Amplitude of a single mode quantized light field in a cavity

Question:
When a single mode light field in a cavity is quantized, how is the amplitude of the field obtained? Is the dependence of the amplitude on parameters like cavity volume and frequency of radiation obtained from purely dimensional considerations?
On a broader note, where does the physical input in dimensional analysis come from? In other words, how do we know that the dependence of a quantity on various parameters obtained from purely dimensional considerations is physically true?
Answer: I assume that you refer to a form of the field given by:
$$ E(z,t) = \left(\frac{2\omega^2}{V\epsilon_0}\right)^{1/2} q(t)\sin (kz).$$
The solution from Maxwell's eqs. is
$$E(z,t) = A q(t)\sin (kz),$$
where $q$ satisfies the eq.
$$ \frac{d^2q}{dt^2} + c²k^2q = 0.$$
Now, you must be able to get the magnetic field $B$, and use the fact that the total energy of the field is
$$ H = \frac{1}{2}\int dV\left(\epsilon_0 E^2 + \frac{1}{\mu_0}B^2\right) $$
If the integration is performed on a box of sides' length $L_x, L_y, L_z$, then, if you choose $A$ as shown in the first equation of this pots, you end up with
$$H = \frac{1}{2}(\dot{q}^2 + \omega^2q^2),$$
which resembles the energy of a harmonic oscillator of unit mass. The amplitude $A$ is picked like that because of this reason I believe.
Answer: I think a great answer to your question can be found in this lecture from 16:49 to 19:45. To reproduce Dr. Lukin's derivation, we want the amplitude associated with a single cavity photon to be equal to the energy of that photon, which we know to be $\hbar \omega$. The time-averaged energy $dU$ in a differential volume $dV$ due to an electromagnetic wave with amplitude $E_0$ is given by
$$dU = \epsilon_0 E_0^2 dV$$
Where the usual factor of $1/2$ that comes from taking the time average of the electric field is cancelled by the (identical) energy contribution of the magnetic field. So the energy of a photon with amplitude $E_0$ in a cavity with effective volume $V$ is
$$\epsilon_0 E_0^2 V = \hbar \omega$$
which gives for the amplitude associated with single photon
$$E_0 = \sqrt{\frac{\hbar\omega}{\epsilon_0 V}}$$
I suppose as a result you would say that if you have $N$ photons in a cavity mode the field amplitude is $E_0N^{1/2}$ to give a final energy of $N\hbar\omega$.
So although this is not a rigorous quantum field theoretic derivation, I think the thought process is a little more careful than pure dimensional analysis.
Regarding your second question, dimensional analysis provides a sense of the natural orders of magnitude for quantities involved in a problem, I don't think anyone would say it does more than that.


SPOnline - Follow a site programmatically on behalf of another user

Question: I have a site in SharePoint Online that I would like all users in my organization to follow.  I have been trying to use CSOM, Powershell and Rest to achieve that with no success - I can only do that for the current user.
Ideally, I would like the process to run every night and set all new users and users who "un-followed" my site to follow it again. Is it possible?
Answer: It's not possible as per this documentation:
Common programming tasks for following content in SharePoint
To quote:

Create an instance of a manager object in the context of a specified
  user

CSOM: not implemented
JSOM: not implemented
REST: not implemented
SSOM:  SPSocialFollowingManager (overloaded)

You can only follow for the current user using client-side calls. Unfortunately only SSOM allows managing following for other users.
(With Groups and the Microsoft Graph you are also out of luck as the docs for addFavorite explicitly states it's for the current user only.)


Can the permutation group of a matrix be generated by the subset of row permutation and column permutation?

Question: Consider a 3*3 matrix
$\left( \begin{array}{*{20}{c}}
{{m_{11}}}&{{m_{12}}}&{{m_{13}}}\\
{{m_{21}}}&{{m_{22}}}&{{m_{23}}}\\
{{m_{31}}}&{{m_{32}}}&{{m_{33}}}
\end{array} \right)$ and its permutation group, which consists of all the possible permutation over its element such as $\left( {{m_{22}},{m_{33}}} \right) \cdot \left( {{m_{11}},{m_{21}},{m_{32}}} \right)$. I wonder whether all these permutations can be expressed as a composition of several row permutation and column permutation? If so, does it still hold when we consider a non-square matrix?
Comment: If I understand your question correctly, the answer is no.  As you have defined it, there are $9!$ elements in your permutation group, but there are only $6$ permutation matrices.
Answer: Of course not, for the following reason. When applying either a row or a column permutation, then the set of products
$$m_{1\sigma(1)}\cdots m_{n\sigma(n)},\qquad\sigma\in\frak{S}_n$$
is unchanged. On the contrary, if you apply another permutation of the entries, this becomes false. For instance, if you apply the transposition $(m_{22},m_{33})$, then the product $m_{12}m_{21}m_{33}$ becomes $m_{12}m_{21}m_{22}$.


What should I do if I received a mail from company about offer letter by post but a week passed out I haven't received yet?

Question: The company has written in mail that I will be receiving their offer letter within 2-3 days by post but no letter received yet and my onboarding is scheduled on 25 August.
Answer: Pick up the phone and ask them what's happening.
It is entirely appropriate to directly call the HR department, preferably whomever you have previously been in contact with there. This is part of their job.
Comment: It is entirely appropriate to directly call the HR department, preferably whomever you have previously been in contact with there. Thus is part of their job.
Comment: without an explanation, this answer may become useless in case if someone else posts an opposite opinion. For example, if someone posts a claim like _"Don't pick up the phone and avoid asking them what's happening._, how would this answer help reader to pick of two opposing opinions? Consider [edit]ing it into a better shape, to meet [answer] guidelines
Comment: Big company or no, the HR department *is* responsible for sending out offer letters and other employment-related documents. (The hiring manager is not.) An alternative is to call the hiring manager and to ask that *s/he* call HR. You need that letter in your hand if "onboarding" is a week hence. (They might decide not to trust the mail and to have the letter ready. Maybe they can e-mail you a PDF ...)
Comment: As it is a big it company I don't think it is right to direct call the hr


How many Character Points for a Psionicist to buy proficiency with Restricted Weapons like Swords?

Question: In skills & powers for AD&D 2nd edition there's a rule that says

"Character Class Restrictions Non-warrior characters are normally
restricted by the rules dictating the types of weapons they can
acquire proficiencies for. However, by spending extra character points
for a weapon proficiency slot, a character can purchase a proficiency
that he would otherwise not be able to possess. A rogue or a priest
can acquire a proficiency for a weapon that is normally restricted to
a warrior's use. One additional character point must be spent when the
slot is purchased—the slot will cost 4 points, instead of the 3 rogues
and priests usually pay. A wizard can spend 2 additional character
points to become proficient in a weapon normally allowed to a priest
or a rogue. If the wizard wishes to become proficient with a weapon
that is normally limited to fighters, however, he must spend 3 extra
character points."

And a Table that lists penalties for non proficient and familiar:

Warrior –2 –1
Wizard –5 –3
Priest –3 –2
Rogue –3 –2
Psionicist –4 –2
Nonclassed NPC –4 –2

As a DM I am wondering what is a good Cost for Psionicists to use Class Restricted Weapons, as their penalties do not line up either with the Rogue/Priest (5 points total penalty) or the Wizard (8 points total penalty).
The cost theoretically appears to be 3 points if we think of them as Wizards, but they use d6/+2 for hit dice, and use Rogue THAC0, so should we use the 2 point penalty instead?
Additional notes: There are psionicist kits like Shinobi and Janissary that use restricted weapons.
Trivia: the 1st edition Psionicist in Dragon 78 was allowed "swords" but not Two-handed swords.
Answer: Aha! it took a while but i found the answer in Skills & Powers in the Psionicist Section
the cost is 4 CP "Proficiency in a single weapon from the warrior list"


wallet will not sync it has been a week and I still do not have my money in the wallet. It took out my account

Question: I have video proof that this has been going on for a week I have restarted the computer multiple times and have yet to receive any money into my account. I do not even have a wallet number.
Comment: Hello, welcome to Bitcoin.SE! In order for someone to help, you'll need to provide more info, such as: what software are you using? What sort of errors are you receiving from the software? What do you expect to happen, and what actually happened? Where did you buy bitcoin from? Do you have a transaction ID for a transaction that was sent to you? Etc
Comment: I am using a Windows laptop, using Bitcoin Core Wallet. When i went to the bitcoin website to buy bitcoin it said i needed a wallet, therefore I downloaded BCW. It has now been six whole days and I have yet to see the deposit I made to the account to receive money. I do have a transaction ID from blockchain and it shows that all was approved, now I am just not seeing it in the wallet. As of now I want to know why for six days the BCW app on my laptop has shown that it is syncing for six days and has yet to give my money?
Answer: You are using wallet-software that requires you to download the whole Bitcoin ledger before it can read your balance. The Bitcoin ledger (the blockchain) is currently about 210 gigabtyes, so you would need a very fast internet connection to download it in just six days. If you don't have 210 gigabytes of spare storage, your laptop's disk will fill up before it gets to the part of the blockchain with your transaction and you'll not be able to see it.
You may therefore want to consider importing your address into some other wallet-software that's able to read a remote copy of this enormous ledger. Such wallets are very popular, but ultra security-conscious people do like to have their own copies of the blockchain, and only you can decide how much security you need.
In the meantime, any blockexplorer web site will tell you whether or not your coins have been sent to your address. If they're at your address, you have nothing to worry about other than the minor inconvenience of either waiting for the blockchain to download or importing your address into some other wallet-software before you can spend your coins.
It will make more sense if you remind yourself that bitcoins are not stored in wallets. Wallets are just tools for drawing up & signing transactions (and they also let you view your balance). The coins themselves exist only on the blockchain.
EDIT
Putting together the information in your two comments, it sounds like you've already transferred your coins to an address controlled by you (check your address on a blockexplorer web site). If so, your predicament is like having a faulty debit card (i.e. the money's in your account but you can't see it). I suggest you study how to export the private key for the address that you used. Once you've exported it, you can import it into a lightweight wallet that doesn't insist on downloading the whole blockchain. Personally I would use Electrum, but other lightweight wallet software is available. Electrum has a create new wallet wizard, with an option to "Import Bitcoin addresses or private keys", where you should be able to paste your exported private key.
Private keys are valuable to you and to anyone who hacks into your computer, so you want to proceed carefully. Consider disconnecting your computer from the internet during the export/import process. You don't need the internet for key management chores.
Comment: Thanks so much for this advice. How exactly do I transfer the funds from blockchain to another account say for example, "Upaycard"?
Comment: I've added to the answer above. Obviously, if you can't see your coins in a block explorer website, my edits don't help much, but your earlier comment says you have a transaction ID, so I'm assuming the transaction (you can look *it* up too, on a block exporer!) shows your coins moving to *your* address.


