WinHttpRequest on Mac OSX Excel 2011 VBA

Question: There is no WinHttopRequest in "References" on Mac version of Excel 2011.  I have tried following approaches which I have seen in other posts:
Set HTTP = CreateObject("MSXML2.ServerXMLHTTP")

which gives me a '429' runtime error: ActiveX controller can't create object.
Is there a way to use WinHttpRequest or something similar on Mac Excel? I have had no luck with query tables either, and want to avoid that solution.
There should be a simple http GET solution to this problem I would think. Just cant find it out for Mac Excel.
I am trying to get data from Yahoo Finance api url:
Dim URL As String: URL = "http://finance.yahoo.com/d/quotes.csv?s=" & Symbols & "&f=snl1hg"
Dim HTTP As New WinHttpRequest
HTTP.Open "GET", URL, False
HTTP.Send

I know this works on windows, but I am using a Mac.  Please advise. Thanks!
Comment: Did you find a solution for this? Facing a similar case. and I also want to avoid QueryTables
Answer: You can use QueryTables in place of the HTTP Get call (WinHttopRequest), which evidently is not supported by Mac Excel 2011. The code below worked for me - enter the tickers in column, starting with A2, enter yahoo finance tags (i.e. a,b, r, n) in row starting with B1.
You can assemble the URL to call YF for the csv, then use a QueryTable to make the call and paste the results in your worksheet.
Code working on Mac Excel 2011:
Sub Yahoo_Finance_API_Call_MacExcel2011()
    Dim head As Range
    Set head = Range("A1")

    Dim wb As Workbook 'In the event that you'll use different workbooks
    Dim src As Worksheet 'In the event that you'll use different a source worksheet
    Dim tgt As Worksheet 'In the event that you'll use different a target worksheet

    Set wb = ThisWorkbook
    Set src = wb.Sheets("Sheet1")
    Set tgt = wb.Sheets("Sheet1")

    'Assemble Symbols for API Call
    Set rng = Range(head.Offset(1, 0), head.Offset(1, 0).End(xlDown))
    For Each cell In rng ' Starting from a cell below the head cell till the last filled cell
        Symbols = Symbols & cell.Value & "+"
    Next cell
    Symbols = Left(Symbols, Len(Symbols) - 1) ' Remove the last '+'

    'Assemble Tags or API Call
    Set rng = Range(head.Offset(0, 1), head.Offset(0, 1).End(xlToRight))
    For Each cell In rng ' Starting from a cell to the right of the head cell till the last filled cell
        tags = tags & cell.Value
    Next cell

    'Build URL
    URL = "TEXT;http://finance.yahoo.com/d/quotes.csv?s=" 'Use TEXT to collect API data below
    URL = URL & Symbols & "&f=" & tags

        'Range("A1").Value = URL 'This will output the assembled URL in a1 for QA if need be

    'Call API
    With tgt.QueryTables.Add(Connection:= _
            URL, _
            Destination:=Range(head.Offset(1, 1), head.Offset(1, 1).End(xlDown)))

        .RefreshStyle = xlOverwriteCells
        .TextFileParseType = xlDelimited
        .TextFileCommaDelimiter = True
        .BackgroundQuery = True
        .TextFileCommaDelimiter = True
        .TablesOnlyFromHTML = True
        .Refresh BackgroundQuery:=False
        .TextFilePromptOnRefresh = False
        .TextFileStartRow = 1
        .TextFileParseType = xlDelimited
        .TextFileTextQualifier = xlTextQualifierDoubleQuote
        .TextFileConsecutiveDelimiter = False
        .TextFileTabDelimiter = False
        .TextFileSemicolonDelimiter = False
        .TextFileSpaceDelimiter = False
        .TextFileTextQualifier = xlTextQualifierDoubleQuote
        .SaveData = False
    End With
End Sub

Answer: I do not believe Macs have anything equivalent to MSXML.ServerXMLHTTP. A suggestion from another stackoverflow thread is to use QueryTables. In summary the thread suggests:
With ActiveSheet.QueryTables.Add(Connection:="URL;http://carbon.brighterplanet.com/flights.txt", Destination:=Range("A2"))
    .PostText = "origin_airport=MSN&destination_airport=ORD"
    .RefreshStyle = xlOverwriteCells
    .SaveData = True
    .Refresh
End With

Comment: I'm looking for a more direct, http get solution. Are query tables really the only way to get data from yahoo finance in excel on a mac? That seems unlikely.
Comment: @brno792 Were you able to find an optimal way to do this?


Lattice gauge and spin network

Question: I see the similarity between the Lattice Gauge and Spin Network.
(For example, both theories depict the node part as quantum (the latter is explained as spin).)
Are there any other mathematical, physical similarity between the two theories?
Comment: Being more precise, can you firstly define your "lattice gauge theory" and your "spin network"? Does Dijkgraaf-Witten gauge theory counts as your lattice gauge theory? Does Levin-Wen model counts as a spin network for your definition?
Comment: Yes , both does count as part of the theories .
Comment: General answer is fine actually . In any case , both theories seem similar .
Answer: This question is already quite old, but since it was modified recently, let my provide a further answer for other latecomers to this question.
There is a relation between lattige gauge theory and spin-network states, namely the latter one form a basis of the Hilbert space of lattice gauge theory. Let me be a little bit more explicit.
First of all, a spin-network corresponding to a (compact) Lie group $G$ can abstractly be defined to be a triple $(\gamma,\rho,i)$ consisting of the following data:

A finite directed graph $\gamma=(\mathcal{V},\mathcal{E})$ with target and source map $t,s:\mathcal{E}\to\mathcal{V}$.
A map $\rho$ assigning to each edge $e\in\mathcal{E}$ an irreducible representation of $G$.
A map $i$ assigning to each vertex $v\in\mathcal{V}$ an intertwiner of the type
\begin{align*}i_{v}:\bigotimes_{e \in\mathcal{T}(v)}\mathcal{H}_{e}\to\bigotimes_{e^{\prime}\in\mathcal{S}(v)}\mathcal{H}_{e^{\prime}},\end{align*}
where $\mathcal{H}_{e}$ and $\mathcal{H}_{e^{\prime}}$ are the vector spaces corresponding to the representations associated to the edges $e$ and $e^{\prime}$.

Now, spin-networks appear naturally in the discussion of gauge theory on a graph. For this, consider a (compact, connected, finite-dimensional) Lie group $G$ as well as a principal $G$-bundle $P$ over a $d$-dimensional manifold $\mathcal{M}\cong\mathbb{R}\times \Sigma$, where $\Sigma$ is some space-like hypersurface representing space. Furthermore, let $\gamma=(\mathcal{V},\mathcal{E})$ be a finite directed graph embedded in $\Sigma$. Now as usual in lattice gauge theory, we discretize a connection $1$-form of our principal bundle (i.e. a gauge field in physics terminology) by assigning a group element $g\in G$ to each edge. Hence, the space of discrete connections is given by
$$\mathcal{A}_{\gamma}:= G^{\vert\mathcal{E}\vert}$$
A gauge transformation can be described via a group element $h_{v}$ living on the vertices of $\gamma$ and acts as
$$g_{e}\mapsto h^{-1}_{t(e)}g_{e}h_{s(e)}$$
on some group element $g_{e}$. Hence, the set of gauge transformations is given by
$$\mathcal{G}_{\gamma}:= G^{\vert\mathcal{V}\vert}.$$
Using the (normalized) Haar measure on each copy of $G$ in $\mathcal{A}_{\gamma}$, we get a well-defined Hilbert space $L^{2}(\mathcal{A}_{\gamma}/\mathcal{G}_{\gamma})$, which can be identified with the subspace of $L^{2}(\mathcal{A}_{\gamma})$ consisting of gauge-invariant functions.
Now, the main point is the following. As discussed at length in arXiv:gr-qc/9411007, one can show, using the theorem of Peter-Weyl and some tricks, that there is the following isormophism:
$$L^{2}(\mathcal{A}_{\gamma}/\mathcal{G}_{\gamma})\cong\bigoplus_{\rho\in\Lambda^{\mathcal{E}}}\bigotimes_{v\in\mathcal{V}}\mathrm{Int}\bigg(\bigotimes_{e\in\mathcal{S}(v)}\mathcal{H}_{\rho_{e}},\bigotimes_{e\in\mathcal{T}(v)}\mathcal{H}_{\rho_{e}}\bigg ).$$
where $\Lambda$ is the set of all (equivalence classes of) irreducible and unitary representations of $G$ and where $\mathcal{H}_{\rho}$ denotes the representation space corresponding to some $\rho\in\Lambda$. Using this as well as the above definition of spin-networks, it is clear that spin-network states actually span the Hilbert space of gauge theory on a graph.
Answer: You ask if there are mathematical similarities between the two.  In fact there is a mathematical structure called a gauge network that accommodates both lattice gauge theory and spin networks.  It was introduced by Marcolli and van Suijlekom in
https://arxiv.org/abs/1301.3480
(it was published in the same year you posted this question).
Here is a rough description: a spin network is in some sense a discrete version of a spin manifold ("spin" manifold being one whose global structure allows for spinors). A lattice gauge theory is a discrete version of a manifold with a gauge theory defined on it (a 'principal G bundle over a manifold' in jargon).
A gauge network attaches more general algebraic data than either of these two theories to the nodes and edges of a graph.  Both spin networks and lattice gauge theories are special cases of gauge networks.  So if you work with the gauge network structure, you are in the right discrete setting to start talking simultaneously about gravity and gauge theories.

Lattice gauge and spin network


How to display facebook profile picture and name of my site visiter using PHP(without authenticating)?

Question: I just want to display the username and profile picture of my blog visiter near the commenting area.I want to do without any authenticating with FB,Is there any way to do this ?? Am i totally wrong ??
Comment: Sorry for my poor language :(
Answer: There is no way of doing it without the user authenticating. Getting the name and profile pic from FB requires authentication. The user must give you permission to access their information via the FB SDK.
You can read more about it here: https://developers.facebook.com/docs/sdks/


