, would you have accepted this edit? örs Is that what you mean? What would you do with it Michael? Would you be uncomfortable with the British usage "I have got"?
Comment: @MichaelPetrotta Please note above comment. I don't know how to notify several people in one comment.
Comment: @örs Please note above comment to Bart.
Comment: That looks perfect to me, David. It significantly improves the post (lots of edits fix one or two issues, and leave a number of others untouched) without damaging it (that happens sometimes). Your edit was approved in 13 minutes, much faster than the average of 68 minutes Tim noted, and the quality of the edit may account for that (or it could have been time of day, phase of the moon, ...)
Comment: @MichaelPetrotta Did you notice someone took away one of my z's! :) I checked the spelling of whizz kid with a very experienced American colleague and his instant response was one z, so I'll just leave it now.
Comment: Honestly, I'd never seen "whizz" before I saw it in your post. I immediately thought it was a typo or spelling error, and that's why I removed it. Now, *"I have got"*: I recognize (or should that be "recognise"?) that as a UK variant, so I wouldn't have touched that. Over on this side of the pond, we've grown more tolerant in the past 250 years...
Comment: One of the toughest things in TEFL is to strike a balance between listening to your native-speaker instinct and respecting a well-documented standard usage which doesn't sit right. I am very uncomfortable with the verb pattern "recommend  to  ", but have to teach it to my upper-intermediate students.
Answer:
What is involved in the review process

As for the review process itself, I can only comment on how I do it.
If the edit clears up all the issues I see within the question/answer, I'll accept it. If the edit leaves serious problems within the content then, even if the edit itself is good, I might reject it. Usually I go to the question myself, improve it and judge how much time it has taken me to do so. If this is significantly more than the other editor will have spent, I might just mark his edit as "not helpful". If you glanced over a minor issue, no problem. I'll correct it and mark your edit as helpful. Moral of that story: Don't just fix one minor thing, fix the entire content.
With regards to code edits, things become slightly more complex. Editing code in the question itself is usually a big nono, apart from something like indentation corrections. If I feel your edit changes the question or might actually hide the OPs problems, I'll reject it.
Code edits in an answer I only handle when I'm confident of their correctness or appropriateness. If edits of such content are so significant that I can't judge it with a simple review, I might leave it alone, or reject it along the lines of "should have been an answer instead".
Whatever you do, don't make the fact that you've edited a question, part of the question itself. If your edit is brilliant, but you leave something like "Edited by me for clarity" or if you even add a whole separate section with an "edited version" of the content, I will reject it. (Though I have the rare nice day where I actually go to the question to remove the fluff you added). Edited content should appear as if written by the OP.
As for the comments you received on only editing those questions which won't be closed outright anyway, please do so. When the edits appear in the review queue, I am not filtering them out anymore based on the quality/appropriateness of the question itself. So if you edit something which should not be on this particular Q&A anyway, I will review it. The review queue is often full as it is anyway, so please don't add to that by editing bad/off-topic content.

typically how long might it take?

With regards to the average time taken between the suggested edit and final acceptance or rejection, I have no hard data. I have handled edits mere seconds after they were suggested, but have also seen some which hung around for several hours (though this is less likely with the new review queues). If the suggested edit is substantial enough that its evaluation requires specialist knowledge in a topic, it might take some time. If the edit is not merely a spelling correction (or something similar) then I usually stay away from them if they are not in my area of expertise. Such edits might take a little longer to get evaluated.
Comment: There is no hurry at all in accepting an answer. Leave it open for a while, see if you get some more responses. If you do, great. If you don't, have a look at how others value my contribution. For all intents and purposes I might be out of my mind. :) Whenever you feel that all has been said and nothing more will be added, you make the judgement call whether you want to accept my answer as correct or helpful.
Comment: That's a very thorough answer.  Now, I'm uncertain about the etiquette of accepting the answer.  You yourself say that this is only your own opinion so maybe I should wait for another sample answer?  Do you have any advice about how you accept an answer? Part of my reason for saying I had edited the question was out of politeness to the person who asked the question.  Now I understand the review process better I realise that's not necessary or useful.
Answer: What I've taken away from this discussion and Bart's excellent answer
I don't know if an answer is the right place for the following summary, but I'm exploring this whole process so here goes in case it helps other newish collaborators like myself:

