I bought a domain from Arvixe without hosting, how can I set up email?

Question: I bought a domain from Arvixe, without any hosting plan. In the domain settings, I didn't find anywhere to set up email accounts. Do I have to buy their hosting plan to have email accounts?
Comment: Short answer. Very likely. E-mail does not just come from the sky. It may be that your host or another can handle e-mail for your domain. There are services specifically designed for e-mail only. As well, many host options can allow this too. Go seek these out. We cannot make recommendations here. But this should be easy to find and set-up.
Comment: Thanks. I contacted their customer service and was told that emails do not come with a domain only purchase. You have to buy their hosting plan
Answer: Buying a domain entitles you to your own www.example.com. Many domain providers will let you keep your domain name for a year. The option you must have with your purchase is the ability to assign an IP address to your domain so that a server is contacted when the domain is requested on the internet. The IP address association with a domain name can be easily demonstrated through the use of a command-line utility PING.
As for the rest of the services such as email, Its best that you rent a server with email software installed on it. If you have enough money and you want a full stack, you could go for a dedicated server with a cPanel/WHM or equivalent software installed then you can have tons of options including email address options, FTP, subdomain creation, logging and more.
Your server provider must provide you with at least one usable IP address you can associate your domain name to.
Once all that is done, give abut 24-48 hours for the domain to be setup correctly across the entire internet.
Comment: "Many domain providers will let you keep your domain name for a year."  What does that mean?  Is there some process by which they may let you keep it for less than the year you paid for?
Comment: I believe theres a possibility that one could run into a company that allows one to rent a domain which lasts less than a year. It's always a good idea to check complete terms and conditions of any service used to make sure everything is alright before purchase.
Answer: The key issue is getting your domain into the Domain Name System so that the internet knows where different services are (mail, website).
If you buy hosting it comes with DNS services.  If you don't want hosting, some companies will include DNS when you buy a domain, other hosting companies will sell you DNS only hosting.  Once your domain in in the DNS you then have plenty of options for email e.g. gmail, a mail account elsewhere, your own mail server.
Answer: Yes, unless promised "free email with domain name registration" you do need hosting services to have @domain.com email service.
Answer: No. You won't need to buy email. Just use some free email for domains provision
If you wish to use emails on domain for your personal use and or for a small number of people, you wont't need to spend on professional email or any kind of hosting if you wish to use it for a few accounts or limited number of space.
There are free email for domains providers like

Pawnmail -
Which allows 2 GB Storage forever free for any number of accounts you wish to create. It allows us to login through a roundcube webmail interface or we can setup an email client.

Zoho for domains -
Which provides 5 GB each for upto 10 user accounts in a domain.

The steps you need to follow for both the above or any other service are

Setup MX Records in your DNS
Create accounts in the control panel
Wait for dns propagation
Login using the webmail or setup an email
client.


In quantum mechanics, is $|\psi\rangle$ equal to $\psi(x)$?

Question: So I'm going through my notes and I think I've confused myself. We often imply
$$
|\psi\rangle \to \psi(x)\\
\langle\psi| \to \psi(x)^*
$$
for instance when we talk about eigenvalue equations we interpret
$$
\hat{H}|\psi\rangle =E|\psi\rangle
$$
as simply
$$
\hat{H}\psi(x)=E\psi(x)
$$
but I don't understand why we say $|\psi\rangle \to \psi(x)$ because if that was the case then
$$
\langle\psi|\psi\rangle=|\psi|^2
$$
when clearly
$$
\langle\psi|\psi\rangle=\int_{-\infty}^\infty |\psi|^2 dx = 1
$$
I'm obviously missing something simple, could anyone point out where I'm going wrong?
Comment: @user7971589 Typically, when I have seen that arrow used in this context, it is explained in the following way: $x\to y$ means that $y$ is the representation of $x$ in a particular (previously-specified) basis. So, for example, if we're working in the position basis, we can say that $|\psi\rangle \to \psi(x)$, $\hat{p}\to -i\hbar\frac{\partial}{\partial x}$, and $\hat{x}\to x$. In the momentum basis, we can say that $|\psi\rangle \to \psi(p)$, $\hat{p}\to p$, and $\hat{x}\to i \hbar\frac{\partial}{\partial p}$.
Comment: What does the arrow $\to$ mean? Do you think you could make the statement $|\psi\rangle\to\psi(x)$ mathematically precise? What is the exact relation between $|\psi\rangle$ and $\psi(x)$?
Comment: @user7971589 This is a general point, but whenever you're learning something and you think something fuzzy like "I think X isn't Y, but X can be replaced with Y?", that should be a mental stop sign to go back to the basics and figure out what's _really_ going on! It's best to stop these misconceptions the moment they appear.
Comment: Not sure to be honest, in my head I treat it as a equality but I don't think it actually is. I meant it as $|\psi\rangle$ 'can be replaced with' $\psi$ to answer questions
Answer: To start, the kets are vectors, which means if we want an explicit realization of them, we would need to write them with respect to some basis. The first basis most people see is the position basis, where the basis kets are the states of definite position. Then, an arbitrary state $|\psi\rangle$ can be written as
$$ |\psi\rangle = \int_{-\infty}^{\infty}dx \ \psi(x)|x\rangle,$$
where $\psi(x)$ is the familiar position space wavefunction. This should help explain why the last line has the integral in it, it comes from this superposition of states of definite position.
Now for rewriting $\hat{H}|\psi\rangle = E|\psi\rangle$, the exact same thing happens, but the reason there aren't any integrals is because the $\hat{H}$ in $\hat{H}\psi(x) = E\psi(x)$ is written as an differential operator acting on position space now, rather than an operator that acts in some way on an arbitrary ket. This might be a slight abuse of notation, but it is usually clear from context what basis the Hamiltonian is written in in a given equation.
Comment: Just to add, the position basis is not always used. $|\psi\rangle\to\psi(k)$ is also very common, where $k$ is the wavevector and is tantamount to working in momentum space. This is why for general things, $|\psi\rangle$ is preferred, as it exists without any regard to a specific basis.
Comment: Thankyou this helped a lot. To clarify then, in a different basis would the definition be $|\psi\rangle=\int_{-\infty}^{\infty}dk\psi(k)|k\rangle$
Comment: Yup! That's correct
Answer: The state of a quantum system is an element of a vector space (a Hilbert space, specifically). The notation $|\psi\rangle$ is used to label a particular vector in this space. Operators such as $\hat{H}$ map states in the Hilbert space to other states in the Hilbert space, so $\hat{H} |\psi\rangle = |\phi\rangle$, where $|\phi\rangle$  is another state in the Hilbert space. Some vectors might be eigenvectors of an operator, in which case we might write $\hat{H}|\psi\rangle = E |\psi\rangle$ (where $E$ is just a regular number, the eigenvalue).
If we are considering a wavefunction that describes the real-space motion of a particle, then we might be interested in the spread of the wavefunction over space. The Hilbert space in this case is infinite-dimensional. A natural choice for the basis vectors in which to expand $|\psi\rangle$ is the position eigenstates $|x\rangle$, where $x$ is any position in real space. $|\psi\rangle$ can always be expressed as a superposition of different basis states with different amplitudes in each basis state:
$$
|\psi\rangle = \int dx \ \psi(x)|x\rangle
$$
where $\psi(x)$ is now a function that maps positions to some complex amplitude.
We can formally relate $\psi(x)$ to $|\psi\rangle$ as follows: take the inner product of the expansion of $|\psi\rangle$ in the position basis with any particular position eigenstate $|x'\rangle$:
$$
\langle x' | \psi \rangle = \int dx \ \psi(x) \langle x' | x \rangle = \psi(x')
$$
where we rely on the fact that different position eigenstates are orthogonal (ie., $\langle x' | x \rangle = \delta(x-x')$).
So $\psi(x) = \langle x | \psi \rangle$. That is, the amplitude of the wavefunction at a position $x$ is given by the projection of the state $|\psi\rangle$ onto the basis state defined at position $x$, $|x\rangle$.
Answer: This is something that confuses many students learning about quantum mechanics.  It is important to understand the concept of
different 'spaces'.  Physicists deal with many different spaces: the usual 3D real space we live in, and many other abstract 'spaces'.
The Hilbert space of a particle is the 'space' of all possible states the particle can be in.  It is NOT the same 3D space that the particle actually 'moves around in'.  The Hilbert space also obeys the usual laws of linear algebra, but it is infinite dimensional.
In particular, the dimensionality is equal to the number of points in space, and the set of basis vectors is $\{\,|x\rangle\,|\,x\in \mathbb{R}\}$.  The meaning of $|x\rangle$ as a state is "the particle is located at x".  $|x\rangle$ is also a basis vector of the Hilbert space; note that there is one for every $x$.  So $|2.77\rangle$, $|\sqrt{3}\rangle$, $|\pi\rangle$, and $|42\rangle$ are all basis vectors, as are an infinite other ones.
The important equation is actually $\langle x|\psi\rangle = \psi(x)$.  This is the 'proper' version of your $|\psi \rangle \rightarrow \psi(x)$.  The vector $|\psi\rangle$ lives in Hilbert space and is the abstract mathematical object representing the state of the particle.  The equation $\langle x|\psi\rangle = \psi(x)$ essentially says that the scalar product of $| x\rangle$ and $|\psi\rangle$, or, 'the component of the vector $|\psi\rangle$ along the $|x\rangle$ direction' is equal to some number which we call $\psi(x)$. Because of the infinite dimension, you can take all of these scalar products (i.e. all the components of $|\psi\rangle$ along all the possible 'directions') and piece them together into a continuous function, the wave function.
Answer: Technically speaking if $\vert \psi\rangle\to \psi(x)$ then $\langle\psi\vert\to \int \,dx\, \psi(x)^*$. The integral symbol guarantees that glueing a bra and a ket together, $\langle \phi \vert \psi\rangle$, is a scalar product and a (generally complex) number, gives the same as the number computed as $\int dx \phi^*(x)\psi(x)$.
More specifically, $\psi(x)$ is really $\langle x\vert\psi\rangle$, as per this question. Using $\langle x\vert\psi\rangle =\psi(x)$ allows to interpret $\psi(x)$ as the component of $\vert\psi\rangle$ along the basis vector $\vert x\rangle$.  If you buy this then you can also express $\vert \psi\rangle$ in the momentum basis, where $\psi(p)=\langle p\vert\psi\rangle$.
There are even states - like spin states $\vert \pm\rangle$, for which there is no "spatial" wavefunction, i.e. $\langle x\vert +\rangle$ doesn't make sense as the spin degree of freedom does have physical dimensions.
The Dirac notation is thus a way of emphasizing the vectorial nature of states, which can be combined and multiplied by scalars just like vectors can.  It also emphasizes that expressing $\psi(x)$ or $\psi(p)$, i.e. the wavefunction in position or momentum space, is basically a choice of basis, much like choosing to express a vector in spherical or Cartesian coordinates.  Indeed, to go from one basis to the other we need the transition formula $\langle x\vert p\rangle \sim e^{-i p x/\hbar}$.
Answer: It's actually not the case that $\langle \psi | = \psi(x)^*$. A more accurate statement would be as follows.
$$\langle \psi |\cdots  := \int_{-\infty}^\infty dx\ \psi(x)^*\cdots $$
This bra $\langle \psi |$ can thus be thought of as a function that maps ket states $| \phi \rangle$ to a complex number.
$$|\phi\rangle \overset{\langle \psi |}{\mapsto} \langle \psi |\phi \rangle :=  \int_{-\infty}^\infty dx\ \psi(x)^* \phi(x) \in \mathbb C$$
Answer: When the state space is the projectivized Hilbert space of $L^2$ functions on $ℝ^1$, $|\psi\rangle$ is by definition  the name of the equivalence class represented by the function $\psi$.  $\langle\phi|\psi\rangle$ is  by definition the same thing as $\int|\phi(x)\psi(x)|$ (when the state space is the projectivized Hilbert space of $L^2$ functions on $ℝ^1$".).
Comment: I don't understand what you mean by "in this context" here, because it leads to a very strange notion of "by definition."
Comment: @probably_someone: "in this context" means "when the state space is the (projectivized) Hilbert space of $L^2$ functions on (in this case) ${\mathbb R}^1$".


