Where can I access American Mathematical Monthly problems given an index?

Question: I don't know if this is the appropriate website to ask, so I understand if this post gets closed. I want to explore (and maybe solve) some of the currently-unsolved problems submitted by readers on The American Mathematical Monthly. The solution submission page has them listed, but I don't know how I can easily access, say, problem 12412, because I don't even know what edition it's in. How can I do this? Thanks.
Comment: Related: [Is there an index for solutions to American Mathematical Monthly problems?](https://mathoverflow.net/q/27271/)
Comment: You can access issues older than three years ago at https://www.jstor.org/journal/amermathmont
Comment: I understood what you meant, but "currently unsolved" might be a slightly misleading way to describe these problems: my understanding is that when people submit the problems to the AMM, they generally provide solutions to them.
Comment: You may see the  Monthly problems in the *American Mathematical Monthly* (the journal).  Your technical library should have it.  Members of the Mathematical Association of America are able to get personal on-line subscriptions to it.
Answer: Unfortunately, I don't think there's any particularly easy way to find a specific problem given its index number, but let me summarize some of the comments (and add some of my own) in a community wiki answer.
The American Mathematical Monthly is published electronically on the Taylor & Francis website. You can often get a free first-page preview of the Problems and Solutions section, so one approach is to manually step through each issue and click on the preview. For example, you mentioned Problem 12412; using this manual method, I found it in
Problems and Solutions, Volume 130, Issue 8. In this case, the preview page actually provides the full text of the problem in question.
Even if this method doesn't yield the full text of the problem, it should let you narrow down which issue the problem appears in. After that, it's a matter of tracking down the actual content. Options include:

Becoming a member of the Mathematical Association of America. I don't know what your status is, but for example, a student membership is $35/year.
Going to your local college or university library. I don't know how convenient this is for you, but this will likely allow you to get access to the content for free.
Finding someone who is willing to share the content with you. This might be a math professor at the aforementioned university, or someone you can find online. Here is one website which invites people to share.

For content more than three years old, you should be able to use JSTOR, but it's rare that a problem that old will still be "live."
Comment: I already have institutional access to the latest issues with Taylor & Francis online (which itself is accessed through MathSciNet), so that's yet another option for people in the future looking at this post!


How to send some data to Mapper class (running on data in a HBase database)

Question: I need to send some information for mapper jobs running on nodes in HBase. I already have defined the data as static member in the class but it seems when the mapper is being run on the other nodes, the data is not being transferred to the nodes. Is there any way to do that?
public class GetResultFromVerticesIDMapper extends TableMapper< ResultVal, IntWritable > {

    public static HashMap<Long, Boolean> vertexIDsHashMap;
    public static int                    nResultComponents;

...
}

Answer: I did this through sending parameter to the configuration:
Configuration conf;
conf.addResource(new Path(hbaseConfigurationFile));
conf.set("simplificationParameters", simpParameters.toString());

and getting them into mapper class as follows:
_simplificationParameters.fromString(context.getConfiguration().get("simplificationParameters"));

Answer: Above way you mentioned is one way..
Other way, if you send System properties then it will be set to context automatically.
while running the program you can pass parameters as system properties (with -D option) or accept input and set them using System.setProperty("simplificationParameters","user_entered_input")
you can take it back in mapper or reducer like the same way you did it in your answer... i.e. context.getConfiguration().get("simplificationParameters")


Merging 2 datasets

Question: In the given task, they've provides us with 2 datasets namely the test & train dataset. So, I was wondering if we could merge those 2 datasets into 1 data-frame and delete the duplicates. Would you advise me to follow this approach or is it going to have an adverse effect like overfitting?
Answer: Test data should never be used in the training process. The reason why there are test data ist that you test your model performance against data the model has never seen before. By doing so, you "simulate" a situation in which the trained model is "in production" so that you can get a good idea of how well it works.
In most cases a model is able to reproduce what it has seen during training. So model performance will usually be "rather good" when you test the model based on data it has seen during trianing. However, showing "new" data (not seen during training, i.e. the "test data") usually gives a good idea how well the model generalizes (with "new" data).
Train your model on the training data only, make a prediction on both data (train/test), and compare both results. I guess you will find different results in terms of model performance.

Merging 2 datasets


How to show that the Laplacian is not closed?

Question: Let us define the Laplace operator $A:D(A)\subset L^2(\Omega)\to L^2(\Omega)$ by setting $Au:=-\Delta u$ and $D(A):=H^2(\Omega)$ (without any boundary condition), where $\Omega$ is a bounded domain with smooth boundary. From Weyl's lemma, we can infer that this operator is closable. However, the proposition 4.1 in the paper, titiled "Boundary value problems for elliptic partial differential operators on bounded domains", shows that $A$ is not closed. The approach there is not direct and has to use the concept of "quasi-boundary triple".
I just wonder if there is a simple, direct method to show that $A$ is not closed.
Comment: In the one-dimensional case of $\Delta$ on a finite interval $[a,b]$, it is easy to see that there is a continuum of (proper) self-adjoint extensions beyond $H^2$, depending on choices of boundary conditions.
Answer: In the multidimensional case the domain of the maximal operator associated to
$-\Delta$ in $L^2(\Omega)$ is given by
$$
D_{\max}:=\{f\in L^2(\Omega):\Delta f \in L^2(\Omega)\},
$$
and here $\Delta f$ is understood in the sense of distributions. This space is larger than $H^2(\Omega)$ as e.g. all harmonic functions in $L^2(\Omega)$ belong to $D_{max}$, but these
harmonic functions are not necessarily in $H^2(\Omega)$ (since their weak derivatives may not be square integrable).
Now equip $D_{max}$ with the graph norm $\Vert f\Vert:=\Vert f\Vert_{L^2}
+\Vert\Delta f\Vert_{L^2}$.
It is known that the space $C^\infty(\overline\Omega)$ is dense in $D_{max}$
and hence also $H^2(\Omega)$ is dense in $D_{max}$ with respect to the graph norm, but $H^2(\Omega)\not=D_{max}$ by the argument above. The maximal operator associated to $-\Delta$ coincides with the adjoint of the minimal operator defined on $H^2_0(\Omega)$ and hence the maximal operator is closed.
It follows that the closure of the Laplacian defined on $H^2(\Omega)$ is the maximal operator and that the Laplacian defined on $H^2(\Omega)$ is not closed.
In the one-dimensional case the situation is different: If $\Omega=(a,b)$, then the maximal domain $D_{max}$ coincides with $H^2((a,b))$, and hence
in this case the Laplacian $-\Delta=-\frac{d^2}{dx^2}$ defined on $H^2((a,b))$ is closed.
Comment: Thank you very much. The argument is fine and acceptable. But could you give an example of a distribution that $u\in D_{max}$ but not in $H^2$? I need to explain to my students. So, an easy one is appreciated.
Comment: I think a harmonic function with the form $\frac{1}{|x-1|^\alpha}$ may work. Thank you very much
Comment: Does this answer contradict the comment by @paulgarret under the question?


Can a Battle Master fighter continue to use maneuvers after they have spent all their superiority dice?

Question: Can a Battle Master fighter continue to use their maneuvers after spending all of their superiority dice? Or can they not do any more maneuvers until they finish a short or long rest?
Comment: @CédricAntunes Please don't use a comment to answer the question. If you have a confusion about why the question is being asked, use a comment to say what exactly is confusing, but without just answering the question. Thanks!
Answer: It is not clearly stated, but Superiority dice are required to use Maneuvers.
The rules say:

When you choose this archetype at 3rd level, you learn maneuvers that
  are fueled by special dice called superiority dice.

emphasis mine
Maneuvers tell you to "expend one superiority die" or similar wording as part of the description of the effect, but it does not make it clear that the maneuver is contingent on the expenditure in the wording. This could be better edited. However, the general wording, and the textual reference that says that maneuvers are "fueled" by superiority dice, make the intention fairy clear.
Answer: No, they can't use up more maneuvers when they're out of dice. And yes; they have to take a short or long Rest in order to get the dice back to then be able to do maneuvers.
After reaching 3rd level in the Fighter's class and choosing the Battle Master subclass, you're given 4 Superiority Dice to use as maneuvers (You get one more upon reaching level 7 and level 15). These can be used in a variety of interesting and tactical ways, but after expending it you will be forced to wait until your next short or long Rest in order to get these back.
This doesn't sound like a problem to me, in comparison to the Bard's Inspiration Dice. A bard gets a total of their Charisma modifier in dice, which he can use in a few ways, but upon use, the bard will have to wait until a long Rest to get these dice back for the day.
(Keep in mind that the Fighter is only allowed to use one maneuver per attack.)
Answer: ... under combat superiority it says "maneuvers are fueled" by the special die.
BUT
in the description of the maneuvers themselves rather that saying " When you hit a creature with a weapon attack, you expend one superiority die to...." It says "When you hit a creature with a weapon attack, you CAN expend one superiority die to...."
the word can suggest to me that you do not have to expend a superiority die...
Especially when you look at the parry maneuver that specifically says, "can use reaction and expand one superiority dice" which indicates for this to work you must use a superiority dice.
Same wording structure for Rally Maneuver as well as riposte
Comment: Welcome to rpg.se! Take the [tour] and visit the [help] when you get a chance. Unfortunately this answer is incorrect "you can [...] to" means you must do the first thing in order to do the second. Thanks for participating and happy gaming!


