What is the shortest distance between electron and positron before they are annihilated?

Question: I just want to know how does an electron felt the presence of a positron before they are converted into energy? Also how does the electron tell if it is positron or proton if this makes any difference?
Comment: I didn't know electrons have feelings. Might as well be more cautious, not to hurt their feelings...
Comment: The question is posed in nontechnical language, which makes it unclear whether there is any well-defined sense in which it can be answered. Depending on how the question is construed, the answer might be basically "Bohr radius" of positronium, which is on the order of 0.1 nm.
Answer: In QED, the basic annihilation process is described by a $\bar{\psi }{{A}^{\mu }}{{\gamma }_{\mu }}\psi $ interaction term, in which the electron $(\psi )$, positron $(\bar{\psi })$ , and photon $(A)$  field operators act at exactly the same point.  However, there are small, higher-order corrections to this term, which arguably make it non-local.  (This is fancy technical jargon for John Rennie's caveat, when interactions become significant.)
This is the process that plays in ${{e}^{-}}{{e}^{+}}\to {{\mu }^{-}}{{\mu }^{+}}$ .  Some different happens in ${{e}^{-}}{{e}^{+}}\to \gamma \gamma $, where the incoming electron and positron are annihilated at two different points.
The idea that two particles must have zero separation to annihilate may seem surprising in classical terms, but not in quantum mechanics, where particles of definite momentum have undefined positions.
Answer: I don't think there is a simple answer to this.
The problem is that in quantum electrodynamics the particles we call electrons and positrons are the states described by the quantum field in the limit of negligible interactions i.e. when the particle is too far from any other particles for any significant interaction to occur. In this limit the particles are described by the free field states i.e. Fock states.
The problem is that when interactions become significant the states of the quantum field are perturbed away from the free field states. We can calculate this perturbation, using (unsurprisingly) perturbation theory, to calculate scattering probabilities, but we don't actually know what the states are. So in the annihilation process it isn't really the case that there is one electron and one positron present because the state of the quantum field cannot simply be described as an electron state and a positron state. If we insist on trying to describe it using the free field states then we have to conclude there are other particles present as well i.e. the virtual particles.
The point of all this is that during the annihilation process the electron-positron separation isn't well defined so it doesn't make sense to ask how close the two get before they are annihilated.
Answer: As a greasy handed experimenter I have a simple answer to this. Look through all the high-energy, exclusive data you have available on $e^+ + e^- \to 2\gamma$ (or possibly $e^+ + e^- \to \mu^+ + \mu^-$), pick the event with the largest squared four-momentum transfer $Q^2 = -t$ (or invariant mass $s$), and use that to characterize a length scale
$$ L = \frac{hc}{\sqrt{-t}} \quad\text{or}\quad \frac{hc}{\sqrt{s}}\;,$$
which you declare to be the best experimental answer to date.
I assume the winning event is to be found somewhere in the LEP II datasets, but I have no idea what the answer might be.
However, there is no theoretical upper limit on those experiment observables and therefore no theoretical lower limit on the distance (short of the Planck scale). Finding such a lower limit would mean either a discovery that the charged leptons are not fundamental or that you've hit the scale of a more fundamental underlying theory (string theory, super-symmetry, etc.)


Ruby Text Analysis

Question: Is there any Ruby gem or else for text analysis? Word frequency, pattern detection and so forth (preferably with an understanding of  french)
Answer: the generalization of word frequencies are Language Models, e.g. uni-grams (= single word frequency), bi-grams (= frequency of word pairs), tri-grams (=frequency of world triples), ..., in general: n-grams
You should look for an existing toolkit for Language Models — not a good idea to re-invent the wheel here.
There are a few standard toolkits available, e.g. from the CMU Sphinx team, and also HTK.
These toolkits are typically written in C (for speed!! because you have to process huge corpora) and generate standard output format ARPA n-gram files (those are typically a text format)
Check the following thread, which contains more details and links:
Building openears compatible language model
Once you generated your Language Model with one of these toolkits, you will need either a Ruby Gem which makes the language model accessible in Ruby, or you need to convert the ARPA format into your own format.
adi92's post lists some more Ruby NLP resources.
You can also Google for "ARPA Language Model" for more info
Last not least check Google's online N-gram tool. They built n-grams based on the books they digitized — also available in French and other languages!
Comment: Many thanks for your answer, I'll check those resources. But both of the answers tend to encourage me to manage my own routines. Maybe downsize them a bit.
Comment: to create a reliable statistic about n-grams, you will need one or many very large training corpus(es) of text data.... e.g. the colleciton of all WSJ articles for a given period of time could be such a corpus.  Processing such large amounts is very time consuming. I love Ruby in general, but for this task a dedicated C-tool is probably better suited.   Once you have the Statistics accumulated, you can use the resulting N-Gram in a Ruby Program - that's memory intensive, but not time intensive.
Comment: those training corpuses are typically domain specific.. make sure you have such text data available in large quantities, otherwise your language model will be over-fitted .. that means that it does not generalize to new data and is basically useless..
Comment: My requirements are not so harsh. I just require a basic analysis of text. I could do it like, most used word, syllabic count, statistical comparison, etc. I don't require context, semantics or anything. Just a basic understanding of vocabulary to be able to identify plurals, simple orthographic errors, words similarities and such.
Comment: And patterns. Like idioms or grouped words like "United States", "République française" (did I mention french?) You comments are very helpful, thanks.
Answer: The Mendicant Bug: NLP Resources for Ruby
contains lots of useful Ruby NLP links.
I had tried using the Ruby Linguistics stuff a long time ago, and remember having a lot of problems with it... I don't recommend jumping into that.
If most of your text analysis involves stuff like counting ngrams and naive Bayes, I recommend just doing it on your own. Ruby has pretty good basic libraries and awesome support for regexes, so this should not be that tricky, and it will be easier for you to adapt stuff to the idiosyncrasies of the problem you are trying to solve.
Like the Stanford parser gem, its possible to use Java libraries that solve your problem from within Ruby, but this can be tricky, so probably not the best way to solve a problem.
Comment: Yeah, I saw the Java thingies while searching. Looks interesting but heck, I'm a Ruby fanboy ^^ Hoped that there would be some simple stuff that would save me the time needed to develop simple analysis, statistics and so on. Thanks for your answer.
Answer: I wrote the gem words_counted for this reason. You can see a demo on rubywordcount.com. It has a lot of the analysis features you mention, and a host more. The API is well documented and can be found in the readme on Github.

Ruby Text Analysis


Compiled modules support for openWRT

Question: In order to add driver for my wireless adapter,
Bus 001 Device 002: ID 0bda:8172 Realtek Semiconductor Corp. RTL8191SU 802.11n WLAN Adapter
I am building openWRT for my GL.iNet AR150. The source I use is
gl-inet/openWRT
According to this page, my card RTL8191SU should be supported by the driver rtl8192su. However,  after I chose rtl8192su in make menuconfig and building the whole openWRT, I cannot find the module in /lib/modules. Even more confusing is that, under /lib/modules/ there are two folders, 3.18.84 and 4.9.102, while the kernel I'm running is 4.9.102.
Using find I can locate the pkg of rtl8192su but not the ko files. As a sanity check, I also looked at another driver I selected in menuconfig, rtl8187, which is found but in the /lib/modules/3.18.84 folder instead of 4.9.102 folder. Did I miss something obvious?
Thanks.
Update:
Now the adapter can be recognized.
[  545.455019] usb 1-1: new high-speed USB device number 2 using ehci-platform
[  545.651473] r8712u: register rtl8712_netdev_ops to netdev_ops
[  545.655903] usb 1-1: r8712u: USB_SPEED_HIGH with 4 endpoints
[  545.675639] usb 1-1: r8712u: Boot from EFUSE: Autoload OK
[  546.508983] usb 1-1: r8712u: CustomerID = 0x000a
[  546.512163] usb 1-1: r8712u: MAC Address from efuse = 08:10:79:56:a8:90
[  546.518852] usb 1-1: r8712u: Loading firmware from "rtlwifi/rtl8712u.bin"

but I am having trouble to connect it to a known network. After ifconfig wlan1 up I can see it's running. However, when I connect it using iwconfig wlan1 essid "_networkID_" key s:_password_ no IP is assigned. Is there anything I can try with to find out what the problem is?
Answer: You should assign an address ip to your interface and set up the default gatway:
iwconfig wlan1 essid "networkID" key s:password

or:
iwconfig wlan1 essid "networkID" key password

Then:
ifconfig wlan1 192.168.1.xxx netmask 255.255.255.0 up
route add default gw 192.168.1.1

Compiled modules support for openWRT


What kind of tire will fit a Columbia touring 26 inch bicycle

Question: I have a girls 26 inch Columbia touring bicycle, its old. It needs new tires, on the tires is the size of 26 x 1 3/8 inch. Yet, all tires at Walmart are too small to fit the rim. I even went to a 27 inch and it was way too big. My question is (and I have searched extensively and found no answers) Will a Schwinn tire fit this bike? Or, will those 700mm fit.  Thanks
Comment: Tires with the 26 x A B/C style sizing require a replacement of the [exact same size](http://sheldonbrown.com/tire-sizing.html#fractiondecimal).  Only if the size is 26 x A.B can you substitute one width for another.
Comment: In 26 inch tires with non-decimal width specifications, pretty much every width is a different rim diameter (bead circle diameter), and a different diameter requires a different rim.  A 26 x 1 is 571mm, a 26 x 1 1/4 is 597, a 26 x 1 1/2 is 584.  Check out the "Fractional Sizes" table in the link.  26 inch tires with decimal widths, on the other hand, all fit a 599mm rim.
Comment: I didn't realize that. Looking at the linked article, I don't see the discussion of why A B/C sized tires require replacement with the same width. Can you say more?
Answer: The tire brand alone won't help. Over the years bike makers and tire makers have invented an amazing range of sizes. It keeps us on our toes…
I'd guess that the tires you found at Walmart were 26 x 1.375 (or some other decimal size). You'd think that 1.375 and 1 3/8 inches would be the same. Alas, they are not. This question, Why does a decimal sized 26" tire fit my rims but a fractional sized 26" tire doesn't?, explains in more detail.
What you need to do is to find a tire with a fractional size (or look for a tire marked 35-590 or 37-590 – this is an ETRTO/ISO tire size, the numbers are the width (35 mm or 37 mm) and the "bead seat diameter" of the rim (590 mm). Most tires will be marked with this number and it will let you be sure you're getting the right size.
Your local bike shop will probably be able to get these for you, or if you'd rather order online one place that carries a good selection is Harris Cyclery. The link will take you to their selection of tires.


