How to remove skull, which obscures the vision?

Question: Each time you kill someone a skull appears, that obscures your vision for a second. How to avoid it and make the targets behind it visible?
Anything would be good, for example, remove the skull, make it more transparent, make it different color (so at least I can see the outlines of the enemy).
Comment: Is this really a big deal? The skull shows for a quarter of a second at most, and verifies that the target is dead. In the screenshot you provided, it looks like that person is already been killed, the red outline just hasn't been removed yet.
Comment: I fail to see where the intent of question matters, or whether it's a "big deal" or not. The question was asked, an answer was given, let's move on with our days.
Comment: @MageXy, first of all, why is it important if it is big deal or not? I just want it to be removed. Second, it is not a quarter of a second, it is longer than mccree shoot cool-down (0.5 sec). The image illustrates how after killing one enemy (killed enemies disappear immediately) you can hardly find a head of another enemy.
Comment: @Frank, erm? "Can hardly find = can't find". You can remember its position, but you can't see it at the time when your weapon is ready to shoot, so you must to shoot blindly.
Comment: If you can find another opponents head in that timeframe, what's the issue? It obviously doesn't interfere.
Answer: There is no way to change the UI in that way currently and no planned feature to do so. Right now the only way you possibly could have the skull not show up is by turning off the entire UI with Alt+Z but this would cause far more problems by having literally no HUD.
Answer: There is currently no way to do so.


Editing Excel without opening and closing by Excel Engine

Question: Is there a way to read Microsoft Excel file without opeining it?
Opening and closing excel files are very slow operations indeed.
In my program I want to edit 17 excel files at once and save them.
Please help
Answer: By "opening an Excel file" and commenting that it's very slow you are presumably opening it in Excel - which you shouldn't be. Find yourself a library for working with Excel files, there are plenty around, some third party, some by Microsoft, and work with the files but not by automating the program. The later formats (2007 onwards) use (zipped) XML, which can be read/modified directly.
Search StackOverflow for existing questions on Excel and you'll find hundreds/thousands.
This one looks a promising link, gleaned from here - https://stackoverflow.com/questions/170652/whats-the-best-net-library-for-parsing-and-generating-excel-spreadsheets
http://www.carlosag.net/Tools/ExcelXmlWriter/
Answer: I assume you are opening your files using Excel interop, which is slow. If your files are Excel 2007 or above, you can create or edit files using the free Microsoft Open XML SDK. It takes advantage of the fact that Office documents in versions 2007 and above are essentially zipped xml files, and provides an API to simplify working with these documents, without requiring Excel itself.


Mahalanobis distance fixed at root 2

Question: I have this data set of 12 input variables with a size of over 9000. Wanted to find any outliers using mahalanobis distance. As a central point, I decided to take mean and median of all observations, getting two central points. When I calculate the distance between the centre and datapoints using scipy, I get a uniform value of root 2 across all points. I even tried by implementing the distance formula in python, but the results are the same. I have also checked every step, including the inverse covariance, where I had to use numpy's pinv due to singular matrix . 1) Is it because of mean/median, what should be the central point ? 2) Is it saying something about the data set, most data are unique. 3) Is there something I am failing to understand. I am a stats amateur, so any insight is appreciated.
Answer: Barring programming errors, that would mean that $$(x_i - \mu)^T\Sigma^{-1}(x_i-\mu) = 2$$
for all points, which seems highly unlikely. It is also surprising that the inverse does not exist (unless some of those variables are exact linear combinations of the other), as you have more that 9000 observations and only 12 variables.

Mahalanobis distance fixed at root 2


does ifft of higher order equal upsampling signal

Question: I have a signal with length 2000, and am taking an fft by 2048, after I perform analysis and some operation on signal in frequency domain I tried to convert it back to time domain. I received a signal length 2048, looking similar to what I would expect but not the same. I know that taking higher point fft simply generates more samples of the DTFT we would expect from the original signal, it does not add new information. However, when I take ifft my data length increases, I am assuming it must time domain interpolation between signal and output is same as if I would upsample my expected 1000 point signal (the operations I perform can be ignored, in my case if this is true for simply taking higher fft and then ifft of signal as is, then its true for me). I am not sure if this assumption is true, as I have yet to grasp how the IDFT works when length in frequency domain changes. I am performing analysis using MATLAB and am aware of the sampling rate and time. Can anyone please share some insight on if my assumption is true and any mathematical insight into it or point me to some freely available resource?
Answer: The longer FFT simply pads zeros to the original time domain signal. The inverse FFT should then be the original signal with zeros appended to the end.
The OP is correct that taking a longer FFT interpolates samples in frequency, and specifically provides more samples of the Discrete Time Fourier Transform. Why this occurs is clearer by observing the difference between the DFT and DTFT formulas as given below:
DFT:
$$F(k) = \sum_{n=0}^{N-1}x[n]e^{j k \omega_nn }$$
Where $\omega_n = 2\pi /N$
DTFT:
$$F(\omega) = \sum_{n=-\infty}^{\infty}x[n]e^{j \omega n }$$
For $\omega$ continuous from $0$ to $2\pi$ (for unique span)
Notice the two key differences: The time increment n in the DTFT extends to plus and minus infinity, and the frequency domain variable $\omega$ is a continuous function. As we add zeros to x[n] in the DFT, we approach the DTFT and ultimately for the same $x[n]$ all results of the DFT as more and more zeros are added will be samples on the continuous DTFT function.
In general related to this it is worthwhile to understand a universal property of all the Fourier Transforms related to signals that repeat or not (periodic or aperiodic) and continuous or sampled signals:
If a signal is continuous in one domain (time or frequency) it must be aperiodic in the other domain. And the counter-case is also true; if the non-zero values of a signal are discrete in one domain, it must be periodic in the other if we allow the domain to extend toward infinity. We see that immediately with the premise of the Fourier Series Expansion in that any single-valued analytic waveform over a finite interval can be represented by a discrete and infinite number of frequencies; adding together all those frequencies in the time domain over an infinite time duration will result in a periodic waveform.
With that observe the Continuous Time Fourier Transform (CTFT) as shown in the slide below. Regardless if the waveform itself is zero valued over a portion of the interval, the key point is the Transform itself uses a time axis that extends from minus infinity to plus infinity, and for that reason the frequency domain MUST be a continuous function: Frequency resolution is the inverse of observation time. If we have infinite time to observe a stationary signal, then we will have infinite frequency resolution. So the CTFT is of a signal in time that is continuous and aperiodic. Therefore the result in frequency is aperiodic and continuous.

Now compare this to the Fourier Series Expansion (FSE) as shown in the slide below, of a time slice of the same time domain signal. As previously explained, if the time is extended beyond the time interval from $0$ to $T$ the resulting waveform will be periodic. The FSE is specifically done over the interval from $0$ to $T$, but periodicity is implied as the exact same result in frequency would occur at the non-zero values if time was extended to minus and plus infinity. The frequency axis for the FSE is discrete since the results are defined for $n \omega_o$ with integer n. What also would occur if time was extended to infinity as explained for the CTFT is the frequency axis would become a continuous function, but the new interpolated values between the original samples would all be zero valued. The FSE of a signal in time from 0 to T is the same as a signal that is periodic in time over that same period. The signal in time is continuous and periodic, therefore the result in frequency is aperiodic and discrete (non-zero values are discrete).

Now compare this to the Discrete Time Fourier Transform (DTFT) as in the slide below for the same time domain waveform.  The only difference from CTFT is that the signal is now sampled in time. Thus the signal that is now discrete and aperiodic in time will be periodic and continuous in frequency: The same frequency result in the CTFT becomes periodic in the DTFT.

Finally the Discrete Fourier Transform (DFT) combines both effects of the FSE and DTFT; time is discrete as in the DTFT and restricted to a finite interval as in the FSE, with implied periodicity in that we can equally extend time to plus or minus infinity without changing the non-zero results in frequency. Thus the DFT for the same time domain waveform as the cases above would appear as in the slide below. The signal that is discrete and periodic in time would be periodic and discrete in frequency.
Comment: Thank you , this helps immensely, was really looking for the part "The inverse FFT should then be the original signal with zeros appended to the end." I was only wanting to find out why does it mathematically equal appending zeros to original signal after bigger than length ifft.
Comment: @vikassajanani do you see how this answers that or is that part still not clear? See how in the last two slides comparing the DTFT to the DFT where the DTFT is the DFT with infinite zeros appended. So as you append zeros, you increase the number of samples in frequency until it ultimately becomes continuous. So if zero padding creates the interpolated spectrum, reversing that interpolated spectrum gives you back the original (zero-padded) signal.
Comment: yes, I see it now, it just became very clear, thanks for the amazing explanation. These slides are wonderful resources.
Comment: @vikassajanani Thank you! They are from a course I teach that is coming up online if you are interested (lots more like this): http://ieeeboston.org/event/digital-signal-processing-webinar/ that focuses on providing a unique intuitive understanding of DSP with but beyond the math.


Is it always true that a non-adiabatic reversible process is an isothermal process?

Question: In a Carnot Cycle, a reversible isothermal process is a non-adiabatic reversible process. Is it always true that a non-adiabatic reversible process is an isothermal process?
Comment: Does this answer your question? [Does an isothermal process always imply an adiabatic process?](https://physics.stackexchange.com/questions/630064/does-an-isothermal-process-always-imply-an-adiabatic-process)
Answer: The word isothermal means "at constant temperature", and hence in an isothermal process
$$\Delta T=0\ \ \ \ \ \ \ \text{Isothermal Process}$$
In an adiabatic expansion, there isn't any net heat flow and we have
$$\delta Q=0\ \ \ \ \ \ \ \  \text{Adiabatic Process}$$

Is it always true that a non-adiabatic reversible process is an isothermal process?

No! Consider an example of the process which is not an adiabatic process nor an isothermal process. Consider a process in which the state of system changes from $A(p_i,V_i,T_i)$ to $B(p_f,V_f=V_i,T_f)$ which is called an isochoric process. In such a process,
$$\delta Q=C_VdT\not= 0,\ \ \ \ \ \ \ \Delta T\not=0 $$

Everything taken to be reversible!
Answer:
Is it always true that a non-adiabatic reversible process is an
isothermal process?

