$(1+x^M)^{1/M}$ need to be able to calculate any order derivatives vs. $x$ and $M$ for $x\ge 0$ and $M\gt 2$

Question: cannot delete my own question, so I try to overwrite it instead...
Comment: This is to be used for a compiler. Thus the max. order of derivatives depends on the input source code written by the user.
Comment: Do you have an idea what the maximum orders of derivatives needed will be?
Answer: One approach is to use the Faà di Bruno formula:
$$ \big((1+x^M)^{1/M}\big)^{(n)} = \sum \frac{n!}{m_1!1!^{m_1}\cdots m_n! n!^{m_n}} (1/M)^{\underline{m_1+\cdots+m_n}}(1+x^M)^{1/M-m_1-\cdots-m_n} \prod_{j=1}^{n} \big(M^{\underline{\phantom{,}j}} x^{M-j}\big)^{m_j}, \qquad M\notin\mathbb{Z} $$
Here $a^{\underline{b}}$ is the falling factorial, and the sum is over all $n$-tuples of integers $m_1,\ldots,m_n\geq0$ such that $\sum_k km_k = n$. This formula is likely to be unstable for very large values of $n$.
A decent alternative is to evaluate the contour integral
$$ \frac{n!}{2\pi \mathrm{i}}\int_\gamma (1+z^M)^{1/M}\frac{dz}{(z-x)^{n+1}}, $$
where the contour $\gamma$ is a circle of some radius $R$ surrounding $x$, $z = x+Re^{\mathrm{i}t}$, with $R$ chosen such that all the branch points of $(1+x^M)^{1/M}$ in the complex plane are outside the circle. The choice of $R$ affects the stability of this formula. This is a straightforward Fourier integral, so one can directly use an existing library to compute it, and one can also evaluate many derivatives at once efficiently using FFT.
Comment: I don't see how the first method avoids calculating the derivative vs. M at x==0. The problem is, that I always end up with a log(0.0)...
Comment: @GrapschKnutsch In the first formula, when $M$ is an integer, $x^{M-j}$ should be replaced with $x^{M-j}[M\geq j]$ because it's really $(x^M)^{(j)}$ (see the full formula). At $x=0$, one just has $0^0=1, 0^k=0$. Not sure where you get $\log$. When $M$ is not an integer, the function is not infinitely differentiable at $x=0$, anyway, so the formula has to break down. The second (Cauchy integral) formula is usually slower, but more robust, and doesn't need such tinkering.
Comment: "the function is not infinitely differentiable at x=0": Does this mean, that this function should not be part of a transistor model or it should not be calculated a x=0 when being part of such a model?
Comment: @GrapschKnutsch You lost me there, I have no idea what you mean. What model? Can you perhaps create a separate question about what you actually want to do?
Comment: This function is part of some famous transistor model (a mathematical description of electrical currents as a function of voltages) used in simulation of electrical circuits. Can I assume, that this function is not suitable for this purpose, as it is not physical.


How do circuits work where from the touch of a button on the wire of a headphone, you can pause or play music?

Question: I am creating a project for my engineering course. I ve decided to make headphones that have the ability to pause music when taken off from the ears and vice versa. I will be using long lever switches as well. How will i be able to create a circuit where pressure is applied to the ear pads from wearing it and causing the music to play or stop. Please help me. I don t know too much about electronics and i thought that it would be a bit of a challenge.
Answer: There is no general answer here, so the answer is: it depends.
Some headsets have an extra wire to the phone and pressing a button shorts that wire to ground via a resistor. If you want 2 buttons, use 2 different value resistors, a circuit in the phone detects which resistor is connected so then it can deduce which button you pressed.
On a wireless headphone or headset there could be a wireless connection to the phone, for example bluetooth. So there the switches are connected to the bluetooth chip in the headphones.
A simple remote via cable could be made by just shorting the audio signal line to ground. That will give a "click" in the sound though. Again some circuit in the phone is needed to detect that the signal line is grounded because you pressed the button.
Comment: Check for presence of 60Hz power waveforms near the ear-pads. The human's head will provide those waveforms


Who are the next-generation (Cursed Child) Hogwarts teachers/staff?

Question: In Harry Potter and the Cursed Child, we know that very few teachers remain at Hogwarts from Harry's days there. Who are the new Hogwarts teachers? I do not own a copy of the book, so I can't look directly. I know that Neville Longbottom now teaches Herbology. Who are the rest of the new teachers/staff?
Answer: Prime timeline - The teachers we know about are

McGonagall- Headmistress.
Neville- Herbology.
Madam Hooch- Flying.
Hagrid- Groundskeeper, Care of Magical Creatures (he isn't mentioned as a teacher in Cursed Child but in "19 years later" he is.)

In the second timeline

Hermione is added as DADA.

In the third timeline we only know about

Snape - Potions
Umbridge- Headmistress.

All timelines - teachers that probably stayed

Professor Binns
Flitwick.
Comment: I've played with the formatting a little in order to make the answer more readable. You might also want to consider adding some supporting quotes.
Comment: I haven't read the play script yet but what is this about timelines? Is time travel/splitting timelines a major plot point of the play?
Comment: @TylerH Yes. The main plot of the play is having next generation Hogwarts students exploring (and interfering with) alternative timelines in the past.
Answer:
Headmistress-Professor Minerva McGonagall
History of Magic-Professor Cuthbert Binns
Muggle Studies-Unknown (possibly Arabella Figg or Arhtur Weasley)
Charms-Professor Filius Flitwick
Flying-Madam Rolanda Hooch
Potions Master-Professor Horace Slughorn
Care for Magical Creatures-Professor Rubeus Hagrid
Herbology-Professor Neville Longbottom
Divination-Professor Firenze and Professor Sybill Trelawney
Defense Against the Dark Arts-Professor Hermoine Jean Granger-Weasley
Astronomy-Professor Aurora Sinestra
Arithmancy-Professor Septima Vector
Ancient Runes-??? (No idea!)
Alchemy-??? (wasn't even mentioned in the first place!!)
Comment: Hi, welcome to SF&F.  What are your sources for this?
Answer:
History of magic-Professor Binn
Muggle Studies-Professor Charity Burbage
Charms-Professor Flitwick
Flying-Professor Hooch
Potions-Professor Horce Slughorn
Care of Magical Creatures-Professor Hadrig
Herbology-Professor Hannah Abbot [Neville's wife]
Divination-Professor Trelawney
Defence Dark Art Magic-Professor Amycus Carrow

Read Harry Potter and the Cursed Child u will get it
Comment: Hi, welcome to SF&F.  Can you provide any references for these?
Comment: Didn't Burbage run into You Know Who in DH?
Comment: I have read Cursed Child, but it's been a while. Also, I believe that Charity was fed to Nagini in Deathly Hallows. Welcome to the site tho! A general rule here is to cite references/source material when offering an answer, and research first. That's alright though, everyone had a first time!


Chaos and integrability in classical mechanics

Question: An Liouville integrable system admits a set of action-angle variables and is by definition non-chaotic. Is the converse true however, are non-integrable systems automatically chaotic? Are there any examples of non-integrable systems that are not chaotic?
Comment: @AngusTheMan You need to make sure what definitions are behind the terms you're using, to elaborate: when you say "integrable", do you mean the integrability of the system as defined by Liouville? In which case, if the hamiltonian system with $n$ DOF does not exhibit at least $n$ global first integrals of motion, all in involution (Poisson commuting), then the system is not Liouville integrable. For such non-integrable system, neighbouring bundles of trajectories in phase space, spread exponentially in time w.r.t. one another, which in turn means unpredictability in longterm, so it is chaotic.
Comment: @AngusTheMan You're welcome. The Lyapunov exponent is a possible measure of sensitivity to initial conditions, more specifically, given 2 trajectories with initial separation $\delta x(0),$ the separation at time $t$ is given by: $|\delta x(t)| \approx e^{\lambda t}|\delta x(0)|.$ The exponential divergence of paths here is characteristic of a classical *chaotic* system, the used formalism is irrelevant. As to why it is exponential, is another discussion, in fact in very long times it is not exponential anymore. All this is related to ergodic properties of chaotic systems.
Comment: What is the precise definition of *chaotic* you are using?
Comment: @ACuriousMind I was in particular thinking of deterministic chaos, where the slight change in initial conditions leads to a drastic change in evolution, but would look for guidance on this topic! :)
Comment: @Phonon This comment really helped me, thank you. Would the exponential decay you refer to be the Lyapunov exponent? Is this a general property of the Hamiltonian formalism that it is exponential separation? (e.g. why not linear or quadratic etc)?
Comment: Do these notes help you? http://www.srl.caltech.edu/phys106/p106b01/topic3.pdf http://massey.dur.ac.uk/sag/download/Chapter3.pdf
Answer: The key point here is that, any dynamical system that is not completely integrable will exhibit chaotic regimes1. In other words not all orbits will lie on an invariant torus (Liouville's torus is the topological structure of a fully integrable system), in principle a chaotic system can even have closed stable periodic orbits (typical for regular/integrable systems) for some initial conditions, the set of such conditions has measure zero (meaning the states on that orbit are only reachable from other states of the same orbit).
In order to get yourself acquainted with such concepts, I suggest looking into 2D dynamic billiards. These models are of great interest because their dynamics are solely defined by the shape of the boundary, circular, ellipsoid, stadium etc. Now an interesting example to showcase here would be the oval shaped boundary (note circular and ellipsoid billiards are regular because of their symmetries):

In the above image (by Tureci, Hakan, et al. 2002), on left you see the poincaré map2 of the 2D oval billiard (with specular reflection), and on the right you see 3 examples of different regimes of the system. This is a perfect example showcasing a system that admits only locally integrable regions. Case a) corresponds to a quasi-periodic orbit, only marginally stable. Case b) shows a stable periodic orbit surrounded by a stable island and finally case c) corresponding to the entirety of densely dotted regions of the map, is indicative of chaotic motion. For further reading, I suggest looking into some of the articles on scholarpedia, and of course not to miss this fantastic review by A. Douglas Stone.

1For example all non-linear systems that are not Liouville integrable (as explained in comments). Note that linear systems can always be solved by exponentiation. But that said one must be wary of distinctions between solvability and integrability.
2These maps are obtained by choosing a poincaré section, and finding the intersection of trajectories in phase space with this section. Such maps allow for a representation of the evolution of any dynamical system, regardless of the dynamics involved. For more intuition, see here.
Comment: @AngusTheMan Crudely put, yes :)
Comment: For fun, here's a [clip of a pool variant](http://www.bbc.com/news/magazine-34015430) with an elliptically shaped table, where aiming and potting is still perfectly possible, thanks to its integrability ;)
Comment: Thank you for a great answer! So the phase space of all but the simplest systems is actually a mix of different regions (integrable +non-integrable). The patch relevant being determined by the initial conditions?