Tarski Fixed Point Theorem Counterexample?

Question: I have a quick question regarding Tarski's fixed point theorem. It states that for all order-preserving $f:X\to X$, the set of fixed points must be a complete lattice. I was wondering how that would apply for this function:
$$f(x) = \begin{cases}0 & x\in\left[0,\frac12\right] \\
x & x\in\left(\frac12,1\right]\end{cases}$$
so that $f:[0,1]\to[0,1]$. But the set of fixed points would be $\mathcal{F} = \{0\}\cup \left(\frac12,1\right]$. However, $\mathcal{F}$ is not a complete lattice, as the subset $\left(\frac12,1\right]$'s infimum is $\frac12$ but $\frac12\not\in\mathcal{F}$. I'm not sure what I'm misinterpreting in either the statement of the theorem, why my counter does not apply, or if I'm misunderstanding the definition of a complete lattice. For the record, $\mathcal{F}$ does have a max and a min fixed point ($0, 1$ namely), but the statement is related to complete lattices and hence my question. Thanks!
Answer: In $\{0\}\cup(\frac12,1]$, $0$ is a lower bound of $(\frac12,1]$. It is also the only (and hence greatest)  lower bound of $(\frac12,1]$ in $\{0\}\cup(\frac12,1]$.
Comment: So a complete lattice in $\mathbb{R}$ need not be subcomplete as a sublattice of $\mathbb{R}$? Thanks!
Comment: @bof Ok, so set of fixed points of $f(S)$ itself is a lattice with the order inherited from $\mathcal{P}(X)$? Also, briefly ignoring completeness, in $\mathbb{R}^n$, are there subsets of $\mathbb{R}^n$ that are lattices in its own right but not a sublattice of $\mathbb{R}^n$? Thanks so much!
Comment: @bof Also a follow up. Let $\mathcal{F}$ be the set of fixed points of $f(S)$. Then take $X',X''\in\mathcal{F}$, then their join and meet are no longer the union and intersection of them, but rather the "smallest" subspace that contains them both and the "largest" subspace that both of them contain, respectively. Is that correct?


When the aircraft tires are "worn to limits", should it be recorded as a pireps or mareps?

Question: How should be correct recorded when the aircraft tires are "worn to limits" - as a pireps or mareps (the data is reflected in the reliability report)
Comment: Welcome to Av.SE.  I'm unfamiliar with both of those terms in the context of reporting aircraft defects, and I suspect that will be true of most people here on this forum. Without more context such as what operation or organization you're referencing, this question is likely to be closed as 'unclear what you're asking.'
Answer: A "PIREP" (pilot-report) is an in-flight report of weather or other things that other aircraft should be aware of that ATC may not know.
I'm not sure what MAREP you are referring to, the only one I can find is a "marine report" which wouldn't apply here, so I assume you mean "maintenance report".
If you are talking about a commercial aircraft, this would be noted in the aircraft log and the pilot would call maintenance to have the tire changed. This may go in the "maintenance report" you are talking about.
If you are talking about private aircraft, this would be called a "Squawk" and either the pilot would remove the tire and have it changed, or the aircraft would be brought to the closest maintenance facility for service.


Does my character know he's ill while incubating a disease?

Question: My character was bitten by a Dire Rat and through accidental meta-gaming reasons, I know he's incubating Filth Fever. I want to ask the Paladin in our party to remove it before it can affect my stats, however, I don't really know if my character would know he's ill before the stat damage occurs. Is there any ruling for this?
Comment: @Feathercrown: This question is about 3.5e, not 5e. (Also, [don't answer in comments](https://rpg.meta.stackexchange.com/q/6533/33569).)
Comment: @ChumpNicholson Yes, that works in fluff, however I'd like to know if there is a ruling for that anyway.
Answer:
Disease
When a character is injured by a contaminated attack, touches an item
smeared with diseased matter, or consumes disease-tainted food or
drink, he must make an immediate Fortitude saving throw. If he
succeeds, the disease has no effect—his immune system fought off the
infection. If he fails, he takes damage after an incubation period.
Once per day afterward, he must make a successful Fortitude saving
throw to avoid repeated damage. Two successful saving throws in a row
indicate that he has fought off the disease and recovers, taking no
more damage.
These Fortitude saving throws can be rolled secretly so that the
player doesn't know whether the disease has taken hold.

Emphasis mine, that last line only makes sense if the character doesn't know.
But, of course, even if the character doesn't know that he's incubating a disease for sure, he could suspect it. Dire Rats are animals, if the character (or any member of the party that witnessed the attack) has ranks in the skill Knowledge (Nature), he could know that these creatures have an infectious bite, and that you could have been affected, and suggest healing magic just as a preventive measure.
Comment: You could guess there is a chance you have caught something but remove disease must surpass a DC check against the infection. The DM could note the success/failure of the roll but not tell you whether it was removed. Could work well for character development, is the disease really gone? Perhaps that is hanging over the character, they wake one morning after sleeping outside - is that ache in your joints just from camping and the damp air or is the infection slowly creeping up on you?
Comment: @LioElbammalf [Remove Disease](https://www.d20srd.org/srd/spells/removeDisease.htm) doesn't have to surpass any check, you may have mistook it with the [pathfinder version](https://www.d20pfsrd.com/magic/all-spells/r/remove-disease/).
Comment: you're right, I did get them mixed up, thanks for picking up on that. Would have been a good idea otherwise! :)


Why are vanilla OTC options are quoted in delta and vol?

Question: Why do we quote options in delta bid-ask & volatility bid-ask & why not it is quoted in terms of option premium?
Comment: Anyone trying to answer can use a real vanilla option quote to demonstrate...
Answer: Both the delta and volatility are a one-to-one mapping from the strike and premium respectively, so it does not harm to quote the latter in terms of the former. Now, does it any good? As options are often delta-hedged immediately, quotation in terms of delta allows to do so easily, cf. Wystup (link):

This allows market participants to ask various partners for quotes on a 25-Delta call, which is spot independent. The actual strike will be set depending on the spot if the trade is close to being finalized.

Quotation in terms of volatility allows to compare how over- or underpriced options on different strikes are, as it makes the in-the-money part of the premium be accounted for.
Fitting option pricing models to the volatilities makes more sense as well, as several stylized facts (e.g. smile, smirk etc) are only pronounced in this representation.


What does "setContentView" really do?

Question: In my main activity, I would like to have it set up, so that I first get met by a contentView just showing a background and some text. After X seconds, I want to change to my other view (GLSurfaceView).
This is obviously something I am doing completely wrong.
This is how I've imagined it could've been done (it's all in the onCreate method):
setContentView(R.layout.main);

try {
    Thread.sleep(10000);
} catch (InterruptedException e) {
}

viewer = new Viewer(this);

setContentView(viewer);

Where layout Main is what I want to show at the beginning and Viewer is my GLSurfaceView class.
What happens is that it just goes black for 10 seconds and then it starts loading the objects I've got that is shown through OpenGLES.
There's nothing wrong with the layout Main, since it works if I just erase the lines under where the Thread.sleep takes action. Though, nothing happens before the Thread.sleep is over...
With that said, my questions are following:

Why is the contentView not changing until after Thread.sleep is done?
What would be an appropiate solution to what I want to achieve?
Answer: I'm assuming this in your onCreate() and thats why you are seeing nothing.
The way I would implement this is to start a thread using AsyncTask sleep in the doInBackground and in the onPostExecute set up the new view.
Answer: Don't make sleep the main thread(UI thread).Use a threads,AsynkTask or TimerTask for that type of works instead.
Answer: You're not sleeping the UI thread in the way you think you are.
The simplest thing for what you're looking to achieve is to separate the views into separate activities and let Android handle the transition between the views. It adds another file to your codebase, but it's fairly straightforward. Let's say your initial, plain view (R.layout.main) is for a SplashActivity activity, and your post-splash view goes into PostSplashActivity. Then you could do something like this:
public class SplashActivity extends Activity {
    private static long DELAY = 10000; //milliseconds;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);

    new Handler().postDelayed(
            new Runnable() {
                @Override
                public void run() {
                    Intent postSplash = new Intent(SplashActivity.this, PostSplashActivity.class);
                    SplashActivity.this.startActivity(postSplash);
                    SplashActivity.this.finish();
                }
            }, DELAY);
    }
}

This will draw your R.layout.main layout, and then puts a startActivity call for your PostSplashActivity on the message queue and tells the queue to wait DELAY milliseconds to execute it.
Answer: It seems like you are making the main thread sleep. This may be why the code is running tell after.
Answer: It sounds like you want something like a splash screen. I like to think of these as separate to the following screen, so always use a separate activity rather than calling setContentView twice. You'd still need to sleep in a thread.
Just personal preference though...

What does "setContentView" really do?


what is difference with WCF and other web services?