Reaction of borax with NaOH

Question: How does borax  react with NaOH?
My textbook says that sodium peroxoborate is formed. I couldn't find the reaction anywhere else.
Can anyone verify as well as explain it?
Comment: This is nonsense.
Answer: The following reaction schemes are adapted from [1, pp. 76–77].
Amorphous borax $\ce{Na2B4O7}$ forms tetrahydroxoborate in concentrated sodium hydroxide solution:
$$\ce{Na2B4O7 + 7 H2O + 2 NaOH → 4 Na[B(OH)4]}$$
Upon fusion sodium metaborate is formed:
$$\ce{Na2B4O7 + 2 NaOH →[\pu{700-750 °C}] 4 NaBO2 + H2O}$$
In order to form peroxoborate, a peroxide should be introduced in the first place, e.g. by using hydrogen peroxide (see also Preparation of Sodium Peroxoborate):
$$\ce{Na2B4O7 + 2 NaOH + 4 H2O2 + 11 H2O → 2 Na2[B2(O2)2(OH)4] * 6H2O}$$
References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)
Answer: Andselisk has given a more satisfactory answer for this question, which has been accepted. Yet, I'd like to give some interesting facts and view about borax, if some readers are interested.
The name borax is derived from the Persian word borak, meaning white. IUPAC name of borax is sodium tetraborate decahydrate ($\ce{Na2B4O7.10H2O}$). Also note that, after analyzing its crystal structure, borax is more properly formulated as $\ce{Na2[B4O5(OH)4].8 H2O}$ and given its systematic name 2,4,6,8,9-pentaoxa-1,3,5,7-tetraborabicyclo[3.3.1]nonane-3,7-diolate, sodium salt, hydrate (1:2:10), accordingly.
When dissolved in water, borax is subject to completely dissociate, and then,  hydrolyzes to orthoboric acid ($\ce{B(OH)3}$) and $\ce{ OH-}$ anions, yielding a $\mathrm{pH}$ of about 9.13 according to the equation:
$$\ce{ Na2[B4O5(OH)4].8 H2O (s) -> 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 8H2O (l)}$$
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 5H2O (l) <=>  2Na+ (aq) + 4B(OH)3 (aq) + 2 OH- (aq)}$$
Therefore, borax reacts with strong acid solutions. For example:
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 2 HCl (aq) + 5H2O (l) -> 4B(OH)3 (aq)  + 2Na+ (aq) + Cl- (aq)}$$
The liberated $\ce{B(OH)3 (aq)}$ acts as a Lewis acid on $\ce{ OH-}$ ions from autoionization of water, and hence, in equilibrium with water:
$$\ce{ B(OH)3 (aq) + 2H2O (l) <=> B(OH)4- (aq) + H3O+ (aq)}$$
Thus, it can also be back-titrated with alkali solutions such as $\ce{NaOH}$. However, note that the boric acid is a weak monoprotic acid ($K_\mathrm{a} \approx 6.4 \times 10^{-10}$). Therefore, it cannot be directly titrated with standard alkali for analytical purposes. Yet, by the addition of an organic polyhydroxy compound such as 1,2- or 1,3-diol, it is converted to a much stronger acid, which can be titrated using phenolphthalein. This is due to the complex formation between hydrated borate ion and either 1,2- or 1,3-diol (see illustration in the diagram):

It is noteworthy that the combination of these two reactions is commonly used by some analytical chemists to standardize their acidic/basic solutions.
Now, consider the following two reactions:
$$\ce{ 2Na+ (aq) + B4O5(OH)4^{2-} (aq) + 5H2O (l) -> 2Na+ (aq) + 4B(OH)3 (aq) + 2 OH- (aq)}$$
$$\ce{ 4B(OH)3 (aq) + 8H2O (l) <=> 4 B(OH)4- (aq) + 4 H3O+ (aq)}$$
If you add these together, you'd get:
$$\ce{ B4O5(OH)4^{2-} (aq) + 9 H2O (l) -> 4B(OH)4- (aq) + 2 H3O+ (aq)}$$
To cancel $\ce{2H3O+ (aq)}$, add $\ce{2NaOH (aq)}$ to both side. Now, you'd get:
$$\ce{ 2Na+ (aq) + 2OH- (aq) + B4O5(OH)4^{2-} (aq) + 5 H2O (l) -> 2Na+ (aq) + 4B(OH)4- (aq) }$$
Now, compare the equation Andselisk wrote in his answer for the borax reaction with strong alkali solutions. For example, $\ce{2NaOH (aq)}$:
$$\ce{ Na2B4O7 + 7H2O + 2 NaOH -> 4NaB(OH)4 }$$
Both are acceptable ionic and condensed formations, respectively.
Sodium peroxoborate ($\ce{ Na2[(OH)2B(O-O)2B(OH)2]}$), on the other hand, can be prepare by the oxidation of sodium metaborate ($\ce{NaBO2}$) using hydrogen peroxide ($\ce{H2O2}$):
$$\ce{2NaBO2 + 2H2O2 -> Na2[(OH)2B(O-O)2B(OH)2]}$$
Sodium peroxoborate can be precipitated out of water as hexahydrate: $\ce{ Na2[(OH)2B(O-O)2B(OH)2].6H2O}$. Sodium metaborate ($\ce{NaBO2}$) can be prepared by heating borax and $\ce{NaOH}$ together:
$$\ce{ 2NaOH + Na2B4O5(OH)4.8H2O  -> 4 NaB(OH)4 + 3H2O -> 4 NaBO2 + 11 H2O}$$
Comment: Did you write thus answer based in memory or from a source? I would like to read the source if possible, to study more(if possible).
Comment: It was not one source. There are few of them. I made a Google search under few key words including the title of the question. Trick is you need to be in a campus computer to be able to see all the references. Sorry on that part.
Comment: What does a campus computer change? I have never used one.
Comment: They have campus' subscribed journal access, of which you can't have with your personal computer at home. That's all.

Reaction of borax with NaOH


How to store a Christmas pudding to make it drier?

Question: I have prepared a Christmas pudding but it came out too wet to our taste. What is the best way to store it to make it drier but without spoiling it?
Answer: A storage method that will dry out your pudding will essentially be staling it. Starches retrograde (lose moisture) fastest between 17F and 46F (-8c and 8c). Unfortunately, this not only causes the starches to release moisture, they also revert to their crystalline form which can make them unpalatable. If you still want to do this, the refrigerator is the perfect place to keep it in this temperature range, but it will still take some time to lose moisture.
Alternately you could speed evaporative moisture loss by storing it somewhere warm and dry (a low oven or over a pilot light?) Again, this will probably make your pudding dry in an unappealing way as the outside will dry faster than the interior.
Comment: Thanks. I've chosen to store it in the fridge at +2C. the texture and flavor definitely improved after being stored overnight.


How to collect Bikurim practically?

Question: So we know that Bikurim was collected by tying a reed to the first fruit of a particular field when it would be found budding.  We also know that Chazal say to give one 60th of one's produce as Bikurim.  The question is how to do this properly.  It would seem necesary to know how much fruit the field will yield in advance how to then know how many fruits to mark off.  Am I missing something?
Comment: Cool question! May we merit to need this knowledge practically soon.
Comment: Amen Isaac.  Do Jews actually keep Bikurim in Eretz Yisroel today?  Not the part about bringing it in a basket to the Beis HaMikdash - but do Jews mark off the fruits and set aside 1/60th like is done with Terumah?
Comment: @chaimp No. Bikkurim only applies when there is a Temple around, unlike Terumah.
Comment: Similar: http://judaism.stackexchange.com/q/45374
Answer: Experienced  farmers can estimate in advance the Yield for the Year and there is usually only slight variation so he does know how much fruit the field will yield in advance .
Comment: So, when Chazal said 1/60th, they were saying this only for experienced farmers?  Even so, no matter how experienced somebody is at farming there is no way to know how much rain there will be and all sorts of other conditions.

In Pirke Avos Chazal say clearly to not estimate with Ma'aser.  I know that this is Bikurrim, but it is hard to understand why that concept would not apply here.

I thank you for the plausible answer, but I think there must be something more to it.
Comment: your right,all people refered to as farmers and  can tell if it will be a good year this much will come out if not this much there are taught growing up a big problem with city slickers (as I am) lerning Zeraim is they dont know the Facts on the ground.
Comment: Is the same done for Teruma and Ma'aser?  Are the preemptively calculated?
Comment: @chaimp They are not because they cannot be separated until after the produce is picked, unlike Bikkurim.
Comment: Thank you for the response 4 years later! :) I totally forgot that I even asked about this and it was nice reading this exchange again.

How to collect Bikurim practically?


Rashi Script in Aruch Hashulchan

Question: In my edition of the Aruch Hashulchan, there is often a paragraph that has Rashi script written inside brackets (see pic). What is the Rashi script? Is it the mechaber or someone/something else? I couldn't find any info on this in the opening pages of the Oz VeHadar edition of the Aruch Hashulchan.
Comment: It would seem from the one in the next sif (21:3) as well as many others that it is the Mechaber himself.
Answer: This exists in all editions of the Aruch HaShulchan. See for example the first page of this older edition.
Perhaps you couldn't find anything on this because unexpectedly the introduction to the work is in Choshen Mishpat. This is because it was the first section of Shulchan Aruch that he wrote.
There, note 8, he explains that he put these pieces in square brackets (and a different font), because they're not specifically to tell you the final halacha. Rather, they're his novella in understanding of the gemarra or rishonim. To not get bogged down with these novella he put them in brackets so you could skip them.
Comment: @Bachur613: Also, as someone who learns Arukh haShulchan Yomi, a warning... He rarely explains enough for you to fully understand those novella. Rather, you often have to see the gemara or rishon inside to know what he is talking about. Contemporary publishing would probably use footnotes, as it's clearly "extra material" and not assumed to be the core text everyone would read. (That said, I try my best...)

Rashi Script in Aruch Hashulchan


Test connectivity to NAT Gateway over firewall.

Question: I created a VPC in aws and added a NAT Gateway so my instances in the VPC can consume other resources on the internet. Some of these resources belong to a 3rd party and are behind  a firewall for which my NAT Gateways IP addresses have to be added so that I can successfully connect. They claim to have added my ip addresses but they are unable to verify because in the true nature of the NAT Gateways they are not able to ping/telnet my NAT Gateway IP's with a response. Is there anyway for them to verify if they can establish a connection to my NAT Gateway IP's?
Comment: Oh ok, thanks @Michael-sqlbot . I'll delete my previous comment as it's incorrect. NAT Gateways are an outward communication mechanism, mostly for accessing resources on the internet, so any testing you need to do needs to be for your end.
Comment: @Tim all the testing i do on my end shows that they havnt configured properlly. But sadly that is not enough for them.
Comment: @Tim that isn't accurate.  NAT Gateways have a single EIP.  It doesn't change.  https://docs.aws.amazon.com/vpc/latest/userguide/vpc-nat-gateway.html
Answer:
Is there anyway for them to verify if they can establish a connection
  to my NAT Gateway IP's?

Nope, they won't be able to connect to your resources. You must connect to them from inside of your VPC.
However are you positive that your routing through NAT actually works? For VPC NAT you generally need at least 2 subnets:

DMZ (or public) one that has IGW (Internet Gateway) attached and where the resources have Public or Elastic IP address. The default route in the Route Table 0.0.0.0/0 points to the IGW.
That's where you set up your NAT Gateway.
Private subnet where the default route 0.0.0.0/0 points to the NAT Gateway and the resources (instances) don't have public IPs.

If you've got these 2 subnets create an EC2 instance in the Private subnet and try to connect to internet, e.g. do curl http://ifconfig.co - if it comes back with your NAT gateway's Elastic IP your routing works. If it times out it's not properly configured and you'll need to dig some more into it.
Only when you get your general internet access working then you can work with your 3rd party on making sure their is set up correctly.
Update
To verify on their end they will have to set up network traffic logging for your NAT IP and verify that

traffic arrives from you to their external interface, and
that traffic is being passed through their firewall to reach the target internal system.

In Linux they can use e.g. tcpdump to monitor the traffic, on Cisco or other HW routers they'll have their own tools. It's up to them to verify that your traffic is being passed through.
Comment: @bubunyo did the answer help? If yes please accept it to reward the responder with reputation points. Thanks :)
Comment: i had done and confirmed all of these. I can confirm by doing `curl http://ifconfig.co` on private instances inside the VPC that the right NAT GW ip are being used. My issue is if there is any way for them to verify from their end.
Comment: @bubunyo updated the answer with some hints re the remote firewall. You'll need to colaborate with them to verify that.
Comment: @bubunyo did my answers help? If yes please accept it as a reward for the time spent helping you. Thanks!