Behavior of infinite convolution products

Question: I was wondering if there were some general results about repeated convolution products. So far, I have only dealt with single convolutions (for instance for constructing dense subspaces in $L^p$, or when working with Fourier series or Fourier transforms), but I was wondering, given a function $f$ such that $||f||_{L^1} \leq 1$, then $|| f \star f||_{L^1} \leq ||f||_{L^1} ^2 \leq 1$, and so on. Being that the $L^1$ norm is always bounded, all the functions obtained by repeatedely convoluting $f$ with itself should be well defined, at least for almost every $x$. The questions is: what can I say about this succession? Also, is this problem somehow "useful" (I mean if it is of some relevance in some scientific applied field, like for example the one-time convolution that lets us write solutions of physical problems, like heat) or is it just a useless mathematical object that we pure mathematicians love so much?
Answer: Of course there's no such thing as an infinite convolution - if your question is to make any sense we have to recast it as a question about limits of finite convolutions. And then we have to ask in what sense the limit exists.
We never get anything interesting as a limit in $L^1$ norm:

Suppose $f\in L^1$ and let $f_n=f*f*\dots*f$. If $g\in L^1$ and $||f_n-g||_1\to0$ then $g=0$.

Proof: The Fourier transforms satisfy $(\hat f)^n\to\hat g$ pointwise. Hence $\hat g(\xi)=0$ or $1$ for every $\xi$;  since $g\in C_0$ we have $\hat g=0$, hence $g=0$.
You see similarly that the norm  limit simply cannot exist unless $|\hat f|<1$ everywhere.
It is possible to get interesting things from convolutions of measures, if we take the limit in the weak* sense. For example, let $$\mu_n=\frac12(\delta_{1/2^n}+\delta_{-1/2^n})$$and $$\nu_n=\mu_1*\dots*\mu_n.$$Then $\nu_n$ converges weak* to $\frac12$ times Lebesgue meaure restricted to $[-1,1]$.

Exercise Take the Fourier transform and derive an interesting infinite product for $\sin(\xi)$.

Edit: Come to think of it, what I say above is misleading, giving the impression that there's a crucial difference between weak convergence and norm convergence here. Of course that difference is going to make a big difference, but in fact the real reason for the difference bewteen the two cases presented above is that in one case we're talking about the convolution of $f$ with itself, while in the other we're talking about the convolution of infinitely many different measures: We can have $||f_1*\dots*f_n-g||_1\to0$ in nontrivial ways, while on the other hand we cannot have $\mu*\dots*\mu\to\nu$ weakly except in certain trivial cases (I'll leave it to you to work out what the trivial possibilities are, looking at the Fourier transform, and assuming that our flavor of weak convergence implies pointwise convergence for the transform.)
Comment: Yeah thanks, I wrote it in a hurry and made a huge mistake. I edited the post though. Thanks for the examples!
Comment: Also, what do you mean by weak convergence? My definition involves scalar products, but we are in $L^1$ so we don't have any
Comment: The Banach-space dual of $C_0(\Bbb R^n)$ is the space of complex measures on $\Bbb R^n$; accordingly if $\mu_n$ is a sequence of complex measures then $\mu_n\to\mu$ in the weak* topology if $\int fd\mu_n\to\int f\,d\mu$ for every $f\in C_0$. If all the $\mu_n$ have support in a fixed compact set then $\mu_n\to\mu$ in this sense implies that $\hat\mu_n\to\hat\mu$. There are other sorts of weak convergence, defined by saying $\int f\,d\mu_n\to\int f\,d\mu$ for other classes of functions $f$; one popular version is to consider bounded continuous $f$, which also implies that $\hat\mu_n\to\hat\mu$.
Comment: Oh I see what you mean now. Indeed I was thinking about a different thing. I actually don't know much about measures in general (always used Lebesgue so far), but I think I got the hang of it more or less


How do I extract the v, r, s values from a signed transaction?

Question: If I have a signed transaction such as 0xf86c098504a817c800825208943535353535353535353535353535353535353535880de0b6b3a76400008025a028ef61340bd939bc2195fe537567866003e1a15d3c71ff63e1590620aa636276a067cbe9d8997f761aecb703304b3800ccf555c9f3dc64214b297fb1966a3b6d83 (taken from https://github.com/ethereum/EIPs/blob/master/EIPS/eip-155.md), how do I go about extracting the v,r and s values from this without using some external library?
Answer: This part of the signed transaction contains the v, r, and s fields:
25a028ef61340bd939bc2195fe537567866003e1a15d3c71ff63e1590620aa636276a067cbe9d8997f761aecb703304b3800ccf555c9f3dc64214b297fb1966a3b6d83

We can go through each byte (One byte is two characters in the above transaction)
The values are:
v=0x25=37

r_length=(0xa0-0x80)=0x20=32 (bytes)
r=0x28ef61340bd939bc2195fe537567866003e1a15d3c71ff63e1590620aa636276

s_length=0xa0-0x80=32

s=0x67cbe9d8997f761aecb703304b3800ccf555c9f3dc64214b297fb1966a3b6d83

The encoding of an Ethereum transaction is called RLP and you can find the find it definition in the Ethereum Yellow Paper.
Comment: Brilliant, thank you. Just one more question: How did you know which part of the transaction contained the v,r, and s fields?
Comment: Nevermind, I got it - it's because v is 1 byte, r and s are 32 bytes. Plus the 2 bytes indicating the length of r and s that makes 67 bytes, which is the length of the string you posted. Thanks!
Comment: Yes. But the RLP encoding also shows that the v value has a length of one byte since values from 1 to 127 are simple encoded as their byte value; values above that has a value indicating the length of the number in bytes followed by the actual value. Also, the three last values of a signed transaction are always v, r, s in that order.


How to identify codec (AAC vs ALAC) used in an m4a file

Question: Given that the m4a format is merely a container, and can contain music encoded with a lossy (AAC) or lossless (ALAC) codec, is there a simple way to determine which codec was used for a given file?
Answer: When I use the file command on an .m4a file, I get back a generic response of "ISO Media, Apple iTunes ALAC/AAC-LC (.M4A) Audio" for both AAC and ALAC files. I can think of two other approaches off the top of my head that don't require either the heavy hand of iTunes or installing new software :

Open the audio file in question with QuickTime Player. Select "Show Movie Inspector" from the 'Window' menu (or press ⌘-i) to see the file's size, data rate, and the codec used in its creation.
Use the afinfo (audio file info) command in Terminal.app to get a more information than QuickTime Player provides. Here's some sample output:

$ afinfo /Users/admin/Desktop/01\ The\ Guns\ of\ August\ 1.m4a

Num Tracks: 1 Data format: 2 ch,  44100 Hz, 'alac' (0x00000001) from
16-bit source

4096 frames/packet

Channel layout: Stereo (L R)

estimated duration: 18702.930431 sec audio bytes: 1006059541

audio packets: 201367 bit rate: 430332 bits per second

packet size upper bound: 12334 maximum packet size: 12334

audio data file offset: 48

optimized audio

824799232 valid frames + 0 priming + 0 remainder = 824799232

For any readers unfamiliar with using the terminal, it's quite simple, and you won't break anything. Open Terminal, and type in--or cut and paste from below--the word afinfo followed by a single space:
afinfo

Next, drag an audio file over to the Terminal window and drop it in. You will see the file's path get appended to the afinfo command. If dragging the icon caused the Finder to become the active process, you'll need to click on the Terminal's window to wake it up before the final step of pressing enter to run the command. If you like, you can get information on more than one file at a time by dragging and dropping multiple icons instead of just one. When you're finished, just quit Terminal.app as you would any other program.
Answer: I wanted to do this in order to identify which m4a files in a folder were ALAC and which were AAC. This is easily done to an entire folder like this:
afinfo ./*.m4a |grep -B 4 alac
which gives a result like:
File:           Track 09.m4a
File type ID:   m4af
Num Tracks:     1
----
Data format:     2 ch,  44100 Hz, 'alac' (0x00000001) from 16-bit source, 4096 frames/packet

To just get the filenames, I used this:
afinfo ./*.m4a |grep -B 4 alac |grep File: |sed 's/.*\///'
resulting in:
Track 09.m4a
Comment: @fd0: That wouldn't work, but I think this would work better since it would only output the files that were alac: `for file in *.m4a; do if afinfo "$file" | grep -q \'alac\'; then echo "$file"; fi; done`
Comment: @jjj - Yes, your solution works. You might add the `-b` option to `afinfo` to limit the text output- `afinfo -b FILE | ....`
Comment: A lot of grepping and piping goin' on. Wouldn't `grep -l alac  /path/to/file` work for you?
Comment: @fd0 Good catch!
Answer: I ended up looking for this type of question
as i needed to to look at the top level of my music hard drive to identify
which ones were alac so i could convert them to compressed m4a(aac) for better cross platform capabilities.
I wrote a function that looks recursively at the current folder ( i.e. if you need to go deeper than beyond the working directory you ran it on )
if you need to change the depth, the -maxdepth option can be added to the find command in $m4a_files , as well as specifiying paths to not look into, like i have with ! -path "*.app*" .
I used echo to print the command to a variable and run it via eval because I had trouble here and in the past with other command line tools ( like qpdf ) iterating the output of find into another tool via a for loop.
allu () {

    # alac lookup

    echo
    echo "looking up m4a files"
    echo

    # bypasses paths with .app in their name so if you run it on some place like ~/Downloads
    # and you have an app downloaded there it won't look in it

    m4a_files=$( echo $( find . -name "*.m4a" ! -path "*.app*" | sed "s|.*|\"&\"|" ) );

    # syntax afinfo $m4a_files |grep -B 4 m4a |grep File: |sed 's/.*\///'

    echo 'checking for alac'
    echo

    # looks alac files up
    file_lookup_merger=$( echo "afinfo" $m4a_files "|grep -B 4 alac |grep File: |sed 's/.m4a*\///'" ) ;

    # uses eval to return the list of files it found *with srnd. quotes*
    eval $file_lookup_merger |sed 's/File:           //'| sed "s|.*|\"&\"|";

}

Edit: Come a day later after i wrote the above, I wanted to write a simpler, more cross-compatible verison of this that worked on both mac and linux so I wrote one using Exiftool by Phillip Harvey https://exiftool.org/
( homebrew install )
https://formulae.brew.sh/formula/exiftool#default
# finds alac files

( find . -name *.m4a -exec exiftool -T -audioformat -filepath {} \; )|grep alac

# prints list of alac files for use as a string separated by spaces

echo $( ( find . -name *.m4a -exec exiftool -T -audioformat -filepath {} \; )|grep alac |sed 's.alac..g'|sed "s|.*|\"&\"|")|sed 's." /."/.g'

Answer: If you go to iTunes > View > View Options > Show Columns > Kind it will include a column that tells you if the file is Apple Lossless or AAC


Can you add a wild card to a completed book?

Question: In Hand and Foot, after completing a melded book, can you add a wild card to it?  For example, once you have completed a book, but have a wild card left in your hand, can you add the wild card to a previously completed book, in order to go into your Foot or to go out?
Answer: According to Pagat.com:

A meld cannot contain more than seven cards, and a partnership is not allowed to have two incomplete melds of the same rank, but if you complete a pile, you can then start another meld of the same rank.

So not only can you not add a wild card, you cannot add any card. You'll need to attach it to another meld in progress, or make a new one.


