What is covered by CDIC savings deposit insurance?

Question: In the unlikely event that a financial institution fails in Canada, what is covered by the Canadian Deposit Insurance Corporation?  What is the limit of coverage?
Answer: CDIC covers deposits at financial institutions...typically banks and trust companies...insurance companies and co-ops have their own version of it.
The limit is $100,000 per account.
This can be in things like bank accounts, GICs, term deposits, etc.
Terms of up to 5 years are covered.
You can have more than one account at an institution and have them both insured.  For example, an open or non-registerd account is insured separately from a RRSP account, and if you have a joint account, that is also separate.
In addition to the above, you would be separately insured if you had deposits at different institutions, for example, 2 different banks.
Comment: Welcome Joanna!  Does President's Choice Financial and CIBC count as different banks?  PCF services are provided by CIBC.
Comment: A clarification: I believe the limit is per *type of account in the name of one depositor*, not just *per account*.  So 2 plain bank accounts (whether chequing or savings) in the **same** person's name would *together* qualify for $100,000 protection, not separately.  And as Joanna has mentioned, RRSPs and joint accounts qualify as separate types of accounts with their own separate protection.
Comment: Zephyr, to answer your question: http://www.cibc.com/ca/cibc-and-you/to-our-customers/dep-reg-info.html  I think the answer used to be "different banks" (see note at that page 'Products **formerly** issued by Amicus Bank') but now the PCF products appear to be offered by CIBC directly.  Amicus bank apparently underwent voluntarily dissolution; see http://www.gazette.gc.ca/archives/p1/2005/2005-09-03/html/misc-divers-eng.html.
Answer: The CDIC provides an insurance cap of $100,000 per depositor within each member institution. If you have two accounts in your name at the same institution or at different branches of the same institution, these accounts together are only insured up to $100,000. For example, if you have two accounts, both in only your name, at the same bank, with a total deposit value of $110,000, you're only insured up to $100,000 on those deposits, even though you have two accounts.
Investopedia describes the six categories of accounts that the CDIC uses. Each of these account types is protected by a separate, $100,000 cap.

Single accounts under one name (this is the category I describe above)
Joint savings
Savings held in trust for another person
Savings held in registered retirement savings plans (RRSPs)
Savings held in registered retirement income funds (RRIFs)
Savings used to pay realty tax on mortgage payments

Joint accounts

Joint accounts are insured separately from each account holder's personal account. For example, Tom and Tara have a joint account worth $100,000, and Tara also has an account in only her name with a balance of $50,000. In the event of bank failure, Tara would receive $50,000 from the account in her name, and the couple would receive $100,000 from the joint account.

Trust Savings

Deposits that are held in trust for another person are insured independently of any additional accounts in the name of either the trustee or beneficiary. So, if Tom and Tara also have a $40,000 term deposit held in trust for their daughter Tina, CDIC would pay another $40,000 in addition to the $150,000 for their other accounts.

Registered Plans

Registered plans include deposit investments like RRSPs and RRIFs. Not all registered plans are eligible. Investments must mature in no more than five years, and mutual funds are not covered.

Mortgage Tax Payment Accounts

Mortgage tax payment accounts are accounts that depositors use to pay realty taxes toward mortgage properties held at a CDIC institution. As with the other categories, funds in these accounts are insured separately from other savings.
Comment: Thanks. So if I had 100k in one bank and 5k in another bank and both banks go under, I am insured for 105k, not 100k. Because you said it is per member institution
Comment: @Kaushik Yes, assuming a) your accounts are at different *banks*, not just different branches of the same bank, and b) both banks are member institutions.


Graphics Card Problems

Question: I've just bought a new gaming PC. I installed Windows 8.1 (64bit) on my PC as that is the minimum requirements for most games. However my driver file says my graphics card is for a 32 bit system so I basically can't run games. Any ideas how to fix this?
Comment: What graphics card do you have? Basically you need the 64bit driver from the manufacturer, if there is one available. I can post a link to one once I have the graphics card information.
Comment: I have a GeForce GTX 1050 Ti 4gb. I dont have a clue how to get the drivers. Im new to all this, i would appriciate any help :)
Answer: The driver to download for your card would be located at: http://www.nvidia.com/download/driverResults.aspx/113442/en-us.
When running the installer, you can select "Custom (Advanced)", then check the box that says "Perform a clean installation" to remove the previously installed 32bit driver. You can also custom select which items to install with the new driver.
Comment: No problem! If you run into any problems getting the updated driver to work just reply.
Comment: Yep - glad you got it working!

Graphics Card Problems


Comma notation in universal qunatification

Question: Is the expression $\forall a,b,c \in M : \varphi(a,b,c)$ equivalent to $\forall a \forall b \forall c : (a \in M \land b \in M \land c \in M) \rightarrow \varphi(a,b,c)$ ?
Comment: Yes. That is so.
Answer: The formula $\forall a \in M : \varphi(a)$ is syntactic sugar for $\forall a \ (a \in M \to \varphi(a))$ and $\forall a, b, c \in M : \varphi(a,b,c)$ is syntactic sugar for $\forall a \in M \ \forall b \in M \ \forall c \in M : \varphi(a,b,c)$ which according to the former syntactic sugar means $\forall a \ (a \in M \to \forall b \ (b \in M \to \forall c \ (c \in M \to \varphi(a,b,c))))$ which is logically equivalent to $\forall a \ \forall b \ \forall c \ (a \in M \wedge b \in M \wedge c \in M \to \varphi(a,b,c))$.
Answer: Yes, the former notation is an abbreviation of the latter. We often even leave out the commas, so you would see $\forall abc \in M : \varphi(a,b,c)$.
A similar abbreviation is used for the existential quantifier:
$$
\exists abc \in M : \varphi(a,b,c)
$$
means
$$
\exists abc : a \in M \land b \in M \land c \in M \land \varphi(a, b, c).
$$


How to compile to form a DLL

Question: I have sample project codes. When I compile it inside Visual C++, it produce as EXE. I would like to know, how can I change the final output to DLL form. Means that, when I compile, it become DLL.
Answer: In a nut shell, a dynamic link library (DLL) is a collection of small programs, which can be called upon when needed by the executable program (EXE) that is running.
E.g Suppose, it would be if the program (exe) needs to get the free space of your hard drive. It can call the DLL file that contains the function with parameters and a call function. The DLL will then tell the executable the free space. This allows the executable program to be smaller in size and not have to write the function that has already exists
You can check  the required steps here : How to Create a Simple Win32 DLL ?
Answer: Compiling it as a DLL is comparatively easy, namely you set the "Configuration Type" in Configuration -> General in the project properties to "Dynamic Link Library".
However this isn't going to accomplish much unless you also export the entry point(s) into the DLL.
Answer: When you create a new project choose "Class Library" in the New Project dialog:

How to compile to form a DLL


What to call an "online illustrator"?

Question: Is there a term to refer to an illustrator who mainly posts his or her works online? I mean, a lot of illustrators started out this way, so I am curious if there's a term that we can use to refer to them.
Comment: How do they post their art? Facebook? Their blog? I think more details are needed for a good answer.
Comment: Sorry for the ambiguity, but in my native language we just call them "internet illustrators" regardless of the websites or social media they post their arts.So I was wondering if English has a term for this.
Answer: If the artist is creating art for websites as a job, the most common terms are "web graphic designer" or just "graphic designer", although the latter is also used for people who work with print media.
For artists who publish their own personal works online, you might try "web artist", "internet artist", or "net artist" which both return quite a few examples of usage in Google results for those terms.

What to call an "online illustrator"?


Why doesn't the earth's surface collapse onto itself?

Question: From what I know the sun (or any other star) nuclear fission keeps it from collapsing under its own gravity until the fuel runs out, if that's so why doesn't the earth surface like the tectonic plates collapse to the earth's core? is it because of similar processes that happen in the earth's outer core (nuclear fission)? and if so what would happen if they would once stop?
Comment: A.K, I'm sorry but Earth science questions are off topic here. You could ask it on Physics.SE or wait for the Earth Science proposal to hit beta.
Answer: For the same (an more prosaic) reason milk skin floats over boiling milk - because it is lighter than the layer below. During the formation of the planet it went through a process called Iron Catastrophe, were a large parcel of the denser materials like iron and nickel sunk to the core, leaving the lighter materials close to the surface.
Nice graph with relative densities follow:

Fission contributes to keep our internal engine super-hot and well-oiled, allowing for our metallic core to spin and generate our magnetosphere. Eventually we'll run out of heat, the outer core will solidify and stop spinning.


Simple Azure Push Service for Windows Phone

Question: I have an Azure project and want to be able to send push notifications to windows phone (7 and 8). I already did this tutorial: http://www.windowsazure.com/en-us/develop/mobile/tutorials/get-started-with-push-wp8/?fb=de-de
But it didnt helped because I have to send broadcast messages.
I would be nice if there is a Microsoft Push Notification Service where I can register my Azure Service and my Windows Phone App. Then my Azure Service simply could send a message to the MPNS which would broadcast it to all registered windows phones. I mean something like Windows Azure Service Bus Hubs. Unfortunately this Service Bus Hubs dont work with windows phone actually.
Is there something else I could use or am I forced to create such a service all by myself. For example as a WCF Service.
Are there other possibilities? What's the best way to do this? Is there a tutorial?
Answer: Per the MSDN documentation for Notification Hubs, "[s]upport for Android and Windows Phone will be added soon.".