Editing should be unobtrusive in an effort to help the questioner express the question in a clearer more useful way. The editor should not draw attention to him/herself.
Wait patiently for a response from a reviewer.  They seem to work pretty fast.
If you want to make public suggestions or clarify something etc, then add a comment or, ultimately, an answer, not an edit.
You don't need to be over-polite.  As you can see from my initial response to MichaelPetrotta I took his editing a bit personally.  Since this is a collaborative venture, that's inappropriate. So, just edit and use the standard tool to explain what you've done if necessary.  Don't add justifications, excuses or credentials to the body of the question. If the questioner doesn't like having his/her question edited, we've been well warned this is the wrong forum for him or her.
Don't waste the reviewers' time by editing a question that really isn't useful no matter how well your editing improves its expression.

Levels of Intervention
There seem to be many layers of editing that different contributors are comfortable with.

Minor typos and spelling mistakes.
More major problems with syntax and word choice that render the question difficult or unintelligible.
Higher levels of concern about style, elegance and expression that help to polish this forum and make it a useful long-term repository of good questions and answers.
Last and clearly most important: editing the code samples and technical points needs to be done lightly and with great care. Attempt with caution.

I will be attempting levels 1 and 2 for the time being, especially with obviously non-native speakers. I have no great ambitions to get involved at levels 3 and 4.
Comment: I have a suspicion that the easier it is to tell an edit is "good" the quicker it's likely to get reviewed.


Being asked for credit card information for updates on apps already purchased

Question: I have purchsed the apps & most I have are free but I'm being asked for credit card information for apps that are being updated.  I'm not being charged for the updates, am I?
Comment: Your credit card information is probably out-of-date. In other words, the card linked to your account is probably an old invalid/expired card.
Answer: Whenever you use your Apple account they will try to validate the information in your account. If there is missing address information you will be prompted for it. If you haven't agreed to the latest Terms and Conditions you will be prompted to agree. If you have credit card information on your account Apple will try to validate it. If the information is out of date, you will be prompted to update it even when installing a new free application or upgrading old applications.
You aren't being charged for updates.


Xcode 10 update issue with Alamofire objectmapper

Question: I updated my iOS version to 12.0.1, when I run my code in Xcode version 10 getting errors like  > Invalid redeclaration of '<-' (Alamofire objectmapper).
when I go back to Xcode version 9.4 it works fine with swift 3.3 but it not running in my updated version 12.0.1 iOS, it showing error like "This iPhone 6s is running iOS 12.0.1 (16A404), which may not be supported by this version of Xcode." Please respond. Thanks in advance
Comment: Thanks for responding , i upgrade the latest one still get Invalid redeclaration of '<-',
Comment: for ios 12 or above will work only on xcode10 if you want then need to upgrade almofire latest version of it.
Answer: I used pod update AlamofireObjectMapper in Xcode 10. iOS 12 and higher are only built by Xcode 10. You must use Xcode 10 and update AlamofireObjectMapper to the newest version.
Hope this help!
Answer: Copy the entire line below and update podfile. Should be good
pod 'AlamofireObjectMapper', '~> 5.0'


What does the solution of a PDE represent?