Hadoop passing variables to mapper and reducer

Question: I'm a complete beginner with Hadoop.  I've built Word Count, and I'm fairly sure I understand the basics, but I'm having trouble extending that to an actual problem.  My (modified) code is below:
for (Item i : set) {
    for (Item j : set) {
        Score s = score(i, j);
        renderer.render(s);
    }
}

I'd like to use Hadoop to distribute this.  I can write a Mapper and a Reducer, but I don't know how to pass the set to the Mapper and the renderer to the Reducer (or if that's even the idiomatic way to handle this).  I also feel like I need to write my own Writable to handle passing the pair (i, j) between the Mapper and Reducer, but I don't know the best way to do this.  Any help would be appreciated.
Answer: What you are doing is essentially a cartesian product of set with itself.
You would probably need to implement a custom input format.
Here is an example of a generic Cartesian Product job: https://github.com/adamjshook/mapreducepatterns/blob/master/MRDP/src/main/java/mrdp/ch5/CartesianProduct.java
You can see the same logic you have done above to the input paths at line 67-77: https://github.com/adamjshook/mapreducepatterns/blob/master/MRDP/src/main/java/mrdp/ch5/CartesianProduct.java#L67-L77


Person/Group that claims all the work but can't deliver because they are overloaded

Question: Is there a phrase or expression to describe a person or group that claims all of the work/tasks but can't deliver because they are so oversubscribed?
Comment: I'm looking for a way to describe the behavior and the consequences of their behavior.
Comment: I'm attempting to characterize this behavior as a reason for not meeting a deadline.  Where should the blame lie if one party bites off more than they can chew and leaves the others with little responsibility?
Answer: If I am not wrong, then you have requested an expression for a ventripotent coveter(s) who avowed to discharge all the work they(he) voluntarily requested to accomplished. But, now they(he) are(is) sitting in the stockpile of work and can't deliver due to their inefficiency. I regret, that I don't have a word or a phrase for this situation. Though, I have suggested three idioms, if you may prefer.
NOTE: Initially, I thought-out a word Sisyphean and I thought to share it with you. According to Oxford dictionary, it means: Denoting a task that can never be completed and a person/group doing such task can be called Sisyphus. But unfortunately, Greek mythology does not allow us to use Sisyphus for the situation you have described. Sisyphean task, according to mythology is a repetitive and never ending task. But, somehow Oxford Dictionary uses Sisyphean in a liberal sense. So, if we go with the flow, then, Sisyphus can be used for the person who does a task that can never be completed. It doesn't say anything about the task was alloted to them or they(he) vowed to do it by themselves(himself). But, according to Greek mythology it was a curse by God to Sisyphus. So, yes it was a given repetitive task by God to Sisyphus.
I can recall few idioms for this situation. Those three idioms are:
1) bite off more than they can chew: (Source: theidioms.com )

try to do more than one is able to do
undertake a promise one cannot accomplish
accomplish attempt to do something which is hardly achievable
to start or promise something to do more than one can accomplish

2) eyes are bigger than one's stomach: (Source: Merriam-Webster)

(used to say) that someone has taken more food than he or she can possibly eat

3) thou art like the Epicure, whose belly is sooner filled than his eye:
Explained beautifully by John Lyly in the book Euphues: the Anatomy of Wit' and 'Euphues and His England
Answer: I would say that those people or groups have overburdened themselves:
From overburden:

[Merriam-Webster]
: to place an excessive burden on
// She overburdened me with work.
// Why overburden yourself when people are offering to help?

Note that although I think this is the most natural word, it doesn't actually necessitate that the work can't be done because there's too much of it for the people to handle. To make that explicitly clear, you'd have to say something like:

They overburdened themselves to the point of not being able to complete their work.

As an expression, you could say that they have bitten off more than they can chew:

[Collins]
If you say that someone has bitten off more than they can chew, you mean that they are trying to do something which is too difficult for them.
He bought the old hotel but soon realized he had bitten off more than he could chew.
Answer: You state that they take all of the work. The person or group represents a "monopoly" in the trade of that service. They set the price and the schedule because they are apparently the only ones providing the service. A single person or a company can monopolize business services or goods.
https://www.thefreedictionary.com/monopoly
https://www.thefreedictionary.com/monopolize


Adjoint representation of the Lorentz group

Question: Is it possible to construct an adjoint representation for the Lorentz group?
Comment: Would [math.se] be a better home for this question?
Answer: I suspect the impeccable WetSavannaAnimal answer is too abstract and general to satisfy the OP. Let me vulgarize it a bit. The Lorentz group has 6 generators, so your adjoint rep will be a 6-dimensional rep, a set of 6x6 matrices, which satisfy the same Lie algebra (commutators) as the conventional 4x4 matrices transforming the fundamental (x,y,z,t) 4-vector. As an aside, note the curvature form (parity inv. 2-form) of the Lorentz group is also 6-dimensional.
Recall the Lorentz group may be written in a pretty form, that is rotations $J_i\equiv  −\epsilon_{imn} M_{mn} /2 $   and boosts, $K_i\equiv M_{0i}$, so that
$[J_m,J_n] = i \epsilon_{mnk} J_k$ , $[J_m,K_n] = i \epsilon_{mnk} K_k$ , $[K_m,K_n] = -i \epsilon_{mnk} J_k$.
Might further note the important simplification  $[J_m +i K_m,   J_n −i K_n]$,  which permits reduction of the Lorentz  algebra to su(2)⊕su(2) and efficient treatment of its associated  representations.
Now consider the six 6x6 matrices with the 3 J+iKs on the upper left 3x3 subspace of these 6x6 ones, and the 3 J-iKs on the lower right block,
spanned by the 4,5,6 indices. Keep the indices of the upper left block  to be the usual 1,2,3; and rename the indices of the lower right block from 1,2,3 to 4,5,6. The commutation relations are then manifest, and the structure constants $f_{mnl}$ are sparse, basically $\epsilon_{ijk}$ for
indices (1,2,3) or (4,5,6), and zero otherwise.
So, then, as you probably learned from SU(3), these very structure constants f provide your 6x6 matrices in the adjoint, with one of the 3 indices (taking 6 values) specifying the particular generator of the Lorentz group represented.
Comment: As you explained, the adjoint rep should be $6$ dimensional. Using the usual labelling of the irreps $(j_1,j_2)$ , is the adjoint $ (1,0) \oplus (0,1) $  (which is 6-dimensional, but reducible) or $(1,1)$, which is 9-dimensional...? From your descriptions it sounds like it is indeed the $ (1,0) \oplus (0,1) $ rep, but everywhere I've found that the $(1,1)$ rep is the spin 2 rep. (c.f. https://en.wikipedia.org/wiki/Representation_theory_of_the_Lorentz_group#Common_representations)
Comment: @JakobH ,  I am not sure why you are quoting WP for something it does not state: it certainly does not state, or imply, that the (1,1) rep is the adjoint of this group. The Lie algebra has only 6 generators, so, of course, the dimension of its adjoint rep is 6, not 9. Where on earth ("everywhere"??) did that 9-d rep come from? It is only the rep of the energy-momentum tensor, no?  The Lie algebra itself is "reducible", **su(2)** ⊕ **su(2)** , if you *really* wished to abuse this language!
Comment: Yes, unfortunately, for whatever reason, this extremely long Wiki article does not mention the adjoint rep.  My flawed logic was the following: in gauge theories of gravity with a gauged Lorentz group the gauge bosons live in the adjoint rep. Naively these gauge bosons should be gravitons and a common statement is that these have spin 2. The Wiki article states that the $(1,1)$ rep is the spin $2$ rep and this led to my confusion that the $(1,1)$ rep could be the adjoint rep.
Comment: Oh, OK, *far too naively*. The actual gauge fields of the tangent space Lorentz group are the *spin connections* $\omega_\mu^{ab}$ and these are 6 vector fields, as the Lorentz indices *a,b* are 4 each, so 4x3/2=6.  They eventually end up getting dressed up into metric components through Vierbeine... Their field strength is a half-dressed Riemann tensor... But that is another question...
Comment: [Spin connection](https://en.wikipedia.org/wiki/Spin_connection).
Comment: Thanks for pointing me in the right direction! Again I find a bit strange that in the Wiki article and elsewhere there is no mention of the rep these gauge fields transform under... (To particle physicist almost everything is representation theory,  but GR people seem to rarely talk about these things)
Answer: Every Lie group has an adjoint representation. I'm not sure what definition you come at the adjoint representation from, but here's the fundamental one which I'm sure you'll see is always meaningful.
Think of a $C^1$ path $\sigma:[-1,1]\to\mathfrak{G}$ through the identity in a Lie group $\mathfrak{G}$ with $\sigma(0) =\mathrm{id}$ and with tangent $X$ there.
Now think of a general group member $\gamma \in\mathfrak{G}$ acting on that path so that $\sigma(t) \mapsto \gamma^{-1} \sigma(t) \gamma$. This too is naturally a path through the identity and has a transformed tangent $X^\prime$ there. This transformation is linear. We say the group "acts on its own Lie algebra $\mathfrak{g}$" in this way and write the transformation $X\mapsto\gamma^{-1} X \gamma$. This is just notation, but in a matrix Lie group it is also a literal matrix product. Perhaps, then, less confusingly we write $X\mapsto {\rm Ad}(\gamma) X$, where the linear transformation ${\rm Ad}(\gamma)$ wrought by $\gamma$ on the Lie algebra is now a member of $GL(\mathfrak{g})$, the group of invertible linear transformations of the Lie algebra $\mathfrak{g}$ thought of as a plain vector space. The association
$$\rho:\mathfrak{G}\to GL(\mathfrak{g});\;\rho(\gamma)= {\rm Ad}(\gamma)$$
is a homomorphism as is readily shown. This is sometimes called the Adjoint Representation of the Lie group. Through the big Ad Adjoint representation $\mathfrak{G}$ is mapped to a new Lie group, this time always a matrix Lie group, a subgroup of $GL(\mathfrak{g})$.
Now, then, we can look at
$${\rm ad}(X)\stackrel{def}{=}\left.\mathrm{d}_t {\rm Ad}(e^{t\,X})\right|_{t=0}$$
This too is a linear operator on $\mathfrak{g}$, although in general not an invertible one. Indeed, you can show without too much strife that:
$${\rm ad}(X) Y = [X,\,Y]$$.
So in fact the big Ad adjoint representation induces a homomorphism of Lie algebras $\rho^\prime:\mathfrak{g} \to {\rm Lie}(\rho(\mathfrak{G}))$. This too is a homomorphism of linear spaces and moreover a homomorphism that respects Lie brackets. It is thus a Lie algebra homomorphism and it is also called the adjoint representation (of the Lie algebra). I like to quaintly call it little ad adjoint representation.
Now here for me is one of the most beautiful equations around:
$${\rm ad}([X, \,Y]) = [{\rm ad}(X),\,{\rm ad}(Y)] = {\rm ad}(X) . {\rm ad}(Y)-{\rm ad}(Y) . {\rm ad}(X)$$
This is a restatement of the fact that little ad respects Lie brackets. But it is also a form of the Jacobi identity in disguise. Wow! That's the REAL meaning of the Jacobi identity: it's there so the adjoint representation of a Lie group, clearly a very basic and fundamental thing, induces a homomorphism in the corresponding Lie algebras that respects Lie brackets. Everything's exactly as we would expect and so, if you're ever designing a Universe, that's why you must remember to throw the Jacobi identity in! Write a note for yourself now so you don't forget! Now my LaTeX skills aren't up to drawing a commutative diagram from memory, so I hope you can see there is a pretty neat and simple one.
Just another couple of interesting facts. The kernel of Big Ad is the centre of the the Lie group. The kernel of little ad is then the centre of the Lie algebra. So if the Lie group is simple, i.e. contains no normal Lie subgroups, then there cannot be a continuous centre to annihilate by the homomorphism. Same is true if the group simply has no continous centre for reasons other than simplicity. So there are no bits of the Lie algebra that get wiped out. The original Lie group and the image of big Ad then have exactly the same Lie algebra. If further there is no discrete centre in a simple Lie group and if the group is connected, then the Lie group and the image of the adjoint representation are the same Lie group.
Okay. So now let's specialise to the Lorentz group. There is no continuous centre, by inspection of the commutation relationships. Therefore the Lie algebra of the image of big Ad is exactly the same as the Lie algebra of the Lorentz group. Nor is there a discrete centre in the Lorentz group. $\ker(\rho) = \{\mathrm{id}\}$, so, by the homomorphism theorem, the image of $SO(1,3)$ under the big Ad adjoint representation is the Lorentz group itself.
Comment: I always thought that the Christoffel symbols are in the adjoint representation of the Lorentz group. They are the required Gauge connection and are thus in the adjoint representation.
Answer: Every group has an adjoint representation. If you know differential geometry then defining this is straightforward. It is a map:
$Ad: G \rightarrow Aut(G')$
Where $G$ is a group and I write $G'$ for its Lie algebra.
More precisely, it is defined via the conjugation map:
$Cj: G \rightarrow Aut(G)$
So $Cj(g): G \rightarrow G$
And we set $Cj(g).h := g^h = hgh^{-1}$
So taking the tangent bundle gives
$TCj(g): TG \rightarrow TG$
And so the tangent space at the identity is:
$T_eCj(g):T_eG \rightarrow T_eG$
But the tangent space at the identity of any Lie group is simply its Lie algebra. Hence we have, $T_eG = G'$ and so the preceding is:
$T_eCj(g): G' \rightarrow G'$
And we set:
$Ad(g) := T_eCj(g)$
It turns out that the adjoint map for $SU(2)$ is simply the double covering map of $SO(3)$.