Not found "Create Add new " option in the Current Employer field if the org name typing

Question: We're already apply " Access API Permission " to anonymous user for creating Add New option organization to the current employer field but still not works. Civicrm-4.7. And added public automcomplete ext. to the current employer field. I go through the documentation https://github.com/TechToThePeople/publicautocomplete and added code in the civicrm.setting.php
Answer: From the documentation:

If you want to force the user to submit only a value from the list (or
  leave to leave it blank), thereby preventing the user from creating
  new organization records, set the require_match config option to TRUE
  (it defaults to FALSE):
global $civicrm_setting;
$civicrm_setting['eu.tttp.publicautocomplete']['require_match'] = TRUE;

This means, if you want to allow the user to create new organization records, you would omit this line, or set it to FALSE.
Also note: If you omit the line or set it to FALSE, you won't actually see a "Create New" option; you'll just be able to type in the name of an organization that's not in the list of options.
EDIT:
Regarding the potential for typos and duplicates, that will always exist any time you allow the user to enter a new organization name. CiviCRM will  follow the unattended dedupe rule for organizations when deciding whether to use an existing org or create a new one.
Answer: As we understand it, this default is an either-or. Either it is set to FALSE, i.e. allows user to input anything into the field, including misspellings (?)
-or-
user may choose from a restricted list (group) "submit only a value from the list (or leave to leave it blank), " i.e. forced to leave.
We would want it as a conditional. If the user cannot locate their employer, then they have option to "create new" organization (and create the ensuing relationship in civi) H
Comment: This looks like a follow-up to my answer. If so, it's better as a comment to that answer than as a new answer. Please see my edit there about typos and dedupe rules.
Comment: We're getting the message we need "50 reputation" to add comment above (?) re : duplicates can you clarify what is meant by "unattended" vs. "unsuperivsed" dedupe rule for orgs. right now, the extension appears to be working yet it creating a new org each time we select the correct Employer. Tried adding a new unsupervised rule based solely on name not working tho. H
Comment: https://civicrm.org/blog/spidersilk/understanding-civicrm-dedupe-rules


If a slime tube hardens when exposed to air, why doesn't it harden when I pump it?

Question: A question for the philosoraptors :v
Comment: I thought tube sealants hardened when rapidly changing pressure, from the 50+PSI inside your tube, to the ambient pressure outside your tube?   If it was purely "drying" in the air then the sealant wouldn't seal quick enough to fill the puncture ?
Comment: Voting to leave open - its pretty clear the question is "why doesn't slime harden in the tube?"
Answer: For the same reason a salt-water solution in a closed half-filled bottle will not evaporate - even though it is in contact with some air.
Open the lid - and eventually the water evaporates to leave just salt.
The slime solution is in a "balance" with the air in the tube - since it is a "closed" system. When the tube is punctured the slime flowing out of the puncture is no longer in a "closed" system and evaporates / dries in the open air.
Takes me back to Chemistry and reaction kinetics (yuck) .. but anyway that's my take on it.
Comment: The whole process is working under pressure - which *may* have a bearing on the speed the of the process. Additionally, the rapid throughput of air passing through the puncture provides a larger than normal head space over the slime solution for evaporation to occur - so it dries far more quickly.
Comment: That's reasonable, but how does the slime/sealant seal a puncture on an inflated tube quick enough to retain the air?  If it has to wait for the sealant to dry, it could take too long, and you have a flat.
Answer: According to the name-brand Slime supplier's own article on The Science Behind Slime Tire Sealant
:

Slime repairs flats with a mechanical seal, meaning physical particles actually plug up the hole. There is no chemical reaction at the puncture site.
These physical particles are a combination of long and short fibers, as well as rubber particles (Those little black bits you see mixed in with the green). We call this special mixture of particles Fibro-Seal technology.

So the sealant's liquid is just a water-based carrier to help the particles move to where they are needed. It doesn't dry out (much) in your bicycle's tube for the same reason it doesn't dry out in the bottle, just because there's not really any place for the water to go once the air in the tube gets to 100% humidity. And based on the Slime company's own description, it never really needs to "harden" so much as it just ends up clogging up any holes.
Hypothetically one might even be able to re-hydrate old Slime if it dries out (unless there's other volatile parts of the mixture that go missing too); the manufacture does not provide instructions for this and would probably rather you buy more instead!
Think more "leaves in a gutter downspout" instead of like any sort of glue or epoxy. Without water the leaves just blow around, but when rain rushes along the gutter the downspout can get "sealed off" by the junk it brings with it.
Comment: Good points.  Welcome to SE!     I'm not sure about "rehydrating" else existing puncture-filling plugs in a tyre might re-dissolve, or perhaps they are and continuously turn over at a microscopic level?     And I do wonder if other products work the same, like Stan's or Caffeelatex etc.


How do fish communicate?

Question: Fish, as far as I know (of course there may be exceptions), don't have any vocal organs, despite the fact they are proven to be social. So how do they manage to communicate with each other? On land, organisms like the honey-bee have their "tail-wagging dance" and ants have pheromones, so what about fish?
Answer: Fish also produce sounds; the difference is that they are put in different positions by the body to sound. The sound produced by fish is about 20Hz and therefore is difficult for humans to hear. However, these sounds can be recorded and reproduced using instruments.
Of course, some fish also rely purely on action to exchange sounds. Some fish use ultrasonic communication, similar to dolphins (although dolphins are not fish).
Some fish transfer information by changing body colour. These fish are usually located in the tropical shallow water. Some fish pass through the water with their fins flapping to communicate courtship information. A few, mostly deep-sea fish, release pheromones.
Comment: What do you mean by " they are put in different positions by body to sound"? Also, it would be really great, if you can provide any reference to substaniate your answer>>>
Comment: can you provide any reference to substantiate your answer???
Comment: Please provide some sources for the information you have in your answer.
Comment: means thier  "tail-wagging dance" will make some noise
Answer: There are a variety of fish species that produce sounds, including many species of Bocaccio (i.e. rockfish). This article is a great resource for learning more about 'fish choruses', a phenomenon where many fishes produce noise during dawn/dusk hours, much like the dawn chorus we're more familiar with for birds.
One of the most famous examples of fish vocalizations is probably the plainfin midshipman fish (listen here!). In this case, the hum is used to attract females for mating.
As for how these sounds are produced, this interview with a scientist at Scripps Institution of Oceanography says
"Fish make sounds by oscillating their swim bladder, or by rubbing the bones where their fins attach to the body, similar to how crickets make sound by rubbing their legs".
Comment: You should also check out Dr. Rodney Rountree's website. It looks old but has lots of useful information about fish sounds.  http://www.fishecology.org/soniferous/soniferous.htm
Answer: I second MZiegenhorn's answer and Xavier Mouy's comment that a variety of fish species produce sounds, and I'll shamelessly refer you to our new website FishSounds for an inventory of published research (updated up to the end of 2020 - 2021 update coming soon) on fish sound production, and some example sound clips.
Fishes don't have vocal chords, but they can produce sounds in a variety of ways. In Ingrid Katz's words: "Fishes are known to produce sounds using five mechanisms: 1) swimbladder pulsations, 2) stridulation, 3) hydrodynamic movements, 4) substrate, body and tendon vibration, 5) air release. "
Regarding yazhuo's answer about fish sound production mechanisms, you can check Eric Parmentier's work, such as for example this book chapter:
Parmentier, E. and R. Diogo (2006). Evolutionary trends of swimbladder sound mechanisms in some teleost fishes. Communication in Fishes. F. Ladich, S. P. Collin, P. Moller and B. Kappor. Enfield, NH, Science Publishers: 43-68.
Now that I mention a book chapter, the entire book would be an excellent source to answer the original question, if you are up for the reading.

How do fish communicate?


meaning and usage of "что ты"

Question: What is the meaning of "что ты" when person uses it during conversation like this:
Это твой папа?
Нет! что ты! Это мой дядя.
Does it mean What are you saying? or You're saying the wrong thing!?
And another question: can I use it with "Вы" in a formal situation?
like: Это ваш папа?
Нет! что вы! Это мой дядя.
And will it not be rude?
Answer: "Что ты" (or "что вы") means "you're wrong" or "your suggestion is over the top".
Saying this is definitely not rude. Whether it's actually polite depends on the context. For an innocent mistake, it's probably not very polite. For a guess that was meant as a compliment, "что вы" is perfectly fine.
Ex.: "Это Вы сами сочинили?" - "Что Вы!" ("You composed that yourself?" - "Of course not!")
Comment: good point on a context of compliment
Answer: Это твой папа?
Нет! что ты! Это мой дядя.
I would say that "что ты/вы" is a colloquial version of "you're mistaken". Both ты and вы can be used. It sounds normal to me.
Answer:
It means the person is wrong.

Yes you can. It's not rude, but in my opinion, it's not particularly polite either to the extent that it's impolite to point out strangers' mistakes. But it's acceptable.
Comment: *it's not polite to point out their mistakes to strangers* - So, taking the dialog above - when a strangers think your uncle is your father, it's not polite to correct them? Personally I see nothing impolite in "- Скажите, эта морковка по 30? - Нет, ну что вы! Всего по 15."
Comment: you misread what i'd written, i think it's not particularly polite (which still doesn't amount to IMpolite) **to point out** mistakes to strangers, it's enough to share factual information without stressing the fact of a mistake, especially when it's inconsequential... it's because when you make the fact of the error known it may be construed to mean "are you blind/deaf?" or "are you an idiot?" which are spelled out in rude style of speech, while polite speech is all about tact and avoidance of even a semblance of offensive statements as much as possible
Comment: I'd argue things like "Что вы" or "Нет, конечно" stress the mistake in a negative sense. Simultaneously it may also mean "You're mistaken, and your question in fact could offend me, but no, I'm not offended as it's ridiculous to assume you asked your (whatever silly) question w/o a reason and with no good intentions".  After all, since "что вы" usually goes as a response to a question, the very fact of the question implies the asking person is ready or even wants to be corrected. And personally I would be offended more by Robot-like-sterile-SJW-friendly binary compressed answer.
Comment: that's right, the very fact of asking a question means that a person isn't sure, so what's the point of additionally stressing their mistake? they kind of preemptively admitted it by asking the question... or you're missing my point, correcting doesn't require letting them know they're wrong by saying что вы, the actual answer speaks for itself
Comment: i don't know the origin of the expression but it seems like a shortened version of a more verbose one, for example it's a part of a very rude expression "Что ты несёшь?" - "What the hell are you talking about?". not saying this is the origin but the actual origin could have also been impolite
Answer: According to D.I.Kveselevich, что ты! is used to express negation, disagreement, warning: what!; what's got into you?; what a thing to say!; why, no, really; no fear, not me (him, her, etc.); whatever has come over you?
The more formal вы removes impoliteness (if any) completely.
Answer: "ты" is similar to the archaic English "thou" as a singular form of address,
Example:

Thou hast not half that power to do me harm  As I have to be hurt
(Shakespeare).

