Does "garçon" mean male waiters only, not female waitresses?

Question: Does garçon mean (male) waiters only, not waitresses? I can't find a site which addresses that question, though etymologically, and in French, it means "boy".
Comment: @user3293056 - calling out "garçon" for a waitress would just sound ridiculous. But you can do it, of course.
Comment: To make the question more interesting, and not one about the word's definition, you could edit your question and ask what do Americans and native speakers call female servers or waitresses. By the way, even the terms "waiter" and "waitress" are being replaced by the gender neutral term "server"
Comment: **MODERATOR NOTE:** Please stop editing in typos, misspellings, and erroneous tags. Continued edits of this nature risk the post being locked for a content dispute.
Comment: Comments are not for extended discussion; this conversation has been [moved to chat](http://chat.stackexchange.com/rooms/60714/discussion-on-question-by-user3293056-does-garcon-mean-male-waiters-only-not).
Comment: In order to capture a server's attention does one call out "garcon", "server", or "waiter/waitress" ? (I can't find that French squiggly symbol on my laptop). PS The question has nothing to do with grammar, but with meaning and usage. The term is, defacto, a French loanword so those tags {meaning, etymology, loan-words, french} appear totally appropriate.
Answer: Garçon is used only for male waiters:

A waiter in a French restaurant or hotel.

Origin:

French, literally 'boy'.

(Oxford Dictionary Online)
Garçon

From c. 1400 as "young male servant, squire, page." Meaning "a waiter" (especially one in a French restaurant) is a reborrowing from 1788.

(Etymonline)
Comment: @user3293056 The cedilla is the correct spelling.
Comment: @user3293056 - Garçon is used for male waiters,  whether you like it or not.
Comment: you mean ***only*** used for male waiters? could you please stop using the cedilla, i seem to have wronged you in a former life. quoting etymology won't cut it, sorry Josh
Comment: At least in the USA, "Garçon" is not currently used for any servers at all, whether male or female, except as a (somewhat rude) joke.
Comment: that's not definitive, kinda replies to a different question, and i've already read it
Comment: @user3293056 - nothing personal, it is just a question of usage.
Answer: I'm not French, so I'm forced to rely on the net, but my sources tell me that garcon is no longer used, as it is disrepectful and condescending
Instead, for instance, this source suggests serveur/serveuse instead.
Francophones are encouraged to comment.
Comment: While serveur/serveuse might be the current French vernacular, I don't believe that either word would be used in an English-speaking context.
Comment: Yes, I can confirm that the term is less and less used in France in this specific usage, sounding now a bit arrogant, when it's not tongue-in-cheek or "played" snobism. And, for the record, I never heard it used in France to refer to a waitress, it's reserved to waiters.
Comment: I think if you are summoning a male waiter in France these days you would probably say "monsieur!" (at least that's what I do) as I have also been told that garçon is no longer the thing.  Whether you refer to a female waiter as Mme or Mlle, I don't know. Probably on the basis of her apparent age as you would in any circumstance (and put your foot in it).
Answer: Unfortunately, it is quite hard to find a dictionary entry that explicitly says the waiter must be male. The OED comes quite close, even though the wording technically separates "boy" and "waiter":

garçon, n.
  a. A boy, serving-man, waiter; in English use chiefly a waiter in a French hotel or restaurant.

You're right to point out that etymology doesn't cut it, not strictly. After all, garçon in this sense was "reborrowed" as early as 1788, very shortly after the seeming neutral sense "server" (as opposed to "male child") began to appear in French (TLFi entry 1c).
That said, it's worth noting that all nouns have a gender in French and garçon in French could only be used of males at every stage of its development. The fact that the English borrowing is used in French settings according several dictionaries' entries supports the idea that it would behave in conformity.
I'll also add my experience to that of everyone else who feels that garçon is reserved for male waiters. I think this feeling is also influenced by the widespread knowledge among English speakers of the modern French meaning "boy"; I expect that someone hearing "garçon" for the first time will usually receive that translation.

By the way, the cedilla is not optional, even though it's a pain to type. On Macs, the shortcut is Alt/Option + C. On PC, hold ALT and type 1 3 5 on the numpad.
One reason it's not optional is that it changes the sound. English speakers would generally be temped to read garcon "gar-kon" instead of the correct "gar-son" if they happened across it.

As the other commenters note, the most up-to-date answer would be that it's used of neither male nor female waiters, being seen as something of a snobism or, at best, outdated. @WhatRoughBeast's links are helpful.
As we can see from this Ngram, "garçon" is by far the least used (and a search for garçon individually shows a peak at the turn of the 20th century, followed by a tortuous but definite decline). Several times as common are waiter and waitress, and the gender-neutral server has seen a massive spike in the last two decades.
Comment: Downvoter, please justify your reasoning. This is an answer offered in good faith, and any improvements would be appreciated. Thanks!


Scattering of blue wavelength and red wavelength in our atmosphere

Question: I've read the reason sky appears blue is because blue wavelength is being scattered by the gas molecules, dust particles etc. Thus, because of this scattering, we are basically being bombarded with pure blue wavelength the most from the atmosphere in all directions.While during sunset, the blue wavelength is scattered to a maximum and we only see the least scattered lower wavelengths of yellow and red hues. But if blue wavelength is scattered to a maximum then shouldn't our eyes only perceive more blue colour?  I know there's a mistake in my reasoning but I just wanted to clear this off in my head. Thank you
Comment: Have you seen [Wikipedia](https://en.wikipedia.org/wiki/Diffuse_sky_radiation)?
Answer: It's not quite correct to say that blue light is scattered to a maximum at sunset and other wavelengths are scattered the least. As far as Rayleigh scattering is concerned shorter wavelengths are always scattered more than the longer wavelengths and this amount stays the same at all hours of the day.
Each particular wavelength of light experiences the same amount of scattering per distance traveled regardless of the time of day, and shorter wavelengths scatter more than longer wavelengths.
But what is different is that at sunset the light must travel through more atmosphere to reach your eye. That means, EVERY wavelength experiences more chances to be scattered, however much each wavelength scatters by. But since the blue scatters more than red, by the time it reaches your eye, so much blue light has been scattered (perpendicular to your line of vision)  along the way that almost none is left to actually enter your eye.
Your eye only detects light that actually travels in a line to enter your eye. When the blue light is scattered, it is scattered in all directions Let's say the 6 sides of a cube, for example. One of those direction is directly away from your eye, four are perpendicular to your line of vision, and only one is actually towards your eye. And each time that light scatters in 6 directions, the scattered light scatters again in 6 directions over and over again; Each time sending more of the light that initially was traveling on a line straight to your eye in directions away from your eye.
Answer: First, during day the sun is over our head so think light have to travel less distance and on entering our atmosphere it is scattered and as violet is having shortest wavelength it is scattered max. (not blue) , but as our eyes are more sensitive to blue color so we see the sky blue.
Second, during evening/ morning the sun rays have to travel larger distance. So, in doing so the rays having shorter wavelengths are  left behind or does not reach the earth but light or rays having larger wavelengths(e.g. red, orange) reach earth's atmosphere and get scattered and we see the sky reddish or orangish.


Trigger to apex class

Question: I want to transfer a trigger code into apex class, and the trigger code happens to have trigger.isinsert and trigger.new attributes. Can you suggest me how to do it?
Comment: These two can be the helpful links that can be referred: https://salesforce.stackexchange.com/questions/61237/how-can-i-move-the-following-trigger-logic-into-an-apex-class https://salesforce.stackexchange.com/questions/41591/help-with-moving-a-case-trigger-to-a-class
Comment: This question really does need additional details for anyone to be able to provide more than general advice. From other comments, it appears that you may want to transfer your code into a Visualforce controller. That would be good information to include in your question, as it does change the approach that you'll need to take. Including the code for the trigger you're trying to transfer into a class would also help (and any code for the class that you already have/tried). You can [edit] your question to include these details.
Comment: What it boils down to, in the end, is that we have no idea what the code you're looking at is. Right now, we're basically driving a car blindfolded, and you're asking us to drive you somewhere. We could try to make some turns, but being blindfolded, there's a high chance we'll end up crashing into another vehicle, a building, or maybe driving off of a bridge. You need to remove our blindfold, and you do that by _including as much detail as you can_. What is the code you're working with? What are the details about the problem you're trying to solve?
Comment: With the information you've provided so far, the only thing I can say is, "then why don't you simply remove the trigger context variables from your code?" If your rebuttal to that is "ok, but then my code doesn't compile" or "but how will I know which records to insert/update?" _how do you expect us to know that if you don't provide **details**?_
Answer: These variables are accesssible in the Trigger's Context.
You can refer below links, which explains using a handler along with a Apex Trigger.
https://developer.salesforce.com/forums/?id=906F000000093cKIAQ
http://www.infallibletechie.com/2014/05/trigger-handler-controller-in-salesforce.html
Comment: I don't want to call it from a trigger.. I just want to create a method, instead of a trigger.
Comment: @Trishagni Well, if you don't plan on using a Trigger, then where do you plan on invoking your method ? A batch class ? VF Page ? API ?
Comment: @JitenderPadda: I have a controller class for that object. I want to implement the logic in a method in that class instead using a trigger.
Comment: @Trishagni Trigger context variables can be used outside of a trigger, so long as the piece of code using a trigger context variable was called (directly or indirectly) as part of a trigger. Additional caution is due when using trigger context variables outside of a trigger body though.
Comment: @DerekF: Thanks for reverting back, the problem is solved .   i am really new in salesforce and started coding about 3 months back, i am still learning, sorry for not providing you with all the info!
Comment: @JitenderPadda: thankyou for reverting back.

Trigger to apex class


Will I need to move app updates to SD card in Android 4.3?

Question: If I move a certain app to SD card (from App Info page), what happens when Play Store installs that app's update? Will the update automatically go to SD card, or will it go to internal storage and I'll have to move it to SD card again?
Answer: From my experience with my Droid after this functionality was added to Android, updates "go" to where the app is. That is, I did not need to re-move the app to the SD card after an update. As far as I know the functionality for that has not regressed. (Actually, I don't think much work has been done on it at all. Google would like for phones to just have internal storage.)
That said, there's probably more chance of something going wrong with an update in that scenario. You may find occasionally that you have to move an app back to internal storage to update and then move it back to the SD card. Or possibly re-install. But it's probably rare.
I don't currently have a device with 4.3 and an SD card, so this is a bit of conjecture based on my experience with an older version of the OS. Caveat emptor.
Answer: The Google Play error <code>-18</code> indicates an app update fails to write to SD (due to some left-overs from a previous installation of the same app; see e.g. in the app description of VitalPlayer, this post on XDA, or try this customized Google-search for 11.900 XDA-results on the same topic). The recommended solution is to move the app to internal storage, trigger the update again, and then optionally move the app back to SD.
From this I conclude: the app gets updated in the selected storage.


