What information is there about the Doctors family

Question: I am in the process of catching up on the full set of Doctor Who series. I have started with 9 and am finishing the current series before I cycle back to the original episodes and Doctors. In "the doctors daughter" a daughter is manufactured by a machine out of cells from the doctor. In that episode he mentions that she was not his first child. Does anyone know more information about his previous families?
Answer: The Doctor's family is a complex issue. It largely depends on what you consider canon or not. The First Doctor had Susan who was refered to as his granddaughter. This implies that he had at least one child but this child is never seen nor directly referenced in the series. The expanded universe has a complicated and controversial take on Time Lord reproduction involving devices called Looms. This is covered in the novel, "Lungbarrow".
Comment: There have been a couple episodes in the new Doctor Who series where he mentions off hand that he had a family, but they were all gone now.  I don't recall him ever getting more detailed than the fact of their existence.
Answer: The Doctor's father was Gallifreyan and his mother was Human. His first wife was Gallifreyan (we assume, as Susan grew up on Gallifrey) and he had at least one child, who then had Susan. Susan married David Campbell, and they had children together. We know that before the Time War he was a father and a grandfather, but afterwards, he was neither.
Comment: Sources? Any sources?


How to ensure that a custom emacs macro operates on region?

Question: Suppose I define a custom macro in emacs in my .emacs file. Is there a way to ensure that it works only on the region I marked in a buffer before executing my command, without going into details of the macro definiton?
Answer: I don't think this is possible at the level of macros, it's going to depend on the commands inside the macro. That being said, macros have gotten more flexible since I learned Emacs, so it's possible that I'm missing something.
Emacs offers another possibility to work on part of a file: narrowing. Select the region you want to work with, and press C-x n n (narrow-to-region) to narrow the buffer to the selected region. Now the part outside the region is hidden, and all commands will affect only the visible part. Press C-x n w (widen) to go back to working on the whole file.
A feature that's often convenient together with narrowing is indirect buffers. Run M-x clone-indirect-buffer or C-x 4 c to get a second buffer editing the same file. The two buffers have the same content, and saving either saves the file, but the buffers have separate regions, modes, and visibilities, so you can use narrowing in one and keep the whole file view in the other.
Comment: @student Yes, all the narrowing-related commands are on the `C-x n` prefix.


Java Applets with GWT

Question: I read all over the internet about GWT compilation of Java code into JavaScript.
I have an old applet code that I would like to translate into JavaScript. I am kinda lost in going from the typical applet functions (init, start stop, run .. etc) to GWT entry point class. Also, my applet is actually a game that heavily uses graphic package which I am not sure how that will translate by GWT
Does anyone have any pointers or sample code to guide me through this ?
Answer: That's not going to work. GWT emulates only a very small subset of the JRE and provides a library of widgets and a RPC infrastructure for communicating with a server amongst other things. GWT is NOT a tool for porting Java applets or anything like that.
If you want to port your game to the modern web platform, I recomment using PlayN. It also has bindings for GWT.

Java Applets with GWT


What are common UDP usecases?

Question: Can anyone tell be where to use the UDP protocol except live streaming of music/video? What are default usecases for UDP?
Answer: UDP is also good for broadcast, such as service discovery - finding that newly plugged in printer.
Also of note is that broadcast is anonymous, you don't need to specify target hosts, as such it can form the foundation of a convenient plug-and-play or high-availability network.
Answer: UDP is stateless and is good for applications that have large numbers of clients connecting to a server such as time servers or DNS.  The fact that no connection has to established and maintained reduces the memory required by the server.  There is no handshaking involved and so this reduces the traffic on the network.  On the downside, if the information transferred requires multiple packets there is no transmission control to ensure that all packets arrive and in the correct order - but in games packets lost are probably better than late or disordered.
Answer: Anything else where you need performance but can survive if a packet gets lost along the way. Multiplayer games come to mind, for example.
Answer: A very common use case is DNS, since the overhead of creating a TCP connection would by far outweight the actual payload.
Additional use cases are NTP (network time service) and most video games.
Answer: I use UDP to add chat capabilities to our applications. No need to create a server. It is also useful to dispatch events to all users of our applications.
Comment: and if message is lost? chat looks as a perfect case for tcp
Comment: I've never seen a lost packet using UDP. Also, using UDP does not requires a known address. For instance, an application can listen to a certain port and in order to send a message to all users just broadcast it using UDP. We have been using this approach for many years without any problem. Mainly we use it in local networks.

What are common UDP usecases?


Will I need kanji in Japan

Question: So I'm trying to learn Japanese since there is a slim chance that I will go to Japan for a tournament next year,I want to know if it's necessary to learn kanji just for a few days in Japan??if you are a Japanese or live in Japan do you need to know kanji so you can go to stores and stuff??
Answer: You won't really need Kanji if you're only staying in Japan for a couple of days. Perhaps an online translator would be enough. But when living in Japan, I would defenitely recommend you to learn kanji. Some words are pronounced the exact same way where you would have to "guess" the meaning of it using the words around it. As a native, it's also very much easier to read long sentences with kanji rather than without them. (Kind of like reading english with no spaces or punctuation)

Will I need kanji in Japan


How can I measure amplitude of a photon and an electron wave?

Question: I want to know how is amplitude measured for a photon and an electron wave
Thanks
Comment: What do you mean by the amplitude of a photon or an electron? Are you talking about the amplitude of their wave function at some value, are you thinking of them as waves and asking about the amplitude of that wave? There isn't enough clarity to answer this question meaningfully. Voting to close for now.
Answer: In mainstream physics, both the photon and the electron are point particles. The photon does not have an amplitude. It has energy $E=hν$ where $ν$ is the frequency of the classical electromagnetic wave built up by a large number of photons.
See how in the double slit experiment one photon at a time, the points accumulating show  the interference pattern of the classical wave.

Figure 1. Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1'000, and 500'000 frames

The random points on the left are photon footprints. The interference pattern on the right analyzed will give the frequency and thus measure the energy of the photon.
The similar experiment one electron at a time, shows the footprint of the electron , and the accumulation shows that the wave nature of the electron is in the probability space, as seen by the interference pattern that appears when adding up many same momentum electrons.

Buildup of interference pattern from individual particle detections

This is the way to measure the wave nature of the probability of measuring an electron (which is a point particle as seen when falling on the slits one at a time).
Comment: Really nice answer.


Question about dynamic window approach?

Question: I have a my mobile robot and plan to use the dynamic window approach to collision avoidance. I have read the paper ,but have one inequality i can't derive it.

could you tell me? thanks!
Answer: This falls out of the basic equations of motion for a body under constant acceleration.  The simplest thing is to start with this equation:
$$
v^2 = v_o^2 + 2ad
$$
Where we are starting at some velocity $v$, and decelerating at $a$ (or $\dot{v}$ in your equation) to a stop.  (So $v_o$ = 0).  The distance traveled, $d$, must be less than the distance to the obstacle, $dist(v,w)$.
Comment: Hi, Ben. Thanks for your reply. But, how to explain the angular velocity w? Further more, once the mobile robot start to slow down, the trajectory must be changed.
Comment: The same equations apply to angular velocity.  You decelerate at some constant angular acceleration, and your angular velocity must be small enough so that you will not collide with the object before you stop.  I believe these equations assume the trajectory is constant and not changing.  They are only used for verifying a given trajectory is admissible.

Question about dynamic window approach?


Generating data structures by parsing plain text files

Question: I wrote a file parser for a game I'm writing to make it easy for myself to change various aspects of the game (things like the character/stage/collision data). For example, I might have a character class like this:
class Character
{
public:
    int x, y; // Character's location
    Character* teammate;
}

I set up my parser to read in from a file the data structure with syntax similar to C++
Character Sidekick
{
    X = 12
    Y = 0
}

Character AwesomeDude
{
    X = 10
    Y = 50
    Teammate = Sidekick
}

This will create two data structures and put them in a map<std::string, Character*>, where the key string is whatever name I gave it (in this case Sidekick and AwesomeDude). When my parser sees a pointer to a class, like the teammate pointer, it's smart enough to look up in the map to fetch the pointer to that data structure. The problem is that I can't declare Sidekick's teammate to be AwesomeDude because it hasn't been placed into the Character map yet.
I'm trying to find the best way to solve this so that I can have my data structures reference objects that haven't yet been added to the map. The two easiest solutions that I can think of are (a) add the ability to forward declare data structures or (b) have the parser read through the file twice, once to populate the map with pointers to empty data structures and a second time to go through and fill them in.
The problem with (a) is that I also can decide which constructor to call on a class, and if I forward declare something I'd have to have the constructor be apart from the rest of the data, which could be confusing. The problem with (b) is that I might want to declare Sidekick and AwesomeDude in their own files. I'd have to make my parser be able to take a list of files to read rather than just one at a time (this isn't so bad I guess, although sometimes I might want to get a list of files to read from a file). (b) also has the drawback of not being able to use data structures declared later in the constructor itself, but I don't think that's a huge deal.
Which way sounds like a better approach? Is there a third option I haven't thought of? It seems like there ought to be some clever solution to this with pointer references or binding or something... :-/ I suppose this is somewhat subjective based on what features I want to give myself, but any input is welcome.
Comment: You might want to read up on Dependency Injection; while the specifics would be different from what you're dealing with, the issues should be similar enough that you may gain some insight.
Answer: When you encounter the reference the first time, simply store it as a reference. Then, you can put the character, or the reference, or whatever on a list of "references that need to be resolved later".
When the file is done, run through those that have references and resolve them.
Comment: Typical two-phase parsing, akin to a compiler/linker system. Good, standard way to go.
Answer: Well, you asked for a third option.  You don't have to use XML, but if you follow the following structure, it would be very simple to use a SAX parser to build your data structure.
At any rate, instead of referencing a teammate, each character references a team (Blue team in this case).  This will decouple the circular reference issue.  Just make sure you list the teams before the characters.
<team>Blue</team>

<character>
    <name>Sidekick</name>
    <X>12</X>
    <Y>0</Y>
    <teamref>Blue</teamref>
</character>

<character>
    <name>Sidekick</name>
    <X>10</X>
    <Y>50</Y>
    <teamref>Blue</teamref>
</character>

Comment: This would certainly work for the example I've given, but the example was an oversimplification of what I really want to do, and there are some cases where decoupling two objects isn't really feasible. Or, at least, it would make it harder for the person making the file to be parsed.
Answer: Personally, I'd go with b).  Splitting your code into Parser and Validator classes, both operating on the same data structure.  The Parser will read and parse a file, filling the data structure and storing any object references as their textual names, leaving the real pointer null in your structure for now.
When you are finished loading the files, use the Validator class to validate and resolve any references, filling in the "real" pointers.  You will want to consider how to structure your data to make these lookups nice and fast.
Answer: Will said exactly what I was about to write. Just keep a list or something with the unsolved references.
And don't forget to throw an error if there are unsolved references once you finish reading the file =P
Answer: Instead of storing Character object in your map, store a proxy for Character. The proxy will than contain a pointer to the actual Character object when the object is loaded. The type of Character::teammate will be changed to this proxy type. When you read in a reference that is not already in your map, you create a proxy and use the proxy. When you load an character which you already have an empty proxy in the map, populate it with your newly loaded character. You may also want to add a counter to keep track of how many empty proxy you have in the map so you know when all referenced characters have been loaded.
Another layer of indirection....it always make programming easier and slower.
Answer: One option would be to reverse the obligation. The Map is responsible for filling in the reference
template<T> class SymbolMap // I never could rememeber C++ template syntax
{
   ...