How to recognize USB to RS232 null modem cable

Question: I have been trying to establish connection between my PC and uC development kit over serial line. There is a standard serial port on the development kit. In oposite there is only USB ports on my laptop. So I have used USB to serial cable for the connection. The problem is that I am not sure whether it is a so called null-modem cable. Can anybody more experienced tell me how to double check it? Thanks in advance.
Answer: It would be very unusual for it to be a null-modem cable. A USB to serial converter normally allows you to add a serial port that is connected in, and works in the same way as the original COM ports fitted to desktop PCs. You would use a null-modem cable to connect, for instance, two PCs together.
If you would need a null-modem cable to connect your development kit to a desktop, you would also need one to connect to your USB converter.
Answer: I manufactured a serial device for 20 years.  I have never seen a USB-SERIAL cable with null modem.  Look in your System => Device Manager for Ports COM and LPT.  It should be there if the driver is installed. The title or the driver may give more information.  Most of these cables use the Prolific PL2303 chip.
If you are having problems, its most very likely a driver issue.  The serial port and Windows was 97% of my tech support issues.
Try installing the Prolific Driver, the odds are good it will work.
IF that doesn't work Win 10 does a good job of finding the drivers automatically.
It is most likely NOT a null modem cable.  It will be a DTE. Not a  DCE.  IBM designed the PC to be a terminal to their big iron so made it a data terminal rather than data circuit-terminating equipment.  The USB cable's serial port is the same as if the PC had a serial port.
A development board made to work with a PC should be a DCE device using a straight through cable and not requiring a null modem unless the designer is an idiot.
The pinout will look like this: (To PC means to USB)
Comment: Thank you for your reaction Misunderstood. I have checked the Device Manager but unfortunately there is no information about the used chip.
Comment: See my updated answer
Comment: Misunderstood, do you think that my problem is wrong driver?
Comment: I don't think you have a driver installed.  In device manager do you have an Unknown USB device?  What happens when you plug in the cable?  There should be a notification from Windows.
Comment: There is USB Serial Port (COM4) item under Ports (COM & LPT) in my Device Manager. I have also tested the connection via PuTTy and crossover connector at the end of the usb to serial cable. I have written some characters and immediately I have received them back into the terminal window. So I suppose that the driver is alright. I have doubts concerning with the cable. I think that my cable isn't the null-modem cable.
Comment: @Steve No it is not a null modem, updated my answer.
Comment: So my cable is the right one. The only place where the problem could be is my software for MCU in the development kit. Thank you very much for your help Misunderstood.
Answer: I usually don't bother to figure which way it is, I just connect it and if it doesn't work reverse term 2 and 3 on serial end.
Comment: What about DSR and RTR?  And RTS and CTS?


Moment calculation

Question: Consider image below. The weight of the fire-fighter is 840 N. What is the torque of the fire-fighter's weight about P and what is the value of the force C which cancels out the torque?
Comment: Please check out the FAQ, specifically, "However, we do not allow certain kinds of questions:

"Do my homework"-type physics questions
"A 4kg ball is traveling at 8m/s in the x direction, how do I find...""
Comment: to add to what Tobias said, you should look at some standard resources first. Don't ask us to do your problem for you, but ask about the specific concept that confuses you.
Comment: http://physics.stackexchange.com/faq
Comment: all we _will_ (possibly) give to you are hints, but first you have to provide some details what you have tried so far. In case you don't have a clue where to start, look at wikipedia's entry on torque, and don't forget to use good ol' pythagoras
Answer: The answer to both your questions are hinted at or can be worked out from the equation: (Horizontal_Force_C) x 12m = (FF_weight) x 5m.  You get this equation by summing up the moments around P and setting that sum = 0.  You could pick any number of points about which to sum these moments, but point 'P' is the obvious choice for ease of setting up the problem and performing calculations.

Moment calculation


How to disable automatically generated comments in node, but enable programmatically created comments?

Question: I decided to programatically print comments un Views using PHP footer. The problem is that they now appears twice (my code + default). When I disable comments in content type settings, both disappear. How can I disable the default comments and keep my comments printed by php code? I tried to edit node.tpl file. but there is no section responsible for showing comments there (I use Garland).I do this because I am using a complicated views structure and I would like the comments to be displayed below the node, but not at the end of the whole view. Any advice appreciated.
Answer: Configure comments to be show in separate page, and you'll be able to show them with your view. Also, you've to hide the comment link if you do it.


Just upgraded to a newer version of Photoshop, how do I get Lightroom to recognize it?

Question: I had PS5.1 when I installed LR4 and now I have CS6, but LR4 still opens everything in CS5.1. How do I make it open everything in CS6?
Comment: I have licenses for CS5.1 and LR4 under the old licensing model whereas I acquired CS6 under the new CC subscription model. Should I ever cancel my CC membership I would like the ability to revert back to LR4/CS5.1 like I had it yesterday before I purchased CC.
Comment: Why not just uninstall CS5.1? Any particular reason why you would want to revert to the old version?
Answer: Assuming Windows OS, if you are comfortable editing your registry:

open registry (using regedit command)
navigate to HKEY_LOCAL_MACHINE\SOFTWARE\Microsoft\Windows\CurrentVersion\App Paths
look for a key called "Photoshop.exe"
if it is there, it will probably contain a path to CS5.  Edit the entry and paste in the path to your CS6 executable
if it is not there, add it using the information in the following

http://helpx.adobe.com/x-productkb/multi/edit-photoshop-command-missing-photoshop.html
Alternately, uninstall CS5 and then reinstall LR4 and it will pick up CS6.
Comment: Just wanted to say thank you to MikeW. I used your solution and it worked a charm. Thanks!
Answer: Apparently one solution is to have your computer lockup and have to hard reset causing LR4 to see CS6 upon startup. I'll leave this open for a more solid answer, preferably one that includes how to revert, or select, the version (or install) of PS that should be used in case I want to go back to an old version at some time.
Comment: Had you rebooted your computer?  Most likely the file association wasn't changed until reboot.  I imagine it just uses whatever has the association registered in Windows, though that is a bit of a guess.
Comment: I'm on a Mac, but my guess is that LR checks for PS on startup and uses the latest one it finds. Because LR was open when I installed CS6 it probably didn't find it until I reopened LR because of the crash.
Comment: even on Mac, it's probably still done off file association, though they may look for installed versions of Photoshop.  In general, file associations would probably be easier though as the OS already provides it and it includes the path to the executable.


Logistic regression where dummy dependent variable is heavily clustered at '0'

Question: As the title suggests, the distribution of my DV is as follows: 0=4,230; 1=41. There are 9 primary IVs I'd like to regress it on as well as 7 CVs. Something tells me that the skew in my DV will compromise any test of significance/effects. Is this true? And, if so, how should I proceed in this situation? Thanks in advance for the help!
Comment: It may be worth your while to read [this thread](https://stats.stackexchange.com/q/291371/7290) & my answer there.
Comment: Ah, I see: 4,230 is 4230
Comment: In addition to the gung's link and the answers below please search "complete or quasi-complete separation in logisic regression"
Comment: Could not understand your question due to some strange things. 1) What made you call your binary DV specifically "dummy"? (was there any nominal variable? how is it used?). 2) What does fractional `0=4,230` mean? 3) What is CV? (covariate?) Please edit the Q to make everything clear.
Answer: With only 41 1's and that many independent variables it seems quite possible that you'll have degenerate data (for example, all levels of an independent variable might only have 0's as responses).
Even if the model can be fit, the variance around the parameter estimates will probably be very large, the estimate of that variance will be suspect, and confidence intervals may not be reliable.  You'll have a high chance of a type II error.  Essentially you might not have enough data to find any significant effects.
Comment: `all levels of an independent variable might only have 0's as responses` How can that be?
Comment: I actually went ahead and ran the model. There were several significant effects. But I suppose beyond the p-values, the results will largely be uninterpretable?
Comment: I think you may have misunderstood me. My IVs are all normally distributed ordinal and continuous scales.  It's the dependent variable that has mostly 0s.
Answer: First, your main problem is not going to be with the significance tests, but with bias in your estimates. The root of the problem, as I also mention below, is not so much the "skew" in the distribution of your DV, but because of the very small number of 1s.
As far as I know there are a couple of ways of dealing with rare events. One would be at the sampling stage, where you retain all $y_i=1$ and then sample from your $y_i=0$ so that it's not a rare event any more. How many 0s you would need to sample, could be left up to cross-validation, especially given that you only have so few 1s. Another approach is to run some sort of penalized logit (the Firth model), which also deals with the separation issues you'll probably face.
A good (semi-recent) review of this issue and discussion of solutions can be found here. The canonical discussion in my discipline is here.
Finally, a similar issue has been discussed in a previous post, but I'm not considering yours a duplicate, because of the extremely small number of 1s you have. I think this is a very important distinction, because other than any other potential problems associated with rare events, you also potentially have very limited variation in the Xs associated with the $y_i=1$.