No. A non-adiabatic process is any process that allows heat transfer between a system and its surroundings. Many processes, including isobaric and isochoric processes, involve heat transfer.
Hope this helps.
Answer: A) Non adiabatic reversible process means:

$q$ is not equal to 0

Reversible process i.e the change in state is not at equilibrium with its surrounding while the process is happening, unlike irreversible process.
In simple words, no integration is needed.

B) Reversible Isothermal process means:

$\Delta T = 0  $ is the only speciality.

Using the first law of thermodynamics i.e $\delta U = \pm  q  \pm W$

$q = c \cdot m \cdot ΔT$  where $c$ is specific heat capacity.
For B, we can write  $$\frac{nf \cdot R \cdot \Delta T }{2}= q \pm
W\text{ or }pdV $$
Since it is clearly written in the definition that $q \ne 0$, none of the variables in the formula of $q$ can be equal to $0$. Therefore, $\Delta T \ne 0 $ for a reversible isothermal process.
For A,
$$\frac{nf \cdot R \cdot \Delta T }{2}= q \pm W\text{ or }PdV $$
It states that for this condition that $\Delta T = 0$. Therefore, $0 = q \pm W$.
A and B do not match each other's conditions since $\Delta T $ of both the processes are not same.
No, non adiabatic reversible process $\ne$ Reversible Isothermal process.


Add a "request CW mode" button to the Ask Question page

Question: Since only moderators have the power to make a new question CW, please add a "request CW mode" button to the Ask Question page, to make it easier for users to request that their questions become CW.
Comment: There's nothing the community moderators can do about the content of the Ask Question page. So I'm migrating this to Meta.SO. On the other hand, note that you can always flag your own question for moderator attention with the request to make CW.
Comment: +1 to this request, though it should really be on meta.SO (ordinary mods here can do nothing about this).
Comment: @Akhil: I have no idea whether or not the current CW setting is a network-wide or a site-specific preference. I do know that at any case Jeff reads here a lot.
Comment: No, no, no, no, no, no, no, no, *no*. The checkbox was taken away so people would *stop even thinking about* CW for questions.  See the two bullet points at the bottom of waffles's answer.
Comment: @Aarobot: the CW feature continues to be widely used on Math Overflow.  It is very useful, for instance, as a middle ground between distinctly acceptable and distinctly unacceptable content, as well as for asking questions which probably do not have a single correct answer.  The users at math.SE are missing this feature.  Simply repeating "no" a bunch of times to our request seems obnoxious to me: *in our community, this is a feature we found to be helpful*.
Comment: Having said that, I don't understand why we shouldn't just be asking for the "make this question CW" box to be reinstated, possibly with a modest minimal rep for use.
Comment: @Pete: The use of CW as a designation for borderline questions is *exactly why the feature was taken away*. Nothing but respect for you folks at MO, you should run the site how you see fit - but here on SE we've simply concluded that the feature does more harm than good.  Either a question is acceptable or it's not.  If it's not, close it.
Answer: Hmmm ... not following how this would be better than this.

If you feel so strongly that a question you are asking should be CW I think you should ask yourself why.

Why do you think you deserve no rep for this question?
Why do you think this question even belongs on the site?
Comment: I would add as the first bullet: Why do you want everybody to be able to edit your question?
Comment: @Aarobot ... I thought about that, but now with the "pending edits" system - this become arcane. "Why would you want low rep users to edit your question without the edit being reviewed?" ... Weird thing to want


Tool for analyzing DNS config

Question: I have made dns using bind. Can someone recommend me some tool that I can paste the code in to test for errors?
Answer: If your BIND master configuration file is, for example, /etc/bind/named.conf,
then you could use the named-checkconf tool that comes with modern versions of BIND (Debian-based distributions may package it separately into bind9utils package) to check it for errors:
named-checkconf /etc/bind/named.conf

If your configuration is complex and has several included files, this command will show all the configuration assembled into a single listing:
named-checkconf -p -x /etc/bind/named.conf | less

If you want to test any configured DNS zones, you can use:
named-checkconf -z /etc/bind/named.conf

There is also named-checkzone for checking the zone files more specifically.
You might also find in your distribution's list of available packages a separate dlint utility that can be used to check your DNS server for common problems. This utility works by contacting the DNS server and examining the specified DNS domain (and any subdomains, if they exist and you haven't specifically forbidden it).

Tool for analyzing DNS config


making instant yeast rise higher/more

Question: I make a relatively quick oatmeal bread (with bread, whole wheat flours and old-fashioned oatmeal); it calls for instant/fast rise yeast.  I am wondering how to make it rise a bit more than it does....
Comment: Welcome Tovah, we need more information to help. Please edit your question to include the recipe and instructions.
Comment: The whole wheat flour and oatmeal will make a bread more dense than a bread made from all white flour.
Comment: A pinch of ginger might help : https://cooking.stackexchange.com/q/12084/67
Answer: That depends whether it still needs to be 'quick'. You can autolyse your flour, which will help the bread to rise. https://www.kingarthurbaking.com/blog/2017/09/29/using-the-autolyse-method This is just adding water to the flour - no yeast - then adding the yeast later in the process.
However this will add time.
You can possibly increase temperature, i.e. use a warm place. Essentially your yeast converts sugar into carbon dioxide and alcohol (which burns off). The carbon dioxide gas makes the loaf rise.  Your yeast species probably works fastest around 32-35C.

(SC in that graph)
A lower temperature will mean slower activity, so everything else being equal 1 hour at 32C will give much more gas than 1 hour at 25C. Fast rises can lead to voids in the loaf, so that's something to bear in mind.
If your bread has no added sugar (you don't mention this), then the yeast will not have much to work on. Your options here are to add (more) sugar or to add a flour improver containing amylase, which breaks down the complex carbohydrates in flour into sugar. Other flour improvers contain ascorbic acid (vitamin C), which helps strengthen the gluten network.
Answer: One thing that limits the effective rise is if you knock any air out of the dough late in the process. This means you should avoid transferring it between containers after the final rise. This is especially true for high hydration doughs. That's a problem if you rely on baking in a preheated dish.
Answer: One thing is if you have yeast, give it 30 minutes to settle in water before adding to the dough. Give it a chance to proliferate.
Comment: This won't make it proliferate - you need to add a small amount of sugar to the water to get the proliferation to happen. You can also do it by adding flour, such as when making a sponge before making the bread, or other nutrients such as vitamin c and soy protein.

making instant yeast rise higher/more


How to copy formulas from one spreadsheet to another?

Question: If I select say 10 rows in Google Spreadsheets and use ctrl+c and then create a new spreadsheet and try to paste all the data it only moves over the values but not the formulas. Is there any way to paste all the data including the formulas? This feature works only if I paste it in the same spreadsheet. Thanks!
Comment: Yes, I have tried. This feature only works on the same page, if I open any other spreadsheet it doesn't.
Comment: Have you tried right click > Paste special > Paste **formula** only
Comment: Found a way to make it works.
Answer: I know this is a few years after the initial question, but I found a quick(ish) workaround that requires a couple steps outside of google sheets..
I tried the 'Show formulae' method as this was a good way to copy formulae in bulk, but also ran in the same issue as above.  Here's what ended up working best for me:

I copy the formulae (I do this in one column batches) to a word processor.
Then I use a little strategic find and replace, in my case, replacing the " ( " symbol in this chunk
of formulae with " ('[sheet name here]'! " so that all the formulae
in the text I copied is now amended with the proper sheet
referencing.  Depending on your formulae, this step could change for you, but chances are there should be some similarities here between cells to take advantage of with F&R instead of doing it manually one-by-one.
Once I have my block of proper formulae, I copy the whole thing, select the group of cells in the column I originally
copied the formulae from and paste the whole block of text from the
word processor.  This pastes each formula in its own cell since
they're separated by line breaks, all correctly attributed!
One thing to note is that if the cells you're pasting into are merged,
this will undo that merge since google sheets generates new cells with
the text data rather than pasting values into the existing cells.  The
formatting should remain though.
Answer: First you need to show formulas :

Then copy and paste values only.
Comment: Wow! That's awesome, it really works! Thank you, you are a lifesaver! I have no idea how you managed to come up with this idea.
Comment: There's only one problem. Let's say one of the cells in a row has a formula =sum(B66,B64). Normally, if I copy and paste a row that contains such a formula, it automatically takes the coordinates of a new row and B66 and B64 become B1 and B2 or whatever. If I do the way you suggested, it doesn't.


Using BLE, how fast can I get ~50 bytes of data from the master to a slave? How does that change when increasing the number of slaves present?

Question: In personal project of mine I'm looking to have a system where a number peripherals need to have provided to them, in as real-time as possible, small (~50 bytes max) amounts of information from a central controller. I was considering using low power Bluetooth for this and I'm wondering what kind of performance I could expect. How many times per second could I send this 'packet' of data if there was only one peripheral? What about if the number increases to, say, ten?
Answer: You are going to be limited by what is known as the Connection Interval.  This defines how often the Central (host) communicates with the Peripheral.
BLE uses a frequency-hopping scheme; two devices each send and receive data from one another on a specific channel, and then meet on a new channel sometime later.  The time between the each hop is defined as the Connection Interval.
There can be a maximum of six packets (four for iOS) sent per connection interval, and each packet can have up to 20 bytes of payload.  So in your case you would use up three packets, which would fit in one connection interval.  You would waste the remaining packets.
According to the BLE specification, the allowable range for the Connection Interval parameter is from 7.5 ms to four seconds.  With a minimum Connection Interval of 7.5 ms, you could send theoretically 133 connections per second. So your throughput would be 133*50*8 = 53.2 kb/s. This doesn't count the initial time to establish an initial connection while scanning for Peripheral BLE devices (which will be sending out Advertising packets on a periodic basis).  Also, if an ACK has to be sent back to acknowledge receipt of the packets of data, this will require another Connection Interval, and thus halve the data rate.
Note: the Apple iOS guidelines limit the minimum connection interval to 20 ms, not 7.5.  So you would only have a maximum of 50 connections per second.   In that case the throughput would be 33*50*8 = 20.0 kb/s.
Here's some more information about BLE data rates.
Comment: @PlasmaHH Thx for the edit.  I try to proofread my stuff, but some things just get by once you've looked at it too many times.
Comment: Thanks for this information! How does the situation change when I include more than one slave? Do I have to "establish the initial connection" every time I switch between the slaves, thus consuming even more Connection Intervals, or am I set, once I'm set?
Comment: @Sanuuu No, you shouldn't repeatedly have to establish a connection; from what I've read (I haven't tried this myself), you can have multiple instances of a connection object, and just loop through them.  The exact syntax will vary depending on what BLE stack you are using.