What is the wizard proficiency for the Spell Save DC

Question: What is the wizard proficiency for the Spell Save DC, is it the armour proficiency or what.
Comment: Welcome to the site! When you get a chance, please take the [tour]. I think it would make the question more clear if you were more specific about which passage of the PHB was unclear to you.
Answer: Each class gives one stat called proficiency bonus. It starts at 2 and increases from there. It is added to a lot of things like attack rolls, skills you are proficient with and your spell save DC*.
For example see the wizard class here.
The spell save DC is 8+ proficiency bonus +casting stat modifier (int for the wizard)
*Among other things
Comment: Thanks! that was extremely helpful, now I can finish my character.


In the Bernstein-Vazirani algorithm, what is the use of the second Hadamard gate?

Question: In the Bernstein-Vazirani algorithm, what is the use of the second Hadamard gate? What happens if I remove it? Would the algorithm works fine? I read something about it closing the interference.
Answer: After application of the oracle but before the application of the second Hadamard gates, the register is in the state:
$$\frac{1}{\sqrt{2^{n}}}\sum_{x=0}^{2^n-1} (-1)^{f(x)} |x\rangle.$$
Although the phase $(-1)^{f(x)}$ is correct, the the probability (the squared amplitude) is uniformly distributed over each basis state.  Thus without the second Hadamard gates you would not learning anything about $f(x)$, and your algorithm would not work.
The second Hadamard gates do indeed "close the interference".
Comment: what does close the interference mean?
Comment: I copied that from your question; I'd not heard the term before but it seems like a nice way to say that the second Hadamard gates allow for constructive and destructive interference of the phases.


What's the goal of Adam and Claudia?

Question: What's the goal of each Adam and Claudia?
It seems to me that Adam would like to end time which sounds like he would like to break the cycle while Claudia would like to keep it?
Answer: Adam wants to destroy the knot. Claudia wants to save her daughter, Regina.
Adam wants to destroy the knot, wiping out both Adam's and Eva's worlds. He plans to do this by killing the pregnant Martha from Eva's world and her unborn child (The Origin / The Unknown) by using the energy of the apocalypses of both worlds on her.
Adam [talking to Eva's world Martha in S03E06]:

What's growing inside you, the origin, must die.
But it can't be killed by normal means.
It's born of both worlds, and the energy of both worlds is necessary to destroy it.
The apocalypse in my world, and the apocalypse in your world.
This here is the end.
A machine that not only crosses time, but also space.
It focuses the energy of both apocalypses onto this one point.
Your son only exists because the matter exists.
And now he will die through the matter, and all of us with him.
None of this... will exist anymore.
Both worlds will extinguish one another. Absolute annihilation.

Adam's plan, however, doesn't work.
Claudia's goal, on the other hand, is to save her daughter, Regina. She learned about the original world and that it is the only world where Regina doesn't get cancer.
Claudia [talking to Eva's world Claudia in S03E07]:

I still remember exactly what she said.
"If everything goes right... Regina will live."
I've thought about that all these years.
I just can't believe that by that she meant her suffering was to repeat over and over forever.
There must be a way to untie the knot without destroying all life in it.
A way for Regina to live.
Really live.
I think neither Eva nor Adam know that path.
But I'll find it.
In my world or yours.

Claudia [talking to Adam in S03E08]:

I've spent 33 years looking for answers in your world and in hers.
I've tried to put together the pieces of the puzzle.
To understand how everything can be reborn from the same family tree over and over again.
Until I realized that we're not all part of the knot.
Both worlds are a cancer that must have grown from something else.
If you remove it, you destroy all that was born of it, but you keep everything alive that already existed in the origin world.

Claudia [talking to Tronte about Regina in S03E08]:

She isn't part of the knot.
She will live.

Claudia [talking to Adam in S03E08]:

Everything I did, I did for Regina.
All these years, I've been searching in both worlds for a way for her to live.
A way to break the chain of cause and effect.
Both in your world and in Eva's world, I've tried to understand how everything is connected.
Really connected.
Until I finally understood that our two worlds were borne of another.
That Regina will always die in both worlds.
That she can only live in the world from which everything was created; the origin world.

Claudia [talking to her younger self in S03E08]:

If everything goes right... then Regina will live.

Quoting the Dark official website's articles on Jonas Kahnwald / Adam and Claudia Tiedemann (emphasis mine):

SEASON 3 EPISODE 7 - A NEW PLAN
Progress with the time travel portal is slowly moving forward. The setbacks start to take their toll physically and Jonas is forced to accept one after another. He is no longer worried about preventing the apocalypse - he wants to destroy the knot and wipe out the world. Jonas has become Adam.

SEASON 3 EPISODE 7 - THE END OF THE ORIGIN
Adam holds Martha from Eva's world captive. He knows she is pregnant with the origin of the time traveler family tree. He ties Martha up in the time travel portal in the former nuclear power plant and directs the forces of both apocalypses on her, killing her and her unborn child. Adam believes this will destroy the knot and end his own existence, but his plan doesn't work. He's still alive.

SEASON 3 EPISODE 8 - CLAUDIA'S EXPLANATION
Old Claudia arrives and explains to Adam that his attempt has failed due to quantum entanglement. Eva found a way to change the course of events, meaning there is a second pregnant Martha out there that will give birth and preserve the knot. She explains that there is another way to break the cycle: by changing things in a third world, the original world where everything began.

SEASON 2 EPISODE 2 A VISIT FROM HERSELF
In the summer of 1987, Claudia receives a visit from an older woman in her office The woman claims to be Gretchen's owner, which is true seeing as the woman is indeed an older version of Claudia. She was the one who brought Gretchen from 1953 to 1986. The woman shows Claudia the time machine and explains that she must stop a man called Adam if she wants to save her daughter Regina.

SEASON 3 EPISODE 7 THE SELF-MURDER
Claudia eventually realizes that neither Adam nor Eva have plans to save Regina so she takes matters into her own hands. When she meets Claudia from Eva's world, she shoots her and takes her golden time travel sphere. From then on, she pretends to be Claudia from Eva's world. Eva gives her the blueprints for the time machine that the clockmaker will later build.

SEASON 3 EPISODE 8 - THE ORIGINAL WORLD
Claudia tells Adam about the existence of a third world - the original world - which has been broken into two parts due to an unfortunate event. She tells him that it's possible to undo the event, ensuring Adam and Eva's worlds would never exist. Claudia is sharing this information in her own interest, because the original world is the only one in which Regina doesn't get cancer.
Answer: Adam wants to destroy the knot. He figures that the only way to do it is by killing the pregnant Martha using the energies of both worlds. However, for this, Adam needs everything to proceed in an untouched manner so that he can kill Martha at the right time, once she has conceived. But when he does kill Martha, nothing happens to the knot because an alternate pregnant Martha is alive and well thanks to Quantum Entanglement.
Claudia wants to end the knot too. But given the cancerous bootstrapped nature of both worlds, she figures there must be a third normal world which gave birth to the two knotted worlds. She knows she cannot convince Adam because he believes his plan is gold. So she waits for him to fail. Once he does, and is now prepared to listen, Claudia tells him that he needs to send Jonas to get Martha and go to the Origin World to destroy it. Adam agrees.
Adam's World actually has two Claudias and that is discussed here - How did Claudia come back from the dead?
Answer: Dark is a very complicated show and hard to understand, but as I can recollect, what Adam wants is for the cycle to remain the same and Claudia is opposite of him and wants the next cycle to finally break.
Comment: You're right about it being complicated and Adam/Claudia's intentions not being clear, but there are two quotes of Adam where he outright states he believes this is the last cycle (s02e06) and that the knot can be undone (s02e08). He may be lying, but it currently doesn't particularly seem like Adam wants it to continue exactly as it is either. Based on some hints, we (the viewer) may have projected "good" Claudia vs "bad" Adam more than is actually established.
Answer: The question is currently unanswerable as the show has been sufficiently vague about what's going on, with additional hints of ulterior motives and things not seeming what they are; enough so that you simply cannot tell right now.
Adam has been talkative but we cannot know if he's been truthful. He has repeatedly stated that he wants to break the time loop, and that he believes the "next" loop (i.e. the one after the events we witness) will be the last one. He's been mostly painted as the villain - but there have been subtle suggestions that Claudia's position is not as do-gooder as the show would have us believe - Adam mentioned this to young Jonas in 1921.
Claudia seems to be driven almost solely by wanting to counter Adam's plans. That doesn't prove she's the good guy in this conflict though. She's adamant and not much is known about her plans. Whatever Adam's true intentions are, it seems reasonable that Claudia is aware of them and wants the opposite. But even that isn't fully clear - maybe Claudia doesn't know Adam's true intentions and they may want the same thing but not realize it (I don't think so, personally, but it's not been disproven yet by the show).

What's the goal of Adam and Claudia?


When can we say a transaction is committed in mysql?

Question: I know the transaction can ensure that all the queries are committed or rolled back, but what is the point at which we can say ok, the transaction has been committed and all changes are durable?
I think it is when commit flag is written into redo log (take mysql as an example), is it right?
Answer: For all intense and purposes, the transaction is commit - then commit statement is complete. At that point all changes are durable and will survive reboot even if the data is technically still in "dirty pages" and not in the table's pages.
When dirty pages become white? That depends on the RDBMS and each one has a lengthy discussion in documentation about this. For MySQL the rules for this are described here: https://dev.mysql.com/doc/refman/8.0/en/innodb-buffer-pool-flushing.html
Comment: Good answer, but it applies to a single (non-replicated) server.  Old replication is async.  Semi-sync is well,...  Galera involves the gcache.  Etc.  In those situations, then "maybe".
Comment: But there is another problem: when can we say commit statement is complete from mysql server's view? Notice: I know it is hard from client's view, but i think it is a determinate action for server, after executing a certain c++ statement for example.