   /// fill in target with thing name
   /// if no name yet, add it to the list of thing that will be name
   void Set(T& target, std::string name);

   /// define name as target
   /// go back and fill in anything that needs to be name
   void Define(T target, std::string name);

   /// make sure everything is resolved
   ~SymbolMap()
}

that won't interact well with value/moving semantics but I suspect that not much will.


Can I have 2 boatlift motors (rated for a total of 22 Amps), on a old 20 amp circuit if I replace the circuit breaker with a 30amp breaker?

Question: I replaced my 2 boatlift motors with new ones that are each rated for 15amp starting & 11amp continuos,, this takes a total of 30amp starting & 22amp continuos & after a few seconds they trip my 20amp breaker, so I was thinking of just replacing the breaker with a 30amp instead of having to replace the wiring running thru my attic all the way to my boatlift.
Comment: Can you live on the boat? Because you're going to burn your house down.
Comment: Ideas of this type is why there is an electrical code.  Plug in a toaster and kettle, trip the breaker.  Put in a bigger breaker and the house burns down.  Check what wires you do have first, 12 gauge limits you to 20 amps.  Might be lucky and have larger gauge because of the distance.
Comment: Your mentality of "just change to a bigger breaker" is dangerous. With that mentality, why even have circuit breakers at all?   That said, is the boat lift **the only thing** on the circuit and do the motors have onboard overvoltage protection?  What does the boat lift documentation say?  Can you change to smaller/the correct motors?   Do you really need to use both boat lifts at the same time?
Comment: @manassehkatz - That's what I had to do before re-wiring my dock.  Start one motor, then the other (no choice there since the switches were mounted on the covers).  They would both run for maybe 20 seconds, then the single 15A breaker would trip.
Comment: *You chose* to replace the motors with larger motors; perhaps after you replace the breaker with a larger breaker while not also using larger wire (and pulling a permit) and find out after your house burns down that choosing to do that invalidated your fire insurance coverage (it's practically arson...) you'll learn a thing or two about making choices.
Comment: Hi again, so what would be my best inexpensive way to fix this problem? Right now I have to run one boatlift motor at a time, if I run both I trip the breaker,  it would be very expensive to run heavier wires from the panel in the garage to the dock, maybe 80- 100 feet away.
Comment: I bet your motors are drawing more than 11 amps when running.  At 22 amps total draw, it should take a long time for the breaker to trip, I think.
Comment: @SteveSh If they're started close to the same time then they'll both draw 30A to start = 60A = relatively fast trip. One at a time would run OK (though still over the rated limit) and start one, wait a few seconds, start the other would *probably* squeak by, but just barely.
Comment: You think it would be expensive to run new wiring. Imagine the cost of replacing the dock and/or house due to fire!
Answer: Only if you like electrical fires.
The reason that we use breakers in the first place is to prevent wires from being dangerously overloaded.
Compare the wire for your cell phone charger to the wire for your electric oven, electric dryer, air conditioning, or other large electrical device.  Notice that the phone cord is super small and the oven cord is pretty chunky?  That extra thickness is because the oven uses a lot more power than your phone.  If you tried to splice your phone cable into your oven's wiring, and turned on the oven, the phone cable would overheat instantly and quite likely melt and catch fire.
The difference between a 20A circuit and a 30A circuit is not quite as dramatic, but the same principle still applies: to carry more current, you need thicker wires.  That's just physics, there's not much wiggle room there.  Your breaker is tripping right now to prevent a similar melt-and-catch-fire scenario from happening within the walls of your house.
There are a few possibilities that could work with the existing wire:

Can you replace your motors with 240V versions?  At 240V instead of 120V, you can get double the power for the same number of amps!  At 240V, you'd have a 4800 watt limit instead of your current 2400 watt limit, using the same 20-amp-rated wire and a double-pole 20A breaker.  You'll need to replace all of the outlets on this circuit with the proper 240V outlets as well.
Did the guy who originally ran the cable put in a thicker gauge than necessary?  (Extremely unlikely due to the cost reasons you've already discovered, but couldn't hurt to check.)  If you can post photos of the markings on your existing wire, someone can
help you identify if it's thick enough or not.  Note, however, that you must be able to determine for certain that the wire is oversized along the entire length of the run - otherwise, that one smaller section will have the same melt-and-catch-fire problem.  Note: see my edit below, there may be problems with this solution.

If neither of those options check out, then running new wire is going to be your best bet.
EDIT: A 30A circuit may be out of the question, period.  You cannot have NEMA 5-15 or 5-20 receptacles (aka "normal 120V receptacles") on a 30A circuit, it's a violation of the instructions which are part of the UL certification.  (You care about UL certification because if something goes wrong and the insurance company finds out, they might reject a claim because you used it in an illegal manner.)  If your boat lift motors are hard-wired (no receptacles involved), they may have a point in their instructions which may state that it is protected by a 20A or maybe even 15A breaker.  If that's the case, then your only option is to run a new, separate circuit for the second motor, keeping the first motor on the existing circuit.
If these motors plug into a receptacle, or if there are other receptacles on this circuit, or if the motors require a 20A or 15A breaker to protect them, then you'll need to run a new 15A or 20A circuit to one of the motors (and using the existing circuit to power the other motor).  On the bright side, the cost for 14 or 12 gauge cable for the new circuit will be less than the 10 gauge you were looking at initially.
Comment: Interesting part about 240 V motors.  Some motors, I think, allow you to change from 120 V operation to 240 V operation just by swapping some wires inside the access cover of the motor.
Answer: No, because you probably only have 12 gauge wire running from the breaker to the boat motors.  12 gauge wire needs to be protected with a 20 A (max) breaker.
I had a similar situation, but I had two circuits out to the dock.  So I was able to run another UF cable out to the motors and have each motor on it's own breaker and feed from the house.
Here's a earlier post very similar to you situation GFCI's on BLACK/RED dual 15-amp circuits.
Answer:
Right now I have to run one boatlift motor at a time,

Well yeah, because you enlarged the motors without thinking about electric service.
I really don't quite understand your situation.  Do you have two boat lifts for 2 boats, and want to run them both at the same time for some reason? Don't.
Or do you have 1 boat lift with a motor on each end of the boat? If so, hacking the machine to use a larger motor was a serious blunder. It isn't faster; motor speed is the same. The machine may not be electrically able to handle the additional power, and it could fry stuff.   If you fit larger motors because your boat is too heavy, you may be physically overloading the boat lift also.  You should really consult with the factory or an engineer on any modifications to the boat lift.

Hi again, so what would be my best inexpensive way to fix this problem?

Install the correct motors.  If you haven't thrown out the old motors, many motor shops can rebuild motors.
Every alternative solution is going to involve either installing a battery system out there, or trenching new conduit to install a new cable (I assume the original cable is direct burial, since you wouldn't be complaining if it was in conduit).
Hmm, battery.
The battery answer doesn't sound so bad costwise.  To understand why, you need to learn the difference between power (flow of energy in that moment) and energy (total work done).  Power is how fast you are using the energy.
The boat lift takes a lot of power, but only for a very short time which means it uses very little energy. Suppose it uses 2600W for 1 minute. Thats 2600 watt-minutes or 43 watt-hours. Well heck, a typical golf cart battery is 1000 watt-hours.
So you don't even need new ones, just hit a golf cart battery seller and ask for trade-ins that can't make it 18 holes anymore. Get them cheap.
A 100W battery charger will replenish the battery from 1 boat lift cycle in a half hour. Your circuit can handle 100W.
I would want 2 or preferably 4 batteries given the high surge load. 12 volts or 24 volt system.
The only trick is the AC inverter.  It will need to be quite large to start both motors. But it only needs to run when the boat is lifting.
Oh and by the way, did I mention? If you come back from boating and the AC power has failed, you can still lift your boat.
Comment: maybe swich to DC motirs instead og getting an inverter.
Comment: The battery+inverter part sounds overly complicated.  I've never lost power to run the lift in over 12 years.  If that happens, just tie the boat up to the dock.
Comment: @SteveSh Maybe, but OP's criteria don't forbid complicated and do forbid costly.  "cheaper than trenching a new circuit" is plainly stated, and I also infer "cheaper than replacing motors again".


Is "completely" in "completely equal" a redundancy?

Question: Is "completely" redundant in "Women and men are completely equal"?
Comment: It's not completely redundant.
Comment: Yeah, it's got two Ls and two Es.
Comment: We can say two things are "mostly equal" or "usually equal" or "almost equal".  To contrast "equal" with that, we may say "completely equal".
Comment: Possible duplicate of [die out completely?](https://english.stackexchange.com/questions/378720/die-out-completely). See also [Is 'very crucial' redundant?](https://english.stackexchange.com/questions/70523/is-very-crucial-redundant)
Answer: The human mind, and therefore human language, doesn't seem to like to deal with (binary) absolutes. Even though linguistically we may see many binary or absolute definitions, in usage speakers will tend to qualify even the simplest absolutes.
For example, strictly speaking, something is either wet or it is not, Yet nobody bats an eyelid at the idea of something being a bit wet or completely wet.
Some people take it to extremes. I have once had a teacher (a language teacher even) who regularly was looking for "more optimal solutions". Needles to say, another teacher (of Latin) cringed every time she heard it.
However, in general, absolutes are often not meant as absolutes. As a computer programmer I have noticed people don't always take kindly to a black/white, true/false absolute binary approach to language. If you want to get examples, try answering "yes" the next time someone asks you if you would like tea or coffee.
As for equality, mathematically it's absolute. Two things are equal, or they are not. However, in the real world mathematics holds less sway.
It is very common to compare "how equal" men and women are in different countries, for instance. I can't think of any native speaker thinking it strange when someone claims women are less equal to men in some countries than they are in others.
Because such absolutes are so often not treated as absolutes, it doesn't harm to specify an absolute when you actually mean it as such, as in completely equal.
Answer: Yes.  The word "equal" requires there to be no difference between the two things being compared.
However, it is common (especially in colloquial English) to emphasize a comparison ("exactly the same," for example) to underscore the idea that the speaker will brook no discussion about possible gray areas.