Can an EU member state recall or change their nominated commissioner mid-way through a term?

Question: Can an EU member state recall or change their nominated commissioner in the European Commission, mid-way through a term?
Answer: No, a commissioner is nominated by a member state but s/he is not a representative of that state. They are not supposed to get instructions, be recalled or be otherwise influenced by their state of origin. The same is true for judges in the EU court of justice, who are named by a member state for a fixed term mandate but approved (and potentially revoked) by the court itself.
The most senior representatives of EU member states at the EU level are the permanent representatives, who prepare Council meetings.
Answer: Commissioners are nominated individually by the member states, but then the European Parliament confirms them as a group.
Comment: That's a procedural detail.
Comment: @Relaxed: No, it's more than that. Since the EP confirmed the group, member states cannot change the membership of that group. Either the EP would have to individually confirm replacements (violating the group confirmation) or the EP would not confirm replacements (opening a major loophole).
Comment: @MSalters That's nonsense. In fact, the exact same situation arises when a commissioner resigns, which happens regularly. The real reason member states are not allowed to recall their Commissioner is simply that this is the way the institutions are intended to work, and not merely a side-effect of the procedural details of the confirmation by the European Parliament.


Using QR algorithm to compute the SVD of a matrix

Question: How to use the QR algorithm to compute the SVD of a matrix $X\in R^{m\times n}$? Is there any algorithm for doing that?
Answer: The SVD can be obtained by computing the eigenvalue decomposition of the symmetric matrix
$$\begin{align}
\begin{bmatrix}
0&X\\X^T&0
\end{bmatrix}
&=
\begin{bmatrix}
U&0\\0&V
\end{bmatrix}⋅
\begin{bmatrix}
0&Σ\\Σ^T&0
\end{bmatrix}\cdot
\begin{bmatrix}
U&0\\0&V
\end{bmatrix}^T
\\&=
\frac1{\sqrt2}
\begin{bmatrix}
U&-U\\V&V
\end{bmatrix}⋅
\begin{bmatrix}
Σ&0\\0&-Σ
\end{bmatrix}⋅
\frac1{\sqrt2}
\begin{bmatrix}
U&-U\\V&V
\end{bmatrix}^T
\end{align}$$
The eigenvectors have the form $\begin{bmatrix}\pm u_k \\ v_k\end{bmatrix}$ with $u_k$ and $v_k$ being the left and right singular vectors, with eigenvalues $\pm\sigma_k$ and some zeros to fill the dimensions.
Since the Hessenberg form of symmetric matrices is tridiagonal, heavy simplifications are possible. These simplifications lead directly to the Golub-Kahan algorithm.
Comment: So for finding the svd of X, we first find the Hessenberg decomposition of (XX') (let's call it H) , then using QR iteration, Q'HQ is a diagonal matrix with eigenvalues of XX' on the diagonal. Q is the matrix of eigen vectors which is equal to U of svd. then we can find V. Am I right?
Comment: You can do that, but it is a numerically bad idea, since in $XX'$ the condition number of $X$ gets squared. What Golub-Kahan does is to apply the QR algorithm to the block matrix above, but using its structure to avoid doubling the dimension. The first step is to decompose $X=U_0BV_0^T$ where $B$ is a bi-diagonal matrix and $U_0$, $V_0$ orthogonal matrices composed of Householder reflectors. Then the usual bump-chasing is performed.
Answer: Quoting the wikipedia article on SVD:

The left-singular vectors of $M$ (i.e. the columns of $U$) are eigenvectors of $MM^T$.
The right-singular vectors of $M$ (i.e. the columns of $V$) are eigenvectors of $M^TM$.
The non-zero singular values of $M$ (i.e. the non-zero diagonal elements of $\Sigma$) are the square roots of the non-zero eigenvalues of both $M^TM$ and $MM^T$.

The QR algorithm finds eigenvalues and eigenvectors of square matrices. $M^TM$ and $MM^T$ are square matrices. There might be a better way, I don't know, but this is the naive, obvious way.
Answer: My answer to this is the following (see the MATLAB snippet below). You make QR-decomposition for A and then repetitively take the R matrix, transpose it and apply QR-decomposition to R'. Evidently, QR applied to the upper-triangular R gives the matrix R again producing nothing new but if you apply QR to the lower-triangular R' and keep doing it again and again you'll see that the resulting R will converge to a diagonal matrix with singular values on its diagonal. Whereas if you multiply all the orthogonal Q-matrices obtained in the process you'll get U and V matrices. See the code example:
<code>   N = 15; % number of iterations
   R1 = A;
   U = eye(size(A));
   V = eye(size(A));
   for i = 1 : N
      [Q1, R1] = qr(R1);
      [Q2, R2] = qr(R1');
      R1 = R2';
      U = U * Q1;
      V = V * Q2;
   end
   S = R1;
</code>
This is a manifestation of the known QR algorithm for finding eigenvalues of a symmetric matrix but applied to a more general situation. Notice that here the matrix A need nor be symmetric and still we obtain both S and U and V together.
Comment: Hi Dan, is this a known method? I just thought of a similar idea over in a discussion thread in MO https://mathoverflow.net/a/436944/297 and googled to see if anyone else had come up with it, and this answer was the only hit.
Comment: Well, I've given the link to wikipedia article describing the similar method but it seems a bit different than mine. I didn't find any mention of this particular algorithm either. I invented it from my head.


How to get CustomDocumentProperties using Excel Interop?

Question: The below code is used to get the custom document properties for Excel workbook.
var xlApp = Globals.ThisAddIn.Application; // This works in VSTO Excel Add-in
var xlApp = new global::Microsoft.Office.Interop.Excel.Application(); // This doesn't work anywhere
xlApp.Visible = true;
global::Microsoft.Office.Interop.Excel.Workbook workbook = xlApp.Workbooks.Open(file, false, true, Type.Missing, Type.Missing, Type.Missing, Type.Missing, Type.Missing, Type.Missing, Type.Missing, Type.Missing, Type.Missing, false, Type.Missing, Type.Missing);

global::Microsoft.Office.Core.DocumentProperties properties = workbook.CustomDocumentProperties; // Exception occurs here
global::Microsoft.Office.Core.DocumentProperty property = properties["propertyname"];

The first 2 lines are references to the Excel Application. One obtain the reference from VSTO add-in internals, the other is a regular new Application().
When using the Application from VSTO internals, the code run fines without any problems. But when using new Application(), the workbook.CustomDocumentProperties line throws InvalidCastException:

Unable to cast COM object of type 'System.__ComObject' to interface
  type 'Microsoft.Office.Core.DocumentProperties'. This operation failed
  because the QueryInterface call on the COM component for the interface
  with IID '{2DF8D04D-5BFA-101B-BDE5-00AA0044DE52}' failed due to the
  following error: No such interface supported (Exception from HRESULT:
  0x80004002 (E_NOINTERFACE)).

I am trying to make it to work on a C# winforms project without VSTO. A lot of examples and tutorials use new Application() for Excel interop, but I noticed that Microsoft.Office.Interop.Excel.Application is an interface, so using new on interface is actually strange to me. How can I create a proper Application that can get the CustomDocumentProperties?
Reference Assemblies I am using:

Microsoft.Office.Interop.Excel v2.0.50727 Version 220.127.116.11
Microsoft.CSharp v4.0.30319 Version 220.127.116.11
Answer:
I  noticed that Microsoft.Office.Interop.Excel.Application is an interface, so using new on interface is actually strange to me.

That is strange indeed, but by design. The Excel.Application interface is decorated with the CoClass attribute telling the actual class to instantiate on 'instantiating' the interface. More about it here.

But when using new Application(), the workbook.CustomDocumentProperties line throws InvalidCastException:

Strange indeed again. I have experienced some issues myself using document properties. It seems that the actual class returned differs from the spec, so I moved to use dynamic in order to prevent type casting issues.
So instead of this:
Microsoft.Office.Core.DocumentProperties properties = workbook.CustomDocumentProperties;

Use:
dynamic properties = workbook.CustomDocumentProperties;

Comment: Not as far as I know. Actually a lot of the interop classes are using `dynamic` for this exact reason.
Comment: Changing both `DocumentProperties`  and `DocumentProperty` to dynamic actually works! But not sure if relying on `dynamic` is a good practice? Is there no way to ensure the correct type that the code is dealing with to prevent further possible runtime problems?
Comment: I find it comical that the Microsoft Web pages on this generate the noted error, thanks Patrick Hofman!
Answer:
How can I create a proper Application that can get the CustomDocumentProperties?

There is no need to create a new Excel Application instance if you develop an add-in. You should use the Application property provided by the VSTO runtime:
var xlApp = Globals.ThisAddIn.Application; // This works in VSTO Excel Add-in

But if you develop a standalone application which automates Excel, in that case you need to create a new Application instance by using the new operator:
var xlApp = new global::Microsoft.Office.Interop.Excel.Application();

Use the late binding technology (Type.InvokeMember) for getting or setting document property as the How To Use Automation to Get and to Set Office Document Properties with Visual C# .NET  article suggests.
Answer: I've had a same problem. Today it has been resolved. There is a different approach to derive the results. The question and its' answer is at
How can I read excel custom document property using c# excel interop
Here is my implementation.
public string CheckDocProp(string propName, object props)
        {
            Excel.Workbook workBk = Globals.ThisAddIn.Application.ActiveWorkbook;

            object customProperties = workBk.CustomDocumentProperties;
            Type docPropsType = customProperties.GetType();
            object nrProps;
            object itemProp = null;
            object oPropName;
            object oPropVal = null;

            nrProps = docPropsType.InvokeMember("Count",
                BindingFlags.GetProperty | BindingFlags.Default,
                null, props, new object[] { });
            int iProps = (int)nrProps;

            for (int counter = 1; counter <= ((int)nrProps); counter++)
            {
                itemProp = docPropsType.InvokeMember("Item",
                    BindingFlags.GetProperty | BindingFlags.Default,
                    null, props, new object[] { counter });

                oPropName = docPropsType.InvokeMember("Name",
                    BindingFlags.GetProperty | BindingFlags.Default,
                    null, itemProp, new object[] { });

                if (propName == oPropName.ToString())
                {
                    oPropVal = docPropsType.InvokeMember("Value",
                        BindingFlags.GetProperty | BindingFlags.Default,
                        null, itemProp, new object[] { });
                    return oPropVal.ToString();
                    break;
                }
                else
                {
                    return "Not Found.";
                }
            }
            return "Not Found.";
        }

Usage:
object docProps = wb.CustomDocumentProperties;
            string prop1 = ExistsDocProp("<CustomProperty>", docProps);


What are the provisions for mentioning the trademark or trade name of source code orobjects licensed in open source software for proprietary software?