Question: I'm confused with WCF and other web services (such as asp.net ASMX,.net Remoting),can anybody tell me what is difference with WCF and the others and when should I use it, thanks!
Answer: WCF is a communication stack that allows services to be exposed over HTTP (like ASMX) and TCP (like Remoting) as well as Named Pipes (which is really an intra-machine cross process call), MSMQ and with .NET 3.5 REST.
It allows this because it's decoupled the communcation parts of the service away from the business logic. All you need to do is decorate your service classes, methods and DTO's with the appropriate contract attribute ([SeriviceContract], [OperationContract] and [DataContract] respectivly.)
This had the benefit of being able to write a service once, and allowing many different kinds of clients to consume the same service (i.e. Java clients can use HTTP, .NET clients can use TCP, legacy can use MSMQ, etc.).
WCF will still allow you do use all the features of each transport, including security, transactions, reliable messaging, etc., but you need to use some care. Not all features work on all transports, and you need to design accordingly. WCF allows you to specify in your contract which features are required. This prevents somone from trying to expose your service in a way that does not support the required feature set (i.e. if your service requires transactions, the WCF runtime will not allow the service to be accessed via a basic HTTP endpoint).
WCF is also extensible via custom behaviors (which influence how the WCF runtime works) and custom channels (which control how WCF services communicate with the outside world.)
WCF has a bit of a learning curve compared to ASMX, but the benefits ABSOLUTLY out weight this learning curve.
Hope that helps.
Comment: In WCF, DataContractSerializer can access non public member so it requires full trust while XML Serializer(in web services) doesn't. This should be used with care as fully trusted code accesses all resources on your machine.
Answer: WCF is communication library that is superset of both .NET remoting and "old" ASMX web service and successor of both of these libraries.
WCF web services have much better support for WS-* standards and have less problems with interoperability.
Basically, you should use WCF since .net Remoting and ASMX might be considered legecy (and, if I recall correctly, some .NET remoting bugs were only fixed in WCF and never in .NET remoting itself)
Answer: I think it is fair to say that WCF replaces ASMX and remoting.
You can achieve everything you could with ASMX and remoting and more with WCF, but you have  a lot more capabilities and generally a lot more control over what's going on.
So I believe that, if you can, you should be using WCF.


When server is booting high load?

Question: From CSF statistic my server load in 30 days is never higher than 3, I have 8 core cpu and 3 sites with 10 000 unique visits per day.
When I reboot my server, and I log in into console, when he is loading cpanel, httpd, mysql etc I check always TOP command and server load is increasing up to 10, then he start to decrease to normal about 0.60.
Why server load is high when server is booting?
This is not the issue, just the question.
Answer: Because when your server starts, all these daemons perform their initialization and health checks. MySQL for instance might check all its tables and apache needs to load and initialize all modules.
Comment: You are right!
Thanks!
MySQL is using CPU on boot I checked now.
Answer: Load is the average number of waiting processes. When there is a lot of disk IO for example, your load will go higher because more processes are waiting their turn. Whereas when your processes are performing operations on a faster medium (RAM) the IO should be much lower and therefore the load should be lower.
When something like a database starts up it is often loading a lot of data from the disks and thus generating higher than normal load averages.
Answer: Also, since you have CSF you also have a process that gets started by CSF @ boot. The md5sum process checks the hashes of files causing a high iowait thus increasing the load of your server.

When server is booting high load?


How is the input not floating when there is a pull-down resistor?

Question: I am wondering why the input to D2 will not be floating when the switch is open. If the switch is open, wouldn't it act like a mini antenna and provide a random input to the D2 pin?
Comment: Because R1 sets the impedance to ground at D2 i.e. for all reasonable values of R1, the input can be considered to be grounded when the switch is open.
Comment: You are right it does act as an antenna to input D2, but it's almost certainly a poor one and you shouldn't worry about it. I'm sure the Arduino people have a program to measure an input, run it and see what you get. That's the point of having an Arduino.
Answer: I guess you are getting the doubt because of confusion that the moving part of the switch is opened and this is making you think that it is floating and so D2 must also be floating. No no no. The moving part of the switch and so D2 are at zero volts.
D2 is input. And GND is zero volts. There is no current through R1. So no potential difference across R1. So the voltage at D2 is also zero when the switch is opened.
In other words, Voltage at D2 = Voltage across R1 + GND voltage = 0 + 0 = 0 Volts.
Comment: So if D2 isn't connected to GND, D2 would be floating. Correct?
Comment: @Flux yes, in general, it is the case, D2 is floating. But remember that in some microcontrollers, few pins can be programmed as internally pullup or internally Pulldown mode. For eg, if the pin is programmed in internally input pulldown mode, the " D2-R1-GND" connection in your attached pic will be inside the uC. In this internally pull-down mode it will not be floating. You need to refer the datasheet of the device you have to know if it can be programmed to input pulldown mode.
Answer: The term "floating" refers to an input or tristated output that has nothing at all connected to it, so any tiny current can cause it to take on pretty much any voltage within limits.
The input is solidly connected to +5 V when the switch is closed. When it is open it is connected to 0 V through R1. If the value of R1 is appropriate for the situation there will be no problem. Note that some situations (long wires, noisy environment) may call for R1 to be quite low or for additional filtering (hardware or firmware).
One way to evaluate this is to imagine a current being injected into the input. Say it takes 1.8 V for the input to be recognized as a 1 when it should be zero. Then a current of less than 1.8 V/R1 will not cause problems. So if R1 is 4.7 kΩ, it can deal with 380 µA of current, which is quite a lot.
If R1 is 1 MΩ then 1.8 µA could cause the input to be interpreted incorrectly. As well as making it noise sensitive, it's possible that leakage on the PCB or out of the chip could contribute some or all of that current.
So why not make R1 100 ohms or something like that, you may ask. It wastes power when the switch is closed if you make the value unnecessarily low.
Usually pull-up resistors of a few kΩ to 10 kΩ are okay in most situations, but if battery power is used you may want to consider 50 kΩ or 100 kΩ. In extreme situations, perhaps with a bit of filtering, 10 MΩ may even be acceptable. For example, if you want long life in a permanently-on circuit running from a lithium button cell.
Answer: A pull-down (or pull-up) is a little like a spring on a (physical) gate.
If you push on the gate, it'll open. The spring is weak enough that you don't even have to push very hard to open it. Nonetheless, it's strong enough that as soon as you quit pushing on the gate, it'll swing closed again. It'll then stay closed until something actually pushed on the gate again.
A pull-up or pull-down acts like that spring. You want a large enough resistor that a pin can "push" it to the opposite value easily--but as soon as it quits driving (pushing) a value, the resistor ("spring") will pull it back to the value you've chosen, and it'll stay there until something drives it to the other value again. It won't just sit half-open, swinging in the breeze, so to speak.
Answer: As always in engineering (as opposed to more theoretical fields like physics and theology ;)), it depends.
Any input will have a certain leakage current. Multiply this current by a sufficiently large resistor value and the input voltage becomes large enough to be detected as a specific value which may cause problems. So, there is a tradeoff between resistor value - smaller is better in terms of fixing the open condition low, while bigger is better in terms of minimizing power dissipation, as Spehro has pointed out. Which is more important depends entirely on your situation and values.
And, yes, the loop formed by the wires and resistor can act as an antenna. For large resistors or long wires, and high-impedance inputs, pickup on external wires can be a problem. You need to look up the subject of electrical shielding. In fact, back many moons ago I worked on a system which had a switch with an incandescent bulb in it. Pushing the button would turn on a function, and this function turned on the indicator bulb. Since both the switch and bulb were in the same unit, their wiring was tightly bundled together. The switch input picked up the current spike associated with the bulb turn-on and thought the signal meant that the switch had been pushed again, so it then turned itself off. This is not how it was supposed to work.
Answer: Because D2 is connected to ground, and the input you receive is the reference voltage. If it's connection to ground was interrupted, then you would have an antenna.


The function runtime is unable to start. Microsoft.Azure.ServiceBus: Value for the connection string parameter name '@Microsoft.KeyVault

Question: I have an Azure Function with a ServiceBusTrigger in it. Now I have set the value of ServiceBusConnectionString in application settings. I am basically taking the connection string value from the Azure KeyVault. But I am always getting an error as below, whenever I run my Azure Function after I did this changes. Any help is really appreciated.
Comment: Why bother using KV if you're going to put the plaintext secret in App Settings?
The app should read the secret from KV at start.
Comment: The point is that app settings are all stored in one place and can be re-used across multiple services - I do the same myself. Once the new app settings service is out of preview then I'd agree with you, but until then this is a setup I use myself
Answer: I fount out what was causing this issue. Apparently there was a typo in the App Settings value. I was missing the SecretUri= in the value. So, instead of @Microsoft.KeyVault(Secret URI with version), we should use @Microsoft.KeyVault(SecretUri=Secret URI with version).

So after doing that changes, the error was gone. Hope it helps.


Keyframe all bones in armature at once?

Question: How do I keyframe the location, rotation and scale for all bones in an armature at once?
Answer: Select the bones you want to keyframe, then change the mode from Object Mode to Pose Mode.

Then, once you are satisfied with editing your bones, press I and then click on LocRotScale.

This will add a keyframe for the location, rotation and scaling for all selected bones. I suggest you read this.
Answer: Select all bones in pose mode,
Press I and T to insert keyframes for LocRotScale (i.e. location, rotation, scale)

Keyframe all bones in armature at once?


google pixel 2 phone rear camera not working suddenly

Question: i am using google-pixel 2 phone since  2017. phone not able to start camera app  and app getting crashed when i tried to open rear camera . front camera is working fine .i contacted google support team they said that try to clear app cache and data. i cleared camera app cache,data but no luck. how do i get my phone rear camera to working condition ? pleas help me out to get it working. thanks
Comment: I faced the same problem and I actually used beta versions just to have their latest taste. If you do the same then it's okay because of the bugs and beta versions aren't stable. If nothing helped at all then first try to delete the data from about or secodly you can restore your phone to stable version.
Comment: thanks @SurajitDas . i am also using beta versions . how can i restore to stable version?
Answer:
Before downgrade to the stable version of android you've to keepfew
things in mind.

You've to backup your data (because flashing erase your data permanently)
You've to download the stable image file of your device
You've to download Android SDK Tools (don't have to download Android Studio, Command Line Tools will be enough to get your jobs
done)

Now the process of flashing into stable version will take some time and follow the rules below.
1. Turn your developers mode on (by clicking the phone's build number
multiple tiles).
2. Enable USB debugging, unlock OEM and then connect your phone with you PC.
3. Unzip the SDK Tools and then open command promt inside that folder.
4. Unzip the image's zip file into the SDK Tools' folder.
5. Unlock your phone's bootloader, and accept the unlocking by clicking the volume up button of your phone.
6. Go to the fastboot mode and then flash all the image files using "flash-all" command.
7. After successfully flashing to the stable version don't forget to lock the bootloader (if you don't do experimental stuffs then it'll be better to keep the bootloader locked).
8. Set your phone up.
And now everything will work fine and will be stable after setting up your phone with your credentials.