Having trouble understanding part of `'runtimepath'` documentation

Question: I get the basics of the runtimepath as explained in :h 'runtimepath'.  Except for the remark at the end
for Neovim (nearly identical in Vim):
    You can put a directory before $VIMRUNTIME to find files which replace
    distributed runtime files.  You can put a directory after $VIMRUNTIME
    to find files which add to distributed runtime files.

What I don't understand is the sense in which putting a directory before $VIMRUNTIME replaces distributed runtime files. I thought that putting something before in $RUNTIMEPATH just affects the order in which its contents are sourced/loaded.
So could this just be poor phrasing in the documentation or am I missing something?
And in this case, I interpret 'distributed' as meaning 'coming packaged with vim when you installed it' instead of the sense of 'distributed computing' or 'distributed version control'.
Answer: It does affect order loaded; but, most of $VIMRUNTIME files check certain "did_X" guards and bail out. So if earlier files set those flags, $VIMRUNTIME files skip themselves.


Can we dump database of db-synch and use it for initialising a fresh db-synch instance?

Question: Db-sync takes about 48 hours to follow the tip, starting from a fresh installation fully. Is there a way to improve the sync time? Is it possible to dump the database and transfer it to a fresh installation?
Answer: You cannot dump the PostgreSQL directly, as cardano-db-sync depends on other states as well.
You need to use the cardano-db-tool to take and restore the snapshot.
Answer: You can use the official snapshots, provided by IOHK. Just restore them after setting up the database.
They are usually linked in the release description on Github: https://github.com/input-output-hk/cardano-db-sync/releases
You will find a link to the list of snapshots, grouped by schema version.
As of today, the current schema version is 11 and the list of snapshots are listed here: https://update-cardano-mainnet.iohk.io/cardano-db-sync/index.html#11/


How to reduce the indexing of a website on the google?

Question: I have to build two websites: one for the staging and another one for the production. and both are hosted on the server, but when I search on google with the website name, both websites are comes together, and the staging one comes before the production website, I just want that when I find the website by the name only the production one will come, I do not want to show my staging website to visitors. Both websites are built in the next.js. There would be any way to handle to SEO of the production website or reduce the rank of my staging website.
Comment: Can't you just ban google bot in your robots.txt file
Comment: Do the production and staging sites use different host names (different domains or subdomains)?
Comment: But I have already hosted the website and with the name- XYZ-staging.com and xyz.com, so now how what I can do for this ...
Comment: Does it also affect the production site ??
Answer: Long-term solution
(Assuming you are using Apache. Equivalent solutions exist for other web servers.)
Any of these solutions is fine:

Use htpasswd to only expose the website to users with the credentials.
Use the htaccess to limit access to the staging site to a specific IP range.
Use the htaccess to limit access to a specific user agent. Use a browser plugin to change your user agent to this specific one.

These three solutions are all good for privacy and security. You will probably not want anybody to just click around on your staging site. And all three solutions prevent Google bot from crawling your site.
Pick the solution that best fits your use case. If you do not have a static IP address or have a lot of coworker and stakeholders who need access to the staging site, some of these solutions might not be very convenient for you.
Short-term solution
Since Google already indexed your site, you should first do this:

Authenticate your staging site in Google Search Console.
In Google Search Console, manually remove the URLs that Google Bot has crawled so far from the Google Index.
Add the robots-noindex meta tag to all documents on the staging site. That means the next time Google crawls them, they will be removed from the index.

In theory, step 3 alone would be enough. Steps 1 and 2 just help to speed up the process.
Once Google has no more pages of your staging site in the index (or only has them in the index without any content, saying "No information is available for this page."), activate one of the long-term measures.
If none of the long-term solutions work, you could, in theory, stick with having robots-noindex everywhere on the staging site. Google would never show the site in search results. However, your competitors or hackers might find it.
Additional concerns

Make sure the robots-noindex from the staging site never make it to the production site. You would - temporarily - lose all your Google rankings.
Make sure there are no links from your production site to the staging site. You should crawl the production site with a tool like Screamign Frog and make sure all links, canonical, hreflang, alt, etc., always go to the production site and never to the staging site.
Comment: It sounds like part of the question is how do you use the same code for both production and staging while keeping Google back just off just the staging site


For a coaxial rotorcraft, Do horizontal and vertical stabilizers in the tail really make much of a difference?

Question: Can we design a coaxial rotorcraft without tail and tail boom completely? What might be the consequences?
Answer: The horizontal tail surface is there to generate downforce that holds the nose up in cruise. Without it, the faster you go, the more the nose points down as the machine's body wants to "trail" so to speak behind the tilted rotor disc.
The vertical tails are to provide a weathervaning effect to make the machine passively point ahead in forward flight without requiring a torque control input.
Take those surfaces away, and you are stuck with pointing down at an unpleasant angle while cruising, and having to constantly work the pedals to keep it pointed straight. A fly by wire yaw stabilization system working the pedals in the background could probably take care of the yaw part, but I don't know how you could deal with the pitch part without a stabilizer surface.


Unable to load Linux after Windows installation

Question: I installed Linux OS on the PC. After that, I installed Windows 7. Now, I am unable to load Linux again. How to install Grub in Windows 7?
Answer: You'll have to boot to an Ubuntu live CD to reinstall grub.
From the tutorial located here:

Mount the partition your Ubuntu Installation is on. If you are not sure which it is, launch GParted (included in the Live CD) and find out. It is usually a EXT4 Partition. Replace the XY with the drive letter, and partition number, for example: sudo mount /dev/sda1 /mnt.
sudo mount /dev/sdXY /mnt

Now bind the directories that grub needs access to to detect other operating systems, like so.
sudo mount --bind /dev /mnt/dev &&
sudo mount --bind /dev/pts /mnt/dev/pts &&
sudo mount --bind /proc /mnt/proc &&
sudo mount --bind /sys /mnt/sys

Now we jump into that using chroot.
sudo chroot /mnt

Now install, check, and update grub.
This time you only need to add the drive letter (usually a) to replace X, for example: grub-install /dev/sda, grub-install –recheck /dev/sda.
grub-install /dev/sdX
grub-install --recheck /dev/sdX
update-grub

Now grub is back, all that is left is to exit the chrooted system and unmount everything.
exit &&
sudo umount /mnt/dev &&
sudo umount /mnt/dev/pts &&
sudo umount /mnt/proc &&
sudo umount /mnt/sys &&
sudo umount /mnt

Shut down and turn your computer back on, and you will be met with the default Grub2 screen.
Answer: The easiest way to ensure a dual booted machine is to load Windows first, and then load Linux.
This ensures that GRUB will see all the drives and mount points. On first boot, GRUB will load, and then you can then arrow down and choose which OS to boot. Your Windows boot drive will be displayed last in the listing, behind Linux.
Even better, to avoid rebooting your machine to get from one OS to the other, load Windows, download VMware Player (it's free), and then load Linux as a virtual machine (the other way around, is OK too, however the Linux version of VMplayer is a little tricky, so you might want to use VirtualBox).


How to define relation of "bigger" and "less" and not run into a circularity?

Question: Note: In the post by number I mean real number.
What is a definition of the relations "bigger" and "less" ? The definition I saw in my book was "$a>b$ if $a-b$ is a positive number" and "$a<b$ if $a-b$ is a negative number". But the problem with this definition is number $a$ is positive if $a>0$ and similarly with negative number and this definition seems to be kind of circular. So how can we define " less" and "bigger" ?
Comment: There are various ways around it.  The first way is to avoid defining $>$ and $<$ in such a way that it makes any reference to positive or negative numbers (*for example using dedekind cuts*).  Another way is to avoid defining positive and negative with any reference to $>$ or $<$.  The set of positives, $P$, can be defined as the set satisfying the properties: i) $x,y\in P\implies x+y\in P, x\cdot y\in P$, ii) $x\in P, y\notin P\implies x\cdot y\notin P$, iii) $x,y\notin P$ and $x\neq 0,y\neq 0\implies x\cdot y\in P$, and iv) $1\in P,0,-1\notin P$.
Comment: @Acccumulation: I don't believe JMoravitz claimed to be able to do that.  They merely said they could define Dedekind cuts without using *positive and negative numbers*, which is much easier since Dedekind doesn't care about zero in particular.
Comment: What the set of numbers you are working with?
Comment: @Ennar Real numbers, I will note that in the post.
Comment: Are you sure that this was meant as definition? How are the reals constructed?
Comment: You might be interested in looking at an axiomatic development of the real number system, such as the presentation in Spivak's book Calculus.
Comment: @littleO OK, thanks, I will look at it.
Comment: @Ennar I am not very familiar with the construction of reals.
Comment: @JMoravitz Can you give a definition of "Dedekind cut" that makes no reference to an order?
Comment: @EricDuminil Sorry, what are you talking about ?
Answer: The axioms for the real numbers provide one way to settle this issue. The axioms are broken into two portions: the field axioms concerning algebraic properties of the binary operations $+$ and $\times$; and the order axioms which require the existence of a total order $x<y$ satisfying three axioms:

if $x < y$ then $x+z < y+z$ for all $z$;
if $0 < x$ and $0 < y$ then $0 < x \times y$.
The completeness axiom.

You can then define the positive numbers to be all numbers $x$ such that $0<x$.
On the other hand you don't have to be satisfied with the axiomatic approach to the real numbers themselves, because the real numbers can be constructed from the rational numbers, which can be constructed from the integers, which can be constructed from the natural numbers, which satisfy Peano's Axioms. This is a long drawn out story, but the salient point for your question is that in Peano's Axioms for the natural numbers there is no need to even mention the concept of positive or negative numbers, but on the other hand the order relation on natural numbers can be defined like this (addition and multiplication having already been defined):

Given natural numbers $m,n$ define $m < n$ if there exists a nonzero natural number $k$ such that $m+k=n$.

Once that task is accomplished, one embeds the natural numbers into the integers and defines order similarly:

Given integers $m,n$, define $m<n$ if there exists a nonzero natural number $k$ such that $m+k=n$.

Then one embeds the integers into the rationals and defines order on the rationals:

Given rationals $r=\frac{a}{b}$, $s=\frac{c}{d}$ where $a,c$ are integers and $b,d$ are nonzero natural numbers, define $r < s$ if $ad<bc$.