Can a project be "too good" for a Master's?

Question: Pretty straight forward. I am starting to apply to grad school and I was discussing possible projects with some (grad student) friends of mine. When I mentioned something in particular one of them said "that might even be too good for a Master's, sounds more like a PhD..." Then he also went on to say that I could use part of a project like that for a Master's and then use the rest to finish it off with a PhD. For some time I thought about this, then I decided to ask it here.
So is there such a thing as a project that is too good for a Master's? And if so, should my future advisor generally know where this "boiling point" is?
Note: if you need more info about my major of study or the specific sub-field, leave a comment and I can make an edit.
Comment: The most important criteria for a project for you, both PhD and Masters is that you complete it and can graduate. A too good project is likely something that has a low probability of completing within a Masters which is shorter than a PhD.
Comment: "Too good" is euphemistically for unrealistically ambitious, a pipedream in plain English; and yes, such projects exist.
Answer: From my reading of the conversation between the two of you, he meant 'too good' in one of two ways:

"Too Good" as in the topic will yield many high-quality results, potentially many publications, and opportunities for continued/extended research.
"Too Good" as in requiring too much time to complete in the typical Master's degree time period, and actually should be understood as "Too advanced" or "Too in-depth".

In answer to your second question 'will future advisor know the boiling point?' Yes. That is what they are there for, to help guide the scope of your project to a Master's level.
Answer: I don't see a problem here.
If the project is too good, then
(a) you will not be able to finish it in the timeframe of a Masters, in which case you can continue working on it for your PhD, OR
(b) you will finish it for your Masters, in which case you have a great Masters thesis, and then you can tackle something else for your PhD.
You should not kill yourself to achieve case (b).  If you find that the problem is too difficult, just do case (a).
Finally, be aware that other graduate students may not be able to gauge the difficulty very well.  Only you will know once you have begun working on it.
Comment: (a) can be *very*, very bad if it's not a project for which one can report meaningful intermediate results. On what basis would the masters degree be awarded?
Comment: @ff524 that is always a possibility with a research project.  You can't know ahead of time if you will have any success, even with a problem that seems easy on the surface.  But even if no substantial progress is made, you can demonstrate a good knowledge of the background material, previous work in the area, and describe your own (failed) attempts.  This can produce a passable thesis.
Comment: Yes, we don't know everything ahead of time, and we can do some damage control if we turn out to have misjudged a problem; but we use our experience and judgement to identify problems that have a high probability of having something meaningful to show for the work after N years. I would definitely discourage the OP from taking on a thesis project if he knows *a priori* that there is a very small chance he will get meaningful results in the time given. There's a difference between "we can deal with this outcome if it happens" and "I don't see a problem, no need to try to avoid this outcome".
Comment: In my experience many master students propose projects that would take two or three PhD project to complete. It is the job of the advisor to help students come up with more realistic plans, and they can do so as they have the experience the master students lack. So it is a huge problem which can and should be prevented.


Intersection multiplicity of projective plane curve and tangent line

Question: Show that a line $L$ is tangent  to a projective plane curve $C$ at a smooth point $P$ if and only if $I_P(C \cap L) \geq 2$. If $P$ is a point of multiplicity $2$, and $L$ is tangent of multiplicity $2$ at $P$, show that $I_P(C \cap L) \geq 3$.
Comment: I proved with using dimension. But question is seen to be proven with simply geometry. Geometric picture..
Comment: Can you give your proof using dimension?
Answer: Let $P=(0,0)$ and tangent line is $L:x-y=0$. (We shall assume them with using natural transformations.)
Now write $C(x,y)=F_0(x,y)+F_1(x,y)+F_2(x,y)+...$ where $F_i$ are homegenous polynomial degree $i$.
If $F_0(x,y) \neq 0$, this means $P$ is not on the curve. If $F_1(x,y) \neq L$ then, $L$ is not a tangent. Hence we can put $x=y$ and eliminate $F_0$ and $F_1$. Then we get at least a term with $x^2$. This proves first statement.
In second, $F_0(x,y)=F_1(x,y)=0$ since multiplicity of $P$ is 2 and $F_2(x,y)$ need to equal $L^2$ since $L$ is a tangent of $P$ with multiplicity 2. Hence we can put $x=y$ and eliminate $F_0$, $F_1$ and $F_2$. Then we get at least a term with $x^3$. This proves second statement.
Comment: I think that this is really geometric proof.


Add -al to verb to make a noun

Question: Do you have any examples of nouns that are formed by adding -al to a verb?
I can think of one example (rental), but would like to have a few more. Thank you for your help!
Comment: BillJ - Thank you for that information!
Comment: By the way, how do you claim to know that the noun "rental" is formed from the verb "rent"?  The word is 700 years old, so its etymology isn't clear.  It could have just as well come from the noun "rent" being made into the adjective "rental," as in "rental property," which then got back-formed into the noun "rental," meaning "rental property."
Comment: Benjamin Harman - That occurred to me, and that's why I'm asking if anyone has any other examples. Barring no more examples, I would then venture to guess that it went from adjective to noun, instead verb to noun.
Comment: Few -_al_ nouns have been created since the nineteen century ("referral" appears to be one), and it is questionable whether the suffix is still productive. Long-established ones include "arrival", "denial", "refusal", "removal", "trial".
Answer: There are a fair number of deverbal nouns ending in -al: examples include arrival, disposal, dismissal, denial, betrothal, approval, appraisal, avowal, betrayal, trial. You can find more examples by looking through the results of this search on the dictionary aggregator site OneLook for nouns ending in al.
The Wiktionary entry on the suffix -al points out that in some cases, -al nouns coexist with -tion or -sion nouns from the same verb, as with proposal and proposition.
Describing the verbs that take -al as a suffix
A number of linguists have been interested in figuring out rules about which verbs take -al as a noun-forming suffix. It has been observed that the noun-forming suffix -al only attaches to verbs that are stressed on the final syllable. There are some more complicated conditions that have been hypothesized to apply to the use of this suffix; you can see some discussion in "Against Phonological Conditioning of WFRs", 1992, by Anna Malicka-Kleparska, or in "Phonotactic Properties of Morphological Units", by Edmund Gussmann and Bogdan Szymanek, in Morphologie: Ein Internationales Handbuch Zur Flexion und Wortbildung, Volume 1.
Malicka-Kleparska, citing Marchand 1969:237, says that trial and rental ought to be regarded as loaned words, rather than as words that have been formed in accordance with the English derivational system. These two words do not conform to a generalization that says that -al attaches only to Latinate verbs with the structure prefix + root (p. 435 in Phonological Investigations, edited by Jacek Fisiak and Stanislaw Puppel). According to Malicka-Kleparska's analysis, the stress pattern of -al nouns is not a condition on the use of the suffix, but only a predictable result of the morphological structure of the verbs that it attaches to (p. 437).

Add -al to verb to make a noun


How to get image from node

Question: Lets assume that i have a complex texture not just single image, it has  noise texture nodes i have a complex image output so here's the problem how do i get that complex image from nodes.
Comment: I'm sorry, I can't understand what you're asking based on what you've written. Are you able to expand and/or add images to explain the problem?
Comment: i want to export the the image i mixed with nodes for my unity3d project
Answer: The easiest way is to bake the procedural texture into an image:

Add a plane into the scene, UV unwrap it and assign the procedural texture to it through emission shader (renders the fastest).
In UV/Image editor create a new texture of desired resolution.
Add texture node into the shader (unconnected but selected) and load the texture into it.
Bake the emission to the texture. Save texture as you desire.
Answer: You could apply your node(s) created texture to an evenly lit plane. Then set up a camera parallel to the plane and then render out an image file.

How to get image from node


How does a Station in WLAN gets list of other associated STAs with the same AP?

Question: There are many STAs(Stations) connected with a single AP. My question is how does a STA knows about other STAs that are connected to same AP. For example in Windows when we click Network icon we get list of other Computers which are connected to same AP. So basically which frames are exchanged between a STA and AP to get list of other connected computer to same AP?
Answer: It's the same as on a wired Ethernet LAN. There's nothing wireless-specific.
Windows generally uses SMB browsing and NetBIOS Name Service and Link-Local Multicast Name Resolution.
UPnP has some discovery mechanisms too, which I think involve putting a weird HTTP GET into a UDP message and multicasting it.
The IETF standard is ZeroConf, which Apple calls Bonjour. It uses Multicast DNS (mDNS).
Cisco devices use the Cisco Discovery Protocol (CDP).
A protocol called Link-Layer Discovery Protocol (LLDP) has its vocal fans, but I'm not sure how widespread it is. Unlike most other discovery protocols, with LLDP you can discovery the topology (logical map) of your network, but I don't think LLDP is as good at discovering what application-level services are being offered by each device.
Many other discovery protocols have come and gone, and many are still around, but only used by one company or product or protocol family.
Well, to be fair and complete, I should mention that Wi-Fi Direct (the Wi-Fi Alliance's way of doing 802.11 point-to-point without an AP, as used by the Wi-Fi Alliance's "Miracast" wireless screen mirroring technology) does have a wireless-specific discovery protocol. As I recall, Wi-Fi Direct-capable devices spend some amount of time sending out special beacons and responding to special probe requests with special probe responses. Other Wi-Fi Direct-capable devices can find them by doing a channel scan much like how a normal STA finds an AP.


What kind of scam did someone claiming to be from UPS attempt?

Question: I received a phone call early yesterday morning from an unknown number. I answered it and someone asked "Is this bitmaker?" (they had my real first name). I said yes, and then they said they were from UPS, and that's all the information they needed and hung up. Googling the number says that other people have received the same thing, and it isn't a real UPS phone number. What was the angle of this scam?
Comment: It's possible they simply wanted a recording of your voice saying "yes".
Comment: @chepner Could you elaborate on this in form of an actual answer? Without any context, your comment is really just causing panic without providing much useful information or actionable advise.
Comment: @chepner this reminds me of a very popular scam in Switzerland where they are a real company offering some useless product like spam protection or w/e. They start with some innocuous yes/no type of questions and slip a "...and you would like to order product X" where (often elderly) people just keep saying yes. A month later they get a bill and if they ask why the recording is played back. Saw dozens of those cases in consumer magazines :). Probably not the case here unless they are going to do some editing where they use your 'yes' with a fake question.
Answer: I suspect someone has a list of names and phone numbers and they are going through them to see which ones are still valid. Two possible reasons for this could be:

If you recently applied for a job, or applied for an apartment, etc, online, it could simply be that this service is verifying the phone number you supplied is accurate. This could help weed out fake applications.
A recently validated list of names and numbers is much more valuable to telemarketers. It could be that the telemarketers themselves are validating the numbers, or the validated list is going to be sold to telemarketers in the near future. The idea is that when you answer a call and someone asks for you, the call is likely to last longer than a cold call, and this increases the chances of a sale. I would not be surprised if in the near future you receive one or more calls from someone asking for you by name, followed by a sales pitch. When you get the call, be strong, my friend.