Question: So I took a course in PDE's this semester and now the semester is over and I'm still having issue with what exactly we solved for. I mean it in this sense, in your usual first or second calculus course you are taught the concept that your second derivative represents acceleration, your first derivative represents velocity and your fucntion is some of position function. So in the world of PDE's what does the function U(x,t) really capture? Your derivatives model rates of change of a phenomena in a broad sense, but what does the soultion model? So for example the solution of the Heat equation is your U(x,t) function modelling temperature? or or what is the solution to the wave equation modelling? Are these some sort of "position functions"? I ask because I'm going over Green's equations and the relationship with the Dirac function and I'm trying to wrap my head around what is being described. I understand the steps towards a solution, but it is the description of what is occurring which is faltering and essentially is the important part behind why we do mathematics.  Also is there some sort of visual representation of these solutions?  Thanks
Answer: What the function you get in the end represents depends on your problem. One thing to note is that a certain equation, just expressed as a PDE, might appear in several different application problems.
E.g. the heat equation can be used to model e.g. the temperature over time in an object, but can also be used to model financial things (more examples can be found on the Wikipedia page for the heat equation).
The mathematical problem for PDEs is how to get a closed-form solution for a PDE or how to integrate the PDE numerically in a computer, etc. What it is used to model is more of an engineering problem (or rather, the problem of finding what equation describes a certain phenomena is engineering). This might lead to a disconnect between what things mean in the "real world" and how to get the mathematical results that you can do something with.
That said, I often find it very useful and motivational to find in which engineering or scientific problems certain mathematical things appear. In my experience, this is often glossed over in mathematics classes, and focused too much on in engineering classes.
I would encourage you to try to find application examples of the things you mention outside the mathematical sphere and see how they arise there. I think it is good to distinguish between things that model the physical world and things that are just mathematical tools to solve the mathematical problem, so you do not try to fit some purely mathematical thing in some ill-fitting physical simile.
Answer: They can represent nearly anything, it depends on what you are modelling with the differential equation. In general, the solution of a differential equation is simply a function.
Using the example of the heat equation, the solution $u(\mathbf{x},t)$ is the temperature at position $\mathbf{x}$ at time $t$. See for example the figures in this link.
Answer: Let's just use the heat equation as an example. The solution of the heat equation $u(x,t)$ can be used to describe the temperature at any position $x$ of a thin rod, at any time $t$.
See the following picture:

Suppose a rod was heated at the middle point and we measured its temperature afterwards. The temperature would be like this. And this could be the solution of a heat equation.
Answer: The solution is just a function. That's what you're aiming for at the end. E.g. in the 1D wave equation, you get $u(x,t)$. You know what a function is right? For the solution to the wave equation, you plug in the position on the x axis and the time and $u(x,t)$ will tell you the amplitude of the wave at that position at that point in time. That's all there is to it really.
Answer: There are tons of situations where the solution of a PDE discribes the behaviour over time:
1) pendulum
2) (radio active) decay
3) mass-spring-damper systems
4) inductance-capacitance-resistance systems
5) heat distribution over time
6) water level in a bucket with a leak
7) Temperature of a heated house
8) (electromagnetic) waves


Simple, secure API authentication system

Question: I have a simple REST JSON API for other websites/apps to access some of my website's database (through a PHP gateway). Basically the service works like this: call example.com/fruit/orange, server returns JSON information about the orange. Here is the problem: I only want websites I permit to access this service. With a simple API key system,  any website could quickly attain a key by copying the key from an authorized website's (potentially) client side code. I have looked at OAuth, but it seems a little complicated for what I am doing. Solutions?
Comment: HTTP Basic authentication over HTTPS.
Answer: You should use OAuth.
There are actually two OAuth specifications, the 3-legged version and the 2-legged version.  The 3-legged version is the one that gets most of the attention, and it's not the one you want to use.
The good news is that the 2-legged version does exactly what you want, it allows an application to grant access to another via either a shared secret key (very similar to Amazon's Web Service model, you will use the HMAC-SHA1 signing method) or via a public/private key system (use signing method: RSA-SHA1).  The bad news, is that it's not nearly as well supported yet as the 3-legged version yet, so you may have to do a bit more work than you otherwise might have to right now.
Basically, 2-legged OAuth just specifies a way to "sign" (compute a hash over) several fields which include the current date, a random number called "nonce," and the parameters of your request.  This makes it very hard to impersonate requests to your web service.
OAuth is slowly but surely becoming an accepted standard for this kind of thing -- you'll be best off in the long run if you embrace it because people can then leverage the various libraries available for doing that.
It's more elaborate than you would initially want to get into - but the good news is that a lot of people have spent a lot of time on it so you know you haven't forgotten anything.  A great example is that very recently Twitter found a gap in the OAuth security which the community is currently working on closing.  If you'd invented your own system, you're having to figure out all this stuff on your own.
Good luck!
Chris
Answer: OAuth is not the solution here.
OAuth is when you have endusers and want 3rd party apps not to handle end user passwords. When to use OAuth:
http://blog.apigee.com/detail/when_to_use_oauth/
Go for simple api-key.
And take additional measures if there is a need for a more secure solution.
Here is some more info, http://blog.apigee.com/detail/do_you_need_api_keys_api_identity_vs._authorization/
Comment: Here's an Internet Archive link if someone is still interested (I still thought it was):