Citrix XenApp - Understanding

Question: Can you please help me with my understanding of xen app. Here is my understanding of citrix xenapp - When any virtual application ( such as word ) is started from xen app server, it basically creates a tunnel between the user machine and the xen app server. The xen app server, creates a session for the user and starts streaming the application or hosts the application inside that tunnel / terminal . The user is basically logged in terminally to that session. When the terminal session is establised, it operates through wfc32.exe ? Where does the ICA.exe come into picture in this case. What is WFICA.EXE ? Where is the PNAgent.exe ( on the xenapp server ? )
Answer:
Some statements to help clarify things

Citrix (XenApp) is based on terminal services.  The user is logged onto a terminal services session (think RDP).  This is displayed in steps [5-7].
The web interface is exactly that, something that is presented in a web browser.  You may be aware the application shortcut can also be in other locations.  These are populated / managed by the PNAgent.exe

The user's start menu
The user's system tray

The .ica file, is simply a text file with an authentication ticket to allow the user to connect to the target terminal server.  ICA is a separate protocol to RDP, and on your client wfica.exe interprets those ICA files and establishes the connection and everything else that's hapenning client side for that application (screen renderings, mouse events, keyboard events, shortcuts, etc...)
Comment: Great.. Thanks alot for the information provided. Couple of questions what about wcfrun32.exe and pnagent.exe ?
Comment: okay got it , pnagent.exe is part of Web interface. What about wcfrun32.exe , I dont see wfICA.exe in my machine. ( I see wcfrun32.exe ), I assume this must be new version then ?

Citrix XenApp - Understanding


Transferring Pictures and phone numbers

Question: How do I remove pictures and emails from my Nokia android phone. And how do I copy my address book to my new Samsung Galaxy S5 phone?
Answer: To transfer any file/picture/etc.. between your phone and PC, you'll need to turn Developer Options on and enable USB Debugging.
Some Android devices already have Developer Options listed in the Settings menu. Most, however, need a certain "key combo" to bring it out of hiding
1) On your phone, navigate to Settings 》About Device (Phone) 》and tap your device's Build Number 7 consecutive times. A caption will display at the bottom of your screen, stating how many taps are still needed & when you're considered a "Developer".
2) Once Developer Options is turned on, navigate back to Settings 》Developer Options 》and enable USB Debugging.
Developer Options is located at the bottom of your Settings, above "About Phone"
3) After enabling USB Debugging, plug your device to your PC via USB cable.
4) Swipe down the notification bar on your phone 》tap USB Options  》and verify your device is set at MTP. If it isn't, change it to MTP.
WALA! You are now able to swap files, pictures, etc.. from your phone to your PC and vice versa.
Ensure proper connection between your phone and PC by clicking your PC's Start Menu, then navigating to the "Devices & Printers" section. Your phone will register as either an MTP Device or your phone's model number if properly connected. If not, it'll say a name you won't recognize and have a yellow triangle with an exclamation point (!) in the middle.

Transferring Pictures and phone numbers


Intel HDMI-1 disconnected Xrandr

Question: I did a fresh install of Lubuntu 20.04, I had to use "graphic safe mode".
My monitor is showing the boot screen / BIOS.
If I add nomodeset to GRUB, it works with 800x600 resolution. It's a touchscreen with built in PC from RC Stars.
I installed TeamViewer when using "nomodeset". The monitor is completely black but I can reach the computer through TeamViewer.
Answers for this questions I've found after a day of googling doesn't help and the answers is quite similar.
sudo lspci -v
00:02.0 VGA compatible controller: Intel Corporation HD Graphics 5500 (rev 09) (prog-if 00 [VGA controller])
    DeviceName:  Onboard IGD
    Subsystem: Intel Corporation HD Graphics 5500
    Flags: bus master, fast devsel, latency 0, IRQ 45
    Memory at f6000000 (64-bit, non-prefetchable) [size=16M]
    Memory at e0000000 (64-bit, prefetchable) [size=256M]
    I/O ports at f000 [size=64]
    Expansion ROM at 000c0000 [virtual] [disabled] [size=128K]
    Capabilities: [90] MSI: Enable+ Count=1/1 Maskable- 64bit-
    Capabilities: [d0] Power Management version 2
    Capabilities: [a4] PCI Advanced Features
    Kernel driver in use: i915
    Kernel modules: i915

xrandr
Screen 0: minimum 8 x 8, current 1024 x 768, maximum 32767 x 32767
eDP1 connected primary 1024x768+0+0 (normal left inverted right x axis y axis) 0mm x 0mm
   1024x768      60.00*+
   1024x576      59.90    59.82
   960x540       59.63    59.82
   800x600       60.32    56.25
   864x486       59.92    59.57
   640x480       59.94
   720x405       59.51    58.99
   640x360       59.84    59.32
   512x384       60.00
DP1 disconnected (normal left inverted right x axis y axis)
HDMI1 disconnected (normal left inverted right x axis y axis)

xrandr --output HDMI1 --auto --right-of eDP1
Doesn't do anything

uname -a
5.8.0-45-generic #51~20.04.1-Ubuntu SMP Tue Feb 23 13:46:31 UTC 2021 x86_64 x86_64 x86_64 GNU/Linux

sudo xrandr --addmode HDMI1 1920x1080
xrandr: cannot find mode "1920x1080"

How do I troubleshoot to find out the problem? Windows installation works, but none of the Ubuntu distros/versions I've tried has worked
Answer: Have you tried updating your pc?
press alt+ctrl+F3 and login do sudo apt update && sudo apt upgrade
Also have you downloaded the drivers if not that might also be the problem.
Comment: I have upgraded. I removed and reinstalled xserver-xorg-video-intel

Intel HDMI-1 disconnected Xrandr


What Pagan elements were adopted in christianity?

Question: I have commonly come across the belief that "newer" religions integrate existing customs and traditions into their own. So what traditions and customs in Christianity can be traced back (with the help of proper references) to Paganism?
Comment: Christianity is the result of the coming into the world of Jesus Christ. It is not a 'reinterpretation' of previous speculations. This question is not clear as to whether it is seeking to know what _corruptions_ entered into the true faith in the early centuries of the Christian Era or whether the question is seeking to deny the essential truth of the coming of Jesus Christ.
Comment: Possibly celebrating birthdays
Comment: At its core, Christianity is a major (re)interpretative overhaul of ancient Judaism, which itself represents a major (re)interpretative overhaul of [ancient Canaanite paganism](http://en.wikipedia.org/wiki/Ancient_Canaanite_religion); as the new religion later spread to non-Jews, it sought to apply the same (re)interpretative mechanism to these newly encountered faiths as well; furthermore, Judaism itself already shared many concepts or ideas in common with non-Jewish belief systems, being itself a descendant of pre-Abrahamic paganism.
Comment: Christmas, for one thing. And Easter and Halloween, at least in that both are celebrated by Christians although they originated in pagan religions. Other than that, elements such as concepts and titles (like that "Pontifex Maximus") have been adopted from pagans.
Comment: There is also no evidence in the Bible that God's servants celebrated birthdays but it does mention an Egyptian pharaoh (Genesis ch. 40) and a Roman district ruler (Matthew ch. 14) as doing so. In both cases someone was also killed on that day.
Comment: @Lucian - Any evidence  for your statements ?
Comment: @dezkev: Start with the New Testament, and, whenever a prophecy is invoked, go back to the original Old Testament passage, and read it in context; you will soon notice that they are mostly [typological](http://en.wikipedia.org/wiki/Typology_(theology)) (Matthew 2:15 vs. Hosea 11:1, Matthew 1:23 vs. Isaiah 7 & 8, etc.) or allegorical in nature (Galatians 4:22-31). Then go on to early Christian writings, and notice that a similar approach is employed for (re)interpreting pagan myths or beliefs along Christian lines; or simply read a scholarly work on the history of religions.
Comment: @Lucian - thanks for your response. I would start with the OT itself,  that is the basis of Judaism and Christianity. While many pagan beliefs were similar to the OT cultus (like animal sacrifices) they were given by Jehovah Himself and endowed with new meaning and significance.  Besides the OT cultus  predates  paganism- see Genesis Ch. 4  - Cain and Abel were bringing sacrifices to God!
Answer: What Pagan elements were adopted in christianity?
Many pagan elements were adopted in the Early Church. Some of elements that are claimed to be of Pagan origin are not true, while many are of a dubious nature and can not be clearly proven.

Pagan influences on Christianity
The early Christians adapted many elements of paganism.: Ancient pagan funeral rituals often remained within Christian culture as aspects of custom and community with very little alteration.: A type of song sung at death, the ritual lament, is one of the oldest of all art forms. As soon as death was immanent, the ritual began, then came the "struggle of the soul" and prayer for the dying. John Chrysostum gives a vivid account of the dying soul seeing angels and demons - "account books in hand" - struggling against each other in a contest for possession of the dying person's soul. Macarius of Egypt (fourth century) writes of such a contest which is only resolved by the intervention of the person's guardian angel - which is roughly parallel to Plato's daimon.
Pagans and Jews decorated their burial chambers, so Christians did as well, thereby creating the first Christian art in the catacombs beneath Rome. This art is symbolic, rising out of a reinterpretation of Jewish and pagan symbolism. Christian piety infused the symbols with its own fresh interpretation. Christian art had something fundamentally new to say as it gave visual expression to the conviction that the human soul can be delivered from death to an everlasting life. Neither Judaism nor any pagan religion had previously made such a claim. "The Jewish faith puts little emphasis on immortality, and pagan beliefs about the afterlife were vague, uncertain, and sometimes dismal".
While many new subjects appear for the first time in the Christian catacombs - i.e. the Good Shepherd, Baptism, and the Eucharistic meal - the Orant figures (women praying with upraised hands) probably came directly from pagan art.: Pagan symbolism in the form of Victories, cupids, and shepherd scenes are scattered throughout the catacombs. Jewish and pagan use of sheep and goats, birds in a tree or vine, or eating fruit, especially grapes, seven steps leading up to a tomb, a pair of peacocks, the Robe of sanctity, the reading of scrolls, are all found in pagan art and adapted in the Christian art to express the hope of immortality in Christian terms. Pagan sarcophagi had long carried shells, and portraits of the dead often had shells over the head of the dead, while some put a shell over a grave. Christians and Jews adapted the convention, identifying it with another symbol - the halo. For the Christians who made the catacombs, these symbols were necessary to convey their message.
Many previously pagan holy places were converted to Christian use. In 609 Pope Boniface IV obtained leave from the Byzantine Emperor Phocas to convert the Pantheon in Rome into a Christian church, a practice similar to that recommended eight years earlier by Pope Gregory I to Mellitus regarding Anglo-Saxon holy places, in order to ease the transition to Christianity. According to Willibald's Life of Saint Boniface, about 723, the missioner cut down the sacred Donar's Oak and used the lumber to build a church dedicated to St. Peter. Around 744, Saint Sturm established the monastery of Fulda on the ruins of a 6th-century Merovingian royal camp, destroyed 50 years earlier by the Saxons, at a ford on the Fulda River.