Finally, one embeds the rationals into the reals and defines order on the reals:

Given reals $x \ne y$, define $x < y$ if there exist Cauchy sequences of rational numbers $(r_n)$, $(s_n)$ converging to $x,y$ respectively, such that $r_n < s_n$ for all $n$.
Answer: It all depends on how you define the real numbers. The important thing is that you can define "positive" without any reference to $<$.
In the Dedekind cuts approach, where a real is a pair $(A,B)$ of subsets of $\mathbb{Q}$ such that $A\sqcup B = \mathbb{Q}$ and $\forall a\in A, b \in B, a<b$, then a real $(A,B)$ is said to be positive if and only if there exists $a\in A, a>0$ (note that this $>$ is the ordering on $\mathbb{Q}$, so it's not circular).
In the Cauchy sequences approach, where a real is an equivalence class of Cauchy sequences in $\mathbb{Q}$ (the equivalence relation is induced by the ideal of sequences converging to $0$), then a real $[(q_n)]$ is said to be positive if there is $\epsilon >0$ (again, this is in $\mathbb{Q}$ so no circularity here) and $n$ such that $\forall k \geq n, q_k \geq \epsilon$ (again, in $\mathbb{Q}$). One easily checks that this doesn't depend on the choice of representative of the class $[(q_n)]$, and hence yields a sound definition.
There are of course other approaches to defining the real numbers, and they all come with different ways of defining positivity.
What matters is that the ordering always comes from the ordering of $\mathbb{Q}$, which has been defined before, so it's never circular. Now one might ask how the ordering on $\mathbb{Q}$ is defined. And once again, this depends on the definition of $\mathbb{Q}$.
For instance, defining $\mathbb{Q}$ as $\mathbb{Z}\times \mathbb{N}^*$ modulo the equivalence relation $(a,b) \sim (c,d) \iff ad= bc$, then you can say that the class of $(a,b)$ is positive if and only if $a>0$ (this is in $\mathbb{Z}$, so no circularity here). One again checks that this doesn't depend on the representative, and so this is indeed well defined. Once you have the definition of positivity you can define an ordering as you mentioned in the beginning of your post.
It remains to define an ordering on $\mathbb{Z}$, and similarly this can be done without circularity by using the ordering on $\mathbb{N}$ (of course it also depends on the definition of $\mathbb{Z}$ one is using).
The last step is defining an ordering on $\mathbb{N}$. Obviously this also depends on the definition of $\mathbb{N}$ , but it's most usually defined as the first infinite ordinal (existing because of the axiom of infinity) and so it comes with a "natural" ordering : $\in$. Indeed, in this definition, $0 := \emptyset, 1:= \{0\}$ and more generally, $n+1 := n\cup \{n\}$ and so $n=\{0,1,...,n-1\}$, so $<$ coincides with $\in$. Now $\in$ (membership) is a primitive notion in set theory and so we don't define it.
Comment: With the $\mathbb{N} = \omega$ approach, we do define *equality* via the axiom of extensionality.  OTOH, the Peano axioms consider equality primitive and have to define $<$ in terms of successorship or addition instead.  It's all a matter of perspective.
Comment: @Kevin : well it depends on the person you're asking. Some take $\in$ *and* $=$ to be primitives of set theory, and some indeed define $=$ via extensionality. Similarly for Peano, some authors take $<$ to be a primitive; I don't know whether there's an overwhelming consensus on these matters
Answer: $a\ge b$ if there is a number $c$ with $a=b+c^2$
Answer: The positive numbers are those numbers which are the square of some other non-zero real number. The negatives are the rest (except zero).
Comment: Then how you prove that you can't express -1 as a square of real number ?
Comment: @ЮрійЯрош This depends on the axioms you are given and what is your definition of $-1$. I would say $-1$ can be defined as the additive inverse of the multiplicative neutral element (if these terms are defined). If your axioms are strong enough to prove that $1/x$ is never the same as $-x$, then I can prove it. But as I said, I don't know your axioms.
Answer: You can't.
An order an a set is any relationship with the symbol "$<$" so that 1) at most one the following its true: $a < b; b < a$ or $a=b$ for any $a,b$ in the set. 2) if $a < b$ and $b < c$ then $a < c$.
For example for $\mathbb N$ we can define $a < b$ if $a$ is a non-trivial multiple of $b$.  1) Given $a, b \in \mathbb N$ at most $a = b$ or $a = k*b; k > 1$ (as $a < b$) or $b = k*a; k > 1$ (so $b < a$).  Only one, or none, of those can be true. and if $a < b$ so $a = k*b$ and $b< c$ so $b = m*c$ then $a=(mk)*c$ so $a < c$.
....
But, for the one we know and love,  we define the rationals $\mathbb Q$ to be an "ordered field" which means:
For any $a,b\in \mathbb Q$ there are two operations: multiplication and addition so that $a+b \in \mathbb Q$ and $a*b\in \mathbb Q$.
A1) $a+b = b+a$.
A2) $(a+b) +c = a+(b+c)$.
A3) There is an element called $0$ so that $a+0 =a$ for all $a\in R$.
A4) For every $a$ there is an element called $-a$ so that $a + (-a) = (-a) + a = 0$.
M1) $ab = ba$
M2) $a(bc) = (ab)c$
M3) There is an element called $1$ so that $a*1 = 1*a = a$ for all $a$.
M4) If $a \ne 0$ then there is an element called $\frac 1a$ so that $\frac 1a*a = a*\frac 1a = 1$.
D) $a(b+c) = (ab) + (ac)$
O1)There is a "total order" so that for $a,b$ exactly one of the following is true: $a < b; b < a$ or $a=b$
O2) If $a < b; b< c$ then $a< c$.
O3) If $a<b$ then for any $c$, $a + c < b+c$.
O4) If $c > 0$ and $a < b$ then $ac < bc$.
With those axioms we can prove all sorts of stuff, the most pertainent being:
Claim:  $1 > 0$.
Pf:  $1 \ne 0$ so either $1 > 0$ or $1 < 0$.  If $1 < 0$ then $0 < 1 +(-1) < 0 + (-1) = -1$.  Then $(-1)*(-1) > 0$.
Lemma: $0*a = 0
Pf:  $0*a = (0+0)*a = 0*a + 0*a$.  So $0 = 0*a + (-(0*a)) = 0*a + 0*a + (-(0*a)) = 0*a  + 0 = 0*a$.
Lemma: $(-1)*(-1) = 1$.
Proof $(-1)*(-1) + (-1)=(-1)*(-1) + 1*(-1) = (-1 + 1)(-1) = 0*(-1) = 0$
$ (-1)*(-1) =(-1)*(-1) + 0 = (-1)*(-1)  + (-1) + 1 = (-1)*(-1) + 1*(-1) + 1 = (-1)(-1 + 1) + 1 = -1*0 + 1 = 0+1 = 1$.
So we have $(-1)*(-1) = 1 > 1$ which is a contradiction.
So $1 > 0$.
From there you can  determine which rational numbers are greater than $0$ and which aren't.
Answer: The definition of less-than via positive numbers is only circular if you in turn define the positive numbers in terms of less-than. But that is not the only way to define the positive numbers. Another way to define then is as follows:
We define a positive cone as a set $P\subset\mathbb R$ with the following properties (the exact axioms vary, but apart from the fact that sometimes $0$ is included in the positive cone, all those definitions are equivalent):

$P$ is closed under addition and multiplication (that is, if $a\in P$ and $b\in P$, then also $a+b\in P$ and $ab\in P$).
If $a\in P$, then $-a\notin P$.
If neither $a\in P$ nor $-a\in P$, then $a=0$.

A number $a$ is then called positive if $a\in P$.
Note that this definition does in no way refer to a relation "less than" or "greater than". Therefore it is now in no way circular to define $a<b$ as "$b-a$ is positive", that is, $a<b :\iff b-a\in P$.
For the real number, we want also have completeness, which can also be expressed using the positive cone. We call a set $S\subset\mathbb R$ upper-bounded if there exists a number $b\in\mathbb R$ so that $\{x-b|x\in S\}\cap P=\emptyset$. A number $b$ with this property is then called an upper bound. Now $b$ is called a least upper bound if there is no $c\in P$ such that $b-c$ is also an upper bound. Completeness now demands that every upper-bounded set has a least upper bound.
One can indeed easily show that with this definition, you get a strict total order with the additional properties that one demands when defining "$<$" directly.

Transitivity: If $a<b$ and $b<c$, then $a<c$.
Proof: $a<b$ means $a-b\in P$, $b<c$ means $b-c\in P$. But then, $a-c = (a-b)+(b-c)\in P$ because $P$ is closed under addition.
Trichotomy: For any $a,b\in R$, exactly one of $a<b$, $a=b$ or $b<a$ is true.
Proof: Either $a-b\in P$, then by definition $a<b$. Otherwise, either $b-a\in P$, then by definition $b<a$. Otherwise since neither $a-b\in P$ nor $b-a=-(a-b)\in P$, then by definition of $P$, $a-b=0$, that is, $a=b$.
If $a<b$ then $a+c

Proof: $(a+c)-(b+c) = a+c-b-c = a-b \in P$
If $0<a$ and $0<b$ then $0<ab$
Proof: $0<a$ means $a = a-0 \in P$, same for $0<b$. Because $P$ is closed under multiplication, $ab\in P$, therefore $0<ab$.
$b$ is an upper bound to $S$ according to the definition above iff it is an upper bound according to the usual order definition.
Proof: Be $b$ an upper bound as defined above, that is, $\{x-b|x\in S\}\cap P=0$. Then we have:
\begin{align}
\{x-b|x\in S\}\cap P=0
&\iff \forall x\in S:x-b\notin P\\
&\iff\forall x\in S: x=b \lor b-x\in P\\
&\iff\forall x\in S: x=b \lor x<b
\end{align}
But that is exactly the definition of an upper bound for an ordered set.
$b$ is a least upper bound according to the definition above iff it is a least upper bound according to the usual order definition.
Proof: Assume $b$ is an upper bound. Then it is a least upper bound according to the above condition iff there exists no $c\in P$ (that is, $0<c$) such that $b-c$ is also an upper bound. Note that the equation $x=b-c$ can always be solved uniquely both for $x$ given $c$ and for $c$ given $x$. Moreover $0<c\iff -c<0 \iff x=b-c<b$. Thus the condition is fulfilled iff there exists no $x<b$ that is also an upper bound. But that is exactly the condition for $b$ being a least upper bound in the usual order definition.
Completeness according to the definition above is equivalent to order-completeness
This is evident from the fact that the definitions of upper bound and least upper bound agree, and the formulation in terms of those concepts is the same.


windows volume shadow copy recover entire volume

Question: For those that don't want the long story, I'll start with my questions:
Has anyone done a VSS entire volume revert of a large volume (over 8TB) before?  And if so, how long should I expect it to take? If it matters, assume the difference between the snapshot and current state is in the 500GB - 1TB worth of change.  Does windows set the volume up as inaccessible as it's performing the revert?  Is there anyway to tell whether VSS is actually still involved in reverting the volume?
For those interested in the background:
I have a Windows 2012 server that hosts the backups of my VM infrastructure.  Power got cut while running a backup job to it and the underlying datastore holding those backups got corrupted.  The backup software would no longer work with the repository.
There is a VSS job on that volume that snapshots nightly.  So figured no big deal.  I mounted the VSS snapshot and started the recovery of the backup repository.  Unfortunately, the backup repo is relatively large (over 4TB) and the restore process was looking like it was going to take about four days to finish.  I let it run overnight.
Today, I check in on things and the restore of the files is still running fine.  But being the patient person that I am, I decided that I didn't want to wait another couple of days and I might as well just revert the entire volume.  So, I went in to the properties of the volume,  shadow copy config, selected the shadow copy I wanted to revert to and went with Revert.
The Revert dialog came up warning that was irreversible.  I said ok.  The revert dialog went away a few seconds later and now trying to access the volume, I get an access denied message and properties of the volume shows it as 0 bytes.
It makes a certain amount of sense that Windows would take the volume offline and make it inaccessible as it was doing the revert.  But it's been about 30 minutes now and I don't know if it's doing anything or whether at this point the volume is a lost cause.  Checking the resource monitor I do see storage activity on the volume that is inaccessible but am unsure of a way to verify if it's activity due to the revert process.
I actually still have the shadow copy of the volume I'm trying to restore mounted (not actively accessing it but the data in the snapshot is still accessible through it).  Is it possible that's causing a problem with the revert?  I don't want to unmount it if unmounting it won't help the revert. As worst case, I could spend a week or so copying it somewhere else.
So back to my questions:

Is this revert behavior normal? Does windows mark the drive as inaccessible as it reverts?
Should I expect the revert process to take a long time?
Can I tell if Windows is still involved in reverting the volume?
Will having the snapshot that is being reverted to mounted cause a problem with the revert?
Answer: The volume came back online early this morning.  Just in case someone else comes across this:
The revert behavior appears to be normal.  Windows will set the volume inaccessible (access denied) and the volume will appear as 0 bytes while in the process of reverting.
The process can take a long time.  It took close to 20 hours in my case.  Not sure if it's a function of how many changes occurred after the snapshot but would guess that's likely the case.  Not going to do a test to verify.
Windows Resource Monitor will show activity on the volume being reverted-  seems like there should be a better way to monitor it but resource monitor works.  It's what gave me hope that it will still in the process of reverting.
Having the snaphot mounted didn't prevent it from being reverted.