"Are they the same?"
"I'm sure they're the same."
"You're sure?  Because I'm not sure they're the same."
"Look, they're exactly the same!"

Benefitting Andrew Leach's comment: the mathematical symbol for "almost equal" is ≈.
Comment: However, since two quantities can be "almost equal", grades of equality are possible.
Comment: Not really.  Using the world of mathematics as a reference, something "almost equal" is identified using a squirlly equality symbol: **≈** (rats, I'll put that symbol in my answer so you can see it).  While colloquially we regularlly presume "almost equal" or "equal enough," from a definitional standpoint, "equal" and "almost equal" are not equal.


How do I create spiked shading in Illustrator

Question:

I am trying to create shading similar to this, where the shading is spikey, how would I go about doing that in illustrator?
Comment: There isn't really any shortcut. While using tools such as custom brushes, blends, etc. May help a little bit. In the end one merely *draws* it. Software, can't yet, replace actual artists.
Comment: You will pretty much have to draw the shapes and then fill them with the shading/ highlight colors you want- If you started with the right raster image it could be possible to create the shapes you want using Illustrators image trace (with great skill).
Comment: Just out of curiosity what do you expect the answer to be?
Answer: Yes as the comments said, the only way yet is to Draw Spikey! But in my opinion the best Tool is Pencil Tool.

Make a copy in front of your main object (here you have petals).
Then start editing them by redraw the boundaries of the object inside and shape your desired illustration (make sure "Edit selected path" is enabled in Pencil Tool Options (double click on tool icon to bring options)).
*** The step two is the key to quality, it depends on your experience on using a Mouse (like me) or a Pressure Pen.
Recolor your redrawn object lighter or darker for highlight or shadow effect.

*** For more details repeat the steps above for created new shadow or highlight objects.
Comment: I dont buy the argument that pen tool is the best tool, it is cerainly fast but nowhere as clean as the secon picture.
Comment: @joojaa If you mean pencil tool, I agree.
Comment: @Wolff right my bad
Comment: .. I would **agree** that the pencil tool **is a good tool** for this, provided it's configured well. The Pencil tool offers **a lot** more natural variety in the drawing than the other tools will, and it's faster overall. Other tools tend to lend to too much "digital uniformity" in my experience.
Answer: You really do not want especially little. The spiky patterns have 2 roles:

They present the curvature of the assumed 3D surface. To do it they must be plausible projections "like they were drawn on a curved surface"

In big picture they also present the brightness variations, the shading, as you have already said.

Image engravers met the same problems as soon as image printing plates became popular. Deciding and carving the wanted curves which fulfill both demands and were possible to make needed and still needs specially talented persons who refined their skills years. Drawing in a computer what you want is substantially easier because you can UNDO and retry one place as many times as you want. Engravers are productive only it they get it right straight in the fly.
As others have said much earlier, you must draw the patterns. Programs can give some shortcuts, but the taperings, curved forms and placements of the spikes must be decided by you. Let's try an example. I make it in Inkscape, because it has a little more flexible path drawing and editing tools (=pen, node tool) than Illustrator, at least my legacy version of Illustrator.

In the left there's an imagined curved surface. In the right I have drawn a complex path (=Bezier curve) which I believe to be good. The black spike pattern was drawn by clicking with the pen. It was like a saw blade.Straight line segments were pushed easily to curved with the node tool.
Its drawn bigger than the actual surface for good reasons:

if it happens to be otherwise good one can easily clip the extras off by taking an intersection
the extras drawn over the board will be needed later when blurring is used to check does the black shape generate a good shading in the big picture (=when one doesn't pay attention to individual spikes) The extras prevent transparency leaking from outside.

But is the curvature any good? To see better the black shape must be clipped to the size of the red surface and then one must fix the spikes, if needed.
In Inkscape one cannot adjust a shape under clipping mask. For that reason I made a group which contains only the black shape. That group got a duplicate of the red shape as a clipping mask:

The black shape can still be selected in the objects panel and it can be tweaked with the node tool. A couple of tweaks was made for less wrong curvature and to get one node on the left edge.
Does it make a good shading when watched as a part of bigger composition? It can be checked by blurring it temporarily. As said, the extras beyond the edge of the red shape prevent blurring to bring transparency from outside:

This is the effective shading caused by my black shape. The blur radius must be so high that the forms of the spikes fade and so low that no edge lightening happens over the surplus black area beyond the edge of the red shape.
If it's good, remove the blur. The complexity can be reduced by releasing the clipping mask, ungrouping and making an intersection between the red shape duplicate and the black shape:

Tweaking the curvatures of the spikes one by one is tedious. Illustrator has a way to draw arcs by clicking. It's the main content of one earlier answer (by Mr.Kerr) and its shown there perfectly. Another earlier answer used the pencil - that's a perfect method is one's hands and eyes are up to the task. Both methods need some real drawing skills when one creates a new original drawing.
Totally different way way to get curvatures easily is to use Envelope Distortion (Illustrator). Less drawing skills can be enough with it, but the ability to see the right forms is still a must.

In the top left I have a simple saw blade pattern and a rectangle.
In the top right I have use the rectangle as an envelope distortion top object. The saw blade and the rectangle are selected and Object > Envelope distort > Make with top object is applied. The same could be got by scaling and using Envelope Distort > Make with Mesh > 1 x 1.
In the bottom the corners of the generated distortion mesh are tweaked with the direct selection tool to 2 different distortion versions.
In theory one could increase the mesh complexity by adding more rows and columns or by using a more complex envelope top object. That unfortunately becomes easily unmanageable, so use simple envelopes. For me making a bigger one than 2 x 2 mesh has always been an effective way to get a reason to think again "why in the hell I didn't try this in 3D"
One can create the spike pattern curvature and the base shape curvature at the same time with envelope distortion. An example:

In the left there's a group of flat looking shapes. In the right they have got envelope distortion. I'm not claiming that a good looking flower scene where this could belong to exists, it only demonstrates what's possible.
Important to know:

The darkest red part covers actually also its both neighbors as a backround shape. That's because exactly fitting seams would leak the white background due the antialiasing and heavy bending easily doubles it due the roundings in the math.

One may wonder how in the hell could I decide beforehand the form of the pink part in the flat version so that it makes the hanging edge in the distorted version. I guessed it. To stay in truth it didn't succeed, but envelope distortion fortunately allows editing the flat shapes and seeing the distortion result in the fly. I deleted the not so well guessed flat shape and placed to the left a duplicate of the fixed version.
Comment: Using the Envelope Distortion is a great way to tweak the spikes easily and create a very geometrical shading
Answer: I'd use the Curvature Tool  for drawing zig zag lines with perfect curves.
Basically, hold down Alt for corner points, and the curves should take care of themselves.
Example

Carry on drawing all the lines, making sure to overlap the lines where they go over an edge

Once you have finished, select everything and make separate pieces using the Shape Builder tool. Hold down Alt and click to delete any overlapping lines. And finally fill the pieces of the design with the desired colour.


How to know if I can't solve an equation with "standard" methods?