Some claim that Christmas a replacement of the feast of Hanukkah or of the Invincible Sun. But there is no evidence that Sol Invictus was observed anywhere outside of the borders of the Roman Empire. Earliest record of Sol Invictus being celebrated on the 25th of December is from 354 A.D. There is seems to be little evidence to suggest it was celebrated before 4th Century in Rome. But we know that the Syriac Christians were already celebrating Christmas (although it was considered a minor feast leading up to Epiphany) in the last week of their season of Subbara, which falls on December in the Gregorian Calendar. This practice dates from at least the mid-4th Century.

Festival of Dies Natalis Solis Invicti
The Philocalian calendar of AD 354, part VI, gives a festival of natalis invicti on 25 December. There is limited evidence that this festival was celebrated before the mid-4th century.
The same Philocalian calendar, part VIII, claims that the Lord Jesus Christ was born eight days before the calends of January (that is, on December 25).
Since the 12th century, there have been speculations that the near-solstice date of 25 December for Christmas was selected because it was the date of the festival of dies natalis solis invicti, but historians of late antiquity make no mention of this, and others speculate Aurelian chose December 25 to shadow early Christian celebrations already on the rise.

Saturnalia was not celebrated on December 25, so why all the fuss about it being link to Christ's birth.

The date of Saturnalia
One very popular misconception about Saturnalia is that it was celebrated on December 25th. This claim has been repeated a million times on the internet, but it is simply not true. In truth, Saturnalia was originally a one-day festival that took place on December 17th, but the holiday was so popular and beloved that, by the first century BCE, it had been extended into a seven-day long festival, which began on December 17th and continued until December 23rd.
The emperor Augustus (ruled 27 BCE – 14 CE) tried to shorten Saturnalia to only three days of celebration, lasting from December 17th to December 19th. Nonetheless, most people continued celebrating after the holiday officially ended, leading later emperors to restore some of the days of celebration that Augustus had removed. The emperor Caligula (ruled 37 – 41 CE) is reported to have extended the festival to five days, lasting from December 17th until December 21st.

Some claim that the title Queen of Heaven that Catholics employ towards Mary the Mother of Jesus is of pagan origins. This claim can not be historically proven.

Queen of Heaven was a title given to a number of ancient sky goddesses worshipped throughout the ancient Mediterranean and the ancient Near East. Goddesses known to have been referred to by the title include Inanna, Anat, Isis, Nut, Astarte, and possibly Asherah (by the prophet Jeremiah). In Greco-Roman times, Hera and Juno bore this title. Forms and content of worship varied. - Queen of Heaven (antiquity)

The term "queen of heaven" appears in a context unrelated to Mary. The prophet Jeremiah, writing c. 628 BC, refers to a "queen of heaven" in chapters 7 and 44 of the Book of Jeremiah when he scolds the people for having "sinned against the Lord" due to their idolatrous practices of burning incense, making cakes, and pouring out drink offerings to her. This title was probably given to Asherah, a Canaanite idol and goddess worshipped in ancient Israel and Judah; and not Mary the mother of Jesus.

Queen of Heaven is a title given to the Virgin Mary, by Christians mainly of the Catholic Church and, to a lesser extent, in Anglicanism, Lutheranism, and Eastern Orthodoxy. The title is a consequence of the First Council of Ephesus in the fifth century, in which Mary was proclaimed Theotokos in Greek, a title rendered in Latin as Deipara or Mater Dei, in English "Mother of God".
Biblical basis
In the Hebrew Bible, under some Davidic kings, the gebirah, the "Great Lady", usually the Mother of the King, held great power as advocate with the king. In 1 Kings 2:20, Solomon said to his Mother Bathsheba, seated on a throne at his right, "Make your request, Mother, for I will not refuse you." William G. Most sees here a sort of type of Mary.
A statue of the Assumption of Mary typically crowned with 12 stars. A reflection of the biblical image in Revelation 12. Statue by Attard, Malta.
In the New Testament, the title has several biblical sources. At the Annunciation, the archangel Gabriel announces that [Jesus] "... will be great, and will be called the Son of the Most High; and the Lord God will give to him the throne of his father David. He will rule over the house of Jacob forever and his reign will be without end."(Luke 1:32) The biblical precedent in ancient Israel is that the mother of the king becomes the queen mother. Mary's queenship is a share in Jesus' kingship.
The Catholic Church views Mary as the woman clothed with the sun in Revelation 12:1–3: "A great and wondrous sign appeared in heaven: a woman clothed with the sun, with the moon under her feet and a crown of twelve stars on her head. She was pregnant and cried out in pain as she was about to give birth. Then another sign appeared in heaven: an enormous red dragon with seven heads and ten horns and seven crowns on his heads." The Church accepts Revelation 12 as a reference to Mary, Israel, and the Church as a three-fold symbolism through the Book of Isaiah and affirms Mary as the mother of Jesus as the prophetic fulfilment described in Revelation 12 (cf. Isaiah 7:14, 26:17, 54:1, 66:7).
In the Hebrew Bible, the term "queen of heaven" appears in a context unrelated to Mary. The prophet Jeremiah, writing c. 628 BC, refers to a "queen of heaven" in chapters 7 and 44 of the Book of Jeremiah when he scolds the people for having "sinned against the Lord" due to their idolatrous practices of burning incense, making cakes, and pouring out drink offerings to her. This title was probably given to Asherah, a Canaanite idol and goddess worshipped in ancient Israel and Judah. - Queen of Heaven

Mysteries of Isis influences on Christianity are of a very dubious nature. The question of whether they influenced Christianity is controversial and the evidence is unclear.

Mysteries of Isis
Some aspects of the mysteries of Isis and of other mystery cults, particularly their connection with the afterlife, resemble important elements of Christianity. The question of whether they influenced Christianity is controversial and the evidence is unclear; some scholars today attribute the similarities to a shared cultural background rather than direct influence.
Possible influence on Christianity
The mysteries of Isis, like those of other gods, continued to be performed into the late fourth century CE. Toward the end of the century, Christian emperors increasingly restricted the practice of non-Christian religions. Mystery cults died out near the start of the fifth century. They existed alongside Christianity for centuries before their extinction, and some elements of their initiations resembled Christian beliefs and practices. As a result, the possibility has often been raised that Christianity was directly influenced by the mystery cults. Evidence about interactions between Christianity and the mystery cults is poor, making the question difficult to resolve.
Most religious traditions in the Greco-Roman world centered on a particular city or ethnic group and did not require personal devotion, only public ritual. In contrast, the cult of Isis, like Christianity and some other mystery cults, was made up of people who joined voluntarily, out of their personal commitment to a deity that many of them regarded as superior to all others. Furthermore, if Isiac initiates were thought to benefit in the afterlife from Osiris's death and resurrection, this belief would parallel the Christian belief that the death and resurrection of Jesus make salvation available to those who become Christians.
Some scholars have specifically compared baptism with the Isiac initiation described by Apuleius. Before the early fourth century CE, baptism was the culmination of a long process, in which the convert to Christianity fasted for the forty days of Lent before being immersed at Easter in a cistern or natural body of water. Thus, like the mysteries of Isis, early Christian baptism involved a days-long fast and a washing ritual. Both fasting and washing were common types of ritual purification found in the religions of the Mediterranean, and Christian baptism was specifically derived from the baptism of Jesus and Jewish immersion rituals. Therefore, according to Hugh Bowden, these similarities are likely to come from the shared religious background of Christianity and the Isis cult, not from the influence of one tradition upon the other.
Similarly, the sacred meals shared by the initiates of many mystery cults have been compared with the Christian rite of communion. For instance, the classicist R. E. Witt called the banquet that concluded the Isiac initiation "the pagan Eucharist of Isis and Sarapis". Feasts in which worshippers ate the food that had been sacrificed to a deity were a nearly universal practice in Mediterranean religions and do not prove a direct link between Christianity and the mysteries of Isis. The most distinctive trait of Christian communion—the belief that the god himself was the victim of the sacrifice—was not present in the cult, or in any other mystery cults.
Bowden doubts that afterlife beliefs were a very important aspect of mystery cults and therefore thinks their resemblance to Christianity was small. Jaime Alvar, in contrast, argues that the mysteries of Isis, along with those of Mithras and Cybele, did involve beliefs about salvation and the afterlife that resembled those in Christianity. But they did not become similar by borrowing directly from each other, only by adapting in similar ways to the Greco-Roman religious environment. He says: "Each cult found the materials it required in the common trough of current ideas. Each took what it needed and adapted these elements according to its overall drift and design."
Answer: I'm fairly sure this answer presents a view that will be very unlike the others, and will generate many objections.
So to save time, I'll simply give a brief summary of the idea.

Some answers will say that the early Roman Church adopted many pagan practices, giving them Christian names, in order to attract new converts (syncretism).
For instance, many statues and other depictions of Isis and Apollo were simply renamed to represent Mary and Jesus, and many local festivities were allowed to continue, but in the spirit of Christianity rather than their pagan origins.
But what really happened is the exact opposite.
The existing pagan church wasn't absorbed by Christianity.
Instead, the pagan church applied Christian nomenclature to its own practices and beliefs in order to suppress the true Christian religion, which was growing rapidly and seen as a threat.
Those that continued to follow the original biblical religion were branded as Judaizers and persecuted.
Eventually paganism, rebranded with biblical terms, became what the world now considers to be Christianity.
For 2000 years, true Christianity has survived only as small isolated groups.
Almost all of what is now called Christianity is pagan in origin.

Compare:

the fish-head hat worn by Dagon against the mitre.
statues of Isis against statues of Mary, including as a madonna and child.
images of Apollo against those of Jesus.
Mithras and sun-worship against Christmas.
Constantine's "In this sign you shall conquer" against a sundog phenomonon.
Representations of the sun against various forms of the cross.
Ancient fertility rites against Easter celebrations.