Children's SF: young pilot flies spaceship into meteor storm, patches damage to ship (spoilers of another book)

Question: Reposting a search I have going on Goodreads, in search of more suggestions:

Age group: children (upper primary / elementary school - maybe 10-12yos).
Genre: SF
Date: 1950s to 80s (read it late 80s / no later than 1990).
Origin: unknown

Plot details (so far as I can remember):
A young spaceship pilot on some urgent mission flies his craft into a meteor storm / asteroid belt / bunch of space rocks of some kind. The first few bang off the ship harmlessly enough, but one then punctures it, causing air loss. He hurriedly applies a flexible patch, but it is too small and is sucked / blown out into space. He applies a bigger one, with much the same result. He eventually gets one big enough to hold.
I have no idea about anything else that may have happened in this book.
The space setting and simple idea (flying across space on some kind of rescue mission) makes me think this could be quite old (the hopeful era of space of the 1950s / 60s.)
It is also quite possible that this book comes from a series.
Progress so far (contains spoiler of another book):
I thought of this some time ago, but held off on posting it while trying to think of more details. I also thought it might come from an extended series (like Tom Swift,) which would make searching a somewhat massive tsk.
I came across something similar in First on the Moon by Hugh Walters, but it is not quite right. I am fairly confident that the pilot in the book I am looking for was solo, while the protagonist in FotM is accompanied by a Russian rival-turned-friend. More certainly, the solution is not the same (the two companions in FotM use their hands to plug the holes.) It also alludes to a flexible patch that pilots could use, perhaps meaning that this episode appeared in another book in the series. Unfortunately, the series runs to 20 books, so I haven't gotten very far.
I also had a look at Trapped in Space by Jack Williamson, but this doesn't match, either. There's a hole in a spaceship, but it's caused by a laser cannon, while the patch job is done with a spray gun.
Answer: I think that may be the short story "Rock Pilot" by Harry Harrison, first published in 1965, but it's many decades since I read it and I don't have access to a copy to check if the details match.
Comment: Thanks Mike.  I can't get a description of the story, and the two anthologies it appears in are hard to find, so might take a while to check.


Is it possible to clone a game, and make modifiactions to it?

Question: I basically want to know whether or not it is possible to copy a mobile game and make it into a new game by modifying it . I want to be able to access , moderation tools and basically alter, things like characters, events, in ap purchases etc. Alter it to make it different enough from the original and make it available to other users. The game I want to copy is no longer in the play stores and I want to change that.
Answer: Is it technically possible? Perhaps. You could disassemble the APK, but when it relies on any server-sided components, you would have to clone those yourself. But taking a game executable made by someone else, doing modifications to it and then releasing it yourself is a copyright violation. It does not matter if the game is still being maintained or even if it's still available on the market. Copyright expires decades after the death of the author. It also doesn't matter if you changed a bit of stuff to make it more original. If there is just a tiny bit that remains from the original, you are violating the copyright on that.
"But what about all the other people who are doing that and their ripped games stay up?"
There can be a variety of reasons why the copyright owner did not take actions yet:

They are not yet aware of the violation
They lack the legal awareness of their rights and how to enforce them
They don't have the time or money to deal with it
The violator and/or the platform they release on are in some country where the copyright owner can not reach them
They don't care

But whatever the reason might be, you can not count on the same reason to apply to you.
So when you like a game and would like to create your own version of it, then you have to start from scratch. You can not reuse any assets or code from the original. You have to recreate them yourself. For more information about how closely you can stick to the original without getting into legal hot waters, check out the question "How closely can a game legally resemble another?".


Can I still win if I take knight while the opponent has two pawns with king remained?

Question: I play as white. I have a rook and queen and 5 pawns while the opponent has a knight and two pawns. I hope I can win this game. But is it legal to take the knight in next move? Then the opponent will have only two pawns with the King. Can I still play after that and win the game?
Comment: I don't see the position so this cannot be answered. Can you paste the FEN into your text at least? Or a pic would be better.
Answer: It's perfectly legal to take his knight and his two pawns, then promote all your five pawns, and mate him with rook and six queens. Watch out for stalemate, he must be able to move his king without going into check at all times.
There are no rules saying you can't take any more of your opponent's pieces.
Comment: "There are no rules saying you can't take any more of your opponent's pieces." Pins can prevent it. Although this is not going to happen in the OPs description of material. It cannot be said whether it is legal to take *any* pieces without seeing the position. I think this should be down voted. Sorry.
Comment: There are no rules _specifically_ saying you can't take any more of your opponent's pieces. Of course you can't take a piece if, for example, it leaves you in check. That sort of exception isn't relevant to the OP's question.
Comment: It's not clear what is relevant to the OPs question since we can't see the position. :)
Answer: A quick answer is that it is legal as long as you are not in check and you will not be in check immediately after making the move and you are moving your pieces according to the rules. The only risk I can see with grabbing the knight is that you can cause a stalemate and the game will be drawn. A stalemate is where your opponent cannot make any legal moves and at the same time the opponent's king is not in check.


Valums Ajax Upload: Canceling a file saved in the input

Question: I need to be able to cancel the file that is saved in the hidden iframe file input.  The problem is if you select a file and try to upload and get an error (for instance if you haven't filled out all the information in the form yet) if you try to upload the same file again it just doesn't work, the onSubmit won't fire ever.
I added a custom cancel function to the AjaxUpload object.  I am aware they have updated their script to a new version however what we have works fine and when I took a look at their new stuff it is completely changed in the state of how it works and what it offers.  The function is as followed:
<code>        cancel: function () {
            this._clearInput();
            this.iframe.src = "javascript:'<html></html>';";
        },
</code>
This should have done the trick however it is not working as desired.  Calling myUploader.cancel() isn't erroring and it's not working.
I just need a simple hack or work around to get this to work.
Any ideas?
Thanks
Answer: Found a workaround by using jquery to search for the input and clear it out.
$('input[type=file]').val('');
for those that have the same issue.


Accessing custom settings from a managed package using the Salesforce Platform license

Question: Can we access (read, create, update and delete) custom settings from a managed package via Apex code? The person manipulating custom settings will be using Salesforce Platform license, not Salesforce license. We can make it work with Salesforce license. In our case, we get the permission denied error.
Answer: I think your class must be written using 'with sharing'. If class is not written 'with sharing' then it will run in system context and will have full access to do curd operation on Custom Setting.


Defining SLI / SLO for ETL and Reporting Application

Question: All,
We're just started on SRE journey and trying to define SLI / SLO for our application.
It is an ETL application where 1. feeds (e.g. start of day, end of day data feeds) comes from various upstream and gets loaded with some transformation. 2. Once feeds get loaded there are some jobs which process the data and populate other tables 3. based on point 1 and 2 the data is made available for downstream applications and users.
In this case, we considered User experience as,

Data availability for downstream application at specific time for specific region
Data availability for Users at specific time for specific region

We defined time for above two points by when the data would be required to be available. Based on this, we created SLOs for us that in case we have to make sure we delivery data at right time we would need at least 1.5 hrs to investigate, reprocess the feeds i.e.
SLA for downstream/users - 1.5 hrs --> this we considered as SLI.
On Measuring the same, we capture how many times we are unable to get the feeds processed by SLO timings and put % i.e. 95% of the time we should be able to deliver data before SLO time.
Is this the right approach to define SLI / SLO?
If we look at most of the documentation / videos etc. on SLI / SLO they are more focused on micro service responses and measuring them on success / failure or latency etc. but couldn't find something around ETL or Reporting applications.
I might be wrong in above approach hence looking forward to an expert advice to get more understanding.
Thank you in advance for your help.
PS: couldn't find tag for SRE or any practices near it so have tagged question with DevOps.
Answer: The terms "SLI," "SLO," and "SLA" have precise meanings that apply across the spectrum of scale, domain, and abstraction. Although most literature focuses on microservices, that's because microservices are "hot" right now. To understand the concepts more fundamentally, look at the last words in each acronym:

An indicator is a measure. It's something that you look at, a piece of data that can answer a question. "How fast are our responses?" "How many errors are occurring in the ETL process?" "What is our cache hit ratio?"
An objective is a goal. This is where you want to be. They can be aspirational (e.g. we currently have 20 errors per day in the ETL process, but our goal is 5), or they can be steady-state (e.g. our response time is 200ms and it cannot fall below 250ms). There is no enforcement or liability tied to these goals.
An agreement is a legally binding statement of intent. If an agreement is broken, financial (or even legal) penalties can be on the line. For example, if you agree to an annual SLA of five nines of uptime (99.999%), a single outage of more than five minutes will blow your SLA and may constitute a breach of contract.

Hopefully that helps.


SDL sprite animation problem

Question: I'm currently drawing my character from a single file, and my sprite class handles it taking the file path as an argument. Now, I need to add a simple animation, cycling between two frames. The problem is that I have the other animation frame in another file, not in the same file as the first frame. How can I handle the animation without loading and unloading the two files and without merging the files into one?
Comment: An example of your sprite sheets would be helpful.
Comment: @Grey They're just 16x16 sprites starting from the origin, each in their separate files.
Answer: If you want to keep it with your current system, you should make a Sprite class with an array of frames (SDL_Surface/etc) that you load at the beginning of the game. Have the renderer choose the correct image based on time or some other constraint. Also, I'd advise you use sprite sheets as there is less overhead and they are generally easier to work with (More encapsulated). And definitely do not load/unload the files during run-time as that would be a huge slowdown.
Comment: yes. for reference, my engine (`carnage engine`) does sprite animation by regrouping frames in the same image and displaying subrects at different times. code source is available: http://sourceforge.net/projects/carnage-engine/
Answer: As CobaltHex said, using a sprite sheet would make this much less complicated, and can speed up your program overall. By creating two different SDL_Rect objects that correspond to the coordinates and dimensions of the two animation sprites you want to use, you can create two options for your what you want a character/entity to look like.
After that, simply using a timer and implementing a switch statement or if-else statement can create a neat looping animation, perhaps for an idle animation, or for a running animation, etc. Take this advice with a little caution though: I'm using this method in my game, and I'm a total newbie to the gamedev community, so I'm sure there are better methods out there.
<code>int initialTime;

while(!flag)
{
    initialTime = SDL_GetTicks();
    int phase = (initialTime/250)%16; //16 divisions, each .25 seconds long
    if(phase%2 == 1) /*apply surface #1*/;
    else /*apply surface #2*/;
}
</code>
So, technically, you could have 16 frames in our idle animation, or change around the length of each frame, whatever you want.

SDL sprite animation problem


Commerce cart form alter to do an availability check

Question: I have some nodes I use as rentable objects, and I want to check availability on the Commerce Cart page (made by Views, the one you see at /cart). I also need to check it a last time before the user is sent off to payment, but as for the views Cart form, what function actually handles the quantity updates? I can see that the edit quantity field has my pre-update value as default_value and the new value as value, but which function actually sets that on submit?
Comment: Is this node has product attach to it, which you send for checkout ?
Answer: I'm not sure what you're asking about quantity for, so I'm going to set that question aside and answer the broader question of confirming availability before purchase. In Drupal Commerce, the product pricing system serves a two-fold purpose:

Determine the price of a product for the current user.
Determine if the product is no longer purchasable.