https://web.archive.org/web/20140708081525/https://blog.apigee.com/detail/do_you_need_api_keys_api_identity_vs._authorization
Comment: LOL: the link to blog.apigee.com is broken and shows "Access Denied: You are not authorized to access this page." But it's from 2009 anyways, so not sure how relevant the info was...
Answer: If someone's client side code is compromised, they should get a new key.  There's not much you can do if their code is exposed.
You can however, be more strict by requiring IP addresses of authorized servers to be registered in your system for the given key.  This adds an extra step and may be overkill.
I'm not sure what you mean by using a "simple API key" but you should be using some kind of authentication that has private keys(known only to client and server), and then perform some kind of checksum algorithm on the data to ensure that the client is indeed who you think it is, and that the data has not been modified in transit.  Amazon AWS is a great example of how to do this.
I think it may be a little strict to guarantee that code has not been compromised on your clients' side.  I think it is reasonable to place responsibility on your clients for the security of their own data.  Of course this assumes that an attacker can only mess up that client's account.
Perhaps you could keep a log of what ip requests are coming from for a particular account, and if a new ip comes along, flag the account, send an email to the client, and ask them to authorize that ip.  I don't know maybe something like that could work.
Answer: Basically you have two options, either restrict access by IP or then have an API key, both options have their positive and negative sides.
Restriction by IP
This can be a handy way to restrict the access to you service. You can define exactly which 3rd party services will be allowed to access your service without enforcing them to implement any special authentication features. The problem with this method is however, that if the 3rd party service is written for example entirely in JavaScript, then the IP of the incoming request won't be the 3rd party service's server IP, but the user's IP, as the request is made by the user's browser and not the server. Using IP restriction will hence make it impossible to write client-driven applications and forces all the requests go through the server with proper access rights. Remember that IP addresses can also be spoofed.
API key
The advantage with API keys is that you do not have to maintain a list of known IPs, you do have to maintain a list of API keys, but it's easier to automatize their maintenance. Basically how this works is that you have two keys, for example a user id and a secret password. Each method request to your service should provide an authentication hash consisting of the request parameters, the user id and a hash of these values (where the secrect password is used as the hash salt). This way you can both authenticate and restrict access. The problem with this is, that once again, if the 3rd party service is written as client-driven (for example JavaScript or ActionScript), then anyone can parse out the user id and secret salt values from the code.
Basically, if you want to be sure that only the few services you've specifically defined will be allowed to access your service, then you only option is to use IP restriction and hence force them to route all requests via their servers. If you use an API key, you have no way to enforce this.
Answer: All of production of IP's security seems produces a giant bug to users before getting connected. Symbian 60s has the fullest capability to left an untraced, reliable and secure signal in the midst of multiple users(applying Opera Handler UI 6.5, Opera Mini v8 and 10) along with the coded UI's, +completely filled network set-up. Why restrict for other features when discoverable method of making faster link method is finally obtained. Keeping a more identified accounts, proper monitoring of that 'true account'-if they are on the track-compliance of paying bills and knowing if the users has an unexpired maintaining balance will create a more faster link of internet signal to popular/signatured mobile industry. Why making hard security features before getting them to the site, a visit to their accounts monthly may erase all of connectivity issues?  All of the user of mobile should have no capability to 'get connected' if they have unpaid bills.  Why not provide an 'ALL in One' -Registration/Application account, a programmed fixed with OS, (perhaps an e-mail account) instead with a 'monitoring capability' if they are paying or not (password issues concern-should be given to other department). And if 'not' turn-off their account exactly and their other link features. Each of them has their own interests to where to get hooked daily, if you'd locked/turn them off due to unpaid bills that may initiate them to re-subscribe and discipline them more to become a more responsible users and that may even expire an account if not maintained. Monthly monitoring or accessing of an identified 'true account' with collaboration to the network provider produces higher privacy instead of always asking for users 'name' and 'password',  'location', 'permissions' to view their data services. IP's marked already their first identity or 'finding the location of the users' so, it's seems unnessary to place it on browsers pre-searches, why not use 'Obtaining data' or 'Processing data.'
Comment: Perhaps you might want to restructure your answer into blocks to improve readability.