Are there "unsociable" irreps? (Definition inside)

Question: If, with a bit abuse of notation, $U\notin R\bigotimes{R}\bigotimes{R}\bigotimes{...}$ (i.e., regardless how long you clebsch up $R$, $U$ won't appear in the expansion), I call $U$ unsociable with respect to $R$ (in the group $G$). Clearly everything else is unsociable with the $1$ irrep, since (duh) $1\bigotimes{1}\bigotimes...=1$. For finite groups there can be nontrivial (ahem) cases: For $G=C_{3v}$ (the symmetry group), $A_2\bigotimes{A_2}=A_1$ and thus $E$ is unsociable w.r.t. $A_2$.
But what is the deal with Lie groups? E.g. is the defining irrep of $E_7$ unsociable w.r.t. the adjoint one?
Comment: I've heard (but not understand) that cuspidal irreps (of say G(F_q)) are those which you cannot get from tensoring principal series... So might be cuspidal are unsociable with principals... I would be happy if some expert can comment more on this...
Comment: @Alexander: As Burnside's theorem shows, your recollection is not quite correct.   Essentially, the cuspidal characters of a finite group of Lie type are elusive because they don't show up in the ordinary process of induction from easy characters of parabolic subgroups.   This is what prompted the more sophsiticated approach of Deligne-Lusztig.
Comment: @Hauke: Be careful about the language "defining irrep" here, which isn't helpful. Whether the notion of "unsociable" will be helpful I won't try to predict.
Comment: @Jim Humpreys You are right, thank you ! I misunderstand the claim. It is claimed that cuspidal does enter decomposition of induced reps coming from tensor product of subgroups... http://www.imsc.res.in/~amri/html_notes/notesch4.html#x7-260004.1 "4.2. Cuspidal representations.
The cuspidal representations of GLn(Fq) are those which are disjoint from all representations of the form π′∘ π′′, where π′ and π′′ are irreducible representations of GLn′(Fq) and GLn′′(Fq), where n = n′ + n′′ and n′ and n′′ are both positive."
Comment: I mean cuspidal does NOT enter...
Answer: It follows from a theorem of Burnside/Steinberg that if G is a finite group then the tensor powers of a module V contain all irreps if and ony if V is faithful for G.
Comment: Or, to put it another way: an irrep $U$ occurs in some tensor power of $V$ if and only if $\ker(V) \leq \ker(U)$. In Isaacs' character theory book, this theorem is attributed to "Burnside-Brauer", by the way.
Comment: http://en.wikipedia.org/wiki/Faithful_representation
Comment: Burnside did it over C and R. Steinberg (no relation to me) did it for semigroups over any field. Rieffel extended it to bialgebras.
Comment: THX! So it's even in the Wiki? :-)
In any case, since by accident I found LiE (http://www-math.univ-poitiers.fr/~maavl/LiE/form.html) I never have to annoy MO with
nagging questions for practical details of Clebsch-Gordan series again. Or so I hope :-)
Answer: This is meant as an extended comment (sometimes correction) to things said in various answers and comments.   As a public service I'll try to fill in some of the history and references.  I recall some of this coming up previously on MO (exercise: find it).
In the framework of representation (or character) theory of a finite group $G$ over $\mathbb{C}$, Burnside led the way in his 1911 treatise (Chap. XV, Thm. IV): given a faithful representation of $G$, all irreducible representations (or characters) occur in some tensor power.    A proof of this is given in the 1962 Curtis-Reiner book (32.9); a later addendum to that book refers also to the slightly later work of Brauer and his student Steinberg.   Note that faithfulness is an obvious necessary condition here, not emphasized in most sources.
Steinberg, Complete sets of representations of algebras, Proc. Amer. Math. Soc. 13 (1962), 746-747, offered a simpler proof of Burnside's theorem.  Then Brauer, A note on theorems of Burnside and Blichfeldt, Proc. Amer. Math. Soc. 15 (1964), 31-34, provided his own version of Burnside's theorem with a number of nice refinements: limiting how many tensor powers are needed, limiting the size of the characteristic 0 field, commenting on what remains true in prime characteristic.   Both of these papers are available online at the AMS website.
Meanwhile, in his 1946 book Theory of Lie Groups (Chapter VI), Chevalley treated compact Lie groups and essentially provided a variant of Burnside's theorem.   But here it's essential to use both the faithful representation and its contragredient (dual).   The proof is in the context of the ring of representative functions, Tannaka duality, Peter-Weyl theorem.
In a later short note, G.I. Lehrer provided a much more algebraic proof of Chevalley's theorem (again for compact Lie groups):  Produits tensoriels de grepresentations de groupes de Lie, C.R. Acad. Sci. Paris 275 (1972).   He also refers to Steinberg's paper.
From compact Lie groups it's not so far to the finite dimensional representation theory of a semisimple Lie algebra or algebraic group in characteristic 0.  But Burnside's theorem and its variants take their sharpest form for groups, whereas a single Lie algebra may belong to simple Lie or algebraic groups or the same Lie type but different isogeny type.    This conceals the requirement that one start with a faithful representation.
Comment: Solution 1 to exercise: http://mathoverflow.net/questions/10126/reference-for-this-theorem-in-representation-theory

Solution 2: http://mathoverflow.net/questions/18194/faithful-representations-and-tensor-powers
Comment: @Jim, M. A. Rieffel, Burnside's theorem for representations of Hopf algebras, J. Algebra 6 (1967), 123–130 proves the analog for finite dimensional bialgebras. Both Steinberg and Rieffel work over any characteristic because they don't rely on characters.

Steinberg really proved his result for semigroups (not necessarily finite). He showed if V is a faithful module for the semigroup S, then the tensor algebra of V is faithful for the semigroup algebra of S. If S is finite one immeiately deduces each irreducible is a constituent of a tensor power.
Comment: @Benjamin: Excellent solutions (but the question won't be on the exam).
Answer: Let $G$ be the simply connected group of type $E_7$. Then its center is $\mu_2$, the cyclic group of order $2$. Of course, it acts trivially on the adjoint representation, hence the same holds for all components of all tensor powers of the adjoint. But the action of the center on the defining representation is nontrivial, hence it is unsociable with the adjoint.
Answer: The proof of Burnside-Brauer adapts easily to give the following result for Lie groups:

Let $G$ be a compact semi-simple Lie
group and let $R$ be a faithful irrep
of $G$. Let $U$ be any irrep. Then $U$
occurs in $R^{\otimes M}$ for some
$M$.

Some preliminary comments:
Since $G$ is compact, the Lie algebra of $G$ is the direct sum of a semi-simple Lie algebra and an abelian Lie algebra; so the semi-simple hypothesis just says that there is no abelian summand.
Since $G$ is compact,  we may assume that $R$ is a unitary representation, by the standard averaging argument. Since the representation is faithful, we may thus view $G$ as a subgroup of a unitary group. Let $n= \dim R$ and consider $G$ as a subgroup of $U(n)$ from now on.
Let $Z$ be the center of $G$. By Schur's lemma, $Z$ acts on $U$ and $R$ by scalars; let $\psi_R$ and $\psi_U$ be the characters by which $Z$ acts on $R$ and $U$. Since $G$ is semi-simple, $Z$ is discrete. Now, $Z$ lies in the center of $U(n)$, which is $S^1$, so $Z$ must be a cyclic group; say $Z \cong \mathbb{Z}/m$. Let $\psi_U = \psi_R^h$. What we will actually be showing is that $\mathrm{Hom}(U, R^{\otimes (h+mN)})$ is nonzero for all sufficiently large $N$.
Proof sketch: Let $\mu_G$ be Haar measure, normalized so that $\int_G \mu_G=1$. Then
$$\dim \mathrm{Hom}(U, R^{\otimes (h+mN)}) = \int_G \chi_R^{h+mN} \overline{\chi_U} \mu_G$$
where $\chi_R$ and $\chi_U$ are the characters of $U$ and $R$.
Since $\chi_R$ is the character of a subgroup of $U(n)$, we have $|\chi_R(g)| \leq n$, with equality only when $g$ is of the form $e^{i \theta} \mathrm{Id}$. Such $g$'s are precisely the elements of $Z$. So, for large $N$, the integral is dominated by the contributions from small neighborhoods of the points of $Z$. For $z \in Z$, we have $\chi_R(z)^m = n^m$ and $\chi_r(z)^h \overline{\chi_U}(z) = n^{h+1}$. If you work out the asymptotics of the integrals, using the method of steepest descent, you get the contribution from each $z \in Z$ is of the form
$$c n^{mN} N^{-\dim G/2} (1+O(N^{-1/2}))$$
for some $c>0$.
In particular, for $N$ large, all these contributions are positive and $\dim \mathrm{Hom}(U, R^{\otimes(h+mN)})$ is nonzero. $\square$
Commment 1 I'd love to know whom to credit for this trick. I came up with it in this blog thread and used it again here but I'm sure it's not original to me.
Comment 2 If $G$ does have an abelian factor, life is harder. The obvious counterexample is that, if $G$ is $S^1$, the irrep $R$ is $\theta \mapsto e^{i \theta}$ and $U$ is $\theta \mapsto e^{-i\theta}$, then $U$ is not a summand of $R^{\otimes n}$ for any $n \geq 0$. More subtly, let $G = U(3)$, let $V$ be the standard three dimensional irrep and let $R=\mathrm{Sym}^2 V$ and $U = \bigwedge^2 V$. Then $U$ is not a summand of $R$, and the matrix $e^{i \theta} \mathrm{Id}$ acts on $R^{\otimes n}$ by $2n \theta$ and acts on $U$ by $2 \theta$, so $U$ is not a summand of $R^{\otimes n}$ for $n>1$.
Comment 3 If $G$ is a complex semi-simple group, the same result holds for $G$ by the standard nonsense relating complex and compact representations.
Comment 4 If $G$ is a real semi-simple group, life is more confusing. For example, let $G = SL_3(\mathbb{R})$, let $V$ be the standard three dimensional rep of $G$, and let $R = \mathrm{Sym}^3 V$. Then $G \to GL(R)$ is injective, but the corresponding complex representation $SL_3(\mathbb{C}) \to GL(R \otimes \mathbb{C})$ is not. One should be able to formulate a statement here with a bit more care, but I'm not going to do it.
Comment 5 I have no idea what the answer is if $G$ is not semi-simple.
Comment: Instead of assuming $G$ semisimple you can assume (after assuming WLOG that $R$ is unitary) that the image of $G$ in $\text{Aut}(R)$ is contained in $\text{SU}(R)$. Then $R^{\ast} \cong \Lambda^{d-1}(R)$ where $d = \dim R$ and your argument from the other thread works.
Comment: @David: Note my reference to Chevalley, where the dual representation is also used.   Your approach will need this extra ingredient at some point.  Concerning your Comments 4 and 5, I'm unaware of further work in those directions (especially the last).
Comment: Putting in the dual rep will allow you to discard the semi-simple hypothesis (and replace it by either reductiveness or compactness), see my answer to http://mathoverflow.net/questions/58633
Comment: I agree that looking at $R^{otimes m} \otimes (R^{\vee})^{\otimes n}$ is more useful than looking just at $R^{\otimes m}$, but its not what the OP asked for.