How to rotate login screen in elementary os Freya

Question: I've been attempting to find a solution to this for a few months now but I can't seem to find anything. I have two monitors, one rotated clockwise.  At the login screen the rotated monitor does not rotate the login menu to match. When logged into Elementary the monitor is correctly configured. Is there a way to get the login screen to be rotated as well?
Answer: Try to use arandr programm which export bash script which on executing configure your screens. When you have "the script" for configuration you must execute "script" after the moment the xinit is executed.


Evaluating the gradient at any given point and the directional derivative.

Question: I seem to have some confusion concerning evaluating the gradient for a given point. The question is subtle so you have to follow the history. Lets assume we  know the gradient definition and are given some point. Just as an example. Lets say f(x,y) = ln(xy^2. And the point  we are interested in is (1,1,0).  So now a standard question would be to determine the direction of the steepest ascent from this point and the maximum rate of increase. Easy enough. Calculate partial derivatives and evaluate at given point to get direction and then take the norm of the gradient to get the maximum rate of increase which is a scalar. So far so good.  Initially When you calculate the gradient it does not give you a specific value but a vector with x and y components  which form a gradient field. Correct? Then AFTER you evaluate it at the point ( 1,1 ) sure enough you get the direction of the maximum increase in steepness and the maximum value is the norm. First note: I don't see how the evaluation can be anything else because only ONE possibility exists at the point. The direction at that point can only be one vector and the norm can only be one value.  Similarly if you define the directional derivative as the gradient doted ( that is the dot product ) with the unit vector u in direction u which equals the  norm of the gradient times the cosine of the angle between the gradient and the direction of the unit vector u  it is at maximum steepness when cos of the angle is 0. So the maximum steepness for the directional derivative is when u is in the same direction as the gradient. Again my confusion is that it can't be anything else for the gradient at the point. No other choice is available so why bother to calculate the directional derivative at all? The vector field has all the information you need. Subtle question but driving me crazy.  The directional derivative appears to me to be useless. Thank you.
Comment: I'm not sure I'm understanding the question.  The directional derivative in a given direction at a point gives the instantaneous rate of change of the function at that point in that direction, and it's true that one of those directions will be the gradient itself.  It's true that all the information you need is to calculate the directional derivative is contained in the gradient, but sometimes (in real-world applications, e.g.) you want to extract that information so that it can be put to use.  Taking the dot product with the unit direction vector is the way of extracting that information.
Comment: i accept your comment. Thank you
Answer: To answer your question:
Yes, the vector field contains the information but only in an implicit form.
The calculation of the directional derivative is the means by which the information is made explicit. Thus the directional derivative is very important.
Comment: I accept the answer but the dot product is a cludge, ad hoc. Once you calculate the gradient which is done with partial derivatives there are ***NO*** degrees of freedom for the maximum and minimum slopes , they are what they are at the given point therefore knowing the cosine angle for the direction derivative is completely useless as I pointed out in my original question. It is what it is at that given point.


The role of variance of the distribution plays in Bayesian inference

Question: Given prior $ \mu \sim \mathcal{N}(\mu_0, \tau^2) $, likelihood $ X_i | \mu \sim \mathcal{N}(\mu, \sigma^2) $, we know the closed-form solution of posterior is $ \mu | X_1, X_2, \ldots, X_n \sim \mathcal{N}(\mu_n, \tau_n^2) $ with $ \frac{1}{\tau_n^2} = \frac{1}{\tau^2} + \frac{n}{\sigma^2} $ and
$ \mu_n = \frac{\frac{1}{\tau^2} \mu_0 + \frac{n}{\sigma^2} \bar{X}}{\frac{1}{\tau^2} + \frac{n}{\sigma^2}} $. It can be observed that given the same number of observations $n$,the larger the $\sigma$ is, the larger the variance in the posterior is. I find it hard to get an intuition of this, because I feel fitting the same number of observations (i.e., MLE) should have the same level of certainty regardless of how spread out the observations are. In other words, I feel it is easy to understand as the number of observations gets large, the uncertainty in posterior diminishes, but the uncertainty should be independent of the sample variance (obviously I am wrong but I want to have the intuition why that is the case).
Answer: Think about a simpler problem: estimating the mean $\mu$ in a frequentist setting.
It is sensible to use the MLE, $\bar X$, as our estimator. How accurate is this on average? We know that $\bar X \sim \mathcal{N}(\mu, \sigma^2/n)$ so the mean squared error is $\sigma^2/n$, i.e proportional to the population variance.
This is quite a fundamental point. It is easier to estimate a population mean when the population variance is low, because (on average) your observations will be closer to this mean. For example, if the effect of a medical intervention is very similar for all patients (low $\sigma$), you don't need a large sample to estimate the mean effect accurately.
Comment: Hey thanks for the explanation. I agree the intuition for the sample variance on the sample mean is clearly explained by your illustration, but mean is just one specific case of the parameter. For example, it is still not clear how will the error on the sample variance estimator change w.r.t. the true variance.


"it would take me 1–2 seconds"

Question: or "it would took me 1–2 seconds"?
Comment: Nice to see the en dash used correctly.
Answer: "It would take me 1-2 seconds" is correct. It takes the infinitive.
Answer:
Once in the past, it took me one or two seconds.
Previously, it has taken me one or two seconds (at most …).
So far, it has taken me one or two seconds (and may take longer to complete).
Currently, it takes me one or two seconds (every time, to complete).
In the future, it will take me one or two seconds.
If I had done it yesterday, it would have taken me one or two seconds.
If I did it tommorrow, it would take me one or two seconds.
Comment: So, "took" never follows any auxiliary in standard English.
Comment: not too sure about 'standard' English - but I remember growing up in the north of England with people saying things like "it could have took me forever"
Comment: Yeah worth noting that HorusKol is right and that's common in many northern English dialects - but it's not "real" academically
Answer: If you are wanting to indicate how long it would have taken you to complete a task that has already been completed:

It took Stephen 10 seconds to get the right answer, but it would have taken me 1 or 2 seconds.

"it would take me 1–2 seconds"


Is there an effective decision procedure for determining whether a truth-tree of first-order logic is infinite?

Question: Reading Tomassi's Logic, he says that the truth-tree method is effective in establishing the validity of a valid sequent of first-order logic, and the invalidity of some invalid sequents. However, trees for some invalid sequents will just go on to infinity, and so "we may never be able to establish mechanically that an invalid sequent... is actually invalid" (p. 362). This led me to wonder whether there is an effective decision procedure for determining whether a tree is infinite. Having seen a few infinite trees, they seem like something that a computer should be able to detect (although I don't know much about computing so I haven't much faith in my intuitions here!). And if there were such a procedure, wouldn't first-order logic then be effectively decidable, since given any sequent, we could see whether the tree it produces is infinite or finite, and if it's infinite, then the sequent is invalid, and if it's finite, we can use the tree to see if the sequent is valid or invalid.
Comment: Yes, it would make it decidable. But order logic isn't decidable, so there is no such procedure.
Comment: I think it may help to consider an example where the infinite-ness of the relevant tree is non-obvious. Let $s(\overline{x})=t(\overline{x})$ be your favorite Diophantine equation; reinterpreting that equation as a formula $\varphi(\overline{x})$ in the language $\{0,1,+,\cdot\}$ of rings, we have that that Diophantine equation has a solution in $\mathbb{N}$ if and only if $\exists \overline{x}(\varphi(\overline{x}))$ is a consequence of the semiring axioms (the "only if" direction isn't quite obvious, but note that every semiring admits a homomorphism from $\mathbb{Z}$). (cont'd)
Comment: Let $\rho$ be the conjunction of the finitely many semiring axioms, and consider the status of $$(\star)\equiv\rho\rightarrow\exists\overline{x}(\varphi(\overline{x}))$$ in case the original equation $s(\overline{x})=t(\overline{x})$ has no solutions over $\mathbb{N}$ but non-obviously so. The resulting tree will be infinite but not obviously infinite (think by analogy with the "look for a solution" process, which of course never terminates ...).
Answer: Two points.

Start off a tree with the premises and negated conclusion of the argument you are interested in. There is indeed a systematic procedure which will either generate a closed tree (showing the argument is valid) or generate a tree with an open branch from which you can read off a valuation which makes everything on the branch true and hence shows the argument you are interested is invalid. But there is no general method of deciding whether this systematic procedure will indeed produce an open branch. As @spaceisdarkgreen says. (And it is provable that there can be no such procedure.)
But if you are interested in tableaux methods, Tomassi's book won't take you very far (it's far from a top recommendation). For better options, and alternatives to tableaux systems, see Beginning Mathematical Logic Chap 3, downloadable from https://logicmatters.net/tyl
Comment: Congrats on 50k ^_^


Add Team Foundation Server from Visual Studio 2012 Express

Question: I have a signed up for Visual Studio Team Foundation Service. I have setup a new Team Project. When I click the "Open new instance of Visual Studio" link, nothing happens. So, I figured I would try to add the team foundation server manually. So, I tried adding the following URL as my team foundation server. The preview shows: http://mydomain.visualstudio.com:8080/. When I attempt to add it, I get the following error:
<code>TF31002: Unable to connect to this Team Foundation Server: http://mydomain.visualstudio.com:8080/.
</code>
I also tried connecting to this URL http://mydomain.visualstudio.com:8080/defaultcollection with the same error. What am I doing wrong?
Answer: try http://mydomain.visualstudio.com:8080/tfs. This format is what worked for me