And what could be more blatant than the Egyptian obelisk that forms the centrepiece of St. Peter's Square at the Vatican.
A few years ago I wrote an illustrated summary of this: A Very Short History of Nominal-Christianity.
It is totally lacking in references and other scholarly attributes, but it might give you a few ideas that you can research yourself.
Comment: ***"Almost all of what is now called Christianity is pagan in origin."*** And I always thought it was Jesus Christ.
Comment: Your only source is most inaccurate. Just one simple point: It relies heavily on images as proof. At the Council of Nicaea, bishops did not wear mitres. That tradition came about some 500 years later to say the least (about the year 1000, even then the form of mitre was very different).


how do I save the extra files that come in a dmg?

Question: After I download a DMG to my Mac and install the apps that come in it, where should I keep the miscellaneous other files (like Readme and license terms) that come in the DMG? I want to keep them, but where, and how do I keep them associated with their app? Keeping the entire DMG just for its little misc files seems wasteful of storage. What is the normal practice?
Answer: Personally, I create a folder within the Applications folder with the same name as the app installed from the DMG and put the extra files in that folder. That way, they're always side by side and I can easily find them.
That, of course, is just personal preference. There is no strictly correct way of doing that.
Comment: It might if the app has some integrity checks on itself. Also, some upgrades completely replace the app package. You may also want to verify that the app doesn't already includes those files internally in the about or help menus.
Comment: Would it break things if I  put them in app's package contents itself?


Should I pronounce is ksenia or zenia?

Question: Xenia is from what I understand a directly borrowed word from Greek. It didn't come through french first or anything, so I see no reason to pronounce is 'z'enia. Why not pronounce as it would be pronounce in Greek: 'ks'enia
Comment: Because pretty much everyone in the US pronounces it "zeen-e-uh".
Comment: @HotLicks I've noticed this, that is why I ask
Comment: Well, that's how it's spelled, and when presented with an unfamiliar word most English speakers will attempt a phonetic pronunciation.
Comment: @HotLicks but how should it be pronounced in English?
Comment: What does the dictionary say?
Comment: 'z'enia, but the definition I'm looking for isn't there. Also,
Wikitionary has the name Xenia literally listed as being spelt Ksenia.
Comment: There are maybe 8 dictionaries online if you Google "define xenia", and none of them have the definition you're looking for??
Answer: We don't pronounce xenia with /ks/ because it would violate English phonotactics (the way sounds are put together in English words) and graphotactics (the partially regular relationship between sounds and spellings), and English speakers generally prioritize these above fidelity to Greek pronunciation.
English has its own sound system which does not allow words to start with /ks/. Imported words are generally changed to fit into this system. Furthermore, in this particular case, the Greek pronunciation is almost irrelevant, because there are only a few loanwords from Greek that English speakers got by listening to Greek speakers (e.g. gyro/gyros; even that often has spelling pronunciations). Most Greek loanwords in English, especially in technical fields, are derived from the written word (in particular, the Latin transcriptions of Greek words).
The Oxford English Dictionary says the word xenia, in the sense "A supposed direct action or influence of foreign pollen upon the seed or fruit which is pollinated," was taken into English from Latin xenia.
Most people pronounce rare Graeco-Latin words like this based on their spelling. The standard way to pronounce the letter "x" at the start of a word in English is /z/. The standard way to pronounce the letters "ia" at the end of a word is /iə/ or /ɪə/ (depending on the accent). The standard way to pronounce the letter "e" in the context eCia (where "C" stands for any single consonant letter that represents a single consonant sound) is /iː/. Put them together, you get /ziːniə/ or /ziːnɪə/. Most English speakers don't know what the pronunciation is in Greek; many of them don't care, and others (like me) don't see any reason why we should try to pronounce recent loan-words from Greek exactly like in the source language when there is a familiar pattern that has already been established by older Greek loan-words. The pronunciation of words already changes over time anyway; what's so bad about it changing in the transition between different languages?
Personal names are a different story. They are associated with individuals who are probably used to hearing a particular pronunciation. So English speakers often try to approximate foreign sounds in personal names. The extent to which people are capable of and willing to doing this varies, though.
Comment: Plus, it's the way they pronounce it in [Xenia, Ohio](https://en.wikipedia.org/wiki/Xenia,_Ohio).

Should I pronounce is ksenia or zenia?


I live with ten people at a time. How many total housemates should I have gone through to expect that two of my housemates would share a birthday?

Question: I live in a house with ten people. I've had about 50 people live here since I moved in. Nobody has ever shared the same birthday before, but a new housemate shares a birthday with a current housemate. What were the odds of that happening? How many people would have to cycle through a house of ten people for us to expect that two of the housemates share a birthday at some point?
Comment: Hint:  given $10$ people the probability that all the birthdays are different is $\frac {365\times \cdots \times 356}{365^{10}}\approx .8831$ under the usual sort of assumptions.
Comment: Note:  to get a precise answer I expect you'll need to clarify the problem.  I'm assuming (possibly incorrectly) that you essentially want to look at groups of $10$ and are wondering how many such groups you'll need to look at before the probability that you'll see a group containing matching birthdays is at least $95\%$, or some number like that.  But you should edit to say what, exactly, you had in mind.
Comment: Alternatively, if you imagine a system along the lines of "every X months one person leaves and a new person arrives"  then, given that the prior group had no match, then the probability that the next group will have  a match is $\frac 9{365}$.  But, really, you should clarify what you had in mind.
Comment: @lulu I can impose that person leaves the house and is replaced by a new individual every month. Does that clarify the problem?
Comment: Not really...the problem doesn't really refer to the time involved, just the number of replacements.  In my posted solution I made the simplifying assumption that only one person is replaced at a time.  Definitely changes things if you allow more complex replacements.
Answer: I don't see how this is different from the birthday paradox, see Wikipedia for instance. On a group pf 23 people, there are around 50% chances that  two of them share a birthday.
With the cycling of people the problem seems easier, ie. when a new roommate arrives, you can easily compute the probability of him/her sharing a birthday. But then again, if you live with 10 people and expect 13 people to arrive in the next years, you have 50% chances of that happening. If the house accepts no more than 10 people at a time, then you should see the closed formula of the birthday paradox.
Comment: I agree with you that the problem is ambiguous.  The way I read it, it differs from the standard Birthday Problem in that you are only concerned with the $10$ people living in the house at one time.  That is, it doesn't matter if the new guy matches someone who moved out a while back.  Of course, different interpretations of the problem are possible and I do not know which reading was intended.
Comment: The wrinkle is that people move out. I only live with ten people at any one time.
Answer: I don't think the problem is entirely clear, but perhaps it makes sense to model it this way:
We suppose that the house starts with $10$ people.  Then, at intervals, one person at a time leaves and is replaced.  How many such replacements do we expect to have before the probability that at one moment the house will contain a birthday match is at least $95\%$ (or whatever threshold you care to specify).
Of course we assume that each birthday from $1$ to $365$ is equally probable and that each birthday in your sample is independent of all the others.
To do it, let's compute the probability that we have seen no match after $N$ replacements.
The probability that there was no match in the starting group is $$\frac {365\times \cdots \times 356}{365^{10}}\approx 0.883051822$$
Given that, the probability that the first replacement does not create a match is $$\frac {365-9}{365}\approx 0.975342466$$
Thus the probability that we have seen no match after $N$ replacements is $$0.883051822\times (0.975342466)^N$$
We set this to $.05$, our confidence threshold, and take logs to solve.  With this choice of threshold we get $\boxed {N\approx 115}$ So you need a lot of cycling.
Note:  if you have had $50$ roommates in total, that means $N=40$.  With that choice, the probability that there will have been a match at some point is about $.67$, so it isn't particularly surprising.
Comment: I think this is a good model. Alternatively, we might "expect" two people to share a birthday once the probability passes, say, 50% (since it's at least more likely to have happened, than not). I believe we get the same $n = 23$ as the birthday problem, in this case...
Comment: @pjs36  Yes, I agree.


error "Cannot add TextRun in TextRun." when generating docx file in CiviCase

Question: I have customized CiviCase for a client who is using it to manage project grants. Things have gone pretty well overall. There are many custom fields and several related document templates in use. "Print/Merge Document" works fine when generating a pdf, but often fails when generating a docx file.
The error that is generated on screen is simply "Cannot add TextRun in TextRun.", but looking into the CiviCRM log shows a little more detail and that the error is in the 3rd-party phpoffice code (which I've never heard of before today).
<code>    $Fatal Error Details = array:3 [
  "message" => "Cannot add TextRun in TextRun."
  "code" => null
  "exception" => BadMethodCallException {#12247
    #message: "Cannot add TextRun in TextRun."
    #code: 0
    #file: "/var/www/sitename/wordpress/wp-content/plugins/civicrm/civicrm/vendor/phpoffice/phpword/src/PhpWord/Element/AbstractContainer.php"
    #line: 266
    ...
    /var/www/sitename/wordpress/wp-content/plugins/civicrm/civicrm/CRM/Utils/PDF/Document.php:90 {
       CRM_Utils_PDF_Document::html2doc($pages, $fileName, $format = [])
    ›   $section = $phpWord->addSection($pageStyle + ['breakType' => 'nextPage']);
    ›   \PhpOffice\PhpWord\Shared\Html::addHtml($section, $html);
    › }
</code>
The error only occurs when one of the custom fields of type "Rich Text Editor" is populated. All of the simpler fields seem fine. I'm not familiar with this code and so am not really sure whether this is a bug (and out of my power to fix) or some glitch in my custom fields that should be corrected.
Any thoughts?
UPDATE: Converting the rich text field type to multi-line text didn't fix the problem. Accented characters were converted to html entities and paragraph tags were added (or maybe kept) in the filed conversion. However, deleting the content and re-adding it as UTF-8 text with the accents into the multi-line field did prevent the error. I guess there is a bug in the phpoffice code with respect to html entities. Changing the field type is not a fix, but I may be able to live with it if there is no better solution.
Running CiviCRM 5.61.0 on WordPress 6.2.2 on Debian Bullseye
Comment: PDF generation is a bit of a dark art and it sounds like you've identified the source of the error as non-UTF-8 non-ascii characters. Your 'rich text field' in combination with wordpress would be responsible for how accented text gets encoded when submitting to CiviCRM, so that's where I'd look for a better solution.
Answer: If you're feeling brave - we're using version 0.18.3 (Feb. 2022 release) of PhpWord in Civi.  If you know how to use Composer, you could upgrade to PhpWord 1.1 (May 2023 release).
If you don't know Composer, you can also probably delete the phpword directory in <code>/vendor/phpoffice</code> and unzip the new one in there.
Hopefully this is obvious, but don't do either of these on a live site! Use a test site.