How can I effectively research for a high-fantasy setting?

Question: When you write in a modern day setting, you research the culture, climate, location and history of that setting. You can do the same for historical fiction or low-fantasy settings based on real places.
When you write for a high-fantasy setting none of this information exists. The history, culture, environment and peoples are entirely your own creation. How can you effectively use research to help create interesting high-fantasy settings?
Some methods I have used include:

Read a lot of other works in the genre. Borrow the things I like.
Base parts of the setting on real world cultures or locations.
Do no research and make it up as I go.

The problem with the first two methods is that they often result in unoriginal/boring settings that have been done to death. The third method is a lot of work and often ends with plot holes or inconsistency in world-building. Any advice people can give on how to use research to improve my high-fantasy settings is useful.
Comment: It all (imho) boils down to two pathways: 1) Do research to get creative inspiration; 2) Do research to validate/substantiate your creative concept.
Comment: I'm asking because those two pathways imply a very different kind of research, really a two different questions. I can't answer #1, but may be able to make a few suggestions for #2.
Comment: @Alexander Both would be valid answers to this question. If you can support it and it's about research techniques to improve a fantasy setting it's a good answer.
Answer: I think for option 2, the problem is if you research the same things everyone else researches, your fantasy will reflect that, and thus have...

unoriginal/boring settings that have been done to death

Can you research other cultures?  For example, in answering someone's question about fire/ice and elements, I found this site https://fireupwaterdown.com , and from there, let's look at this post: https://fireupwaterdown.com/2015/10/14/dragons-a-force-of-nature/
Dragons aren't always fire, some are wind or rain.  I think part of what made The Never Ending Story by Michael Ende stand out to me was the Luck Dragon -- instead of a typical European Dragon, Ende brought in a more eastern conception.
A cultural-sensitivity reader/advisor is a good idea -- you want to be careful you're not fetishizing the "exotic" or doing things that are done to death in other places.  (Killing your particular superhero's girlfriend may make sense for the story as it is, but DO note that it's part of a larger pattern of killing superhero's girlfriends.)
But picking a few elements from another culture and having them be baseline in your fantasy may allow great new combinations.  I highlight that word, because I feel that most originality is about mixing the right things together in a new way.
Black Panther, besides being part of the Marvel (fantasy) cinematic universe, also postulated an African nation without a colonialist past.   The designers used elements from the real countries of Lesotho (blanket design) and Congo (rare minerals). Each tribe within Wakanda was based on details from real, sub-Saharan nations.  Things were mixed, but with thought and specifics, not a random mish-mosh. (I'm white & American, so I may be missing many nuances, of course.  I'm an optimist though, and assume that this was all respectfully done.)
So my advice for research is to go farther afield than you expect.
Answer: When making up a fictional story setting, there is always something that you have seen or read somewhere else that led to inspiring you for making that place up.
The points that you have discussed are true. If you want inspiration, you need to look somewhere for ideas. You might not take everything. Maybe you like the scenery someone has described, but their culture seems boring to you. You might like another culture described by someone else.
Make points of what interests you. Assemble them together. Put in your own imagination. Make a mix of other's and your ideas.
That's the beauty of fiction. There is no wrong as long as it is well explained.
And what you said about plot holes is true, but once you write the first draft, you can always get a few test readers to read your writing and point out what seems off to them. In that way, you can keep filling up the holes until there are no more left. :)
(I'll always suggest to let someone else test read your work instead of doing it yourself, because you rarely find the blunders that you have made in your work, as your brain has made up excuses for them to be correct already.)
Answer:
Base parts of the setting on real world cultures or locations.

Unless you have near-infinite time to do your worldbuilding, you're probably going to keep coming back to our world.  So start there!  Not in the sense of "modern-day Earth with the serial numbers filed off", but in the sense of specific examples that are already fairly detail-rich.  For example:

If you need to show a language, think about what broad properties you want it to have, look for a known language (could be historic) that has many of those properties, and study that as a baseline, then adapt.  Instead of choosing between a deep dive into constructed languages or a dozen phrases with no clear grammar structures, adapt something that already has vocabulary and grammar and sounds to your liking.
If you need a physical setting, learn from ones that already exist.  Your story is set in the molten lava fields of the salamanders' homeland?  What details can you glean from volcanoes here on Earth, both when they erupt and when they simmer for decades without erupting?
If you're developing a new culture, what are its defining elements and do you know any cultures that share some of those elements that you can study?  How do people in nomadic hunter-gatherer tribes manage childcare?  What consequences do you see in the economies of nations that are always at war?

It's fun to reason out things like this, and I can suggest a site that can help with everything from orbital mechanics to centaur clothing, but you probably don't have time to work it all out for yourself, and there are probably some analogues already here in this world that you can look to for inspiration and some implementation details.
Answer: Warning: this answer links repeatedly to TV Tropes. Don't let this happen to you.
Since all three methods you mentioned have drawbacks, I'll try to suggest tweaks to them.
Let's start with reading the genre, then borrowing what you like. This amasses facts of the form "when it comes to X, story Y does it in way Z". Apart from your concern that this is boring or unoriginal, reading the works themselves to get these facts can be very inefficient. Now, writers should read widely, of course; but this is not, IMHO, one of the reasons to do it (at least, not if by "read" you mean "read works" rather than "read about the works", an underappreciated item in the writer's toolkit).
Instead, go through these pages' examples regardless of medium and genre, Ctrl+Fing for these keywords; and rather than borrowing these ideas wholesale, think about what assumptions you've made they challenge, then see what that helps you come up with. That'll hopefully make it much more original, much more interesting. The best litmus test you'll have access to is whether it interests you.
I contest that real-world settings have to be boring or unoriginal, but it depends on which ones you pick. Guess what: we all (think we) know ancient Rome. There have been so many cultures in history, even in one region over time. The world's empires alone will give you all the fodder you need for something unlike anything you've ever read. So will the world's religions. I know I keep mentioning Coco, but I learned so much from that film about both certain afterlife beliefs and real Mexican culture.
Having said all that, one culture may not be enough for you. You're trying to write a story that makes internal sense, and cultures often don't. Greek myths, for example, are incredibly inconsistent. (Usually if I look up one of their deities' parents, I discover sources differ.) Cut and change what doesn't work, and borrow from a second culture, real or imaginary, if you have to.
For the third option, work smarter, not harder. If your creative focus is on characters first, and you reveal only what you have to about their culture for their decisions and misadventures to make sense, you'll have your best chance of having a first draft that makes sense. Now, let me be clear: you can come up with a Hemingway-style iceberg early on, maybe even spend the first few pages fleshing it out. You will excise it, just as you'll excise the occasional inconsistency you encounter in a read-through. Nine times out of ten, you'll only need to delete or rewrite one sentence or paragraph to fix it anyway.
Even fantasy is, to some extent, like realty unless noted. Sometimes you'll realise the people in your world couldn't do A the way we do because of B; on other occasions, you'll realise they would in fact do C like us, because people aren't (too) stupid or evil (or smart or good, depending on the parallel).
Above all, think about the tech and values of your culture. Have they legalised same-sex marriage, and did they read about the ruling on their smartphones? Feel free to move forward or backward relative to our timeline, of course. Fix what bugs you about the conventions of the genre.
Answer: If you are writing via planning and worldbuilding, your creative process goes like this:
idea => concept => detailed plot/setting
Research can be (and should be) used in every step of the process. Your methods #1 and #2 seem to imply that this research should predate the "idea" step and should be a source of inspiration. I can not really suggest anything there. Anything that gives the writer new ideas is a good thing, but this is highly personal.
A different kind of research is needed when an idea is developed into a concept, and then concept serves as a stage for specific plot developments. Obviously, you would want to avoid any holes in concept and in plot, and likely want to avoid any striking similarities with existing stories. So, after you have your idea, #1 question is
Have this been done before?
If you are well familiar with your genre, chances are that you immediately can pinpoint other works where something similar had been done. Anyways, it would be good to do some Google or TV Tropes search to see how original your idea really is. The outcome could be different:

Looks like my idea is very original. Hooray!
Other people have done similar things, but my idea is really
different;
My idea is not new at all. Duh!