Question: I'm particularly fascinated by transcendental equations whose posses closed form solutions and when I pose some of them to my friends or teachers I heard a lot of "You can't solve this in closed form" or "You can't solve this with ordinary functions" and I always thought: when and how can I be sure that this is the case ?
In other words suppose that I have an equation in one variable involving some exponentials and polynomials or even trigonometric functions etc. 99% of the time if the solution is nontrivial no one can get to an explicit formula but only to an approximation. But shouldn't we actually have to prove that ordinary functions is not enough to solve such an equation ? I mean may there is a solution which is 3 pages long but contains only common functions.
So what I want to know is:
1) If I have an equation how do I know that I can't solve it using any finite composition of polynomials, exponentials, trigonometrics and their inverses ?
2) If I'm in the case in which I can't and I decide to add one new function to the list (take Lambert's as an example) can I apply a similar reasoning to know when I can use this new function?
clarification for 2) take as an example the equation $xe^x=1$ and let the Lambert's function represent its solution: $x=W(1)$ if then I want to solve $x^2e^x=1$ I don't need to introduce another function to my list, the solution still can be expressed in terms of Lambert's function: $x=2W(\frac 12)$ but if I then approach $xe^x+x=1$ I need another function? How can I prove/disprove this?
Comment: There is a simple algorithm: try to solve it with standard methods, fail, lie down, try not to cry, cry a lot.
Comment: If you do not even have transcendental functions, only a polynomial, you still might not have a closed-form solution in radicals. The usual explanation goes something like this: "Galois theory mumble mumble." Even the criteria to tell which polynomials above degree 4 have closed-form solutions in radicals and which don't are complicated. Is the situation better if we allow a few transcendental functions to be part of "closed form" but open the possible equations to the same kinds of transcendental functions? I don't know, but it doesn't sound good.
Comment: @DavidK "Galois theory mumble mumble" is a wonderful phrase and I am pretty sure I have said it a bunch of times.
Comment: @DavidK you pretty got the point, actually I heard that there is a new method to solve equation with something called "ultra-radicals" or "hyper-radicals" sry but I don't remember the exact name.
Comment: I am not even sure whether the given problem is decideable in general.
Answer: The problem you pose stems from an implicit false dichotomy introduced from the various meanings of "solve".
Any equation can be solved, whether polynomial, linear or transcendental. The algebraic process of solving usually involves algebraic transformations which preserve or do not preserve the equivalence of the solutions to the original one. When it does not preserve it, care is taken to examine superfluous or extraneous solution.
The last step, not always but usually, involves inverting a given function. The problem is that the domain of functions which are algebraically invertible (i.e. the inverse has a closed form which allows direct calculations) is somewhat more limited than the domain of functions which are generally invertible.
Suppose, for example, I have reached the step and need to solve:
$$e^z+z^2+1/z+z^8=0$$
Solving for $z$ amounts to inverting $f(z)=e^z+z^2+1/z+z^8$. The existence of such an inverse is always guaranteed by the Lagrange Inversion Theorem, so away from critical points, there always exists an $f^{-1}(w)$, such that:
$$w=f(z)\Leftrightarrow$$
$$z=f^{-1}(w)$$
Call this $f^{-1}$ "myCrazyInverse(w)". It exists and is well defined in a neighborhood of a good solution, provided the solution is not a critical point of $f$.
This function can now be approximated, using numerical methods, provided it is fairly smooth locally and therefore the original equation can be solved to the user's $\epsilon$ satisfaction.
The inverse of this particular example either has a closed form, or it doesn't. If it does, we dump this example to the list of "alebraically" solvable examples and are done with, since the inversion can be done manually. If it doesn't, it may take a while for mathematicians to hold-on to this function in the mathematical memory of things, depending on usage.
For example, it's fairly useless to classify the inverse of this $f$ in the official archives of mathematical functions, because its use has been very rare. Have you seen it before? Chances are nil. It is a very rare kind of function (the inverse) and functions are usually classified in order of some use.
The most notable (and fairly late) example you may probably already be familiar with, is the Lambert $W$ function. Its use has become somewhat "standard" lately, not because it just "solves" the transcendental equation $z\exp(z)=w$, rather because mathematicians have encountered it often-enough in various branches of science, such as astronomy, quantum physics and optics, away from mathematics themselves, which guaranteed a fairly frequent usage.
If you didn't know its properties, it would be fairly useless, too. The following implication (as an algebraic step) is actually valid:
$$z\cdot \exp(z)=y\Leftrightarrow z=W(y)$$
but away from knowing actual ways to calculate values $W(y)$, it is as moot a step as the inversion above with "mycrazyinverse".
It just so happens that this function is of use in many fields, hence the step of the inversion of $z\exp(z)=y$ has been included in the allowed algebraic transformations which solve all the related transcendental equations and provide a "closed form" solution.
Even under the above notes, many mathematicians still doubt the inclusion of this inversion step, given the multivalued nature of (most crazy) inverses and I actually sympathize with them. In view of the multivalued nature of some maps, it can be argued that such steps are valid algebraically only under specific qualifications, which preclude semantic mathematical nonsense, such as:
$$z\cdot\exp(z)=y\Leftrightarrow z=W(y)$$
The correct step is qualified by the branch index only and denotes an equivalence of sets:
$$z\cdot\exp(z)=y\Leftrightarrow z=W(k,y),\,\,\,k\in\mathbb{Z}$$
You can wittness the resulting confusion in answers which use the Lambert $W$ function to solve equations, as a wrong answer: $x=W(whatever)$. Whenever this function solves an equation, it always specifies infinitely many solutions, whether real or complex, because of the branch index. It's never a single value, whether the teacher wants only real solutions or not.
Addednum (for clarification for #2)
For some reason I didn't see that last question, so I am adding a reply here.
The Lambert $W$ function enjoys certain algebraic properties, which allow it to solve many transcendental equations, such as close variants of the original $z\cdot \exp(z)=y$, like $z^3\cdot \exp(z)=y$.
All these are exactly the equations which through some valid algebraic manipulations can be eventually be brought to the form:
$$f(z)\cdot\exp(f(z))=y\,\,\,\,(1)$$
for some $f$ and some $y$, since the solutions can then be described as:
$$f_k(z)=W(k,y)\Leftrightarrow z_k=f^{-1}(W(k,y)),\,\,\,\,k\in\mathbb{Z}$$
Maple contains an algorithm which does exactly that in each given case, but the algorithm's success alone is not a proof, because the algorithm has to stop sometime.
Fortunately, correct algebraic syntax makes for a solvable algebra/group (by definition), so you can be sure that if some equation can be factored into form (1), it will be factored in $O(g(n))$ time, where the latter is the factoring time of the algorithm and this gives you deterministic certainty with a very small fail rate $\epsilon(g(n))>0$.
Now push $\epsilon\to 0$ and you've proved it.
Answer: "Closed form" means expressions of allowed functions (Wikipedia: Closed-form expression). If an equation is solvable in closed form depends therefore on the functions you allow.
A general method for solving a given equation $H(x)=0$ is to apply the compositional inverse $H^{-1}$ of $H$: $x=H^{-1}(0)$. In general, $H$ and $H^{-1}$ are correspondences. But often it is possible to split the problem into subproblems where $H$ and $H^{-1}$ are functions. For applying this method, $H$ and $H^{-1}$ have to be known. That means they have to be in closed form.
For the elementary functions, there is a structure theorem that can help to decide if a given kind of equations of elementary functions can have solutions that can be expressed as function value of an elementary function.
The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in [Ritt 1925] Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 answers which kinds of Elementary functions can have an inverse which is an Elementary function.
Risch also proved this theorem in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J. Math 101 (1979) (4) 743-759.
You can also take the method of
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.
Two methods for simpler transcendental elementary equations are given in [Lin 1983] Ferng-Ching Lin: Schanuel's Conjecture Implies Ritt's Conjectures. Chin. J. Math. 11 (1983) (1) 41-50 and in [Chow 1999] Chow, T.: What is a closed-form number. Am. Math. Monthly 106 (1999) (5) 440-448.
If a given function can be decomposed into compositions of algebraic functions and other known Standard functions than $\exp$ and $\ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope to prove such a generalization of Ritt's theorem for this class of functions.
If you can find a representation of your given function $H$ that is a composition of functions that doesn't contain a multivariate algebraic function and if the inverses of all member functions of this composition representation are known, you can apply the single inverses in opposite order and gets the whole inverse in closed form.
For equations solvable by Lambert W, see my answer here: Algebraic solution to natural logarithm equations like $1-x+x\ln(-x)=0$ and my comment here: https://mathoverflow.net/questions/289248/what-are-the-most-general-methods-for-solving-equations-in-closed-form-with-lamb.


Magnetic Flux of a lone solenoid and coaxial solenoids

Question: If you run a current through a solenoid, does it have any magnetic flux on its own of does it need an externally produced magnetic field for there to be any net flux through the solenoid? The physics problem I am doing consists of 2 coaxial solenoids. One is inside of the other. When calculating the flux through the inner solenoid should I only account for the magnetic field produced by the outer solenoid?
Comment: When you run a current through a solenoid, it generates magnetic flux on its own.
Answer: So the inner solenoid has no current thru it?
You also need to describe the relative juxtaposition of the two windings.  Does the outer winding completely cover the inner?


Computer won't connect to network after coming out of suspend

Question: So I am new to Linux and I am not very good at CLI and much prefer gui instead. But I have 3 issues that might all be stemming from my suspend mode. When I close the lid of my laptop it goes into suspend mode. When I resume work I usually cannot get internet (wired). I usually have to restart to get a connection. Also lately I will lose my connection randomly while using my computer ( this has just started with in the last 2 days). The other is Cinnamon usually crashes after I resume from suspend. Any ideas? I have been trying to scour the internet with no avail. Thank you.
Answer: Try running this code after coming out of suspend mode for resuming the network services:
<code>sudo service network-manager restart
</code>


What are the best ways to maximize int score?

Question: I am playing a level 10 alchemist, and because alchemist needs as much as he can for int (bombs per day, bomb's damage, craft alchemy, etc.), i want to maximize it. I need the cheapest items i can get, as i almost have no money, as well i need it to well. Any extracts that can do that is great too. As well feats and discoveries.
Comment: Please edit the question to say exactly what you're asking: best, cheapest, or both (if both, with what priority, or with what constraints); and what important details of your situation are. Currently the question is too broad. It may not be too broad with more details, but we can't keep guessing at the details.
Comment: A good question would show your research.  I assume you know about some sources of INT bonuses, right?  Please list them so people aren't doing busy-work.
Comment: So do you want to know the *best* ways, or the *cheapest* ways? These are probably not the same.
Comment: My only problem is that our dm gave us less then 200 gp which is bad for level 10 so i tried to tel him this ia nothing but he didin't listen. I am not sure if we will earn money, but i'll try to convince him for more money
Comment: Kind of. I am an elf so i have currently 20 at int. Is there any cheap item/extraxt that can help me? I know there is contagen, but it lowers str to high int and my str is 12 so..
Answer: Headbands of Vast Intelligence, Tomes of Clear thought, and Races that grant a bonus to intelligence are the most common boosts to intelligence.
The headbands cost 36,000gp for +6 Enhancement Bonus to intelligence, but only 18,000 if you happen to have someone in your group with access to Craft Wondrous item and Fox's Cunning. You don't need access to Fox's Cunning if you take a +5 Penalty to the crafting DC. Tomes of Clear thought are exorbitantly expensive 137,500gp for a +5 Inherent Bonus.
Races that get a +2 Racial bonus to intelligence are:

Human
Half-Elf
Half-Orc
Elves
Drow
Ratfolk
Samsarans
Sylph
Wayangs
Blues (From Psionics Unleashed)
Answer: There are a number of methods, many already listed, to boost Intelligence.
Here are a few, some that are extreme longshots(As in GM FIAT only with no way for PCs to normally obtain):

Headband of Vast Intelligence  +2/+4/+6(or variants such as Headbands of Aerial Agility, Mental Prowess, Mental Resilience, Mental Superiority, Ponderous Recollection, Shifter's) Type: Enhancement

Tome of Clear Thought 1-5 Type: Inherent

Ioun Stone(Scarlet and blue) Type: Enhancement

Stat Increase at levels 4, 8, 12, 16 and 20 Type: Untyped

Spells/Extracts are useful for damage and in the form of Crafter's Fortune Craft Alchemy checks

Racial Bonuses, See Sandwich's answer.