Does dropout require multiple passes of the same data set, as a sort of ensemble method?

Question: I'm a bit confused about dropout -- on one tutorial, it was described as basically an 'ensemble method' of sorts. This implies that you might need to create an ensemble of networks. Is this the case, where you would need to get a consensus from the ensemble at the end? Or is dropout run over one single network?
Answer: Dropout is applied over one network. Sometimes (like with non-dropout networks) you will run your data through it multiple times before it converged and this number will be a bit higher on average with dropout but it is one network. Per layer you have a dropout probability during training and during testing/prediction you use the full network without dropout by multiplying the weights by (1-p) where p is the dropout probability. This is because during training (1-p) of the nodes are actually used.
Why this is related to ensembles is that every training instance is basically trained on a different network, by randomly dropping out nodes, forcing it to learn specific things using different nodes because it will not have all the nodes available at all times. It is not a traditional ensemble in that you combine multiple networks, just during training it acts a bit like it.
Comment: Ahhh.. So it's not that an instance is discarded at some node and is gone for good. Instances just skip certain nodes, all in one network only.
Comment: Exactly, this way all types of relevant information is captured by multiple nodes but in different ways, which enables somewhat of an ensemble-like behaviour.


Could A Planet with Fiery Rain Survive?

Question: The Idea
I'm creating a planet where rain is small drops of fire, and anything in its way will be burnt. (No idea how this happens, and the technicalities don't bother me, either)
Every night the wind blows in dust, which covers everything in a thin layer. Anything that is covered by this dust is protected from the fire rain. The dust doesn't stick to the ashes of anything that was already burnt by the rain.
Unless there's wind that moves the dust, or something becomes exposed midday that wasn't exposed overnight, I figure that most of the outdoor world - from trees to houses to roads - would be safe from the rain.
Points of Interest
My main concern is the people on this planet. If someone is walking outside and it starts raining, unless they have somewhere to run, and a way to extinguish whatever fire may have already started on their clothing from the drops, I'm not sure they have much of a chance of survival.
There is no electricity, most of the world is green, full of forests and trees, and communities all small and sparsely spread out. (Unless for some reason this wouldn't work under such circumstances)

The Question
Could people live like this? How many would still be alive, after hundreds of years on this planet? What measures could they take to protect themselves?
I am aware that this question is very general and could be taken in any direction - from living deep underground to creating shields made out of dust. More than anything else, I'm interested in hearing if life with this rain is realistic. I'm also interested in hearing ideas on how to overcome the struggle of the rain and what measures could be taken against it.
 A question that seemed similar but didn't quite answer my question for various reasons: How to survive a 24 hours of heavy lava rain?
Comment: If the trees are covered in dust, how do they photosynthesize?
Comment: Well, isn't it somewhat important? What do the people eat? Can there be any non-burrowing animals? What fumes are released by the burning droplets? Will a torrential fire-rainstorm just cook everything through radiant heat? (probably!) Can you get local oxygen depletion that can cause asphyxia? I mean, maybe you don't care about any of these things, but without knowing the list of things you don't care about, forming an answer is tricky. I'm not going to downvote or VTS, but you might wanna think about some of this stuff.
Comment: Hmmm good question. Didn't think of that... Honestly, the details of how the nature world works bother me less. I'm sort of taking the idea section as fact and trying to make sense of everything from there. Like, no idea how it works. It does, so I won't touch it. Now what about everything else?
Comment: One last thing, @User613. You've misused the [tag:reality-check] tag. It's a common failure because people don't bother reading the tag wikis and jump to conclusions about what a tag does based only on its name. It is ***NOT*** for the purpose of asking, "is X realistic?" Curiously, the shortest-and-sweetest summary of the tag comes from the [help/on-topic], "Reality Check: Provide the relevant details of your world and a situation and we will make sure it remains internally consistent." Which is not what you did. For these reasons, I removed the tag.
Comment: Looks like you're describing [Dragonriders of Pern](https://en.wikipedia.org/wiki/Dragonriders_of_Pern), by Anne McCaffrey. Although in that case, the rain is not literally made of fire, but "a mycorrhizoid spore that voraciously consumes all organic material, including humans and their crops".
Comment: I believe Iain M Banks's "Player of Games" features a planet with a weird fire-based ecology. I am not sure if it has fire rain though.
Answer: Make it so
You are unbothered by the premise that every day it rains fire.  We are encouraged to let that be.  What about the rest?
The rest is your story.  This approach reminds me of midcentury absurdism, or Kafka's Metamorphosis.  There is nothing in the story about how or why he became a big beetle.  He just did, ok.  Then the rest of the story is about what comes next.  The central premise is not examined.
In some ways this is like zombie fiction.  Zombies, yes, yes.  They do not bear rigorous examination, the zombies.  The zombie apocalypse is a contrived scenario within which one can tell human tales: suffering, triumph, good and evil.  Zombies are just context.
So too your rain of fire.  It rains fire.  Why would someone go out in it?  Did they forget?  Do they wrap their lives around the fire, like the characters work around the monsters in Bird Box?  Do they ignore the fire and then run afoul of it over and over, never learning?
Your rain of fire might be a metaphor.  There are some societal issues one cannot mention, even in a WB stack answer, because some people are wound so tight and so set in their beliefs they will immediately downvote.  But you could rename such an issue "rain of fire" and then explore it at your leisure because the zealots are concrete thinkers and they will never catch on.  Some people will, and they might think.
Comment: I can't see another answer.  Given a) magic fire rain without explanation; b) magic dust-that-protects-from-fire-rain-and-does-not-block-sunlight; and c) no actual water rain to sustain all that green life (since all rain is fire) the only conclusion is that it's all unexplained magic for authorial reasons.
Answer: Who cares if it's realistic. Reality is boring, which is why fiction exists. How could people live?

Your plants have evolved to withstand the fire. There are forests on Earth that expect periodic forest fires to cause seed germination. So you have a precedent if you're worried about "realism." If you don't like them being that fire resistant, make long-term growth (like trees) have bark that's fire resistant and have short-term growth (like flowers) have a super-fast growing period so when the fire hits, the world recovers. Their seeds use the energy from the fire to germinate.

Your animals have evolved to avoid the rain! They favor digging burrows with the animal kingdom's version of a sewage P-Trap and a secondary tunnel to keep the air from being consumed while the water burns itself out in the P-Trap. Larger animals might be less abundant, but they'll favor caves or learn to build large versions of upside-down bird nests to keep the rain out (something akin to Beaver dams). Or they have super-dense fur with a non-combustible oil that allows the water to flow off so no burn is felt by the animal. Heck, if birds can survive -20 degree winters (they're outside my window!) then it's believable they can survive the burning rain until it rolls off their penguinesque feathers.

Your humans learned how to adapt that animal fur to create (for lack of a better term) rain coats. They may look a bit like Fremen Stilsuits, but the point is to allow them to walk around in the rain. And they can always keep an eye on the animals, who would have developed a fine sense of knowing when the rain is coming. And their houses may favor sod, adobe, or stone (stone and clay shingles exist!) to keep their houses dry (if not cool).

So, I think your story is not only cool, but well within suspension-of-disbelief. And on this site, keep in mind that most of the time what you should be looking for is suspension of disbelief... not "realistic." Just my opinion.
Comment: @TheRubberDuck Using the Real World to demonstrate that an idea can achieve suspension of disbelief is the backbone of this site. When people ask for "realistic" they're almost always asking for "scientifically bullet-proof." Worse, it causes our users to get into the habit of thinking that an answer is only viable if the answer really is scientifically bullet-proof. Frankly, if the worst thing you have to complain about my answer is that I whined about the OP's use of the word "realistic," I did OK.
Comment: You start by saying "Who cares if it's realistic?", but your bullet points make an effort to provide explanations based on the real world. Maybe your point that "most of the time what you should be looking for is suspension of disbelief" should be the focus.
Comment: I just think the idea in your last paragraph would be a better opener than a closer. The comment wasn't a "complaint" that you need to defend against. Particularly, based on
"Using the Real World to demonstrate that an idea can achieve suspension of disbelief is the backbone of this site," it seems we're in agreement "that 'most of the time what you should be looking for is suspension of disbelief' should be the focus". But it's not currently the focus; it's an afterthought.
Answer: Life uh, finds a way. This planet is not Earth and it seems that this rain isn't a "just happened now" thing. If it is life is screwed, so let's assume it isn't.
If there is life on this planet at all then it has evolved to survive the fire rain - that's what evolution is, after all. I can imagine many ways this is possible. For one, perhaps life has evolved to cover itself in large sacks of water-filled vesicles that are replenished from groundwater and are sacrificial. This is a huge metabolic problem (it takes a lot of energy) so life that lives on the surface is almost surely going to be either very small or very good at pumping lots of water into itself.
Animals evolved underground and stayed there for the most part. Those that came up have ways to deal with the fire rain. Either they can very rapidly burrow when the rain comes or they themselves are able to withstand the fire. I can imagine a few ways this happens: there are so many, and they are so small, they they just die but enough get missed by the raindrops that they survive. Or they can just handle the heat - some extremophiles on Earth can do that, but I don't know about anything big. Finally, they grow scales or something that can handle the heat for some time. We have some evidence of species on Earth using metalic salts in their exoskeletons, so perhaps these creatures also evolved something similar. Perhaps they eat asbestos...
Answer: Yes, this is realistic
According to your alternate-reality tag, you have defined new climate systems. Things that don't stay still and get covered in dust, will get burnt. It sounds like you've got everything … covered?
But this is really as far as a world building consultation can take you since you are the master of this domain. You may want to continue with a new question and reality-check a solution to your world, such as the decision to exist in caves (that would be realistic, except they have no electricity so I don't know how they would function).
As a suggestion, while I am here:
Raincoats?
You have a fireproof material (the dust), so glue that onto a raincoat.
Answer: Does it ever rain water? If no, then you may need just a brief explanation for the water cycle. One possibility - evaporated water, instead of going high into the atmosphere to form clouds, stays relatively low as mist, while the higher levels are dominated by the clouds that rain fire. Fog, mist and dew could all be part of the environmental protections against everything burning to the ground, alongside the dust and some gentle evolution.
On the topic of evolution; A lot of plants and animals on earth show adaptations to wildfires, and might be fun to play around with. The cones of many pine and spruce trees, for example, are designed specifically to open up under high heat, revealing their seeds only after a fire has cleared ground for a new tree to grow. So, a lot of cones and hard nuts, a lot of birds and burrowing animals (flight and digging being common adaptations for avoiding fires) who eat cones and hard nuts and each other.
As far as human adaptations are concerned, the possibilities are endless. Trying to predict and even manipulate the weather has always been a part of human civilization, and could be neatly nestled into a world like you've described. Soothesayers and what not would try to tell when and where the rains would occur, tanners and tailors would learn how to use dust in the construction of clothes with varying success, and probably everyone would know the best ways to put out a fire.
On a larger scale, if we're talking about the move from neolithic people to early civilization, I don't think rains of fire would stop people from creating cities. It might stop them from ever getting much further along with their cities than certain developmental eras, but that could be in your favor as a storyteller.
Again, the water cycle is central. If it ever rains rain, human agriculture is probably more or less the same, though regularly interrupted. If we're talking about fog and mist instead, humanity's core crops will be affected - probably a focus on rice, or something similar? The whole idea of submerged farming, like you see in rice paddies, seems to work well with the concept of fire falling from the sky. Maybe even coastal societies that survive mostly on fish and seaweed, who I imagine are largely unaffected by the fire rain.
Either way, I think that the overall concept does feel realistic enough for an audience to go along with it. As long as we know what's happening with the water, rains of fire aren't a bridge too far.