Disclaimer: this is conjecture only.


Does it make sense to preprocess (normalise or standardise) this data for GAN?

Question: I'm working on a project where I have a dataset for a dynamical system (pendulum) containing a trajectory, energy cost and corresponding control actions (See below). I'm using a generative adversarial network (GAN) in a supervised settings to train it as a regression model. The trajectories are the labels, the control actions are the targets. During test time the GAN takes the trajectory concatenated to noise (per literature on conditional GAN's) as input and outputs the required control actions. This works ok, but not as good as expected. It seems that it is not able to learn the dependency between the labels and the data as well as I had expected. My guess is that it has to do with the labels and their domain.
I was always under the assumption that it does not make sense to normalise or preprocess my data, but now I'm second guessing.
The control actions (lambda and mu) are always between -1 and 1. The cost is always between 0 and 1. The trajectory (two initial and two final states) is not constrained like that except that the initial condition lies between -pi and pi. The data is generated by feeding random control actions on random initial states into a dynamical model resulting in a final state.
I always thought that if I were to normalise/standardise my data it would break the dynamical dependencies between the variables but some people around me have suggested otherwise.
Note: t1 and cost are indeed equal and are supposed to be
Are there suggestions on how to preprocess the data below?
Quick recap: Left 4 columns are the targets, right 4 columns are the labels.
Answer: A GAN is a type of ANN, and therefore it makes sense to normalize: https://stackoverflow.com/questions/4674623/why-do-we-have-to-normalize-the-input-for-an-artificial-neural-network
Whenever you normalize data, you can just keep the statistics used to transform it to return it to its original state. Doing this makes normalizing data scale across nearly any machine learning process.


What is the "unambiguously correct" pictogram for database storage?

Question: I want to draw a mockup design of a set of microservices forming a web application. I need to represent a database storage in it. I would like to know if there is  consensus on which pictogram is accepted as representing the storage of the data in the database (not the database application itself).
Comment: "Correct" may not be the best word, in your title. Perhaps "standard" is more correct.
Comment: I take the point @RadLexus - the word is rife with connotation. perhaps a better adjective would be  "unambiguous" ?
Comment: Yes, that's better :) It would only need to be unambiguous *in your context*, that is, if it appears in between several other icons to store other data, say, plain text or images. I guess you'll have to judge that by yourself.
Answer: Well, from what I know the most used icon for data storage is:

[ image source ]

[ image source ]

[ image source ]
You could possibly get away with a cloud storage icon:

[ image source ]
..or a combination of the two:

[ image source ]
Comment: Thanks for the reply. That's close to what I'm looking for - except I want it to explicitly convey that it's SQL-like data. So the generic "storage" pictograms are not going to cut it. My feeling is that http://i.stack.imgur.com/Hf7gW.png comes closest ?
Comment: @brucellino That does come first. I've added another icon for SQL at the top of my answer. Bottom line is, that what you should be going for.


do I need gfci outlets for a storage room receptacles

Question: I wired a storage room and run one 20 amp circuit for some outlets and to power the water heater and furnace does this circuit have to be gfci protected?? Or can I install regular receptacles
Answer: If this area is an unfinished area of a basement or crawl space then GFI protection is definitely required for ALL 120V, 15A & 20A receptacles. No exceptions, other than for fire or burglar alarm systems.
The furnace must be dedicated and should NOT have a cord and plug.
2011/14 NEC

422.12 Central Heating Equipment
Central heating equipment other than fixed electric space-heating
  equipment shall be supplied by an individual branch circuit.
Exception No. 1: Auxiliary equipment, such as a pump, valve,
  humidifier, or electrostatic air cleaner directly associated with the
  heating equipment, shall be permitted to be connected to the same
  branch circuit.
Exception No. 2: Permanently connected air-conditioning equipment
  shall be permitted to be connected to the same branch circuit.

If the water heater has a typical 120V cord and plug you have no choice but to use GFI protection.
Answer: I started this as a comment on Speedy Petey's answer and ran out of room. He's completely correct.
I was just going to comment that GFCI protection can be implemented with GFCI circuit breakers in the panel, in which case you would install regular receptacles, and protection extends to everything downstream from the GFCI device on the same circuit. The same applies if you install a GFCI receptacle and connect additional regular receptacles to the load side of the GFCI. You only need one GFCI for the whole circuit.
So you need to run at least two circuits and maybe three. One of them will be hard-wired into the furnace. The other(s) will feed the receptacles and you might want a dedicated (or mostly dedicated) circuit for the water heater, depending on how much electricity it uses.
For example, this point-of-use 120V water heater draws 1,500 Watts. That's 12.5A on a 120V circuit. That is below the 80% continuous load limit on a 20A circuit (it's above 80% on a 15A circuit), but there isn't a ton of headroom and it's jut one device. Plug something like a space heater or hair dryer into the same circuit when the water heater is on, and you will trip the circuit breaker.
Comment: I think the rule with water heating equipment, is that it can be on a shared circuit if it uses less than 50% of the capacity of the circuit.  But as always, I could be mistaken.


cohomology of the loop space of a Lie group: divided power structure

Question: Let $G$ be a complex simple simply connected algebraic group and $Gr_G$ be the corresponding affine Grassmannian (it is well known that $Gr_G$ is homotopy equivalent to the base loop space $\Omega K$ of the maximal compact subgroup $K\subset G$). Let $H=H^*(Gr,k)=H^*(\Omega K,k)$ be the cohomology ring where $k$ is a field of characteristic $p>0$. The computation of $H$ by Yun and Zhu in their paper "Integral homology of loop groups..." shows that the commutative $k$-algebra $H$ has a divided power structure, at least when $p$ is not too small.
Can this be seen from first principles? In particular, is it related to Cartan's divided power structure on homology of a commutative simplicial $k$-algebra?
Comment: I wouldn't expect a simplicial k-algebra structure on the cochains of $\Omega K$... I think Steenrod operations are an obstruction to having such a thing (right?) and those are certainly nontrivial in this case. On the other hand, I think this might be an instance of Koszul duality in characteristic p (if that's any use to you).
Answer: Let me make the following remarks regarding divided power structures and cohomology of loop spaces. Let us fix a field $\mathbb{F}$.

1) Le me recall the Eilenberg-Moore spectral sequence. Let us suppose that we have a fibration $$F\rightarrow E\rightarrow B$$ such that $B$ is $1$-connected, $H^k(B;\mathbb{F})$ and $H^k(E;\mathbb{F})$ are finite dimensional $\mathbb{F}$-vector spaces for any k. Then we have a convergent spectral sequence
$$E_2^{p,q}=Tor^{H^*(B;\mathbb{F})}_{p,q}(H^*(E;\mathbb{F}),\mathbb{F})\implies H^*(F;\mathbb{F}).$$
In the case of the path fibration $$\Omega X\rightarrow PX\stackrel{ev_1}{\rightarrow} X$$
we get a convergent spectral sequence
$$E_2^{p,q}=Tor^{H^*(X;\mathbb{F})}_{p,q}(\mathbb{F},\mathbb{F})\implies H^*(\Omega X;\mathbb{F}).$$
2) Behind the Eilenberg-Moore spectral sequence you have the Eilenberg-Moore quasi-isomorphism
$$C^*(F;\mathbb{F})\simeq Tor^{C^*(B;\mathbb{F})}(C^*(E;\mathbb{F}),\mathbb{F})$$ in order to take into account of the multiplicative structure given by the cup product we have to be very careful and use the $E_{\infty}$-algebra structure on the singular cochains $C^*(-;\mathbb{F})$ and play with $E_{\infty}$-torsion products. This is very well explained in M. Mandell's paper "E_{\infty}-algebras and $p$-adic homotopy theory" (Topology,  2001, have a look at section $5$).
3) When one can replace the dg-algebra $C^*(X;\mathbb{F})$ by a commutative dg algebra i.e. when it is quasi-isomorphic to a cdga $A^*(X)$ the Torsion product
$$Tor^{A^*(X)}(\mathbb{F};\mathbb{F})$$
is a cdga and by results of Cartan (Cartan's seminar 1954-1955, exposé $7$) when $char(\mathbb{F})>0$ its cohomology has a divided power algebra structure. The relationship between this divided power structure and the one coming from the homotopy ring of a simplicial commutative algebra is explained in B. Richter's paper "Divided power structures and chain complexes."  Alpine perspectives on algebraic topology, 237–254,
Contemp. Math., 504, Amer. Math. Soc., Providence, RI, 2009. The relationship is explained thanks to the Dold-kan correspondence and the classical Bar construction.
4) Last point, when can we replace $C^*(X;\mathbb{F})$ by a cdga? I know at least three cases: when $X$ is formal over the field $\mathbb{F}$, when the field is of characteristic zero, in the Anick's range if $X$ is $r$-connected and $dim(X)\leq r.char(\mathbb{F})$.

Conclusion: for any $1$-connected finite CW-complex $X$ the cohomology algebra of $\Omega X$ with coefficients in a field $\mathbb{F}$ has a divided power structure provided $char(\mathbb{F})>>0$. This divided power structure is inherited from the one coming from the classical Bar construction and is of simplicial nature.
Comment: That's great, but can you please also provide a reference or a brief explanation on the "Anick's range" part of (4)? Does the obstruction to formality vanish for degree reasons in that case?