Get Page Type (Category/Tag) Using AJAX

Question: I want to know if the opened page is a category/tag page. From the JavaScript file, I can pass only the permalink of the page. Is there a way I can find the page type?
Answer: One thing you could do is set a variable in the archive.php like . Or in any template do if (is_archive()) { /* set the variable */ }.
And then in the JS file do if (typeof isArchive != 'undefined') { } but there are probably way more elegant solutions that I'm unaware of.
This also probably assumes you load your scripts in the footer now that I think about it.
edit: of course you could also conditionally load your JS only on the archive pages and woudln't need to test anything in the javascript itself.
Comment: I am aware of both of these solutions, but neither of them are so elegant to use in a premium theme. I want do the check via AJAX. Reasons:

1. I don't want to output script tags unnecessarily in the code.

2. I want to have a single JS file.
Comment: Fair point, and I'm too interested in this if someone posts a decent solution :) Or how to check tag/category/post format what so ever.
Answer: The only way I could accomplish this is by using the wp_localize_script function. The function is quite straightforward and lets you simply send anything from server side to your script.
This also outputs the variables inline, but this is the best WordPress allowd in this situation.

Get Page Type (Category/Tag) Using AJAX


How would a finite state transducer with no output function work?

Question: Say I have a finite state transducer, 6 tuple FST with a set of states, input symbols, output symbols, transition function, a start state, and an output function. If I combine the output function into the transition function and create this mutated FST called iFST, where the output happens on the "edges" and just one symbol is outputted each step. Would this new FST still be able to compute the same functions as a normal FST?
Answer: It seems your notion is called a Moore machine (output function in the nodes) whereas the other variant is called Mealy machine (output on edges).
This is what wikipedia states:

"Every Moore machine $M$ is equivalent to the Mealy machine with the
  same states and transitions and the output function
   $G(s, \sigma) \rightarrow G_M(s)$, which takes each state-input pair $(s, \sigma)$
  and yields $G_M(s)$, where $G_M$ is $M$'s output function.
However, not every Mealy machine can be converted to an equivalent
  Moore machine. Some can be converted only to an almost equivalent
  Moore machine, with outputs shifted in time. "

In another terminology, the transitions of a FST are of the form $(p,a,b,q)\in Q\times \Sigma\times\Delta\times Q$, where, $\Sigma,\Delta$ are the input/output alphabet of the transducer: from state $p$, reading $a$ from input, move to state $q$, writing $b$ to output. Sometimes this is called a Generalized Sequential Machine, and the more general definition of a FST allows  $Q\times \Sigma^*\times\Delta^*\times Q$ instead.
Now the transitions of a Moore machine, which in wikipedia formalization [but with my symbols] has a transition function $T : Q \times \Sigma \rightarrow Q$ and output function $G : S \rightarrow \Delta$, can be written in the above FST format like $(\,p,a,G(a),T(p,a)\,)$, for every $p\in Q$, every $a\in\Sigma$.
I do not know what a "normal" FST is. I prefer output on edges (like the input) to make input/output more symmetric.
Comment: The issue may have to do with $\varepsilon$-transitions resp. non-determinism in general, which allows us to pick different edges but not necessarily states. That said, "with outputs shifted in time" seems to be a restriction not typically considered in automata theory; there you'd only be interested in the resulting language.
Comment: I am not sure if this is the same. I was talking about completing removing the output function. If I am not mistaken Mealy machines still have an output function. And by a normal FST what I meant is having an output function. Could I combine the output and transition function and create a "iFST" and have it be able to compute the same functions as a FST


How axial precession affects the night sky

Question: I am working on the model which can visualise how the night sky change over time (from observer's point of view). I was thinking how axial precession affect positions of stars and planets on the night sky, but after hour of straight thinking I can't put it together. I use as reference point of view the north pole. Lets say that axial precession is circular (following circle) and we are looking at the sky at the same part of the year. If I understand it correctly, the stars/planets will move in two ways - in circle and down from the original position. So in the quarter of axial precession cycle, stars which were originally on the norhtern sky, are now on the eastern sky (I read that axial precession proceed in the opposite way as rotation of the planet) and sky should appear as if planet doesn't have axial tilt (which should make planet without seasons) - in different words, angle between star's original position and current position is the same as axial tilt. In the half of axial precession cycle, stars which were originally on the norhtern sky, are now on the southern sky, angle between their original position and current position is double of axial tilt. That's my understanding of axial precession. But apparently (from what I read) axial precession just shifts seasons (and doesn't stop them in the quarter of cycle) and I also run some simulations in Stellarium and stars with closer declination to the equator don't change their declination as much as stars near the north pole. I really don't know what I am missing.
Answer:
How axial precession affects the night sky

For  layspeople, there are no discernible effects whatsoever. Even if they could magically live for millennia.
For the kind of nerd who is properly represented in media such as the tv show The Big Bang Theory, the main observable effect on the night sky is that the polar star changes every few thousand years. It was Alpha Draconis back at some point when ancient Egypt was not ancient. It is Polaris now, and will be Gamma Cephei when Valve releases Half-Life 3. All stars in the Northern hemisphere seem to circle around whatever star is the current polar one. There is no true southern polar star, so observers there wouldn't notice much regarding this.
The secondary effect is that the dates for equinoxes and soltices changes a little bit over time. As in approximately 13,000 years from now (both past and future), summer in the Northern hemisphere would be around December and the Northern winter would be in the middle of the year. The effect on the night sky are the dates when night are longer or shorter.
Also because of the above, the position of the Sun relative to the stellar background has a similar cycle. 13,000 years from now, you'll have to be born between May and June to be a floricking Sagittarius.
Over such vast spans of time, all stars other than those eligible to become polar change positions more due to their own trajectories around the Milky Way than because of Earth's movements.
Comment: @Bedlasky if "you know that", you should have mentioned it in your question. People can not magically guess the depth of your pit of knowledge.
Comment: I know that. Stars will move around their trajectories in my model. I need something that works for cca 10 000 years. With this model I can basically track celestial events in any moment of history of civilaziton. Not mention this model can be used for any world.
Comment: *"For layspeople, there are no discernible effects whatsoever:"* Depends on whether those lay people do or do not look up to the sky occasionally or not. For somebody who lives at mid-latitudes in the northern hemisphere, I would say that seeing Orion up in the sky in mid-summer would count a striking discernible effect. (In Romanian there is a semi-well-known poem by Geo Bogza: *No ship has ever returned from the South Seas or from the Capricorn such a pure and elegant frigate as Orion when he returns in autumn; over green forests his white light never did shine, nor over meadows of hay...*)
Answer: What axial precession does is move the celestial north pole on a circle with a radius equal to the axial tilt. For Earth, a complete precessional cycle is completed in about 26,000 years. The entire starry sky appears to revolve 360° eastwards $\rightarrow$along the ecliptic$\leftarrow$ during a precession cycle. Year after year, the vernal equinox, aka the first point of Aries, appears to move westwards by about 50″ per year, or by about 1′ every 71½ years.
Saying that the vernal equinox appears to move westwards with respect to the stars is the same thing as saying the stars appear to move eastwards with respect to the vernal equinox.
For somebody who is at least a little bit familiar with the night sky, this has two very noticeable effects:

About 13,000 years from now, stars which right now are visible in summer (because they culminate at night) and not in winter (because they would be in daylight) will be visible in winter and not in summer; and vice-versa.
For people who live at mid-latitudes, the best example may be the striking constellation of Orion. Everybody who is even remotely familiar with the night sky can recognize it. Right now, for most Europeans, or Asians, or North Americans, Orion is a winter constellation, not visible in summer; but 13,000 years from now it will be a summer constellation, not visible in winter. (Of course, for Australians Orion is a summer constellation, and 13,000 years from now it will be a winter constellation...)
Of course, the effect is gradual, so that about 7,500 years from now Orion will be visible in spring in the northen hemisphere, roughly from mid-winter to mid-summer, but not in autum.
The twelve constellations of the Zodiac are, by definition, on the ecliptic. About half of them are visible in summer, and the other half are visible in winter. As the precession cycle progresses, about every 2,200 years one zodiacal constellation moves from the summer-visible group to the winter-visible group, and one moves from the winter-visible group to the summer-visible group.
This is the dawning of the Age of Aquarius... When the vernal equinox is in Aquarius this means that the constellation is visible in northern hemisphere autumn and southern hemisphere spring, and invisible, because it would be in daylight, in northern hemisphere spring and southern hemisphere autumn.

The position of the north and south poles in the sky changes quite fast on historical scales.
For example, about 2,300 years ago, the Greek explorer Pytheas of Massalia complained that there were no stars around the north pole. (In his days, the closest star from the north pole was β Ursae Minoris, sometimes called Kochab, which is a bright star, but was almost 10° away from the pole.) The north pole came within 5° of our current North Star, α Ursae Minoris, only about 1,000 years ago, and 1,000 years from now it will again be more than 5° away from it. (The closest approach will be around the year 2,100, when Polaris will be less than half a degree away from the north pole.)
For a quick reference, the full Moon is about half a degree across.