connect to openvpn using username and password

Question: Is it possible to connect to an openvpn server using a username and password. I read that you need a config file on ubuntu, but I don't have a config file.
How can I connect to my openvpn server using credentials?
Answer: Yes it's possible.
To do this, you have already an OpenVPN server installed, and the user created in the server.
The easiest openvpn client is network-manager. If you are using Ubuntu run:
aptitude install network-manager-openvpn
restart network-manager

Now click on the network-manager applet, select configure VPN, and setup a new open-vpn connection.
Set the gateway to you server
Set the type to Password
Point your CA to a copy of your server's ca.crt and everything should just work
Attached is a simple client configuration file that will work.
Edit it to match your server's settings where appropriate.
You will need this and your ca.crt in the same directory.
On linux my file is called /etc/openvpn/client.conf
##############################################
# Sample client-side OpenVPN 2.0 config file.
# for connecting to multi-client server.
##############################################

# Specify that we are a client and that we
# will be pulling certain config file directives
# from the server.
client

dev tun
proto udp

# The hostname/IP and port of the server.
remote my-server-2.domain 1194

# host name of the OpenVPN server.  Very useful
# on machines which are not permanently connected
# to the internet such as laptops.
resolv-retry infinite

# Most clients don't need to bind to
# a specific local port number.
nobind

# Try to preserve some state across restarts.
persist-key
persist-tun

# Certificate Authority
ca ca.crt

# Username/Password authentication is used on the server
auth-user-pass

# Verify server certificate by checking
# that the certicate has the nsCertType
# field set to "server".  This is an
# important precaution to protect against
# a potential attack discussed here:
#  http://openvpn.net/howto.html#mitm
#
# To use this feature, you will need to generate
# your server certificates with the nsCertType
# field set to "server".  The build-key-server
# script in the easy-rsa folder will do this.
ns-cert-type server

# Set log file verbosity.
verb 3

# To start the openvpn client, simply type:
# openvpn --config /etc/openvpn/client.conf

That's it.
Answer: On the server side you need a line like:
plugin /usr/lib/openvpn/openvpn-plugin-auth-pam.so openvpn

The above will enable the pam plugin and make it to use the /etc/pam.d/openvpn file as config (note: file does not exist by default, you may use 'login' instead of it to validate unix credentials or set up the openvpn one with the authentication method of your choice (ie: google authenticator))
On the client side as mentioned above you should use auth-user-pass in openvpn configfile, or if you use network manager choose "Password with Certfificates" (User + Pass + cert based auth) or the simple "Password" option (User + Pass based auth) on the Identity tab.


Django user logout with remote authentication

Question: I have a website that uses remote authentication via Kerberos. I've set everything up according to the Django documentation (https://docs.djangoproject.com/en/dev/howto/auth-remote-user/). Logging in works great, however logging out doesn't end the session. What can I do to ensure the user is fully logged out? Closing the browser window entirely does seem to work, but users need to have the ability to log out without closing the browser window.
Answer: When user logs out from django, django session is destroyed, but user is still logged to Kerberos, and next request will automagically create new session again. You can't do anything about it in your django application. User have to log out from Kerberos himself, because Kerberos might be used to authenticate user in other services/applications at the same time.
Answer: Use HttpResponse to return status 401.
<code>def logout(request):
    return HttpResponse(content, status=401)
</code>
This will log the user out from Kerebos.
Credit: I saw this solution elsewhere related to Basic Authentication.


US Visa DS 160 and appointment

Question: I recently completed the DS 160 form and sent it to an agent to assess it, the agent didn't know I had already created my profile in Cgi federal for an appointment and paying visa fees. So, he created another one in my name through which he gave the info sheet to pay the visa fees. The one I created I only stopped at visa class and it doesn't have all information including passport and all of that, it only has my full name in it. What should I do? Would that be a problem?
Thanks
Answer: This should be fine.
At some point I had two accounts with all DS-160 details filled in because I've lost access to the earlier one.
Make sure you reference correct application number and don't mix them up.

US Visa DS 160 and appointment


Difference between heteroskedasticity and overdispersion

Question: Are both terms equivalent? They seem very similar to me. Or does one imply the other?
Answer: No, they are not equivalent. In fact, they are quite unrelated.
Heteroskedasticity is when variance differs between "situations". For instance, in a regression task, the variance of the residuals may be larger or smaller, depending on whether a particular predictor has a certain value or not.
Overdispersion is when variance is greater than the expectation.
For instance, in OLS we assume homoskedasticity (and if we suspect heteroskedasticity, we should use special tools). However, we make no assumptions on equi- or overdispersion. In fact, since we assume that all residuals have the same variance (per the homoskedasticity), while the conditional mean can be any value at all, we might have all of under-, equi- or overdispersion for different observations if they just differ in their conditional means enough.
Conversely, a Poisson regression assumes equidispersion: each observation's conditional variance should be equal to its conditional expectation. But since we are in a regression framework, the expectations will vary with the predictors... and therefore, so will the variances. Thus, in a Poisson regression, we will usually have heteroskedasticity. But it's not a problem here, because Poisson regression handles this automatically.
I work in retail forecasting. Retail time series usually exhibit heteroskedasticity (sales increase during promotions, but so does the variance of sales), as well as overdispersion. (In terms of theory, the negative binomial distribution should be a good description.) However, there are a few products that vary very little and move very fast, like milk, where we might have underdispersion (and still heteroskedasticity, e.g., because of weekday patterns). Or if products habitually go out of stock - maybe the baker consciously plans to go out of stock so they don't have stock left over at the end of the day: censored demand - we will often again have underdispersion.


How to set the same number of datapoints in the different ranges in correlation chart

Question: I am beginner in working with machine learning. I would like to ask a question that How could I set the same number of datapoints in the different ranges in correlation chart? Or any techniques for doing that? . Specifically, I want to set the same number of datapoints in each range (0-10; 10-20;20-30;...) in the image above. Thanks for any help.
Comment: What is the reason for wanting to do that?
Comment: In my actual case, after getting the correlation chart, it somehow got the saturation and overestimation. I asked for my supervisor and he just said to me for doing that. I have no ideas how to do. If you have any suggestion or guide me to do that, I strongly appreciate.
Comment: there are ways to make the data less dense depending on the language.  you can also do this manually by binning the output data. e.g taking the midpoint of each of the ranges e.g 0-5=2.5.    That would reduce the density but not sure how to exactly same number in each bin.  that would take a little programming.
Comment: Have you tried something like that before? Or just an example in Python, JavaScript or R.
Comment: I've edited my real chart case.
Answer: You can bin your variables to prevent overplotting and make the output cleaner.
Here is an example from StackOverflow:
https://stackoverflow.com/questions/16947210/making-binned-scatter-plots-for-two-variables-in-ggplot2-in-r
This may not be exactly what you need since you did say you want the same number of datapoints in each range (or bin).  You would have to add some code if you wanted that exact format.


Importing LaTeX table code into LyX

Question: I create tables in a spreadsheet program, convert them to LaTeX code and then use an ERT to import into my LyX document.  When I paste the LaTeX code the entire thing turns out very ugly looking. The table, however, works fine but its impossible to edit within LyX due to the ugly formatting when pasted.  Is there a way to import a LaTeX file into a table float instead of inserting an ERT in LyX?
Comment: Try to (1) export your LyX document (including only the ERT table) to `Latex (pdflatex)` and then (2) to import the resulting `.tex` file back into LyX. If the table is not too complex and you are lucky, LyX will transform it into a LyX table that then may be Copy&Pasted to your original document and edited there with LyX's table editor.
Comment: @alex Is it possible posting any snapshot?, please.
Comment: this is my least favorite thing about Lyx. I've come up with another (unsatisfactory) workaround: the table pastes just fine if there are two new-lines between every line (it collapses one of them). So when I really need to edit the table I copy-paste to a text editor, add a second newline, then paste to LyX.
Comment: I personally am always copy-pasting `R` output to `TeX`, so this solution may be too narrow, but building on my comment above, the following (inspired by [this](http://stackoverflow.com/questions/12148770/changing-the-color-of-negative-numbers-to-red-in-a-table-generated-with-xtable) solution to different question) works: `cat(capture.output(xtable(your_table)),sep="\n\n")`
Answer: If the table (LaTeX code) is saved in a file table1.tex it is possible to include it into the float (or elsewhere) in LyX by using "Insert > File > Child Document..." with "Include Type: Input".
For editing the table in file table1.tex, you can right-click on the field "Input: table1.tex" and choose "Edit Included File...". A text editor should open up displaying table1.tex.
Comment: If you like this way better, you might like using instant preview, which will show what the table will look like in LyX when it's rendered to PDF (if this doesn't make sense, try it to see): First make sure instant preview is turned on. Go to Tools > Display and  "Instant Preview" should be "on". Then highlight the child document inset (the grey box with the tex "Include: yourfile.tex") and go to Insert > Preview.
Answer: If it's a simple file (simple in terms of simple LaTeX. It can be as wide or as long as you want), you can use the tex2lyx program. The easiest way to do this is to go to File > Import > LaTeX (plain). Then copy that table and paste it into your main document.