Origin of Nuclear fusion energy

Question: What causes the mass of the two reacting nuclei to convert to energy and if it is for a purpose then why is it released. What is the source for this cause (if there)?
Comment: Does this answer your question? [Nuclear fusion energy origin?](https://physics.stackexchange.com/q/681043/)
Answer: Neutrons and protons consist of quarks, and when a neutron and a proton are squeezed tightly enough together, the quarks in each begin to interact and cause them to attract one another. The resulting bound state represents a lower energy configuration that what existed before, which means that when the binding occurs, energy is released, and the bound particles remain stuck together.
This in turn means that every possible bound state of neutrons and protons has a characteristic binding energy associated with it, and the biggest change in energy between the unbound and bound states happens when hydrogen is converted into helium.
Comment: Can you explain why does the "resulting bound state represents a lower energy configuration"?
Comment: Imagine the attractive force between the proton and neutron as a stretched spring, drawing them together so as to allow the stretched spring to relax. in its relaxed state, the spring is in a lower energy state than it was when it was stretched out.
Comment: Wow , thanks but can you also explain if a nucleus is bombarded with a neutron , doesn't the neutron bond with the other protons due to the strong force. If yes , then how does fission work. I know this is out of topic but curiosity is...Anyway , thank you for your pains.
Comment: Kindly answer my question above
Comment: fission is a completely different animal. Using the strong force, you can load a nucleus with extra protons and neutrons until the mutual repulsion of all those protons becomes cumulatively strong enough to overwhelm *nuclear force* and thereby tear the nucleus apart purely by *electrostatic forces*.
Comment: But why do the particles have less energy when brought together , does it have any scientific explanation?
Comment: ????????????????

Origin of Nuclear fusion energy


Login screen resolution

Question: So I just installed Ubuntu 16.04 on my Dell XPS 13. When I get to the login screen it starts at 3200x1800 for a quick second, then I get a way lower resolution like 800x600 but as soon as I login it adjusts to 3200x1800. How would I get 3200x1800 permanently on my login screen? I didn't have this issue on 14.04.
Answer: force your desired resolution in grub
sudo nano /etc/default/grub

Find this line
#GRUB_GFXMODE=640x480
change it to
GRUB_GFXMODE=3200x1800
then rebuild the frames and update grub
echo FRAMEBUFFER=y | sudo tee /etc/initramfs-tools/conf.d/splash
sudo update-initramfs -u
sudo update-grub

tell us if this worked.
**EDIT:
if the problem is about the login screen resolution only, read here
Answer: If people have the issue on the latest Ubuntu 22.04
The answer is indeed
sudo cp ~/.config/monitors.xml /var/lib/gdm/.config/monitors.xml

Make sure to create the file ~/.config/monitors.xml by making any change in Settings > Display
Create the /var/lib/.config folder

This tool is also nice to customize your login screen (background, ...)
https://realmazharhussain.github.io/gdm-settings/
The button in Display > Apply current display settings will basically do the above copy, but only if you have done the above steps
Display in Login Manager Settings
Comment: Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer).
Answer: Hiting ESC in GRUB and entering videoinfo showed that GRUB doesn't support all the screen resolutions available in gnome-control-center display so I used a workaround from this answer:
sudo cp .config/monitors.xml /var/lib/gdm3/.config/

Reboot to see the effect
This workaround was tested in Ubuntu 18.04 if you use version 17.04 or lower (LightDM) either upgrade to version 17.10 or above or install gdm3 sudo apt install gdm3 select your preferred resolution in gnome-control-center display then sudo dpkg-reconfigure gdm3
Select gdm3 if prompeted copy monitors.xml as shown above then reboot
Answer: FOR Q4OS ONLY Not Ubuntu, Although... Could be Similar.
I Search all over, and nothing worked. And then I thought how I saw two connected monitors using the command "xrandr -q". And I thought how about Unified Outputs(Both Screens being the same, like a mirror setting.) I definitely had one screen connected. My issue was that Q4OS asked me to select my Monitor setup when I first install, but I didnt answer it at the login screen, I answered it after, when it asked again. AND I didnt want to reinstall Q4OS so I check the display settings and attempted to change the resolution of the second monitor, And I saw the second monitor was disabled(But xrandr -q told connected) with a max resolution of 1024x, AND I know I had LVDS (all-in-one PC screen Capability) disabled in bios, so I went to BIOs and enabled the LVDS setting in my BIOs video. And when I booted up the Login Screen was full 1920X1080, already. When I logged in, the main screen was wrong, so I saw nothing, no dock. I had to right click and then run a command, and then type display and then click unified outputs or merge the two monitors dragging them to each other with mouse(or just click Unified Outputs). As soon as you see the OS GUI Graphic Monitor output setup graphic prompt, click unified outputs, the two screens with same desktop, or else it wont save, when restarted. Probably only for Q4OS.
My Fix Run Down:
Terminal and then type command "xrandr -q"
Look for more then one Connected Monitor outputs Mine was HDMI and DP.
If more then one Continue to read further.
*Right click on desktop > Run a Command > Type "Display" > Find "Displays" from the drop down menu.
*Now you should see two boxes indicating monitors. You can either drag one monitor to the other one merging OR you can simply Click "Unified Outputs" right under it, second button over from the right side. IF THE RESOLUTION HAS A MAX LESS THAN THE MAIN DISPLAY continue reading.
*Check if you have LVDS or any built in Screen system > Go to bios and enable this for Resolution support! > Once enabled > Boot OS.
*With the new login resolution already fixed by the enabling of LVDS.
*Log into the OS, and then you might have to right click > run command > type "display" and do that setup we did prior.
*When the screens are unified once again, you should see the OS GUI Graphical Setting prompt, NEEDED Quickly select "Unified Outputs" (Two monitors with the same desktop), dont let it fade away or you'll have to reboot again and right click and command and then display settings all over again.
I hope this helped someone!

Login screen resolution


Hydrogen extraction efficiency of newish nano particle AA battery

Question: The internet is ablaze with the new nano particle based extraction method of converting water to hydrogen. In 2002/2003 when I was 16 there was a similar craze about a lawn mower motor which supposedly ran on water. Indeed there were even conversion kits on the internet for vehicles. They didn't work because of a fundamental inefficiency of the conversion. Will this new nanotechnology improve the process so that cars can run on converted water or does there still exist a limit?
Comment: Is your question one of economic feasibility?
Comment: No physical like does a theory so limit the reality of the presentation
Answer: The article you refer to is about the electrolytic splitting of water.
A 100% efficient electrolytic cell would require a voltage of about 1.23V to split water, but for various reasons a simple electrolytic cell requires about 1.48V. The difference between the voltages is called the overpotential, and it increases the amount of power needed to split the water because the power required per unit of hydrogen produced is proportional to the cell voltage. The excess power goes into heating the hydrogen and oxygen produced, and in this case it means that simple cells are about 83% efficient at converting electricity into hydrogen.
Catalysts can be used to increase the efficiency, and indeed platinum based catalysts can be used to reduce the overpotential and make cells with near 100% efficiency. The problem is that platinum is expensive. The result from the Stanford team is that a much cheaper nickel based catalyst can achieve the same efficiency as platinum. The paper is here, but note that it is behind a paywall.
If the catalyst proves to be stable enough then it will be useful for electrolytic production of hydrogen, but the improvement in efficiency isn't going to change the world overnight. It still takes a lot of power to electrolyse water so it's only feasible when cheap electricity is available.
Comment: Let's add, just for completeness, that no improvements in water electrolysis will make hydrogen a good transportation fuel. This is linked to a number of issues, like the storage problem, the low energy density, and the missing oxygen (the fact that the atmosphere only has 20% oxygen limits the thermodynamic efficiency of fuel cells, again, which compounds the electrolysis efficiency problem).
Comment: @caseyr547: I've amended my answer to respond to your edits.
Comment: @CuriousOne For what it is worth, hydrocarbon and other fuels also share the inefficiency handicap posed by the less than 20% oxygen content of the atmosphere.


What is the best way to check my RAID HDDs from windows?

Question: I have a file server (2003) and I want to be able to check if the HDDs are good or not. They've been running for over 3 years.
Answer: Try THIS or THIS, basically something that looks at your server's SMART counters - if you'd let us know what your server and/or disk-controller is we might be able to help more.
Answer: If your server has SMB software you have to use that to monitor the predicted failure status. Dell, HP, etcetera should have a download on the support site where you can download the management software.
If no SMB but you are using hardware raid you'll have to reboot during a maintenance period to check the array status in the raid controller.
If no hardware raid or SMB then you can use HDTune, Speedfan, CrystalDiskInfo, or any of dozens of other programs to monitor the smart data.


Block/allow access to certain PHP pages

Question: I have a number of PHP scripts that are meant to be included into other PHP scripts, but otherwise should not be directly accessible to anyone via a Web browser. How do I block people from accessing these files directly over the Internet? Is this possible using htaccess? Would it be better to put all these scripts into a sub-folder first? Also, what is the correct HTTP error/status code to display to visitors in this case?
Answer:
Would it be better to put all these scripts into a sub-folder first?

Yes. Then create a .htaccess file in that subdirectory with the following (mod_authz_core) directive:
Require all denied

(Assumes Apache 2.4+)
This will return a 403 Forbidden response for anyone trying to access anything inside that subdirectory.
Even better, put these PHP "includes" outside of the document root and they are inherently protected from direct access without doing anything else.

UPDATE: There are a handful of files that I may want people to access

In that case, you can use negative lookaheads (regex) in a <FilesMatch> container to create exceptions for the files you wish to allow access.
For example:
<FilesMatch "^(?!(foo|bar|baz)\.php$).+$">
Require all denied
</FilesMatch>

All files, except for foo.php, bar.php and baz.php will be blocked. (?!...) is a negative lookahead assertion, so whatever regex is contained in this (ie. (foo|bar|baz)\.php$) cannot match for the expression as a whole to be successful.

UPDATE#2: Checked my server and Apache Version 2.2.23

If you are still on Apache 2.2 (EOL January 2018) then you will need to change the Require all denied directive with the following:
Order deny,allow
Deny from all

Comment: Yes, what kind of exceptions? Certain files? Certain clients? Time of day? - Depending on the type of exception required you may need to change _how_ you are blocking access. (But why do you need an "exception" for something that should only ever be accessed server-side?)
Comment: In that case you can use a `` container with a _negative lookahead assertion_. I've updated my answer.
Comment: I've updated my answer for Apache 2.2. (FYI Apache 2.2 reached end-of-life at the end of 2017. Further, version 2.2.23 was released in June 2012. The latest version on the 2.2 branch is version 2.2.34 which was released July 2017.)
Comment: Is it possible to make a few exceptions to the rule?
Comment: There are a handful of files that I may want people to access, not sure yet. Thanks.
Comment: Crap. I checked my server and it has Apache Version 2.2.23. I'm on shared hosting so there is not likely to be any upgrade soon.