How do I get a specific outfit at the start of GTA Online?

Question: I notice that changing your lifestyle also changes your features / clothing. I was playing around with what hours I spent on everything after finding a good outfit, but even though I set the hours back at the same amount (or close to), the outfit is something else. Is there a guide to what numbers will give what outfit? or a way to manually change outfit?
Comment: If you're wondering about hats and sunglasses: they are chosen at random whenever your outfit changes and you can choose them after defining your lifestyle.
Comment: For me, it looked like that your outfit changes based on what stats you place. I kept adding a few points and removing a few points from one of the options and noticed that it chose one of several different outfits but eventually came back to the one I wanted. I think it's based from all of your stats though.
-kuma
Comment: @kuma I think it's actually based on whatever one's the highest - it switched whenever a different one became the highest.
Answer: Having had another play around with the character creator it seems that although the outfit is based on the various stats you've chosen, it also picks - at random - between 4 or 5 outfits.
You can test this yourself:

Set all your stats how you'd like them - your character will change into an outfit.
Pick a stat (let's go with Legal Work) and remove all the hours you put in. Hopefully, if you had enough in the first place you'll see your outfit change to something else.
Put the same amount of hours back into Legal Work.
Chances are you'll have a different outfit to the one before.
Using this process I was able to cycle through the outfits available to my character and pick one I preferred.

I can't find any definitive list of outfits but this should help finding the one you want.
Answer: You can definitely start with a suit, I had to show up when I put stats like going to legit work and sleeping but then I couldn't replicate it.


Why was the veal q removed from HNQ?

Question: I see that the veal q was also removed from HNQ https://skeptics.stackexchange.com/posts/49953/revisions
Why?
Answer: Similarly, the reasons here explain that we're trying to curb extraneous mod efforts.
Differently, there's quite a few flags mostly along the lines "upset the OP doesn't seem genuine", on the question and the OP's self answer and even some comments. We also opted to lock it for a day, since it's becoming a "content dispute" that might motivate a meta post of its own along the lines "what does the community expect us to do when this happens?".
The volume of activity that needed mod intervention on the post wasn't terribly burdensome, but people are flagging the OP's question, answer, and comments as "spam" and "other" with messages that he's trolling, etc. To be clear, we're on the same page with not really liking his answer and belligerence in accepting the evidence in the other answers, but he's not really breaking any rules, so we're at a loss what the flaggers expect.

Why was the veal q removed from HNQ?


Can a measurement be 100% accurate? What about counting things?

Question: I have never heard about any experiment performed which is 100% accurate (Sorry,if there is any).By the term accurate I mean measurements exactly equal to the theoretical value .Is it every time that an experiment performed is not accurate? Why is it so?
Comment: Do you have any "100% accurate" measuring devices?  How would you confirm that?
Comment: I just counted the number of tick divisions on the face of my watch. I got 60 precisely.
Comment: What does "accurate" even mean in this context?
Answer: Some food for thought:

Accuracy vs. Precision.
In everyday conversation the words Accuracy and precision are often mixed around quite a lot, but in science it is very important to keep these concepts separate. Accuracy refers to whether the statement is True or False, while precision is a measure of the statements detail.
For example, if I were to say that my height was between 1 cm and 100 m this statement would be accurate (it is true), but it is very imprecise. I could also say my height is 1.234274721 cm — this statement is very precise (many significant figures) but is not accurate — I am much taller than 1 cm! Ideal measurements should be both accurate and reasonably precise. So that they tell us stuff that is both true and useful.
I assume by your question you mean "100 % precise" and NOT "100 % accurate", but I will try to answer both questions.
Can I be infinitely Precise with a measurement?
No measurement is infinitely precise. There are many reasons for this. One fundamental reason is that an infinitely precise measurement of a quantity (say the temperature) would be some number T = 23.43284731..... that never, ever ends, so would require an infinite memory to store. There are other reasons such infinite precision is not possible (the finite temperature of the detector for example).
Can I be 100 % accurate with a measurement?
See my statement above where I said my height was between 1 cm and 100 m. Is this measurement accurate? I think it is. What kind of odds would I give on this measurement being true? Very very good ones. But is it "100 %" accurate? What does this statement even mean? Is this the "%" that would be implied by the odds I would offer if I was gambling on this statement? Or is it the statistical chance of me being right (so if you got millions of copies of me the % that did not make a mistake and get it wrong).
How do we know the truth anyway?
Accuracy just refers to whether a statement is true or false. To know if a statement is true you need to know not just the statement, but also the true state of the world. How do you know this? All we can really say for certain about the accuracy of a measurement is whether that measurement is consistent with other measurements or statements we believe to be true. We cannot access the objective truth directly for a comparison.
Answer: You're not right. For example, for every billiard ball sunk into the hole, the experiment result is 1. That is because

The measurable object (the ball with diameter 57 mm) is big in relation to the resolution of the measurement instrument (our eyes with a resolution of ≈ 50 μm)
The setup after some settling time is irreversible, the ball hangs firmly in the net.

But it is very easy to change the setup to get a result with less than 100% accuracy. You will be blindfolded. Or/and the room with the billiard and you will be revolved head-over.
So any experimental setup includes the measurement instrument and a stable over time final state. With which instrument for example could you measure the size of an electron? Or its position? The neutrino is not interacting - or only in very rare cases - with the electron, and the measurement of any deflection is much more difficult. Any photon interacts with the electron, changing its energy content. And the oscillating electric and magnetic field components interact with the electrons field in an unmanageable manner.
Perhaps you know, that the Copenhagen interpretation is so called because

quantum mechanics can only predict the probability distribution of a given measurement's possible results. The act of measurement affects the system, causing the set of probabilities to reduce to only one of the possible values immediately after the measurement.

Scientific progress is progressing in leaps and bounds, and with the increasing complexity of experiments, there is a certain likelihood that the boundaries of unknwledge will shift further.
Answer: That's so because I think the initial conditions and the conditions during the period the experiment is made are always changing and hence we have to take that into account.
During our experiment sometimes(most of the times), we have to take into account many real world problems which can be ignored in theory.
Moreover there are also situations when the instrument or the device limits the maximum possible accuracy achievable. In these cases we are not concerned about $100 \% $ accuracy because the result works just fine enough. However with the right tools, methods and conditions we can go arbitrarily close to $100 \% $ accuracy.
You'll notice that in the results, the numbers which aren't fully accurate or determined with certainty are usually marked with parenthesis "$()$".
However when we go down to the quantum scales which involve sub atomic particles, Heisenberg principle prohibits $100 \% $ accuracy even in theory.
We should keep that in mind :-)
Comment: When you get down to the quantum realm, 100% accuracy isn't possible even in theory - *any* measurement tool you use will change the properties of the object you're measuring. It is not possible, for example, to determine a particle's position and velocity with 100% accuracy at the same time - the more accurately you measure one, the less you know about the other.


Customise Sharepoint QuickLaunch menu per subsite page

Question: Is there a way I can have a custom quick launch menu on a particular page on a MOSS subsite.
I have tried detaching the page layout on on the page this works fine and copys the master page layout code into the aspx file. But it does not copy the code for the Quick launch menu.
I know I can change the Quick Launch items per site but can I change this per page on a site?
Thanks
Nav
Answer: Yes. For this you need to create a new master page with different Quick launch. And go to your page and change its master page.
Answer: I did this by using JQuery to hide the Quick launch panel. Rather than creating a new master page


C# / Tfs-API - Weird Parallel.Foreach error: Collection was modified; enumeration may not execute