Template adjustments such as Advanced, Half-Celestial/Dragon/Fiend, Fey Creature, Foo Creature, Nightmare Creature, Animal Lord Type: Untyped

Racial adjustments, such as finding a Vampire to turn you. A jiang-shi is superior to a western vampire for Intelligence boosting purposes. Type: Untyped

Summoning either a Contract Devil(and making a contract with such) or a Succubus and convincing them to grant you their Profane Gift Type: Variable(At GMs discretion for Contract, of Profane for Succubus)

Leadership feat to get an item cohort take the perfection advancements for inherent up to +5 INT
Answer: The standard intelligence-boosting items and methods are as follows:

Headband of Vast Intelligence (+6 Enhancement Bonus)
Tome of Clear Thought (+5 Inherent Bonus)
Stat increase at level 4 and every 4 levels thereafter (+5 Untyped Bonus)

Additionally, as an Alchemist, you can either take the Cognatogen discovery or the Mindchemist archetype to get an ability that essentially is like Mutagen for mental stats. Through this ability, you can get up to a +8 alchemical bonus to your Intelligence for 10 minutes per Alchemist level.
Please see @Sandwich's answer to get a detailed like of Int-boosting races.
Answer: 1) The custom magic items price guidelines are explicitly to be followed exactly when crafting attribute boosting ion stones, even by a very conservative reading. The formula is

(bonus squared × 1,000 gp × 2)

which makes your attribute scores, including int, limited only by your access to gp.  As a member of a level 10 party, you should have access to vast quantities of gold via things like Fabricate, scrolls of Polymorph Any Object, etc and the exact amount will depend on if you have a rogue in the party and, if not, how much time your wizard is willing to spend making you gp.  Based on experience, you'll probably end up limited more by time than gold, here, unless you have a rogue in which case the bonus is essentially arbitrary.
Unfortunately, in WBL (wealth-by-level) games all your 62000 gp would only net you +5 via this method (+6 if you took any crafting feats), which is very lackluster for the investment.
2) In a WBL game, the Headband of Intellect provides one of the cheapest sources of long-term int boosting, but caps out at +6, while the Crimson Sphere Ion Stone caps out at +12 if the second method of Resonant Power determining is used, but is much more expensive.  A flawed Scarlet and Blue Sphere Ion Stone with the right resonant power will give you +4 Int but -2 Con for 6500, which includes the cost of a wayfinder. A Dull Gray Stone with the right resonant power can grant +2 Int for 525 gp. This last option provides the very best gp/bonus return available, but is generally insufficient for a 10th level character.  The +4 int -2 con is probably what you want at level 10, especially since you have lots of access to physical buffs as an alchemist.
This answer assumes you already have a 10th level character and are trying to optimize for that character, and neglects both the cost of purchasing the right resonant power (~0-1250 gp for a dull gray, ~0-300000gp for a scarlet and blue sphere) and the potential for multiple resonant powers on a singe stone (technically you can get +8 int from a roll of 93-04-04-04 on a Scarlet and Blue Sphere, with a 1:100000000 chance if truly random).
Answer: Two so-far unmentioned methods:
Ageing effects, middle aged or higher can get you +1 to Int. (Along with the other mental stats, at a cost of physical stats)
Also technically levels Dragon Disciple would increase it, but it's a terrible return on investment! (i.e. for the sake of the +2 intelligence boost alone)
The Alchemist's Cognatogen is probably one of the best returns on investments though when your budget is 200gp.


Do the photons emitted along with the gravitational waves take more time to reach earth than the ones emitted after them?

Question: I'm puzzled because the gravitational waves do warp the spacetime locally. They do it continuously as they propagate. So the photons travelling in tandem with these waves must be constantly following a curved path and hence must be travelling more distance to reach earth than 'normal' photons from the same source that travel without these disturbances. So do they take more time to reach earth?
Answer: In vacuum, that is, in absence of matter and electromagnetic interactions, photons and gravity waves follow the same geodesics (if gravitons are actually massless... Let us assume it for we have no strong evidences for the contrary). In their propagation photons are not affected by the ripples and the stretches in space-time caused by the waves, because they are travelling along with them. There are no front crossings.
You can figure this out looking at the photons as a surfer riding a wave; if he moves along with the wave, that is, he's at rest with respect to the wave, and he doesn't perform any trick or strange path, the space he covers is exactly the same as if the sea were calm. He doesn't measure the slope and the curves in the surface of the water, which would constitute and increment in the path length respect to when the sea is calm.
In conclusion, neglecting the electromagnetic interaction of the photons with the interstellar medium, photons travelling in tandem with the gravity waves, and those leaving later from the same source, take exactly the same time to reach us.
EDIT: consider also that the stretching effects of gravitational waves are only transverse (at least in General Relativity) to the propagation direction of the waves. Thus, a light ray travelling in their same direction, won't be effected by any effect of dilatation/compression of the space-time. This is indeed the working principle of laser interferometric detectors: measure the interference of two orthogonal light beams.
Comment: Re "if gravitons are actually massless": one of the results from the LIGO detection is a very strict upper limit to the mass of the graviton ($< 10^{-22}$ eV), actually more stringent than the [current limit](http://pdg.lbl.gov/2015/tables/rpp2015-sum-gauge-higgs-bosons.pdf) on the mass of the photon.
Comment: Thanks for answering. However I do not understand both points. I see what you mean here. But I don't see how a travelling spacetime ripple is different from a bump in spacetime formed by a star. In the second case it seems to curve the spacetime. So do you mean to say that light travelling along this curved path really doesn't travel more distance?
Comment: Regarding the transverse aspect, do you mean to say that if the wave is coming from south to north, then instead of the possibility that the space is compressed in north-south direction, it is the possibility that the g.wave rotates the light beam travelling in the east-west arm to make it out of sync with the other one's polarization to produce the interference? The description given here (https://www.ligo.caltech.edu/page/what-is-interferometer) doesn't mention about this transverse polarization aspect. Did they make it too sugary for public?
Comment: Hi! As regards the first question, you see (high frequency) ripples in the space-time fabric because you are at rest with the source and the waves are travelling towards you at the speed of light. Instead, light rays are travelling in tandem with the gravity waves, so they are at rest with them and they  don't see the effects of such ripples. I think that the surfer analogy could help. According to his point of view, at rest with the wave, he is on a slope (the front of the wave) and he doesn't experience the waving movement that would experience a boat bound at the harbour.
Comment: And the LIGO link you pointed to does in fact tell the same as regards the transverse deformations the waves produce. In the picture at the top of the page ('Basic schematic...') the wave is moving from top to bottom in a direction perpendicular to the plane of the detector. Imagine such plane defined by the two arms of the interferometer, and name them $x$ and $y$. Then the wave that moves towards with the $z$ direction produce ripples in the space(-time) of the ($x,y$) plane, that is the one of the detector.
Comment: I think that the best way to grasp this aspects is looking at the huge amount of videos have been released on YouTube in these days. ;)
Answer: You are right - it would take longer for light (and gravity and anything else) to make the trip during the event than after. But the effect is tiny, as gravitational waves shrink and expand space in virtually equal amounts - and in the linear regime - which the waves are in for virtually the whole trip, the compression and expansion is perpendicular to the direction of motion.
It is only because of the non - linear behaviour of general relativity that there is a very slight overall increase in time. I don't know what the amount of time would be, but something like 99% of the extra delay would occur in say the first million km from the event. So even if the effect was say 1% over the first million km, that would amount to only an extra (light takes 3 seconds to go a million km) 0.03 seconds. Then virtually nothing for the next billion years of travel. Those numbers are for illustration only - I'm sure someone has done the calculation somewhere.
After a million km of travel, the gravitational waves are well into the linear regime and the effect turns from a small effect into a completely ignorable tiny amount.
So 1.3 billion years for light travel without the event, and 1.3billion years + 0.03 seconds for light travelling in the event.
Please note that the 0.03 seconds is nothing more than a complete guess, I would like to see the real number.
Gravitational waves will only look like the neat linear fully transverse waves as one gets out of the intensive emission region.


Enable VT-x in your BIOS security settings (refer to documentation for your computer)

Question: While I was adding a virtual device in my Android Studio, Android Studio is Showing This Error. See This Image:

.
Answer:
shutdown you pc and open bios settings, and enable Virtual Technology-x option and restart your pc.
done.
Comment: How do I open Bios Settings?
Comment: it depend on system.
Comment: which laptop you have? or if you have desktop then which motherboard do u have
Comment: I got these Setting Thanks
Comment: press escape while starting for hp laptop
Answer: In Short -> You must enable VT-x Technology in your BIOS.
Here are the detailed steps:
1- Restore Optimized Defaults (Not Necessary)//Steps to start BIOS
Its better to restore Optimized Defaults before, But following steps are not necessary:

Reboot the computer and open the system's BIOS menu. This can usually be done by pressing the delete key, the F1 key or Alt and F4 keys depending on the system.
Select Restore Defaults or Restore Optimized Defaults, and then select Save & Exit.

2- Enable VT-x Technology in BIOS (Necessary)

Power on/Reboot the machine and open the BIOS (as per Step 1).
Open the Processor submenu The processor settings menu may be hidden in the Chipset, Advanced CPU Configuration or Northbridge.
Enable Intel Virtualization Technology (also known as Intel VT-x) or AMD-V depending on the brand of the processor. The virtualization extensions may be labelled Virtualization Extensions, Vanderpool or various other names depending on the OEM and system BIOS.
Select Save & Exit.

Note: Many of the steps above may vary depending on your motherboard, processor type, chipset and OEM. Refer to your system's accompanying documentation for the correct information on configuring your system.
Test:
Run cat /proc/cpuinfo | grep vmx svm. If the command outputs, the virtualization extensions are now enabled. If there is no output your system may not have the virtualization extensions or the correct BIOS setting enabled.
Detailed instructions can be found Here
Comment: How can I do that in wondows 10
Comment: Thanks Milad But I Cannot vote as I Have less Reputation
Answer: Even i have enable VT-X tech and disabled secure boot. My Android Studio failed to load emulator saying dev/kvm not found.
After doing some research on this. Finally i solved it out.
This issue came after updating HAXM. I found some useful answers. which tell this issue is in HAXM 7.2.0. See this issue on github
Steps to solve:

Uninstall Haxm from SDK manager.
Download previous version of HAXM v7.1.0 from this release page.
Install this HAXM.