While #1 is the most common use case, #2 even has some core representation: if you add a product to your cart and then disable that product as an administrator, the product will be removed from your cart thanks to a default product pricing rule that unsets the price of disabled products. We'll make this process more explicit in the next version of Drupal Commerce, but right now you should understand that if the unit price of a product line item is "unset" or set to a NULL (empty) value, the product will not be considered purchasable.
This will result in one of two behaviors:

For a product in the cart, it will be removed.
For an Add to Cart form, the Add to cart button will be changed to Product not available.

So, in your case, I'd just add a product pricing rule that performed the availability check. If the product is not available, the Add to Cart form will not allow the product to be purchased. If it was already in the cart, it would be removed, so you might use a separate rule to remove unavailable products from the cart and display a message informing the customer what happened. If they were on a checkout page, it would be user friendly to also send them back to the cart page.
Since I don't know how you're checking availability, I'll just add that if you can't do what you need through Rules directly, you can just whip up a simple custom condition to perform the check for ya. There should be plenty of examples in our code and elsewhere.


Elliptic regularity and inhomogeneous Neumann boundary condition

Question: Consider a harmonic function $u$ defined on $D : = \{ (x, y) \in \mathbb{R}^2 | (x, y) \in \overline{B(0, 2)}, y \geq  0\}$, that is, the closed upper half ball centered at $0$ and radius $2$. Let $u$ satisfy the inhomogeneous Neumann boundary condition $\frac{\partial u}{\partial y}(x, 0) = -u(x, 0)f(x)$, where $f(x)$ is smooth and bounded for $|x| \leq 1$. I am looking for a reference stating that $u$ is bounded around $(0, 0)$. Also, what is the biggest set around $(0, 0)$ that $u$ is necessarily bounded on? Thanks!
Answer: The present form of the question is very confusing. First of all, the boundary values are defined only for the flat part of the boundary. Secondly, even on the flat part, the condition given is not a Neumann condition due to the presence of $u(x,0)$ on the right hand side. If $f(x)$ is a constant function, say $f(x)= c$ for every $-2 \leq x \leq 2,$ then it is the homogeneous Robin boundary condition $\frac{\partial u}{\partial y} + c u =0$ if $c \neq 0$ and a homogeneous Neumann condition if $c=0.$
Hope it helps.


Cholesky decomposition for matrix with only one small negative eigenvalue

Question: In my project, I am computing some Quadrature Rule for Truncated Normal Distribution using the python code here (http://people.sc.fsu.edu/~jburkardt/py_src/truncated_normal_rule/truncated_normal_rule.html). This code involves a Cholesky decomposition of some moment matrix. In some cases, it works and gives me the desired nodes and weights, but in many cases mainly when I increase the number of quadrature points, it says that moment matrix is not longer PSD. I checked the eigenvalues of that matrices and the most of the time I get only one negative value of small order (~ 10^(-6)). Is there any trick to fix this and make the Cholesky decomposition works for my case. Thank you.
Comment: I'm not certain there is much you can do without heavily altering your machinery. This is a badly scaled problem in general, so that will need to be incorporated into your method from the beginning, otherwise you should expect this kind of stiffness effect.
Answer: Is it possible to slightly modify your moment matrix so that it is PSD, which will make the computation slightly less accurate but still preserve an approximate solution?
The book "Numerical Optimization" by Jorge Nocedal and Stephen J. Wright, which can be found here, provide some techniques for modifying symmetric indefinite matrices. These methods include: eigenvalue modification (you could flip the sign of your negative eigenvalue since it is of extremely small order), adding a multiple of the identity (adding $\tau I$ to your moment matrix, where $\tau$ is a tiny positive real-valued number and $I$ is the identity matrix of the appropriate size), and the Modified Cholesky Factorization method, which has many variants.


How to display text in a new activity when two radio buttons from two different activities are clicked?

Question: I am using Java and Android Studio. I have one activity with a few options to select from, and once one of them is selected, they click the button that takes them to the next activity. They do the same thing again except this time when the user clicks the button it takes them to a new activity where it displays the results based on the two radio buttons clicked previously. How can I do this?
Comment: The question is how do I fix the view error marked in red?
Answer: You cannot declare a method inside another method. Put a closing brace after onCreate.
When you are opening a new activity, you are doing it with an Intent. Put the values you want to send (bool clicked) the new activity inside the intent
Comment: I think you cannot put more values inside one case. Write 2 cases one after another without a break
Comment: I am not sure how you want to reuse the IDs, but if you want to store the value of the radio button, store it in a private field of the activity.
Answer: You can send some variable through intent and onCreate() method of the second activity, you can use that variable to differentiate your functionality.
This video demonstrates the functionality of the radio button: https://youtu.be/zJG9Prn3vZ0
In this video you can know how to pass arguments to another activity: https://youtu.be/OMCctaE66b8
Hope that helps you to solve your problem.


Do I have to list a country each time I visited it for UK visa application, or can I just list it once to say I've been there?

Question: I am a Canadian. I will be studying in the UK this fall. The question on the visa application specifically asks "Have you visited any other countries in the past 10 years?" It then ask for the date and length of travel. If I have visited a country more than once, do I say so, or do I just record it once?
Comment: You list each visit
Answer: You should state the date/duration of every visit in the past 10 years (or up to what is specified in the latest guidance and/or allowed in the system), even if they were to the same country.
The guidance notes to complete form VAF2 (the paper equivalent of your online Tier 4 visa application form, now only available to North Korea residents) says the following:

7.2 Have you travelled outside your country of residence, excluding the UK, in
  the last 10 years?
Answer Yes/No. You should include travel for study, training, business trips etc.
  When giving this information please provide details of all your trips abroad that you
  have not described in answer to question 7.1.

Hence they are expecting the list of trips, but not merely the list of countries you have visited.
Comment: @uberqe Please can you clarify which part of the answer is an assumption in your opinion? I can try and further clarify it. I totally agree that "other country" should be taken at face value when answering the yes/no part of the question, though with the guidance telling one to provide details of all of their trips, I see it as a harder requirement than "couldn't hurt" to do so, which makes me a bit puzzled...
Comment: That's an assumption. While it couldn't hurt to include all trips, "other country" means what it says.
Comment: The guidance is for the paper version of the form which is not the same as the one referred to in the original question. Many questions on this form are different from the ones on the Access UK website which is not the same as the Visa4UK website for the same type of application. For example, the paper form does not separate trips to premium destinations to make it part of another question. On the online Accessuk version, the applicant is specifically told to list only his/last three "visits" to these. IMO the UK application process is vague and requirements/questions frequently change.
Comment: @uberqe That makes more sense to me, thank you for clarifying. I don't have immediate access to the forms on different platforms but I agree the questions, guidances, and form itself frequently change. I don't think your observation contradicts the essence of the answer to OP's question (i.e. list each visits instead of list of countries if you visit the same country multiple times), but I will try to reword it to direct readers to the latest guidance to make the answer last longer. Thanks again.
Comment: There is no official guidance for the Access UK or visa4UK forms. The instructions (as ambiguous as they are) are embedded in the forms themselves.


can the spring mechanism in a mepla match cabinet hinge be replaced?

Question: The "spring" for this mepla hinge has broken.  It's shown left of the hinge in this photo, "appearing" to be unbroken.  Is there anyplace i can get new spring?
Answer: Try taking the hinge to a home store like Home Depot or Lowe's.  I just replaced  a number of similar hinges and they have a huge selection. You'll need to buy the entire hinge but they are not expensive.
Comment: FWIW, my home derpo/lowes have terrible selections of hinges. The OP might get lucky, but I wouldn't bet on it. My cabinet/door place has an entire aisle devoted to euro hinges.
Comment: I second @AloysiusDefenestrate. Home Depot only has 1 Everbilt hinge for each broad category (35mm, soft/normal closing, overlay type). I bought one of these and the mounting holes are several mm offset from the old Mepla hinge. So if there are plastic dowels inset in the door or cabinet frame, you will need to remove these or drill through them, before Everbilt will work. This might be true for other brands as well, so check the schematics of the distances between the mounting holes.
Answer: The Mepla hinges are no longer made, nor available except (sometimes) on eBay. As one hardware store puts it: "Fortunately, the Mepla hinge isn't in production anymore due to its design flaws. Unfortunately, that means it is hard to find a replacement."
I just needed to replace 30-year-old Mepla cabinet hinges as well, and one had a broken spring part like yours. Ideally, I agree it would be elegant and frugal to buy replacement (sub)components for devices which are "mostly" still in good condition. However, it's really not economical for items this cheap (for the manufacturer, and probably not for consumers). Hinges like this can be bought at box stores like HD for about $12/pair; and I've seen (probably better) products at online cabinet/hardware suppliers for as low as $3.50 each.
Stock availability aside, my 35mm full overlay hinge does not seem to be serviceable. The "spring leaf" part is anchored on four (or more) tiny pins that function like axels, but seem to be capped or rivited at the ends so they can't be removed. In fact, I'm not sure if I can dismantle any more parts to this, additional to the primary parts for adjustment and attachment to the cabinets.
Answer: It's unlikely that you can get a new spring, but the good news is that you'll be able to find a new euro cup hinge. Take the entire unit (including the base plate that's still attached to the cabinet) to a cabinetry supplier. If you don't have one, refer to the Blum catalog and try to match the relevant offsets. (Hint: this is hard, as there are a zillion different combinations available.) Replace all the hinges on the door -- don't mix and match.


Is All for One more powerful than All Might?