"ты" is informal and singular, while "вы" is formal or plural in Russian, just like "you" is in modern English.
Using "вы" informally is silly, if not pretentious.
Это ваш папа? - Нет! Что вы! Это мой дядя. Yes, it translates to "What are you saying?" or "Of course not!", but the person has broken informal and formal rules (rude and pretentious). The second reacts politely yet emotionally сorrectly using 'Вы'.

meaning and usage of "что ты"


Understanding Maxwell's Equations in a Box

Question: Maxwell's equations for $n$ charged particles each with charge $e_j$ are known to be (in cgs) $$\begin{align}
 \nabla\cdot\textbf{E}(t,x)&=4\pi\rho(t,x)\\
 \nabla\cdot\textbf{B}(t,x)&=0\\
    \nabla\times\textbf{E}(t,x)&=-\frac{1}{c}\frac{\partial \textbf{B}(t,x)}{\partial t}\\
    \nabla\times\textbf{B}(t,x)&= \frac{1}{c}\left( 4 \pi\mathbf{J}(t,x)+\frac{\partial\mathbf{E}(t,x)}{\partial t}\right)
\end{align}$$
With $$\begin{align}
 \rho(t,x)&=\sum\limits_{j=1}^ne_j\delta(x-x^j(t))\\
 \mathbf{J}(t,x)&=\sum\limits_{j=1}^ne_j\dot{x}^j(t)\delta(x-x^j(t))
\end{align}$$
Where each $x^j$ represents the position of the jth particle. Now if we restrict ourselves to a box $V$ and impose periodic boundary conditions, that should in principle allow us to expand everything using Fourier series $$\begin{align}
  \mathbf{E}(t,x)&=\frac{1}{|V|}\sum_{k}\vec{a}(t)_{k}e^{ik\cdot x}\\
\rho(t,x)&=\frac{1}{|V|}\sum_{j=1}^ne_j\sum_ke^{ik\cdot (x-x^j(t))}=\frac{1}{|V|}\sum_k\left(\sum_{j=1}^ne_je^{-ik\cdot x^j(t)}\right)e^{ik\cdot x}
\end{align}$$
Then if you apply the electric Gauss's law you get the next equation for the Fourier coefficients $$ik\cdot\vec{a}(t)_k=4\pi\sum_{j=1}^ne_je^{-ik\cdot x^j(t)}$$ This is completely analogous to the equation you find by performing the Fourier transform, but with continuous $k$. However, in this case if we consider the Fourier coefficient for $k=\vec{0}$ this results in $$\sum_{j=1}^ne_j=0$$ That is, the net charge must be 0. Clearly it isn't physically reasonable to only consider systems with net charge 0, so my question is if there is anything wrong with my procedure? If not if the problems has to do with how I naively assumed Maxwell's equations work in these sort of systems?
Comment: I think this is an inherent property of trying to solve Poisson's equation on a compact manifold without boundary (in your case, the 3-torus).  See [the last paragraph of this answer](https://physics.stackexchange.com/a/523242/81133) for some clues.
Comment: @MichaelSeifert why solving Poisson's equation on compact manifold is relevant? We already have the field, the problem is convergence of its Fourier series, no?
Answer: In a compact space, the net charge must be zero, by a simple topological argument: the net charge is proportional to the volume integral of the divergence of the electric field. By Gauss' law, this is equal to the surface integral of the electric field over the boundary, but a compact (periodic) space has no boundary!
Or you can also see it in coordinates: the integral $\int \nabla \cdot \mathbf{E}\, dV$ in a 3-torus is
$$\iiint \left(\frac{\partial E_x}{\partial x} + \frac{\partial E_y}{\partial y} + \frac{\partial E_z}{\partial z} \right)\, dx\, dy\, dz,$$
and each of these terms is zero after applying the fundamental theorem of calculus and the periodic boundary conditions.
Comment: @JánLalinský: For a compact 3-manifold, any closed Gaussian surface divides the manifold into two volumes, either of which can be viewed as the "interior". (This is easiest to see on the surface of a sphere:  does the Equator, a closed curve on the Earth's surface, have the North Pole or the South Pole in its interior?) The flux integrals for these two regions will be identical except that the "outward normal" will flipped at all points;  so the flux integrals for these two volumes will be equal in magnitude but have opposite sign. ...
Comment: This makes no sense - yes we can consider a boundary where the field is discontinuous and the flux integral is undefined, but this does not imply that the flux integral is zero. It is not defined, so it cannot be zero. In other words, the boundary is unusable for Gauss theorem, and this implies nothing about total charge. When we take slightly smaller box than the one used for Fourier series, the flux is defined and non-zero.
Comment: ... Thus, by Gauss's Law the amount of charge in these two volumes must be equal and opposite. This holds even if you use a "slightly smaller box" than the entire volume.
Comment: @MichaelSeifert Thanks, that's a better argument than the one I gave!
Answer: Convergent Fourier series for the Coulomb field may not exist; it is not L2 integrable through the singularity, only L1 integrable and not all L1 integrable functions have convergent Fourier series. If it exists, one cannot exchange summation and differentiation in that series as you have shown that leads to contradiction.
If we use Fourier transform instead, we find that the transform behaves at $\mathbf k = 0$ as $\mathbf k/|k^2|$ so your argument does not work.
Comment: http://users.jyu.fi/~salomi/lecturenotes/FA_distributions.pdf These notes suggest the procedure is justified as a distributional equation, were the fourier series converges for periodic test functions.
Comment: Where in the notes is it shown that the Fourier series for the Coulomb field is the same distribution as the Coulomb field?
Comment: Theorem 2.4.6 shows that a periodic distribution is uniquely defined by it's Fourier decomposition, and 2.4.7 shows that you can exchange the derivative with the summation.


Resulting torque on the cross section

Question:
I have trouble solving part A of the question. I thought the resulting torque would just be 10k * L of the beam.
Answer: Your suggested solution indicates that you seem to be confusing torque with bending moment. This is understandable since they are often used interchangeably in common parlance.
However, they mean very different things in engineering. To understand the difference, it's important to clarify the standard notation for a beam's local axes:

The x-axis is the beam's longitudinal axis. That is, the axis which describes the beam's length.
The y- and z-axes are the beam's transversal axes, describing it's width and height.

So, given that, we can then state that:

Bending moment is the internal stress state along a beam due to rotations around the two transversal (y and z) axes. When you think of a beam or cantilever deflecting downwards due to a load, that beam is under bending moment.
Torque is the internal stress state along a beam due to rotations around its longitudinal (x) axis. When you think of a screw being twisted, a torque has been applied to it, causing torsion.

So, the calculation you expected of $\text{Force}\times\text{Length}$ would give you the bending moment of the beam, not the torque.
However, the general gist of your solution is correct: both bending moment and torque are equal to $\text{Force}\times\text{Distance}$. However, to calculate the torque you need to use another distance.
The question states that the force is "acting on the outer edge of the cross-section in the direction of the symmetry axis." From this we can conclude where on the section the force is applied and in which direction it's pointing:

I personally find "outer edge" to be unclear, but it's fair to assume they mean the left- or right-most point of the section (one of the top corners).
Likewise, "in the direction of the symmetry axis" is also unclear since triangles have three axes of symmetry, but it's fair to assume they mean the vertical one, which therefore means the force is vertical.

So now we just need to calculate the torque itself. The force is vertical and on the left- or right-most edge of the section. Since the force is vertical, the torque is calculated by multiplying it by a horizontal distance. The relevant distance in this case is to the beam's centroid, which runs along that vertical axis of symmetry. Therefore the distance is equal to half the section's side, so $120/2=60\text{ mm}$. Therefore
$$T = F\times d = 10 \times 60 = \pm600\text{ kNmm}$$
I had to put that $\pm$ sign because the problem as stated doesn't give us enough information to know whether the resultant torque is positive or negative. That depends on where the load is applied (left or right-most corner) and the force's direction (positive or negative, though I suppose one could argue it should be in the positive "direction of the symmetry axis"... which itself is unclear since we can't necessarily assume the positive direction is upwards).
Answer: In a thin walled open tube of uniform thickness J is;
$J = \frac{1 } {3 }\times Ut^3  $
t = thickness of the wall of tube.
U is is the length of the median boundary (perimeter of median cross section)
So the rotation is $ \theta = \frac {TL} {GJ} $
$  T = 10kN \times x,\space x \space is\space 1/3 \space of\space triangle\space height. $
$  Height of triangle =3x=60 \times \sqrt 3= 103.92mm \space x = 34.64mm $
$T = 10kN\times 34.64mm $
Torque = force times distance to center of rotation.
So this is the answer to first part  (a).
Comment: Why is $x=h/3$? The force is vertical ("acting in the direction of the axis of symmetry"), so the relevant distance to the centroid is in the horizontal direction, which would make $x=h/2$ (horizontal distance from edge to centroid).
Comment: @Wasabi, the equilateral triangle has three axis of symmetries, perpendiculars from one of the corners to the middle of the base across from it. May be you are thinking of the vertical line through Geometrical Center of triangle but that is not an axis of symmetry.
Comment: Sure it is. It's not the only one, sure, but it certainly is an axis of symmetry.
Comment: no, it is not. when I get back from work I explain. I

Resulting torque on the cross section


Ingress client certificate authenticate requires CA certificate to be stored in secret?

Question: I want to enable client-certificate authentication in my AKS cluster and I have a basic question which I just don't seem to understand. As per the docs, ingress requires the CA certificate to be stored in a secret. My question is: Assuming that I use client-certificates that have been issued by a trusted CA (that's how it works right? CAs issue client-certificates that they sign?), why would a trusted CA give me their CA certificate to be stored in AKS cluster as a secret? Do CAs just hand out their certificates out to public? Isn't that a security issue? (since I can sign client-certificates using that CA certificate)
Comment: If a CA were to hand out a _key_ (or a certificate making another key into a CA), [it would indeed be newsworthy](https://archive.is/o/www.wired.com/2010/03/packet-forensics/). *Packet Forensics doesn't advertise the product on its website, and when contacted by Wired.com, asked how we found out about it. Company spokesman Ray Saulino initially denied the product performed as advertised, or that anyone used it. But in a follow-up call the next day, Saulino changed his stance… "Our target community is the law enforcement community."*
Answer: In the context of X.509, which is used in TLS, we usually use the term "certificate" to refer to a structure which contains, along with identifying information, a public key.  Since this key is public and the certificate is generally available to anyone on the Internet, there's no risk in providing it.  It probably needs to exist in a secret store only because it's a convenient place to load it from, not because it's actually secret.
What you're probably thinking of is the private key corresponding to the public key in the certificate, which you of course won't be able to access.  Publicly trusted CAs usually store the private key in a hardware security module, so even the people running the CA probably can't access it (only use it to sign things).
Answer: Certificates are intended to be public.