Now everything should work fine as before.
Comment: I installed HAXM v7.6.5 directly from the link above and fixed the problem as well.
Comment: I had VT-X enabled too, but I'd still keep getting OP's error. This is the solution that worked for me.
Answer: It was because of docker which uses hyper-v. You just have to remove hyper-v from windows features.
Answer: No restart, no BIOS needed in my case. I have downloaded and installed the latest version of HAXM from the releases: HAXM releases
Answer: Even after enabling Intel Virtualization Technology (Intel VT-x), I got the error message ' dev/kvm not found ' in Android Studio.by making Hyper-V feature off in Windows 8.1 issue was fixed.
here is how to access Windows Hyper-V feature
Control Panel -> Programs and Features -> Turn Windows features on and off.
Answer: Just to make it simple:
The solution is just to enable vt-x or Virtualization Technology in bios, which is under Advanced Tab.  and once it's enabled, the error disappears.
FYI
I had similar issues starting up my Android emulators in Appium studio for my mobile testing, and on top, I had latest bios, which looked so different to the standard one.
So attaching screenshot of my computer bios, but the option should be there on any Bios settings.  Just need to boot the computer, and push Esc or some function key to see the computer bios, and then find the correct option to enable it under Advanced Tab, (most importantly, you may have to scroll down as the option would be down the list)
I left my Hyper-V feature as is, which was enabled though.
Answer: HP computer method:
Make sure your BIOS is updated before changing the settings. If you have an HP computer, they have an HP Support Assistant app you can configure to automatically install BIOS updates. Then follow the instructions on how to update BIOS.
Then you can look up which HP computer for how to change the BIOS in a search engine.
For an HP ZBook, follow these steps:

Restart your computer with the shift key pressed (before you click restart) until a menu appears.
Choose BIOS Setup on the screen (or press F10).
Click on Troubleshoot.
Using your arrow keys in this menu, go to Advanced Options.
Select UEFI Firmware Settings.
Select restart.
It reboots into a Startup menu
Choose BIOS Setup
With arrow keys go to Advanced tab.
Choose the System Options.
Check both the Virtualization Technology (VTx) and the Virtualization Technology for Directed I/O (VTd) boxes.
Go back to the Main tab and at bottom choose Save and Exit.
Computer will restart.
Answer: For Mac users, see this answer:

https://stackoverflow.com/a/47414989/2788608

Essentially, your OS may be blocking Intel from running what it needs to get your AVD going. Go into System Preferences -> Security and Privacy and you should see an option there to enable the Intel processes. Restart Android Studio and you should be good to go.
Answer: I had similar issues and below is how i fixed it:

Restart your PC, when booting up press f10 key to select or open BIOS Settings.
Locate and check / enable Virtual Technology-x option under Device Configuration Settings
Save changes, system restart.
All should be working fine now
Answer: Check "Virtualization" status under performance option in your task manager. If you already have enabled it in your BIOS and still status is "Disabled", then go to bios, disable it and save & exit. Restart or shut down again. Enable it in BIOS again and save & exit. This time you'll see the status changed to "Enabled", It took me 3 attempts (don't know why it took that much time, but finally it worked).
Answer: I guess the same error happened to me after I installed Docker. The solution below worked for me:

Go to Turn Windows features on and off
Uncheck the Hyper-V checkbox

Uninstall Intel Hardware Accelerated Execution Manager via Programs and Features. After the uninstall process, restart your computer (trying to install a new release of HAXM without rebooting didn't work for me).
Go to HAXM GitHub page and download the release. I downloaded HAXM v7.6.5.
Install HAXM

After all these steps, everything is fine for me.
Answer: Intel HAXM is required to run this AVD.
VT-x is disabled in BIOS.
Enable VT-x in your BIOS security settings (refer to documentation for your computer).this error on android studio
I dont no how to do Bios Security


How long does it take to learn German at a B1 level?

Question: I speak Hungarian and learnt English as a second language (both are at a C1/C2 level) and recently developed an interest in German. I tried to find out roughly how long it would take to learn German at a B1 level but the only thing I found were ads for courses. Could someone please tell me roughly how long it would take to teach myself German with 2 hours of self-study a day? Also what good free resources are out there?
Comment: Hi and welcome to Language Learning Stack Exchange. Your question appears to lack a bit of precision and this makes it difficult to come up with an answer. (1) Are you a native speaker of a Germanic language (English?)? (2) B1 and B2 are different enough to justify separate questions. Which of these are you interested in? (3) Two hours of studying (on your own?) is also rather vague, since it is unclear how you study and how much time you dedicate to each skill that is covered by CEFR. Could you please edit your question to make it more specific?
Answer: The Youlden twins, who speak 20 languages, claim that you can learn a language in seven or even five days. Michael speaks 30 languages and learned Hindi at an advanced level in four weeks. The average person doesn't even speak their native tongue correctly. So anything between 5 days and never seems possible.
How long you would take to learn German at a B1 level is impossible to predict. My best guess would be to assume that learning German will take you equally long as learning English.


How to configure windows server 2008 with two networkcards? Clients do not have internet access

Question: I recently migrated my windows 2003SBS to a windows 2008 standard server. The server has two network cards and has the DHCP DNS and AD roles. On the old SBS machine there was a wizard that helped nicely to configure one NIC as the internet gateway and the other to connect the clients. Now I am on my own and failing...
The server can access the internet, the clients can logon to the domain and browse one another's shared files, but the clients have no internet access.
How must I configure this?
Answer: I followed this the other week when helping a friend out, specifically the Dynamic Version with RRAS.
If you have a simple network, you could follow the static path and it takes about 2 - 3 mins to get working.
Answer: It was actually simpler than the article above stated. You need to first add the network policy and access services role, than go to the MMC of Routing and Remote Access and run  the "Configure and enable remote access" wizard. There will be an option to configure internet sharing through NAT that will nicely show the two NICs and offer to create the bridge for you.
Comment: I'd be careful with bridging, as you will need to be getting your IP addresses from your Internet connection and having all machines directly open to the internet.


Finding maximal cardinal independent set given oracle

Question: The problem is given an oracle $O(G, k)$ that would say if graph G contains IS of size k devise an algorithm for finding independent set of max cardinality that makes poly number of calls to the oracle. My attempt has been that first finding the maximal possible size and then try to find the set of that size by removing vertices one at a time. I understand for a given node it either has to be in or not in the set, then I noticed that that there are multiple overlaps between chain of removals and I could devise a DP algorithm of sorts. But I'm just really stuck after that and was wondering if any hint could be given.
Answer: Let $O(G)$ be an oracle that returns the maximum cardinality of an independent set in $G$; you can implement it given your oracle using binary search.
Pick an arbitrary vertex $v$, and form a graph $G'$ by removing $v$ and all its neighbors. If $O(G') < O(G) - 1$ is false, then $v$ doesn't participate in a maximum cardinality independent set, and we try more vertices until we find a vertex that does. If $O(G') = O(G) - 1$, then we put $v$ aside, and continue in the same way with $G'$.
Taking all vertices put aside, we obtain a maximum cardinality independent set of $G$.
Comment: Wow, thank you so much. Is there some sort of methodology for approaching problems like this or is it just intuition?
Comment: This is a very standard search-to-decision reduction. After seeing this one, you will be able to come up with many more.


What (probabilistic models) can only output decisions when they are certain?

Question: I'm basically looking for approaches, models, algorithms for the following situation (a fault diagnosis problem):
I have an input set $\{x_i\}_{i \in \{1..m\}}$ with $n$ binary features of cases (think of "faults" or "alarms" that fired) and $k$ classes. Each case $x_i$ can belong to at least one class and at most $k$ (so I'm dealing with multi-label classification).
Now some relations in the data set are utterly boring/uninformative (say, feature $a$ says "Mechanical Error occurred" and label $b$ means "Mechanical Error fixed").
But even more generally, whenever $x_a = 1$, I see all kinds of labels, i.e., $a$ is not predictive. Put differently, the relation is not "functional".
Other input features $c$ might have a much more "functional" nature, such that whenever $x_c = 1$, I can easily deduce $y_d = 1$.
For instance my training set could look something like this:
$ [0, 1,0] \mapsto \{4, 1\}$
$ [0, 1,0] \mapsto \{2\}$
$ [1, 0, 0] \mapsto \{1\}$
$ [1, 0, 0] \mapsto \{1\}$
So, knowing $[0,1,0]$ is not really informative whereas $[1,0,0]$ tells me (with high confidence) that the label 1 is active.
I'm looking for the latter pairs, a classifier that only extracts meaningful pairs and ignores uninformative inputs.
Could you point me to relevant techniques / keywords? Thanks a lot!
Answer: You can get probabilities for all labels. You can set a threshold for confidence if proba is greater than some value classify otherwise remain silent.


Is there a way to list Django signals?

Question: Is there a way to see which signals have been set in Django?
Answer: It's not really exposed in docs but Signal is just a class that contains a list of receivers which are called on event. You can manually check this list:
from django.db.models.signals import *

for signal in [pre_save, pre_init, pre_delete, post_save, post_delete, post_init, post_syncdb]:
    # print a List of connected listeners
    print(signal.receivers)

Comment: You listed 7 signals, and these are correct. But to get the list algorithmically, `from django.db.models import signals` and then `[s for s in vars(signals).values() if isinstance(s, signals.ModelSignal)]`
Comment: Also, by using that list, I find that `post_syncdb` appears to be renamed to `post_migrate`.
Answer: There's a django app called
django-debug-toolbar which adds a little toolbar at the top of all django served pages providing info related to the backend of the page's rendering, such as how many queries were executed, how much time they each took, etc. It also prints out signals. I don't use signals in my app, so I have never used that feature, but it's there.
Answer: If you want to list only the connected receivers for a specific signal on a specific model, you can look at _live_receivers. For instance, if you want to list the connected post_save hooks for a model named MyModel, you can do:
from django.db.models.signals import post_save
from models import MyModel
print(post_save._live_receivers(MyModel))

I found this approach in the Django source code by looking for how has_listeners works: https://github.com/django/django/blob/3eb679a86956d9eedf24492f0002de002f7180f5/django/dispatch/dispatcher.py#L153
Comment: It's generally a bad idea to use non-public methods. They are marked with '_' prefix for a reason.
Answer: I wrote little command that shows all signal listeners: https://gist.github.com/1264102
You can modify it to show signals only.

Is there a way to list Django signals?


What is the significance of the Jaccard similarity score?