Question: The Ultra Analysis stats give All for One perfect stats in everything, 6/6 S+ for power, speed, intelligence and technique, when All Might only has 6/6 S+ in power and speed, and 6/6 S in intelligence and technique. Is All for One more powerful than All Might?
Comment: related: [How is "One for All" stronger than "All for One"?](https://anime.stackexchange.com/questions/46658/how-is-one-for-all-stronger-than-all-for-one?rq=1)
Answer: Depends, if you take the fact that All Might defeated All for One twice then no, All for One is weaker but still, All For One weakened All Might to the point where everyone knew his skinny form. But if you take their quirks against each other, All For One can actually win (without protag power)
Answer: I think One For All is stronger.
Think of it like this: All For One is definitely weaker like All Might. Supporting this fact is that he still hasn't healed, and is wearing a life-support mask. When they fought the first time, he messed up All Might's stomach really badly. In the second fight, All For One only held him back with his hands.
And when class A and class B were attacked on the campsite, Muscular lost to Midoriya who has One For All. When Midoriya passed by Spinner during the attack, he claimed that Muscular was the second or first strongest in the group.
Dabi claims that Tomura was weak for sending a bunch of rookie villains and that Dabi brought a bunch of hard-to-defeat villains.
Therefore, I think One For All is stronger.
Comment: The question is asking about All For One and All Might, not One For All. Thus, looks like the only relevant point is the 2nd paragraph, but it's unclear who is stronger/weaker from this answer.


Will Brave New World disable my old mods?

Question: I have a couple of mods, mostly custom civilizations, that were designed around G&K (Gods & Kings). Now that the new expansion is out, will installing the new expansion have unintended consequences for my old mods? If so, is there anyway to play with my old mods on the occasion that I feel like it?
Answer: Your old mods will be disabled, but in the options screen of the game, you can change the current expansion to Gods and Kings. That will let you play your old mods when you want to.


Detect if role instance exists in windows azure

Question: Is there a service inside windows azure that I can query a non cached record of role instances to see if a role instance exists, and if it exists, what is its IP address and port?
Answer: Yes that's possible from within your cloud service:
<code>foreach (var role in Microsoft.WindowsAzure.ServiceRuntime.RoleEnvironment.Roles)
{
    foreach (var instance in role.Value.Instances)
    {
        foreach (string endpointName in instance.InstanceEndpoints.Keys)
        {
            RoleInstanceEndpoint endpoint = instance.InstanceEndpoints[endpointName];
            IPEndPoint ipEndpoint = endpoint.IPEndpoint;
        }
    }
}
</code>
If you want to do it from outside your cloud service you'll need to work with the Service Management API: The Get Deployment operation returns configuration information, status, and system properties for a deployment.
Comment: Is Microsoft.WindowsAzure.ServiceRuntime.RoleEnvironment.Roles a cached list?  From my understand this value is cached and is polled every couple of minutes.  So in theory a role instance can get created, do some work and other role instances will still believe that the role instance hasn't even started.  That wouldn't be an ideal situation for me.  I am looking for an entry that would record the existence of an instance before it is loaded into the load balancer.


How can DFT of a two dimensional array be found using program for one dimensional DFT in C?

Question: I have the program four1.c from Numerical Recipes in C to calculate the Discrete Fourier Transform (DFT) of a one dimensional array. I want to calculate the DFT of a two dimensional array using four1.c, how can I do it?
Comment: Why not just use a 2-dimensional FFT library?
Comment: I think @BillBarth is right, why not use something like FFTW?
Comment: Yeah, I'd just use a 2-D FFT library. Hypothetically, one might object to using a GPL'd library. SciPy uses FFTPACK instead of FFTW due to licensing issues, and FFTPACK only implements 1-D transforms, so SciPy implements $n$-D FFTs by composing 1-D FFTs from FFTPACK. (For an in-house tool not intended for distribution, it's all moot anyway.)
Answer: You just take the DFT across each row and then across each column, or columns then rows, it doesn't matter.


ActionScript 2 vs ActionScript 3 performance

Question: Any benchmarks of Integer, Array, Function, Class, etc, performance in these 2 languages (compilers)?
Comparisons would also be helpful.
Answer: Since AS3 runs in a different virtual machine (AVM2) than AS2 (AVM) there have been made significant performance improvements with the introduction of AS3. In general, opcode executed by AVM2 runs about 10x faster than similar opcode running in the old AVM. This of course does not mean that everything in AS3 is 10x times faster than with AS2. With the introduction of AS3 all the display functions, or all the typical Flash API's, haven't increased in speed only the logic and arithmetic part of the code execution process. E.g. applying a blur filter on a visual element is just as fast with AS2 as it is with AS3 however, doing 3D transformations is much faster with AS3. AS3 introduces also some new data types which can increase performance as well (all though that doesn't necessarily always be the case).
Here are some links with information about AS3 performance:

Avoid ints in AS3
Types in AS3
AS3 Performance Tests
Flash AS3 and AVM2: Performance Tuning
Answer: i think you have to be more specific in what exactly are you looking for, i doubt anyone benchmarked and compared everything. also, AS is only the language, but there are speed differences between player versions, platform and browsers too.
but as a general rule, AS3 is almost always faster, sometimes with orders of magnitude.


Should the UX team be organizationally with IT or Product Management?

Question: I'm a UX veteran but new to company that has a Product management team. There is often questions about UX being under Product instead of the IT area.  What are your thoughts? If it helps, my UX team also writes all the User stories.
Comment: What kind of company is it?  IT means very different things in different companies. At technology companies, IT runs the corporate network and organizational tech assets whereas at Fortune 500 companies IT may actually build applications.
Comment: In your company, what does the Product management team do vs IT?
Comment: Welcome to UX.stackexchange. There isn't a clear cut answer to this question. It depends. In many ways UX is it's own entity (think of a Venn Diagram in which it overlaps many departments.
Answer:
Should the UX team be organizationally with IT or Product Management?

Neither. Or rather, both.
UX, ideally, is a team that facilitates product design and development by bringing all the stakeholders together to the table. PM, IT, Marketing, Business, etc.
The fact that the UX team is writing the user stories--however--indicates to me that this company is not unlike a lot of companies in that they really don't understand Agile. :)
User stories should also be an entirely collaborative process. Usually driven by business, but not dictated by them.
(Personal rant: In terms of an org-chart, I find that UX shouldn't be under any particular branch as we're supposed to answer to the end-user--not a particular department head. All-to-often in large organizations when UX is placed under a particular branch of the Org Chart, someone above has undo political influence on how customer needs are addressed.)


How can I install VMWare Player on Ubuntu 12.10?

Question: I need help installing VMWare player on my Ubuntu 12.10 system because I am having problems installing VirtualBox
Answer: Download the player from the VMware website:
https://my.vmware.com/web/vmware/free#desktop_end_user_computing/vmware_player/6_0
The download will come through as a '.txt. file. Once you have the file, open the command prompt and go to the location that you downloaded the file to:
For example if you put it in your 'Downloads' directory you could issue:
cd ~/Downloads/

Then you will want to execute the script to do the install. You can do that with by issuing the following commands:
chmod +x <name_of_file>
./<name_of_file>

Where the <name_of_file> is the name of the VMware player txt file you downloaded.
This should start the installation wizard to install the program.
Answer: virtualbox is easily installable using APT. Just with typing:
sudo apt-get install virtualbox

in a terminal you can install virtualbox. If you are doing so, and still have problems ask another question and include the problem you are facing.


Copying the full path of a file to text file

Question: I want to get the full path of all files in a directory and and the entire path has to be saved in one text file. Can you give me the script for that?
Answer: GNU coreutils contains the realpath program that returns the full path for each argument. The following program prints the full name of all files in the current directory (not expanding symlinks, -s; not recursive):
realpath -s *

Depending on your shell, hidden dot-files will not get expanded. For that, you also need to execute shops -s dotglob.
Answer: You should use readlink (available in GNU coreutils):
readlink - print resolved symbolic links or canonical file names

Syntax:
readlink -f <file-name>

Answer: The shell variable PWD contains the full path to the current directory.
You can use printf '%s\n' to print a list of strings with a newline after each string.
printf '%s\n' "$PWD"/* >somefile.txt

The wildcard pattern * excludes dot files (files whose name begins with a .). If there are dot files, you can use printf '%s\n' "$PWD"/.* "$PWD"/* >somefile.txt; note that this also lists . and ...


How can I determine the necessary joint angles given the position of this UR-3 robot?

Question: I'm trying to perform inverse kinematics on this robotic arm with 6 DoF. However, I'm not exactly sure how to start. I know a technique when the last three joints make up a spherical wrist, however, I'm not sure if this robot fits that category. Because Joint 6's and Joint 4's Z-axis are facing the same direction. However, when Joint five moves the end effector 90 degrees CW, then Joint 6's z-axis would be facing into the page, and then it might be considered a spherical wrist.
Comment: Robotics opensource libraries make this task simple. You do not need to write the IK solver from scratch. To help you with that, please let me know if you have experience with URDF and MoveIt.
Comment: Is there a way to do this via Matlab? @Robotawi
Comment: Yes, @ARJ, I just added an answer considering that.
Answer: Here's a paper that includes the derivation of analytical IK solutions for the UR5, but says that the wrist is not spherical.
There's also an implementation of analytical IK here in the ur_kinematics package for ROS 1. There may be a limitation that these IK solutions only work with the nominal kinematic structure of the robot.
The UR robots store an internal factory calibration that describes the as-manufactured kinematics of the physical robot. This can be important for accurate IK solutions. See, for example the ROS 2 package ur_calibration.
I don't know if using the calibrated kinematics breaks some assumptions in the solution process or not. (For example, no real-world as-built robot has strictly parallel axes for the shoulder lift, elbow, and first wrist joint.)
I agree with other answers that I'd use an existing IK package. I've personally found that numerical inverse kinematics solutions are fast enough for anything I've done with UR arms.
If you do need closed-form solutions, you can take a look at the IKFast library. I also came across this library which uses IKFast to find solutions for the UR arms. It looks like the ur_kinematics ROS 1 package can also use an IKFast solution if it's available.
Answer: It's pretty uncommon to do this on your own/by hand: there are plenty of libraries that will do this for you. In ROS/ROS 2, the most frequently used is MoveIt, and the ROS-Industrial project maintains a configuration for the UR3 (among others). With that setup, you can follow the regular MoveIt tutorials to calculate and achieve joint trajectories/positions.
Here's the relevant ROS Industrial project, and here is the ROS MoveIt tutorials.
You don't mention ROS in your question so if that's not what you plan to use, you might also be interested in (for example) IKPy: provide a joint configuration and it'll manage the computation for you.
If you're trying to understand the underlying algorithm and how they operate, MoveIt uses OMPL under the hood, which has its own extensive documentation that you can dig into.
Comment: OMPL knows nothing about inverse kinematics. These have to be supplied externally (if used at all, typically you only plan in joint space and use only FK).
Comment: ? @SimonSchmeißer Not sure what you're referring to here. I haven't claimed that OMPL knows about the specific IK of the arm being used, I've said that it provides algorithm implementations for MoveIt (which, per the website, is accurate).
Answer: If you are interested in solving the Inverse Kinematics (IK) problem with Matlab, you can use  the Inverse Kinematics Designer tool. This tool helps design an IK solver given a model from Matlab's robot library or URDF robot model.
Fortunately, the robot library already includes a model for the UR3 robot included in your question. Look at the examples part to see all the robot models in the library. If you want to use another robot, check the library or start from its URDF model.
This documentation page will help you start Inverse Kinematics. The three mentioned links are connected. Please read to figure out how they fit together.


can geolocation fields be used to calculate and display distance?