Block/allow access to certain PHP pages


Inheritor dies before trust distributed, then what?

Question: A man, say, Hilbert, buys second-to-die life insurance, then dies. Then one of his inheritors, Daniel, dies. Then Hilbert's spouse dies, making the policy ready for distribution to the inheritors. Is the dead inheritor Daniel still eligible to receive his share—I.e., if Daniel writes instructions in his will regarding what is to be done with his portion of the policy in the event of his early death, will that be effective?
Comment: In Spain, the moment Daniel dies he no longer can inherit. Note that the inheritance happens *at the moment of death*, if Hilbert dies and then five minutes later Daniel dies, then Daniel has inherited and whatever he would have received from Hilbert passes to his state. The fact that the documents had not been signed and the goods have not actually changed ownership would not be relevant. I would be surprised if the USA was different, but I do not know enough to post an answer.
Comment: I'm no expert, hence this is just a comment, but it seems to me that if the policy rules say it only yields up its payout on *second* death, then *Hilbert's* estate - and the disbursement of it - have nothing to do with the policy. Pay *Hilbert's* heirs from what is *Hilbert's*, then when Hilbert's spouse dies, pay the policy beneficiaries according to the *policy's* rule.
Comment: In the event of simultaneous death (or that it cannot stablished who died first), neither Daniel inherits from Hilbert nor Hilbert inherits from Daniel.
Answer: Note: I am not a lawyer; this is not legal advice.
The situation is likely to vary by state.
It is likely to get complicated unless either (a) Daniel is a direct issue (child/grandchild etc.) of the Hiberts, or (b) their trust contains explicit clauses as to what should happen in the event of Daniel dying before the last of the Hiberts. If neither of those apply, I believe that Daniel's portion would revert to the residual estate of (probably) the spouse.
If there is a possibility of such a situation arising, the consensus is that you should explicitly make provision in the trust for alternative beneficiaries should any of the original beneficiaries have died.

Will Daniel Inherit...?
The article What Happens if a Beneficiary Dies? starts with a warning:

Not planning for what will happen if a beneficiary dies can have unexpected and costly consequences. Complications can arise with life insurance policies, payment on death (POD) accounts, transfer on death (TOD) accounts, wills, and trusts.
Source: https://seniorscenemag.com/what-happens-if-a-beneficiary-dies-2/

For the case where the beneficiary dies first (i.e. the OP's situation), it states:

1] If the beneficiary dies before the person making the gift: The law prohibits us from leaving property to a deceased person. When there are no named living beneficiaries, TOD and POD accounts as well as life insurance policies are paid to the decedent's estate, requiring probate.
With a will or trust a gift to a deceased beneficiary will lapse (go away) unless protected under Florida's anti-lapse statutes (discussed below). When a specific gift (e.g. $10,000 or the home) lapses it becomes part of the residual estate (what is left after distributing specific bequests). If the lapsed gift is part of the residual estate, it goes back into the pot to be divided among the remaining residual beneficiaries.

(Note: decedent's estate in the above is that of the person whose will or trust is being distributed, not that of the potential beneficiary).
...Probably Not
So: apart from "anti-lapse" situations (see below), if the (potential) beneficiary is already dead when the settlor of the trust dies (and there are no instructions in the trust to handle this eventuality), the allotment to that beneficiary passes back to the settlor's estate (one of the Hiberts: also see below) and does not pass to the beneficiary's estate1,2.
On the subject of "anti-lapse", the same article notes:

The Florida's Probate and Trust Codes have anti-lapse provisions which provide that a gift will not lapse if the deceased beneficiary is a descendant of a grandparent. In that case, the gift goes to the deceased beneficiary's lineal descendants: first to children and if a child has died to his/her children. However, rather than relying on Florida Law,   it is always best to specify who will receive a gift if the primary beneficiary dies.

As implied by "Florida" in the above quotes, the situation in the US can vary by state, although as Rules Regarding the Distribution of a Trust When a Beneficiary Is Deceased on LegalZoom states:

While each state drafts its own set of trust laws, there has been a movement to create a standard trust code. As of publication date [not specified], the Uniform Trust Code has been enacted in 23 states. The UTC has been endorsed by the American Bar Association and AARP. The UTC offers a good general framework. However, when attempting to determine the rules specific to your situation, review your state laws and consider consulting with a licensed attorney.
Source: https://info.legalzoom.com/rules-regarding-distribution-trust-beneficiary-deceased-22486.html

To Whose Estate Will the Allotment Return
Assuming there are no clauses in the trust dealing with Daniel predeceasing the Hiberts, the portion that would have been allotted to Daniel will return "to the decedent's estate". At first glance, this would then be covered by the terms of the spouse's will (since they died second).
However, it is common (and advisable) for wills to contain so-called "Titanic" clauses: what should happen if two parties (particularly a pair of spouses) die within a short time of each other. The purpose of such clauses is to prevent confusion (=legal wrangling) should, for example, both parties die in a car accident and it is unclear which died first.
Therefore, assuming the Hiberts' wills contained such clauses, if their two deaths occur within the time specified by those clauses, the question of what happens to the trust allocation may end up being decided by the will of the first to die and not the second.
Without an explicit clause, the results will again vary by state, although some have adopted the Uniform Simultaneous Death Act which states that if two people (e.g. husband and wife) die within 120 hours of each other (and their wills contain no special clauses), each is considered to have predeceased the other. More on this can be found in the article Why your estate needs a 'Titanic clause' on CNBC.

1 This mirrors what happens with Wills, according to the (UK) solicitors Howells:

What Happens if a Beneficiary Dies Before the Testator?
If a beneficiary dies between the point when the Will was made and the death of the testator, under this scenario the beneficiary's estate will usually have no benefit from the Will.
If the beneficiary has died before the testator, the benefit is said to have lapsed, although there are exceptions to this rule.
Source: https://www.howellslegal.co.uk/news/post/What-Happens-If-a-Beneficiary-of-a-Will-Dies

(The "exceptions to this rule" as similar to the "anti-lapse" rules above: in the case where beneficiaries are direct issue (children/grandchildren etc.), the allotment normally will go to children/grandchildren of the beneficiary rather than reverting to the residual estate).

2 As the LegalBeagle page notes – assuming no clauses in the trust to the contrary – the allotment will pass to the beneficiary's estate should the beneficiary die after the settlor (whether or not the trust has been processed when the beneficiary dies).


Can I manipulate a Shutterstock vector for a logo?

Question: I am designing a logo for a friend. I have the font and colors done. All I need is an icon for it. I found a vector on shutterstock that would be perfect. I took the vector and used it as a base and re-designed it so it's completely different from the original. Should i be worried about copyright and licencing.
Answer: Yes. You should be worried about copyrights and licensing.
Royalty-free stock imagery is typically set to strictly forbid use in a logo. Shutterstock specifically states that logo usage is not permissible on their site - unless you purchase the rights to the image which can cost considerably more than typical royalty-free costs. You will find this rather common among stock image providers.
Shutterstock's Help Center topic on logo/trademark usage
And be aware, there is no such thing as "edit the image by X amount and it's okay"


Coefficient of determination ($R^2$) and sample size

Question: Is there any relationship between $R^2$ and sample size - does the $R^2$ increase with sample size? And does the adjusted $R^2$?
Comment: Whether the R^2 changes on average with sample size can't be answered without some assumptions about the process you're sampling. Since the X's are fixed in regression, this presents some difficulties in guessing what sampling circumstances you might mean. However, it's possible to specify reasonable circumstances in which the $R^2$ should approach a 'population' value.
Answer: It depends on whether you are interested in $r^2$, the sample correlation coefficient, or the $R^2$ multiple correlation coefficient, used to assess the performance of regressions.
Both $r^2$ and adjusted $r^2$ are negatively biased--that is, the sample values are slightly smaller than the corresponding population value--but the adjusted formula is somewhat less biased. In addition to the sample size, the amount of bias depends on the value, with $r^2$ near zero and one showing the least bias and those near 0.6-0.8 showing the most bias.
Table 1 of a paper by Zimmerman, Zumbo, and Williams (2003) illustrates the bias as a function of sample size and correlation value. Elsewhere in the paper, they show simulation data indicating that the Fisher and Olkin and Pratt adjusted $r^2$ reduce this bias considerably.
There is also a decent amount of work looking at "$R^2$ shrinkage", which is a  related phenomena that comes up a lot in regression-related contexts, but has the opposite sign (it is positively-biased, and adjustments bring it back down). Yin and Fan (2001) have a fairly comprehensive comparison of methods for estimating it, and Page 3/205 has some citations to descriptions of the problem.
Finally, you should be aware that there are lots of methods for adjusting $r^2$/$R^2$ (in fact, there are even multiple ($\ge3$) versions of the Olkin and Pratt adjustment formula floating around, some of which correct for the number of parameters), so it might help to be more specific about whatever you have in mind
Comment: Adjusted $R^2$ < $R^2$. If $R^2$ is negatively biased then adjusted $R^2$ must be more biased, not less. Figure 1 in the Zimmerman et al paper you linked to shows the bias of rank correlation.
Comment: Gah, that should have been Table 1 instead of Figure 1 (which has both Pearson $r$ and the rank correction). Are you sure that adjusted $r^2$ is biased more, rather than less? I thought (one) of the reasons to adjust was to reduce that bias, and Figure 2 (really, not the table!) seems to confirm that.
Comment: Zimmerman et al are talking about $r$, which is biased low and needs to be increased, whereas the OP and Yin & Fan are talking about $R^2$, which is biased high and needs to be reduced.


How to pull an event report that lists how many people bought tickets at the different level of ticket prices

Question: I need a report for an event on Civi that lists how many guests bought tickets at the different levels we had our tickets for sale at. How can I do that? I can't figure out if it's possible with any of the 4 default reports Civi offers for events.
Comment: Hello Andrea. Welcome to CiviCRM Stack Exchange. Please edit your question to include anything you might have tried and the issue you encountered with those possible solutions or questions you have in documentation that wasn't clear. You may want to read up on the documentation for reports in the CiviCRM to get started and then refine your question. https://docs.civicrm.org/user/en/latest/events/reports/
Answer: You will want to use: https://civicrm.org/extensions/extended-reports extension - it gives you access to many more templates - including price set - line item ones.
Answer: we are using a 3rd party PHP-based reporting tool which lets us get reports based on SQL queries on the civicrm database - the native Civi reports are template-based and more brittle IMO. You could run the event participants list, export the results to .csv and then sort the resulting rows by participant fee, or net amount after pulling the .csv into Excel.
 HTH


Kanji changing to boxes in Excel when I switch from Japanese to English input

Question: I'm creating a database that includes a column of Japanese amidst many columns of English, but every time I switch my input language to English, the kanji turn to boxes with the character 字 inside them. I'm not sure if this will continue you when I turn the Excel database into a searchable online database. The problem resolves itself when I switch back to Japanese input, but it's very weird and makes me worried.
Do any of you have any idea what's going on with this?
Answer: The fact that the problem resolves itself when you switch back to Japanese implies that your content is still there, intact.  When you switch to the English language input, Excel is changing its view encoding to a Western character set that doesn't include kanji so it can't display them.  As long as your online search function supports the appropriate Japanese character set, it should work okay.