Operators Conceptual Meanings in Quantum Mechanics

Question: I'm confused about the meaning of various operators in quantum mechanics. I know that $X|\psi\rangle = x\psi(x),$ where $X$ is the position operator, and the projection is made onto the position basis. Additionally, $P|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x),$ where $P$ is the momentum operator and the projection is made onto the position basis. I believe that applying a Hermitian Operator (like $P$ or $X$) is intended for making measurements. What does such measurements physically represent?
Additionally, it seems that the unitary operator $e^{\frac{-iaP}{\hbar}}$ modifies $|x\rangle$ to $|x + a\rangle$ unlike $P|\psi\rangle.$ Why are they different, and how does $P$ generate a unitary transformation in $e^{\frac{-iaP}{\hbar}}$? I believe there's a connection to the Lagrangian (i.e. spacial transformation symmetry), but I'm not sure how it gives that formula. I also don't understand why the unitary operation shifts $|x\rangle$ to $|x + a\rangle$ but the operation $P|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x),$ which seems to retrieve information from the wave.
Comment: One way to deal with the disorganization is to single out *one* question and ask that. Right now you have a lot of text and it is even difficult to see what exactly you are asking! Even if no one answers, the work spent in trying to organize and single out a question is going to help you understand more, if not everything.
Comment: I apologize for my question's disorganization. I shortened my question to make it easier to read. If it's still difficult to understand, I will edit it again.
Comment: Measurements physically represent... measurements ! I.e. the ones you make with instruments $-$ we sometimes tend to forget that physics is an experimental science, even quantum physics, because of the way it is taught.
Comment: $e^{-ia\hat{p}/\hbar} = e^{a\partial_x}$ is nothing else than the translation operator (in the position basis), which you have certainly encountered in your calculus course and whose effect translates $x$, such that $x \mapsto x+a$, hence $e^{a\partial_x}f(x) = f(x+a)$.
Comment: When you write a formula, put dollar signs around the whole formula: `$P|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x)$`, not just around every single symbol that requires them as you did: `P|$\psi$> = -i$\hbar$$\frac{d}{dx}$$\psi$(x)`. When you are talking about a symbol, like $P$, put dollar signs around it: `$P$`. See my edits.
Comment: Three nice answers (as I write), each with a different slant.   If you can grok these you will have a better understanding than most undergraduate graduates.   But don't expect it to be easy going.
Answer: A measurement of the position in quantum mechanics is to have a small detector with the capability to catch a fish in the river in a small volume $V(x)$. There is only one fish in the river for each experiment and the waiting time for detection is unlimited.
If it is a classical fish, the detector is measuring the integral of the density of the static position probability distribution $\rho(x)$ multiplied by the characteristic function $\chi_V(x)$ over one dimension
$$  P(X \in V) = \int_{\mathbb R}  \rho(x) \ \chi_V(x) \ dx$$
$$ E[X] = \int_{\mathbb R} \ (X \rho(x) ) \ \chi(x) \ dx = \int_{\mathbb R} \ (x \rho(x) ) \ \chi(x) \ dx$$
So the position operator in P-theory is defined by the multiplication of the density with its argument.
If it is a very small quantum fish, we replace each operator X acting on the probabilty density of position by the same operator acting on an element of the Hilbert space
$$X \rho(x) = x \rho(x)  \longrightarrow  \psi(x)^* (X \psi(x) )= x |\psi(x)|^2$$
The second fundamental operation is motion. The translation operation on analytic functions is readily provided by the Taylor expansion:
$$\psi(x + dx) =\sum_n \frac{dx^n \partial_x^n}{n!} \psi(x)  = e^{dx \partial_x} \ \psi(x)$$
Here the exponential can be understood as a formal operator exopnential, that has to coincide with the limit of repeated translation by refinement of the partition of  an interval according to
$$\lim_{n\to \infty} (1 + x/n)^n = e^x$$
So its natural to define
$$\left(1 + i (- i \partial_x ) \psi(x)\right) - \psi(x)$$
as the canonical displacement differential in Hilbert space.  The factor $i$ is invented in order to make $-i\partial_x$ symmetric with respect to partial integration over all of space and acting in the left factor.
$$\int_{\mathbb R} \psi(x)^* (- i \phi' (x)) dx = \int_{\mathbb R} (-i \psi(x)')^* \phi (x)) dx  $$ for all vectors in Hilbert space with images of the derivative inHilbert space, too.
This means that the probabilty distribution of the momentum of that single fish in the experiment is always taken as an integral over the whole space. As always, the probability in quantum mechanics is event counting of independent, single particle experiments with destructive, terminating measurements over times long with respect to the dynamics of the system.
Answer: There is a book by Olver called "Applications of Lie Groups to Differential Equations" and I would recommend you read the first chapter to get an intuition as to is happening. This is a Springer Book so you can download the PDF for Free: (https://link.springer.com/book/10.1007/978-1-4684-0274-2)
All of this has to do with infinitesimal generators and the exponential map.
Let's start with the simplest and most fundamental example consider a vector field V, i.e. an object that assigns a vector to each point in your space. Take your space to be a Manifold, i.e. a space that looks locally like Euclidean space.
The important question now is, where do these vectors live?
The answer is that at each point of your space, your manifold M, there is a Vector space called the tangent space at x, $T_xM$, where the vector assigned to x lives.
Now consider the following:
Suppose you have a vector field on M, V, and a curve on M whose derivative at each point is given by the tangent vector assgined by V.
That curve is called an integral curve of the vector field. Now consider the maximal integral curve, or flow of a vector field going though one point, that is, (uniequeness by ODE theory) pick some point x on M, then the flow is just the longest integral curve that passes through x.
This flow is precisely what a commutation relation in quantum mechanics describes, or what the momentum generates.
You mentioned the example of $exp(-iε\hat{p})$, well here you see what this actually is:
Consider simply $exp(ε\frac{\partial}{\partial x})$ acting on a point x.
What is the result, what happens? (Try this out, will give good intuition as to what is happening in general)
You get a translation, $exp(ε\frac{\partial}{\partial x})= x+ε$.
In other words the vector field $v = d/dx$ generated a translation.(The basis of the vector are the basis of the tangent space)
Momentum is given by $-i\frac{d}{dx}$, so $exp(-iεp)=exp(ε\frac{d}{dx})$, which as we saw generates translations.
This is extended far beyond translations, as for exampl time evolution is generated by the Energy, rotations by angular momentum etc.
This is Lie theory, where the power lies locally, in the infinitesimal generators!
By exponentiating the infinitesimal generators you construct one-parameter groups of transformations that can describe symmetries.
These infinitesimal generators form what is called a Lie Algebra, which tells you how the flow of one infinitesimal generator affects the other. This is called the Lie Derivative, it is the true form of the "looking-and-comparing-neighboring-points" derivative, and is based on the pullback, but that is not that important right now.
When talking about vector fields, this Lie Derivative, and the Lie algebra is given by the Commutator!
So now let's put all the pieces together:
The way we think of uncertainty in observables is in terms of their infinitesimal generators and how one affects the other!
In other words, one can replace complicated, and perhaps non linear conditions for the invariance and symmetries of a subset, or functions or any object with a linear condition of infinitesimal invariance or condition for the infinitesimal generators that generate the group whose action we originally cared about!
I hope this helps clear things out a bit. Feel free to ask more question or point out anything I missed!


Pencil Mark On Vintage Poster

Question: How can I remove pencil marks from a Vintage Poster without damaging the poster itself? I have an Original, 1939 New York World's Fair poster, with a pencil mark on it. I'd like to find out how to get the mark off, without harming the poster's artwork.
Answer: This "answer" is in response to HOW to remove pencil marks on paper. It is information based on my personal knowledge and experience as an illustrator, graphic designer, and part-time art instructor.
Important: This answer is NOT in response to how to handle valuable documents or collectable vintage items as noted in the comment to your question by Robert Cartaino. It is a good idea to consult a professional restorer. It might even be better to leave such documents "as is." This answer is not a recommendation to follow for valuable materials.
First: Pencil leads are composed of clay and graphite mixed in different proportions to be practical for their intended use. High clay content resists wear while low clay proportion leaves a darker mark for the pressure used. Leads are graded with a number or a letter H (4) for Hard, B (1) for black, and HB (2½) for the best of both qualities.
Second: Poster printing stock is a complex material of varying amounts of fibre and starch and with different surface textures which affect the paper's "tooth." Tooth refers to the abrasiveness of the paper surface and it's ability to hold a (pencil) mark. There can be a baryta coating on the surface of the kind of paper used for printing the poster. Paper cannot be repaired. Once fibres are broken (by folding or denting), removed or lifted (by scraping), torn, or warped (by moisture) the damage is permanent and cannot be reversed.
Third: There are many different kinds of erasers and techniques for removing marks on paper. Erasers can be hard or soft, stiff or flexible, natural or artificial, solid or semi-liquid. Some can be impregnated with various abrasives (pumice or fibre glass) to scrape or abrade the mark (and the surface). There are simply too many different kinds of materials to review in this venue.
Fourth: After identifying the type of eraser for the kind of pencil mark, the manner of erasing is to avoid working the mark into the paper base. Likewise, avoid damaging the delicate surface of the original document with excessive pressure.
How: Work under the best non-glaring illumination you can find. Clean the work area. Wash your hands. Use a CLEAN eraser. Clean the eraser frequently to avoid re-contaminating the marked area. Work across the mark from one clean side of the piece, across the mark to a clean area. Work a bit at a time working from one end of the mark to the other end of the mark. Some choose to mask (cover) the adjacent clean area with an erasing shield to isolate mark while you work on it.

Pencil Mark On Vintage Poster


New miner warnings and errors, dropping nodes and peers

Question: Im new to mining.  I had several older blades sitting around not doing anything so I thought why not have them doing something.  Anyway these are all CPU based miners.  I understand its not effecient, but that's not why Im here.  I get a lot of what you see in the photo and I have no idea if that's something I'm doing wrong or something wrong with that particular state or block verification.  For the most part everything is "green", but it regurgitates these red and yellows enough to bring me here.  What are these and is the problem on my end?  I suppose that because I can't actually see whether or not this rig is doing its job and sticking rewards in the right place (I guess until it happens), I don't really know how affected I am by this either.
Answer: Those are warning, some other node tried to connect and it failed. From the log entries it seem it is still synchronizing. To test mining with CPU try the ropsten test network.
Comment: I think you got it, more-or-less.  I didn't think to open the ports required 30303 (TCP\UDP) and 8545 (TCP).
Comment: @addohm The p2p port 30303 should be open. Don't open RPC port 8545.


what's the exact meaning of 'of the people'

Question: I listened to a news report and it said she really is of the people when describing the royal wedding news and showing Meghan Markle with crowds of screaming fans. My Korean text book translated the sentence as she is really popular among people. How come of the people means "popular"? What is the exact meaning of of the people?
Comment: It does not mean "popular". It quite literally means of the people, someone who shares their concerns and beliefs, likely someone who comes from the same background, who is accepted to be one of them. One of the people. Popularity is just a possible byproduct of that. Because of course if you're accepted as the representative of a group, you'll probably be popular with that group.
Comment: The people consider that 'she's one of us' (I can't think of a good way to avoid this report/quotative conflation), rather than aloof, remote, stuck up, considering herself above the common man.
Answer: people (n.)  etymonline

Meaning "body of persons comprising a community" first recorded late
  13c. in Anglo-French; meaning "common people, masses" (as
  distinguished from the nobility) first recorded c. 1300 in
  Anglo-French. Meaning "one's own tribe, group, etc." is from late 14c.
  ... Legal phrase The People vs., in U.S. cases of prosecution under
  certain laws, dates from 1801.

and  macmillan:

[plural] ordinary people who are not members of the government or the
  upper classes

and

Government of the people, by the people, for the people, shall not
  perish from the Earth. - Abraham Lincoln

My sense is not popularity ( as she was not until her path intersected royalty) but is of the common people, of my people, of the proletarians, among us 'bread-eaters'.