Importing LaTeX table code into LyX


Sobolev space inclusion

Question: Let $U$ be a bounded open domain in $\mathbb{R}^n$. If $W_0^{k,p}(U)$ is the completion of $C_c^{\infty}(U)$, the space of smooth and compactly supported functions on $U$, with respect to the $W^{k,p}(U)$ norm, does the inclusion $$W^{k_2,p}(U)\cap W_0^{k_1,p}(U)\subset W_0^{\max\{k_1,k_2\},p}(U)$$ hold?
Comment: Note that $W_0^{0,p}$ is just $L^p$, so $W^{1,p} \cap W_0^{0,p}$ is still $W^{1,p}$ and is not contained in $W_0^{1,p}$. PhoemueX's answer shows that the same happens for general $k_1<k_2$.
Answer: No, this is not true in general. To see this, consider the one-dimensional case $U = (0,1) $. Then, if $k_2 > k_1$,
$$
W^{k_2, p} \cap W_0^{k_1, p}
= \{ f \in W^{k_2, p} \,:\, \partial^\alpha f (x)=0 \text { for } |\alpha| \leq k_1 - 1\text { and } x \in \{0,1\} \}
$$
is not a subset of
$$
W_0^{k_2,p} = \{ f \in W^{k_2, p} \,:\, \partial^\alpha f (x)=0 \text { for } |\alpha| \leq k_2 - 1 \text { and } x \in \{0,1\} \},
$$
as can already be seen by considering suitable polynomials.

Sobolev space inclusion


Community bot posts additional bounty

Question: I just saw something for the first time where the user Community posted an additional bounty on this question. As it turns out, this user (Community) is a bot. I thought this was amusing, but I'm also curious how is a background process empowered to do this? What criteria must be met by an answer to trigger the bot to do post a bounty? How big can these bounties be?
Answer: Information systems that associate an owner with all objects in them need to account for what happens to those objects when their original owners go the way of all things.
What happened in this specific case is that someone posted a bounty and then self-deleted their own account.
Whenever that happens, the orphaned bounty is adopted by, or if you would, "re-owned" by the Community user (user id == −1).
This is like how when the parent process exits while it still has running children, those orphaned child processes are adopted by init (process id == 1), so that there are never any unowned processes.
It remains an open question whether the moderation team or the SE team will be manually awarding those bounties before they expire, or whether the default 50% unawarding process will be followed.  My bet is on the latter, but I may be wrong.
Comment: @oerkelens That is something you would have to ask of the self-deleted user, who is no longer with us.
Comment: @tchrist did you mean to use a commonly-used euphemism for "dead" in that comment?
Comment: That does not explain why a bounty was posted on that question, 3 years after it was posted and answered?
Comment: Hello! It seems that you have refunded the *wong* bounty. Please see chat, and "take the biscuit" question.

Community bot posts additional bounty


Nurses in India

Question: In many Hindu serials especially Mahabharata we see female nurses assisting a male doctor my question is do these nurses who exist historically/religiously in scriptures get paid? Some argue women cannot get a job that has a salary as per Shastra, but this is untrue since well I won't say it, but here is the source: [Source] So, to reiterate my question is were nurses present in ancient India and did the role include females who worked for a salary?
Comment: Hospitals didn't exist in India before British rule. And the concept of nursing as a profession didn't emerge globally before the Crimean War. Hence it's absurd to ask whether nurses existed in ancient India.
Comment: There are certain anecdotes about Buddhist nuns performing roles similar to modern-day nurses but such activities are no different from that conducted by the Christian nuns in hospices run by the MoC, so they can't be technically considered as nursing.
Comment: @অনু hospitals didn't I mean during war time and we know medical practitioners did exist correct?
Comment: Yes doctors did exist, but not nurses. Besides, you should know that here we don't Wikipedia as a authority, let alone serials.
Comment: @অনু I know I am just making sure you understood the question correctly.
Comment: @অনু There is a concept of nurse in cikitsāśāstras, as i have discussed in my answer. It's just that the concept is a bit different that West, and the paricaras (which play the role of both nurse & attendant/helper of doctor) are not necessarily female. You can check my answer to this post.
Answer: In order to talk about 'nurses' in the framework of cikitsāśāstras and Indian civilization, some basics regarding cikitsā and its practice as mentioned in śāstras, need to be detailed.

bhiṣag dravyamupasthātā rogī pādacatuṣṭayam /
guṇavatkāraṇaṁ jñeyaṁ vikāravyupaśāntaye //
~ Carakasaṁhitā sūtrasthāna (9.3)

Bhiṣaj (vaidya), dravya, upasthātā (paricāraka), & rogī, constitute the four fundamental pādas of cikitsā, these four upon being guṇayukta lead to the alleviation of rogas.
Here, upasthātṛ is a paricāraka, who is like an assistant, helper or attendant of the vaidya, as specified by Cakrapāṇidatta in his commentary (comm.) Āyurvedadīpikā. The next two quoted ślokas specify the diff. b/w vaidya & paricāraka, by stating their guṇas.

śrute paryavadātatvam bahuśo dṛṣṭakarmatā /
dākṣyaṁ śaucamiti jñeyaṁ vaidye guṇacatuṣṭayam // (9.6)
upacārajñatā dākṣyamanurāgaśca bhartari //
śaucaṁ ceti catuṣko'yaṃ guṇaḥ paricare jane // (9.8)
~ Carakasaṁhitā sūtrasthāna

The four guṇas of a vaidya are-  pūrṇarūpa jñāna of cikitsāśāstras (śrute-paryavadātatvam), extensive practical experience in cikitsākarma (bahuśo-drṣṭakarmatā), dākṣyam, & śaucam. - (9.6)
The four guṇas of a paricāraka are- upacārajñatā, dākṣyam, loyalty/devotion to the doctor (anurāga ca bhartari), & śaucam.- (9.8)
In his comm. Āyurvedadīpikā, Cakrapāṇidatta adds that upacārajñatā means being jñāta on how to prepare yūṣa (soup), rasa (juice), etc., do patients' (rogin) massage, make them sleep, do their sevā, etc. (upacārajñatā yūṣarasādikaraṇasaṁvāhanasvāpanādijñatā). So, the requirements of theoretical expertise of śāstras and practical experience are not made of the paricāraka of vaidya, the paricāraka literally meaning one who does paricara i.e. follows after, and so, paricāraka's first allegiance is to the vaidya s/he is working under, instead of the patient. Apart from dākṣyam & śaucam, a paricāraka must be upacārakuśala, knowing exactly what is involved in tending and attending to a patient. Caraka saṁhitā sūtrasthāna (15.7) further specifies that they (paricārakas) should be able prepare pulses or soups, cook rice, give baths (snāpaka) & massages (saṁvāhaka), lift patients (utthāpaka), help them lie down again (saṁveśaka), grind necessary auṣadhis (auṣadhapeṣakaṁsca). All such paricārakas should have the requisite guṇas of of śīla, śauca, anurāga, dākṣya, and respectful behavior (prādakṣiṇya). They should be upacārakuśala, trained in all necessary karmas (sarvakarmasu paryavadātān), and not averse to any required karmas (sarvakarmasvapratikūlān), and so on.
Suśrutasaṁhitā sūtrasthāna (34.24) states that a paricāraka should have the following guṇas-  friendly/tender (snigdhaḥ), non-reproaching (ajugupsu), balavāna, dedicated in the caring for the patient (yukto vyādhitarakṣana), obedient to the vaidya's orders (vaidyavākyakṛd), and tireless (aśrāntah).  [snigdho'jugupsurbalavān yukto vyādhitarakṣaṇo / vaidyavākyakṛdaśrantaḥ pādaḥ paricaraḥ smṛtaḥ].
There is no norm in Āyurveda unlike modern west that the paricārakas should be specifically female in all cases (i.e. for all patients). Since, most of the directives in cikitsāśāstras refer to the patient in masculine form, the paricāraka too is assumed to be a man. But we can safely assume, that the gender of the paricāraka may most likely depend on the gender of the patient. Usually, where a female patient is specified in śāstras, the paricara (of vaidya) is always female. Thus, a midwife is assumed to be a woman. Suśruta saṁhitā śārīrasthāna (10.8) explains that there should be four female paricārakas (catasraḥ striyaḥ) at the bedside (of the garbhiṇī), who are mature (pariṇatavayasaḥ), not easily upset (aśaṅkanīyaḥ), skilled at giving birth (prajananakuśalaḥ) and whose nails are cut short (kartitanakhaḥ). [tataḥ kṛtopadhāne mṛduni vistīrṇe śayane sthitāmābhugnasakthīmuttānāmaśaṅkanīyāścatasraḥ striyaḥ pariṇatavayasaḥ prajananakuśalāḥ kartitanakhāḥ paricareyuriti].
The dhātrī (wet nurse) is of necessity female. Carakasaṁhitā śārīrasthāna (8.52) & Suśrutasaṁhitā śārīrasthāna (10.25) describe her in some detail. Caraka states in śārīrasthāna (8.52) that a dhātrī should be - belonging to the same varṇa as that of baby (samānavarṇa),  yuvatī (yauvanasthāṁ), viśvāsapātra, rogarahita, without any hīna or more aṅgas, without durvyasanas (addiction/vices), not kurūpa, without nindita svabhāva/veśa, belonging to the baby's deśa, not kṣudra, not one who does kṣudrakarma, from uttamakula, having affectionate disposition towards children, having a living male child, having profuse lactation,  not careless, not sleeping when the baby has done excrements, not patita by dharma or ācaraṇa, kuśala in baby's sevā, śuddha, having dveṣa of aśuddhatā, and endowed with excellent breasts & milk. (nibhṛtāmanāturāmavyaṅgāmavyasanāmavirūpāmajugupsitāṃ deśa jātīyām akṣudrām akṣudrakarmiṇīṁ kule jātāṁ vatsalāmarogāṁ jīvadvatsāṃ puṃvatsāṁ dogdhrīmapramattām anuccāraśāyinīmanantyāvasāyinīṃ kuśalopacārāṁ śucimaśucidveṣiṇīṁ stanastanyasampadupetāmiti). The next ślokas (8.53-57) detail the properties of excellent breasts and characteristics, complications, etc. of breast milk.
In śārirasthāna (8.58), Caraka states dhātrīkarma-  when a dhātrī with madhura & śuddha breast milk is found, then before breastfeeding, she should take a snāna, apply candana paste on stanamanḍalas, wear a clean white dress, and put one of these auṣadhis (upon herself) - aindrī, brāhmī, śatavīryā, sahasravīryā, amoghā, avyathā, śivā, ariṣṭā, vāṭyapuṣpī, and viśvaksenakāntā. Then taking the baby, she should sit facing the east and make him suckle the right breast first. (dhātrī tu yadā svādubahulaśuddhadugdhā syāttadāsnātānuliptā śuklavastraṁ paridhāyaindrīṃ brāhmīṃ śatavīryāṁ sahasravīryāmamoghāmavyathāṃ śivāmariṣṭāṁ vāṭyapuṣpīṁ viṣvaksenakāntāṁ vā bibhratyoṣadhiṁ kumāraṁ prāṅmukhaṁ prathamaṁ dakṣiṇaṁ stanaṁ pāyayet //). Suśruta in śārirasthāna (10.25) adds that while breastfeeding, the baby's face should be towards east while the dhātrī's face should be towards north. (tithau śiraḥsnātamahatavāsasamudaṅmukhaṃ śiśumupaveśya dhātrīṁ prāṅmukhīṁ copaveśya dakṣiṇaṃ stanaṃ dhautamīṣatparisrutamabhimantrya mantreṇānena pāyayet)
As per Kauṭilya's Arthaśāstra (10.3.47), on battlefields and in army camps, women play an auxiliary role and are mentioned with vaidyās (and their paricaras who carry instruments, etc.), being at the rear of army. This is stated as follows-

cikitsakāḥ śastrayantrāgadasnehavastrahastāḥ striyaścānnapānarakṣiṇyaḥ puruṣāṇāmuddharṣaṇīyāḥ pṛṣṭhatastiṣṭheyuḥ //

Cikitsakas (vaidyas) carrying śastras (surgical instruments), yantras (medical apparatus), agadas (medicines & auṣadhis) , sneha (medicinal oils), vastras (bandages), and women in charge of food & drink with the duty of emboldening the men, should be stationed at the rear (of the army).
Kauṭilya's Arthaśāstra (3.13.30) further states -

kāruśilpikuśīlavacikitsakavāgjīvanaparicārakādirāśākārikavargastu yathānyastadvidhaḥ kuryāt / yathā vā kuśalāḥ kalpayeyuḥ tathā vetanaṁ labhet /

The group of those who work in hope (or renumeration) such as kāru (artisans), śilpi (craftsman), kuśīlava (performers), cikitsakas, vāgjīvanas (bards), paricārakas, and others should receive a renumeration (vetana) as others of that type do or as experts may determine.
So, as per Kauṭilya's Arthaśāstra (3.13.30), cikitsakas and paricārakas (including female ones such as dhātrīs) came in the category of āśākārikas, i.e. those who work in the expectation of renumeration/reward, rather than a wage. Jain (1947:179) states the following -

If a vaidya was brought to the home of a patient, he was provided with a hot bath and other comforts. He has to be fed on old rice and was provided with costly utensils. After the examination the vaidya was to be given his proper fees.
There were state vaidyas as well. It's said that once a vaidya did not treat the queen of a certain king properly and so he was put to death. Another physician was addicted to gambling and he didn't attend to his work. His śāstra on cikitsā was stolen by robbers and his surgical instruments rusted. When this was brought to the notice of the king, he stopped his salary.