Do CAs just hand out their certificates out to public? Isn't that a
security issue?

Yes, CAs do hand out their certificates. And no, this is not a security issue. Any CA is considered as trusted only if you know its certificate. In some cases, for manual verification knowledge of fingerprint may be sufficient. But in many cases for trust a knowledge of certificate is needed. Technically, it means that CA is trusted only if its certificate is stored in your trust store. That's making certificates publicly available is not a security issue. It is even vice versa: CAs want that their certificates are broadly known and are accepted (and stored in trust stores) by as many parties as possible.
When you install some operation system, Windows, Linux, no matter what, also certificates of trusted CAs will be installed. When you install FireFox, also certificates of trusted CAs will be installed.
For CAs it is very hard to become trusted and to be included in the install packages of important software, because there is a very strict process for this that requires that CA fulfills many special requirements.
To the secrets: Often applications need passwords to access some external resources. It is unsafe to store such passwords in the application code. Passwords are made available to the application only when application is put to the Kubernetes environment. Kubernetes uses for this a mechanism called secrets. The article you linked uses this mechanism to store certificates. Certificates are not secret. Then why public information like certificates is stored as a secret? This is because the word "secret" is misleading in this context. Better name for "secrets" in Kubernetes would be "credentials" or "authentication data" or something similar.
Answer: A "certificate" is a digitally-signed message which says, in a machine-readable language, "<this public key>'s corresponding private key is authorized for <these purposes>". The language is usually X.509, which stores "these purposes" in the Key Usage bitfield and Extended Key Usage sequence. (If you tried to change either field, it would invalidate the digital signature.)
A "certificate authority" is just any key (or the operator of a key) which computers have been configured to trust such signed messages from.*
If the authority "signs your key for client authentication", they have likely not included keyCertSign in the list of authorized uses; it's probably only been given id-kp-clientAuth. This means that other software which trusts that authority will trust your key to authenticate any client using it directly to negotiate a secured connection, but it will not trust any certificates that your key were to create (even if such certificates only authorized id-kp-clientAuth).

*In this case, you are creating a Certificate Authority, but (obviously) only your cluster is configured to trust certificates from it.

Note: The CA Certificate must contain the trusted certificate authority chain to verify client certificates. … For more details on the generation process, checkout the Prerequisite docs].

If sg1993's Cert Authority (that you generated back during the prerequisite stage) certifies a key for use as a client, server, or anything else, such certificates will only be respected or recognized by software that's been configured to do so (i.e.: that you've installed the CA's key on).

Usually you would obtain this from a trusted source, but for this example we will just create one.
Answer: An initial point to note here is that the documentation you link to is for implementing Client certificate authentication for applications running on your cluster using an ingress resource, not for the AKS cluster itself.
In terms of why they're suggesting you store the information in a secret, there's actually a couple of pieces of information being stored here. First is a CA certificate which, as has been pointed out in other answers, is public information, so no problem there.
The other piece is the private key for your server alongside its certificate. It's that key that is important to maintain the security over, as anyone with that key can impersonate your server.


View file diffs in merge commits

Question: When I open the commit view for a regular commit in magit, I can see the diff for each file. When I open a merge commit, I can only see a summary of the number of lines that were changed. Is there a way to get the first behavior for merge commits?
Answer: When viewing a merge commit or while hovering it in the log, press d r <RET> to open a diff view showing all the changes.
Comment: The reason you don't see any changes is that magit calls `git show` with `--cc`.  See `man git-diff-tree` for a description of how these flags compress the patch output. Your solution diffs against the first parent and is the simplest way of seeing what changes were brought in by the merge.  You may also want to modify the range to diff against the second parent.
Comment: I have not yet accepted this answer because I am hoping that someone knows of a better way, preferably a configuration that changes the default behavior.

View file diffs in merge commits


Make an existing project a framework iOS

Question: I have an existing project. I just like to ask if it is possible to convert it to a framework?   I have tried searching but I cannot find any tutorial on how to do it.
Comment: check [this Link Adding an Existing Project to a Workspace](http://developer.apple.com/library/ios/#recipes/xcode_help-structure_navigator/articles/adding_a_project_to_a_workspace.html). This might help you. http://developer.apple.com/library/ios/#recipes/xcode_help-structure_navigator/articles/adding_a_project_to_a_workspace.html
Answer: Solution for iOS7 or previous
If you want to create a framework in iOS7 you can follow this link. It explain how to create a framework starting from a static library.
http://www.raywenderlich.com/65964/create-a-framework-for-ios
Solution for iOS8
With iOS8 is possible to create framework natively. You can find a good video in the WWDC14 website called Building Modern Frameworks (you'll find both videos and pdf).
Cocoapods
From your question is not clear what you need to do with the framework. If you need to share a lib you can give a look to the cocoapods project. You can use it for both public libraries or privates. I'm using it at work for both the scenarios
Comment: Will creating a Cocoa touch framework using the new Xcode 6 template not work on versions older than iOS8?
Comment: From what i know no. There are solution to workaround this problem like: Create a static library for older iOS and include the it inside a framework for iOS8
Comment: is it really necessary to create a static library? Can't I just add a new framework target to an existing project?
Answer: For iOS I would suggest CocoaPods unless you don't want to distribute the code.
Even in that case you could use CocoaPods. Especially to avoid embedding other third-party libraries or frameworks.
Answer: For such a useful feature it baffles me that Apple has left this feature out of user's hands, unfortunately Apple in their wisdom has deemed that only OSX would get the framework treatment.
iOS frameworks are effectively packaged static libraries which follow a defined structure that XCode recognises - and hence the need for build script (as in Ray's guide) as this doesn't exist in XCode.
Depending on your mileage with XCode (and shell scripts) and it's peculiarities this project is one saving grace.  ios universal framework. You do have to know what you are doing, as it works in only a majority of cases - you may fall into the other 10%.
The documentation could do with an improvement, but the real cruft of the help is here. You do have to install the 'real' and/or 'fake' frameworks, create an initial framework from the templates installed and copy over your files.
Answer: You are able to create a "Framework Target" in a existing project and setup the necessary files in the "Build Phases"
You can find more examples here


Should I make major edits to a question?

Question: I made a major edit to The Fermi paradox then wrote an answer to it. In fact, I rewrote the body of that question entirely. That doesn't seem like such a bad thing to do because the edit didn't become visible until it was reviewed. I actually wrote my answer before the edit became visible and now the other answers were written before the edit to the question by people who probably never would have dreamed of writing that as an answer to the current form of the question. Is it better to write a new question that's related to an existing question but more on topic then mark the original question as a duplicate of it than to make a major edit to the original question? If so, how can I mark it as a duplicate?
Comment: "After all, https://www.youtube.com/watch?v=JHQOX8EVNmE&t=4s shows that bankers were more selfish than they were trusted not to be." the hell does that have to do with the question?
Answer: If your edit makes existing answers invalid then your edit is too substantial. It can be ok to do substantial (as in entirely re-write) edits to a question, but only in cases where the question is going to be closed otherwise and only within the intentions of the original asker (a lot of site scope can be down to interpretation so editing to make some 'more on-topic' can be a bit of a grey area).
As the help center says, edits should be used (emphasis mine);

To fix grammar and spelling mistakes
To clarify the meaning of the post (without changing that meaning)
To include additional information only found in comments, so all of the information relevant to the post is contained in one place
To correct minor mistakes or add updates as the post ages
To add related resources or hyperlinks

In short; don't change the meaning of the question beyond the intent of the asker and don't make substantial edits unless absolutely necessary.
Of course it can be a fine line between inconsequential and too substantial edits. Err on the side of caution and certainly if your edit affects existing answers then you've gone too far.

Is it better to write a new question that's related to an existing question but more on topic then mark the original question as a duplicate of it than to make a major edit to the original question?

You shouldn't intentionally be asking duplicates.
If the current question is OK in meaning and intent but needs editing then edit it (within the guidelines).
If you have enough to add to the question that would make it a new question then ask a new question, but link to the original question for context and explain what more you have to add and why that makes it not a duplicate.
Comment: I see that the change has been undone which is good, but reviewers never should have approved my edit in the first place. Can I ask a new question that's the same as I originally edited that one to be then mark that question as a duplicate of it then move my own answer to the new question?
Comment: I just took another look and your question does look to be different enough to warrant a new question, you don't want to mark it as a duplicate though (as I said you shouldn't be intentionally asking duplicates). You can link to the other question in the body of yours if it gives context but it's not necessary (there's nothing wrong with asking related questions, as long as it's not an exact duplicate)...
Comment: All of that being said, I've never been on Astronomy SE so I'm not sure about scope there, you should ask on [Meta Astronomy](http://meta.astronomy.stackexchange.com/) if you want more specific advice for questions there.


Why does iTunes play the whole album on shuffle?

Question: As the title states, my library is a bit screwed up. For example, when I shuffle my entire library it tends to play some of the albums in its entirety. But this only happens to a few of them. In this case, it's Land of Pleasure & Caress Your Soul by Sticky Fingers and Man on the Moon II: The Legend of Mr. Rager by Kid Cudi. And no matter how many times I keep reshuffling the library, it tends to repeat the whole album everytime it gets to one of the songs. It doesn't even start from the middle, it always begins from the start till the end of the album.
Answer: It does this because of some broken code in the most recent releases of iTunes. They added a new feature to help those of us who want to properly group, and play in order, classical music. If the Work tag is populated then once itunes sees something in that tag it plays all of the items on the album with the same Work tag. Many have used the groupings field in the past and the latest version grabs the groupings info and stuffs it into the new Work field. So if perhaps you had something like SONY (an album label) in the groupings filed now iTunes thinks every track on the album is part of a classical Work called Sony and that you will want to hear the album in order as you would if it were movements of a classical work. Very frustrating and foolish error on their part. One hopes the can fix this in the next release as it makes shuffle play pretty much un-usable if one uses the grouping field as it was originally intended - to group songs in some way like as Jazz.Standards or World.Music, etc.