Which is a Better Introduction?

Question: If I were to introduce myself, which would be better?

はじめまして、あかみです。どうぞよろしく。
はじめまして、あかみです。よろしくお願いします。
How would you introduce yourself? Any better recommendations instead of 1 & 2 ?!

Thanks :D
Comment: 1+2=3. どうぞよろしくお願いします.
Comment: どうぞ よろしく お願い いたします。is more polite
Comment: "(どうぞ)よろしくお願いします" is better in a formal situation, while "どうぞよろしく" (or simply "よろしく") is more natural when you're introducing yourself to your friend's friend. Adding "どうぞ" is optional.
Answer: Both are fine.
For some reason my first text book taught どうぞよろしく but when I went to Japan the first thing I noticed was how rarely, if ever, I heard it. People always said よろしくお願い(いた)します, so I started to do the same.
どうぞ means "please" and I would say it is used more often to make a request into a polite invitation ("please sit down"/どうぞお掛けください).
Grammatically どうぞ still means please when you say どうぞよろしく (you are effectively dropping お願いします) but when making introductions it is something of set phrase and as you know gets translated as "How do you do", "Pleased to meet you".
Answer: While I think  the sentiment expressed in both answers that "both are fine" is generally correct. I think you will hear the します and いたします forms much more frequently, and I recommend using them except in contexts where you are sure the other 「どうぞよろしく。」 construction is preferred.
I'm sure I'll get downvoted if my reasoning is wrong, but my sense is that どうぞよろしく sounds quite informal in a first acquaintance context. Mainly, this is because dropping ending verbs is a feature either of 文語 (which is definitely not normal for meeting people for the first time) or casual speech.
To me どうぞよろしくお願いいたします fits bets when I'm meeting people who are above me or joining a 学会 to which I haven't previously belonged. Used in other contexts where I'm one of the more important people in the room, it might sound overly polite (I think). I use this type when meeting my new coworkers or superiors.
[どうぞ]よろしくお願いします  is an all-purpose workhorse greeting sentence. I use it when I meet equals (by status) or when I am meeting a mixed group of people. So for instance, I would use this to the [事務所]{じむしょ} staff. I also hear this when the chair did his greeting to all of the incoming students.
どうぞよろしく。 is more informal (even with the どうぞ). I use then when at a party doing greetings or when meeting friends of friends for fun.
So if I don't know the layout, I will probably opt for どうぞよろしくお願いします. If I know where I am expected to fit, I will go with one of the others.

Which is a Better Introduction?


Why are Grizzlies in the playoffs and GSW are not, even though Warriors have more victories?

Question: Why are Memphis Grizzlies in the playoffs and Golden State Warriors are not even though Golden State Warriors have more victories? Why was the last game important if it's not the number of victories that matters?
Answer: For the 2020-21 season the NBA introduced a play-in tournament (somewhat similar to those we saw in 2019-20 post Covid shutdowns).
As part of these play-in games, the 7th and 8th seeds of each conference at the end of regular season played each other, where the winner would be ranked the 7th seed for the playoffs. The 9th and 10th ranked teams, would also play each other. The winner plays the loser of the 7th and 8th seed game and would be ranked as the 8th seed of the playoffs.
Here is a picture showing what I described above with more clarity:

In the play-ins, the Warriors finished the season 8th, and played the Lakers who finished 7th. Lakers won that match and clinched a playoff berth as the 7th seed.
Since the Warriors lost, they played the winner of 9th and 10th seed, which was the Memphis Grizzlies. So the Grizzlies and the Warriors played a single game elimination to enter the playoffs. Since the Grizzlies won, then entered the playoffs as the 8th seed, despite finishing 9th in the regular season.
Comment: So there were playoffs to enter playoffs :) Thanks for explaining. Somehow I did not find in articles or news why that last game was so important, but I noticed that somehow in both conferences the last games were decisive. Why do they need to make life so complicated :)
Comment: Haha yes, a mini playoffs for lower seeds they dubbed "play-ins". They're really juicing out for more games due to the covid lost revenue


Closures break induction in correctness proof of interpreter

Question: I'm trying to prove the correctness of an interpreter for a simple extension of untyped lambda-calculus with De Bruijn indices. The interpreter is bounded, i.e. in order to ensure its finiteness it has an additional parameter $n \in \mathbb N$ that decreases with each recursive call. The function returns $\bot$ when $n = 0$.
This is the signature of the interpreter: $\mathcal E : (\mathbb N, Env, E) \to (Val \cup \{\bot\})$
Where $Env$ is the set of environments, represented as lists of values accessed with De Bruijn indices, $E$ is the set of expressions and $Val$ is the set of values:
$$ \begin{aligned}
  E &::= \lambda E \mid EE \mid A \mid B \mid \texttt{if } E \texttt{ then } E \texttt{ else } E \mid x \in Var \mid V\\
  A &::= \textit{integer operations...}  \\
  B &::= \textit{boolean operations...}  \\
  V &::= \texttt{True} \mid \texttt{False} \mid c \in \mathbb Z \mid \texttt{Closure} (\rho, E)
\end{aligned} $$
Soundness theorem
$$\forall e, \rho, n, v . \mathcal E (n, \rho, e) = v \implies \langle e, \rho \rangle \to^* v$$
I.e.: If $e$ evaluates to $v$ under the initial environment $\rho$ and in a finite number of steps, then there exists a finite proof that it does, which uses the rules of the small-step semantics I describe in the "appendix" of this post.
Proof Attempt Sketch
I tried to prove the theorem by induction on $e$.
Base Cases
The base cases are lambdas, values and variables. I managed to prove these cases with ease.

Lambdas: Let $e = \lambda e'$. First off, the case in which $n = 0$ can be eliminated, since it makes the premise of the theorem false (ex falso...). If $n = Sm$, by the definition of the interpreter we get that the value $v$ must be $\texttt{Closure}(\rho, e')$. The case can be shown to work by applying the LAMBDA rule, some properties of the star closure and the VALUE rule.
Values: trivial.
Variables: Use rule VAR-KO to eliminate the case in which the variable is not in the environment. Then, it's a matter of applying the VAR-OK rule and the definition of the interpreter, plus some property of the star closure.

Inductive cases (problem is here!!!)
The most complex inductive case, and the one that is giving me problems, is the case of application: $e = e_1e_2$.
The inductive HPs are:
$$  \forall \rho, n, v . \mathcal E (n, \rho, e_1) = v \implies \langle e_1, \rho \rangle \to^* v \quad \text{(IHe1)} $$
$$  \forall \rho, n, v . \mathcal E (n, \rho, e_2) = v \implies \langle e_2, \rho \rangle \to^* v \quad \text{(IHe2)} $$
By applying these to the definition of $\mathcal E$ and doing some case checking (the evaluations of $e_1$ and $e_2$ cannot be $\bot$ by Hp), from the inductive hypotheses we obtain that for some $\rho'$ and $e'$:
$$ \forall \rho, n \langle e_1, \rho \rangle \to^* \texttt{Closure}(\rho', e') $$
and for some $v'$:
$$ \forall \rho, n, \langle e_2, \rho \rangle \to^* v' $$
At this point we use an auxiliary lemma:

If $\langle e_1, \rho \rangle \to^* \texttt{Closure}(\rho', e')$ and $\langle e_2, \rho \rangle \to^* v'$, then:
$$ \langle e_1e_2, \rho\rangle \to^* \langle \texttt{Closure}(\rho', e') v', \rho\rangle$$

Which is easily shown by induction.
We thus obtain:
$$  \langle e_1e_2, \rho\rangle \to^* \langle e', \texttt{bind} (\rho', v') \rangle $$
and:
$$ \mathcal E (S m, \rho, e_1e_2) = \mathcal E (m, \texttt{bind}(\rho', v'), e') = v $$
But we have no induction hypothesis on $e'$, of the form:
$$ \forall \rho, n, v . \mathcal E(n, \rho, e') = v \implies \langle e', \rho \rangle \to^* v \quad \text{(Fake-Hp)} $$
which would allow us to conclude the proof. Do you have any suggestions on how I should finish the proof I sketched in this post?
Appendix: Interpreter definition and semantics
What follows is a partial definition of the interpreter (leaving out the non-relevant portions..)
$$ \begin{aligned}
\mathcal E (0, \_, \_) &:= \bot\\
\mathcal E (S n, \rho, \lambda e) &:= \texttt{Closure}(\rho, e)\\
\mathcal E (S n, \rho, v) &:= v &\text{$v \in Val$}\\
\mathcal E (S n, \rho, e_1e_2) &:= \begin{cases}
                                  \mathcal E (n, \texttt{bind}(\rho', \mathcal E(n, \rho, e_2)), e) & \text{if } \mathcal E (n, \rho, e_1) = \texttt{Closure}(\rho', e) \\
                                  \bot &\text{otherwise}
                                 \end{cases}
\\
\mathcal E (S n, \rho, x) &:= \begin{cases}
\mathcal E (n, \rho, \rho[x]) & \text{if $\rho[x] \neq \bot$}\\
\bot & \text{otherwise}
                            \end{cases}
 &\text{$x \in Var$}\\
\end{aligned} $$
Now for the small-step semantics: The intermediate values of the computation have form $\langle e, \rho \rangle$ where $\rho$ is the environment of evaluation. The termination values are the elements of the set $V' = V \cup\{\bot\}$, where $$V = \{\texttt{True}, \texttt{False}, \texttt{Closure}(e, \rho), c \in \mathbb Z\}$$