I have not said anything about planets. The planets will move relative to the stars as they always do, precession does not have any effect on this. The movement of planets is not synchronized with our years, so that the position of a planet is relative to the stars will not return to the same point after one year. (Of course, except the Sun, which returns to the same point with respect to the stars after one sidereal year by definition.)
About those equatorial coordinates:
The equatorial coordinate system, right ascension and declination, is equatorial. Axial precession rotates the starry sky around the axis of the ecliptic. As a consequence, the change in equatorial coordinates during a precession cycle is neither simple, not the same for all stars.
The ecliptic coordinate system comes to the rescue. Basically, ignoring the variations in the orientation of the ecliptic and the proper motions of the stars, the ecliptic longitude of a star goes through all the values from 0° to 360° during a precession cycle. For stars which are not very close to the poles of the ecliptic this comes to a regular increase in ecliptic longitute of about 1°23′48″ per century. Again ignoring the variations in the orientation of the ecliptic and the proper motions of the stars, the ecliptic latitude of a fixed star remains unchanged during a precession cycle.
So that one can always convert the equatorial coordinates of a star for a given epoch into ecliptic coordinates, apply the regular progress of ecliptic longitude, and convert back into equatorial coordinates for the target epoch.
Beware that the obliquity of the ecliptic, which is needed when converting between equatorial and ecliptic coordinates will change during a precession cycle. How to account for this depends on the time scale; on short time scales, a few centuries, we have very precise formulas, whereas on longer time scales this is governed by one of the famous Milankovitch cycles, varying between about 22.1 and 24.5 degrees in a cycle about 41,000 years long.
Comment: Thanks for detailed answer! But I am still confused. From observer's point of view, everything on the night sky is shifted by x° (x - axial tilt) as oppose to what would you see from space. Half way through axial precession cycle you see everything shifted by -x°. So in quarter of the cycle everything is shifted by 0°. So Sun is exactly on equator = no seasons. But apparently this don't happen - why?
Comment: @Bedlasky: I do not understand what you do not understand. You do understand that the sun makes a complete revolution with respect to the fixed stars in the course of one year? The obliquity of the ecliptic remains more or less constant, and anyway the changes in obliquity have nothing to do with precession. In **any year** there are two moments when the sun is above the equator, and they are called equinoxes. For the movement of the sun, the only thing precession does is that it makes a sidereal year about 20 minutes longer than a tropic year.
Comment: @Bedlasky: Re-reading your comment... First, there is no privileged point of view. Second, the obliquity of the ecliptic does not change during a precession cycle. (Well, it does change, but only a little and this has nothing to do with precession.) Precession changes the direction in which the axis of rotation points, but it always keeps about the same inclination with respect to the ecliptic plane. Look at [this illustration](https://commons.wikimedia.org/wiki/File:Precession_N.gif) and notice that at any point in the cycle the north celestial pole is 23° away from the pole of the ecliptic.
Comment: @Bedlasky: The oscillation between +23° (called the northern solstice) and −23° (called the southern solstice) happens *in the course of one year*. This is why in summer days are long and nights are short, and in winter days are short and nights are long; this is why some stars are visible in summer and not in winter, and vice-versa. And yes, midway between the solstices are two points, one in March and one in September, when the sun is at 0° with respect to the Equator.
Comment: I looked at that illustration and that basically confirm what I said. In a quarter of precession cycle, north pole is shifted by 23° from current position. So equator must be also shifted by 23°. So Sun is on the equator.
Comment: @Bedlasky: It is shifted indeed, but it is **not** shifted towards the pole of the ecliptic. The distance between the axis of rotation and the axis of the ecliptic is always the same. The pole of the axis of rotation moves **around** the pole of the ecliptic, with the axis of rotation always inclinded 23° with respect to the axis of the ecliptic. The equator is always inclinded 23° with respect to the plane of the ecliptic.
Comment: Ok, but how about stars between? How much are they shifted?
Comment: @Bedlasky: In [ecliptic coordinates](https://en.wikipedia.org/wiki/Ecliptic_coordinate_system) their latitude stays constant and their longiture progresses by about 1°23′48″ per century. In equatorial coordinates that varies from star to star, because it's a rotation around an oblique axis. Need to convert from equatorial to ecliptic coordinates at the start of the epoch, apply the progression of the longitude, then convert back to equatorial coordinates at the time of interest. Pay attention to the value of the obliquity used when doing the conversion -- it varies slowly for its own reasons.


Why are resistors that are shorted and connected to each other considered to be in parallel and not in series?

Question: In the following image, we are finding the Thévenin equivalent resistance after shorting the voltage sources. Why are these two resistors considered to be in parallel? It seems they are connected both in series and in parallel.
Comment: how many nodes are in the diagram?
Comment: The ends are connected together on both sides so they're definitely in parallel. The way it's drawn is just putting you on the wrong foot. Redraw the image with the resistors on top of each other and it'll look more familiar with typical textbook examples.
Comment: "The way they're drawn is just putting you on the wrong foot" **This is done intentionally in most textbooks, to confuse and obfuscate**.  One of the *meta-skills* you must learn to succeed in class is how to physically re-draw a diagram in a way that is *electrically* the same, so you can push things around and thus see what is then obvious.    Whether this is lazy textbook writing, or teaching you an important lesson, I do not know.  I'm sure cr*p like this causes many otherwise qualified people to wash out of school.
Comment: The Thevenin equivalent resistance of what? It matters! Label nodes please for this to make sense.
Comment: I commend the OP for asking for clarification on a result that they have worked successfully. The diagram is the result of a Thevenin analysis. The OP has clearly realized that the resistors appear to be both in series and in  parallel. I really don't think their work is cr*p. This is a well presented question.
Answer: Because these are the same circuit:

simulate this circuit – Schematic created using CircuitLab
Answer:
It seems to be they are connected in both series and parallel.

You are correct.
Amusingly, you are the only one to state that. No one else (as of my writing) has said what you have realized: they are connected in both series and parallel.
If you add a 3rd component across one of the resistors (e.g., an ohmmeter), then the resulting circuit becomes just parallel.
Answer: Slide each resistor around the corner so it is in the middle of a vertical part of net, and you will see that they are directly in parallel.  If they were in series, as it looks like across the top of the drawing, the outside ends of the two resistors would not be connected together around the bottom of the drawing.
Answer: Why the resistors are connected in parallel
Definitely the two resistors are connected in parallel and we can measure their equivalent resistance R1||R2 (in my schematics below) by connecting an ohmmeter in parallel to this network. So they are connected in parallel but with respect to the nodes A and B between which the ohmmeter is connected.

simulate this circuit – Schematic created using CircuitLab
Why the resistors are not connected in series
If the two resistors were connected in series, we should be able to measure their equivalent resistance R1 + R2 by connecting the ohmmeter in series with this network... but it is shorted. So we have to break the connection and insert the ohmmeter between the two ends C and D. Thus the two resistors are connected in series but with respect to the nodes C and D between which the ohmmeter is connected, not with respect to the initial nodes A and B.

simulate this circuit
If you still want to say that they are connected in both configurations, then you can say it that way:
"The resistors are connected both in parallel (with respect to A and B) and in series (with respect to C and D)."
Note: Since CircuitLab does not have an ohmmeter, in the schematics above I have assembled it by connecting a 1mA current source and a voltmeter "in parallel":-) At this value of the current, [mV] corresponds to [ohm] and [V] to kohm. So just look at the voltmeter to read the equivalent resistance.
Conclusion
We can talk about connecting in series or in parallel at least two loads to one source or at least two sources to one load (i.e. when there are at least three elements).
When there is only one source and one load (the simplest case), we can say that they are connected "both in series and in parallel"... but it does not make much sense. I prefer to say they are connected "to each other" or simply "connected".
Answer: It depends on point of view. I suspect that the original circuit is something like this with the load across nodes a and b removed:

simulate this circuit – Schematic created using CircuitLab
For Thevenin analysis:
R1 and R2 are in series from the point of view of the sources while calculating the Thevenin voltage Vab.
As shown in other answers, they are in parallel from the point of view of calculating the Thevenin resistance Rab.
Answer: The resistors are in series if the current flowing in them is the same and only 1 end is common for both resistors.
The resistors are in parallel if the voltage across them is the same and both ends of the resistors are shorted to each other.
Here, both resistor ends are shorted and voltage is the same hence they are in parallel. Connect a voltage source between the 2 terminals shown in your diagram and you will see that current in the 2 resistors is different (so, definitely not series). So, through which terminals you are looking at the circuit is critical.


Why is the mass of small elements taken as $∆m$ in center of mass of a continuous body?

Question: A continuous body has continuous distribution of mass. Doesn't $\Delta m$ mean $m_f - m_i$? But, is the mass Changing? If yes, how is the mass varying? Why is the mass of the small elements in a body taken as $\Delta m$? Why isn't it taken just as $m$ (mass of the small element)?
Comment: $\Delta$ means change in general, not final minus initial. Change can be with respect to anything, for example distance in case of continuous bodies, or time in case of finding speed.
Comment: I think the same symbol (in this case, the delta) can mean two different things, in this case when you divide a rigid body into small parts, $\Delta m$ is just the mass of one of those parts
Comment: For summing up or integration, we divide into number of smaller parts. Suppose mass $M$ having some volume is divided into $N$ parts all having same mass then, $\frac{M}{N}=\Delta M$. More the number of parts, less the error.
Answer: It's a good question.
Normally speaking, we consider a quantity written as $\Delta x$ to conceptually mean a "change in a quantity". For example a "tiny change in time" or a "tiny change in position".
However, the usage in case of center of mass calculation is in a different conceptual sense. We can think of chopping up the body into some chunks with each chunk being of mass:
$$ dm = \rho dV$$
Here I am considering that we have a 3-D body but it could be that we have 2d or 1d as well.
In a way, we can think of the two approaches as actually being the same. Suppose you had a rod of length L lying along the x-axis from $x=0$ to $x=L$. You can imagine a situation where tiny rods of mass $dm$ are placed consecutively with tail of one at head of previous as time passes to reconstruct the original rod.
Comment: So, dm doesn't mean the mass is changing, it just refers to a small mass which has some mass m but the notation used is dm for integration purpose right?
Comment: Great answer, showing how the ∆m can indeed be conceptualized as a change, and sharing the small mass picture with the better notation dm as well.
Comment: "small mass which some mass m" ? Maybe you mean small piece which has mass "m", but yeah more or less @jsivesh
Answer: You are right: it is a confusing notation. Usually it is used to "construct integrals", where it represents a finite-sized small chunk of the body, which is later assumed to go to zero size when a limit is taken.  I prefer to use something like $m_n$ for these "chunk masses", and reserve the $\Delta$-notation for coordinate variables in which your "final minus initial" idea makes sense.
For example, consider a bar of length $L$ which has a changing linear mass density such that if you place one end of the bar at the origin of the $x$ axis, the density is given by $\lambda(x) = \lambda_0 + a x \; (0 \le x \le L)$. What is the mass of the bar?
We start by dividing the bar into a finite number of equal-length chunks, $N$ (which you should think of as something like 8, so you can draw a picture and identify one of the chunks, $m_n$). Then we give an approximate expression for the total mass, $M$, which will become exact once we find a Riemann sum, and take the limit as $N \rightarrow \infty$:
$$
M \approx \sum_{n=1}^N m_n
$$
Notice that I have labeled the 8 chunks $m_1, m_2, \ldots, m_8$.  These are the $\Delta m$ chunks that confused you.  Now express the mass of an individual chunk using the density (at the position of the $n$th chunk, $x_n$) and the length of the $n$th chunk:
$$
M \approx \sum_{n=1}^N \lambda(x_n) L_n
$$
The length of each chunk is $L_n = L / N$, but on our chosen coordinate axis, it is more useful to express it as $L_n = \Delta x_n = x_{n+1} - x_n = \Delta x$.  This lets our approximate expression take the form of a Riemann sum, of which we can then take a limit:
\begin{align}
M &\approx \sum_{n=1}^N \lambda(x_n) \Delta x\\
M &= \lim_{N\rightarrow \infty} \sum_{n=1}^N \lambda(x_n) \Delta x\\
 &= \int_0^L \lambda(x) dx\\
 &=  \int_0^L \left( \lambda_0 + a x\right) dx\\
& = \lambda_0 L + \frac{a L^2}{2}
\end{align}
This idea can be generalized to higher dimensions where either
$$
m_n = \sigma(x,y) \Delta A_n = \sigma(x,y) \Delta x \Delta y
$$
or
$$
m_n = \rho(x,y,z) \Delta V_n = \rho(x,y, z) \Delta x \Delta y \Delta z
$$
and to other, e.g., spherical, coordinate systems where you will need to construct $\Delta V$ in terms of those coordinates.