Question: I have a geolocation field setup on an object and want to be able to display the distance from another location in miles on a Visualforce page. I see with the new geolocation fields you can use DISTANCE in the WHERE clause combined with <, >, and other operators but I'd like to see the actual distance not just filter the results, is this possible without using an outside API?
Answer: You can use the DISTANCE and GEOLOCATION formula functions to create a custom field that will calculate the distance in miles to a fixed location. Then you can use this field in your Visualforce pages as required.
DISTANCE(acme_distribution__c, GEOLOCATION(37.775,-122.418), "mi")

Failing that, you can do the haversine formula calculation yourself in Apex:
public Decimal calculateHaversineDistance(Decimal lat1, Decimal lon1, Decimal lat2, Decimal lon2){
    // Earth's radius varies from 6356.752 km at the poles to 6378.137 km at the equator
    Double radius = 6371.00;
    Double dLat = toRadians(lat2-lat1);
    Double dLon = toRadians(lon2-lon1);
    Double a = Math.sin(dLat/2) * Math.sin(dLat/2) +
        Math.cos(toRadians(lat1)) * Math.cos(toRadians(lat2)) *
        Math.sin(dLon/2) * Math.sin(dLon/2);
    Double c = 2 * Math.asin(Math.sqrt(a));

    double kmToMiles = 0.621371;
    return radius * c * kmToMiles;
}

private Double toRadians(Decimal degree){
    return degree * 3.1415926 / 180;
}

Note that this includes the conversion from kilometres to miles in case you are in one of the three countries in the world that don't use the metric system :)
You may want to work to, say, 4 significant figures, as there are a number of simplifications in this approach.
Comment: Unfortunately the custom field option won't work since I need to display the distance between a selected event and other events so it needs to be more dynamic. I also learned I need the driving distance not 'as the crow flies' distance so I'll need to use an API anyway. It looks like the haversine formula would work for actual distance. It'd be nice if they'd expose the logic that DISTANCE uses in the MATH or perhaps a GEOLOCATION class.
Answer: Yes, this is possible without an external API using the Location class, which can be instantiated with a latitude and longitude or queried from the database. Here's the example from the documentation.
// Select and access the Location field. MyLocation__c is the name of a geolocation field on Account.
Account[] records = [SELECT id, MyLocation__c FROM Account LIMIT 10];
for(Account acct : records) {
   Location loc = acct.MyLocation__c;
   Double lat = loc.latitude;
   Double lon = loc.longitude;
}

// Instantiate new Location objects and compute the distance between them in different ways.
Location loc1 = Location.newInstance(28.635308,77.22496);
Location loc2 = Location.newInstance(37.7749295,-122.4194155);
Double dist = Location.getDistance(loc1, loc2, 'mi');
Double dist2 = loc1.getDistance(loc2, 'mi');

Answer: As of Spring '16 release, you can now use the DISTANCE function in SOQL SELECT clause so you can easily display distance in your query results. Do note, this is the haversine distance that @Daniel Ballinger already mentioned in his answer instead of "driving distance".
From the release notes...

DISTANCE() Support with Location-based SOQL Queries
The DISTANCE function is now supported with location-based queries.
  DISTANCE() now works with location and address fields. You can
  retrieve a specific distance (say, of a store from a contact's
  location) and display it to the user. For example:
SELECT Id, Name, Location, DISTANCE(Location, GEOLOCATION(latitude,
  longitude), 'mi') FROM CONTACT;

Sample Apex Code
// Assuming your accounts have their billing address geocoded
// such that BillingLatitude and BillingLongitude are not null
// then this query finds the nearest 3 accounts to a particular location
Decimal latitude = 10;
Decimal longitude = 10;

List<Account> accounts = new List<Account>([
    SELECT
        id, name, billingStreet, billingCity, billingState, billingPostalCode,
        // Using DISTANCE function in SELECT statement, aliasing field as 'dist'
        DISTANCE( BillingAddress, GEOLOCATION( :latitude, :longitude ), 'mi' ) dist
    FROM
        Account
    ORDER BY
        DISTANCE( BillingAddress, GEOLOCATION( :latitude, :longitude ), 'mi' )
    LIMIT 3
]);

// To display the distance value from SELECT statement
// use the 'get' function instead of dot notation.
// This is similar to retrieving field results from aggregate queries.
for ( Account acct : accounts ) {
    System.debug( acct.get('dist' ) );
}

Comment: Curious to know if anyone tried getting the distances by running a query on Junction object. [Select Id, DISTANCE(Location__r.GeoLocation__c, GEOLOCATION(-37.775,122.418), 'mi') FROM ABLocation__c ORDER BY DISTANCE(EDL__Location__r.GeoLocation__c, GEOLOCATION(-37.775,122.418), 'mi') . Wanted to see the possibility if distances can be retrieved. I saw the query is working fine but not able to retrieve the distances (or maybe don't know how to do it :) ).


Google Sheets Appscript on apply filter event?

Question: I want to execute a script as soon as I apply a filter to my Sheet. I can't find any event in the App script Docs that will allow me to do this, no onFilter event or anything obvious anyway. Does anyone know if this is possible, and if so, how to do it?
Comment: Welcome to [webapps.se]. It's not possible. Ref. https://developers.google.com/apps-script/guides/triggers. If you need further help add more details including what you tried a more specific description of your search efforts (include links to the most relevant post from this site that you have reviewed) as is suggested in [ask].
Answer: There is no simple trigger that runs when a filter is applied, but you can use an installable trigger to detect that. You can test an installable "on change" trigger like this:
function onChange(e) {
  console.log(JSON.stringify(e));
}

To make the code run when the spreadsheet is changed, click the clock icon   in script editor, click +Add trigger and choose Select event type > On change > Save.
Note that the trigger will run every time any change is made to the spreadsheet. You will have to carefully examine the event object and the values returned by the various SpreadsheetApp.getActive* methods to determine whether you want your script to do its thing when it gets run through the trigger. When the event is related to filters, the changeType property of the event object will be OTHER.


Can I hug my girlfriend?

Question: My girlfriend wants to give me a hug and neither of us are married or anything could I hug her back or not? My girlfriend is not muslim and she is able to hug me. Would it be Haram if I was to hug her?
Comment: Does this answer your question? [Can I kiss or hug my friend's wife who is like a sister to me?](https://islam.stackexchange.com/questions/51044/can-i-kiss-or-hug-my-friends-wife-who-is-like-a-sister-to-me)
Answer: You cannot hug or even touch your girl friend before marriage, it is forbidden (haram).

Say, "My Lord has only forbidden immoralities - what is apparent of
them and what is concealed - and sin, and oppression without right,
and that you associate with Allah that for which He has not sent down
authority, and that you say about Allah that which you do not know." https://legacy.quran.com/7/33

We should do our best to avoid doing haram things. But we are humans and may do some haram things and the remedy for this is to do good deeds to compensate for the haram things.

A man kissed a woman and then came to Allah's Messenger (ﷺ) and told
him of that, so this Divine Inspiration was revealed to the Prophet
(ﷺ) 'And offer Prayers perfectly at the two ends of the day, and in
some hours of the night; (i.e. (five) compulsory prayers). Verily, the
good deeds remove the evil deeds (small sins) That is a reminder for
the mindful.' (11.114) The man said, Is this instruction for me only?'
The Prophet (ﷺ) said, "It is for all those of my followers who
encounter a similar situation." https://sunnah.com/bukhari:4687
Answer: AssalamuAlaykum. In addressing your question, it's important to note that in Islam, physical contact between individuals of the opposite gender who are not related is generally discouraged. This includes actions such as hugging. This rule is in place to prevent potential pitfalls that could lead to undesirable outcomes. While your girlfriend may not be a Muslim and may feel comfortable hugging you, as a Muslim, it is encouraged for you to uphold the teachings of your faith. Remember, Islam encourages modesty and respect in all interactions. If you find yourself in an unsure situation, it may be best to kindly and respectfully explain your religious beliefs to your girlfriend.

Can I hug my girlfriend?