Attacked in Markarth, Skyrim

Question: When I enter Markarth Keep and get to the first set of stairs towards the King all the guards become hostile and start attacking me. Nothing happens until I reach this point, not even if I drag my weapon. As soon as I reach the stairs they attack me, with or without my weapon drawn.
I only have an 5 gold bounty, and I haven't threatened anybody.
What's wrong? Have anybody else experienced this?
Comment: Have you recently been working on the Forsworn Conspiracy quest by chance?
Comment: @Katustrawfic having the same answer does not mean that it's the same question.
Comment: possible duplicate of [Can I secretly kill every guard in a town?](http://gaming.stackexchange.com/questions/39361/can-i-secretly-kill-every-guard-in-a-town) The question name doesn't make it sound like a duplicate but if you read the answer you'll see why it's a dupe.
Comment: This was a recent quesetion... Brotherhood quests can affect this so can your level of Vampirism (more specifically, level 4).
Answer: Try committing a minor crime in the city(not the keep) so a guard approaches you in the usual manner. Pay off your bounty or go to jail either way your bounty should reset to zero, preventing this from occurring.
Comment: Worth noting that if you opt to do go to jail, you should remove any stolen items from your inventory, otherwise they'll be confiscated and you'll lose them.
Comment: @Alex Before the game decides to reset a chest I'd put my stolen stuff in "Evidence" anyday. The hardest part about getting it back is figuring out where the jail is.
Comment: "Had your stolen stuff stolen? Steal it back!" :) If I recall chests in your home never get reset, but with Skyrim's bugs anything is possible...

Attacked in Markarth, Skyrim


My dog tears up the towels I line her crate with

Question: Is this a behavior problem, or material problem?  Do I need to change what I line the crate with?  Or is she misbehaving?  She's not in the crate, most of the time.  I put her in there when I take a shower, or go out to the store.  Then she tears the towels to shreds!
Comment: How old is your dog?
Comment: @andrewbuilder - 1 year
Answer: It sounds like your dog is bored, and ripping up the towels is a fun way to pass the time.  You can probably take the towels out, and let her lay on the crate pan.  I doubt the towel adds much in the way of comfort of warmth.  Either way, even if you only put her in there while you take a shower or do another equally short activity, put a toy or Kong filled with treats in there with her.
Answer: The issue is that you put your dog in the crate and then leave.  The dog now associates being in the crate with separation, and now you get the anxiety being expressed as destruction.
Crates should be a positive place for the dog to be - it's their space to feel comfortable and safe.
There's plenty of reading to be had on crating dogs, but start with crating your dog earlier and staying around until she's comfortable and relaxed, at which point you can leave without any fuss.


How to map the deep web?

Question: I was thinking of having a crawler visit all the pages on the dark web. But there's no broad similarity in the way the websites are named.
Answer: Please do not attempt this
The entire purpose of hidden services and such is privacy. You suggestions would destroy that. A search engine to find the supposed to be unfindable? A paradox in itself. Some of these services are hidden for a very good reason, such as humanitarian organization under oppressive regime. Your search of the dark web would help the oppressive regime out a lot
If you really need to access a hidden service, contact the server operator. Some services do already offer something similar, like this
Hidden service are hidden and by design of the tor network should remain that way
Comment: Well, my object is not to really build a search engine. Any website that can be crawled can literally be visited in the first place. I don't see therefore, how this would violate the privacy. My main obstacle is to get the possible seed URLs because a TOR URL can have as many characters as one wants. I have seen variations. Those aren't just a set of 16 alphanumeric characters with the onion registered domain. There are tor pages that do something identical like the `fresh onions` which I can't seem to find all of a sudden.
Comment: @saurabh000345 How is it not a privacy concern? Your doing the bad guys work for them! You are handing over addresses that they may not have known before
Comment: I won't be releasing it.
Comment: @saurabh000345 doubtful. You would not go through this just for yourself. And even if you did, you now provide a single point of access to the data: your computer
Answer: Actually your suggestion is what I started before. The main problem is the seeds. We do not have all the onion addresses. I recommend you to read my project details named OnionHarvester:
OnionHarvester
Comment: That's a pretty basic approach and unfortunately, I can't seem to find any other way either. TOR is slow and harvesting the details of that number of websites or even to check if their server is up would be tedious and would take a lot of time. `Don't understand why someone has downvoted your comment though.` If you've run the code yourself, why don't you share the results you might have got?!
Comment: IN your blog, you had linked your question and there I found out something interesting from the `Ethical TOR Research.` It reads: `*It is not acceptable to run an HSDir, harvest onion addresses, and publish or connect to those onion addresses.*` Could you explain this to me? I wanna cross-check my understanding as this seems something serious.
Comment: Actually TOR does not like to Hidden Services get discovered in any way. And I myself think that real Criminals are not in public sites such as The Hidden Wiki. Therefore a way should be available to getting the addresses: 1) Asking Hidden Service owners such as [Ahmia](https://ahmia.fi) does 2) Running rough HSDirs (which the hit rate of Onion addresses is low) 3) Bruteforcing the address space which is acceptable for now But with Onion v3 (56 chars) it seems to be infeasible!
Comment: @saurabh000345 it means tor hidden service are hidden for a reason, whether illegal or legal. Your search engine break the foundation of them

How to map the deep web?


After updating android studio 3.6.3 and flutter it gives me an error that not find or load main class com.android.sdklib.tool.sdkmanager.SdkManagerCli

Question: <code>'C:\Users\lenovo>flutter doctor --android-licenses
Error: Could not find or load main class com.android.sdklib.tool.sdkmanager.SdkManagerCli''
</code>
I have set environment parameter.
Even installed AndroidSdkTools(obsolete)
here is a log of flutter doctor.
<code>C:\Users\lenovo>flutter doctor -v)

[!] Android toolchain - develop for Android devices (Android SDK version 29.0.3)
    • Android SDK at C:\Users\lenovo\AppData\Local\Android\sdk
    • Platform android-29, build-tools 29.0.3
    • Java binary at: C:\Program Files\Android\Android Studio2\jre\bin\java
    • Java version OpenJDK Runtime Environment (build 1.8.0_212-release-1586-b04)
    X Android license status unknown.
      Try re-installing or updating your Android SDK Manager.
      See https://developer.android.com/studio/#downloads or visit visit
      https://flutter.dev/docs/get-started/install/windows#android-setup for detailed instructions.

! Doctor found issues in 1 category.'''
</code>
Answer: I just managed to resolve the same issue, here is what worked for me:

Go to File - Settings - System Settings - Android SDK
Select the "SDK Tools" tab
Here is the tricky part: for most people with the same problem on the web, you simply need to uncheck the "Hide Obsolete Packages" option on the bottom right corner
Screenshot 1. Then, an "Android SDK Tools (Obsolete)" option will appear Screenshot 2. Check it, click Apply and then OK. Now, you will be able to run "flutter doctor --android-licenses"
For me, number 3 did not work, as the obsolete option did not appear. So I uninstalled the "Android SDK Tools" version which was already installed, the non-obsolete (just unchecked it, clicked apply, etc.). After that, unchecking the "Hide Obsolete Packages" option on the bottom right corner made the "Android SDK Tools (Obsolete)" appear.

Unfortunately, I did not take a screenshot of the last process and now I cannot reproduce it. Generally, the steps above were a bit "trial and error" and included reinstalling Android Studio at some point or installing the "Android SDK Command-line Tools", but I am not sure about their effect.
edit: Sorry, did not see you mentioned that you already installed the obsolete sdk tools, so my answer is not really useful...


is learning rate schedule a hyperparameter?

Question: I believe term 'learning rate schedule' is a certain solution for tuning the learning rate. But at the same time, every parameter evaluating the parameter itself can be called a hyperparameter. So can the learning rate schedule be called a hyperparameter?
Answer: Hyper-parameters are essentially parameters that you have to tune manually and aren't learned.
The schedule can be considered a hyper-parameter, as well as all the scheduler's parameters (patience, decay, etc.) and even the decision to use or not a scheduler can be considered a hyperparameter.
Answer: Learning rate scheduling can be achieved through multiple ways e.g. 1 cycle, power, performance.
Selecting the approach is first Hyperparameter.
Then, selecting the appropriate parameter value for the approach will be another set of Hyperparameters. e.g. eta0, step size etc.


Safe mobile browsing

Question: What actions can I take to make web browsing on mobile phone (android) as safe as possible? My main concern are sites that would install malware (has happened before on desktop computer). On a computer, I would run browser in a sandbox environment (e.g. Sandboxie). Are there similar alternatives for mobile phones? Or are mobile environments more completely sandboxed by nature?
Comment: Android sandboxes by default: https://source.android.com/security/app-sandbox
Answer: Installing malware through websites is not as common as you seem to think. The malicious website needs to exploit a vulnerability in your browser which permits to get root access (quite uncommon).
So, as far as your browser is up to date, only the 0days (rare) could affect you.
But, even with less privilege, the malicious website could still tamper the behavior of your browser (spying you, insert scripts in the visited pages, ...).
Sandboxes can be a partial answer to this threat because they focus on isolating a process from the rest of the system (to avoid a malware spraying everywhere). They do are present in Android but they are not the ultimate protection.
Another way to protect you is to disable script (javascript mainly) in the setting of your browser.
But many websites will be broken as client side scripting is now a huge part of the Web 2.0.
Security experts, who want to keep their computer safe, rely on up to date Linux (android is Linux-based) and avoiding crap websites with malvertising (like porn, download, buzznews, ...).
It is less overkill than disabling the scripts.
Maybe take a look at the brave browser which is designed to limit the websites capabilities on your browser.
Comment: Sandboxes "focus on behavior"? Can you explain that? Sandboxes don't make decisions based on behaviour. They don't make decisions. They just contain. An up-to-date Windows box is also secure. I'm not sure that you can claim that security experts focus on Linux for security. Your list of malvertising sites is odd (what is "download" and "buzznews"?).
Comment: As far as I know sandboxes launch the analyzed target in a controlled environment and they have a evaluation function to determine whether or not the security is degraded at the end. So it is technically not an evaluation of the behavior, it is just an analysis of the result, but I used this term to mark a contrast with the simple signature detection.
Comment: Concerning the fact that security experts rely on Linux for security, the main idea was about the packaging of the Web browser (updated with the system, so no surprise like in older windows with up to date system and outdated browsers) which is present in Android too.
"Download" is about Warez, P2P, and other illegal share of cracked movies/games/programs. But also some websites which act as bank of program but ad some "bloatware"/spyware/... in the installer. Buzznews are websites that aggregate sensational news just to increase their traffic and sell ads
Comment: You are talking about a malware sandbox, not application sandboxing. The OP is talking about application snadboxing.
Comment: Yes correct... my bad. I have to edit my answer
Answer: Android has a lot of security features like SELinux, permission system and sandboxing already to try to prevent exploits and malware.
But as with a lot of security, the user might be the weakest link. It is highly unlikely that a malicious website will take root access or even just exploit your browser without you noticing, but what is much more likely is some shady website downloading a malicious .apk on your phone and tricking you to execute it.
The solution to such attack is disabling installing third-party apps, which is already disabled by default. Also most browsers will warn you against downloading an .apk, so at this point getting infected by such malware would be like willingly jumping into a cage with tiger despite all warnings.
The bigger issue to Android security can be apps from Google Play, which not always are closely inspected and examined and can contain malicious code, so you should only ever install programs that you trust.