Can't implicitly convert System.windows.Controls.Image into System.windows.form.control in WPF

Question: I am totally new to WPF. I am converting application from windows form application to WPF application. windows form application is working properly. I have to create a device for rendering image. for that I Used following code.
//define a render Control
<code>private System.Windows.Forms.Control renderControl;

//get and set rendercontrol

public System.Windows.Forms.Control RenderControl

  {

       get { return renderControl; }
    set { renderControl = value;  }
    }

//PresentParameters

pp = new PresentParameters();

pp.Windowed = true;

pp.SwapEffect = SwapEffect.Discard;

    device = new Microsoft.DirectX.Direct3D.Device(0, Microsoft.DirectX.Direct3D.DeviceType.Hardware, renderControl, CreateFlags.SoftwareVertexProcessing, pp);
</code>
I import this class and used WPF Image as renderControl. But it generate error can't implicity covert from windows.Controls.Image to Windows.Forms.Control. How I can convert? Is there any other way to render or for Creating device?
Answer: WPF controls are not compatible or interchangeable with Windows Forms controls.  The best way to host a Direct3D surface in WPF is to use D3DImage.


Authorization plugin for package installation

Question: Is it possible to allow installation of specific packages/profiles without admin credentials by modifying authorization rights?.I was able to allow installation for all by modifying the installation right in authorizationdb.Is it possible to receive any information about package requesting for installation inside auth plugin so that i would be able to evaluate and allow.
Comment: I will assume you don't mean using admin credentials to delegate admin to a non admin account using something like sudoers file. Please clarify what you seek if my answer is not on the mark. Basically, it looks like you're defining malware installer that doesn't rely on tricking an admin  to authorize the install.
Comment: what i meant was to create a authorization plugin that would allow installation of certain whitelisted apps(which might be identified by publisher) without prompting for admin password.
Answer: What you ask is possible in several ways. We use management frameworks like JAMF Pro or Munki to script and manage arbitrary changes and installations so that no user needs to download / package / install or prompt. These are powerful distributed automation options, but they cost time and salary and experience and licensing.
You can roll your own more minimal installer, Most require a bit of skill if you can't change your installer to just run as non-admin, but they are lighter cost and weight than an MDM suite solution. See homebrew for a great example of an installer that needs admin only at setup and not when installing new code.
If you need admin installer and can't use brew cask to package your apps, a dependable way is to write a daemon that runs as admin. Microsoft Office, Adobe and many other software does this, the installer system runs at launch or as a daemon and gets admin privileges and then uses the OS to run periodically.
This avoids the design intent for the installer to ask for admin at the last moment when installations trigger. Following the design intent is generally more secure and far less work for you and your customers.
Comment: in my case i am supposed to allow installation of any third party application(including the one which require admin cred) by standard users which are whitelisted by admin
Comment: We use JAMF framework for this. Munki is another framework that does this if you like open source solutions. @theappledev
Answer: Missing Intent
Unfortunately, the information needed to evaluate if an installer package is whitelisted is not available during requests to obtain an system.install.* right.
Calls to macOS's Authorization Services do not include details about what the authorization will be used for.
Even if you implement a custom Authorization Plug-in, the authorization request will not mention what package is being installed.

Use plug-ins to extend macOS authorization services to perform authorizations in a new way or to implement a new policy that is too complex to be implemented entirely with the authorization policy database.

What Next?
Here are some options:

File a bug report directly with Apple;
Write a SMJobBless helper to grant administrator rights on-demand; see UpdateCore for an example administrator privileged helper tool that installs software;
Consider third party software that can provide a "self serve" service for your users.
Comment: i have implemented a custom authorization plugin.Is there a way to identify which application is being installed?I tried to access context and hint values but was unable to get details about the application being installed
Comment: No, that information is not available in the request. I suspect Apple's engineers would consider the information a possible security leak. A possible solution from Apple would check for a `system.install.pkg.` before a more generic right check.
Comment: I have extended the answer to mention a `SMJobBless` approach and linked to a sample project.
Comment: is it possible to use helper to identify which app being installed currently and grant right?in my case standard users should be able to install any whitelisted 3rd party app without admin credentials
Comment: Yes, you could wrap up [UpdateKit](https://indie.miln.eu/tool/update/) to handle installation for non-administrators of acceptable packages. The wrapped UpdateKit would need to be installed by an administrator to grant the initial rights.
Comment: ok ill check that.Is it possible to do this for installation of profiles(manually by standard user)?
Comment: UpdateKit handles packages (pkg) files. Anything you can package, you can pass to `installer` via UpdateKit/UpdateCore. You should also investigate [Munki](https://www.munki.org/munki/) before writing your own solution.


Select Rows with a Point that fall within a set of points that form a polygon from outside the DB

Question: Using a front end Map UI, I have selected a polygon and received all the Lat/Long points from it. I want to send this data to a PostgreSQL PostGIS database table that has rows which stores a geometry Point in a column. I need to select all the rows that are inside of the polygon. How do I put the external polygon points into the SQL statement and pull the correct data?
Answer: I found the answer:
<code>SELECT * FROM Your_Table
  WHERE ST_Within(geom,
    ST_GeometryFromText('POLYGON((-104.53798595055618 31.54712467875706,
        -102.48477288296839 31.289207699238165,
        -102.68445701550822 30.006579157844246,
        -104.3434563387951 30.233933113969798,
        -104.53798595055618 31.54712467875706))',4326)
    );
</code>


New sd card doesnt work in my Android Alcatel one touch phone

Question: I purchased a new sd card for my Android phone to increase memory.  Its a 32 gb card, but my phone still says not enough memory when I try to download an app. I tried selecting sd card and it shows 32 gb available but still cant download apps.  Then I switched to internal memory just to see if that worked but it did not.  It shows I have 32 gb of memory but phone says not enough memory. I tried to format new sd card but still getting same message.  HELP!!
Comment: What android version are you on?  Older Android didn't support running apps off of SD cards.
Answer: There isn't any problem with your SD card since it is recognized buy the phone. You know what else? Apps can't be placed on sd card without some help. Phone shows to you message 'not enough memory' because you lack internal memory where apps go when installing. So please rephrase your question. And check out this:
insufficient-memory' tag wiki,
'app2sd' tag wiki,
link2sd tag wiki,
adoptable storage tag wiki
Answer: Some data can be stored on SD cards and others not. This depends on how the SD card is integrated in your device.
A rule of thumb is, that apps should be stored on the internal storage of your phone (often this is the only possibility). While pictures and music can be also located on the SD card.
I can not read this from your posting, but maybe you should try to move such user data manually to your new SD card to make space available on your internal memory.


Create setup without writing to registry

Question: I need to make a setup for my .Net4 application. What I expect about installer is that it will just check if .Net4 Client profile is installed on target machine and if not it will deploy the .Net package and it will extract the app to directory where the setup was started.
I don't want to write anywhere else(app is "roaming" and runs from sd card for ex.) as  copying it to start menu or Program Files or write in to the registry. I just want to be able to run the app.
Can distribute the .Net4 Client profile with the setup(any licence limitations)?
Is it possible? How can I start? Thank you for replies.
Answer: Ok, so it is possible. Just remove any registry and application entries in setup project. Easy.


Tachyon velocity in relation to light speed

Question: tachyonic particle are merely hypothetical particle that always travels faster than light., But if real, how fast would they be?
Answer: This is  like asking "how fast is a massive particle ?". The answer is "it depends". It depends on the reference frame / on the dynamics of the particle.
Depending on the reference frame of the observer, a tachyon could have any velocity that is faster than light. In some reference frame they propagate instantaneously. Different reference frames do not agree on if they propagate forward or backward in time.


How to host large files on Google-App-Engine

Question: I would like my server that runs on Google App Engine to host large files such as audio scripts and images, is it possible to store them as a column in a database? If not, what mechanisms may I use?
Answer: You have two options:

Blobstore (currently available in Java, Python and Go).
Google Cloud Storage (currently available in Java, Python and PHP).
Answer: Blobstore and GCS are most likely what your are looking for.
Both services are not covered by the GAE SLAs however. If you need that kind of reliability promise you're stuck with the GAE datastore.
You can put your files in a BLOB property of a datastore entity and serve it from there. Datastore entities have a size limit of 1MB however.
To circumvent that, you must split and re-assemble your files using multiple entities. There again is a size limit to any GAE response which is 32MB.


Writing in Common Styles

Question: How do I make my writing sound like a college textbook, a newspaper article, a scientific journal, a blog post, etc, professionally?
Answer: Practice pastiches.
Pick examples of the style you want to emulate, and then write something like that.  For instance, take a newspaper article about the opening of a library and write something in its style about the opening of an imaginary school.  Pick a blog entry about visiting parents for the Fourth of July and write one like it about visiting grandparents for Labor Day.  Etc.
Answer: Publishers sometimes have a standard and recommendations on how to write an article for them, as freelance writers submit articles and need to know in what style to write for them.
As there are a myriad of possible styles, I recommend searching for them yourself, (rather than me providing a bunch of links).
This is especially true for news articles, academic papers, and whatever else.
The good way to learn, is to write a draft and then go over it again to match a set of guidelines.

Writing in Common Styles