Op-amp with T network in the feedback path

Question: How do I calculate the equivalent T network resistance of the inverting amplifier below? The answer is 10 MΩ, but I don't know how to calculate it. I'm not looking for just a formula, but an in-depth circuit analysis.
Comment: Just do current mesh analysis with a virtual ground.
Answer: Your writing isn't entirely clear. But since you gave the answer you seek, I think I understand what you meant. You want to know how the output affects current entering the inverting input by way of the T-network. Your goal is to find the equivalent \$R_F\$ (if I do understand you) that would have a similar result as in:

simulate this circuit – Schematic created using CircuitLab
(Above, I've treated two resistors as being the same value, \$R_T\$, to help simplify things a little bit.)
Given that the inverting input can be taken as close to \$0\:\textrm{V}\$, for the purposes here, the question is essentially just asking, "How does voltage at the amplifier output affect current entering the inverting node?"
To figure that out, first figure out \$V_X\$:
$$\begin{align*}
V_X &= \frac{V_O\cdot R_G\cdot R_T+0\:\textrm{V}\cdot R_T\cdot R_T+0\:\textrm{V}\cdot R_G\cdot R_T}{R_T\cdot R_T+R_T\cdot R_G+R_T\cdot R_G}\\\\
 &= V_O\frac{R_G}{R_T+ 2\cdot R_G}
\end{align*}$$
From that, you can easily compute:
$$\begin{align*}
I_X &=\frac{V_X}{R_T}=\frac{V_O}{R_T}\frac{R_G}{R_T+ 2\cdot R_G}\\\\
&=\frac{V_O}{\frac{R_T\cdot\left(R_T+ 2\cdot R_G\right)}{R_G}}\\\\
\therefore R_F &=\frac{R_T\cdot\left(R_T+ 2\cdot R_G\right)}{R_G}\\\\
&=R_T\cdot\left(2+\frac{R_T}{R_G}\right)
\end{align*}$$
The result is \$R_F\approx 10.004\:\textrm{M}\Omega\$

Just a note about T-networks, from my own personal experience with electrometers. (I was experimenting with circuits achieving below \$1\:\frac{\textrm{fA}}{\sqrt{\textrm{Hz}}}\$ input-referred noise levels and quite literally having to buy unpackaged dice and use wire-bonders and stable temps at \$-5\:^\circ\textrm{C}\$ [low, but not so low that my window would ice up] in tiny, sealed quartz-windowed modules in order to get there.)
These T-networks actually generate higher current noise (\$i_N\$) and the voltage divider at the output multiplies input offset voltage, drift, and amplifier voltage noise by the ratio of \$1+\frac{R_T}{R_G}\$. These input specifications are often pretty crappy anyway, so it quite quickly becomes insanely impractical to consider multiplying their already annoyingly high offset and drift (low input current FET opamps are almost always used here) by using a T-network instead of a big feedback resistor.
(Sure, the Johnson noise of the feedback resistor has to also be considered. But it's not nearly as much of a problem once you refer that back to the input as input current noise.)

NOTE to OP:
You wrote:

"Shouldn't the first \$R_T\$ and \$R_G\$ be in parallel, with the
  combination in series with second \$R_T\$? Resulting expression is
  \$\left(R_T\:\mid\mid\:R_G\right)+R_T\$. What is wrong with this
  expression?"

There is nothing wrong with that expression except that it's not shown within its proper context. Let's see where that goes.
Starting at the output and working backward:
$$\begin{align*}
V_{TH} &= V_O\cdot\frac{R_G}{R_G+R_T}\\\\
R_{TH} &= \frac{R_T\cdot R_G}{R_G+R_T}
\end{align*}$$
We now have to add the value of \$R_T\$ leading back to the inverting input node to \$R_{TH}\$ to get the total Thevenin resistance seen by the Thevenin voltage computed above. So, the current into the inverting node due to the output voltage of the opamp is:
$$\begin{align*}
I_{V_O} &= \frac{V_{TH}}{R_{TH}+R_T}
\end{align*}$$
But we are actually interested in the effective \$R_F=\frac{V_O}{I_{V_O}}\$ (the equivalent feedback resistance being determined by dividing the output voltage by the current it causes to enter into the inverting node, as if we had such a resistor there.) So:
$$\begin{align*}
R_F &= \frac{V_O}{I_{V_O}}\\\\
&= \frac{V_O}{\frac{V_{TH}}{R_{TH}+R_T}}=\frac{V_O}{\frac{V_O\cdot\frac{R_G}{R_G+R_T}}{R_{TH}+R_T}}\\\\
&=\left(R_{TH}+R_T\right)\cdot\left(1+\frac{R_T}{R_G}\right)
\end{align*}$$
And now, here, you can see your factor present in the above equation. But notice that it is not alone! There is a second factor there.
We could just stop there. But let's follow through:
$$\begin{align*}
R_F &= \left(R_{TH}+R_T\right)\cdot\left(1+\frac{R_T}{R_G}\right)\\\\
&= \left(\frac{R_T\cdot R_G}{R_G+R_T}+R_T\right)\cdot\left(1+\frac{R_T}{R_G}\right)\\\\
&=\frac{R_T\cdot\left(R_T+2\cdot R_G\right)}{R_G}\\\\
&=R_T\cdot\left(2+\frac{R_T}{R_G}\right)
\end{align*}$$
Which is just the same thing as before.
Comment: Shouldn't the first Rt and Rg be in parallel, with the combination in series with second Rt? Resulting expression is \$R_t||R_g+R_t\$. What is wrong with this expression?
Comment: @sarthak The only thing wrong is that you have *taken a piece without putting it into its fuller context*. If you do that, you get exactly what I wrote. Start with \$V_{TH}=V_O\frac{R_G}{R_G+R_T}\$ and \$R_{TH}=\frac{R_T\cdot R_G}{R_G+R_T}\$. Now, you need to compute the current into the inverting node, \$I=\frac{V_{TH}}{R_{TH}+R_T}=\frac{V_O}{R_F}\$. Therefore, \$R_F=\frac{V_O}{I}\$. What do you find for \$R_F\$?
Comment: Between what nodes are you computing the Vth?
Comment: @sarthak \$V_{TH}\$ is the Thevenin voltage made up of \$V_O\$ and ground and the resistors, \$R_T\$ and \$R_G\$ on the right side of the schematic (not to include \$R_T\$ that ties to the inverting node.)
Comment: @jonk ok now i understand how the feedback resistor is calculated , but i   have another little question why R comp is 51k ? , it should be Ri || Rf thats equal to 100k not 51k
Comment: @isam The center of the T network is essentially at ground. So the inverting input mostly sees two 100k resistors in parallel. Not exactly at ground though, because it is 1% off. So the value is set just slightly higher.
Comment: If \$ R_T \gg R_G \$, then the term equal to 2 can be neglected and the gain is approximately \$ \displaystyle\frac{R_T^2}{R_G}\$.
Answer: The simple answer is visible to the naked eye - another voltage divider inserted into negative feedback loop.
What is the idea?
T-network
If we look carefully at the circuit diagram of the so-called "T-network"...

simulate this circuit – Schematic created using CircuitLab
Loaded voltage divider
... we will notice that the two resistors on the right (R1 and R2) form a voltage divider, and the left resistor (R) is grounded, and can be considered as a load of the voltage divider. So the "T- network" is actually a loaded voltage divider. Usually its output resistance is much lower than that of the load and it can be considered as unloaded; so its gain is approximately R2/(R1 + R2). In the schematic below, I have adjusted (slightly increased) R2 so that the gain is exactly 1/10.

simulate this circuit
Building the circuit
Many times I have had to explain this clever circuit trick, but now I will do it in a more different way - based on it, I will formulate a more general principle of virtual resistance modification. I will do it as usual in several successive steps believing that building the circuit is the best way to explain it.
Real resistance
Let's take a 100 kΩ resistor but pretend we don't know its resistance and decide to measure it experimentally. For this purpose, we supply it with a 1 V voltage source Vs and measure the current I through it. As you can see, the resistance is the same - R = 1V/10μA = 100kΩ.

simulate this circuit
Virtually increased resistance
Now let's apply the clever trick by reducing the voltage applied to the resistor R by a factor of 10. For this purpose, we insert an R1-R2 voltage divider between the supply voltage Vs and the resistor R. As above, we pretend not to notice the divider, and measure the resistance in the familiar way - R = 1V/1μA = 1 MΩ. However, what is our surprise when we see that it is 10 times higher!
The trick here is that we "do not see the divider" and use the previous 1 V voltage again... but the current is now 10 times smaller; so the resistance looks 10 times higher. Really clever trick!

simulate this circuit
Op-amp 4-resistor circuit
After revealing the idea, let's make an op-amp circuit of an inverting amplifier where the resistor R2 is virtually "increased" 10 times by the help of the R3-R4 voltage divider. Here the op-amp is "fooled". It does not "see" the R1-R2 voltage divider and "thinks" that the resistance R2 = 1 MΩ. That is why it increases its output voltage 10 times.

simulate this circuit
Equivalent 2-resistor circuit
So the 4-resistor op-amp circuit above behaves as the classic 2-resistor inverting amplifier with R2 = 1 MΩ.

simulate this circuit
Another point of view
Negative feedback followers
The most basic negative feedback circuits, like negative feedback systems in life, are followers (basically they contain amplifiers but they are made to work as followers). These are the voltage follower and the voltage inverter.
Negative feedback amplifiers
To get them to amplify, we apply a well-known trick of life - disturb them. In an effort to compensate for the disturbance, they become amplifiers. So negative feedback amplifiers are actually "disturbed followers". These are the non-inverting amplifier and the inverting amplifier.
Implementation
In voltage amplifiers, disturbance means voltage attenuation and the attenuator we do it with is a voltage divider. So, to make an op-amp amplifier, we insert a "disturbing" voltage divider between the op-amp output and the input. Indeed, in the inverting amplifier it is a bit more sophisticated (there is also a second input for the input voltage), but it is still a divider.
Disturbed negative feedback amplifiers
When, for some reason, only one disturbance is not enough for us, we can disturb the amplifier with another disturbance (add another voltage divider). Such a "double-disturbed non-inverting amplifier" will contain two cascaded voltage dividers in its feedback network. Accordingly, a "double-disturbed inverting amplifier" will contain cascaded R1-R2 summing network and R2-R3 voltage divider in its feedback network... and this will be an "inverting amplifier with T-network" - a subject of this question.
It is as if in this circuit solution, the inverting and non-inverting circuits are merged into one.
Conclusion
From this point of view, an inverting amplifier with T-network is a "disturbed inverting amplifier".


Google-TV device not found to pair with Google Anymote

Question: In building the Google Android app for Anymote tv control (I've got some Sony Google TVs to work with), I never actually seem to find the TV when I run it - just getting "No Google TV devices found on ".  I've tried it on an actual phone (not just the emulator) and have also tried the TV "connect" pairing button, but it never seems to find anything.  Both the phone and the TV are on the same WiFi network.  Anybody get this to work?
Answer: Does the google-tv-remote app work on your sony google tvs?  If so then I would check the source for that app, found here.  This should allow you to troubleshoot why your device is not found.  The process of discovering devices depends on mDns as found here in the docs.  So there are several things to check in regards to making sure multicast packets are allowed on your network/device.
As a fallback I would also allow manual ip entry in case a user wasn't able to discover the device on their network, the google-tv-remote allows for this as well, I would take a look at that source to get an idea of how that applications handles different scenarios.
And as always, if it's not working, provide some examples of how you are trying to do the discovery as you haven't provided any information(source, logs, etc) for anyone to look at.
Answer: My experience with the sony google tv is that the multicast packets are reliably sent out shortly after the box is started. Paring at all other times is virtually impossible


What does 'Gangnam Style' mean?