Difference between the energy eigenstates and the eigenstates of other physical variables

Question: I am having difficulties understanding the difference between the energy eigenstates and the eigenstates of other physical variables. I was told that if a system is an energy eigenstate at $t = 0$ then it stays in that state until an external perturbation is applied. But if the system of any operator is in an eigenstate that corresponds to a physical observable, it can change without any perturbation.
Why is that the case? Where does this difference come from? Also why does a set of eigenstates of an operator has to commute with $H$ in order to be a stationary state?
Comment: Its simply that $H$ is the operator which appears in the Schrodinger equation, which describes the time evolution of the system
Comment: @BySymmetry sorry but I am struggling to find a connection between $H$ appearing in the SE and stationary states.
Answer: The difference is in the time-evolution.
Eigenstates of the Hamiltonian, which are states of definite energy since $\Delta E=0$ for those states, evolve in time according to
$$
\vert \Psi_n(t)\rangle =e^{-i E_nt/\hbar}\vert\psi_n\rangle\,  . \tag{1}
$$
This follows immediately from separation of variables in the time-dependent Schrodinger equation:
$$
i\hbar\frac{\partial d}{\partial t}\Psi(x,t)=\hat H\Psi(x,t)
$$
and the ansatz $\Psi(x,t)=e^{-i Et/\hbar}\psi(x)$ with $\hat H\psi(x)=E\psi(x)$.  This
 makes the evolution of eigenstates of $\hat H$ particularly simple.
If the state $\vert\Phi(t)\rangle$ is not an eigenstate of the Hamiltonian but an eigenstate of some other operator $\hat{A},$ its evolution is NOT of the type given in Eq(1).  The evolution of this state is more complicated and obtained by first expressing  $\vert\Phi(t)\rangle$ in terms of the $\vert \Psi_n(t)\rangle$'s, for which the evolution is simple:
\begin{align}
\vert\Phi(0)\rangle&=\sum_n \vert \Psi_n(0)\rangle  \langle \Psi_n(0)\vert \Phi(0)\rangle \, ,\\
&=\sum_n\vert \Psi_n(t)\rangle b_n\, , \tag{2}
\end{align}
with $b_n=\langle \Psi_n(0)\vert \Phi(0)\rangle$ determined at $t=0$.  Then
\begin{align}
\vert\Phi(t)\rangle&=\sum_n \vert \Psi_n(t)\rangle b_n \, ,\\
&=\sum_n  e^{-iE_nt/\hbar} \vert \psi_n\rangle b_n\, . \tag{3}
\end{align}
Note that, if $\hat A$ is taken to be the Hamiltonian, then by orthogonality of the eigenstates of the Hamiltonian, Eq.(2) will contain only a single term - say the term where $n=m$  and $E_m$ the energy of the state at $t=0$.
If $\hat A$ commutes with $\hat H$ and $\vert\psi_n\rangle$ is an eigenstate of $\hat H$ with eigenvalue $E_n$, then write $\vert\chi_A\rangle = \hat A\vert\psi_n\rangle$
\begin{align}
\hat H\vert\chi_A\rangle&= \hat H\hat A\vert\psi_n\rangle = \hat A \hat H\vert\psi_n\rangle
=E_n \hat A\vert\psi_n\rangle \, ,\\
&=E_n\vert\chi_A\rangle \tag{4}
\end{align}
meaning that $\vert\chi_A\rangle$ is also an eigenstate of $\hat H$.  In fact, if the eigenvalue $E_n$ occurs only one, $\vert\chi_A\rangle$ must be proportional to $\vert \psi_n\rangle$ since this is the only eigenstate of $\hat H$ with eigenvalue $E_n$.   If, on the other hand $\hat A$ does not commute with $\hat H$, then $ \hat A\vert\psi_n\rangle$ is not an eigenstate of $\hat H$ and so will not evolve "simply" as per (1).
Answer: This "speciality" of Energy eigenstates comes from the definition of energy itself. We know from Noether's theorem that every symmetry of a system gives us a conserved charge. Now if your system has a time-translation symmetry, the conserved quantity associated to that is called energy. In other words, when you look at a system at two different times, the only thing guaranteed to be same in both the measurements is energy of the system.
Now every symmetry also has a generator associated to it (whose eigenvalue gives you the conserved charge). For time translation symmetry, the generator is the Hamiltonian itself and energy is its eigenvalue. That's why a system which is in an eigenstate of Hamiltonian(energy eigenstate) will stay just like that. If any other generator commutes with hamiltonian, that implies that the eigenstates of these two are the same and the corresponding states will also be the stationary states.
Mathematically speaking,
$$H \, |E(0) \rangle = E \; |E\rangle $$
$$|E(t) \rangle = e^{-iHt}|E \rangle \implies |E(t) \rangle = e^{-iEt}|E \rangle$$
where in the last step, the state has changed only by a irrelevant phase.


QGIS Print composer using margin and scale

Question: When using the Atlas feature in the print composer is there any way of using both the Margin around feature and a defined scale?
Comment: Correct me, but don't these options cancel each other out? Either you zoom to freature x + 10%, or have a predefined scale.
Answer: Since the feature used to set the view extent margin is the same feature used to create each atlas view, the scale will be dependent on the feature size. It's thus impossible to set a defined scale for features with greatly different sizes and have a constant margin around them. If you use a defined scale, either some map extents will appear with a wide margin or some map extents will only be partially visible in the composer. If you use margins, the scale has to be different for every page.
If your features are identical in size, it's then a choice to make. Either go with a defined scale if you want nice round numbers, or go with a margin if you need a specific view extent margin around the features.


How to blend in when singing in a group?

Question: When I sing an identical song with a group, my voice tends to get louder than the group; Somehow, I think my voice has blended in with the group, but my group tells me that I am singing louder than all of them. How do I actually hear/notice when my voice gets louder than that of the group's?
Comment: Do other people tend to say that you are too loud overall, especially when you're not singing? If yes, then there's your answer.
Comment: Make sure you can hear all the other vocal parts, and then make sure somehow or another that they can hear yours.
Answer: Is this with or without mics? When someone sings, acoustically, they hear their own voice through the bones in their heads as well as through their ears, so they are well aware of what they are singing. Their ears, though, hear the other people singing, but that sound is only received through their ears. experience tells them when they're singing too loud, or not loud enough. That, or constantly being told where they are in the mix.
Using mics has a slightly different slant, as the mix may be pre-determined by the settings on the amp. used. That and the sound in your own head. And, maybe, the sound coming through the foldback - monitor - system, which may not actually be faithfully copying  the mix on the p.a. even though it should.
The bottom line is that your voice should blend with the others, and not be louder/quieter. You have to sing  in situations like this loads of times in order to discipline yourself that you really don't need your voice to be louder than the others to hear what's going on. The appropriate jigsaw puzzle piece is the same size as all the others. Just because it's important for that space doesn't make it bigger!
Listening (to the overall mix) is of paramount importance in any ensemble situation, and not enough players/singers do it. That's about it.
Comment: Oh... we sing with mics.


how to use public:// in page content

Question: As content author creates a page, she wants to have something like . Is it possible? the goal is to avoid hard-coded drupal installation path; that is she doesn't want to have  on test site and  on production site.
Answer: you can use one of this function
file_create_url($uri);
or
image_style_url($style_name, $path)
Answer: You should always be using relative paths in content, like
That way if you ever change the domain name (for example when you are on the dev environment or the site url changes) you don't have to deal with a lot of headache.
You just need to properly configure your test site to handle the /drupaltest/ part for itself.
The rest of the solutions currently are giving you PHP functions for dealing with file paths, so they won't help you when creating content (PHP really shouldn't be going in your content like that).
To configure to run in a sub-directory
There are a couple of ways you can do that.
You could set up your development environment so that apache is pointing to the subdirectory for the web root for that site.
The least technical way though would be to:

In your site's .htaccess file, uncomment the line that says # RewriteBase /drupal and make it say RewriteBase /drupaltest
Install the pathologic module.
Configure the module so that your input formats use pathologic and add your path as /

The pathologic module is useful anyway as it rewrites links in your content to use path aliases instead of paths like node/234.
Be careful when using path auto across multiple domain names, you might find you need to clear the cache whem moving between the testing & live sites to refresh the cached path related content.
Comment: you are spot on - that solutions with PHP functions are not relevant to my question. And also, that my main goal would be solved if I could just have `src=/sites/...`. The problem, though, that although I have base_url set to .../drupaltest, the path for some reason is evaluated from the root. So, the question that I *should* have asked is "how do I properly configure my site to handle /drupaltest/ part for itself"? Any pointers?
Comment: Oops sorry, I added some info to the answer.
Comment: I was tinkering with .htaccess - but couldn't make it work. So, I ended up creating a link on test site for sites -> drupaltest/sites. Kludgy - but simple
Comment: I have done that same thing at times. It's a nice simple solution.
Answer: It seems like the question is not about PHP. You can use Token Filter module to make global token values available as an input filter.

If you install and configure Token Variable module you can type some thing like that.