Even if your idea is not new, that's not necessarily bad. In fantasy genre the setting is important, but the story is the most important part. You may use cookie-cutter elves, dwarves and dragons if you like. If your story is original, it would be perfectly fine. Conversely, if your story is a collection of well-worn tropes, fresh and original setting may not be enough to save your book.
If your idea is original enough, the next question is
Does this concept make sense?
As you flesh out your world, you need to understand how this novel idea would work out in everyday life. It's easier if the idea is not very novel, because other people have already done the work of rationalizing the concept, or supplying some "standardized" magic to make some non-scientific concepts (like flying and fire-breathing dragons) to work without readers asking too many questions. If your idea is very novel, you need to make it work with the minimum amount of necessary explanations. For example, if levitation magic is widespread in your world, it would be natural for different creatures and people to fly. Otherwise, the reader would be striving for an explanation.
I can recommend to get familiar with Brandon Sanderson's First Law, Second Law and Third Law of Magic. Also, other answers have already recommended the sister worldbuilding site where you can find help polishing your concepts.
After you are satisfied with your concept, you can move to the plot. And here lies the next question:
Do I need to tweak my concept to make the plot work?
Here is the area where (imho) most promising, but less than perfect stories are suffering. This problem is by no way limited to the fantasy genre, but it is particularly common here. You need some magic to move your plot and resolve your conflicts, but you don't want it to look like you are inventing new things as you go just to push the plot in the right direction. Again, I can recommend reading Brandon Sanderson's opinion here. It is Ok to encounter new things while on a journey, but it's not Ok to create some reader's expectations and continuously smash them by introducing more and more new things.
Answer: The main thing with High Fantasy or Super-tech science fiction is internal consistency and analogy.
Since you will be dealing with stuff that has no real world direct matches, figure out how the magic/tech/psionics/etc. (I will just use the word, "magic", to represent all of this) work in your story.  Make sure they are either consistent or purposely inconsistent (with how/why well thought out).  Then think of what kind of world would create that magic or be created by that magic.  How would life be affected by magic or affect magic (or effect magic :-) ).  How would societies be affected?
As a shortcut, you could pick current or historic societies as the base society that you then modify.  If you want the story to be approachable and easy to research (as well as taking the very well tread path) base the society off of medieval Europe or Ancient Rome.  If you want to do a lot of research, pick a group like the Inuit. Then figure out how things will change due to the magic.
Then use real world analogs.
If magic is rare and only has a few wizards throwing fireballs, look at historic sword/spear/bow with canons combat.  If everyone has a couple of one shot ranged spells, look at the musket era.  If combat magic is relatively common, look at WWI.  If everyone has trained dragons, move it up to WWII.
For general society, is magical light common (gaslight era or electricity).  Can magic be used to speed up crafting? then look at the industrial era.
In general, look at the effect of the magic not the method and find effect analogs that you can reference.


I there is way to burn ERC223 tokens like we do to ECR20 token?

Question: but as far as i know  alike ECR20, ECR223 token cant be send to an address without having tokenFallback() do we have another way around to burn the ERC223 tokens?
Answer: Yes, you can burn the token by deleting the allocated token to the particular address and subtracting the equal value from the total supply.
<code> function burn(uint256 _value) public {
require(_value <= balances[msg.sender]);
// no need to require value <= totalSupply, since that would imply the
// sender's balance is greater than the totalSupply, which *should* be an assertion failure

address burner = msg.sender;
balances[burner] = balances[burner].sub(_value);
totalSupply_ = totalSupply_.sub(_value);
Burn(burner, _value);
}
</code>


What are Trojan Asteroids and where are they located?

Question: Asides from the main Asteroid Belt, there are Trojan Asteroids - what are these and where are they located?
Answer: A Trojan Asteroid is defined by the Swinburne University Astro page "Trojan Asteroids" as:

Asteroids sharing an orbit with a planet, but which are located at the leading (L4) and trailing (L5) Lagrangian points.

These are often divided into 'leading' and 'trailing', as per the diagram below (from the source linked above):

Even though the term Trojan asteroid is often associated with the 4800 or so that lead and follow Jupiter, it actually applies to any asteroid that displays this behaviour, associated with any planet.
There are numerous Trojans co-orbiting with Mars and Neptune.  Recently, single Trojans have been found co-orbiting Uranus and even Earth.
Comment: L5 is the Sun–Earth Lagrangian point located close to the Earth's orbit 60° behind Earth.
Asteroid 2010 SO16, in a horseshoe companion orbit with Earth, is currently proximal to L5 but at a high inclination.


Why can "bill" mean (almost) opposite things?

Question: Bill is somewhat of an auto-antonym, since it can mean either a piece of paper which has positive monetary value (i.e. a note), or a piece of paper which has negative monetary value (though it only applies to a certain person).  Maybe I'm reading too much into the usages, but it seems strange that the two meanings of bill are nearly opposite.  Does anyone know why this is?
Comment: Any money owed by one entity is owed to another entity. Possibly the term originated before paper currency became as common as it is today.
Comment: Note that the *note* meaning of *bill* is chiefly American: we would say *five-dollar bill* but not *five-pound bill* (and *five-pound note* but not *five-dollar note*).
Comment: I don't think "autantonym" (aka "contronym") is a particularly useful term. The current case  *bill=money/debt* is like *weather=endure/erode*. Two different meanings of a single root word diverge so much they can be seen as opposites. In other cases, like *cleave=split/adhere*, two different words just happen to have the same spelling. Interesting quirks of language, but I can't see why we need *two* technical terms for something so trivial.
Answer: Without doing too much research into this, the etymology of bill shows that the original usage was more like an official document at first.

"written statement," mid-14c., from Anglo-L. billa "list," from M.L. bulla "decree, seal, sealed document,"

This suggests that different people adopted the word to mean similar, but related, concepts. In fact, government currency is a form of money owed, in that (in the days of the Gold Standard) the government promised you the bill's face value in gold. Now it is more abstract but the terms are related, even if seemingly opposite.
Comment: So, in short, *bill (note):government:me::bill (invoice):me:servicer* -- Is that right?
Comment: @drɱ65δ: yes, that is what I was trying to say. But I don't have something definitive to say if that is an intentional or coincidental overlap.
Answer: In general, this situation can arise for two main reasons:
(1) a word that originally meant a narrower concept diversifies to cover related concepts or specific connotations, and gradually over time the precise meanings of these once strongly related concepts can further diverge over time;
(2) what were once two separate words could merge due to sound changes or other phonological processes.
In this case, as mentioned in the answer above, it appears that effectively situation (1) has arisen: "bill" expanded from its original meaning of "official document" to cover various different specific connotations of "document" today, be it the "bill" put up by a "bill poster", the "bill" you pay for your meal, a "bill" of rights, a "dollar bill" etc.
Answer: A "bill" is an evidence the party A owes money to party B.
The "bill" that arrives in the mail from say, the phone company, is evidence that you owe money to the phone company.
A "bill" in the currency sense, is evidence that the government owes money to you. If you give the bill (instead of writing a check) to the phone company, the government then owes money to the phone company.
(This came about because "money" used to be gold and silver. In the U.S. the 16th Amendment to the Constitution gave the Federal government the authority to issue paper money as a substitute for metal. So (most people turned in their gold and silver and received paper "bills" (IOUs) in return.


Using different metrics for training vs validation

Question: Is there any theoretical justification for using a different metric on the validation set to do model selection than what was using for training? For example, one can train the model using some type of differentiable loss (i.e. log-loss) so gradient descent can be ran, then do model selection using some other metric (i.e. F1 score or average precision). Is it the general hope that that the training metric is a good proxy for the validation metric, or that doing better on one means doing better on the other one?
Answer: Validation metric is selected based on the final goal of the model.
For e.g., in medical image analysis to detect the presence of cancer, false negatives (detecting cancerous images as non cancer) should be very low as it is a disastrous condition, but some false positives (detecting non cancerous images as cancer) is still OK as it will not be very harmful or fatal. So, in this case, we tend to focus more on keeping recall(TP/TP+FN ) high to get high TP when compared with FN. Since recall score is non differentiable (cannot be used as loss function), we tend to incorporate it in our loss function which was used for training the model, for better results. Go through this article: https://towardsdatascience.com/the-unknown-benefits-of-using-a-soft-f1-loss-in-classification-systems-753902c0105d
" Is it the general hope that that the training metric is a good proxy for the validation metric?"
Well, yes. The purpose of training the model is to enable it to perform the desired task on unseen data (evaluated using validation metric). Through training, the model learns what to do on new data (validation data), it will perform what it will learn, so to achieve good results on validation data, training metric should be a good proxy for the validation metric.


Adding the Apple ProRaW profile to Lightroom

Question: I have Adobe Photoshop Lightroom Classic CC 8.2.1 with Camera Raw 11.2.1.
How do I add the Apple ProRaW profile to it?
Answer: For the moment Lightroom do not support RAW files from APPLE. You can check here for reference. Also to open at all these DNG files you need CameraRAW 15
So you should wait for Adobe to release appropriate version (and update Lightroom)


Is the n factor/ valency factor of H3PO4, 2 or 3?

Question: I'm very confused with search results I see on Google and different websites. I think the n-factor of H3PO4 is 3 because three H atoms are already bonded to three oxygen atoms. But some sources say it is 2, and I can't find out why.
Comment: If "these" sources say the $n$ factor is $2$, it may mean that they are speaking of an ion like $\ce{H2PO4^-}$. Can you mention the origin of these sources ?
Comment: It can be 1, if being neutralized to H2PO4-, 2 if being neutralized to HPO4^2- or 3 if being neutralized to PO4^3-.
Comment: Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
Answer: The n-factor of a molecule is not absolute. It can change from reaction to reaction.
If we are talking about H3PO4 then it can have a wide variety of n factors. The oxidation state of phosphorus in this compound is +5.
So it can change to only lower oxidation states(+3, 0, -3) (redox) or not change at all(neutralization reaction.
In the first case(redox) the n factor depends upon the change in  oxidation state. $$(+5\text{ to }+3):\text{ n factor is }2$$
In the second case that is in the neutralization reaction it depends on the basicity of phosphoric acid. $$\ce{ H3PO4\text{ to }H2PO4-}:\text{ n factor is }1.$$


Pre-trained CNN for one-shot learning

Question: I'm currently trying to learn one-shot learning using convolutional neural networks. According to this video, the CNN that I use should have been pre-trained on the MNIST. Why must the CNN be pre-trained??
Answer: Because the layers in the CNNs must be already able to extract features from images.
Normally this procedure takes thousands of iterations to be completed, so if they were not trained we could't have the concept of one-shot learning.
Answer: There is nothing like pre-training on MNIST data.

You can take any advanced CNN architecture and then fine-tune the network on your dataset. For performing the one-shot learning, the need for fine-tuning or training is must as this will tune your model to understand and emphasise only on the images that the model has been trained upon.
To simplify, a network trained on faces will be able to recognise only the facial features over other objects in the images which is must for generating the embedding.

In one-shot learning, the network tries to generate embedding by forward propagation of the image data through the network. The generated embedding must only emphasis on the required part of the image and not onto the useless data in the image data. A pre-trained model in this case is chosen that is capable of selecting the features in a image that are required to be emphasised.
Let's try to understand it through the example Face Recognition system. In this case, we would require to have a model that has been trained with images of faces so that whenever we try to create a new embedding, the model must know that it has to concentrate its weights on face rather than the noise in the image. Thus when the embedding are generated, the part of the face is concentrated and thus the cosine distances between the embedding could be minimized to uniquely identify the closeness among the 2 images.
You can get a better understanding of such a network in this video by Andrew Ng.

Pre-trained CNN for one-shot learning


What is "Section 508" very very simply?

Question: I researched at google and found many confusing, looooong explanations for Section 508! made me even more confused.
very simply, what is this Section 508?
Answer: A fairly simple explanation can be found here; summed up nicely by this snippet:

Section 508, an amendment to the
  United States Workforce Rehabilitation
  Act of 1973, is a federal law
  mandating that all electronic and
  information technology developed,
  procured, maintained, or used by the
  federal government be accessible to
  people with disabilities. Technology
  is deemed to be "accessible" if it can
  be used as effectively by people with
  disabilities as by those without.