Question: Does PSY attribute his fashion to his district Gangnam? I've been hearing people nowadays going 'I have the Gangnam Style'. Are they meaning they have adopted the cultures of Korea, Gangnam?
Answer: "Gangnam" is an area in South Korea, specifically a small area in Seoul, South Korea, home to about 1% of the population of Seoul and is about the size of Manhattan.  It is known for its wealthy, "new money", inhabitants.  (The region has only very recently become extremely upscale, with a gradual upswing in the last half century or so. Before that, there really wasn't much there but farmland.)
How wealthy do you have to be to live in Gangnam?  The average price of an apartment there costs around 3/4 of a million dollars per year.  For reference, the average South Korean household's annual income is about $40,000.
So "Gangnam Style" simply is referencing this sort of elite, "new money", wealthy culture and lifestyle that has sprung up around the region.  As pop critic Kim Zakka, who lives in Seoul, said "Gangnam inspires both envy and distaste.  Gangnam residents are South Korea's upper class, but South Koreans consider them self-interested, with no sense of nobility."
Or, as Park Jae-sang (PSY, who himself was from a wealthy family and grew up close to Gangnam) said, Gangnam residents are seen as "good-looking because of plastic surgery, stylish because they can splurge on luxury goods, and slim thanks to yoga and personal trainers."  Sounds a bit like wealthy people from Los Angeles…
Read more about Gangnam Style here.
Cited: (http://www.todayifoundout.com/index.php/2012/12/what-does-gangnam-style-mean/#VkIY1vQidgXoFUHd.99)
Comment: Rents in Gangnam are obviously high, but I think you've somewhat overstated things. According to [this page](http://ibnlive.in.com/news/gangnam-style-so-where-exactly-is-gangnam/299558-2.html) *The average Gangnam apartment **costs** about $716,000, a sum that would take an average South Korean household 18 years to earn.* But at least they'd *own* the property if they paid that much - it's not "annual rent". For comparison, in the UK overall, houses cost about 6 times the average wage (4 times average *household income*).
Comment: Sorry, I might have overstated quite a bit.

What does 'Gangnam Style' mean?


Are there any double stars that I can actually see orbit each other?

Question: If I had a nice amateur telescope†, are there any multiple star systems that I could observe over a few years or a few decades and actually see the movement of one or both of them over time?
My short human lifespan and limited telescope put heavy constraints on the orbital distance, brightness, and distance from the Sun, so I am guessing that if there are any at all, the number is probably small.
†arbitrarily defined as say 8-inch (20 cm) aperture, with a good set of eyepieces, and a sketch pad.
Comment: There are links to various lists of visual binaries [here](http://www.midnightkite.com/index.aspx?URL=Visual). Ignore the optical doubles, they are just stars that look like they're close together.
Comment: I was just about to link that table from Richard Dibon-Smith's site. :D The trick is to find a binary with separation small enough for the period to be small, but large enough (and close enough to us) so that the 2 stars can distinguished.
Comment: This is a years long project, but can be done.  Sky and Telescope recently had a list of visual binaries with shorter orbital periods, but I can't find a reference right now.
Comment: Does [Algol](https://en.wikipedia.org/wiki/Algol) count? Sure, you cannot resolve the individual stars but you can definitely see the drop in magnitude during the eclipses.
Comment: @PM2Ring there are several more links there, for example [this one](http://www.dibonsmith.com/elements.htm) with some periods $P$ on the order of a human lifetime, maybe it's not so hopeless after all?
Comment: You might want to invest in either a reticle or a camera so you can quantify the relative position (spacing projected onto your image plain) and the changes over time.
Comment: @CarlWitthoft the reticle is a great idea, thanks! For this question I've said "... that I can actually see orbit each other..." to exclude astrophotography. With a cooled camera and a computer drive no doubt there are many more candidates, but that would be "cheating" (for the purposes of the question).
Comment: Thanks for that, but thedimming doesn't count as being able to ... actually see (them) orbit each other?
Comment: different but related: [Perceiving movement of stars](https://astronomy.stackexchange.com/q/11767/7982)
Answer:  Vir (12h 42m, –01° 27′)
Probably Porrima, $\gamma$ Vir, is the best candidate for most observers in the Northern Hemisphere to see changes in a binary orbit, particularly using a small telescope. It is a pair of stars with similar size and visual magnitude, of about 3.6. Their orbital period is about 169 years, but the orbit is eccentric, e = 0.88. They are also relatively close at about 40 ly. Periapsis was in 2005, so the stars are now moving away from each other, but their rate of separation is decreasing. Separation at periapsis was about 0.4 arcsec, so would not have been resolved using a small telescope in 2005. By 2015 their separation was ~2.5 arcsec, and will increase to ~3 arcsec by 2020. I estimated a position angle change between 2015-2020 of ~7 degrees. These changes should be detectable with a 100-200 mm (4-8 inch) telescope.
Since most short period binaries are close together, with nearly circular orbits, and often more distant, they are very difficult or impossible to resolve with a small telescope.
Sirius B (06h 45m, −16° 43′)
As mentioned in @MichaelWalsby's answer, it is also possible to observe the orbit of the white dwarf binary companion to Sirius, the brightest star visible in the night sky. Sirius is only 8.6 ly away, and their orbit has a semi-major axis of about 7.5 arcsec, an eccectricity of e = 0.59, and a period of about 50 years. If this pair were similar in brightness, they would be an easy answer to this question. Unfortunately, Sirius B, or the Pup (as the companion is known to amateur astronomers), is ~10 magnitudes dimmer than Sirius, and usually lost in its glare. It takes a night with excellent seeing, i.e. a stable, non-turbulent atmosphere, especially since Sirius never gets much above 30 degrees elevation at the mid-northern latitude where I live. I have only seen the Pup 4 or 5 times (one view was probable but not certain) over almost 6 decades of observing, and I have never seen it in a telescope with aperture under 300 mm. I know other amateurs that have seen Sirius B in 150-200 mm telescopes, but mostly at lower latitudes. However, by seeing Sirius B at intervals separated by decades, I have observed its polar angle change.
I believe the separation of Sirius B is now over 10 arcsec, and still increasing slightly. So for the next couple of decades, observing it might be a bit easier. In recent winters I have  tried with telescopes from a 120 mm refractor to a 250 mm Dobsonian, and occasionally larger, but still have not seen it for several years. This Hubble photo of Sirius gives some idea why Sirius B is hard to observe in small telescopes.
Indirect methods
Also, many eclipsing binaries are often observed, and their light curves measured. By analyzing light curves, orbital elements can be estimated. However, these indirect orbital observations are  probably stretching the intent of the original question.
Comment: Back when USNO (home of the Washington Double Star Catalogue) was giving tours, the astronomers there seemed fond of that one.
Comment: While editing my answer above to add Sirius B as another alternative, I saw the answer from Michael Walsby suggesting the same. I hope my additional information supports his answer.
Comment: "by seeing Sirius B at intervals separated by decades, I have observed its polar angle change" - that is amazing.
Comment: wow! `+1` very nice answer!
Comment: It's not the best.
Comment: @RobJeffries, Yes, I should have qualified my answer for observers in the Northern Hemisphere. I will edit.
Answer: α Centauri (14h 40m, –60° 50′)
The most obvious visual multiple system, where orbital changes can be observed is Alpha Cen A+B, (together with Proxima Centauri). The A/B system has an orbital period of 80 years, but because it is so close (1.34 parsec), the semi-major axis is a whopping 17.5 arcsecond.
The two stars are currently separated by 5 arcseconds on the sky. This will increase to around 10 arcseconds by 2030, decrease to about 2.5 arcseconds in 2037 and then increase again to about 22 arcseconds over the course of the next 15 years, with a changing relative angle all the time. The progress of this could easily be followed with binoculars or a small telescope.

Orbit Alpha Centauri AB Source

Trajectory of Alpha Centauri B relative to A (fixed to the coordinate origin) as seen from the Earth (inclined ellipse) and face-on (horizontal ellipse). The orbit parameters are taken from Pourbaix et al. (2002). The graph has been created with gnuplot based on data generated by solving Kepler's equation.

Note 1: The time stamps refer to the motion as seen from Earth, i.e. delayed with respect to the true motion by the travel of light by about 4.4 years. North is down, as usual in many astronomical charts.
Note 2: For simplicity, the points refer to steps of 1/80 of an orbit rather than to exact 1-year steps (which would be 1/79.91 of an orbit).
Note 3: The ascending node is the point where B (with respect to A) penetrates the celestial plane and becomes more distant to Earth than A.
Comment: @uhoh I live in Sydney, so Alpha Centauri is circumpolar for me. Being able to see the closest star (not counting the Sun) on any clear night is nice. And Sirius is fairly prominent for half the year. Pity I've never seen Polaris, but you can't have everything. :)
Comment: This seems like an excellent candidate, thanks!
Comment: @uhoh It was the 1st one I thought of, but I figured you wanted stars that are easy to view in the northern hemisphere. ;)
Comment: @PM2Ring oh that's so hemispherist of you ;-) I'm near the tropic of Cancer but tickets past the equator (Indonesia) can be really cheap here; you just have to avoid the volcano eruptions and earthquakes, and get used to flying on those small island-hopping airplanes. But WOW! the skies are dark and the Milky Way is bright!
Comment: @PM2Ring: Trust me, you're in the best hemisphere for astronomy. Double cluster looks great but otherwise, you're not missing much looking at the sky from Australia. I wish I could watch 47 Tuc, Omega Centauri, Eta Carina and the Magellanic clouds every night. Plus, Saturn & Jupiter are desperately low in the European skies now.
Answer: Sirius (α Canis Majoris, 06h 45m, −16° 43′)
You could try Sirius B, which is a white dwarf orbiting Sirius A around their common centre of mass. Sirius is one of the closest stars to Earth, and Sirius B has a 50 year orbital period. At maximum separation the white dwarf is about 30 AU from Sirius A, which should allow you to separate them with a good amateur telescope, but I'm not sure where Sirius B is right now. There should be a detectable change in position within 5 years or less. Needless to say, Sirius is a very btight star, and its companion rather faint, so you won't be able to separate them when they are close together.
Comment: This one certainly won't be hard to find. Thanks!
Answer: Mizar and Alcor (ζ Ursae Majoris, 13h 23m, +54° 55′)
Zeta Ursae Majoris is a binary star system in the Big Dipper constellation that is visible to naked eye. In fact, the ability to resolve Mizar and Alcor is taken as a test of good eye sight.
Interestingly, in the Hindu marriage tradition, immediately after the marriage the couple are asked to look at this binary star system (Alcor is called Arundhathi) as a sign of fidelity.
Comment: Mizar and Alcor are an optical double. They are relatively close together and they might be gravitationally bound, but if they are then they have a very long period, since the distance between them is at least 0.28 light years, and possibly as high as 1.5 light years. Mizar is a quadruple system consisting of two binary pairs orbiting each other. Alcor is a binary system.
Comment: @uhoh It says Mizar A & B have a period of thousands of years. So no, you're not going to see much change over a few decades.
Comment: Thanks for the answer! Are they moving fast enough to see any change "over a few years or a few decades" at most?
Comment: @uhoh sorry I do not know. Actually Mizar A is itself a binary with a period of just 20 days. I hope that helps.
Comment: @PM2Ring am I reading [this](https://en.wikipedia.org/wiki/Zeta_Ursae_Majoris#Stellar_system) correctly? Mizar A and B (both doubles) are separated by 14.4 arcseconds and have a period of 2000 years. So they can be resolved, but in a few decades I'm not going to see any change through an amateur telescope?