Safe mobile browsing


How to create a JavaScript button on the related list in lightning that navigates to a visualforce page?

Question: How to create a JavaScript button on the related list in lightning that navigates to a visualforce page. Since the javascript buttons are not working on the lightning experience so we need a workaround for the same. The existing JS code navigates the end user to visualforce. we need to achieve the same on Lex on of the related list.
Answer: The related List does not support Action either in classic or Lightning. They only support button. You can create button of content Type:-

URL
Onclick JavaScript
Visualforce Page

But, Onclick JavaScript does not appear in Lightning. It only
  comes in Classic.

So you are left with only two options, Either you go for button of content type URL or visualforce page. In your case, better to choose the URL as content type of button because you can easily navigate to your users via it if you don't have complex conditions.
If you have conditions like when to navigate where, choose Visualforce page route.

Note:- You should always choose the button of type List Button to make it appear in related List.

Read more about buttons and Links here:- Create Custom Buttons and Links


Is there any motor which moves slowly and gives strong torque?

Question: I need a strong torque motor which move slow. However, I don't want to use gear train because gears will be damaged easily by the strong torque. Please give me some suggestion.
Comment: A gear train in your transmission moves your car, a gear train in a winch can lift tons. I'll reconsider the "gears will be damaged by the strong torque" :-)  ... And there are worm drives...
Comment: "strong" equates to how much force?
Comment: Gears would indeed be normal for this.  In extreme applications there may be stronger alternate forms.  Another thing to look at could be hydraulics, especially if your ultimate need is not rotary, though that is possible too.
Answer: The cheapest and most common thing you will find is a stepper motor. You can salvage a few of those from old printers and scanners.
And if you use microstep control, you can make it very slow and accurate (below the fundamental step of the motor). The problem with steppter motors is that they consume lots of current regardless of speed.
In any case, there are gears that can take a lot of torque. The gearbox in your car does a good job, for instance.
Comment: Stepper motors generally don't have a great torque. Unfortunately microstepping generate a variable torque, and under load this can leads to big positioning error without a closed loop controller
Comment: Ah, but they do have a large torque for low rpms, when compared to other types of electrical motors. If OP says he doesn't want to use any sort of gears, he's limiting his options. As for microstepping, sure, it has its downsides. But it doesn't change the fact that it does allow you to drive a stepper motor smoothly.
Comment: No other specs were mentioned, so I assume this is for a simple mobile robot project. Of course nobody would use a stepper motor when you need 10kW of power. But then again, if you need a professional application, you won't go around asking on a site like this.
Answer: Did you know about these motors :

I think the industry name is direct drive motors or torque motors , these are AC or DC servo motors that are wound and designed for High torque and low speed . Their main purpose is replacing a geared motors ( higher efficiency , better inertia , faster response , zero backlash ) there main application is rotary tables in 5Axis CNC machines and anywhere slew drives where used previously.
Some manufacturers are :

HIWIN
motorpowerco
kollmorgen
Answer: A gear reduction or pulley reduction would likely be your best option to get increased torque and reduced speed. However, this would restrict the working speed to the exact ratio between the input and output gears or pulleys.  Therefore, the solution would be a variable pitch pulley that would provide a wide range of ratios to arrive at the desired speed. There are variable pitch pulley sets that use a belt or a metal belt. The metal belt would likely not be practical for a custom-made device, but there are currently variable ratio transmissions in some automobiles. Check out the transmission in a new Nissan
Comment: Most electric motors can have their speed electrically varied, and the electrical complexity of doing so tends to hold up better than the mechanical complexity of a variable belt drive.


Can I find a power series representation for any $x$ in the domain just by varying the center?

Question: It is well known that $\displaystyle\dfrac {1}{1-x} =  \sum_{n=0}^{\infty}(x-0)^n$ when $x \in (-1, 1)$. If I want a power series representation  on the interval $(2, 4)$, can I simply find a Taylor series for $f(x) = \dfrac {1}{1-x}$ centered at $3$? Can I continue moving the center around until I cover $\mathbb{R}$ \ $\{-1, 1\}$? If not, why would we ever use a center that is not $0$?
Answer: Yes. Observe that:
$\frac{1}{1-x}=\frac{-1}{2}\frac{1}{1+(x-3)/2}=\frac{-1}{2}\sum_{n=0}(-1)^{n}\frac{1}{2^n}(x-3)^n$, as long as $x\in(1,5)$ since the radius of convergence of the power series is 2. In fact, for any $a\in\mathbb{R}-\{1\}$, You can write $1/(1-x)$ as a power series with center at $x=a$, and the power series converges to $1/(1-x)$ as long as $x\in(a-|a-1|,a+|a-1|)$.
Answer: The sum
$$ \sum_{n=1}^{M} -x^{-n} = \frac{1/x-(1/x)^{M+1}}{1/x-1} = \frac{1}{1-x} - \frac{x^{-M}}{1-x} $$
converges to $\frac{1}{1-x}$ for $\lvert x \rvert > 1$. This is akin to a Taylor series at $\infty$, but is called a Laurent series because it contains powers of $1/x$.
One can find the Taylor series of $\frac{1}{1-x}$ about the point $x=a$ by writing it as
$$ \frac{1}{1-a-(x-a)} = \frac{1}{1-a} \frac{1}{1-(x-a)/(1-a)} = \sum_{n=0}^{\infty} \frac{(x-a)^n}{(1-a)^{n+1}} $$
by again using the geometric series formula (Clearly $a=0$ recovers the original series about $x=0$). This series converges if
$$ \left| \frac{x-a}{1-a} \right| < 1. $$
You therefore find, for example, if $a=3$, the series converges if $-2<x-3<2$, i.e. $1<x<5$. This will work for any $a \neq 1$. The general principle is that the series cannot converge on an interval that includes the singularity at $x=1$. In this case, this is the only obstacle to convergence.
The usefulness of the series expansion generally lies in the ability to approximate the function as closely as we like over a finite interval by only taking a finite number of terms (if you happen to know that $12^2=144$, you can get a pretty good estimate for $\sqrt{145}$, to take a very simple example). Hence expanding about a point closer to points you are interested in provides a better estimate for less cost than a further away point (such as trying to estimate $\sqrt{145}$ based on the series expansion of $\sqrt{x}$ at $100$ instead of $144$: it's pretty obvious you won't get the same level of accuracy without taking more terms).


My icloud photos won't upload onto my new macbook air 2018

Question:
I just got this computer 2 days ago and the number of photos its trying to upload keeps climbing, but the bar won't show it's loading. It did at one point and it seemed to be done after a bit, but then it went back to the beginning. Is this something that will just take time? Also, I can't plug my phone into the computer to upload because it doesn't have a USB port (computer came out a couple weeks ago so the charging holes and stuff are new and different.) All the icloud settings say my devices are connected but it's just not working, anyone have advice?It also keeps pausing to "save battery power" which is annoying and I wish I could turn that off. Help?
Comment: Welcome to Ask Different.  You have a lot of photos and the Mac doesn't upload them all at once.  It's designed to upload them in the background, when you're on power, and when you're connected to wifi.  Just let it go for a few more days.  Be sure you have enough iCloud storage, too.
Answer: You can try this found here:

If your Photos library won't open, or if Photos behaves unexpectedly,
  the Photos library repair tool might resolve the issues.
Follow these steps to use the Photos library repair tool:
Hold down the Option and Command keys while you open Photos on your
  Mac. In the dialog that appears, click Repair to start the repair
  process. You might be asked to type in your user account password to
  authorize the library repair.
The repair tool analyzes the the library's database and repairs any
  inconsistencies it detects. Depending on the size of your library,
  this might take some time. When the process is completed, Photos opens
  the library.


How do I prevent getting an Out of Memory error?

Question: Often, when I join a game, I'm able to play for approximately 15 minutes before the game straight up crashes on me, telling me that there was an Out of Memory error. How do I fix this? I am on PC, not Mac.
Comment: Well, for starters, how much RAM do you have?
Comment: Adding to @Timmy_Jim 's comment, how often do these crashes occur. And when did they start? Any installed programs and or viruses that you're unaware of?
Comment: I use a lap top, and currently not at home. I use Avast and run virus checks quite frequently. I have no clue when they started.
Comment: @PacmanLP - Turn off your Antivirus when playing TF2. AV programs have a habit of going nuts every time a program writes/deletes a lot of files (such as temp files created by games).
Answer: Solution 1
I've been searching around and found the answer on the Steam Forum.
Here is a slight paraphrase / quotation of the answer-er (credits at the end).

To start troubleshooting, we need to make sure your game installation is unmodified.
First, you should delete your tf folder located in either, "C:\Program Files\Steam\steamapps\[your steam username]\team fortress 2" or "C:\Program Files (x86)\Steam\steamapps\[your steam username]\team fortress 2".
If you have scripts, maps, etc that you would like to keep, you can copy those out to a safe place before you delete them.
Next, you'll want to reset all your configuration settings. Deleting the tf folder should do this, but just to be safe, set the -autoconfig flag for Team Fortress 2.
To do this, open your Steam library, right-click TF2, click Properties, and in the General tab, click Set Launch Options. Erase everything that is in there (if there is anything) and type in

-autoconfig

Close these windows, then start up TF2. Don't change any settings, and see if it still crashes.