Comment: It is not necessary to use token filter to add the site url to the start of a url. A properly configured site will handle relative links for you without that.
Comment: That is one way (easier for the end user than entering a token every time they post a link + it handles aliases). Also if you have more control over your server you can point your domain name direct to the subdir.
Comment: What kind of configuration you are talking about? If we move site to subdirectory we have to update all img elements in _body_ field.
Comment: See my answer above http://drupal.stackexchange.com/a/80642/10729
Comment: Did you mean using of _pathologic_ module?
Answer: You can also use
theme_image_style($uri)

if you don't want to use an image style.
https://api.drupal.org/api/drupal/modules!image!image.module/function/theme_image_style/7

how to use public:// in page content


Time dilation when moving toward or away from the earth

Question: For a given speed, is the time dilation of a space traveler different if moving radially toward the earth as compared with moving radially away from the earth?
Is the time dilation of a space traveler moving tangentially to the earth always less than radial motion for a given speed?
Comment: Don't confuse the effects of time dilation with those of signal retardation. The direction of travell affects tha latter but not the former.
Comment: I am only interested in time dilation.
Comment: How ever would the space traveler's clock know whether it was moving toward the earth or away from the earth?  And why would it care any more about this than whether it was moving toward or away from Mars?
Comment: Assuming that the space traveler is at earth's geosync orbit radius moving toward or away from the earth is relevant since gravitational potential is either increasing or decreasing.  Does the direction of movement against the gravitational field affect the space traveler's clock with respect to a clock at rest?  I think that the answer requires considering general relativistic effects.
Answer: Time dilation only depends on the magnitude of the relative velocity between the two observers. Therefore, it doesn't matter if the two observers are moving towards or away from each other.
This becomes especially clear when you realize when we discuss reference frames we are are talking about, well, frames of reference. At this point then there really is no sense of two frames moving towards or away from each other. There is just relative motion between the two frames, and this relative motion is what causes time dilation when comparing time intervals between two events between the two frames.
Comment: See:  https://physics.stackexchange.com/questions/174694/would-special-relativity-predict-time-dilation-of-a-geostationary-satellite-comp  It appears that the direction matters for how the time dilation responds; or does it?  I like your answer, but I would like to reconcile it with the former stack exchange response on this topic.
Answer: Gravitational time dilation is a name given to the observation that the higher clock in a gravitational field runs faster than a lower clock.  It is a poor name.
The observation is between clocks.
The same observation is made between two accelerating clocks, one ahead of the other.  It can be calculated from the times each clock shows to the other observer after the acceleration stops, when they are moving inertially.  Both observers then agree the leading clock shows a later time than the trailing clock.
Answer: There are two separate effects to keep track of here- namely time dilation caused by the relative motion of the space traveller, and time dilation due to gravity.
The first effect is dependent only on the relative speeds of the reference frames of the space-traveller and the observer on Earth- the direction of motion has no effect. You can convince yourself of this if you imagine observations made in deep space where there is no local large spherical body (ie Earth) beneath you- you would have no way of judging direction, and nothing to be tangential to.
The second effect is governed by the space-traveller's relative distance from the centre of the Earth, and so a traveller approaching Earth will experience a gradually reducing gravitational frequency shift, whereas a traveller moving away from Earth will experience a gradually increasing one. The space-traveller moving tangentially to the Earth will experience a variation in gravitational time shift in proportion to the component of their motion normal to the Earth's centre, which will be zero in the case of a traveller orbiting the Earth, but will vary for a traveller on a straight path tangential to the Earth.
These effects have to be taken into account when calibrating GPS systems. You can read an interesting summary here: https://en.wikipedia.org/wiki/Error_analysis_for_the_Global_Positioning_System#Relativity


The use of such a or the article THE

Question: While doing some exercises in grammar I came across this question
- Her boss was impressed with...........good job she has done.
I'm confused between "such a" and "the" as answers
My hunch says "the" but I'm not sure.
Please, can someone clear up this confusion?
Comment: Thanks for the answer. So according to what you have said the article "the" is the correct answer.
Answer: When using "the" in this context there is no implied comparison:
Her boss was impressed with the good job she had done.

When using "such a" there is an implied comparison. The boss is implying that the job is better than expected.
Her boss was impressed that she had done such a good job.

The use of such a or the article THE


Deleted old kernels, now can't boot from hard drive or live-usb

Question: Running Ubuntu 16.04 on HP laptop. New to ubuntu.
When I power on computer, it displays a blank purple screen for a moment. Then black screen with blinking cursor. Then goes to 'Startup Menu'. Pressing 'Enter' repeats this cycle.
From the purple screen I can access the Grub menu by pressing 'Escape'. The only option displayed is "*System setup". Selecting this repeats the cycle described above.
From the 'Startup Menu' I can select 'BIOS Setup', as well as the other standard options. I altered 'Boot Order' to boot from USB device first, before hard drive. I then inserted a ubuntu 16.04 live-usb and restarted. Goes through same cycle above, does not acknowledge the USB.
Last week I deleted some old kernels to free up space on the boot drive, I'm gonna go ahead and and guess I screwed up and deleted too much.
I can provide any additional information needed, just ask.
I've dug around online for a couple hours now and haven't found a solution. If you can help (or point me in the right direction) that would be terrific. Thanks!
Comment: Is [this](https://help.ubuntu.com/community/RemoveOldKernels/Problems) useful to you?
Answer: Usually there is a button to press when booting to show boot order and select where you want to boot from. Usually f8, f11 or f12. Have you tried that? Press it right when the motherboard logo is shown.
If you are able to boot into the live session then it's probably easy to fix with boot repair. There is a guide for installing it here.
It also explains the option of installing boot repair on a disk. Maybe you could try that?


How can I detect if someone is using an ssh tunnel as a proxy to connect to a VPN?

Question: I've got an SSH server setup in a remote location, which I use to tunnel traffic from an android tablet. On the tablet, I have the Cisco AnyConnect app to connect to my company's VPN. We want to be able to detect that a proxy is being used by the tablet. Any ideas as to how I could go about doing it?
Comment: All outgoing traffic goes via VPN? If not how can you detect which path does traffic on different computer is using?
Comment: If you are trying to put a stop to something then I would suggest strict firewall ACLs.  If you don't want a user to do something then limit their ability to do so.
Comment: Do you have administrative access to the users device? What about to the SSH server? It could be possible to restrict access with those, but only if they are managed by you.
Answer: A VPN is just routing with protection. In your case, the "proxy" ensures that traffic goes first from the tablet to some server S, from which it then goes to "the Internet", e.g. to the entry point for the company's VPN. The "proxy" uses SSH but that's just between the tablet and server S; you cannot see it from the company.
What you will see, though, from the company, is the IP address of S. A relatively invasive strategy is to use OS fingerprinting on that machine (basically, scanning who is connecting) to try to guess whether S is indeed an Android system, or some other kind of OS. If you detect that "who connects" does not use Android directly, then you have shown that some intermediate system, a "proxy", is in use.
(Whether scanning machines who connect is legal is a complex subject which depends a lot on jurisdictions. Don't deploy that without taking some legal counsel.)
Now, of course, in all generality, there is no such thing as a "proxy / no proxy" dichotomy. A basic tablet user will connect over WiFi when he can, connecting to an access point which will use NAT, and that qualifies as a kind of "proxy". This already prevents the reverse-scanning explained above from really working (i.e. most of the time, the scan will see the access point, not the tablet itself).
The whole point of using a VPN like Cisco AnyConnect is to make the proxy question irrelevant: the VPN uses cryptographic protection exactly so that the paths used by data packets have no impact whatsoever on security. You should not have to ask yourself whether some SSH or NAT or whatever is in place. Thus, your question is kind of weird.
Comment: I should've provided a few more details. I've got a system that allows/denies access based on where the user is connecting to the VPN from. With the current method defined above, I can use my android tablet in "bad" location, but not be denied access because it looks like I'm coming from the SSH Server, which is in a "good" location. Thus, in order to ensure the integrity of the location detection, I need to make sure that a proxy is not being used.
Comment: @suss23 I agree with the last statement from Tom.  If you are trying to block based on their location, but they are (effectively) using a VPN to protect themselves and connect to a "good" location, then why is there a problem?  This sounds like a win to me.  I think your question effectively boils down to "Can I detect if a user is using a proxy" in which case the answer is "maybe".  There should be lots here about proxy detection though, which is pretty much what Tom was addressing.


Weird 4 tile jump height in Terraria

Question: The problem is I have just a 4 tile jump height instead of the 6 tile in Terraria. It is really  really annoying and I don't know why did this happen. I cant find anything about it at the internet, just keep searching but I can't find anything.
Does anyone know how can i get back the 6 tile jump height? I have the game on steam, 126.96.36.199
Comment: Are you in water and trying to jump out of it?
Comment: If you hold the jump button, do you jump once and then stop, or do you keep jumping repeatedly? Former is the expected result (unless you have the Frog Leg accessory equipped).
Answer: The game has no way to obtain a jump height of 4 blocks, you can see all the possible jump heights in the game in this infographic from reddit user DataNalle.
I would try reinstalling the game from scratch. So long as you backup your save data (located in Documents\My Games\Terraria\ on Windows and Library/Application Support/Terraria on Mac.) you should be fine reinstall/verify your game.

Verify game cache: Right click game on library list > Properties > Local Files > Verify Integrity of Game Cache
Delete local content: Right click game on library list > Delete Local Content...

If none of this works and your not completely adverse to cheating you can use http://yal.cc/r/terrasavr/ to give yourself a jump boost buff/item which my help.
Answer: Maybe try holding the jump, if you aren't already, as that's all i can think of and if you keep holding it as long as you can before you fall, it should go as high as you can jump.
Comment: Yep, i hold down the jump button, which is space, and the character goes up to the fourth block only. Really annoying, jsut bought a game and this happens.

Weird 4 tile jump height in Terraria


2010 Workflow with Designer email plus approval on library column

Question: I am trying to create a workflow in designer where it sends the email to the group managing the library that a new request as been added to the library which is working but then I need to send another email to the created:by that the request was either approved or Denied which is determined by a column dropdown in the library and if denied include the reason based on another column text field. What is the best way to create this?
Answer: You can use the same workflow which starts when an item is added to a library and make sure to check the box 'Start workflow automatically when an item is changed'
Create two steps in your workflow and add a condition for the
1st step to check if the workflow status is NOT 'Approved' or 'Denied' - send an email to the group managing the library.
2nd step to check if the workflow status is 'Approved' or 'Denied' - send an email to 'CreatedBy', you can also add currentitem:Reason field in the body of the email.
Comment: How do I check the status as not approved or denied what command allows me to put a or in the step?
Comment: Ok so when I finish the email to notify them that a request has been submitted to the library I wait for edit menu table to start to contain the column that is a dropdown of Approve and Denied. Then I created the two steps. In the first Step I am running in parallel waiting for the status to contain approved or denied then log in the settings. then in the 2nd step I run the If approved do this else denied do this. which just emails the created:by. and in the case of the denied I did look up the column that the host can edit. Should this work?