Question: I'm making software that will be proprietary software, which contains some source code or objects from several different open source licenses such as BSD 3-clause, MIT, MPL, Apache, and GPL 3.0, can I mention the owner of the code or object , what are the provisions regarding the mention of trade names or trademarks when I distribute, sell or provide services to the software that I make?, I hope I get information and references regarding this problem.
Comment: trademark is separate from copyright, but *most* open source projects are happy to get credit (in fact the copyright license often requires it). Some, like Firefox, require you to change the name if you change the software.
Answer: Most Open Source software is not trademarked. Even if it is trademarked, you can name the software in a descriptive, accurate manner. For example, you can state that your software is built upon library X. Indeed, nearly all Open Source licenses require that you provide such attribution in the documentation provided with your software.
But you should avoid creating the impression that your software was endorsed or created by the authors of those Open Source components. While this is explicit in some licenses (e.g. original BSD license), this follows more generally from personality rights (in some jurisdictions) or from fair competition laws.
In any case, you should read and understand the licenses of the software components that you distribute. In particular with regards to MPL and GPL, you should note that these are copyleft licenses that will require you to give recipients some or all of the source code, which may prevent some business strategies you would have otherwise pursued.


Java String declaration

Question: What is the difference between String str = new String("SOME") and String str="SOME"
Does these declarations gives performance variation.
Comment: See also http://stackoverflow.com/questions/334518/java-strings-string-s-new-stringsilly and http://stackoverflow.com/questions/390703/what-is-the-purpose-of-the-expression-new-string-in-java
Answer: String str = new String("SOME")

always create a new object on the heap
String str="SOME"

uses the String pool
Try this small example:
        String s1 = new String("hello");
        String s2 = "hello";
        String s3 = "hello";

        System.err.println(s1 == s2);
        System.err.println(s2 == s3);

To avoid creating unnecesary objects on the heap use the second form.
Comment: The link you shared explains it beautifully. Thanks!
Answer: There is a small difference between both.
Second declaration assignates the reference associated to the constant SOMEto the variable str
First declaration creates a new String having for value the value of the constant SOME and assignates its reference to the variable str.
In the first case, a second String has been created having the same value that SOME which implies more inititialization time. As a consequence, you should avoid it. Furthermore, at compile time, all constants SOMEare transformed into the same instance, which uses far less memory.
As a consequence, always prefer second syntax.
Comment: For the record, there are scenarios where `new String(String)` makes sense, like if you have a very large string and you're only interested in retaining a small substring. The `substring` methods only return a flyweight view of the original string, so using `new String(hugeString.substring(a, b))` forces a copy and lets the GC reclaim the contents of `hugeString` when it goes out of scope. They shouldn't have made it a constructor, though...
Comment: Interesting case of optimization, indeed, but I wouldn't go this path before having done some profiler checks (so would you, I guess).
Answer: String s1 = "Welcome"; // Does not create a new instance
String s2 = new String("Welcome"); // Creates two objects and one reference variable

Answer: First one will create new String object in heap and str will refer it. In addition literal will also be placed in String pool. It means 2 objects will be created and 1 reference variable.
Second option will create String literal in pool only and str will refer it. So only 1 Object will be created and 1 reference. This option will use the instance from String pool always rather than creating new one each time it is executed.

Java String declaration


How to adjust the edge sharpness for the subdivision surface (in edit mode)?

Question:
After you add subdivision and back to edit mode. you can see there is big distance between subdivision surface and framework surface. How can I adjust the framework scale to make it close to the subdivision surface?
Answer: Use Edge Crease (⇧ Shift+E)
Quote from the blender manual

This edge property, a value between (0.0 to 1.0), is used by the Subdivision Surface Modifier to control the sharpness of the edges in the subdivided mesh. This operator enters an interactive mode (a bit like transform tools), where by moving the mouse (or typing a value with the keyboard) you can set the (average) crease value of selected edges. A negative value will subtract from the actual crease value, if present. To clear the crease edge property, enter a value of -1.


What is the friction force at terminal velocity?

Question: As terminal velocity is inversely proportional to viscosity or fluid friction, then my question is: at terminal velocity, the fluid friction is maximum or zero?
Answer: The terminal velocity of an object falling through a fluid is only inversely proportional to the fluid's viscosity when the Reynolds number is small (viscous drag the dominant resistive force). At large Reynolds numbers, terminal velocity is independent of viscosity (see here). In either case, however, the drag force felt by the object is a function of the relative velocity between the object and fluid. At the instant of release the object has zero velocity so, if the fluid is also still, the drag force on the object is zero at this instant. If the object falls under gravity (in a constant gravitational field) and does not lose mass then the impelling force, it's weight, is a constant. Under these conditions, the drag force increases from zero to a maximum value, equal in magnitude to the object's weight and opposite in direction, at which point the object is in a dynamic equilibrium and moves at the terminal velocity.
Comment: @Yasir Mukhtar If this answers your question, please consider marking it as accepted.
Answer: When you reach the terminal velocity, you can apply Newton's 2nd Law $F=ma$, and since there is no acceleration (the terminal velocity is constant in time) , there must be a balance in the forces, such that their resultant vanishes.
In your case, the force into play are viscosity and the cause of the initial motion (for example a push, or gravity). When you reach the terminal velocity, the fluid friction is calculated simply as follows (all the relations are meant to be evaluated at equilibrium, i.e., after the particle has reached the terminal velocity):
$$F_{term}=-F_{friction}+F_{ext}=0$$
$$F_{friction}=F_{ext}$$
and since the friction force is (in general cases) of the form $-\eta v$, at the terminal velocity:
$$\eta v_{term}=F_{ext}$$
In particular, in the case of gravity as external force:
$$\eta v_{term}=m g$$
Answer: At terminal velocity the resistive forces (drag (you need to push the fluid aside) and friction (the actual friction between the falling body's surface and the surrounding fluid)) are equal to the accelerating forces (in most cases gravity).
Answer: Neither. 'Fluid Friction' can be considered as a drag force on a body moving through a fluid which is proportional to the square of the velocity. At terminal velocity the drag force balances the force of gravity - a dynamic equilibrium. If however additional downward force is increased - such as adding additional mass, the velocity will increase and so will the frictional (drag) force.


What is the temperature of an accretion disc surrounding a supermassive black hole?

Question: What is the temperature of an accretion disc surrounding a supermassive black hole? Is there plasma in the disc?
Comment: [This](http://www.scholarpedia.org/article/Accretion_discs#Extreme_mass_supply_:_.22sub.22_and_.22super.22_Eddington_accretion_rates) is a fantastic resource on accretion disks, and this section in particular may help with accretion rates.
Answer: It depends on the distance from the central body. This gives the temperature $T$ at a given point as a function of the distance from that point to the center ($R$):
$$T(R)=\left[\frac{3GM \dot{M}}{8 \pi \sigma R^3} \left(1-\sqrt{\frac{R_{\text{inner}}}{R}} \right) \right]^{\frac{1}{4}}$$
where $G$, $\pi$, and $\sigma$ are the familiar constants, $M$ is the mass of the central body (and $\dot{M}$ is the rate of accretion onto the body), and $R_{\text{inner}}$ is the inner radius of the disk - possibly (if the object is a black hole) the Schwarzschild radius $R_s$, in which case we can simplify this a little more. So the temperature in the accretion disk is far from constant.
Whether or not there is plasma depends on the exact nature of the disk, the central object and the region around it. For example, a supermassive black hole may have different matter in its disk than that of a stellar-mass black hole. I should think, though, that black holes in binary systems accreting mass from a companion should have plasma in their accretion disks, and supermassive black holes might also have plasma from nearby stars.
Comment: I imagine, it would be more interesting to see this equation in number form.
Comment: @AlexeyBobrick With specific numbers from an actual accretion disk?
Comment: Yes, like something you can on page 9 of your reference.


Papers Please Booth Upgrades

Question: I am aware that at the "end of day" screen one is supposed to be able to buy an upgrade for their booth allowing them to use the space bar to enter inspect mode and use the tab to toggle the stamp bar, however there is no instruction stating where to look to purchase the in-game upgrades. I have searched through the "end of day" screen trying to find the upgrades with no clue where they are. if you found them, don't simply tell me that I can buy them on the "end of day" screen, as I already know that (apologies if I sound rude), but instead explain how to get them at said screen.
Comment: The possibility of a housing upgrade will appear (seemingly) randomly at the end of day screen. Once at that screen, an option will pop up that will let you purchase the flat.
Comment: They are listed at the end of day, in red/grey text with a tickbox like food and heat. You can only get them after quite a few days into the story, though.
Comment: I'm not sure if there is an underlying mechanic to make them appear, e.g. a number of successful processings, number of detainments or similar. It is worth researching though.
Answer: The booth upgrades become available in the end of day screen on set days during the game. There's nothing special you need to do to trigger it, you just need to make sure you saved enough cash to pay for it!
Comment: An upgrade costs [5 credits](http://papersplease.wikia.com/wiki/Booth#Upgrades). You also get a notification at the "end of day" screen that an upgrade is available.

Papers Please Booth Upgrades


SKYPE on ubuntu 14-04 -- how to upgrade it?

Question: SKYPE has upgraded its soft some while ago, and I missed my chance to reinstall it at that point. One of my  comps run ubuntu 12-something, and I guess it is hopeless  now to reactivate SKYPE on it now without upgrading it to a more recent ubuntu release (and I hate doing that), while another one (laptop  Dell Precision 5510) runs ubuntu 14-04, and I hope it would accept latest SKYPE soft. If this is still doable, how do I do it?  Thanks!!!
Comment: Similar: https://unix.stackexchange.com/q/409433/117549
Answer: Skype snap package
You don't need to the download the .deb package for Skype for Linux or install Skype from a PPA anymore. Skype has been released as a native snap package in Ubuntu 14.04 and later.
snap find skype
Name      Version      Developer  Notes    Summary
skype     184.108.40.206    skype      classic  One Skype for all your devices. New features. New look. All Skype.