Thanks to scy1192 from Steam for this paraphrased answer.
Solution 2
Another solution mentioned, was to change your Video Settings: ideally disabling the Multi-Core Rendering settings. To achieve this, navigate to the Option 's 'Advanced' tab - you should find Multi-Core Rendering somewhere among the several other settings, do not mess with them - when you find it, disable it.
When you join a TF2-Match, you will probably notice a significant FPS-Drop depending on whether your computer relies on or semi-relies on Multiple CPU Cores.
This is a side-effect at most, it may affect game-play performance, but it is sure to stop any 'random crashes'.
Solution 3
This ties directly back to Solution 1, in that you need to 'clean up' your files in the tf folder (save the important or essentials to a safe-zone, then delete the contents of the tf folder). If you haven't skimmed over it yet, go back to it! The reason why this is classed as another solution, is that many, many, many people have been referencing this exact process... (should I have given credit?)
Anyway, to Solution 4 (I feel inclined to link it to Solution 3, as 3 is just a referral-call) !
My solution revolves around changing your video settings to the absolute lowest and shutting down any of the computer's background processes. If this doesn't work, all I can say is RIP.
Anyway, I hope that this helps and gets you back to your game. Good luck.


Where will the atmosphere of our Earth go if our sun blows it away completely?

Question: What will happen?Will our atmosphere head to the next gravity dominant body?Will our moon hold a part of our atmosphere or is too weak to hold it?Where all those gases are supposed to go?
Comment: @Chappo saw the article...doesn't says anything about where our entire atmosphere will go if it is blown away..about exosphere i knew it before hand,thanks for being helpful
Answer: The Sun erodes a planetary atmosphere through its solar wind.  Presently, the solar wind passes Earth at 250-750 kps, which is roughly ten times the Sun's escape velocity from earth orbit.  So atoms and molecules of gas eroded by the solar wind are carried right out of the Solar System into the beginnings of interstellar space.
At that point (ca. 100 AU) the solar wind starts interacting with the interstellar medium (IM) which is a very low density plasma. The Sun is plowing through the IM at a speed around 25 kps, and there's a bow shock ahead of it and a long tail "downstream."  The Sun's magnetic field also plays a big role at extreme distance.
It seems likely that some of the solar wind escapes altogether, while some of it piles up out there. (Note that at 100 AU, there's a lot of space and Earth's entire atmosphere and all its water too, wouldn't make a noticeable impact on gas density out there.)
So, best answer is that as Earth's gasses are stripped away, it gets carried out to around 100 AU by the solar wind where it stagnates, some staying there and some leaking off into the IM.


How can I make icons not stay on my dock?

Question: Whenever I quit an app, the icon remains on the dock. I always have to remove it, and I hate having to do it for every app. Is there a way to make it not behave this way?
Answer: A normal app remains in the Dock under two circumstances:

The app is currently running (indicated by a triangle or glowing dot underneath)
The app has "Keep in Dock" enabled

Unless the app is enabling "Keep in Dock", then the reason the apps stay in your dock should be because the app is still running. You can quit apps by using the app menu and choosing Quit, or pressing Cmd-Q.
If the app is one that doesn't automatically quit when all its windows are closed, then there isn't really a foolproof method to force them to quit. This question includes an explanation of why this is so, and a particular answer even gives a small AppleScript that might help.
If the app isn't running (no triangle or dot) and the "Keep in Dock" setting is getting enabled without your doing so, then that is unusual behavior. You might want to check if it's happening on a different user profile; perhaps there's something running causing this to happen, and if so, it's likely limited to just your user account. Did you try using any software around the time it started happening?
Comment: I understand this, but it's not behaving properly. Every time I open (even a new app) an app, it checks "Keep in dock" even though I explicitly uncheck it and reopen it. I  always have to remove every app off the dock. I already do use cmd+q for quitting apps.
Comment: Hrm, that's very strange behavior; can you try creating a new user profile temporarily and see if the behavior persists with that user account? Perhaps there's an obscure setting or third-party program in your profile causing this, and this test will eliminate or confirm that possibility.
Comment: Try to open TinkerTool and restrict Dock to modify it's contents
Comment: @StephenJennings I just created a new profile and the problem did not occur.
Comment: Apparently, re-ordering icons in a dock will cause them to stay.  Haven't been able to find a way around this but figured I'd mention.
Answer: If the above suggestion didn't work you may have to take a deeper dive.

Go into your user Library folder using the Go Menu.  If you are running Lion you will have to hold down the alt key to see it.
Once you are there you go to the Preferences folder and then delete the file named com.apple.dock.plist.
After this restart the machine and this will reset your dock and you will have to set it up again the way you want it.

As Stephen suggested you should also try with a test account. If the issue is present there as well then you could try deleting the com.apple.dock.plist file in the main Library/Preferences folder.  You can get to it by going to the Go menu and then selecting Go to folder ... then type the the path /Library/Preferences.
Answer: Here is the setting that we were looking for... it makes this normal behavior.

macOS Mojave 10.14
Prevent "Keep in dock" - apps would stay in the dock even after quitting with cmd-q


Do neural networks create a separating hyperplane?

Question: https://colah.github.io/posts/2014-03-NN-Manifolds-Topology/ mentions that neural networks learn a representation of the data so as to make the classes linearly separable. What I fail to see is how does a neural network create a separating hyperplane that separates the classes. How do we know that it creates a hyperplane? What's the math behind it?
Answer: Disclaimer: I assume your question refers to multi-layer perceptrons, used as classifiers. But, if you want to engage in hair splitting, there are other neural network architectures and applications which do not create a separating hyperplane.
In a multi-layer perceptron, each neuron first computes a linear function on its inputs and then passes it through an activation function, which is almost always non-linear. All layers except the last are used to perform some kind of non-linear data transformation, with the eventual purpose of making the data linearly separable, or close to linearly separable. The neurons in the final, output layer also compute a linear function of their inputs. Typically, they would also pass it through a non-linear function, like a sigmoid, but this is not necessary for understanding the concept. It suffices to set a threshold on the linear output of the output neuron, and you get a separating hyperplane.
Mathematically: Let us denote by $x_i$ the output of the $i$-th neuron in the last hidden layer, i.e. the last layer before the output layer. Let there be just one output neuron in the output layer and let $w_i$ be the weight connecting the $i$-th neuron to the output neuron. Also, let $w_0$ be the bias of the output neuron. Then, the output neuron computes the linear function:
$$
y = w_0 + \sum_i w_i x_i
$$
Let us say, we classify all observations for which $y < 0$ as "class A" and the others as "class B". In other words, the equation
$$
w_0 + \sum_i w_i x_i = 0
$$
describes the class boundary. This is again a linear function; an equation describing a hyperplane in the implicit form.
Example: Let's say you have two neurons in the last hidden layer, with the outputs $x_1$ and $x_2$. Then, the equation
$$
w_0 + w_1 x_1 + w_2 x_2 = 0
$$
describes a one-dimensional class boundary (a straight line) in the two-dimensional space, spanned by $x_1$ and $x_2$. If you want, you can reformat the equation to get a perhaps more common explicit form:
$$
x_2 = -\frac{1}{w_2}(w_0 + w_1 x_1)
$$
The same works for any number of dimensions.

Update in response to comments:
The above described network, with only one neuron in the output layer, is suitable only for classifying the data into two classes. For more classes, you'll need more output neurons. In practice, people usually use as many output neurons as they have classes. Each output neuron is "responsible" for recognising one class. It should output a high value if it "thinks" that the present observation belongs to "its" class, and a low value otherwise. Again, each output neuron determines a separating hyperplane, with points belonging to "its" class ideally on one, positive, high-valued side of the hyperplane and all other points on the other.
Of course, in such case it is possible for the network to produce ambiguous results, e.g. two or more neurons claiming that a point belongs to their class. Resolving these ambiguities is beyond the scope of this question, but the basic idea is to trust more the neuron with the highest value. People commonly use some kind of normalisation, like softmax, to convert the network output to "class probabilities".
As noted above, it is also common for the output neurons to apply a non-linear transformation to the previously computed linear function. Usual choices are $\tanh$ and logistic function, which are both sigmoid, meaning that they approach horizontal asymptotes as their arguments approach positive or negative infinity. With a suitable encoding of class labels as the values of these asymptotes, the training of the neural network consists of adjusting the weights so that its output, according to some error metric, approaches the true class labels.
Comment: Re: Threshold: Yes, the activation function must be monotonic. Typically $\tanh$ or logistic, in which case you can simply compare the output to $0$ or $0.5$, respectively. It is, however, unusual to use just a single neuron to classify $n$ classes, as this would imply a certain structure among the classes, which is very unnatural. Instead, you'd typically have $n$ output neurons, one for each class ($n-1$ would technically work, too, but $n$ is simpler to interpret for more than two classes).
Comment: Re: Multiple neurons: Exactly, each output neuron would define its own separating hyperplane, with, in case of a perfect separation, members of one class on one side and members of all other on the other side. For two classes you *could* use two output neurons, but their output would be redundant, either identical or complementary. For more than two classes, it is common to use as many output neurons as classes.
Comment: So I guess it's about having a threshold, but don't neural networks classify according to which of the output neurons have the greatest probability? In this case, is it correct to say that the threshold in this case would be 1/n, where n is the number of classes? I'm a bit confused what effect a non-linear activation function has on this, so if you classify based on whether it's greater than a threshold (after applying non-linear activation) should also correspond to a threshold before applying the non-linear activation, the activation must then be monotonically increasing for this to work?
Comment: So if you have multiple output neurons, then for each output neuron, there corresponds it's own hyperplane that classifies whether it's in this class or not? If you have two output neurons/two classes those two hyperplanes are the same?
Comment: say I have 4 classes, and I'm using softmax, I could have output values of [0.45, 0.30, 0.15, 0.10], I thought that having a threshold of 1/4=0.25 would work, but in this case it would classify the point as belonging to two classes. Also as in https://youtu.be/Ilg3gGewQ5U?t=75 values may not sum up to 1, also in this case, a fixed threshold would not work, as you can have outputs[0.1, 0.01, 0.01, 0.01]aslongasone of them is greater than the others.Classification is not dictated by a threshold value but by how a value compares with other values does this make the hyperplane interpretation fail?
Comment: No, softmax is just used to convert the outputs into (something like) probabilities. You'd choose the class with the highest probability. Or, if the values don't sum to one, the one with the highest value. The hyperplane interpretation is still correct, it is just the question where you draw the threshold.
Answer: The final layer can be seen as a generalized linear model (such a a logistic regression) on extracted features.
Consider a basic network with an input layer, a hidden layer, and an output layer. Draw it out and then cover up the input layer. Now it looks like a linear or generalized linear model (such as logistic regression) on some features.
That's what's happening and what the author means: the network uses early layers to figure out the features. Then the network regresses on those features.
Answer: As the other answers have stated, all the layers up to the last layer can be viewed as a complicated way to create features (feature map), with the last layer representing a linear model. The traditional neural network construction with sigmoid activation function corresponds nicely with logistic regression.
The power of neural networks comes from the nonlinearity of the activation function, since stacking linear layers is equivalent to just one linear layer. Theoretical results are various universal approximation theorems, the first by Cybenko (1989) stating that arbitrary width neural networks with only one hidden layer can approximate any continuous function arbitrarily well. However this is an existence result and does not indicate how to actually learn these weights.
More recently (2018), neural tangent kernels (NTKs) describe the learning process of gradient descent in NNs when considering the limit of infinite width. However, as a kernel method, in the unrealistic infinite-width regime this simplifies to a linear model and does not match up with empirical performance of over-parameterized neural networks (Arona et al, 2019).
Answer: I think I may have answered my question.....
Let $x$ be the last hidden layer before the output layer in the neural network
Say I have $n$ classes
Let the multi-class decision rule be simply the class with the highest output value
Mathematically, we classify $x$ as belonging to a class $i$ if $\forall$$j$:
$f(W_ix+b_i) > f(W_{j\neq i}x+b_j)$ where $f$ is a nonlinear activation.
We now solve for the values of $x$ which satisfy this decision rule (i.e. $\{x|decision(x) = i\}$):
If $f$ is monotonically increasing this implies $f(W_ix+b_i) > f(W_{j\neq i}x +b_j) \iff W_ix + b_i > W_{j\neq i}x+b_j$
bringing everything to one side gives:
$\implies (W_{i1}-W_{j1})x_1+(W_{i2}-W_{j2})x_2+(W_{i3}-W_{j3})x_3 +...+ (b_i -b_j) > 0$
so this decision rule does correspond to a separating hyperplane
Edit: it's a hyperplane for fixed $j$, but the inequality must hold for all $j$, so I guess it must be the region that's in common/intersection of all of the hyperplanes.....?
It would be interesting to obtain an explicit expression and solve for the set $x$ that satisfies a decision rule where instead of $x$ being the layer before the output layer, it instead would be the input to a multi-layer neural network