Law for transforming lambdas into closures
$$ \dfrac{}{\langle \lambda e, \rho \rangle \to \langle \texttt{Closure} (\rho, e), \rho \rangle}\label{redlambda} \quad \text{(LAMBDA)}$$
For all values $v \in Val$:
$$ \dfrac{}{\langle v, \rho \rangle \to v} \quad \text{(VALUE)}$$
Laws for variables:
$$ \dfrac {\rho[x] = v}{\langle x, \rho \rangle \to v} \quad \text{(VAR-OK)} $$
$$ \dfrac {\rho[x] = \bot}{\langle x, \rho \rangle \to \bot} \quad \text{(VAR-KO)} $$
Laws for application:
$$ \dfrac{\langle e_1 , \rho \rangle \to \langle e_1' , \rho \rangle} {\langle e_1 e_2, \rho \rangle \to \langle e_1' e_2, \rho \rangle}\quad\text{(APPLY-1)}$$

$$ \dfrac{\langle e_2, \rho \rangle \to \langle e_2', \rho \rangle} {\langle v_1 e_2, \rho \rangle \to \langle v_1 e_2', \rho \rangle}\quad\text{(APPLY-2)}$$
$$ \dfrac{} {{\langle (\texttt{Closure} (\rho', e)) v_2, \rho \rangle \to \langle e, \texttt{bind} (\rho', v_2) \rangle}}\quad\text{(APPLY-3)}$$
Where bind adds $v_2$ to the environment, adding it "on top" of the list that represents it.
$$ \dfrac{}{\langle v\, e_2, \rho \rangle \to \bot} \quad\text{(APPLY-KO)} $$
Comment: And what you're trying to prove is true?
Comment: @AndrejBauer I think it should. The rules and interpreter are fairly simple, and I made them side by side. What's your take on this?
Answer: You must strengthen the theorem to be proved by induction. In the $e_1e_2$ case, the current induction hypothesis just tells you that $e_1$ reduces to a closure, but not anything more about how the enclosed expression reduces. Another hint that something is missing is that in the $\lambda x . e$ case (and the closure case which is hidden under "values"), you don't make use of the induction hypothesis on $e$, which is precisely about how $e$ reduces.
You will have more luck with a proof by induction on a recursive property like the following:
$P(e) = ``\forall n \rho v,\; \mathcal{E}(n,\rho,e) = v \implies (\rho,e) \to^\star v \wedge (\forall \rho' e',\; v = \mathbf{Closure}(\rho',e') \implies P(e'))"$
This creates new problems that you will have to figure out (Now it's too strong for the case of variables, how to weaken it a bit? Is this recursive property even well-defined? (if not, how to make it so?)). Hopefully this gets the ball rolling for you.
Comment: I'm certain your suggestion is on the right track; I did not manage to complete the proof but I feel like this got me closer. Thanks :)


What do the settings on a multimeter mean?

Question: I wanted to measure the resistance in ohms of an apple, so I plugged my digital multimeter into either end of the apple and adjusted the settings up to 200k and it gave me a reading of about 68. Because the meter was set to 200k does that mean I have to do 68 x 200000 to get the number of ohms for the apple or something else? Sorry, I'm new to all this.
Comment: Generally on the 200k range the multi-meter can read up to a maximum of 199. So your reading of 68 means 68k.
Comment: What does your multimeter manual have to say? Usually the range given is the *maximum* value that can be measured.
Comment: @user2943160 it's long gone. Maybe it could be on the internet.
Answer: Despite the fact the manual for your particular meter is missing, there are some fundamental concepts to using multimeters which you should familiarize yourself with.
I would recommend checking out some of the numerous online tutorials, such as this one from LadyAda, about multimeter usage. (Measuring resistance, specifically.)
Alternatively, you may want to acquire or borrow a meter with a manual and read up on it.
Ranges on meters denote the maximum value that the range supports. So 200k means you can measure from 0 to 200 kΩ. All of the readings shown while using that range should be in kΩ.
For example, if you instead set the range to 200 Ω, it would measure from 0 to 200 Ω, in which case the apple at 68kΩ would be out of range. The multimeter would display some sort of message to indicate out of range, but this differs by manufacturer/model.
Comment: Perfect. Explained everything perfectly! Thank you so much!
Comment: For a lot of multimeter, this denotes the maximum range for which the accuracy is guaranteed. They have some percent of over range that they still display, often 10 or 20%


Two different gre scores

Question: so I gave my gre for the second time and got a score of 160 in quant and 146 in verbal. However first time I got 158 in quant and 150 in verbal. Which score do I send to the university or do I send both the scores.
Answer: This is something that you may well ask your prospective universities.  Also depends on which fields you are planning on pursuing.  Second score seems overall better, but your question doesn't have a lot of information to go off of.
I would note that if your program doesn't weigh GRE nearly as much compared to GPA and research experience (let alone making it optional for reasons such as COVID) that this wouldn't necessarily be something worth tearing your hair out for.

Two different gre scores


How did Snape end up becoming the servant of Lord Voldemort in the very beginning?

Question: I am not very much into the Harry Potter books but have watched every part of the movies like a hundred times and couldn't find how Snape ended up becoming the servant of the Dark Lord for the very first time.
Comment: Your username is misspelled: it should be *valar morghulis* :-)
Comment: although this can be a random username but ill correct the spelling after a month :P
Comment: An interesting question, but alas, a dupe.
Answer: We're not told exactly how, but it's likely that he was recruited while still at Hogwarts. Sirius tells us in Goblet of Fire that Snape had a long-time fascination with the Dark Arts, and that he hung around a bunch of future Death Eaters:

Snape's always been fascinated by the Dark Arts, he was famous for it at school. Slimy, oily, greasy-haired kid, he was," Sirius added, and Harry and Ron grinned at each other. "Snape knew more curses when he arrived at school than half the kids in seventh year, and he was part of a gang of Slytherins who nearly all turned out to be Death Eaters."
Goblet of Fire Chapter 27: "Padfoot Returns"

This was a point of contention between him and Lily, in fact; at one point she accuses him of being eager to join up with Voldemort:

You and your precious little Death Eater friends – you see, you don't even deny it! You don't even deny that's what you're all aiming to be! You can't wait to join You-Know-Who, can you?"
Deathly Hallows Chapter 33: "The Prince's Tale"

Whether he joined while still a student, or shortly after graduating, is unknown.


Is this Rubik's cube broken?

Question: I am trying to solve the Rubik's cube in the picture using the layers' method. I used successfully the same algorithm on another cube, but somehow when I reach the last stage it just does not work. Does this configuration look legit (hence, I just need to try harder :) ) to you or is this cube corrupted?
Answer: This position is not solvable without twisting a corner, I just tested it on a cube that I have handy.
If you are unable to twist a corner (as that cube doesn't seem like one that would be easy to do that on), you can pop off the caps of each center piece and loosen the screw underneath.
Then, either twist the corners once the cube is looser (though this could be some trial and error), or you could disassemble the whole cube and rebuild it, that way you can ensure that everything is how it belongs.
Answer: You might need to twist the corner pieces. If you can't, its probably broken.
Answer: The cube has a twisted cube which is the yellow-green-red. Twist it and after that do an easy algorithm:
R' D' R D
Then after the corner is solved, make an U' move and repeat the step 4 times.

Is this Rubik's cube broken?


Creating a script that cleans RAM upon locking the screen

Question: I own an android with cyanogen mod installed and would like to add a script, that cleans the RAM automatically each time i lock my screen.
I seek to achieve that remains of the encryption key are getting overwritten in order to prevent a boot attack.
Thanks in advance if anyone can help.
Answer: Try installing Clean Master. In older version (Not sure about last update) you could select an option to clear RAM when screen was turned off.


Why does "No license field" Warning Disappear when you set "private": true in your package.json

Question: I have a react application that installs its dependencies via yarn install. When I run this command, I get the following No license field related warnings.
warning package.json: No license field
warning firstname.lastname@example.com: No license field

Then I updated, the package.json file's private property to be true. (See NPM docs)
{
  "name": "some-application-name",
  "author": "Keet Sugathadasa",
  "email": "email",
  "license": "MIT",
  "version": "0.0.1",
  "private": true,
  ...
}

Now, all warnings related to No license field disappeared. Why?
Comment: Because a `private` package will not be distributed and therefore doesn't need a license!
Comment: Also, you spelled `license` incorrectly. It's `license` with an `s` not `licence` with a `c`
Comment: @phuzi, Updated `license`. Thanks. I think you answered my question. If you can post it as an answer with a description, to help others as well. I can accept it.
Comment: For what it's worth, [firstname.lastname@example.com](https://github.com/micrum/react-material-dashboard/blob/v0.3.0/package.json) didn't declare any license in package.json (current version does).
Answer: The warning complains about there not being a license field (with an s), you do however have a licence field (with a c), correct the spelling and the warning should disappear.
The reason that setting "private": true works and prevents the warning from appearing is that private packages don't get distibuted and therefore do not need a license.
Comment: Spelling mistake fixed it. And thanks for sharing that `license` field is ignored when `"private": true`.
Answer: UPDATED: FIXED - CD into the right directory, I was in the parent directory
Old issue:
This did not resolve my issue, in the terminal I see:
homefolder@MacBook-Pro LFHome % yarn start
yarn run v1.22.19
warning package.json: No license field
error Command "start" not found.
info Visit https://yarnpkg.com/en/docs/cli/run for documentation about this command.
homefolder@MacBook-Pro LFHome %

Even though I changed the package.json to:
{
  "name": "seventhave",
  "email": "email",
  "licence": "MIT",
  "version": "0.1.0",
  "private": true,
  "dependencies": {
    "@fortawesome/fontawesome-svg-core": "^6.2.1",
    "@fortawesome/free-solid-svg-icons": "^6.2.1",
    "@fortawesome/react-fontawesome": "^0.2.0",
    "@testing-library/jest-dom": "^5.16.5",
    "@testing-library/react": "^13.4.0",
    "@testing-library/user-event": "^13.5.0",
    "react": "^18.2.0",
    "react-dom": "^18.2.0",
    "react-scripts": "5.0.1",
    "web-vitals": "^2.1.4"
  },


Download App Store app on one Mac to install on another?

Question: I am working with a client who lives in the country, and has absolutely poor Internet speeds. Even though the connection is "high-speed wireless" (I believe it's WiMAX based), he routinely receives about 20 KB/sec on a good day, and almost always less. In other words, think a dial-up modem with even higher latency. Normally, all this guy does is some basic Email and web browsing, and he's just the sort who's patient enough to deal with it. Not to mention, monopolies and ISP business dealings mean he's unable to get any better speed to begin with unless he went with a capped LTE link.
Anyway, this guy recently got a Mac, and is now beginning to enjoy the advantages. He just now discovered the App Store, and has been downloading some small apps. Apps that are say 3MB or so will download in 5-10 minutes for him, but as I said, he's pretty much fine with that.
Now, however, he made the "mistake" of buying a game that weighed in at 3.8GB. He's now looking down the barrel of a 8-day solid download, during which (due to latencies and such) he won't really be able to do anything else with his connection.
He asked me if there was any way I could possibly download the app (on my nice 100Mbit connection!) and give it to him on a burned DVD or USB drive. Of course, Windows vets will be used to this method for dial-up dwellers - the traditional "sneakernet". I explained the issues with the App store being intimately connected to the Apple ID, and he said he'd be glad to share his Apple ID password with me and then just change it afterwards.
I do have a spare Mac that could volunteer itself to be temporarily registered to his Apple ID to download the app. The only thing I am not sure of is if downloading the app on my machine, transferring the .app package to a flash drive and copying it to his machine will work.
I would just try it but since we're dealing with "live" data here (i.e. his account) I don't want to do too many "experiments", and also I would like to be reasonably sure before I schedule another trip out there, only to find out it won't work.
I do know Apple's crypto technologies are pretty deep and intense, and it's conceivable to me that each app download is somehow uniquely encrypted, to prevent obvious piracy.
Has anyone else tried something like this, and if so, can it be done?
Thanks!
Answer: I've tried it, and it works, provided that the app's makers don't put in their own special encryption system. There isn't even a limit of 5 computers per account on the Mac App Store (only iTunes and iOS), so you won't have to worry about that.
Edit: There's another question like this here. The answerer agrees with me.