To install Skype in all currently supported versions of Ubuntu open the terminal and type:
sudo apt install snapd
sudo snap install skype --classic

The skype snap package will be updated automatically when updates are available.
After you have tested that the skype snap package is working correctly you can uninstall the old version of Skype that you were using before.
Ubuntu 12.04
Ubuntu 12.04 is no longer getting software updates and does not support installing snap packages, so download the Skype for Linux .deb package instead or else upgrade from Ubuntu 12.04 to a more recent Ubuntu LTS release. Ubuntu 12.04 is an end of life release. To resolve package dependencies in Ubuntu 12.04, you need to read How to install software or upgrade from an old unsupported release?.
Open the terminal and type:
cd ~/Downloads/
wget https://repo.skype.com/latest/skypeforlinux-64.deb
sudo apt install ./skypeforlinux-64.deb

When you install the Skype for Linux .deb package it automatically adds a repository to keep your installation of Skype up to date.
Comment: thanks, will try immediately on 14-04!!!


Some US AAA Corp are borrowing below UST for 10yr paper. What could be the reasons for this?

Question: Any particular situations which can lead to this?
Comment: They are considered to be more financially stable that the US government. They are considered to have lower volatility than US treasuries thus a lower risk premium. They are smaller quantities of the issues and greater demand leading to higher prices and lower yields.
Answer: I can't find any today.  Johnson and Johnson is rated AAA and the 10yr yield is Treasury +29bp.  There may have been a time during the debt ceiling drama where very short dated corporate paper traded better than Treasury bills, since the latter were potentially subject to a payment delay.


Why the ground state energy of bosons are 0 at T=0K ? Does it violate Heisenberg's Uncertainity Principle at 0K?

Question: Bosons obey Heisenberg's Uncertainity Principle but do not Pauli's Exclusion Principle. That's why in Bose Condensation we get a large amount of particles in a single state i.e. ground state at T=0K. But why the ground state energy is zero ? And in this case is Heisenberg valid ?
Comment: Why would it violate the uncertainty principle?
Answer: Bose-Einstein Condensates (BEC) do not violate the Heisenberg's uncertainty principle.
Traps
Usually, BECs are not in free space, but spatially confined by some potential $V(r)$.  The total Hamiltonian is $H = p^2/2m + V(r)$ so your "$E=0 \rightarrow p=0$" reasoning does not hold.
This potential $V$ is usually approximated as harmonic oscillator, $V \propto 1/2 \, m \sum_i \omega_i^2 x_i^2$, which will have a zero-point energy $E_0 = \hbar/2\, (\omega_x+\omega_y + \omega_z) \neq 0$.  So once again, you do not have $E=0$ as per your reasoning.
A BEC in a trap thus has some spatial extent $\Delta x$. It is not an eigenstate of the kinetic energy, and thus also has some spread in momentum $\Delta p$.
Free space
A BEC in free space, $V=0$, would have a flat wavefunction spread uniformly over all space. Since $V=0$, the wavefunction is an eigvenstate of the kinetic operator, $\psi = e^{\mathrm{i}qx} =  1$ where $q=0$, i.e. it's the lowest plane wave. The momentum $p = \hbar q$ is now known exactly.
But while $\Delta p = 0$, the spatial extent of the wavefunction is $\Delta x \rightarrow \infty$, in accordance with Heisenberg.
Answer: According to quantum mechanics, a particle can only have certain well-defined energies (given by solutions to the Schrodinger equation), the lowest of which is called its ground state energy. You can think of a collection of Bosons at absolute zero all being in their ground state- that doesn't mean that they have no energy, but just that their energy can't possible go any lower.
The Heisenberg Uncertainty Principle doesn't prevent a particle from having an exact energy. What the principle means is that a particle cannot simultaneously have exact values for two separate observable properties unless they 'commute'. This is tricky to explain in non-mathematical terms, but I will give you a specific example.
If you have a particle in its ground state, it has a very exact energy, E say, but its position is not well defined because of the spatial spread of its wave function, so you cannot know both the energy and position exactly at the same time.
In principle a particle could have a very tightly constrained position (ie you could be pretty sure where it was), but in that case its wave function would be a narrow spike. The narrow spike would not be a solution to the Schrödinger equation, so the particle wouldn't have a specific energy.
It is worth spending time reading books about this because the underlying principles- about expanding wave functions in terms of orthogonal eigenstates of other operators- are extremely beautiful once you understand them. If I get the time I will try to supplement this answer with a non-mathematical explanation.


crystal report initialize on x64

Question: I have a program that works properly on x86. also it works on systems with visual studio without error even on x64. but on x64 systems without visual studio, occur error when call reports. message exception is :
The type initializer for 'CrystalDecisions.CrystalReports.Engine.ReportDocument' threw an exception.
Source:
CrystalDecisions.CrystalReports.Engine
Stack Trace:
   at CrystalDecisions.CrystalReports.Engine.ReportDocument..ctor()
   at CrystalDecisions.CrystalReports.Engine.ReportClass..ctor()

and message that contain this decription :
An error has occurred while attempting to load the Crystal Reports runtime.
Either the Crystal Reports registry key permissions are insufficient or the Crystal Reports runtime is not installed correctly.
Please install the appropriate Crystal Reports redistributable (CRRedist*.msi) containing the correct version of the Crystal Reports runtime (x86, x64, or Itanium) required.  Please go to http://www.businessobjects.com/support for more information.

my program is based on:

.net framework 4
visual studio 2012
crystal report 13.0.2

and install CRRuntime_64bit_13_0_2 sometimes.
Can anyone help me?
Answer: One possible solution
Go to your application project, right click, and select "Properties"
On the Build tab, find the Platform Target combo box. It probably says, "Any CPU".
Change this to x86 and your project will compile for 32 bit only, thus eliminating the issue.

crystal report initialize on x64


Non-relativistic limit of the Dirac equation

Question: How to recover non-relativistic limit of the Dirac equation
$$\left( i\gamma^{\mu}\mathcal{D}_{\mu} - m \right)\Psi(x) = 0$$
where $\mathcal{D}_{\mu} = \partial_{\mu} + iqA_{\mu}$. I do not assume that the vector potential $A_{\mu}$ is zero, but I have a particle in central potential and magnetic field. During derivation please highlight where does the non-relativistic assumptions is used and how.
I have found some notes here, but I don't understand why can we Taylor expand this expression
$$\frac{1}{E + m -qA^0}\vec{\sigma}\cdot(\vec{p} - q\vec{A}) \approx \frac{1}{E+m}\vec{\sigma}\cdot(\vec{p} - q\vec{A})  + \frac{1}{(E+m)^2}qA^0\vec{\sigma}\cdot\vec{p}$$
around $qA^0 = 0$ and neglect $\vec{A}$ in the second term.
Comment: The reason for neglecting $\vec A$ is that the magnetic field $\nabla \times \vec A$ is a factor of $v/c$ smaller than the electric field $-\nabla A^0$ (for a static vector potential). See also the answer to [Why are magnetic fields so much weaker than electric?](http://physics.stackexchange.com/questions/128512/why-are-magnetic-fields-so-much-weaker-than-electric)
Answer: The essential reason is that the "$E$" in your equation has no non-relativistic limit. This can be seen more clearly by separating out the rest-mass-energy $mc^2$ and writing $E = mc^2 + H$, with $H$ comprising the total of the kinetic and internal energy. The "$mc^2$" part blows up to infinity as $c$ goes to infinity.
On the other hand $H$ has a non-relativistic limit. At the same time, the mass in non-relativistic theory can be treated as the non-relativistic limit of the "moving mass" $M = E/c^2$. So, to get a non-relativistic version of Dirac requires actually extending the Poincaré group from 10 dimensions to 11, by replacing $E$ by $M$ and $H$, treating them as independent.
So, the confusion arises from confusing $E$ and $H$. The shifting from $E$ to $H$ is precisely what is done with the Foldy-Wouthuysen transformation.
In place of the mass shell identity
$$\frac{E^2}{c^2} - ||^2 = (mc)^2,$$
you then have two identities:
$$||^2 - 2MH + \frac{1}{c^2}H^2 = 0, \quad M - \frac{1}{c^2}H = m.$$
If you solve for $H$, you will see more clearly the connection to non-relativistic theory ... and you will also see more clearly what that extra 11th dimension in the symmetry group reflects:
$$H = \frac{||^2}{m + M}.$$
Compare to the non-relativistic case:
$$H = \frac{||^2}{2m} + U$$
where $U$ is the internal energy - i.e. the "kinetic energy" arising from the kinetics of the internal constituency of a body ... except now, it's being generalized and treated as a thing in its own right that potentially applies even to elementary bodies. So, it's the rest frame value of $H$, just as $m$ (the rest-mass) is the rest frame value of $M$. Substituting into the equations for the two invariants leads to the following modified form:
$$||^2 - 2MH + \frac{1}{c^2}H^2 = -2mU + \frac{1}{c^2}U^2, \quad μ ≡ M - \frac{1}{c^2}H = m - \frac{1}{c^2}U.$$
The drop-down from extended Poincaré-11 to ordinary Poincaré-10 then corresponds to setting $U = 0$.
On the other hand, the non-relativistic limit corresponds to setting $1/c^2$ to $0$. The non-relativistic limit of $M$, $m$ and $μ$ are all $m$, the mass, with all 11 dimensions of the symmetry group kept intact.
In Relativity, the following is invariant:
$$E dt - ·d,$$
where $d = (dx,dy,dz)$. If you make the above replacements, you get
$$H dt - ·d + μc^2 dt.$$
To make a meaningful non-relativistic limit out of this, you have to throw in the proper time, which we can denote $s$ and rewrite it as:
$$H dt - ·d + μc^2 dt - μc^2 ds.$$
Then, you can define a new coordinate $u = c^2(s - t)$ - i.e. the time dilation, itself, ramped up by $c^2$ and write:
$$H dt - ·d - μdu.$$
Now, you have a new invariant that you can take the non-relativistic limit for. In place of the Minkowski invariant
$$ds^2 = dt^2 - \frac{|d|^2}{c^2}$$
are now two invariants:
$$|d|^2 + 2 dt du + \frac{1}{c^2}du^2, \quad ds = dt + \frac{1}{c^2}du.$$
This, too, has a non-relativistic limit:
$$|d|^2 + 2 dt du, \quad ds = dt.$$
The 11th generator in the extended Poincaré group is actually the translation on the $u$ coordinate. Since everything else is $u$-invariant, then ... by the Noether Theorem ... there is a conserved quantity. That's $μ$.
In the non-relativistic limit, $μ$ is the "central charge" that arises when you centrally extend the Galilei group to the Bargmann group - which is actually what the correct symmetry group for non-relativistic theory is, not Galilei!
Extend the operator correspondences appropriately:
$$H ⇔ iħ \frac{∂}{∂t}, \quad  ⇔ -iħ∇, \quad μ ⇔ -iħ\frac{∂}{∂u},$$
replace the $$ and $β$ matrices to rewrite
$$δH - · - εμ ⇔ iħ\left(δ\frac{∂}{∂t} + ·∇ + ε\frac{∂}{∂u}\right),$$
for units $δ$, $ = \left(γ^1, γ^2, γ^3\right)$, $ε$, endowing them with a suitably-defined algebra (which will - in fact - be isomorphic to the complexified Dirac algebra), and add in the eigenvalue equation
$$μψ = -iħ\frac{∂ψ}{∂u},$$
for the central charge. Restricting to the ordinary Poincaré group, for Relativity, by setting $U = 0$ and $μ = m$ produces:
$$iħ\left(δ\frac{∂ψ}{∂t} + ·∇ψ + ε\frac{∂ψ}{∂u}\right) = 0, \quad mψ = -iħ\frac{∂ψ}{∂u}$$
which will be equivalent to the Dirac equation. Taking the limit $1/c^2 = 0$, on the other hand, will produce a result equivalent to the Schrödinger equation.
You can also add in a coupling with the electromagnetic potential. In that case, the non-relativistic version will be the Pauli-Schrödinger equation.
Note that shifting from $E$ to $H$ by subtracting off $mc^2$ corresponds adding a phase factor $\exp(±imc^2/ħ)$ to $ψ$, depending on which direction you're going in.
I demonstrate this, in detail - here Morphing The Dirac-Maxwell Equation To The Pauli-Schrödinger Equation and in the opposite direction, here Morphing The Schrödinger Equation To The Klein-Gordon Equation.
Needless to say, none of this was available in during the time Schrödinger devised his equation. He didn't know that Bargmann was the correct symmetry group for non-relativistic theory, rather than Galileo, nor that its natural geometric interpretation is 5D, not 4D; and even up to the present day there is still not a proper understanding that the correct symmetry group for Relativity is not the Poincaré group, but its (trivial) central extension to the extended Poincaré group.