Could A Planet with Fiery Rain Survive?


Overheated Resistor

Question: I am running a 12 volt DC 25 watt 6 ohm resistor to excite the alternator in my car but it is extremely hot to the touch, could I reduce the heat output by adding another resistor of the same value, in either series or parallel and still achieve being able to excite the alternator?
Comment: It might help if you explained why you are doing this.
Comment: don't know about exact mechanism for alternator but using N resistors of value 6N in parallel would cause each resistor to dissipate on 1/N the fraction of original heat, maybe then you can touch one of them if they are well separated?
Comment: With a regular functioning alternator/regulator, you should just attach battery voltage so the alternator can charge correctly. A resistor is only going to gum up the works.
Answer: most alternators use the small amount of residual magnetism persisting within the iron armature to excite and establish current flow in the field coils, without the need for any external excitation source. Can you explain why you are trying to excite the alternator in this way?
Also note the output of the alternator will flow backwards into and out of this excitation circuit and get the dropping resistor really hot unless you have blocking diodes in the circuit. Have you included these?
And finally- remember that the excitation only needs to be applied momentarily, as the armature is first beginning to spin, and then can be disconnected as the alternator self-excites and comes on-line. Are you doing this?
Comment: Alternators don't use "residual" energy to excite them. They use battery voltage, then have a regulator to maintain it. Generators of days gone past used permanent magnets to excite them and generate energy.
Answer: Your current setup is putting 12 volts through a 6 ohm resistor. Ohm's Law say current equals voltage drop divided by resistance. That is 2 Amps of current. Watts equals voltage (12v) times current (2A). That equals 24 watts, the rating of the resistor. To decrease heat you must reduce the voltage or reduce the current.
You can put another 6 ohm 25 watt resistor in series with the first resistor to get 12 ohms. Here is the math: Current equals voltage (12v) divided by resistance (12 ohms). That equals 1 Ampere of current. And watts equals 12 volts times 1 amp current, that is 12 watts.
You now have half the original heat dissipation, and it is spread out over 2 resistors. Wire-wound resistors can be open-wound around a hollow ceramic tube, or they can be cemented into a finned aluminum body.
If the system voltage is 20% higher, at 14.4 volts, then all the current and watt calculations would be 20% higher. So the orignal setup would draw 2.4 amps and dissipate 34.6 watts, and the dual-series resistor setup would dissipate 17.3 watts.
The Ohm's Law for watts is V-squared over R, so 14.4 volts squared over 12 ohms equals 17.3 watts.

Overheated Resistor


Fission fragments interaction

Question: In a nuclear fission reaction, do the fission fragments interact with each other or with other product neutrons and gammas? if the answer is yes then how could we determine the probability of this interactions and what are the conditions for this interactions?
Answer: This is an attempt at a short  answer to the above post, that I more than welcome a critique of, in the hope of learning more about its almost certainly simplistic assumptions.

In a nuclear fission reaction, do the fission fragments interact with each other or with other product neutrons and gammas?

Why would they not interact in all manner of ways? It's initially a small, highly energetic region, more akin to the LHC than a normal commercial, controllable "slow" nuclear fission reactor. Unfortunately I cannot tell you, (because I simply can't find) the percentages of different types or reactions between the various byproducts of the reaction.

If the answer is yes then what are the conditions for this interactions?

The conditions for these interactions is basically,  it's conversion of mass to energy, even on a relatively inefficient basis, in particle creation terms (as most fission bombs are)  but with still enough particles created to give a supercomputer a good workout.
From: Nuclear Weapon Yield

A standard bomb's energy distribution, in the "moderate" kiloton range, near sea level.
Blast 50%,
Thermal energy 35%,
Initial ionizing radiation 5%
Residual fallout radiation 10%

Please don't forget the vital role of computer modelling of such explosions, since the advent of various nuclear test ban treaties.

How could we determine the probability of this interactions

In the same way as we work out the probability of any reactions of this type,  by calculating scattering amplitudes, decay rates, branching ratios and detection of the longer lived particles. The problem is that the number of reactions involved are rather large and will contain a high percentage of possibly erroneous simplifications. Even then, I would imagine the reduced amount (both in number and complexity) of calculations  would  still occupy a lot of computer processing time and resources.

Fission fragments interaction


Row & Column Operation to Determine Rank

Question: While evaluating the rank of a matrix is it permissible to apply row and column operations simultaneously on a single matrix? Most of the books that I discussed use either row or column operation (but not both) to evaluate the rank. May I apply both on a single matrix to evaluate rank?
Comment: One may always apply a sequence of row operations and column operations of a $n \times n$ matrix $A$ to arrive at $I_r \oplus 0_t$ where $r$ is the rank of the matrix and $t$ is the dimension of its kernel. For a more in-depth explanation, see [this answer](http://math.stackexchange.com/a/332945/98077).
Answer: The rank of a matrix is invariant under application of elementary row and column operations.
So the answer is yes: Any mixture of row and column operations may be applied to determine the rank.


Which users context will SOAP apex class execute in?

Question: I have a SOAP apex class which exposes some webservice methods to be calls from my applications backend. This code has some DateTime methods that return the values based on context user's timezone settings. Now how can I know which user's context this code is executing in or if I want to change the behavior how can i change the context user for thee webservice methods ?
Answer: Ideally, you should only ever use GMT methods. Your application should convert all times to GMT, and your Apex Code should only use GMT. This removes all ambiguity, and helps make sure that you're getting the right information back regardless of the time zone settings. The context it runs as depends on the currently logged in user, which you can determine via the UserInfo methods (there's one for the time zone, if you're interested).
Comment: @y.dixit Yes, if that's the session Id you use, that'll be the time zone used.
Comment: Yes.. if I am not wrong getSessionId from UserInfo class is used to set the Sessionheader for a SOAP call. And this is what determines the login user's context for that SOAP call to resolve time related API calls.
Answer: The context user is the account under which the remote application authenticated prior to calling the web service. This is often, but not always, an "Integration" user, that's not used in the UI and may be API only.
REST services can be exposed to the unauthenticated web over a Force.com Site, but even in that case there's a context user (the site guest user).
Note that this is distinct from the question of whether or not the context user's permissions and sharing rules are being applied - there's (almost) always a context user even in system mode code.
You can obtain details of the current context user via the UserInfo class. If you need to run under a different context user, you'd need to change the authentication behavior of the remote calling application, but your code generally should not be dependent on the context time zone.
Comment: That is correct. The session id is linked to a specific authenticated user session.
Comment: That makes sense. In my application wihch is making the SOAP request I get the sessionId and add it as session header in the SOAP headers. I think this is what determines the current user context for that SOAP call.
Answer: It always runs in System Context. This is from the documentation:

Invoking a custom webservice method always uses system context. Consequently, the current user's credentials are not used, and any user who has access to these methods can use their full power, regardless of permissions, field-level security, or sharing rules. Developers who expose methods with the webservice keyword should therefore take care that they are not inadvertently exposing any sensitive data.

Additionally it mentions further on how to address this scenario:

Apex class methods that are exposed through the API with the webservice keyword don't enforce object permissions and field-level security by default. We recommend that you make use of the appropriate object or field describe result methods to check the current user's access level on the objects and fields that the webservice method is accessing.

Remember, you will always need a User mapped to a Profile to be able to call the webservices, so you can utilize UserInfo apis to make the decisions in your webservice methods.
Comment: I know that the code will run in the system context but if you are using DateTime methods for example something like this
    Datetime dt        = Datetime.NOW();
     dt.hour(); // this will return the current hour wrt the user's context that the code is running in.
Comment: So what exactly is your question? Are you saying this returns the current user datetime based on their time zone? Or it does not? And that you want it to return the current user's time? Based on what I understand this will return the current user's time. And as long as you have that information, you can always take any branching logic out of that.
Comment: It doesn't return current user's time it returns time wrt the user in whose context the code is executing. So what I am experiencing is it returns the time wrt one particular user's timezone setting. i wanted to understand how that works ?
Comment: That is what would be the expected behavior. When you say "current user", the context user which is running the webservice is basically the current user for that webservice scope.
Comment: And how is the user scope set for running the webservice. I mean because the web service is called from outside salesforce using WSDL how is the calling user set for each call ?
Comment: You will still have a **User** mapped to a **Profile** to be able to call a webservice exposed by your organization. So you can always fetch those details in your method using the [UserInfo](https://developer.salesforce.com/docs/atlas.en-us.apexcode.meta/apexcode/apex_methods_system_userinfo.htm) methods.
Comment: Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/77323/discussion-between-y-dixit-and-jayant-das).
Answer: Answers posted by sfdcfox, David and Jayant helped me explore more about the issue and understand better.
I think answer to my question was, whenever you do a SOAP call all the code runs in the system context but if your code has any part which does timezone related thing (like DatTime.hour() that is resolved wrt current user's context) then it will use current logged in user's context. Now because it is a SOAP call the logged in user is no different than the user whose username password was used to get the sessionId for that SOAP call. This session id is added as SessionHeader to the SOAP headers which decides the user context.
I hope this answer is not wrong and my understanding about the functionality is correct, if not I will be happy to pull it down from the post.
PS: answering to my own question because I did not find this information anywhere in the documents or or any post. I have formed this answer by collecting information from multiple answers and my understanding.