I want to learn how to make a CPU from scratch

Question: I am new to electronics, I program from time to time and i know the basics of the logic gates. I have used logisim in the past and I was just wondering if someone could tell me about a good online class i can take or a good site to learn more. Thank you in advance
Comment: I suggest that instead of that you get a very basic micro-controller - I found the 8-bit Microchip PICs brilliant in their design, documentation and application notes - and program it in assembler. (1) It's ready made and you can get it working immediately. (2) They have a small instruction set that's easy to learn but very powerful. (3) You will learn how to move data from the accumulator to registers, etc. (4) It's very cheap and there is a vast amount of support. (5) You can build something useful. At least it will give you a very good understanding of basic architecture.
Comment: I apologise, but sadly your question is 'off-topic' because it is likely to be opinion based. Please read the [help] to understand the sort of questions this site answers. Further, a web search might give answers. So it will likely get closed.
Comment: MIT's closest course to this, 6.004, typically mixes theory and practical implementation of a CPU and has its lecture and lab notes online at http://ocw.mit.edu/courses/#electrical-engineering-and-computer-science  They may change which years's version is posted, and it tends to use a different mix of simulation and/or hardware depending on who is teaching.
Comment: This topic is way off scale in terms of a "cover all the basic's" answer. The CPU has a 'state machine' at its core, which drives a 'pipeline', which is the micro-code needed to drive all the other sections. This is where any answer becomes 'fuzzy', as the section options are up to the designer. @transistor has great suggestions.
Answer: I don't know about online classes, but I can tell you what I did when I designed my first CPU. So there are three steps:

Get a good Digital Design / Computer Architecture book. I strongly recommend the "Digital Design and Computer Architecture" from Harris and Harris, and also "Computer Organisation and Design" from Patterson and Hennessy (2nd edition). Of course there are tons of other books, but these are generally accepted as very good ones.
Learn an HDL (Hardware Description Language). I prefer VHDL, but you can also try Verilog
Design your processor on paper. Begin with a relatively small instruction set, and then you can add functionality or upgrade it (pipeline stages, FPU, etc.)

After you design your ISA and your architecture, you can start implementing it on the HDL you like. Then you can "download" it on an FPGA evaluation board and see your CPU running.
Answer: I can recommend this online coursework. You make a 1 bit ALU, then a 4 bit ALU and then an 8-bit ALU. Download Quartus II from Altera and start making a multiplexer, an adder, a decoder and put it together as an ALU and you got a 4 or 8 bit system that can perform your operations. If you get an FPGA you can even execute your own code in your own custom CPU. FPGA is a good way to learn but you can also simulate everything in a program like Qsys in Quartus II.
Answer: I agree with @transistor's comment.
Programming at assembler level might be a much quicker and more effective way to learn initially than trying to start at designing a CPU.
I think designing a CPU would be quite a lot of work to get it to do anything meaningful, and without proper test cases or use cases, the design might be rubbish but there'd be no way to tell. To get the thing to do anything meaningful, it'll also need memory and peripherals.
To put the scale of the task into context, simple 32bit embedded-style CPUs are over 2000 gates. (Based on recent RISC-V comparisons, I'll try to find the link) I believe that might be without a bus interface, or important peripherals like a communication mechanism or any timers. IIRC the size of the 32bit ALU and registers makes a difference, but a usable 8bit CPU is unlikely to be a lot smaller.
Learning to program at assembler first would be necessary anyway, as you will likely want to test your CPU design, so it isn't a waste of energy.
I'd recommend program in C and assembler with ATmega, maybe even an Arduino as they are quite cheap and quite well documented. They important value of ATmega for me is the relationship between C and assembler is good enough that you can write in C and understand the assembler code relatively easily.
Further, ATmega was designed to be a good target for C in embedded systems, and C/C++ are still the favourite high-level language for embedded development. So you would be learning about a relatively modern architecture designed for a high-level language, rather than assembler programming.
Finally, it is not cluttered with lots of mechanisms designed to support full operating systems like Linux, so there is much less to learn.
A key question is what is it you want to do with your CPU?
However, your question is likely to get opinion-based answers, which is not a good ee.se fit.


What do you mean by linear, surface and volume density?

Question: For continuous mass distribution they introduce linear, surface and volume mass density (represented by $\lambda$, $\sigma$ and $\rho$ constants). I just want to know what they mean. Also, in books they say to use linear density for rings, linear bodies; surface density for hollow sphere, plates. So, on what basis do they use these constants on different bodies, like let's say I want to use them on solid sphere so which one should I use?
Answer: For a 1D object, use linear mass density. For example, along the x-axis, one can specify a mass density $\rho(x)$ for every point on the x-axis.
For a 2D object, use surface mass density. For example, on the 2D surface of a sphere of radius $R$, one can specify a mass density $\rho(\theta,\phi)$ for every point on the sphere, where $\theta  $ & $\phi$ are the azimuthal angles.
For a 3D object, use volume mass density. For example, for a 3D solid sphere, one can specify a mass density $\rho(r,\theta,\phi)$ for every point within or on the solid sphere.
Comment: @Ankit The hollow sphere is a 3D object. But the surface of the hollow sphere is a 2D object. It's similar to saying a piece of paper is a 3D object. But if you draw on one side of the paper, you are drawing on a 2D surface.
Comment: Hmm, "on the 2D surface of a sphere of radius R", isn't sphere a 3D shape even if you consider a hollow sphere?
Answer: Linear density means the change in mass or charge per unit length    (1D)
For surface charge density mass or charge per unit area              (2D)
Volume density is charge or mass density per unit volume             (3D)
These are the properties of a material
Answer: It depends on what you want. Are you looking for something along 1D, 2D or 3D geometry? Some exapmles:

2D: Do you want to know the amount of paint that you must use to paint the roof? Possibly certain parts of the woodwork suck in more of the tint and require more paint. You could talk about the paint concentration or surface density being uneven.

1D: Are you making candles, dipping the wick in molten stearine wax, then you might want to know if you kept it down in that bath for too long at some point along the wick. That point might have accumulated a higher concentration. The linear density is not even.

3D: Finally, the most typical one is the 3D version called volume density (often simply called density) which is what you are interested in when stuffing your pillow with goose feathers. It should be even with the same concentration, the same volume density, throughout.


Measuring very small temperature differences

Question: Can one use a thermometer with $\pm$5 mK accuracy to measure a temperature difference of 2 mK (the measurement is near 100 mK temperature on a sample on an ADR)? Using the same thermometer, I am thinking to measure temperature of the sample, heat the sample slightly, measure temperature again, and take the difference. Does the $\pm$5 mK uncertainty cancel out when I take the difference? My thermometer is sensitive enough, my AC resistance bridge is capable of resolving such small temperature differences, but I want to know if the $\pm$5 mK is really an issue here.
Comment: No, it's NOT possible. the $\pm5\:\mathrm{mK}$ is simply the uncertainty on *each* measurement. Taking a difference does NOT make that go away, quite the contrary, in fact.
Answer: I do not think that you will have a problem measuring a temperature difference of $2$ mK but as to the accuracy you will achieve you need to give/get more information about the type and calibration method of your thermometer.
Suppose that the reading you got on your thermometer was $105.32$ mK' (the K-primed showing it was the value you obtained on your measuring device) the $\pm 5$ mK would indicate that the actual temperature is within about $\pm 5 \%$ of your K' value.
Say that the actual temperature was $103.12$ mK which is well within the $\pm 5 \%$
Now using your sensitive thermometer your next reading is $107.49$ mK' then the chances are that the actual temperature if fairly close to $105.29$ mK.
This is where the type and method of calibration of your thermometer comes in.
I have assumed that the reading on your thermometer in the range $103$ mK to $105$ is always about $2.20$ mK too high.
Given that you have a sensitive thermometer it is this factor which will contribute to the error in your temperature difference.
If your readings were $105.32$ mK' and $107.49$ mK' your measured difference is $2.17$ mK' and the corresponding actual temperatures were $103.12$ mK and $105.29$ mK leading to an actual temperature difference of $2.17$ mK which is the same as the value measured on your thermometer.
However it could be that the second actual temperature is $105.36$ mk which means that the actual temperature difference is $2.24$ mK compared with your measured temperature difference of $2.17$ mK which represents an error of about $3 \%$.
Also built into the error is the sensitivity of your thermometer which I have assumed to be $0.01$ mK' and the reproducibility of readings; by how much do readings of the something at the same temperature change over time.
Repeated readings will reduce the random error but will not reduce the systematic error.
Finally I think that the chances of your first measured reading being $5 \%$ too high and your second measured reading being $ 5 \%$ too low over such a small temperature range are small.