Is it acceptable to write that you are getting a "Ph.D. in [area]" (rather than name of department) on social media?

Question: I am starting a PhD in computer science, and have noticed that a number of classmates will write on LinkedIn or their academic websites that they are getting a "PhD in Machine Learning" or "PhD in Artificial Intelligence" and so on. Is this common/acceptable to write "PhD in [Area of Interest/Expertise]" instead of your home department?
Comment: Common/acceptable to who? It appears common within your class, but how much wider are you interested? Also, if you have reason for believing that it might possibly not be acceptable then it may be worth saying.
Comment: @Ian_Fin, an ideal answer will note if this is audience-dependent and factor that in if needed.
Comment: Given the range of possible audiences, and consequently the number of considerations potentially requiring factored in, does this not make the question rather broad? Perhaps a more focused question would be something like "Is it accurate to describe myself as getting a PhD in Machine Learning specifically, rather than just a PhD in Computer Science?"
Answer: I believe it makes sense.
The PhD education is a very narrow education which will give you very specific knowledge regarding a specific aspect of a research area.
Therefore, it makes completely sense to be a bit more specific rather than general. PhD in computer science sounds very broad and not much descriptive.
Comment: Well, i agree with you. I was focusing more on the research area rather than on the title. For me is obvious that if you are still pursuing it you can not use it officially.
Comment: The problem is that the OP's classmates are not PhD _yet_. I think it would be fine if they say "PhD student in AI".
Answer: This should not be a problem.  These people are providing additional information about what they are studying; nobody is going to be confused and think that they are studying under the auspices of a "department of machine learning," or whatever.
Whether people do this is cultural, and it depends on both the field and institution involved.  Personally, I would not phrase things precisely this way on my c.v.  However, technically, my Ph.D. degree is not "in" anything at all; no field was noted on my diploma or at my commencement ceremony.  Depending on context, I might say that my degree was in "mathematics," "applied mathematics," or "mathematical physics."  Which I would use would depend on my audience and what information I was trying to convey.
Answer: It sounds fine to me too. FWIW, I usually phrase it as two separate statements: I'm getting a PhD. My research/thesis topic is..."
Not only are you clarifying what your field of expertise is, you're also preventing misunderstandings. I study artificial life, but I'm in what's called the "Faculty of Engineering", and for all I know that's what will be listed on my diploma. I certainly wouldn't want people to think I was capable of building a bridge!


How do you send a 64 bit meterpreter stager?

Question: All the stagers that are sent seem to be 32 bit. This is fine, until I have to run Mimikatz which on a 64 bit system you NEED to run the 64 bit mimikatz. Since the 64 bit Mimikatz won't run on the 32 bit meterpreter, I'm in need of the 64 bit Meterpreter.
Answer: Just use a 64 bit Meterpreter payload.
For example the windows/x64/meterpreter/reverse_tcp payload.
You can get a 64 bit shell from a 32 bit shell by using the payload_inject module built-in to Metasploit. You can do this by running the following commands after you already have a 32 bit Meterpreter shell: use windows/local/payload_inject  set payload windows/x64/meterpreter/reverse_tcp. Then setting the other parameters (SESSION, LHOST, LPORT) to yours. Then you get a 64 bit Meterpreter shell!
Comment: This didnt work for me. Here is the error i got.
[-] Exploit aborted due to failure: bad-config: Cannot inject a 64-bit payload into any process on a 32-bit OS
Comment: But I was able to migrate to a 64 bit process using the migrate command and I got a 64 bit meterpreter.


Preston not working

Question: Okay I am new to fallout 4 and when I did the mission where sturges he said go and help build sanctuary and talk to Preston but he won't respond and the first ever mission the Preston gives you has gone missing what do I do I don't what to restart ?
Answer: The wiki mentions this being a bug that makes the mission impossible to complete, so unfortunately the only fix is reloading the last save game you made before starting that mission.
Two tips:

in general: save your game regularly, for example before starting a new quest (this applies to all Fallout games: sometimes there are bugs and sometimes you want to try for a different outcome)
apparently the chance of this bug popping up can be reduced by having all necessary materials before starting the quest.

Preston not working


display classes of a namespace in visual studio (C#)

Question: I am a Java programmer, and just starting to use Visual Studio to do C# programming. In Java IDE such as Eclipse, if I do not know the class name in a package, I can just type a dot (.) after a package name, then I will get a list of all the classes in that package. How I can configure visual studio to do the similar thing, say, if I click a namespace name in a file (for example, using System), or add a dot after the namespace, all the classes in that namespace will be displayed somewhere?
Answer: As far as I remember, the same scenario works. Type System. and Visual Studio should show the Intellisense window with its contents. If it doesn't show up you can try Ctrl+Space to force it appear.
If it still doesn't show anything, you might need to add an assembly reference to the DLL that contains that namespace to your project.
Obviously, if the namespace is System you don't need to add any reference, as it is added by default already.
Also relevant: the Object Browser view in Visual Studio.
Answer: If you want to know, first make sure Ctrl + SPACE is not used by other things such as IME. Then execute this shortcut for IntelliSense within the editor. Visual Studio will display all classes you can access in the context. It may not work like Java, but most people like me find it enough to move on.
To better understand each namespace, I am more used to hit F1 on the namespace to launch MSDN.
Answer: The Visual Studio intelligence is good enough to that for u only you have the correct reference added to your assembly. Or you can try Ctrl+Space to force it.


How can i remove liquidity from a token

Question: I'm learning how to create erc20 tokens, and after adding liquidity to my token for test, I Renounced ownership and now my eth is stuck in Uniswap.
Is there a way to remove the liquidity even as I'm no longer the contract owner?
Answer: You don't remove liquidity from a token unless the token is a pool contract.
You can remove your liquidity from the pool as long as you still hold the pool tokens, to burn for the underlying tokens they represent.
You should first try to do that from the web interface if uncomfortable with calling contract directly.
If your token that is in the pool has some strange mechanic or was launched with faulty parameters such as token tax on transfer designed to work with a different router, it is possible you have locked the weth with them in the pool contract.

How can i remove liquidity from a token


Difference between Socket Connection and XMPP Connection

Question: I am new to these two connections. I used to work with HTTP Connection before and now I wanna move to a new type of connection. I searched about connections and got these two types: Socket Connection and XMPP Connection. What is the main difference between them? I wanna use this for a chat application so which type is better recommended? Thanks
Answer: XMPP is a protocol for communication and so is HTTP. Both XMPP and HTTP will internally use socket connections.
You are confused between application protocols and network layer.
Comment: I see, I successfully integrated the HTTP connection. But no successful testing for XMPP connection. Check this POST I explained more my problem.
http://stackoverflow.com/questions/7432407/try-to-connect-to-server-with-xmpp-connection
Answer: Socket is the essential thing of any existing connections there is. If you want to use any connection that has a host point and a port, very likely (unless they write their own), they use Socket (to open a connection on a given port, send a message, and close the connection port) internally.


Could the folding wingtips of a plane (ex. F-35C) be used as a vertical stabilizer?