Does Gaara belongs to Uzumaki clan?

Question: In the series, it is said that most of Red Haired belongs to the Uzumaki clan. But I don't think that's the case for Gaara, but he got red hair. So question popped up into my mind if he's belongs to Uzumaki clan or is it something mother-father-combination thing?
Comment: @W.Are If I recall correctly, I believe the OP has it backwards anyways. Most Uzumakis have red hair, not most red heads are Uzumakis
Comment: Assuming that this statement is true: it is said that most of Red Haired belongs to the Uzumaki clan. It states 'most of the red haired' not 'all of the red haired'. Take note of that difference. Having read the manga, there never was a mention nor hint of Gaara being an Uzumaki clan member.
Comment: @Wondercricket Ah yes. my mistake :)
Answer: While there are no straight facts that confirm this or not, I can only use contextual conclusion to answer this:
No, it doesn't mean Gaara is of Uzumaki Clan because:

If Gaara was of Uzumaki Clan, it would play a big part in the story and Kishimoto would've definitely use it somehow, but neither of Gaara's parents are implied to be from that clan.
Sasori, as well as his father, are from Sunakagure and have red hair so it probably isn't that uncommon among the Sand Village people.
There are other examples of characters with red/reddish hair that are not from Uzumaki Clan, like Tayuya or Karui so there must be more red haired people in the world.

Does Gaara belongs to Uzumaki clan?


Decoupling Capacitor

Question: For power pins of IC we add a decoupling capacitor, suppose a design misses the decoupling cap. Is there anything that can be done on layout to counter the missing decoupling capacitor? My first guess is to increase the width of the power traces.
Comment: If you're modifying the layout, why don't you just add the missing capacitors?
Comment: If you insist on not modifying the circuit (with Dave and Olin I can't see why) you could add a large power/ground plane. But it would have to be VERY large! Or just add a few traces and stop-mask free copper islands that can accommodate an SMD capacitor... look boss, asolutely no circuit diagram changes! (but I'll have to talk to the boys at assembly about ... never mind, you don't want to know)
Comment: Not much can be done, I guess. But in addition to lowering the resitance of the power traces, lowering the supply's output impedance via additional/lower ESR capacitors may also help a little. - Still, you'll probably have "less than optimum" EMI characteristics from the so-created antenna loop.
Answer: If you're still in layout, then you can go back to the schematic and fix the circuit.  You may have to then move parts around to fit the new cap, depending on how tight the congestion is in that area.  Still, fixing it now is far better than applying a kludge later.
Besides, this is one of the things that should be looked for in any design review, and checking this is also something that should fall out of normal layout procedures.  When you go to lay out a part, the first thing you do is look at the power and ground connections.  If you see pins connected to power that don't have a bypass cap, you go back to the schematic and check whether these are power pins or just signal pins permanently tied high.
It's hard to imagine how a missing decoupling cap would go undetected all the way thru layout.  This therefore a non-question.  If you're thinking about decoupling caps in layout, then you'll fix problems there.  Conciously putting in a kludge around missing decoupling caps in layout makes no sense whatsoever.
If you do manage to screw up and miss a decoupling cap, the question then becomes how to do the rework.  That depends a lot on the geometry of the part.  Most of the time soldering a small leaded cap directly between the power and ground pins of the package will be the simplest fix, but this will vary case by case.
Comment: @whats: That doesn't change anything. If you think about this issue in layout, then you'll go back to the schematic to fix any missing caps you find. It makes no sense to think about this in layout, but install a kludge because the cap *might* be missing than to simply check it and make sure. Besides, adding decoupling caps is the first thing you do when connecting a IC in the schematic. If you find yourself leaving off decoupling caps, then you need to seriously fix your schematic drawing process. There is really very little excuse.  This is a really dumb problem to have.
Comment: "It's hard to imagine how a missing decoupling cap would go undetected all the way thru layout." Plenty of people have PCBs made as part of solo efforts, as non-commercial or hobbyist projects.
Comment: Having said all that, you can quite often get away with missing decoupling caps from the board without any obvious problems. The problems tend to arise when you are manufacturing batches of boards and/or when they are subject to extreme conditions. In which case you will start to see "mysterious" random problems in *some* of the boards.
Comment: I think if we use "net ties" in the schematic, it will make it easier to do layout since a signal pin tied high won't have a net name as VCC.

Decoupling Capacitor


debug a big loop in eclipse

Question: I am debugging a Python program in Eclipse.
I have a breakpoint in a big loop and I expect the problem to occur in the (1000+)th round. How can I reach that ASAP instead of clicking resume 1000 times?
Answer: You want to set a conditional breakpoint.  Read through the link I provided for an explanation and example in python.
Answer: Ctrl-R is the equivalent to placing a breakpoint on the cursor's line and clicking the "Resume" button, except that it doesn't litter your system with unnecessary breakpoints.
Is this what you are looking for?
Take a look at this: http://www.clear.rice.edu/comp310/Eclipse/debugging.html

debug a big loop in eclipse


Send ETH using ethersjs without private key

Question: Most tutorials I googled need wallet's private key to send ETH, such as this. How can we send ETH without knowing the private key using ethersjs? So, that user can just call their MetaMask and send.
Answer: You can use the Signer#sendTransaction method after getting the signer of the MetaMask provider:
async function sendEth (receiver, weiAmount) {
  const provider = new ethers.providers.Web3Provider(window.ethereum)
  const signer = provider.getSigner()
  const tx = await signer.sendTransaction({
      to: receiver,
      value: weiAmount
  })
  return tx.hash
}

await sendEth('0x12345...', ethers.utils.parseEther('0.12345'))

Answer: I am not sure I understand what you mean by user can just call their MetaMask and send.
However, in any case, it is not possible at all to send a transaction without having the private key of the sender.
Compare it to real life scenario: Would you think it is secured to be able to use a credit card without knowing its four digit code ? (yes, I am looking at you contactless payment, heresy)
Comment: Well, the user can just sign the transaction in MetaMask (or whatever wallet provider they use) when prompted, which is what was asked. You don't need to know their private key, only their wallet needs to know it...


OSX Multiple isolated desktops

Question: The question is it possible to have X virtual desktops/screens on my OSX, which would be really isolated from each other? Maybe I've very specific use case :-) but I would like to have work desktop and for instance, home. And I don't want to be notified from one to another, I would like to control it by myself. I don't really need this overlapping. Or even further. I would like to have ProjectA, ProjectB etc... with different browsers, apps etc.. I understand that it's easier maybe to have X laptops :-) but it's quite expensive, to be honest. Therefore I'm looking the software related solution.
Comment: Can't you use virtualbox or parallels or such a program to create a virtual work machine?
Comment: Any virtual machine is soooo slow :(
Answer: You can create a new user account. This will keep things separated.
When using your main "superuser" account you will have to remember to choose "install only for this user" rather than "install for all users" if you want some programs to be unavailable from your other accounts. Even if you "install for all users" this would just give your work user the ability to launch the program, like Chrome, but it would not see the history or bookmarks or extensions that you had installed as the home user.
The instructions vary slightly depending on which version of OS X you are using. In the Sierra dialog shown, you open System Preferences then Users and Groups (Accounts on previous versions) then unlock in the bottom left
then the plus sign to add a new user. Make them a regular account, not admin and they won't be able to see documents from other accounts. There - you have the separate desktops you wanted. To switch between desktops use AppleMenu - logout, then log back in as the next user. It's much faster than a shutdown or restart.  Enjoy separating your work desktop from your home desktop from your programming desktop using native OS X tools.
http://www.apple.com/us/search/new-user-account?src=serp
Comment: Thanks for the suggestion, it's an option, sure. But switching between accounts not so fast, but maybe you're right and it make sense to consider it more

OSX Multiple isolated desktops


Macbook and external hard drive : A better solution?

Question: I have an external hard drive. I use windows and macbook both. The problem is, plugging and unplugging from Mac is very much pain. If I accidentally remove it without 'ejecting', Mac doesn't recognise is next time I plug in. This is a very irritating problem and it keeps coming. Sometimes, while moving laptop here and there, connection gets loose and hard drive gets unplugged without 'safely ejecting'. Windows only warns, but recognises it every time I plug it in. Please suggest a solution. I don't wanna do all commandline stuff to make it work each time.
Answer: What do you want to use this drive for? Storing Documents, Backup, etc?
If you have a router with an usb port you could use this to set it up as a network drive that you can automatically connect when you login.
Answer: If one of your computers never moves then you can leave the usb drive connected to it and do a network share like @MSa suggested.
Here is how to share from OSX to Windows
http://www.howtogeek.com/howto/24950/how-to-share-a-folder-in-os-x-with-windows/
And the other way around
https://support.apple.com/kb/PH18709?locale=en_US


How should I repair this metal gate lock?