Question: I understand how to calculate the jaccard similarity , but never quite understood the logic behind why are we calculating it. How does it show the similarity between two sets? What relation exactly does it show?
Can someone throw some light on this?
Answer: The Jaccard index is $J(A, B) = \frac{|A \cap B|}{|A\cup B|}$. The numerator $|A \cap B|$ counts the number of times an item occurs in both $A$ and $B$. The denominator $|A\cup B|$ counts the number of distinct items in total in $A$ and $B$.
Hence $J(A, B)$ is the fraction of items that are shared between the two sets.
Comment: @yogeshagrawal I have used it in feature selection studies. When building a model that includes a feature selection step, how stable can you expect the selected feature set to be? By perturbing the data set (bootstrapping) and repeating the feature selection, I calculate $J(set_k, set_{k+1})$ to get an idea of how many/which of my originally selected features are likely to be selected in new data.
Comment: Please give any usecase where we can implement in term of machine learning.


the relationship between the growth rate of GDP, GDP per capita and the population

Question: I was reading Capital in the Twenty-First Century by Thomas Piketty and it says the growth rate of GDP is equal to the growth rate of GDP per capita plus the growth rate of the population, and I wonder why this is the case.
Mathematically, $Y = y * N$ by definition, and I have no idea how we can get $Y'-1=(y'-1)+(N'-1)$ from the former eqution. Or am I getting this wrong?
Answer: Start by definition of GDP per capita:
$$\text{GDPpc}= \frac{Y}{N} \tag{1}$$
where $Y$ is GDP and $N$ population. Take log of both sides:
$$ \ln  \text{GDPpc}= \ln Y - \ln N \tag{2}$$
Take the time derivative of both sides (note all variables are implicitly functions of $t$ so this is allowed):
$$ \frac{\dot{ \text{GDPpc} }}{\text{GDPpc} } = \frac{\dot{ Y} }{Y } - \frac{\dot{N}}{N} \tag{3}$$
Now solve for growth rate of $Y$
$$ \underbrace{\frac{\dot{ Y} }{Y } }_{\text{GDP Growth}}=   \underbrace{\frac{\dot{ \text{GDPpc} }}{\text{GDPpc} }}_{\text{GDPpc Growth}} + \underbrace{\frac{\dot{N}}{N}}_{\text{pop growth}}  \tag{4}$$
The derivation above is expressed in continuous time, but note $\frac{\dot{X}}{X}$ is just a direct continuous analogue of discrete growth $\frac{X_t-X_{t-1}}{X_t}= \frac{\Delta X_t}{X_t}$ (for some infinitesimal $\Delta$ they are equal for large delta it would only be approximation) so if you want to switch to discrete time you get:
$$ \underbrace{\frac{\Delta Y }{Y } }_{\text{GDP Growth}}=   \underbrace{\frac{\Delta \text{GDPpc} }{\text{GDPpc} }}_{\text{GDPpc Growth}} + \underbrace{\frac{\Delta N}{N}}_{\text{pop growth}}  \tag{5}$$
Answer: In discrete time (in which GDP is released), the statement is only true in approximation. Economists often use this kind of approximation implicitly.
Write GDP per capita as $y = \frac{Y}{N}$.
Then, growth of GDP per capita is related to GDP and population growth:
$$
\frac{y_t}{y_{t-1}} = \frac{Y_t}{N_t}\frac{N_{t-1}}{Y_{t-1}}
$$
So, the gross growth rate of GDP per capita is actually the product of the gross growth rates of GDP and population. However, we can use $\log(1+r)\approx r$ where $r$ is the (net) growth rate and relatively small (say 0.025 year-on-year). So, for a net growth rate $r$ of $x$,
$$1+r = 1+\frac{x_t - x_{t-1}}{x_{t-1}} = \frac{x_t}{x_{t-1}}$$
we obtain
$$
r \approx \log(1+r) = \log\biggl(\frac{x_t}{x_{t-1}}\biggr)
$$
Applying this to the relationship between the growth rates gives
$$
\biggl(\frac{y_t}{y_{t-1}}-1\biggr) \approx \biggl( \frac{Y_t}{Y_{t-1}}-1\biggr) - \biggl(\frac{N_t}{N_{t-1}}-1 \biggr)
$$
where the terms in brackets are growth rate of per-capita GDP, GDP, population, respectively. Rearranging the terms gives you the statement (in approximation).
Answer: If we are considering a long period in which GDP per capita and population both grow considerably, then the growth in GDP over the period will be significantly more than the sum of growth in the former quantities.  Suppose that over a decade, GDP per capita and population both grow by 20%.  Then GDP will have grown by:
$$\Big[\Big(1 + \frac{20}{100}\Big)\Big(1 + \frac{20}{100}\Big) - 1\Big]\times 100 = 44\% > 20\% + 20\% = 40\%$$
However, when considering short periods with small changes, summing the growth of GDP per capita and population gives a good approximation to growth of GDP.  Making a similar calculation with growth in GDP per capita and population of just 2%, we find that GDP grows by:
$$\Big[\Big(1 + \frac{2}{100}\Big)\Big(1 + \frac{2}{100}\Big) - 1\Big]\times 100 = 4.04\%$$
In this case therefore the degree of approximation is only 0.04%.
When growth rates are measured in continuous time rather than over a period, then (as explained in 1muflon1's answer) summing the growth rates of GDP per capita and population gives the growth rate of GDP exactly.


Why it is very rare to find a compound that lacks carbon to be liquid at standard temperatures and pressures?

Question: It is the first time I heard such a generalization, I had no idea. Why being liquid in standard conditions is special to compounds which contain carbon atom? Which property of carbon cause this?
Comment: It is a property of human mind, and not that of carbon. If you hear about a remote island inhabited by two-headed people, will you seek an explanation, or try to verify whether the thing is true in the first place? Water, mercury, H2SO4, Br2, POCl3 all are liquid, despite containing no carbon. Methane, CO2, naphthalene, adamantane, diamond all contain carbon, but...
Comment: @IvanNeretin Hello there, asking here is my way to verify actually :) I searched and couldn't find anything satisfying that's why I asked here. I already know there are liquid compounds that's why I wrote "rare" not "impossible". Because organic compounds form a majority. Thank you anyway.
Comment: @IvanNeretin - while I totally agree with you, the fact remains that organic chemistry sure makes a lot of things that are liquid at STP. Far more than, say, Osmium...
Comment: @IvanNeretin - OK, I'm getting a little silly, but carbon should get a lot of credit simply for ethanol... Cheers!
Comment: @JonCuster True, but then again, organic chemistry also makes far more gases and far more solids than osmium.
Answer: The property of carbon that causes this is the ability to form rings and chains that are stable under atmospheric conditions.
The only reason why 'so many liquids contain carbon' is basically confirmation bias. There are orders of magnitude more compounds known that contain carbon than those that do not. For the longest time in the history of chemistry, inorganic chemistry was restricted to very simple molecules of only a handful of atoms while polymerisation was already a thing in organic chemistry. Furthermore, a large proportion of ionic compounds is inorganic—or, in the context of this question better: carbon-free—and most of these are solids at standard temperature and pressure adding to the confirmation bias.
Thus, by sheer economy of scale the number of known carbon-containing liquids is far higher than the number of known carbon-free liquids.


Stochastic matrix problem

Question:
A (left) stochastic matrix is one which has only non-negative entries, and such that the entries in each column sums up to $1$. Let $A$ be any (general) $2 \times 2$ stochastic matrix.
a) Show that one of the eigenvalues of $A$ is $1$.
b) Show that a general $n \times n$ stochastic matrix also has one eigenvalue equal to $1$.
c) What happens if the column sums are $1$ as above, but the entries may be any (possibly negative) real numbers?

I'm not sure how to start with this question. Should I explain using the Perron-Frobenius theorem? How to show for (a)? Because I think I can use the theorem only for (b), or is there a simpler explanation? And I'm not sure how to answer (c) as well.
Answer: Hint: a matrix $A$ has eigenvalue $1$ if and only if $\det(A-I)=0$.  Now look at $A-I$ and see what you get if you evaluate the determinant by adding the rows.

Stochastic matrix problem


How to add geometry columns to a layer in QGIS 2.16?

Question: I want to add geometry columns to a shapefile. I use the tool called Export/Add Geometry Columns. The tool automatically creates a new layer with the input layer's attributes and the new geometry columns. In previous versions this tool (very usefully) allowed to add (or update) the columns to the existing shapefile instead of creating a new one. How can this be done in the 2.16 and 2.18 versions? I tried by selecting the curved arrow icon labelled ("Iterate over this layer") but the program crashes ("An error occurred while executing Python code").
Answer: You could use the field calculator and the expression:
<code> geom_to_wkt(  $geometry )
</code>
For point data you can create X and Y columns and use the field calculator with:
<code>$x
</code>
and
<code>$y
</code>
Comment: Just make sure you select the 'Text' field option when creating the new column and the above will work fine


How to do version compatibility testing

Question: I have an App ver - 3.8 and minimum supported version is 4.0. If the App is older than 4.0, an update available is shown to the User within the App.
Is there a way to simulate or test this scenario
Comment: I might have missed the point... would you not just load v3.8 and see if the 'update available' message appears?
Comment: Is this some sort of client-server interaction?  You have api-testing as a tag.  Depending on the mechanism that the necessary level is found, mocking is probably an option.
Answer: To test any thing is to simulate the conditions necessary to for the thing to happen and see if the thing happens.
To paraphrase what Dan Wilson said. If you want to see if v3.8 of an app prompts for an update, load v3.8 and see if it prompts for an update.
When you're working with multiple versions it helps to have a standard procedure for setting up the testing environment with with any supported version of your software.
Answer: In any qa services, the terminology compatibility version testing is basically used for checking whether the features of app supports the latest version or not.
So, in order to test it below scenarios can be verified:

Generally if any user is using an old version build then a pop-up should be shown when user open the app using any older version. Verify that the latest version is mentioned in the app once update is complete.
Verify all the basic features of the app are fully functional: execution of smoke or sanity checklist can be done every time the version
Verify all the features that are added in latest build version (it is generally mentioned by developers in their release notes ) in order to make sure that all required changes are available in the latest build version.
Some of the apps have the feature of rolling back to previous build. If this feature is available in your app, then make sure to test this as well. In this scenario, basic test has to be done to make sure all latest changes are not shown again and user is able to move back to previous build anytime.

How to do version compatibility testing


Does the mass distribution matter in (Schwarzschild) black holes?