Which fruits and vegetables should be kept in a fridge, and which outside?

Question: I'm a bit unsure about which fruits and veggies should be kept outside the fridge and which inside. It seems that apples, oranges, and bananas, are normally kept outside, and melons, watermelons, etc are kept inside. How do I know which ones should be kept inside and which should be kept outside?
Also, apart from making sure that fruits are stored on a different rack of the fridge than veggies, are there any other guidelines about storing them? I.e certain fruits that shouldn't be stored together, or anything?
Comment: See my answer here: http://cooking.stackexchange.com/a/15071/4638.
Comment: For the fruits that shouldn't be stored together, see: http://cooking.stackexchange.com/questions/4993/how-should-i-organize-my-fruits-for-storage
Comment: And see my answer here: http://cooking.stackexchange.com/questions/4993/how-should-i-organize-my-fruits-for-storage/5024#5024
Answer: Apples, pears, and many other fruits are kept in cold storage after harvest, and keeping them in the fridge after buying them will help them keep longer.  Pears, peaches, and other fruits that are sold on the green side, should be removed from cold storage a day or two before using, to allow them to ripen a bit more.  Berries should be unwashed and refrigerated, and only washed right before using them.
Only a few fruits and vegetables are hurt by being refrigerated.  Bananas will turn brown, and tomatoes lose flavor.
Some veggies, such as winter squashes, have a long shelf life outside of the fridge, and refrigeration does not really add to that.
Answer: The most suitable range of temparatures for the storage of fruit is about 4-8 degrees Celsius. Fridge temperature varies between 0 and 10 degrees.
Perishable fruits which contain lots of water (eg, strawberries) can be stored at the bottom of the fridge. However, remember that you shouldn't wash them first.
When it comes to citrus, it can easily be stored in the fridge for some time. However, they should be washed before storing.
It's not recommended to keepfin the fridge fruits, which ripens under the influence of temperature, such as apples, pears, bananas. You can, of course, put them into the fridge for a liitle time to prolong their lives, but it's better to store them at room temperature to ripe and became tastier.
If you decide to keep the fruits in the fridge - remember that they must have air flow. Do not keep them in airtight bags or containers.
Answer: Find a grocery store (or grocery store chain) that you trust has done their research (I would think the bigger the better), and see how they setup their displays. This would be especially reliable for items that store always carries (if they get, say, fresh peas once a year, they might not have done the research.)
Always reuse the research others have paid for.
Comment: I have never seen a grocery store which displays fruit or vegetable in a fridge. But there are some which have cooling and wetting systems which keep the produce at a temperature cooler than ambient, but warmer than fridge. Both cases seem to not be applicable to home storage :(
Comment: Rumtscho, you are, of course, right. But though produce coolers are calibrated specifically for produce, the principle is still the same.


Black Screen After RDP Session

Question: Would anyone know why after an RDP session either from my smartphone or from another PC, when I wake my computer and then proceed to log in, I get a black screen with cursor. I am running Windows 10 Pro 64bit, and haven't had this problem before. After the remote desktop session, it lets me get a lockscreen and prompts me to log in with my password but afterwards, either hitting enter or the arrow, I just get a black screen and the cursor. I am not able to hit any shortcuts or keypresses to make anything come up. ctrl+alt+del, ctrl+shift+esc, alt+f4, nothing seems to work. I only have one monitor so it's not an issue with windows not being able to recognize my default display like other Windows 10 users have been having, this seems to be different.
Answer: The issue was actually caused by Nvidia. Their Geforce Experience program was updated recently and their "Share" feature was updated and when enabled, that's when the problem happened. As soon as the feature was disabled on the PC i was remotely logging in to, the issue was fixed.
Comment: Do you mean "Shadow Play"? I've disabled this, but still have the issue.
Comment: I had the same issue and I must say that GeForce Experience is really a bad piece of sw unless on a gaming-only PC.
Comment: This seemed to solve the problem for me a few months ago, but now it is back. I have removed GeForce Experience since they started requiring a login to use the application, so it must be something else.
Comment: This is correct, but as of Apr. 2018 this feature is called "In-game overlay".
Comment: Are you using the app or the same rdp that comes preinstalled in windows accessories? I've had this issue using the app for older OSs.
Answer:
Try to kill your session on the remote server, if this is not possible. Give the machine a reboot if possible.

Next step is to disable Bitmap Caching in your rdp client.

Open RDC -> Show options -> Experience -> Uncheck: "Persistent bitmap caching"
Reconnect.
Comment: That didn't do the trick, and I'm already running the most updated version of Windows 10. Luckily though, I was able to fix it. Seems like the problem was Nvidia's "share" feature. Worked fine until they just recently updated the Geforce Experience program. Disabled it and that fixed it.
Comment: Forgot to mention that rebooting fixes the issue, but it happens every time I remotely access my PC and rebooting is not a permanent fix. And the problem is not with the remote access itself, that works fine. The problem is that after I terminate the session on the client, the desktop I was remotely logging into has the problem. After the rdp session, I'll actually go to the computer itself and log in locally and the issue happens with the black screen.
Comment: Can you provide some more information about the hardware that you're running. And can you ping both machines when the issue occurs?
Comment: Running Windows 10 Pro 64bit on a custom built PC. Sabertooth Z87 mobo, i5-4690, 16GB Ram, 250 GB Samsung Evo SSD. Everything else shouldn't be relevant, but I can give you the info if needed. Also, I can ping the computer with a smartphone within the LAN before and after I log in to the computer locally. Also, issue happens with built-in NIC on mobo and with dedicated Intel gigabit NIC
Comment: Totally forgot the graphics card,  its a Nvidia Reference GTX 770
Comment: Can you try the following?
Start a new RDP client on your desktop but BEFORE you click CONNECT, click the SHOW OPTIONS link, click the DISPLAY tab and set the DISPLAY CONFIGURATION to a low resolution like 640×480.

I've also found this MS page, though i dont know if it is relevant for your OS. Maybe you can give it a try:

https://support.microsoft.com/en-us/kb/3016725
Answer: Late to the show, but for future reference, try disabling shadow play. I had exact same issue on my LAN & WAN.
Answer: This RDP issue goes back to Windows XP, and still effects Windows 10; it seems to occur after having a network connectivity issue when connecting an RDP session.
If the common solutions do not work (fe. ctrl-alt-end): log in as another user.
In other-words, to be able to remotely workaround this issue: have a second account to connect with, in-case the primary account's session breaks during another RDP session.
Answer: I had the very same issue
Turning off fastboot as mentioned here solved it for me.

" This feature only works when you do a SHUTDOWN and then Boot. It
  doesn't effect a RESTART. To turn it off or on... Go to Control
  panel...Power Options, and select Choose What the Power Buttons Do on
  the left. Then select Change Settings That are Currently Unavailable
  near the top center of screen... Lower down on the window, uncheck
  Fast Startup."

Black Screen After RDP Session


mysql cluster - mysqld node failed to connect to mgmt node

Question: I am trying to setup a mysql cluster but was unable to get the mysqld node to connect to the mgmt node.
This is the error I received when I issue the command with -show warnings
<code>    C:\Program Files\MySQL\cluster\MySQL_Cluster\My_Cluster>mysqld –defaults-file=c
onf\my.102.cnf.txt -show
120310 2:49:27 [Warning] Can't create test file C:\Program Files\MySQL\cluster\
ow\des.lower-test
120310 2:49:27 [Warning] Can't create test file C:\Program Files\MySQL\cluster\
ow\des.lower-test
mysqld: Can't change dir to 'C:\Program Files\MySQL\cluster\ow\' (Errcode: 2)
120310 2:49:28 [ERROR] Aborting
120310 2:49:28 [Note] mysqld: Shutdown complete
</code>
I have looked through the web and most answers point to a permission issue. I have setup a "everyone" user group and assigned full control(r/w) to the mysql folder but it still not working. The setup is based on a windows environment.
Does anyone have any ideas to this?
Comment: were you able to fix the issue? I am facing the same problem.
Answer: hopefully the windows getting started guide can help:
http://downloads.mysql.com/tutorials/cluster/GetMySQLClusterRunning-Windows.pdf
Answer: try to start your command prompt with Run as Administrator hope this works.
Answer: Dont install mysql in Program Files, install it in C:/mysql/ and that should fix any permission problem