Question: The latch on our metal gate rusted and snapped off. I'm having trouble finding a latch similar to this one that bolts onto itself rather than being screwed/bolted into the gate. Since I cannot find a replacement latch, can I drill new holes into this gate to attach a different type of hardware, and if so, how should I do this?
Comment: It's hard to tell for sure, but that almost looks like a custom job to fit the square tube with the latch pin at an angle like that. Unfortunately the "where to find..." portion of your question is [off-topic](https://diy.stackexchange.com/help/on-topic) here. "Better off..." is a matter of opinion (and, therefore, also off-topic), and "how should I go about..." is _very_ broad and could get your question closed. Please take the [tour] and read that link to find out more, then [edit] your question to something specific and on-topic, and I'm sure you'll get a great answer or six.
Comment: Just wondering if that latch was the wrong type in the first place.  It looks like it made for bars with the corners pointing out(diamond shape) instead of being square.  Try searching for clamp on gate latches.
Comment: @FreeMan thanks for the guidance - I hope my edits made this question clearer, first time posting here.
Answer: Can you drill holes through the bars to install a different kind of latch? With the appropriate tools, absolutely.
Should you drill holes through the gate to install a different kind of latch? Well, that might be up for debate.
If you choose to do so, you're opening up the metal of the gate posts to the same rust that attacked the latch itself. Only this time, instead of being on a replaceable part, it's a part of the gate itself.
If you choose to drill, you will very definitely want to apply some serious rust protection (a good quality spray paint, applied well and in multiple coats would be a great start) to the hole. If I were to do so, I'd drill a clearance hole (big enough for the new bolt to pass through w/o touching), paint it thoroughly, then attach new hardware with a through bolt & nut on the other side. I might also consider a layer of silicone caulk between all mating surfaces in an attempt to keep water out.
Will you compromise the overall integrity of the gate by drilling one or two holes for mounting a new latch? Not unless the holes are too big.
Could you possibly compromise the long-term integrity of the gate due to rust? Possibly, but you can protect against that.
Comment: There are numerous similar "peg" latches for wooden gates.  I would think that even if a person didn't want to screw through the bars of the gate, a person could simply clamp the peg on both sides of the gate with a metal plate on the other side, perhaps using a spacer in between.


No passwords stored in Webview

Question: I have an ASUS tf101. Have rooted it and installed SQLite Editor. Have found the Webview file. There is listed some web adresses, but under passwords the fields are blank.
I desperately need one of these passwords, but why are they not listed there?
The password for the spesific site was typed in the browser just a few weeks ago.
Comment: Webview file for which app?
Answer: Just use the Forgot Password option for the website. The passwords are encrypted and you can't find them like that.

No passwords stored in Webview


Output of differentiator op-am

Question: I have struggled with differentiator opam operation
From the formula: Vout = -RC.dVin/dt
For the input signal is square wave(or triangle wave), when Vin increases, the dVin/dt should be positive, therefore Vout should be negative, but when i simulated the differentiator on Circuit Maker, it show the plot as in the picture (when Vin increases, Vout is positive).
Anyone explain this to me ?
Comment: i mean, the formula shows that the output signal should be out of phase with the input, for triangle wave, when Vin increases, dVin/dt be a constant and it's greater then 0, then Vout < 0, but when I use circuit maker, when Vin increases, Vout > 0 , that's what I dont clearly understand about
Comment: @NguyênChương The **waveforms** you have are just for a simple RC high pass filter (differentiator). The **equation** you have is for a high pass filter with an inverting opamp. See: https://electronics.stackexchange.com/questions/82314/what-is-the-advantage-of-non-inverting-op-amp-high-pass-filter
Comment: @NguyênChương Your question seems completely clear to me. You expect to see Vout = -RC*dVin/dt. Instead, simulation shows Vout = +RC*dVin/dt. You want to know why. Please post a picture of what you're simulating here, if possible.
Comment: @Annie, i added my simulated circuit and, the first plot shows the input voltage signal, and the 2nd one is shows the output voltage signal
Comment: Oh I know why, it's because I use an ideal opam there ~~ thank you all
Comment: @NguyênChương Using an ideal opamp has nothing to do, the result should have been inverted. It could be that the simulator is wrong, if that is the setup and the result.
Answer: The "in phase" relates only to sinusoidal quantities (e.g. your last row of pictures). For anything else, it's just that, a derivative, plus an RC time constant, where the case need be.
For your 1st example, the derivative of a square wave (the first half, for now) is a pulse with the value and with the duration of the slope. If the square wave is ideal, the slope is infinite, therefore the pulse is infinite -- no such thing in nature. So, the response for an actual RC differentiator to a finite slope square wave will be the voltage at the input, and then, since the square wave has reached the amplitude and it stays flat, and the derivative of a constant is zero, the output tends to go to zero, but the capacitor is still charged, so it will discharge, creating the pulse you see. If the differentiator would have been an active one (opamp), the derivative would have been a pulse with the width of the slope, and its amplitude formed by the time constant multiplied by the dV/dt.
Since the picture you show is a bit oversimplified, here are two examples of a passive and an ideal, active RC differentiator, with a \$\tau\$=0.1ms, in response to a step input with a finite slope of 1V/\$\mu\$s. The passive one:

and the active one (note the amplitude and the duration):

For the second case, you have a signal that is formed by two ramps, a positive and a negative one. The derivative of a ramp is a constant, so you get alternating positive and negative constants, or a square wave.
For none of the cases above you can talk about a phase, since the waveforms are a sum of sine and cosine waves (see this). But in the third case, it's simply the derivative of sin(x), or cos(x). Here you are dealing with harmonic waveforms, so it's safe to talk about the phase. As it happens, for the differentiator, the phase is 90o.
Also, the output is only negative, -RC*dV/dt, if you're inverting it, as in the case of an opamp.
Comment: for the second case, the derivative of a ramp is a constant, as in the picture, Vin is increasing, dVin/dt should be positive, the -RCdVin/dt should be negative constant, but it's positive in the picture, that's what i dont understand, sorry for the unclear question

Output of differentiator op-am


Magsafe light occasionally turns off while plugged in

Question: Every once in a while, the green light on my Magsafe adapter will turn off, but the computer will indicate that it's still running off of AC power. If I unplug the adapter from the computer and then plug it back in, it'll immediately light up again. Is this normal?
Comment: Have you resolved this issue ?
Answer: Here is Apple's troubleshooting documentation: Apple Portables: Troubleshooting MagSafe adapters
I have a late-2011 MacBook Air, and I have never experienced this issue.
Comment: "Another indication of possible strain relief that is not always visible to the eye is intermittent charging or strange LED behavior. You can often verify this condition by wiggling the cable back and forth near the MagSafe connector and observing whether the LED indicator (which turns amber or green) flashes on or off depending on the position of the cord." I tried that and the light did turn back on, so that sounds like it could be the problem...though the cable doesn't show any visible signs of strain.


design workflow in Eclipse for liferay

Question: I want to design workflow using Eclipse for Liferay CE.
Can anybody tell me how can i design kaleo workflow in Eclipse ?
I searched on the internet but could not find any facility for Community Edition (CE) to design our own workflow diagrammatically. Is there any reliable method other than kaleo for creating workflows for Liferay CE diagrammatically??Any help will be appreciated.
Answer: By design, Liferay does not have a built-in UI in CE for creating Workflows, and does not provide a plugin for doing so. It's one of the goodies folks get when purchasing a subscription to EE.
Liferay EE has a built-in Kaleo Designer, and Liferay Developer Studio also has a Kaleo Designer for Java plugin to give you the same functionality inside Eclipse.
None of the free, open-source options (Liferay IDE or Liferay CE) have this feature.
That said, between the information in the docs on Kaleo Designer and the docs on Designing XML workflows, you should be able to design your definitions using XML.
That's probably not entirely satisfactory, but I think it's the best you're going to manage.
Comment: @nagi That's your best bet!

design workflow in Eclipse for liferay


Why does this error occur on boot?

Question: No matter which Linux distro I try to install, I get this error:
end Kernel panic - not syncing: VFS: Unable to mount root fs on unknown-block(1,0)

I tried Debian Netinst and SliTaz on a 32-bit Tablet PC which currently has Windows 10 installed.
Thanks for your help!
Comment: Show us what you were doing: When you installed the OS, When you got the error.
Answer: As it says, it can't mount the root device.
There are many possible reasons for this. One is that you don't load the initrd that is supposed to prepare for mounting the real root.
You should also check the root parameter on the kernel command line. The 1,0 in unknown-block(1,0) indicates that /dev/ram0 should be used. So either you want to use another root or ram0 has not been prepared.

Why does this error occur on boot?


Spell school specialization dnd 4e

Question: How does the new school specialization work in dnd 4E?  Do you get more feats to choose from or does it just give a static bonus or is it something different from either of those?
Comment: are you talking about for the mage class?
Answer: School specialization is a feature of the Wizard (Mage) class that has been introduced by the D&D Essentials line. It is not naively1 supported by the other two Wizard subclasses: the  Arcanist (first one, published in Player's Handbook), and the Bladesinger (published with the recent Neverwinter Campaign Guide).
At 1st level Mages specialize in one out of the (currently) 6 schools of magic. Note that these schools don't directly relate to the 8 schools of previous editions. Mages then get school-related abilities at

1st: Apprentice Mage,
4th: Apprentice Mage in another school
5th: Expert Mage in a school you attained Apprentice Mage in
8th: Expert Mage in the other school you attained Apprentice Mage in
10th: Master Mage in a school you attained Expert Mage in
11th: Action-point-based benefit related to school you attained Master Mage in.

Many powers on the Wizard power list have now a keyword representing their school of magic. These keywords interact with the apprentice, expert ad master features.
Just for allowing a better comprehension, I've listed below the 6 schools and their corresponding apprentice feature.

Enchantment: When one of your arcane enchantment powers allows you to pull, push, or slide a creature, the maximum distance of the forced movement increases by 2 squares.
Evocation: When you roll damage for an arcane evocation power, if any of the dice come up with a result of 1, pick one of them, reroll it, and use the new result.
Illusion: When you hit a target with an arcane illusion power, the target takes a -2 penalty to the next attack roll it makes against you before the end of your next turn.
Necromancy: When you hit at least one target with an arcane necromancy attack power, you gain 2 temporary hit points.
Nethermancy (shadow magic): Creatures hit by your arcane nethermancy attack powers treat enemies more than 2 squares away from them as having partial concealment until the end of your next turn.
Pyromancy: You gain a +1 bonus to damage rolls with arcane fire attacks. This bonus increases to +2 at 11th level and +3 at 21st level. Additionally, your attacks ignore fire resistance.

School keywords are given to old and new Wizard powers in this official article.