Therefore, there shouldn't be any doubt that many vaidyas received payment for their services, and so the paricaras would most likely attain payment too, either from the vaidya or from the patient directly. A third source of payment for the vaidya & paricaras might be grants from external sources (other than patient).
There's a late 11th century inscription on the wall of the temple at the town of Tirmukkūḍal, not far from Kandhi, covers 540 sq.ft., and is 55 ft. long. The Tirmukkūḍal inscription records grants given by Vīrarājendra (i.e. Cola king Rājakeśarivarman) for the support of a Saṁskṛta viśvavidyālaya, vidyārthīs, hostel, and an āturaśālā (kinda-hospital/clinic). The inscription provides detailed information on Vīrāśōḷan "house for the sick" (āturaśālā). The clinic was provided with 15 beds. The vaidya's name was Savarṇan Kōdaṇḍarāman Aśvatthāma-Bhaṭṭan of Ālappākkam, who was a recipient of land grants to support his medical prescribing. He was paid annually 90 kalams of paddy and 8 kāśus, as well as a grant of land, for prescribing medicines to the patients lying in the clinic, for the dāsas attached to the institution and for the adhyāpakas & vidyārthīs of the vidyālaya. In addition to the vaidya, the inscription mentions the following staff (Wujastyk 2022: 21-22)-

one surgeon who received 30 kalams of paddy
two persons for fetching medicinal hebs who were paid 60 kalams of paddy and 2 kāśu; these staff also supplied firewood and prepared medicines;
two paricaras, who received 30 kalams of paddy and 1 kāśu, and attended patients and administered medicines
a barber who received 15 kalams of paddy
there were also cleaners

So, in this case, not only the vaidya but even paricaras of this kinda-clinic attached to the vidyālaya, are being funded by the royalty.
In Mahāsāṅghika Lokottaravāda Bhikṣuṇī Vinaya (pāccatika-dharma 82), there's an account regarding a Bauddha bhikṣuṇī named Candakamātā, who was a vaidyikā.

The Lord was staying at Kauśāmbī. A bhikṣuṇī, Candakamātā, was accepted in the inner household of the king. She was skilful with root, leaf and fruit medicines. With the relatives of the king, ministers, merchants, and [other] excellent families, she established a (basti) for women. She cured fainting fits, offered medical treatments such as black ointment, [other] ointments, emetics, purgatives, sweat-treatments, nasal treatments, and bloodletting. Before she left, she received soft food and hard food. Then the bhikṣuṇīs found out, "This is not a bhikṣuṇī" they said, "This is a vaidyikā." The bhikṣuṇīs informed Mahāprajāpati about the situation. The Lord said, "Summon Candakamātā." She was summoned. The Lord said, "Is it true, Candakamātā, that you make a living by cikitsā-vidyā ?" "Yes, Lord." "This is badly done, Candakamātā. It is not suitable to make your living by cikitsāvidyā." And the lord [ordained]..."Whichever bhikṣuṇī makes a living by cikitsā-vidyā commits a fault requiring expiation."
(Langenberg 2014: 161-162)

As we see, Candakamātā had opened a kinda-clinic where she treated women patients. Although she did not accept direct payment (as that's against Vinaya), she received food from the women she treated. Her fellow bhikṣuṇīs found her cikitsākarma as not aligning with bhikṣuṇīdharma and accused her of acting like a vaidyikā, instead of a bhikṣuṇī. And then, Buddha ordained that bhikṣuṇīs couldn't make their living by means of cikitsāvidyā, or they would commit a fault requiring expiation. But clearly, in this Vinaya story, it's not mentioned that Candakamātā provoked any criticism of laypeople, as her lay patrons welcomed her medical expertise and flocked to her for cures for variety of ailments.
Comment: Very cool, and informative answer, I am wondering what other jobs females did I know this and prostitution is one, but the latter is evil and against chastity I believe.
Comment: @Haridasa Well, even veśyās had a certain respect in pre-Izlamic Indian society, the Indians were not affected by Victorian prudishness then. They aren't seen as 'evil' in Indian tradition.
Comment: I am very curious on this regard I am no liberal, but the complexity of society is beautiful

Nurses in India


prove that NDFA = {< M1,M2 > | M1 and M2 are DFAs such that there is at least one string x that is accepted by neither M1 nor M2} is decidable?

Question: How can I prove that $NDFA = \{ \langle M_1,M_2 \rangle | M_1$ and $M_2$ are $DFA$s such that there is at least one string $x$ that is accepted by neither $M_1$ nor $M_2\}$ is decidable using the fact that $ANFA = \{\langle N \rangle | N$ is an $NFA$ with some input alphabet $\Sigma$, and $L(N) = \Sigma^*\}$?
Answer: We try to decide the language $NDFA$ using a language that decides $ANFA$.
If $x$ is not in neither $L(M_1)$ or $L(M_2)$, then it means that it's not in $L(M_1) \cup L(M_2)$, in other words there is at least one string $x$ which is not in $L(M_1) \cup L(M_2)$, thus $L(M_1) \cup L(M_2) \neq \Sigma^*$ We know that we can (in finite time) construct a $NFA$, $M$ for language $L(M_1) \cup L(M_2)$. Now since $ANFA$ is decidable, $\overline{ANFA}$ is also decidable, so there is a $TM$ that can decide that $L(M) \neq \Sigma^*$, therefor the original language $NDFA$ is also decidable.
Comment: Thanks. I realize I was having difficulty in understanding the neither/nor statement.
Comment: @N.Ali, No problem, glad I could help


How can my non EU girlfriend stay in Italy more than 90 days?

Question: My girlfriend wants to come to Italy with me (an EU citizen). She is not an EU citizen and wants to stay for more than 90 days in Italy. She is 19 years of age and has a grandaunt (who is an Italian citizen) that is willing to let her live in her house, and declare that she is taking care of my girlfriend. What laws, if any, would allow such a person to stay more than 90 days in Italy?
Comment: You might be better off asking in Travel. We can only answer legal questions, they can offer practical advice.
Comment: I don't think they needed practical advice where she should make arrangements or what hotels allow for stays for longer than 90 days, in fact, the question is clear that there will be a place and a care taker there for the girlfriend. The only how is in the realm of law; it is as much a legal question as it gets.
Comment: Maybe you're not familiar with the field of immigration law and relating fields..?
Answer: Unfortunately, there are no EU visa's for non-family members on the basis of a romantic partnership or relationship.
The EU recognizes family members to be facilitated entry and residence as follows:

Article 2
Definitions
For the purposes of this Directive:
1.
'Union citizen' means any person having the nationality of a Member State;
2.
'family member' means:
(a)
the spouse;
(b)
the partner with whom the Union citizen has contracted a registered partnership, on the basis of the legislation of a Member State, if the legislation of the host Member State treats registered partnerships as equivalent to marriage and in accordance with the conditions laid down in the relevant legislation of the host Member State;
(c)
the direct descendants who are under the age of 21 or are dependants and those of the spouse or partner as defined in point (b);
(d)
the dependent direct relatives in the ascending line and those of the spouse or partner as defined in point (b);
3.
'host Member State' means the Member State to which a Union citizen moves in order to exercise his/her right of free movement and residence.

Some Member States, potentially Italy included, recognize domestic partnerships as equivalent to marriages.
Maybe you and her are able to come to peace with this idea that for a longer stay, you two would have to make deeper commitments. Maybe that is a non-question, and can work well if that is the case in Italy.
Other than that merely on this basis, there is not much room to consider.
If there are other facts that may be relevant, for e.g., she may apply to a university and get a student visa; maybe the familial tie you mentioned or other familial ties or ancestry may entitle her to naturalization and citizenship, but these questions vary from state to state; if she's from a widely recognized less democratic country where she was subjected to or has a well-funded fear of persecution as a member of a social, ethnic or "racial" group or for political opinion or faith, asylum may be another option.
Comment: I would need to look further, but my gut feeling is that this would also vary from Member State to Member State, and although I can read Italian, another member will probably be better suited to answer that one. Credo che voi siate insieme qualche moda nel'UE e nel'Italia!
Comment: Thank you for your answer, in the case that she would have to take a course (in uni for example) in order to get a student visa, would it only be valid for the duration of the course? Or could she stay in Italy from prior to the date?
Answer: The easiest way would be to marry your girlfriend, which would qualify her for a spousal visa (in some E.U. countries, this would automatically give her citizenship). Whether or not this is a reasonable solution would depend upon the relationship.


Unable to update from Ubuntu 20.04.6 LTS to Ubuntu 22.04 LTS

Question: I am unable to update from Ubuntu 20.04 LTS to Ubuntu 22.04 LTS.
Getting below info for lsb release and do upgrade command -
lsb_release -a
No LSB modules are available.
Distributor ID: Ubuntu
Description:    Ubuntu 20.04.6 LTS
Release:    20.04
Codename:   focal

do-release-upgrade
Checking for a new Ubuntu release
Please install all available updates for your release before upgrade

What shall I do ?
Answer: Since 22.04 is not a development release, and because 24.04 has not been released yet, simply run the following commands instead.
First, make sure everything is up to date:
sudo apt update
sudo apt dist-upgrade

and then do release upgrade without the -d option:
sudo do-release-upgrade

Also, if you edited /etc/update-manager/release-upgrades to set Prompt=normal then you should set it back to Prompt=lts before trying again.
Comment: `apt dist-upgrade` is essentially the same as `apt full-upgrade` but in any case and unless you have some packages intentionally held back or pinned, this should apply all available updates before the upgrade.


What is the Incoming Radiation on the Earth's Surface?