Question: Heads-up:
I know that it is not possible to add or remove items from a collection while iterating with a foreach-loop, by using ForEeach() in List<T> or Parallel.ForEach(). That is not what I am trying to do.
What I want to do:
I want to iterate over an array of TFS-WorkItems and create a copy of each item.
The code works fine if it is not parallelized.
Whats weird about it:
An System.InvalidOperationException with the error message

Collection was modified; enumeration may not execute"

is thrown if I want to execute it in parallel.
But not always, sometimes the code executes just fine in parallel.
But I couldn't figure out a pattern...
The code:
    public void Clone(string area, string sourceIteration, string targetIteration, bool includeSubIterations)
    {
        WorkItemCollection wisToCopy = getWorkItems(area, sourceIteration, includeSubIterations);
        IEnumerable<WorkItem> wiToCopyList = (from WorkItem mItem in wisToCopy select mItem).ToList();
        internalCloning(wiToCopyList, targetIteration);
    }

    private WorkItemCollection getWorkItems(string areaPath, string iterationPath, bool inlcudeSubIterations)
    {
        if (inlcudeSubIterations)
        {
            return _wis.Query(@"SELECT [System.Id] FROM WorkItems WHERE [System.TeamProject] = '" + _prj + @"'   AND  [System.AreaPath] UNDER '" + areaPath + "' AND [System.IterationPath] UNDER '" + iterationPath + "'");
        }
        return _wis.Query(@"SELECT [System.Id] FROM WorkItems WHERE [System.TeamProject] = '" + _prj + @"'   AND  [System.AreaPath] UNDER '" + areaPath + "' AND [System.IterationPath] = '" + iterationPath + "'");
    }

    private void internalCloning(IEnumerable<WorkItem> cloneBatch, string targetIteration)
    {
        var po = new ParallelOptions();
        // if i put'1' here everything works as expected
        po.MaxDegreeOfParallelism = 4;
        Parallel.ForEach(cloneBatch, po, wi =>
        {
            WorkItem copied = wi.Copy(wi.Type, WorkItemCopyFlags.CopyFiles);
            copied.Links.Clear();
            copied.IterationPath = targetIteration;
            copied.Save();

            copied.State = wi.State;
            copied.Save();
        });
    }

As you can see, the code is pretty straight forward. I'm not even accessing the collection inside of the loop.
I've tried a lot of different things, creating a new List, using Classes of the System.Collections.Concurrent namsepace etc. But I just don't get where the List gets modified.  (Or is it even my list that throws the exception?)
I hope someone can figure this out, because it would improve the execution time drastically.
Cheers.
Edit:
As i mentioned: I already tried creating a new List before iterating. Parallel.ForEach(cloneBatch.ToList(), po, ... --> Same result.
Additionally, I've updated the code so you guys can see where the IEnumerable comes from.
Edit 2:
If i leave the copied.Save() statement out, I still geht the exception
Comment: What does `copied.Save()` do?
Comment: What class is IEnumberable ? Is it possibly self-updating in case new WorkItems get saved?
Answer: I tried using .PartialOpen() as soja noted but was unsuccessful.  However, using .Open() DID work.
List<string> projectList = ConfigurationManager.AppSettings["Projects"].Split(',').ToList();

foreach (string project in projectList)
{
    Uri uri = new Uri(ConfigurationManager.AppSettings["TFSURI"] + "//" + project);

    TfsTeamProjectCollection tfsTeamProjectCollection = new TfsTeamProjectCollection(uri);
    WorkItemStore workItemStore = tfsTeamProjectCollection.GetService<WorkItemStore>();
    WorkItemCollection workItemCollection = workItemStore.Query("SELECT * FROM WorkItems");

    Parallel.For(0, workItemCollection.Count, (i) =>
    {
       workItemCollection[i].Open();
       WorkItem.ProcessWorkItem(project, workItemCollection[i]);
    });
}

Note that WorkItem is a seperate class designed to process a single work item.
Answer: Before running the parallel operation, try iterating through cloneBatch and running the OpenPartial() method on each WorkItem. That's it.
I was having this same problem while walking through the revision history, not modifying a thing. I think that accessing the history must modify the WorkItemCollection somehow, which causes the exception. In any case, running OpenPartial on each WorkItem, even though it takes more time and uses up more memory, fixed the issue for me.