1 The current trend from WotC is to design feats that allow characters from a subclass to get feature of another subclass. A D&D Insider article explains, for example, how to trade a Cleric (Templar) feature for the Cleric (Warpriest) Domain feature. I think something similar would be done for the Wizard subclasses too.
Comment: I think it's possible to choose one school at 1st level and a different school at 5th level, and your 11th-level feature is based on what you picked at 5th. This is based on the character builder, though, not a direct reference to the Essentials books.
Comment: I've re-read the Mage class. The 11th-level feature is based on what you choose at 10th. Also, you have a bit of flexibility in some choices. Going to edit the answer.
Answer: Basically how school specialization works is the following:
At L1 you select a school in which to specialize. You gain the benefit granted by that school to powers with that school's keyword (For instance Evocation specialization gives you brutal 1 on Evocation powers).
As you level up you gain mastery in your chosen school as well as being able to pick up another school.
There are also some feats tied to specific schools, but I'm not sure if they have specialization in the school as a pre-req.

Spell school specialization dnd 4e


What kind of tools are available to package source code

Question: I would like to create a package (devoid of SVN entries, Eclipse settings) from a project stored in SVN and managed using maven. I should be able to create a package which I can re-use as a snapshot at a latter point in time. How should I do this?
Answer: There are multiple ways you can do this. One of the simplest ways is to use scripts (shell/batch/powershell) and do an svn export. This will get a specified revision of the source code without the SVN entries, unlike svn checkout. Essentially, you will be executing the following command in your script:
<code>svn export [-r REV] URL[@PEGREV] [PATH]
</code>
You can add the commands to remove the eclipse setting in your script.
Answer: Assuming your have setup your project with a SCM, maven will ignore the .svn directories and as long as your eclipse properties files aren't within your "src" folders, you won't get those either.  Your -SNAPSHOT versions can be deployed and reused later.
If you just want to have the SVN repository "SNAPSHOT" saved for later, I would suggest just using a SVN "TAG" to save that point in time within the repository.  That can be done using the Maven SCM Plugin
At that point you can always checkout the tagged version and continue from that exact revision.
Answer: You can use the maven source plugin to do this.


Why does ionization increase from Li to Be?

Question: If ionization energy decreases from $\ce{N}$ to $\ce{O}$ due to the pairing of electrons (causing electric repulsion and greater potential energy) in the $\mathrm{2p}$ orbital in the $\ce{O}$ atom, then why doesn't ionization energy decrease from $\ce{Li}$ to $\ce{Be}$ as electrons are paired in the $\mathrm{2s}$ orbital?
Answer: The s-orbitals have a much greater degree of penetration than the p-orbitals, so they experience the nuclear charge more strongly.  This electrostatic attraction is sufficiently greater than the electrostatic repulsion between the two electrons that the ionisation energy increases.
Because the p-orbitals do not penetrate through the s-orbitals to the nucleus as much there is less electrostatic attraction.  By placing a second electron in a p-orbital the repulsion is now more significant compared to the attractive force, and the ionisation energy is less than expected.
However, each new element also increases the nuclear charge, which in turn increases the attraction that each electron has to the nucleus.  When the next p-orbital becomes doubly occupied this stronger attraction once again dominates, increasing the ionisation energy.


Can linear combinations of any Gaussian random variables be independent?

Question: Suppose that $X=[X_1\; X_2]^t$ is Gaussian vector. My question is whether $U=a_1X_1+a_2X_2$ and $V=b_1X_1+b_2X_2$, where $a_1b_2-a_2b_1\ne 0$, can be independent Gaussian random variables, if $a_1a_2b_1b_2 \neq 0$.
Comment: I believe the answer is yes.  Somebody posted an example in a question a couple days ago.  I believe it was $X,Y,Z$ are independent standard normals, then $Z-Y+X$ and $X+Y$ are independent.  You can prove it easily using the bilinearity property of covariance.  That's not exactly what you are looking for because it involves a combination of three instead of two, but I imagine an example using two is not hard to come by
Answer: If $X_1$ and $X_2$ are independent normal random variables, then at least after normalizing the random variables to have equal variance (for simplicity) you can choose any two linear combinations $a_1 X_1 + a_2 X_2$ and $b_1 X_1 + b_2 X_2$ such that $a$ and $b$ are orthogonal, i.e. $a_1b_1 + a_2b_2 = 0$, and the resulting linear combinations will be independent. You can check this by computing the covariance between the two linear combinations, noting that $X_1$ and $X_2$ are independent and have equal variance. For example, you can choose $a_1 = 1, a_2 = 1, b_1 = 1, b_2 = -1$. This satisfies your conditions on the coefficients.
Answer: Let $X=[X_1\; X_2]^t$ be a jointly Gaussian vector and $K_X=E\{XX^t\}$ be its covariance matrix.  Since any covariance matrix is a positive semi-definite, it can be decomposed as $$K_X=W\Lambda W^t,$$
where $W=[w_1|w_2]$ is a unitary matrix and $\Lambda $ is a diagonal matrix. Then, since $w_1$ and $w_2$ are orthogonal, i.e., $w_1^tw_2=0$, we can prove that $U=w_1^tX$ and  $U=w_2^tX$ are uncorrelated because
$$E\{UV\}=E\{UV^t\}=E\{w_1^tXX^tw_2\}=w_1^tE\{XX^t\}w_2=w_1^tK_Xw_2=\Lambda.$$
On the other hand, we know that if two jointly Gaussian random variables are uncorrelated they are independent. Therefore, if $X_1$ and $ X_2$ are jointly Gaussian, one can choose  $[a_1 \; a_2]=w_1^t$ and $[b_1 \; b_2]=w_2^t$. Note that the elements of $w_1$ and $w_2$ can be non-zero in general.

Therefore the answer to the question is YES!


Islamic View of Veganism

Question: Is it haram to be a vegan? I'm currently a vegetarian, simply because I don't enjoy meat as much as I did before. I don't think that we really need meat to survive, and all this mass production is just horrible! For the animals and for our body, those animals are mostly sick because of their poorly living in a way too small cage and they basically transfer their illness to us, it's just horrible. Same with the egg and dairy industry. I would enjoy those products if people weren't so cruel and kept the animals in a good way. In the dairy industry as an example they get the cow pregnant and once the cow gives birth her baby is taken away from her just to get killed. I just hate how humans use animals as a material; and like I said, I just strongly disapprove of the mass production, I want my body to be healthy and I don't think that stuffing my body with poison will do. Is my thinking wrong? And would it be okay to be vegan?
Answer: Well of course, I think most people would agree with you however(as in how animals are treated in the current day), however, you have to remember that Halal Organic meats are in-fact available where the welfare and well being of the animal is a high priority. Also consider the fact that the requirements for the meat to be Halal is a must and therefore, there is an assurance that the animal is healthy before death.
I really do recommend you watch this video, it quite informative and will pretty much answer your question: https://www.youtube.com/watch?v=WZaW_P-szCU

Islamic View of Veganism


Data sharing between multiple salesforce orgs

Question: What are the option available for data sharing between multiple salesforce orgs? How does the admin monitor two orgs ...any magic control avilable
Answer: You could also use Salesforce to Salesforce feature where you can set up a publish subscriber model and whenever you enter data into one organisation it will be shared with the other organisation . It's out of the box functionality, easy to configure. You can find more information here : https://help.salesforce.com/HTViewHelpDoc?id=business_network_intro.htm&language=en_US
Answer: I know 3 solutions which I used, might be there are another ways to do that.

Informatica Cloud. Read more about it. I guess that you can search a similar products via AppExchange.
Create your own set of scripts for migration data between org using Apex Data loader console interface, your preferable shell (cmd/powershell for Windows, bash for Mac and Linux) and Cron
Develop your own Data synchronization manager inside the Salesforce.com using Apex & web services which can be installed to your orgs via a managed package.

I do believe that if you have pretty simple data schema than the second way is preferable, but for more complicated cases you have to choose between 1st and 3rd options.
Answer: You might also want to have a look at the Cross Organizational Data Sharing pilot. However, the 'spoke' orgs will only have ready-authorization.
Answer: Bedrock Data has a pre-built Salesforce.com connector that can be used to connect multiple Salesforce orgs. Standard Salesforce objects (accounts, contacts, opportunities, leads, tasks, cases, users) can be connected and de-duplicated across org's.
Bedrock Data is also commonly used to connect Salesforce to other sales & marketing tools.  For more info see https://www.bedrockdata.com/bedrock-integrations#.featured
Comment: Are you affiliated with this product? Please take a moment to read [How to not be a spammer](https://salesforce.stackexchange.com/help/promotion).


How to maintain per-node instance state?

Question: My question is about maintaining per-node instance state without the use of Properties. I though that standard Python instance variables would work, but it looks to me like a different node instance is invoked for each execute(). I see the use of properties in the existing nodes, but is there a way that I can use Python dictionaries instance variables instead? Thanks.
Answer: Generally it is not a good idea to store the state of a node in Animation Nodes. The execution of a node should not depend on the executions before. This is also important because nearly all nodes have to work inside of loops - so they can be executed many times with different input in the same node tree execution.
Beside that I agree that it is sometimes useful to store some information (cache data etc.) which cannot be stored in Blenders properties (bpy.props). You cannot use the python dictionary of a node as you noticed. However, it is possible to create a dictionary in the module (file) that contains the node. You can use the identifier of a node (myNode.identifier) as key in the dict. Animation Nodes ensures that each identifier only exists once. The disadvantage of this approach is that the data will not be stored in the .blend file. For eg caching that is no problem though.
This is a 'problem' that is caused by Blender. Animation Nodes (or other addons) cannot change this behavior. Personally I don't care that I can't use the instance dictionary. Maybe it is even better the way it is because so we cannot confuse the data that is stored in the .blend file and other variables.
Comment: I understand about nodes being stateless, but in fact I am caching a result that I think is currently wasting many processor cycles by recreating every evaluation. As far as caching in the module, can you provide an example of one of the current nodes that does something similar? Thanks!
Comment: Sometimes you don't have to implement caching yourself. The Invoke Subprogram node has a simple caching mechanism in the advanced settings. Here is an example where I used caching (not with the identifier though): https://github.com/JacquesLucke/animation_nodes/blob/master/nodes/text/text_file_reader.py
Comment: Store the dictionary on a class variable. Duh, that was obvious. Thanks again and thanks for all your efforts. Mike