Question: In the rare event of the vertical stabilizers being ripped off and the rest of the plane surviving, could folding wingtips (given they are far enough behind the CoM to create a stabilizing moment) be used as a "replacement" for the vertical stabilizers?
Comment: Not really folding wingtips but the [Rutan VariEze](https://en.m.wikipedia.org/wiki/Rutan_VariEze) uses that concept.
Comment: @sophit - As did the B-70. But those were foldable in flight, by design. https://aviation.stackexchange.com/questions/22745/why-werent-the-xb-70s-wingtips-fixed-in-the-down-position
Comment: @WPNSGuy: yes, very good example indeed  Even if in the case of the Valkyrie there was more than just increasing lateral stability.
Answer: If the vertical wing tips have aerodynamic centers that are aft of the center of mass, there will be some yaw stabilizing effect by the tips.
Problem is, the effect will be quite minor, the arm being so short, and without vertical stabilizers, the overall aerodynamic center of the entire body is going to be forward of the center of mass, because there is more fuselage area ahead of the center of mass than behind it.
The turned up tips having so little "leverage" they will likely just make the thing somewhat less unstable in yaw, but not positively stable. They are just too far forward to make more than a minor difference.
Remember, it's just a big weathervane pivoting about the center of mass. Look at a side profile without vertical fins, and imagine where the overall center of the finless profile presented is, relative to the center of mass (assume the center of mass is about 1/4 of the wing chord aft of the leading edge). If the total profile footprint is still forward of the center of mass, the weathervane, or airplane, will want to switch ends and point the other way. Then it will be stable in yaw, just going backwards.
Comment: Agreed. unless the tips were as far back as the vertical stabilizers, then they would not provide as much stability as the vertical stabilizers for sure.
Comment: When you put floats on an airplane, they move the yaw axis neutral point (aerodynamic centre) forward, reducing yaw stability, b/c the floats have more surface area ahead of the CofG than aft. That's why you usually see little finlets or ventral fins added to the tail when floats are installed. On airplanes that DON'T include fins as part of the float kit,you will notice the decrease in yaw stability when floats are installed. Nose wanders in bumps, and adverse yaw with aileron increases.
Answer: Well, no. There is no control authority or code to tell the wingtips to act as a replacement for the vertical stab(s) which have been blown off.
Additionally, you can't fold the wingtips up in flight.
Might they act as sort of a yaw stabilizer? Maybe. But not with any steerable control from the cockpit.
If the system had been designed to do this, like the XB-70, then yes. Since it was not, this will not work. Especially if trying to compensate for having the existing vertical stabs "blown off".
Comment: *"you can't fold the wingtips up in flight"* - Citation required. Why not?
Comment: Yeah I was thinking more along the lines of yaw stabilizer. Just something to make you not immediately go into a spin.
Comment: @Wyatt - Whatever caused the RUD (Rapid Unplanned Disassembly) of the vertical stabs might have an influence on stability.
Comment: haha yeah you're probably right. Just was wondering if in a rare case if the wingtips could provide stability.
Comment: @Richard - I believe it is controlled by a weight on wheels switch, much like munitions release. No weight on wheels, no fold. Also, only the C model has folding wings. The -A and -B do not.
Comment: I feel it is possible that folding the outer panels in flight would have a greater effect on the stability in pitch than in yaw - or in roll, for that matter, supposing that it changes the effective dihedral.
Answer: I agree with the diagram by Mins, however, creating a source of drag at the wingtips would create a yaw force in the correct (rearward) direction with ample 'arm'.
The MU-2 uses drag-brakes instead of ailerons to induce both yaw and bank, however, in general, it is frowned upon to induce drag as part of maneuvering as this creates unexpected stalls in low-speed turns. i.e. MU-2.
The wingtip design would not require the wingtips to pivot upright, just remain as-is while inducing drag, such as a split trailing edge as speed brake, just like the Space Shuttle vertical fin did.
Answer: More or less what others already said, but with a picture...
Yaw control/stabilization is a rotation around the yaw axis, created by a force acting perpendicularly to its arm (left below).

In your suggestion the vertical tip would create a left/right force (right above), applied with an arm which is parallel (the line between the surface and the center of gravity. This results in:

The force is not amplified (arm parallel to the direction of the force).

The resulting moment tends to translate the aircraft left or right, it has no yaw action.

No mentioning the lack of crew control, the reduced lift of the wing, the strength of the unlocked tips, etc.
Comment: ah, I thought that might be the case. I was thinking that sometimes swept wings will have tips aft of the CoM, creating a stabilization effect if the tips fold up.
Comment: @Wyatt: You're right on the principle, this is like a twin tail, however an inefficient one, the arm is short and the area of the tip is small, limiting the torque that can be created. The wingtip area is also subject to a vortex, limiting lift.


Are there benchmarks available for Openbox vs Metacity resource usage?

Question: I'm reading a number of sites on making ubuntu faster/more lightweight (e.g. this guide, and many mention replacing Metacity with Openbox. However, in 12.04 on an Inspiron e1405, I only see Metacity taking 11MiB and (fairly consistently) 0% CPU. Am I missing something? Seems like I would be optimizing zero to replace something that uses that few resources
Answer: About metacity, I love it very much because of the Ambiance theme. But if compared to openbox, I have to admit that it is quite heavy. Openbox uses only 6.6MB of RAM with 0% CPU on my laptop. So, if you are going for speed, give openbox a try.


Young double slit experiment without first slit

Question: What if I take a laser pen and direct it towards the middle of two wide slits? Will I get interference pattern or just two lines? Do I need to scatter photons before they hit the slits to observe interference pattern?
Answer: Yes, you will see interference pattern, provided the slit widths and their separation are appropriate. Look here. If laser beam width is greater than the separation of the slits, the beam may split and might form interference pattern. It is to be noted that particles of appreciable size (coming in the path of laser) will also scatter light according to Tyndall effect.
Answer: The reason incoherent light is passed through a slit is to make it coherent . The light from a point source is forced to be coherent, the incoherent part is scattered off, or absorbed, at the opaque surface where the slits are formed.
Laser light is by construction coherent so no first slit is needed.
Scattering also often results in coherent waves and that is why we [may see interferences]  with street lights; even see them in light patterns in sunlight .1 This video at about 4 minutes shows interference patterns from street lights.


Gymnastics Lunge Position, Hand position

Question: When doing a Lunge position in gymnastics, which position is the hands? I see it sometimes a) pointed in as below, b) or other times straight ahead.
Answer: Its best to have your arms in the position described in a).
This is because a handstand is all about balancing on your hands.
Standing in the position of b) will mean that your chest and torso will be arched. The handstand will also be arched.
Most people's head will be sticking out, so the handstand will be arching.
Your body, if it isn't arched, will be dead straight, and when your hands hit the floor, your hands are far away from the front foot. If the hands are away from the foot, it is harder to balance and the kick up will need to have more energy.
If we were doing a), the body is more or less curved in a dish shape (in the picture the gymnast should have a more round body) and therefore the head will be in, making sure that the body has a less percentage of arching over.
Also, it is easier to control where your hands are placed because you can quickly stretch out if needed or close in if needed.


Difference in acquisition of active vs. passive language usage skills

Question: By passive and active, I mean understanding language when used vs. using it yourself, like in the terms active/passive vocabulary. My experience learning languages is that active skills are much harder to acquire than passive skills. I'm trying to explain this to someone who is learning a second language for the first time, who wants to focus on active language usage and under-emphasizing passive usage, but I think they are having difficulty because this seems backwards in terms of goals. Are there any good written explanations or academic work on the differences between active vs passive skill acquisition, or on different learning strategies for people who want to focus on one or the other?
Comment: There is no such thing as "passive language usage skill". One can know things in a language passively as opposed to actively.
Answer: In my opinion, the most convincing layman explanation of why passive language skills are usually superior to the active ones is that many people can read and understand Shakespeare, but very few can write like him.
If I am not mistaken, I have seen it in Linguistics of Teach yourself series, which I would strongly recommend to language learners trying to go a bit beyond the layman level.
Comment: Please, we don't need debates in comments (deleted), especially rude ones.
Comment: Reading is not a passive activity. What exactly is: passive language skills? In one's own language, one generally has passive **vocabulary**.
Comment: @Lambie By *passive* I mean understanding rather than production (*active*). Thus, reading and listening are passive, whereas writing and speaking are active skills. These four skills are the basis of CEFR evaluations (although sometimes they also add *understanding language structures* as the fifth).
Comment: No, reading and listening are **not passive**. Rebecca, I was not rude. I merely made a point.
Comment: I'm sorry but reading and listening are most definitely not considered passive activities in language learning.
Answer: There are four skills in language learning, and teaching!
In "order of appearance", they generally are: listening, speaking, reading and...writing. There are people who only learn grammar and how to read. Like certain scientists.
Last I looked at this, none of those is passive.
The CEFR says this:

The CEFR describes what learners can do across five language skills:
Spoken Interaction, Spoken Production, Listening, Reading and Writing.
For all five skills at each level, there are sets of detailed 'Can Do'
statements. By dividing Speaking in two, the CEFR focuses both on the
learner's production and their ability to take part in conversations
and discussion. So, for example, under Spoken Interaction there is
information about Turntaking: a Basic A2 learner Can use simple
techniques to start, maintain or end a short conversation, whereas a
Proficient C1 learner Can select a suitable phrase to preface their
remarks appropriately in order to get the floor, or to gain time and
keep the floor whilst thinking.