Question: Why is it that the incoming radiation is spread over an area of a disc = πr^2 rather than half the surface area of the earth?
Answer: The Earth intercepts an amount of solar radiation equivalent to that falling on a disc with its same radius, facing the Sun, but the Earth itself is (roughly) spherical, so that radiation will spread over half its surface area.
Answer: The shadow of the Earth indicates the light area that it absorbs. The shadow is a disk. Remember that light that strikes at an angle (like sunlight that strikes the Earth at dawn or dusk) is less intense.
Comment: If that's the case then πr^2 must be an approximation. Even if the shadow is a circular disc, that does not mean that the light is being distributed on one.
Answer: First assume the light source is far enough away that all light rays are parallel. Then the shadow would be the disk you refer to. This means that that much light struck the Earth. However, that light is not distributed evenly over the lit half of the Earth. (That is why we have seasons).


Make a portable device based on Raspberry Pi

Question: I would like to use RP to create a portable Unix device, something like Nintendo DS. Has anybody have information about the monitor and the power I need to use to make RP as a portable device?
Comment: why on earth would you make a portable Pi? get an android phone, iphone, or maybe ipad or cheap chinese android pad, that already has all the bells and whistles -- touch screen, ARM processor, HDMI output, plenty of memory -- and start playing with it instead of pointlessly spending your time on this very time consuming endeavor.
Comment: @lenik I'm a software guy and want to learn about electronics, and to motivate myself I need to make it portable! Where is the fun on buying full packaged products?!
Comment: @lenik well the only limit is your imagination. Robotics comes straight to mind for portable pi projects or what if you wanted to make a solar powered rover for example or some kind of remote monitoring system?
I'm currently working on some wearable electronics which obviously will need to be portable.
Answer: For monitor/screen information please read this : http://elinux.org/RPi_Screens
You can find information which screen devices are supported. For example this guy uses 1.8" TFT LCD screen.
http://www.whence.com/rpi/
For power supply i suggest USB portable batteries (power bank). I'm using Yoobao YB642 11300 mAh battery with my Rpi. http://www.yoobao.com/en/product.asp?b_id=143&a_id=295
Comment: @HeatfanJohn 11300mAh / 700mA = 16 hours approximately, or significantly less if you plug a lot of peripherals
Comment: I'm using Raspberry w/ BerryBoot (OS disk is USB stick), also there is a USB WIFI adapter plugged to connect network. So with that accessories it can run about 12-14 hours.
Comment: How long can you run your Pi on the YooDao YB642?
Answer: In terms of the screen depending on your need you could have the pi connect to a phones local wifi hotspot and then shh into it with that phone (this works lovely on my iphone i am sure other platforms will have simialr apps)
+1 to using the usb portable batteries. you can get some pretty serious ones now that look similar to an external HDD in terms of size and weight that will power it for hours (probably outlast your phones battery life if that is the method of control you choose)
Answer: Your goal is similar to a project of building laptop with RPi so you can learn from it.


Where is the energy involved in osmosis?

Question: Osmosis creates pressure on the side of the membrane with higher concentration. But where does the energy for this come from?
Answer: All things naturally tend towards equilibrium, moving from high energy to low energy. Osmotic pressure is the force that helps achieve osmotic equilibrium, so it is really just a manifestation of that natural tendency. You don't really need energy to create osmotic pressure, the osmotic pressure will be present until equilibrium is reached. It comes from the higher concentration itself, not some other force. Osmosis is a sort of 'gradient-driven' process, which comes about through entropy.
https://courses.lumenlearning.com/introchem/chapter/osmotic-pressure/
https://simple.wikipedia.org/wiki/Osmosis
Comment: I think this answer skirts around and shows an unawareness of chemical potential energy, might be exactly what he is looking for (or at the very least, related and worth mentioning).
Answer: Consider a U shaped tube with a membrane in the middle, permeable to water but not to salt. It is partially filled with water. The height of the water surface on both sides of the tube will be the same, as they both feel the same atmospheric pressure.
Now we add salt to the left side of the tube, which fully dissolves (and it can't go through the membrane so it all stays there). We observe that the water level on the left rises and on the right side it falls. So there was a net movement of water molecules from right to left.
This movement went against gravity, as it lifted some of the solution, which is now weighting down on the rest of the liquid, on top of the atmospheric pressure already there. Yet the solution stays risen.
Such net movement, implies an attraction between water and salt. Salt dissolves mostly because of entropic effects (rather than, say, forming more stable bonds). Once dissolved, the water molecules are attracted to the charges of the ions, which are stronger than the mere dipole moment other water molecules have.
It is this attractive force that pushes water to the left, and causes the solution to rise. In other words, this attraction supplies the necessary energy to make the solution rise.
Comment: No. If we use sugar to simplify (no dissociation) the water will *not* move from right to left as the pressure due to the *water* remains  the same on both sides.  The side with the sugar  will have higher pressure because that side of the membrane is being hit by the sugar  molecules as well as by the water molecules. In equilibrium that extra pressure is equal to the weight of the added sugar. With dissociation there are extra factors, but they are complications which obscure simplicity of the basic osmosic mechanism.


Identically zero multivariate polynomial function

Question: Let $p$ be a prime and let $F=\mathbb{Z}/p\mathbb{Z}$. Can a nonzero multivariate polynomial $f\in F[x_1,....,x_n]$ such that $\mathrm{deg}_if< p$ for all $i=1,\ldots,n$ be identically zero as a polynomial function $F^n\to F$, where $\mathrm{deg}_if$ stands for the degree of $f$ in $x_i$?
Comment: No. Induct on n.
Answer: No.  The phenomenon of polynomials over fields which are not the zero polynomial but evaluate to the zero function is studied in loving detail in $\S 2.1$ of these notes: in particular, the result you want is Theorem 8.
I also agree with Qiaochu's comment: it would be possible to make much less of a production of this by simply inducting on the number of variables, beginning with the basic fact that no polynomial with coefficients in an integral domain can have more roots than its degree (except the polynomial which has all coefficients zero, if you take the convention that this polynomial has degree $-\infty$).  I have my reasons for the more elaborate treatment given in my notes: in particular, an at least somewhat careful analysis of the situation leads directly to a proof of the Chevalley-Warning Theorem.
Comment: The link to the notes is dead :(
Comment: Great notes! If I may pose a question (even though the original one was answered four years ago): in the proof of Proposition 6, why is it that $Q(t_n)$ has infinitely many roots in $R$? I suppose you could write $Q$ out like $Q(t_n)=\sum_kP_k(t_1,\ldots,t_{n-1})t_n^k=P(t_1,\ldots,t_n)$, but if $P(\alpha)=0$ for all $\alpha\in R^n$, how does that imply that $Q(x)=0$ for all $x\in R$?
Answer: Hint $\ $ By induction the leading coefficient of $\rm\ f(x_n) \ $ is nonzero for some  $\rm (a_1,\cdots,a_{n-1})\in  F^{\,n-1}.\,$ Evaluating $\rm\:f\:$ at this point yields a univariate polynomial $\rm\ 0 \ne \bar f(x_n)\in \mathbb F_p[x_n]\ $ of degree $\rm< p.\, $  Some $\rm\,a\in \Bbb F_p\,$ isn't a root of $\rm\,\bar f,\,$ by a poly $\ne 0\,$ over a domain has no more roots than its degree, and $\rm\:\deg \bar f < p =|\Bbb F_p|.\,$ Therefore $\rm\, 0\ne \bar f(a) = f(a_1,\ldots,a_{n-1},a),\,$ hence $\rm\, f\neq 0.\ \ $ QED
Remark $\ $ In fact the above root-bound property completely characterizes integral domains: the ring $\rm\: D\:$ is a domain $\iff$ every nonzero polynomial $\rm\ f(x)\in D[x]\ $ has at most $\rm\ deg\ f\ $ roots in $\rm\:D.\:$ For the simple proof see my post here, where I illustrate it constructively in $\rm\: \mathbb Z/m\: $ by showing that, given any $\rm\:f(x)\:$ with more roots than its degree, we can quickly compute a nontrivial factor of $\rm\:m\:$ via a $\rm\:gcd.\,$ The quadratic case of this result is at the heart of many integer factorization algorithms, which attempt to factor $\rm\:m\:$ by searching for a nontrivial square root in $\rm\: \mathbb Z/m,\:$ for example, a square root of $\,1\,$ that is not $\:\pm 1$.


How much does the global structure of a Lorentzian spacetime restrict the metric? And vice versa

Question: E.g. if I know that my topology is that of a hyperboloid, how much freedom do I have left for my choice of the metric?
And the other way around: if my metric is some conformal factor times the unit hyperboloid how much freedom do I have left in my choice for the topology? Am I forced to conclude that the topology of this (Lorentzian) spacetime needs to be that of a hyperboloid?
Comment: Certainly the latter statement is at least partially wrong--your spacetime could be a manifold-with-boundary, or have disconnected subcomponents, for instance.
Answer: Let a smooth manifold $M$ be given, namely a suitably non-pathological topological space along with a smooth structure.  A metric $g$ on $M$ is any symmetric, nondegenerate $(0,2)$ tensor field on $M$ having constant index.  The pair consisting of $(M,g)$ is then called a semi-riemannian manifold.
So suppose you hand someone the a topological space like a hyperboloid $H$, and suppose that someone turns this hyperboloid into a smooth manifold by equipping it with a smooth structure, then you can still choose any metric on the hyperboloid you wish and make it a perfectly acceptable semi-riemannian manifold, so needless to say, there's a whole lot of freedom.
Note, however, that (especially in physics literature on, say, AdS/CFT) a hyperboloid $H$, as a smooth manifold, is often defined as an embedded submanifold of some higher dimensional smooth manifold $N$.  Moreover, as it often happens, this higher dimensional smooth manifold already has a specified metric $g_N$.  In this case, there is a particularly natural choice for the metric on $H$; it's called the pullback metric or the induced metric on $H$ depending on who you're talking to.  The pullback metric is defined as follows.  If $f:M\to N$ is the embedding through which $H$ was defined (in other words $H$ is just the image of $M$ under $f$), then the pullback metric is usually denoted $f^*g^N$ and is defined as follows:
$$
  (f^*g^N)_p(v,w) = g^N_{f(p)}(df_p(v), df_p(w))
$$
for all $p\in M$ and $v,w\in T_pM$.  Intuitively, the pullback metric is precisely the notion of distance inherited by $H$ from $N$, a smooth manifold that already has a notion of distance defined on it.  I can't remember the last time I encountered a situation in the (high energy theoretical) physics literature in which someone considered an embedded submanifold with a metric other than the induced metric.
Now, for the second question which is basically the reverse of the first, we need to be careful.  When we use the word "metric" as defined above, implicit in the use of this terminology is the existence of a smooth manifold on which the metric is defined, so in this sense, the reverse direction doesn't make much sense.
However, I think (and correct me if I'm wrong) you're second question can be reformulated as follows:

For each $\mu,\nu = 1, \dots, n$, I have a matrix $g=(g_{\mu\nu})$ of functions $g_{\mu\nu} = g_{\mu\nu}(x_1, \dots, x_n)$ defined on some open subset of $\mathbb R^n$ that is nondegenerate and has constant index, then does there exist a realization of $g$ as the local coordinate representation of the induced metric of some embedded submanifold $M$ of some ambient manifold $N$?  If so, is this realization unique?

If this is not a correct reformulation of your question, then let me know in the comments below.
I don't know the definitive answer to these questions, but I strongly suspect that the answers are yes and no respectively.  If such a realization exists, then, as Jerry Schirmer basically points out in his comment, you can just, for example, chop $M$ into two, disconnected pieces to obtain another realization.
Comment: Your correctly rephrased my question. To avoid the point @JerrySchirmer raised, I would like to put the additional assumptions that the manifold M is without boundary and connected. Would any one know the answer to this more restrictive question?