Think of it in terms of two clocks.
If both clocks tick at exactly the same rate but show different times on their dials then the difference between the times at the start and after 2 days will stay the same.
If however one clock ticks faster than the other then the difference in times on their dials at the start and after two days will not be the same.
Comment: I am using a Ruthenium Oxide thermometer (calibrated by manufacturer) and a Lakeshore AC resistance bridge. Details at http://www.lakeshore.com/products/AC-Resistance-Bridges/Model-372/Pages/Cryogenic-Temperature-Sensor-Performance.aspx. The specs say the device can resolve 3.7 $\mu$K at 100 mK with this thermometer. 4 mK overall accuracy is what got me to post here...
Answer: Yes, of course you can do it.   The 'accuracy' is from a calibration, after all, and your temperature-difference determination establishes a short-term new calibration.  If the apparatus were to have hysteresis (the meter pointer is sticky), or if there were interfering signals (the power supply ripple dominating
an electrical measurement over a short time), those could cause difficulties
that would interfere with your intention, but most measurement apparatus is
well controlled for those kinds of errors.
The key here, is that you are resolving differences below the 'it holds this
calibration for a year at a time' accuracy.   If you can resolve them, with
a repeatable measurement, those
differences ARE measured, in every sense of the word.
Comment: Agreed on your point on short-term calibration. The thermometer is a NTC resistor (Ruthenium Oxide). I guess the limitation to resolving the temperature difference then becomes how sensitive the thermometer is (how much its resistance changes with temperature, at a particular temperature) and how fine a resistance change the bridge can resolve. Right?
Comment: Yes, resolution does set a limit.   There are outlying possibilities of unknown error sources that are not under control (at millikelvin levels, your signal is rather tiny), but it isn't likely that light level or barometric pressure or magnetic field (or other ambient conditions) would affect a resistor.
Answer: Assuming the accuracy reported does not include any nonlinear factors, it refers to the inaccuracy in a single measurement, and will not vary much over the time of your experiment, then you can increase precision by integrating multiple measurements. Signal (the temperature) will add proportional to the number of measurements and noise (the inaccuracy) will add proportional to the square root of the number of measurements. Also, it is typical that inaccuracy will be fairly correlated between measurements and imprecision will be substantially less and uncorrelated. It sounds like you care a lot more about precision than accuracy--so that should help tremendously.
Answer: In general it's a bad idea to take two absolute measurements and subtract them from each other to find a difference that's comparable to the uncertainty in the measurements; the fractional uncertainty in the difference is much larger than in either measurement.  In computer science the problem is sometimes called catastrophic cancellation, but the problem is essentially the same if the imprecision is due to physical uncertainty rather than numerical truncation.
Whether your $\pm5\rm\,mK$ thermometer uncertainty is a systematic uncertainty or a random uncertainty is a question you can probably already by looking at the stability of your data when you're not changing the temperature.  The extent to which you can subtract adjacent measurements depends on the random uncertainty, which causes independent measurements of the same quantity to be uncorrelated.  If the uncertainty is systematic, it's okay to compare adjacent measurements that are larger than the random error.
If what you want is the difference in temperature between two heat reservoirs, you want a differential measurement.  For instance, the operating principle of the thermocouple is a voltage difference between dissimilar conductors that depends on the temperature difference between their two junctions.  In your case, perhaps your cryostat could contain a fairly large thermal mass controlled (i.e. by some pid-driven heater) to be near the target temperatures for your sample.  Put a thermocouple junction on your sample, the reference junction at your reference temperature, and use a sensitive ammeter to measure the relatively large changes in the small current driven by the temperature difference between your sample and your reference.


Botnet callbacks to coin mining sites - how to act?

Question: I found that one user is making connections to coin mining related domains. Is there any way to stop this activity, with respect to that user?
Comment: I assume by your tags that you are an analyst in a SOC and you saw in the proxy logs that someone's machine made connections to bitcoin mining related domains?
Answer: It depends on the context of these connections, but things that you could do are:

Identify whether these connections related to someone just browsing about coin miners (e.g. with intention to install one) or are they made by the actual software/malware (which means that the user has it installed and running). If the user was just browsing the links but didn't install anything, this activity itself is not a threat;
Take a look at the domains' categorization. Does it seem suitable? The names depend on the vendor you use, but if it's something like "Business", "Internet services" or no category then they are likely permitted even though they maybe shouldn't be permitted. In this case you should inform the vendor about the false category, they will recategorize the domains and next time these connections will be blocked. Possibly a local block for particular domains can be locally enforced, from your company's side if you report this activity;
If it is determined that the user has a miner installed, uninstall it or reinstall the whole host (depends really on your company's policy how to act here).

There is a pretty broad spectrum of possible actions in this case, depending on the context, but these things above are the usual.
Answer: Sinkhole the domains.
Either at the DNS server level or in the user's hosts file, make it resolve to 127.0.0.1.
Also, be advised: more and more sites (even legit ones, sadly) are running Javascript bitcoin miners at pageload. The big one that's making waves lately is coinhive, and it seems malware authors are also starting to incorporate it.


npm ERR! notarget No matching version found for jasmine-core@~2.9.0

Question: I created a new angular application. But while doing npm install I am getting the below error. Is anyone came across this kind of issue?
Here is my Package.json content:
{
  "name": "my-exp",
  "version": "0.0.0",
  "license": "MIT",
  "scripts": {
    "ng": "ng",
    "start": "ng serve",
    "build": "ng build",
    "test": "ng test",
    "lint": "ng lint",
    "e2e": "ng e2e"
  },
  "private": true,
  "dependencies": {
    "@angular/animations": "^5.1.2",
    "@angular/cdk": "^5.0.4",
    "@angular/common": "^5.0.0",
    "@angular/compiler": "^5.0.0",
    "@angular/core": "^5.0.0",
    "@angular/forms": "^5.0.0",
    "@angular/http": "^5.0.0",
    "@angular/material": "^5.0.4",
    "@angular/platform-browser": "^5.0.0",
    "@angular/platform-browser-dynamic": "^5.0.0",
    "@angular/router": "^5.0.0",
    "core-js": "^2.4.1",
    "hammerjs": "^2.0.8",
    "rxjs": "^5.4.1",
    "zone.js": "^0.8.14"
  },
  "devDependencies": {
    "@angular/cli": "^1.6.4",
    "@angular/compiler-cli": "^5.0.0",
    "@angular/language-service": "^5.0.0",
    "@types/jasmine": "^2.6.3",
    "@types/jasminewd2": "~2.0.2",
    "@types/node": "~8.0.47",
    "codelyzer": "~4.0.0",
    "jasmine-core": "~2.8.0",
    "jasmine-spec-reporter": "~4.2.1",
    "karma": "~1.7.0",
    "karma-chrome-launcher": "~2.2.0",
    "karma-cli": "~1.0.1",
    "karma-coverage-istanbul-reporter": "^1.2.1",
    "karma-jasmine": "~1.1.0",
    "karma-jasmine-html-reporter": "^0.2.2",
    "protractor": "~5.2.0",
    "ts-node": "~3.3.0",
    "tslint": "~5.8.0",
    "typescript": "~2.6.1"
  }
}

THis is the error
npm ERR! code ETARGET
npm ERR! notarget No matching version found for jasmine-core@~2.9.0
npm ERR! notarget In most cases you or one of your dependencies are requesting
npm ERR! notarget a package version that doesn't exist.
npm ERR! notarget
npm ERR! notarget It was specified as a dependency of 'jasmine'
npm ERR! notarget
npm ERR! A complete log of this run can be found in:
Comment: why u install so old angular version for new project? reinstall global angular cli and generate actual version
Answer: jasmine-core is a peerDependency it is not automatically installed from npm 3 on wards. You need to install it manually.

Install locally: npm install jasmine-core --save-dev
Check to make sure the installed version is the same as the version in package.json. Update the one in package.json if necessary.
Do npm install.

OR if you are not using karma-jasmine in your project, simply remove jasmine-core from devDependencies in your package.json and do npm install. Build should be successful with a warning as in below:
npm WARN firstname.lastname@example.com requires a peer of jasmine-core@* but none is installed. You must install peer dependencies yourself.
Answer: fix Package.json to:
"karma-jasmine": "~3.3.0",

"karma-jasmine-html-reporter": "^1.5.0"

and then run
npm install --force


What is it that makes public wifi so insecure?

Question: What exactly is it that make public wifi so insecure?
Is it because about anyone can connect to the network and that my device will be visible to people just as it is visible to other devices in my home network?
And what is it that's supposed to make 3g/4g connections more secure?
Answer:
Many public access points have no security at all. This means anyone within range (and with the right antennas and amplifiers that can be several hundred meters) can read all traffic.
Many of those that do use WPA2. Anyone with the password can decrypt all traffic.
Even if your traffic was safe from other users do you trust the owner of [random cafe] not to read or mess with your traffic?
With a device like a WiFi Pineapple you might not even be connected to the network you think you are.

3G/4G wise data traffic is encrypted with a per user key. Although it has its own problems - see my answer here.
With any network you do not fully control you should use a good secure channel over the top - for web access either HTTPS/SSL (remote server identity verification and hides content but not who you are talking to) or a VPN.
Answer: Public WiFi is just a open network.
Becouse  anyone in a network can manipulate traffic inside that network (see is there any security threat with open wifi connection)
You are basically trusting all other clients in the network,
not just all wifi clients but all network clients. like:

Wired clients.
Access Points.
Routers.
Switches.
Modems.

all of these could be used to record or manipulate the data you send and receive.
Since you do not have any legal contract with the other parties you can not file a legal claim against any of them if there is any serious misuse (the main reason 3G/4G differs, aside from the per channel encryption as @hector already describes).
So Basically WiFi is as secure as its components are, since in a public / shared WiFi connection you have no control over the components and the components can be added by any party, it has an intrinsic hostile network state (you can not trust anything about the network or what you send / receive on it).
the primary way to safegeurd yourself from manipulation of your actions is employing a VPN connection to a known secure endpoint. This does not close the door for directed attacks to your device though and (if applicable) you should employ firewalls and other techniques to safeguard yourself.
3G/4G does not have the same threat as the network normally does not have a way to transmit data to a device without the device initiating a connection first. (this is a setting done by the Mobile company though not a given network property).
TL;DR
Public wifi is bad for security because there is no known secured part, and any part not known safe is considered insecure.
3G/4G is safer than Wifi because the way they are normally setup is in such a way to protect customers (and carriers).


Does Gnome implement its own ssh-agent?