Question: Is it possible that from the same initial mass different black hole radius will be created due to different mass distribution during black hole creation? If mass is concentrated more on the outside bigger event horizon will be created? If mass is concentrated more in the center smaller black hole will be created?
Answer: No, it is not possible.
This is the whole essence of the so-called no-hair theorem: that all previous details concerning the object from which the black hole formed has no bearing whatsoever on the properties of the final object, except for mass, electric charge and angular momentum. And also, that these quantities enter only as global values, not with their initial distributions.
Notice that the impact of the theorem is even wider than you seem to imply: not only it does not matter which initial distribution these three quantities had (for instance, you might have some matter corotating while some fraction of matter is counterrotating, but it does not matter so long as you have the same total angular momentum), but also matter might be made of leptons or baryons, charged could be made of different number of positively or negatively charged particles, or of mostly neutral particles plus only one sign of charge.
Also, the no-hair theorem states that this property is not the result of some special assumption or of highly symmetrical initial conditions, but that instead it is the necessary outcome of any history of the matter going into the black hole, regardless of any parameter save for the global values of M, Q, J.
Comment: The distribution of matter will affect how much energy is lost to gravitational radiation during formation of the hole, and in that sense will affect its mass and radius, though this doesn't seem to be the sort of effect the original question had in mind.
Comment: So is event horizon always created spherical even if it is created during collision of two heading from opposite direction massive stars?
Comment: @PawelWelsberg Yes, because the total angular momentum vanishes. If the total angular momentum is not zero, the event horizon is flattened at the poles.
Comment: And, yet, there is a non-trivial discussion about the black hole information paradox as raised by Bekenstein, I believe. Is it truly lost, as the classical equations predict, or does quantum gravity save the day, and black holes do have microscopic degrees of freedom (fuzzball?), as string theory seems to predict?
Comment: Too many questions in too little space. My reply dealt with GR. As for quantum effects, though many interesting discussions are present, there is of course no definite prediction about any of these topics because of the well-known lack of a suitable theory.
Answer: Going out on a limb here, if string theory is right, and one can make microscopic black holes in an accelerator, there should be gravitational analogs of Stern-Gerlach (and more general spin polarization) experiments for these microscopic black holes. They could also have measurable excited states leading to decay spectra.
For macroscopic black holes, of course, the thermalization timescale would be unmeasurably short, and they will, just like MariusMatutiae said, not return any useful information about their past beyond these mentioned conserved quantities. That, of course, is no different from nuclei, atoms and molecules.


The negative integral meaning

Question: Whenever I take a definite integral in aim to calculate the area bound between two functions, what is the meaning of a negative result? Does it simly mean that the said area is under the the x - axis, in the negative domain of the axis?
Comment: Yes, this is simply that. Definite integrals compute the signed area between the curve and the $x$ axis. So, the signed area can be positive, negative or zero. For example, integrate $\sin(x)$ between $0$ and $\pi$,  $\pi$ and $2\pi$,$0$ and $2\pi$.
Comment: Note that $\int_1^0 1\, dx=-1$, so integrating a positive function in the wrong direction also yields a negative result.
Comment: @Trey what is the reasoning for the bounty for official sources? This may help us understand what you're going after.
Comment: take a look here https://math.stackexchange.com/questions/1316529/why-does-an-integral-change-signs-when-flipping-the-boundaries
Answer: Not always.
The definite integral of a positive function (that is a function over the $x$-axis) gives a positive area. This is,
$$\int_a^b f(x) dx \geq 0$$
for a function such that $f(x) \geq 0$ when $a<x<b$.
On the other hand, the definite integral of a negative function (that is a function under the $x$-axis) gives a negative area. This is
$$\int_a^b f(x) dx \leq 0$$
for a function such that $f(x) \leq 0$ when $a<x<b$.
Now, the problem comes when you have a function that goes for a while over the $x$-axis, and for another while under it. For example, let's assume that goes over the axis on the interval $[a,c]$ and under it on the interval $[c,b]$. That is
$$ f(x)\geq 0 \text{ when } a<x<c, $$
$$ f(x)\leq 0 \text{ when } c<x<b. $$
Then the integral can be split in two as
$$\int_a^b f(x) dx = \int_a^c f(x) dx + \int_c^b f(x) dx = A_1 + A_2,$$
where $A_1\geq 0$ and $A_2 \leq 0$.
Then, depending on which of the two areas $A_1$ and $A_2$ is "bigger" the result of the definite integral will be positive or negative. That is,
$$ \text{If } |A_1|\geq|A_2| \text { then } \int_a^b f(x) dx \geq 0 ,$$
$$ \text{If } |A_1|\leq|A_2| \text { then } \int_a^b f(x) dx \leq 0 .$$
To sum it up, a negative definite integral means that there is "more area" under the $x$-axis than over it.
Answer: Let us begin with something very important: Integration is wonderful! No, let us discover some of its wonderfulness. At first, as stated and as it is known, the definite integal of a Riemman-integrable positive function $f:[a,b]\to\mathbb{R}$ over an interval $[a,b]$ represents the area enclosed between the curve $f$ defines on the plane, te $x'x$ axis and the lines $x=a$ and $x=b$. which results to the fact that, if $\displaystyle\int_a^bf(x)dx$ is negative, then, at least a "large" enough part of the abovementioned area lies under the $x'x$ axis. All this has been said up to now.
However, as integrating is - in a way - a more generalised way to sum - uncountably many - numbers, it is expected that some of the summation properties should be "transferred" to integration. In deed, to define $f's$ average value, $\overline{f}$, we use the following integral:
$$\overline{f}=\frac{1}{b-a}\int_a^bf(x)dx$$
So, we do exactly what we would do if we have to find the average value of the numbers $x_1,x_2,\dots,x_n$:
$$\overline{x}=\frac{1}{n}\sum_{i=1}^nx_i$$
See, how $n$ is the "length" of the "interval" over which we are summing. Also note that the sign of $\overline{f}$ is independent of $\frac{1}{b-a}$, so,
$$\overline{f}<0\Leftrightarrow\int_a^bf(x)dx<0$$
and
$$\overline{f}>0\Leftrightarrow\int_a^bf(x)dx>0$$
So, if the integral of a function is negative then, it is expected, should one choose a point $y_0\in f([a,b])$ uniformly, that $y_0<0$ - of course, "expected" means "expected" and not "for sure".
A more interesting meaning of a negative integral comes from physics. Let $F:[a,b]\to\mathbb{R}$ be a function with $F(x)$ being the force applied to a mass when it is at position $x$. So, if $F$ was a constant over $[a,b]$, then, as we know from school physics, it work is given by the formula
$$W_F=F\cdot(b-a)$$
Now, imagine that $F$ is not constant over $[a,b]$. So, let us "cut" $[a,b]$ to elementary small pieces of length $dx$ over which $F$ is "almost" constant - of course this is a rather intuitive approach, however it is very useful to explain the idea of the formula that will be given below. Then we have that, on these "pieces" the work of $F$ is given again by the formula $W_F=F\cdot dx$. Now, summing all these "partial works", we get that:
$$W_F=\sum F\cdot dx$$
It is clear that the, mathematically, correct formula for this is - imagine that we are now letting $dx\to0$:
$$W_F=\int_a^b F(x)dx$$
So, thinking of every function $F$ as the force applied to a mass while it is moving through $[a,b]$ gives a clear interpretation to any integral, positive or negative; it is the work of $F$ during that movement. So, that integral being negative, it means that we make more effort to "stop" this mass, than to support its movement.
I hope these were helpful! :)
Comment: the physics example is astonishing!
Comment: I know. I was also very impressed when first met this approach of finding a force's work over an interval. In a similar way, you can find many interesting formulas, as, for instance, for the length of a curve/function over an interval etc.
Answer: This area thing is for kids. What you have to grasp is the general idea of an integral.
We are given a solid $B\subset{\mathbb R}^n$, $n\geq1$,  and a function $f:\>B\to{\mathbb R}$ modeling some variable (geometrical, physical, etc.) intensity on $B$. The integral $\int_B f(x)\>{\rm d}(x)$ then encaptures the total impact caused by $f$ on this $B$ in terms of  a single number. Thinking about this idea, and the properties this total impact should naturally have, brings us to the following setup: Partition $B$ into a large number  of tiny subdomains $B_k$ $(1\leq k\leq N)$ and choose a sample point $x_k$ in each $B_k$. Then the quantity we have in mind is approximately given by
$$\sum_{k=1}^N f(x_k)\>{\rm vol}(B_k)\ ,$$
so that we should define
$$\int_B f(x)\>{\rm d}(x):=\lim_{\ldots}\sum_{k=1}^N f(x_k)\>{\rm vol}(B_k)\ ,$$
whereby the $\ldots$ under the $\lim$ will have to be filled with lots of legalese.
In the simple case of an $f:\>[a,b]\to{\mathbb R}_{>0}$ the function $f$ models an "area intensity" over the interval $[a,b]$, and the integral then is the "area under the curve".

The negative integral meaning


what's analogy learning algorithm ? differences from induction learning algorithm?

Question: I am a computer engineering student and trying to get the idea behind all these Artificial Intelligence Concepts and applications. I know little theoretically about machine learning and some high level brief introduction of artificial intelligence as a whole and neural network. What i am interested in is knowing the similarities or difference between the concepts: Deep Learning, Genetic Algorithm/Programming, Artificial Neural Networks and Machine Learning. How do they relate or how they help each other? What are their particular applications?
Comment: This is a very broad request, covering a large range of topics. Reading a suitable textbook is probably the best way to get this kind of introductory information.
Answer: In a nutshell, (statistical) machine learning is a certain approach to artificial intelligence. Traditional artificial intelligence incorporates hard-coded rules which are written by humans. Statistical machine learning, in contrast, has the computer learning the rules by itself using statistical means.
There are several paradigms used for statistical machine learning, such as support vector machines, artificial neural networks, and genetic algorithms. Recently a certain type of artificial neural networks, deep neural networks, has come to the fore due to unprecedented performance. Machine learning using deep neural networks is known as deep learning.


How to Create Line Break on Google Maps Engine Lite

Question: I am editing a map in Google Maps Engine Lite and when I add a marker and description, it doesn't let me make a new line. Enter saves what I wrote, and Shift + Enter and Ctrl + Enter don't work.
Answer: This question has stumped me for a while and finally I found a solution on the official Google forum. The solution is that you need to press Alt + Enter.
Update: Ctrl + Enter seems to work now as well.
Answer: @Ynhockey is correct, alt + enter gives a full paragraph break (2 lines).
For a single line break (next line), press Alt + Command + Enter