Cambridge.
Another site:
speaking, listening, reading and writing
Passive vocabulary, which is first about one's native language not an L2/

By Richard Nordquist Updated on August 21, 2017 A passive vocabulary
is made up of the words that an individual recognizes but rarely uses
when speaking and writing. Also known as recognition vocabulary.
Contrast with active vocabulary.
According to John Reynolds and Patricia Acres, "Your passive
vocabulary is likely to contain more words than the active one. One
way to improve the range of the vocabulary in your own writing is to
try to transfer words from your passive to the active vocabulary"
(Cambridge Checkpoint English Revision Guide, 2013).

passive vocabulary
Generally speaking, you have to have a very high level to have a passive vocabulary in an L2.
The only way to learn a new language is the hard way. You have to listen and repeat a lot. You have to memorize certain things. You have to work very hard. Just like a baby, the first thing you experience is listening, otherwise you don't know how to "say it".
Usually, listening and then speaking go together. For example, doing a situation exercise with a dialog such as:  "At the Office" or "At the Train Station". Those will be grouped together in a conversational way, the vocabulary and grammar you are apt to find at those places. **You can only talk about being at the office if you have heard it being talked about in a learning exercise. You can't speak ex nihilo.
Speaking for some people is harder than reading. This is an individual situation. But most people hear and then speak. And you can only speak after a lot of hearing! :)
Answer: IMHO (from personal learning experience), it's more complex:
An active vocabulary sufficient to express some complex ideas can still by way too limited to understand native speech.
In addition, native speakers of any language I encountered often speak quite fast, use colloquial expressions, sometimes even dialects, making the understanding task difficult.
So, in more than one language, I've experienced some stage where I was able to express my thoughts (even if sometimes clumsy), but had big troubles understanding the replies.
Answer: First of all, I agree with you on the fact that active skills are much harder to acquire than passive skills. Just think of a baby who is exposed to his/her parents' native language for the first time. It takes some time before he/she is able to create correct sentences, not to mention the complex ones. However, at some point they understand what people are telling them, even if they aren't able to speak well themselves.
Here it is another example that comes to my mind. If you watch lots of movies or TV series in a foreign language with subtitles on for a long time, at some point you will start to get accustomed to the sounds of the language in question (even without being able to understand what people are saying) and you could also start to recognize some recurring words! This is because you passively listened to people talk in that language for some time.
However, whereas passive language skills could be very useful, acquisition of active skills is on a different level. In fact, it requires an actual effort: the effort that is put into learning about the structure of a language without which speaking it becomes difficult, if not impossible. For example, it is not enough to hear how a word is pronounced by native speakers: you have to try to pronounce it many times so that you'll memorize it. In the same way, when you encounter new words or grammar structures, you have to try to understand and memorize them first, and then try to use them. This is what it means to make an effort into learning and acquire active skills.
Comment: There is passive knowledge of/in a language but there is no passive knowledge skill.
Answer: You could explain it this way:
There are degrees of knowing a word, phrase or pattern. These degrees go from recognition, to effortful understanding, to effortless understanding, to limited/inflexible/dependent production, to gradually broader, more flexible, more independent production. Language learning is in itself a developmental process & it is impossible to skip parts of that process. Analogy: Don't try to run before you can walk.
The consensus among SLA researchers is that learners need large amounts of comprehensible input & that learners need to engage in interpreting (making meaning from) it as the necessary but insufficient first stage of acquisition (I prefer the term language development to acquisition).
By interpreting (making meaning from) language, I mean understanding the content of the messages rather than the language code. In other words, the messages remain the same whether the language code is English, German, Russian, or Chinese. That's the meaning we're aiming for so that learners can make form-meaning connections, where form is the language code & meaning is the content of the messages.
BTW, "use", i.e. listening, speaking, reading & writing, in applied linguistics is never referred to as passive because interpreting (making meaning from) language is an effortful & active set of processes.
I hope this helps! :)


Install PHP Curl on Windows and Using it on Laravel 5.4

Question: I am using Laravel 5.4 on Windows 10. How to install PHP Curl on Windows and using it on Laravel?
Answer: Usually all you need is finding php.ini, find line with curl for example:
;extension=php_curl.dll

remove ; from the beginning of the line so it should look like this:
extension=php_curl.dll

Restart your webserver. Extension should be then available.
Be aware you might have multiple php.ini files (one for command line, one for web server), so make sure you are editing the correct file.
Answer: Use the following steps to install curl:

Open https://curl.haxx.se/dlwiz?type=bin in a browser.
Select your operating system in the dropdown box: either Windows /Win32 or Win 64. Click Select!
For Win 32, choose whether you will use curl in a Windows Command Prompt (Generic) or in a Cygwin terminal (cygwin).
For Win 64, choose whether you will use curl in a Windows Command Prompt (Generic) or MinGW (MinGW64). Click Select!
If required, choose your Windows operating system. Finish.
Click Download for the version which has SSL enabled or disabled
Open the downloaded zip file. Extract the files to an easy-to-find place, such as C:\Program Files.

Testing curl

Open up the Windows Command Prompt terminal. (From the Start menu, click Run, then type cmd.)
Set the path to include the directory where you put curl.exe. For example, if you put it in C:\Program Files\curl, then you would type the following command: set path=%path%;"c:\Program Files\curl"

NOTE: You can also directly copy the curl.exe file any existing path in your path

Type curl. You should see the following message: curl: try 'curl –help' or 'curl –message' for more information This means that curl is installed and the path is correct.
Comment: Ok. I already tried it and success in testing. Now, how I use it in laravel?


ViewsLifeTime is not updating instantly?

Question: I got a counter to show how many visitors visit the page in my site. Users are visiting the website , but there is no change on the counter. It is the same like no one clicked it. Has it got a timer job? and what is the default frequency?  Or how can change the value of counter instantly?
Answer: It is actually two timer jobs: A timer job to create the analytics and a timer job to import these analytics to the event store.
The first timer job is called "Analytics Timer Job for Search Service Application" and it is set to whatever schedule you set it to.
The second timer job is called "Microsoft SharePoint Foundation Usage Data Import" and on default runs in a 10 minute interval.
Once these two timer jobs execute your analytics events can be found within search. Bear in mind that there is also Tail Trimming, if there are less than 3 events, they are not tracked or visible.


About "Flow Network based Generative Models for Non-Iterative Diverse Candidate Generation"?

Question: I am reading the Yoshua Bengio et al, Flow Network based Generative Models for Non-Iterative Diverse Candidate Generation. After read this paper, I wonder how to establish a GFlowNet if I don't know all the States?
That is, I only have goals, such as maximum likelihood estimation or some generation models similar to MCMC. How can I initialize the GFlowNet nodes when I don't know all the states in advance?
Comment: This paper might provide you with some intuition: Generative Flow Networks for Discrete Probabilistic Modeling https://arxiv.org/pdf/2202.01361.pdf
Answer: As in deep reinforcement learning, one does not need to literally hold the state space in memory explicitly (i.e. we don't need to "know all the states"). Rather, what is used is a function approximator that maps a description of a state to a prediction. In DRL that would be for example $Q_\theta(s,a)$, or $\pi_\theta(a|s)$.
In this paper, the flow $F_\theta(s,a)$ is what is being approximated, with a deep neural network. All that is needed is thus a description of the current state $s$ that can be given to a neural network (e.g. an image). In the paper, a graph is used as an input to a graph neural network. These graphs are not to be confused with the state space DAG, which in this work is a graph of (molecular) graphs, where an edge represents a graph edit.
Disclaimer, I am the first author of this paper.

About "Flow Network based Generative Models for Non-Iterative Diverse Candidate Generation"？


Possible scenarios for Unix domain sockets interruption

Question: When using TCP & UDP sockets, there are many scenarios which cause connection interruption (slow connection, network reset, etc).
Is there any possible situation which an unix domain socket automatically disconnects or interrupts because of an external reason?
Should developers consider possible interruption in their codes?
Comment: Without even considering the full question, a developer should _always_ consider and expect the unexpected, including interruptions. So the simple answer is yes, expect to handle such a case, even if it's not _supposed_ to be possible. Errors and bugs happen. And finding them is much harder when you do not write your code from the start to be prepared for them to happen.
Comment: Murphy's Law applies to programming. Anything that can fail, will fail, usually at the least convenient time.
Comment: @ArkadiuszDrabczyk and where that possibility comes from ?
Comment: Thanks for your suggestion guys, but my question is more than just "consideration".
Comment: It's always possible that server will terminate unexpectedly, no
matter what type of socket or IPC mechanism is used.
Answer: It's always possible that server will terminate unexpectedly, no
matter what type of socket or IPC mechanism is used. It can happen for
many different reasons, for example:

it may crash because of software bug, for because due to
segmentation fault or failed assertion

it may eat too much memory and can get killed by Linux OOM killer
(it's very easily reproducible in virtual machine)

it may be killed erroneously by another user. Linux is multi-user
system and it's possible that someone could accidentally killed your
process, say they wanted to type kill 1112 but typed kill 1111
instead

Should developers consider possible interruption in their codes?

Yes, they always should. Read manpage of the function you're going to
use and read all possible values of errnos that the function can set
and always prepare for the worst.


Blender 3d view port rotation issue

Question: I have a model under construction - this is ok but I've done something to the way it relates to other objects in the 3d view.
With just the model in the view all appears to be fine when I rotate the view in any direction.
If I add two new objects to the scene and rotate the view, the new objects move as expected in relation to each other but the model always appears to be in front of the other two objects - it never passes behind the other two objects.
If I duplicate the model, the two models behave as you would expect but both are always in front of the others in the scene.
I think I've set some parameter on the model but I can't see it!
Comment: It sounds like you've turned on x-ray for your object.
Comment: Welcome to the site.This sounds like a pivot point issue. I'd suggest doing a search of Blender Stack Exchange using the search parameter "Pivot point", and review the questons and answers shown by the search.
Comment: I spent hours yesterday playing with the pivot point - wasn't that. But it was x-ray!!!! thanks
Answer: It sounds like you've turned on X-Ray display mode for your object.  You can toggle X-Ray mode under the Object properties tab > Display Rollout.

X-Ray display draws the object in front of all other objects (which don't have X-Ray enabled).  It is most often useful for armatures to keep them from getting lost inside the meshes they control.

Blender 3d view port rotation issue


How can i monitor ARP packet on WiFi network?

Question: I want to monitor packet in my network.  It said that ARP packet request broadcasted all over network.  But why i only see my arp request packet and reply packet to the gateway?
Comment: What are you using to monitor?  You need to configure your wireless device to run in "monitor " mode. It's equivalent to promiscuous mode for an Ethernet sniffer.
Comment: Did any answer help you? If so, you should accept the answer so that the question doesn't keep popping up forever, looking for an answer. Alternatively, you can provide and accept your own answer.
Answer: I don't know how do you monitor ARP packet but I think what you see is correct. Here are the reasons:
First, you ARP request packet will be broadcast to LAN by switches. Your computer only needs to send one ARP request packet and the broadcast is done by switches. So the monitor shows your ARP request packet as an indication for this step.
Second, the gateway router sent the ARP reply packet to you (unicast because the switch is self-learning) by the switches. So the monitor only shows ARP reply packet from the gateway router as the indication for this step.
Answer: You can add a TAP between your access point and the switch. Of course you can set a monitor session on all Wi-Fi interfaces. I think it's the simple way to monitor packets transferred by your network. You will be able to see ARP because you AP by default just pass through all packets except beacons.
Comment: There are quite a few frames besides beacons that a AP will not forward.