Question: For a long time I've wondered why when I use ssh on MacOS I need to use ssh-add to add keys to the ssh agent, whereas when I use ssh on Fedora I don't. The agent seems to automatically load keys in my ~/.ssh/ directory. The question pops up from time to time, and I usually research a little until I uncover something new, and satisfy my curiosity.
Some time ago I set up a Fedora server locally, and noticed that there I had to use ssh-add too. At the time I decided to create a config file where I specify the identity file for the only host I connect to, which is github.
Some time later I saw the same behavior on a remote server I set up on rackspace. This server uses fedora as well. However, on that server I need to start the ssh agent, as well as manually add the keys to it every time I want to use ssh to connect to github. For some reason I never set up a config file there, and today I had another one of those moments where I just want to figure everything out.
I think today's been one of the longest research sessions. On my previous research session I learned that Gnome, or more accurately X, starts the ssh agent, and adds keys to it. At that time that was enough to satisfy my curiosity, but today I really wanted to figure out how it does that, and why MacOS does not.
I spent some time reading the Xsession script, and the scripts that it includes as it is executed, and I confirmed what I had read on some sources: X starts the ssh agent and passes the gnome session script to it. So, basically I confirmed what I learned before about X starting the ssh agent. This is the reason I have to manually start the agent in the rackspace server, because that does not run X. I guess the local Fedora server does it's own set up at start time to run the ssh agent, but I haven't looked into that.
However, it wasn't very clear to me how the Fedora ssh agent got a hold of the existing ssh keys, while the MacOS agent does not. I assumed both OSes used the same ssh-agent from OpenSSH, but at this point I'm thinking they do not.
Reading this question Adding SSH keys to ssh-agent on demand I saw that the single answer there mentions that an agent may do something fancy to add keys to the agent beforehand. And it goes on to specifically mention that a desktop environment implementation may have keys in some wallet. This made me think gnome has its own agent implementation.
At this point I want to make a parenthesis to mention that in MacOS I've added a key to my local fedora server to the keychain, and this effectively loads the key in the ssh agent, and unlocks it. I did this some time ago, and it made me think that Fedora was doing something similar, where it had keys in some kind of keychain, and it was loading them that way, so today I started looking into some type of keychain in fedora. I eventually learned about gnome-keyring. Also, the question I linked above mentions gnome-keyring-daemon, and how it launches an ssh-agent
Back to today's research, I started looking into the gnome source. Specifically, the gnome-keyring project. In that project there is an ssh-agent, that seems to have pre-loading functionality for keys (https://gitlab.gnome.org/GNOME/gnome-keyring/-/tree/master/daemon/ssh-agent). This is where my question arises: Does gnome have it's own ssh-agent implementation, and that is what Fedora uses? And if that is the case, what happens to the OpenSSH implementation in Fedora systems?
UPDATE: Yesterday, after posting this question I learned my Fedora system (desktop, not server) is actually using Wayland. I had seen in the about section of the settings that the Windowing system was set to Wayland before, but it wasn't clear to me that this meant X was not being used. This may be obvious to many, but I'm not an expert on any of this. It would also seem that Wayland is not starting the ssh-agent in the same way X does, but rather on demand when it is actually first needed (How do I set a default ssh-agent timeout under Gnome).
It's also been pointed in the comments that my question may be a bit pointless, but I want to get confirmation of my suspicions from someone who actually knows, and understands the inner workings of fedora and ssh-agent precisely because I'm not an expert, and I don't want to just accept my possibly wrong/incomplete findings.
Comment: I don't understand the point of the question. You have found the source code of GNOME's implementation of ssh-agent functionality and now you're asking if it has such an implementation? "what happens to the OpenSSH implementation in Fedora systems?" - you're free to use it if you want.
Comment: Yes, it does. It's quite convenient if you use GNOME Keyring/Seahorse. It also has a GPG agent implementation, IIRC.
Comment: Thanks. You do have a point, but I want someone who knows this stuff to confirm whether gnome does implement, and use it's own version of ssh-agent.
Comment: Thanks. I'll continue researching this topic a bit more.
Comment: Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.
Answer: I found the confirmation I was looking for in the Gnome Keyring SSH Agent wiki page. There it specifically states that the ssh-agent included with gnome-keyring automatically loads files in the ~/.ssh directory that have a corresponding .pub file.
A nice resource, if you want to read more about keyring is this ITnext article.
Comment: Here is where the magic happens https://gitlab.gnome.org/GNOME/gnome-keyring/-/blob/master/daemon/gkd-glue.c#L51

Does Gnome implement its own ssh-agent?


Proof of a graph problem on degree sequence

Question: Prove that:
order of natural number $d_1 , d_2 , \ldots , d_n$ as descending is a ordering graph if and only if the order was sorted as descending $d_2-1, d_3 -1 , \ldots ,d_d -1,\ldots ,d_n$ be ordering graph.
$d_d$ in sequence is d  $d_1$.
(arrangement=order).
Comment: I'm not sure, but I think you are trying to state the [result of Hakimi for vertex degree sequences](http://en.wikipedia.org/wiki/Degree_(graph_theory)#Degree_sequence).
Answer: Hint: If you know that $d=d_1\ge d_2 \ge \dots \ge d_n$ are degrees of vertices of some graph, what can you say about degrees of the graph where the vertex of degree $d_1$ is omitted? Can you go in opposite direction, too? (I.e. can you obtain from a graph with degrees $d_1-1,\dots,d_d-1,d_{d+1},\dots,d_n$ a graph with degrees $d_1,d_2,\dots,d_n$?)
(I assume that the interpretation of your question given in this comment is correct and that the question is about Havel-Hakimi theorem.)


Use stochastic process to express solution to Laplace equation in the whole space

Question: Consider the Laplace equation in $\mathcal{R}^3$
\begin{equation}
\Delta u = f, ~~~\lim_{x\to \infty} u(x) = 0.
\end{equation}
Here we assume $f$ is a smooth, compactly supported function. Of course, $u$ can be explicitly solved with the Green function. I am considering if we can use a stochastic process (like Brownian motion) to express its solution? I know the solution can be expressed as an expectation of a stochastic process (with a stopping time) if we consider a bounded domain with a certain boundary condition. In this case, the boundary seems to play a quite important role. Can we have a similar result for the whole space case?
Comment: In [these notes](http://math.stanford.edu/~ryzhik/STANFORD/STANF227-10/notes227-09.pdf) (James Nolen, PDEs and Diffusion Processes), it appears what you are looking for is treated in Chapter 4.
Comment: Thank you for the note @WillieWong! I found in Theorem 4.1.3 there is a further condition for the representation: the Poisson equation should be like $\sum_{i,j}1/2a_{ij}u_{x_ix_j} + \sum_jb_j(x)u_{x_j} - c(x)u = f(x)$, with $c(x) > c_0 > 0.$ It seems the strict positiveness of $c(x)$ is required for the representation. Do we have to use this condition if we simply consider the Laplace equation?
Answer: If $f(x) / (1 + |x|)$ is integrable, then the solution $u$ is equal to the Newtonian potential of $f$:
$$ -u(x) = \frac{1}{4\pi} \int_{\mathbb R^3} \frac{f(y)}{|x - y|} \, dy . $$
And the Newtonian potential kernel is the occupation density of the Brownian motion:
$$ \frac{1}{4\pi} \, \frac{1}{|x|} = \int_0^\infty \frac{1}{(4 \pi t)^{3/2}} \, e^{-|x|^2 / (4 t)} dt . $$
Thus,
$$ -u(x) = \mathbb E^x \int_0^\infty f(B_t) dt , $$
where $B_t$ is the 3-D Brownian motion (with covariance matrix $2 \operatorname{Id}$).
Comment: Thanks a lot! What if we consider a slightly more general equation $b(x)\cdot\nabla u + \Delta u = f$ with a suitable vector field $b(x)$? Can we get a similar conclusion in this situation? Do we need more structure condition for $b$ to make the argument true (for example, $b(x)$ is a divergence free vector field).
Comment: If there is a Markov process $X_t$ with infinitesimal generator $L = b(x) \cdot \nabla + \delta$, then its potential operator $U$, defined by $U f(x) = \mathbb E^x \int_0^\infty f(X_t) dt$, is *roughly* the inverse of $-L$. Detailed statements require extra care, and this belongs to the field of probabilistic potential theory. Off the top of my head, I do not have a good reference, but, especially if you are willing to read more about it, you can try Dynkin's two-volume book *Markov processes*. Although very old, it is still a very good read. (Of course, there are more excellent resources.)
Comment: Many thanks for your suggestions! They are really helpful!
Answer: consider its boundary condition $lim_{x\to{\infty}}u(x)=0$, to express its inhomogeneity as a hyperbolic term, by inverse function theorem. thus not adequate to define a fast variation function. since stochastic process require a nonzero term under variation two, to make this wave be more frequently, even to make its vibration not ever disappearing.
then define an Ito calculus to represent its vibration, $df=\frac{\partial{f}}{\partial{t}}dt+\frac{\partial{f}}{\partial{x}}dB_t+\frac{1}{2}\frac{\partial^2{f}}{\partial{x}^2}d(B_t)^2=(\frac{\partial{f}}{\partial{t}}+\frac{1}{2}\frac{\partial^2{f}}{\partial{x}^2})dt+\frac{\partial{f}}{\partial{x}}dB_t$, action as a centralized source at both positive and negative side, between $y=(u-\epsilon)t$ and $y=(u+\epsilon)t$ , spanned at a distance of $\Delta{s}$, represented by its non-convergent part.
then define a mandatory function by orthogonal condition, such that $\delta(x-x_0)+c\phi_n{(x)}$. to handle its inhomogeneity boundary by generalized green function. satisfied space symmetry and time reciprocity, and the condition of compact support and smoothness in variation theory.
for example, since $\partial{f}$ is both $t$ and $B_t$'s derivation, are equivalent under the scalar derivation $dB_t$ at order two. so this time's green integral can be expressed as $\int[G(x,t;x',0)$$\psi{(x')}-\phi{(x')}$$\frac{\partial{G(x,t;x',t')}}{\partial{t'}}\vert_{t'=0}]dx'$, correspond to the reciprocal of time.
next, to express its space symmetry as a green integral be similar to the form :
$a^2{\int{[v(t')}}$$\frac{\partial{G}}
{\partial{x'}}\vert_{x'=l}-\mu{(t')}$$\frac{\partial{G}}{\partial{x'}}\vert_{x'=0}]dt'$
is it helpful? thank you.


Is it possible to recover a deleted gist on Github?

Question: I created a a gist on Github and then created a new one after making changes in it. But I accidently deleted the updated one. So I wanted to know if it's possible to recover it. Thanks for helping.
Comment: Have you tried contacting GitHub support?
Comment: @John Doe – Did Github support help you recover a Gist? I'm in the same boat and would love an answer..
Answer: Go to github contact page here and email them saying you need to recover your deleted gist and also mention the link or the description of the deleted gist.They will recover it for you.
Comment: have you tried this before? I've also deleted some Gists by accident and was wondering if they can help to recover it.
Comment: It has been long I posted this but I think I was able to recover the deleted gist successfully. It is just that the github guys need some reference to the gist.
Comment: how would you have a reference if you deleted quite few? Is there a way to do it without?
Answer: I just successfully had a gist undeleted.
You will need to contact support: https://support.github.com/contact?tags=rr-restore

Subject: Undelete A Gist
Body: For best results, write them a nice note, include the link to the gist you deleted.

They undeleted mine within like 10 minutes of submission, though it states it can take up to 5 days.
Comment: Thanks! It worked for me. But I got the reply back in 6 hours and you do not even need to provide the link of the gist