Same procedure for each image in Photoshop: how to speed up the production?

Question: Id' like to know if exists a way to speed up the production in Photoshop in the following case:
I have several images and I have to apply for each one of them the same adjustment. Doing it manually is very time consuming.
So, is there a way to automatize this process?
Comment: yes . use actions https://www.youtube.com/watch?v=NvdtFoetEmc ! or batch processing https://www.youtube.com/watch?v=JcXwhrvHu3w !
Comment: @DesignPhoenix that's an answer, why is it in the comments?
Comment: This is quite a broad question (check [ask]) and it has already been asked  (e.g. https://graphicdesign.stackexchange.com/questions/29927/creating-action-using-photoshop) please use the search function in the website and [edit] your question if you need some more specific help.
Comment: actually there is many asnwers to similar questions! seems i'll just post those links then
Answer: Here are a few posts from GDSE for making an action and batching images by action.
And here are YouTube resources for making an action and batching images.
Comment: Nevermid, I did it and it was extremely simple! Thanks!
Comment: yes you can practically record everything but actions have limitations like they don't understand layer order. if you click on second layer then it'll recorded with name instead of layer order . it'll all be cleared once you get habitated of actions :D
Comment: Thanks for your exhaustive reply. Does this technique works even if I have to work with multiple images, and not only with filter and adjustments? For example, I have 3 RGB images and I want to copy the G channel of the second image into the first one, and the B channel of the third one into the first one.


Ubuntu & Silverlight

Question: I have a business that runs 11 computers that only require 2 bits of extra (i.e. other that the os) software to work - a browser and MS  Silverlight I am currently running on XP but time is running out on that, windows 7 is supported to 2020 and therefore I face a bill of over £1200 to upgrade.
Is there anyone out there that can provide me with a disc image of Ubuntu and Silverlight already installed and working OR if it is not possible please let me know and if possible how much?
Answer: It's possible thanks to Erich Hoover and his pipelight team. You can make yourself Ubuntu ISO after installing silverlight. You can use Ubuntu-Builder to make yourself USB runnable ISO images.
Pipelight can run flash, shockwave, silverlight and unity player through wine FDS-Team | Pipelight - Installation
Comment: ATM I'd suggest using Ubuntu 14.04. I had problems installing Pipelight on 15.10, but on 14.04 it just works (test with Firefox).

Ubuntu & Silverlight


BCP neither gives results nor outputs anything when using valid statements but it does throw errors when passing invalid parameters

Question: I have to use bcp command-line tool to export data from an SQL Server database to a file in a Red Hat server.
I am (apparently) using valid statements but bcp is not producing any kind of output/results.
However, when I execute statements with missing or invalid parameters it displays the respective error.
I am looking for the reason of this issue (e.g. defective installation, bad usage of bcp, lack of permissions or any other known conflict) and how to fix it.

bcp statement:
bcp fully_qualified_table_name out ./data.txt -c -S server -U user -P password

bcp usage:
usage: /opt/microsoft/bin/bcp {dbtable | query} {in | out | queryout | format} datafile
  [-m maxerrors]            [-f formatfile]          [-e errfile]
  [-F firstrow]             [-L lastrow]             [-b batchsize]
  [-n native type]          [-c character type]      [-w wide character type]
  [-N keep non-text native] [-q quoted identifier]
  [-t field terminator]     [-r row terminator]
  [-a packetsize]           [-K application intent]
  [-S server name or DSN if -D provided]             [-D treat -S as DSN]
  [-U username]             [-P password]
  [-T trusted connection]   [-v version]             [-R regional enable]
  [-k keep null values]     [-E keep identity values]
  [-h "load hints"]         [-d database name]

bcp version:
BCP - Bulk Copy Program for Microsoft SQL Server.
Copyright (C) Microsoft Corporation. All Rights Reserved.
Version: 11.0.2270.0

SQL Server version (SELECT @@VERSION):
Microsoft SQL Server 2012 - 11.0.5058.0 (X64)
   May 14 2014 18:34:29
   Copyright (c) Microsoft Corporation
   Enterprise Edition: Core-based Licensing (64-bit) on Windows NT 6.3 <X64> (Build 9600: ) (Hypervisor)

Distribution:
Red Hat Enterprise Linux 6.7 (KornShell).

Invalid statements with respective error message (examples).
bcp THAT_TUB_ACE.oh_nerd.table_name out ./data.txt -c -S sr._bear -U you_sr. -P pass_sword

    SQLState = S1T00, NativeError = 0
    Error = [unixODBC][Microsoft][ODBC Driver 11 for SQL Server]Login timeout expired
    SQLState = 08001, NativeError = 11001
    Error = [unixODBC][Microsoft][ODBC Driver 11 for SQL Server]A network-related or instance-specific error has occurred while establishing a connection to SQL Server. Server is not found or not accessible. Check if instance name is correct and if SQL Server is configured to allow remote connections. For more information see SQL Server Books Online.
    SQLState = 08001, NativeError = 11001
    Error = [unixODBC][Microsoft][ODBC Driver 11 for SQL Server]TCP Provider: Error code 0x2AF9

...
bcp fully_qualified_table_name ./data.txt -c -S valid_server -U valid_user -P bad_word

    bcp fully_qualified_table_name out ./data.txt -c -S valid_server -U valid_user -P bad_word
    SQLState = 28000, NativeError = 18456
    Error = [unixODBC][Microsoft][ODBC Driver 11 for SQL Server][SQL Server]Login failed for user 'valid_user'.

SUMMARY.
The objective is to generate a datafile using the following syntax (or similar):
bcp fully_qualified_table_name out ./data.txt -c -S server -U user -P password

The facts are:

When running a valid bcp statement there's nothing in the window at all (no output) and no datafile is created.
I cannot use option -T (trusted connection using integrated security) for bcp so I have to specify the server, user and password.
Tried queryout option on a very simple small table already but still no luck.
Credentials are valid, I successfully tested them using sqlcmd like the following: sqlcmd -S server -U user -P password -Q 'SELECT * FROM really_small_table'.
The bcp statements under "Invalid statements with respective error message (examples)" section of this question are just examples of invalid statements to show that bcp actually does something but giving the expected results.
Comment: I don't know anything about bcp, but reading the usage statement in your Q, is seems to me that the `-c` option wants a "character type" which you don't provide.
Comment: Hi @JeffSchaller , there are some options to specify the kind of data **bcp** is dealing with...I am originally intending to use a format file but that adds complexity to the statement so in order to keep it simple I included `-c` option which will make **bcp** to store the exported data as **SQLCHAR** ( a [native file storage type](https://msdn.microsoft.com/en-us/library/ms189110.aspx#Anchor_1) ), no matter if it's whether an integer, a date or any other type. So basically by specifying `-c` you get rid of any datatype consideration and simply store the data as delimited strings.
Answer: What I have found on my Debian platform is that the Microsoft ODBC installation process put libodbc.so into /usr/lib64. This isn't an expected location for libraries and bcp cannot find it. But instead of telling you that it can't find a necessary library it simply exits.
Solution for Debian. Your situation should be similar.

Find where the installation procedure has put libodbc.so:
find / -type f -name libodbc.so

Add its directory to a new file /etc/ld.so.conf.d/odbc.conf. I've invented the name, and on my system /etc/ld.so.conf.d is included from /etc/ld.so.conf:
# Required but not documented by MS ODBC installation for SQL Server
/usr/lib64

Run ldconfig to update the list of library locations
Retry the bcp command

You will need to be root to perform steps 2 and 3. I assume that's no big deal given that you've managed to install the SQL Server libraries.
Comment: Hi @roaima ... all I can do is up-vote your answer because (after long communication and bureaucratic procedures) the Unix admin team didn't follow up the issue so unfortunately it will remain unsolved :( I couldn't agree more with your original (unedited) answer: this is `wild!`. I appreciate the efforts though :)
Comment: @HectorTorres no problem, I quite understand! (It was a nicely-researched question, by the way.)


What constitutes legitimate debate on a posted question and what constitutes original research

Question: Recently I asked a question and then based on one of the answers I got, proposed a solution to the question I asked. At which point I was told that "original research" was out of scope for the Philosophy SE.
At which point does speculation about a question constitute just that, and at which point does it become original research?
Answer: To my mind there should be zero "debate" conducted on the main page, which is not intended to host discussion or conversation.
If you are trying to start a debate, discussion or even informal conversation about some point that is being raised, it may help to keep in mind that this is much closer to the intended use of the chat space.
If you are able to specify a bit more about the particular details of the problem you're encountering it may be possible to be a bit more specific.
In general I'd just gently remind community members that questions on the mainpage are not prompts to open general discussion. Any "speculation" you are tempted to embark upon should therefore be sharply focused around elucidating specific philosophical works and excerpts. Citations should be provided where possible to help ground commentary in terms of very particular contexts.