What are Next-to-Leading-Order (NLO) corrections?

Question: I'm currently studying Particle Physics and HEP and this acronym is omnipresent. I know it means next-to-leading-order but, what is exactly the physical meaning of LO and NLO?
Comment: This question is difficult to answer in its generality since you already know the only thing that is true in general: NLO means "next to leading order". The *physical* meaning of such contributions depends on what exactly you are computing, could you perhaps give a more concrete example you wish to have explained?
Comment: Have you learned about quantum field theory and Feynman diagrams yet?
Comment: Just tree level
Answer: You say you know about Feynman diagrams, at least at tree level. So you might have seen loop diagrams, i.e. diagrams that contain a loop of internal lines (see e.g. https://en.wikipedia.org/wiki/One-loop_Feynman_diagram).
Now, the whole point of Feynman diagrams is to expand the phyisical quantity we are interested in as a power series, $$\sigma= a_0 + \alpha\cdot a_1 + \alpha^2 \cdot a_2 +\dotsm$$
For this to make sense, the expansion parameter $\alpha$, which usually is (related to) some coupling constant, needs to be small -- for example, in QED, $\alpha\approx 1/137$. Then, the higher-order terms, i.e. those with higher powers of $\alpha$, are suppressed, and the lower-order terms dominate (note that I'm glossing over quite a number of issues here to get the basic picture across).
In other words, the lowest-power of $\alpha$ with a nonzero term $a_i$ gives the most important contribution, called leading order, and the other ones are higher-order. In particular, the next one is the next-to-leading order (NLO), contribution and so on (for example, this paper computes NNNNLO contributions: https://arxiv.org/abs/hep-ph/0610143).
This notation is used in other fields as well: Whenever you have a approximation scheme with a most important term and successively smaller corrections, it makes sense to speak of leading-order, NLO etc.
Comment: Usually the series is in terms of $\alpha$ (because vertices occur in pairs once you square the amplitude). That's sort of tangential to your point but I wonder if it might be better to use $alpha$ as the expansion parameter instead of $g$ because that's what we usually see in practice. Just a minor thing and probably not worth an edit on its own.
Comment: @DavidZ: you're right, I was a bit sloppy there. Changed.


Isometric map tiles with Openlayers or similar

Question: I wish to show a map layer in a browser based tool such as Openlayers or Leaflet however my tiles are projected isometrically.  Has anybody done this before?
Answer: This map by mapbox does it the best way I've seen Bike Share Map.

That tilts the whole map, though, and you might just want to render isometric tiles. I have no idea how they did it, I assume it's a combination of WebGl, magic and hard work.
The Minecrafter plugin for a Minecraft server does isometric tiles, but it's just a normal Leaflet map with fancy rendering. You could achieve the same thing by using building footprints and their height, for example.

Final example is a little leaflet plugin that extrudes stuff, I have an example here with real data Leaflet 3d Buildings. Code is here: maps.gcc 3d buildings.
Answer: I guess you intend something like http://maps.osm2world.org/
The wiki page should give you some guidance on how to render the tiles:
http://wiki.openstreetmap.org/wiki/OSM2World/Maps
Answer: Andre's solution works if you want to reproject the tiles, however I wished to retain my isometric projection.  Here's a way using leaflet:
Instead of creating a single layer, add an imageOverlay layer for each tile.  When the page is loaded or moved, check map bounds and add/remove layers as necessary.


How to create a boot CD

Question: How to create a boot CD for a dual-booting system? Just consider I am having Windows and Ubuntu, GRUB is the boot loader. I just want to create a dual-boot CD for the same operation.
Comment: What should the Boot CD do?
Comment: And does it have to be a CD or could you also use a USB stick?
Comment: anything is fine .. But i just want to creat the boot disk for dual
Answer: I asked a similar question on LiveCD's. You could make one yourself, if that does what you need.
If you could also use a bootable USB stick (much easier than a CD perhaps), look at this question.

Quoted from Josh Hunt on How to create a live CD with your own applications:

For linux, try Reconstructor to create your own Ubuntu live CD with programs of your choice:

Reconstructor is a Live CD creator for Ubuntu Linux.
It uses the Ubuntu Linux Live CD as a base, and then allows customization of boot screens (usplash), gnome settings, and software (you can also use the chroot environment to make other changes before creating the live cd).
Reconstructor uses the solid Ubuntu foundation, and allows for extensive customization. For example, create a custom Live CD with blender, inkscape, etc. included for a friend in graphics, or simply use econstructor to re-brand your environment (wallpaper, fonts).

From what I have heard, it is dead simple. Just run
<code>sudo python reconstructor.py
</code>
from the folder where you extracted it and follow the nice, pretty prompts. You will need either a live Ubuntu CD or an .iso for one.
Comment: Oh how very nice of you :)
Comment: Never hurts to quote a good answer ;-)

How to create a boot CD


How to use backspace in terminal in ubuntu 13.10..?

Question: I am using ubuntu 13.10.I cant able to use backspace in terminal. While using backspace it just passed through the letters.Please help me...
Answer: You can change this behavior by editing the terminal profile.

Go Profile Preferences, either by:
Right-clicking in the terminal window and choosing Profiles >> Profiles Preferences
Or from the menu: Edit >> Profile Preferences
Go the the compatibility tab
Find Backspace key generates
Select ASCII DEL

This should solve your problem.
Answer: If by "passed through the letters" you mean the cursor goes back over top of the previous character without removing it, then your computer simply can't render the terminal properly. If you type a letter after that, it should replace the letter it was hovering over.
The problem should be resolved by closing (or logging out) of the terminal and opening (or logging back in to) the terminal.
The only other option I can think of is somehow your keyboard layout is incorrect and thinks that the backspace key is the left arrow key.
Answer: The cursor has got 2 states. One looks like a box and the other is an under score. It seems the current state of yours is the box. Simply press the Ins key (Insert key) in your keyboard and the cursor transforms into a line. It should work as normal.
P.S: The Insert key is a keyboard key located on the majority of all computers that allows text to be inserted. When insert is enabled, the text inserted in a document will not overwrite any other text. However, when disabled, text will be overwritten.


Is JRE installed in Windows and Linux?

Question: Do Windows XP, Vista, Windows 7 and all Linux distributions have Java JRE installed so end-users don't have to install it to run Java applications?
Answer: No, not by default (at least for XP and Ubuntu which I know about).
You have to download and install the Sun version if you want it. It may be that some Linux distros include the GNU Java runtimes.
Comment: To add to this: Mac OS X *has* a Java VM installed by default, however, it's maintained by Apple and changes versions only along with the operating system. No other options there but at least you can rely on *some* version of the JRE to exist.
Comment: Apple includes updates to the JVM without a full minor OS version upgrade.  It typically lags behind the sun version though.
Answer: Not by default, but it's quite easy to install it, and if you want your users not to think about it, you could use launch4j which automatically proposes them a link to the latest JRE.
Comment: Many distributions does not allow any arbitrary user to install programs.
Answer: Many Linux distributions include some form of Java, but you cannot be CERTAIN that a suitable JVM is present for your application.
For Ubuntu installing "sun-java6-jdk" is enough to get all you need as a Java deployer.
Answer: Since XP, Windows does not include a JRE (and prior to that, the included JRE was Microsoft's version, not the official Java version, so it always lagged behind and had its own issues).
Many Linux distributions come with a JRE installed, but again it's not the official Java version, since that's closed-source.  The latest versions of Ubuntu, openSUSE, and Fedora (among many others) use the JRE from openJDK.  It works for a lot of Java programs, but not all of them.  A lot of applets and Java programs will crash under it.  You can install the official version in Linux from either your distribution's package manager or from the Java website.
In some Linux systems (including Ubuntu), you may need to manually set the official JRE as default in order for the system to use that one without you specifically telling it to do so every time.  In Debian-based distributions (including Ubuntu) and openSUSE (and likely others), the command to do this is sudo update-alternatives --config java and to enable the offical JRE plugin in the web browser is sudo apt-get install sun-java6-bin sun-java6-plugin.

Is JRE installed in Windows and Linux?


Log directory access on a disk

Question: Is there a way to monitor all read and write operations on a directory and its files in SBS 2011 (Windows Server 2008 R2)?
Answer: The SysInternals Process Monitor tool is able to log all disk access.
Getting any useful information out of this requires that you know what you're be looking for, though. The raw access log is somewhat of a firehose. :)
Comment: I'm about to check it out in a few minutes, but will it allow me to specify a directory to watch?
Comment: It's been a while since I've used it, but I belive you can set filters on the log, so you'd be able to filter out the results to just what you're looking for.
Comment: Played around with the filter, now it seems to be showing me exactly what I need it to. Anyone who gets in this folder now is toast :)
Answer: You can also use the built-in Access Auditing group policy settings on Server 2008 ( http://technet.microsoft.com/en-us/library/dd772630(WS.10).aspx ).
In the Group Policy editor, go to Computer Configuration | Windows Settings | Security Settings | Advanced Audit Policies - Local Group Policy Object | Object Access and enable Audit File System events (Success and Failure).
Then browse to the folder you want to monitor with Windows Explorer, Properties - Security - Advanced - Auditing and add Everyone \ Full to monitor all access.
Any attempt to access the specified folder will result in Security Event Log messsages like this:
<code>An attempt was made to access an object.

Subject:
    Security ID:    YOURDOMAIN\USERID
    Account Name:   USERID
    Account Domain: YOURDOMAIN
    Logon ID:       0x5057c

Object:
    Object Server:  Security
    Object Type:    File
    Object Name:    C:\temp\share
    Handle ID:  0xbf8

Process Information:
    Process ID: 0x12fc
    Process Name:   C:\Windows\explorer.exe

Access Request Information:
    Accesses:   READ_CONTROL

    Access Mask:    0x20000
</code>
You could go as far as automating alerts by attaching a task to the specific event in the Event Viewer to send an email, launch a program, etc when the event occurs.
Comment: This seems like a much more robust solution, especially since Process Monitor won't show anything if the server reboots for any reason.

Log directory access on a disk


Can you actually play pizzicato with regular bowed notes ( at the same time )?

Question:

In the beginning of this video it looks like the player is actually pizzicatoing with the left-hand and playing regular notes on a different string, is that what is happening? Is that even a thing?
Comment: I guitar we call it Hammer Pulloffs. One of left hand's fingers would pluck the string while another might be holding a note. Given how Pag was a guitarist too and took a LOT of inspiration from spanish guitar, thats most probably where it comes from.
Comment: Yes.  Difficult but part of becoming a skilled player
Answer:
is that what is happening?

It certainly is.

Is that even a thing?

It wasn't before Paganini came along. He was the best violinist in the world in his day. Initially his problem was that extant composition for the violin wasn't sufficiently difficult to highlight his superiority. His solution was to compose music that, to begin with, only he could play.
Paganini went on to become the early 19th century equivalent of a modern rock star, going on tour where he was payed fantastic sums of money to perform and with modern style female groupies.
Today, however, when there are more violinists alive than have died through all the ages, pretty much all the top soloists are expected to and can play his compositions. So, yes, simultaneous left hand pizz and arco are a thing at elite levels.
Comment: Not to disagree here, but on a side note, the baroque instrument _baryton_ was regularly played with bowed notes and pizzicato at the same time. The baryton is a kind of bass viola da gamba, but with many added sympathetic strings running under the fingerboard, which can also be plucked by the left thumb {the neck is hollow underneath) while bowing the playing strings.
Comment: I had to do that on the cello part for Villa-Lobos "Jet Whistle"