This page also offers another good, simple explanation:

Section 508, in simple terms, requires
  that when Federal Agencies develop,
  procure, maintain or use electronic
  and information technology,
  information and data should be
  accessible to people with disabilities.

Obviously there's lots more involved in this process, but you already knew that. Hopefully this clarifies things a little bit!
Answer: From A Website for Lawyers:

What is Section 508 Accessibility Law?
Section 508 of the Rehabilitation Act
of 1973, is a US law requiring the
federal government to make all goods
and services-including web sites and
intranets fully accessible to people
with disabilities. It identifies
specific standards for Internet and
Web accessibility, which are often
used as a basis for evaluating whether
or not Web sites meet accessibility
requirements.
Answer: I gave a talk about this at the last PyCon:
http://djangocon.blip.tv/file/3261238/
508 compliance is about making your site accessible to everyone, no matter their handicap. While much of 508 is focused on people using screen readers, you also have to consider those that are physically handicapped (and may have trouble mousing over tiny areas), or those that are deaf (and would love to play your Flash game, but can't hear the beeps required to win!).
Answer: The idea is the following: unless your product is fully Section 508 compliant you may be unable to your product sell to the US Government organizations.

What is "Section 508" very very simply?


How do you beat Endogeny?

Question: I am playing Undertale at the moment, and I am doing the True Pacifist Route, (Post-Genocide). I am up to the true lab, but I am stuck on Endogeny, the dog amalgamate. Is there some sort of trick that I am missing? I do not know how to dodge the attack where the dog turns into some kind of rocket.
Comment: Once you've completed a Genocide run, you can't undo the effects in-game. You'll want to look here to get back to a pre-Genocide state: https://gaming.stackexchange.com/questions/241785/can-i-ever-play-undertale-again-normally You might as well finish your run before you do that, though, and the actual question you're asking here is still answerable separately.
Answer: There are two ways to spare Endogeny:
Just like the fight with greater dog, there is an order of ACTs you have to do:

Beckoning
Petting
Playing
Petting
Petting (again)

After these you will be able to spare him.
Dodging the rocket attack can be difficult but is doable. When he begins a charge, he will go for your last position so try to strafe away from him diagonally and try to avoid getting stuck in a corner or against a wall.
It is not easy but with this you can minimize getting hit to maybe 1 or 2 times.
Bringing some chips from the vending machine or any other healing items will help you survive.
I spoilered the other way to not take away from exploring the hidden things that make exploring Undertale so fun.

 The other way is to use one of the following items:
 The Hush Puppy, which can be obtained in the MTT Resort at the last door to the right. The person will trade it for a Hot Dog. These can be bought from Sans at Sans' Hot Dog Stand in Hotland.
 The Stick, the weapon you begin the game with. When kept in the inventory, you can use it like an item in any of the dog fights to instantly pacify them.
 Both of these items will pacify Endogeny and you can spare him.
If you wish to achieve the yellow text for the Amalgamates in the credits, you still have to do the ACTs as described above, but any of the two items will make this easier as Endogeny will not attack anymore when pacified.

How do you beat Endogeny?


Representation suitable for reconstruction of a tree with bounded degree

Question: I am dealing with reconstruction of molecular graphs for which unlabelled rooted trees with maximum degree 4 are fair approximations. In particular, I would like to encode a small tree (assume number of vertices is less than 20) and degree is upper bounded by 4, such that we can reconstruct the tree form the encoding, decoding is not necessary to be unique but low degeneracy is expected. Currently, I am given the number of vertices at various distance from the root, say $d = [d_1, d_2, ..., d_k]$ where $d_i$ is the number of vertices at distance $i$ from root. However, the number of tree satisfying this condition is very large. I am looking for other representation (as an alternative to $d$) of tree which will be more suitable for reconstruction, may be approximately. I would prefer the representation which is hierarchical in the sense that, first few information are more constrained and informative, and others are gradually less so. For example, here $d_1$ gives complete information on number of nodes at level 1, there is slight degeneracy in constructing level two nodes from $d_2$, etc. My focus is on approximate reconstruction rather than canonical encoding.
Comment: It's quite difficult to provide an answer without knowing what information is available. For example, if the data for these comes from some sort of experiment, only certain kinds of data may be available. I may suggest a representation that uses the ratio of leaves to inner nodes for arbitrary example, but it may be unavailable.
Comment: @DavidCummins these graphs are molecular graphs. First I need to code a molecule preferably into a single number and then reconstruct, with low degenearcy, the graph from the number.
Comment: What is the problem with, for example, having the vertices of the tree labelled and simply storing the parent of each vertex?
Comment: The representation should be invariant of labeling, and should preferably result in a single real number or vector of dimension at most tree height.
Comment: What do you mean by degeneracy? Did you mean [degeneracy](http://en.wikipedia.org/wiki/Degeneracy_(graph_theory)) in graph theory? In this case degeneracy of tree is one, even if your molecule structure is planar, its degeneracy is five. I think you mean bounded maximum degree not degeneracy. Also what's wrong if you just make a complete binary,... tree. I mean the structure that you already know, then the only information that you need is the number of nodes.
Answer: I'm still not sure that I completely understand the problem domain, but here's a few suggestions:

The depth of each leaf.
The difference between min and max number of children at each depth.
The number of leaves below each node.
The difference between the depths of the left and right subtrees for an in-order traversal.
The weights required to build a Huffman tree of the same structure.


Bluetooth Issues in 18.04 LTS

Question: I am using Ubuntu 18.04 LTS and after the recent upgrade to the final release on April 26, 2018, I found that my laptop (Dell Inspiron 1440) was unable to connect to my Bluetooth speakers using a generic USB Bluetooth dongle which was earlier working perfectly in Ubuntu 17.10 Any help in this regard would be highly appreciated.
Comment: https://bugs.launchpad.net/ubuntu/+source/linux-firmware/+bug/1756824 Solution/Workaround in link above.
Comment: Please edit the question to include results from terminal for `lsusb`
Answer:
Install "synaptic".
Search synaptic to ensure "bluez" is installed.
Use synaptic to search and install "blueman".
reboot.
If no icon in task bar, search for "bluetooth manager" in your apps, and start it. Follow the screens to connect your device.
Comment: This is what I was going to recommend. I had to use the same steps to get an Apple Magic Mouse to work with 16.04. The built-in Bluetooth control applet is not versatile enough, but the BlueZ replacement works well.

Bluetooth Issues in 18.04 LTS


Do Non-Jews have the rule of Safek D'Oraisa L'Chumra concerning the 7 mitzvos of B'nei Noach?

Question: Do Non-Jews have the rule of Safek D'Oraisa L'Chumra (if one is unsure about whether a Torah prohibition exists, He must assume the more strict behavior), concerning the 7 mitzvos of B'nei Noach or does this rule apply only to Jews?
Comment: @aBochur If a non-Jew has a piece of meat, and he is not sure wheter it was separated from the animal while the animal was still alive or after it was killed. Must he act strictly because of the rule of  Safek D'Oraisa L'Chumra and refrain from eating it as if it were Aiver-Min-Hachai [A limb from a live animal], which is forbidden by one of the 7 Mitzvos of Bnei-Noach?
Comment: What would be a good example of this?
Comment: @RibbisRabbiAndMore +1 you might wanna edit that info the question
Answer: I saw this discussed in Ateres Shmuel (by Rav Mordechai Rosenbaum) to Kesubos 15b ד"ה אם רוב עכו"ם .
The gemarra there says an abandoned child that is found in a city of mostly non Jews is treated as a non Jew, to the extent that we can directly feed him non kosher food (Rashi). The Ateres Shmuel asks why didn't Rashi just say the child can feed himself non kosher?
He says quoting Rav Avraham Gurwicz that even if the city was 50% Jews 50% non Jews, the child can be lenient and eat non kosher. This is because saffeik doraisa lechumra according to the Rambam is deRabannan, and only a definite Jew has to listen to the Rabbis. Not sure his source for that last part. But we see from here that a definite non Jew doesn't have to worry about saffeik doraisa lechumra. So for example, if he has a saffeik if his food is maybe eiver min hachai, he can eat it (assuming the saffeik is 50-50).
The Ateres Shmuel considers saying that even according to those who say saffeik doraisa lechumra is a doraisa, this questionable non Jew can still be lenient. However, he brings the Rash's opinion that if the city was 50% Jews 50% non Jews, the child can't eat non kosher. He presumably holds saffeik doraisa lechumra is a doraisa. We would then have to conclude either (1) a non Jew has to follow saffeik doraisa lechumra, and in the above case they'd have to worry the food is eiver min hachai, or (2) at least a questionable non Jew has to be stringent, but maybe a definite non Jew not.
Now this assumes the saffeik is 50-50. If there is a majority saying it is muttar, there is a discussion in Achronim if non Jews can rely on the rule of rov (majority). If yes, no problem. If no, then your question will also apply.
Comment: This only works like the Rambam. What about the Rashba?
Comment: @DonielF I addressed his opinion. The Ateres Shmuel would have said even according to the Rashba non Jews aren't obligated to be stringent, but the Rash is unclear if he's a counter-proof. I hope to research this more when I can.
Comment: This is not how I understood the question. I thought the person was addressing more a case like: A non-Jew goes to a restaurant, and he sees they offer crab claws. He cannot know  if the crab was killed before the claw was removed, or if it is eiver min hachai.Does he have to avoid the crab claw because safeiq deOraisa lechumerah? Not a safeiq whether one is a Ben Noach and thus whether one of the Jew-specific 613 apply.
Comment: @MicahBerger me too. You learn the answer from my case, since what's the chiluk? He either has suffik lechumra or not.
Comment: I saw the Mahache Chaim and Maharam Shik address this. Bli neder I'll update this with their insights
Answer: The Rambam Malachim 9 says that since Bnei ketura and bnei Yishmael intermarried, they're both obligated in circumcision misafek. We see you say safek deoraisa by non jews. It's also very logical to say so because the Rashba said that the Rambam said his source for safek Deoraisa likula is learned from the fact a safek mamzer is permmited likhal. It's a derivation that would only apply to jews. Without this derivation, logic says safek deoraisa is lichumra. Everyone agrees logic says safek deoraisa is lichumra, therefore it should apply to non jews as well.
Comment: I saw the Mahache Chaim and Maharam Shik address this. Bli neder I'll update my answer with their insights as what you're saying isn't so simple


Add a product description field to "Edit Selected Quote Line Items" page

Question: My requirement is to display the product description field in "Edit Selected Quote Line Items" page which opens after you go to Quote object-->Related List Tab-->Quote Line Items List-->Add Product-->Select a Product-->Click next. The page which opens after this is the one in which I wish to add a new field (product description).
You can change the fields visible in this page via standard by going to page layout of Quote Line Item Object-->Quote Line Item Layout-->Edit Multi Line Layout-->Select which all fields you wish. Here the issue is that I do not know how to add fields from another object to this page. I understand since this functionality is not possible through admin and I have to proceed via development, but I do not know how to proceed.
I have shared some screenshots below to give you a wider understanding:
Answer: You can consider a cross object formula field from the Product Object field