Vector bundle and its definition

Question: Take this definition of vector bundle: its a triple $(V,\pi,M)$ where $V$ is a set, $\pi:V \rightarrow M$ a sutjective map such that for every $m \in M$ the set $\pi^{-1}(m)$ has the structure of real vector space o dimension $r$ and $m$ a real differential variety of finite dimension with a structure of maximal atlas on $V$. Pay attention that i don't assume that $V$ has a topological structure.
Using this assumptions, can i make $V$ a topological space? For example saing that a subset of $V$ is open if its image with $\pi$ is open in $M$.
Comment: If $r$ is finite, then you can canonically topologize $V$ so that $\pi$ is continuous and for each $m$ the inclusion $\pi^{-1}(m)\hookrightarrow V$ is continuous (assuming the usual topology on a finite dimensional vector space). More precisely, consider the largest topology that contains all $\pi$-preimages of open sets and such that all $\pi^{-1}(m)\hookrightarrow V$ are continuous.
Comment: @Peter Franek The topology that you define is the following: $A$ is open in $V$ iff $A=\pi^{-1}(B)$ where $B$ is opened in $M$. Right? But every elements of $V$ is a linear combination of $r$ vector $v_1,...,v_r$. So if an element $w=a_1 v_1+...+a_r v_r$ i can think at $w$ as the n-tuple $a_1,...,a_r$. With thiss identification can i put on $V$ the euclidian topology saing for example that $A \in V$ is opened iff $\alpha(A)$ is opened in $\mathbb{R}^r$?
Comment: No, if you define open sets to be just $\pi$-preimages of open sets in $M$, that's not enough.  You want more open sets. For example, you want the complement of *any* closed set in any fiber to be open. Therefore, you need to add as many open sets so that the inclusion of the fibers into $V$ are continuous. But I'm not sure if this is the right track at all. (In most books I have seen there is a local triviality condition that immediately generates a topology)
Comment: @ Peter Franek because in your comment i cannot understand what do you mean when you say the largest topology of all preimages of open sets ..... How can i define it? As intersection of biggest topology?
Comment: I thought about a topology generated by all $\pi$-preimages of open sets in $M$ and all sets $U$ such that the intersection of $U$ with each fiber is open (in the vector space). But now I'm not sure if it makes much sense.. will think about it.
Comment: i think that you want to find the topology that makes inclusion continuous application
Comment: (Yes, but under the topology I propose, all fiber inclusions as well as $\pi$ are continuous)
Answer: This kind of approach (have the topologisation of manifold descend from a differential atlas) is followed in quite a rich-in-detail way by the Italian book Abate, Tovena, Geometria Differenziale (which I presume you can read). There you can also check the correctness of your requirements. To quote some lemmas and definitions:

page 62 (on the topologisation of manifolds):

A $n$-dimensional atlas $\mathcal{A}=\{(U_\alpha,\varphi_\alpha)\}$ on a set $M$ induces a topology in which a subset $A\subseteq M$ is open if and only if $\ \forall\alpha\ \ \varphi_\alpha\left(A\cap U_\alpha\right)$ is open. Moreover, it is the only topology allowing all the $U_\alpha$ to be open in $M$ and all the $\varphi_\alpha$ to be topological embeddings.

page 134 (definition of vector bundle):

A vector bundle of rank $r$ on a manifold $M$ is a manifold $E$ provided with a surjective differential map $\pi:E\to M$ such that:

$\forall p \in M,\ \pi^{-1}(p)$ is endowed with a structure of $r$-dimensional vector space.
$$\forall p \in M\ \exists\ U\ni p\text{ open set },\ \exists\ \chi:\pi^{-1}(U)\to U\times \mathbb{R}^r\ \text{ diffeomorphism s.t.}\\ \forall p\in U\ \pi_2\circ\chi|_{\pi^{-1}(p)}\text{ is a linear isomorphism } \pi^{-1}(p)\to \pi_2\left(\{p\}\times \mathbb{R}^r\right) $$
  $\pi_1$ and $\pi_2$ being the projections on the factors of $U\times \mathbb{R}^r$

page 136 (on the conditions determining a vector bundle)

Let $M$ a manifold, $E$ a set, $\pi:E\to M$ a surjective function, $\mathcal{A}=\{(U_\alpha,\varphi_\alpha)\}$ an atlas of $M$ and $\chi_\alpha$ maps $\pi^{-1}(U_\alpha)\to U_\alpha\times \mathbb{R}^r$ such that:

$\chi_\alpha(\pi^{-1}(p))\subseteq \{p\}\times\mathbb{R}^r$
$$\forall U_\alpha\cap U_\beta\neq\emptyset\ \exists g_{\alpha\beta}:U_\alpha\cap U_\beta\to \text{GL}(r,\mathbb{R})\text{ differentiable map s.t. }\\\chi_\alpha\circ\chi_\beta^{-1}:\left(\alpha\cap U_\beta\right)\times\mathbb{R}^r\to\left(\alpha\cap U_\beta\right)\times\mathbb{R}^r\text{ satisfies }\\\chi_\alpha\circ\chi_\beta^{-1}(p,x)=(p,g_{\alpha\beta}(x))$$

Then, $E$ admits exactly one stucture of rank-$r$ vector bundle on $M$ such that $\{(U_\alpha,\phi_\alpha,\chi_\alpha)\}$ satisfies the condition of the point (2) in the definition.

So, as you see, your idea is right in some sense, but the requirements to induce a vector bundle seem quite sharper. The book I'm quoting takes its time (quite the time, actually) to thoroughly deal with these technical lemmas, which seems to me to be useful to your purpose. Of course, I assume it's not the only one in the world.
Added: I don't think I did, but I might have made some mistakes in copying and translating the text.
Added 2: Specifically: let $E, M, U_\alpha,\varphi_\alpha,\chi_\alpha$ as in page 136. Let $\rho_\alpha:\pi^{-1}(U_\alpha)\to \varphi_\alpha(U_\alpha)\times\mathbb{R}^r$ defined by $\rho_\alpha(x)=(\varphi_\alpha\pi_1\chi_\alpha(x),\,\pi_2\chi_\alpha(x))=(\varphi_\alpha\pi(x),\,\pi_2\chi_\alpha(x))$.
$\mathcal{B}=\{(\pi^{-1}(U_\alpha),\rho_\alpha,\varphi_\alpha(U_\alpha)\times\mathbb{R}^r)\}$ is an atlas on the manifold $E$, therefore it induces one and only one topology as of the result in page 62.
Comment: I'm italian so i can read the book in the original language. The traslation is right thanks.
Answer: What is missing from your definition is not only the topological structure of $V$, but the much more important locally-trivial character: there is a covering of $M$ with open sets $U$ such that $\pi^{-1} (U) \simeq U \times \mathbb R ^r$.
Comment: Alex, can you prove that the above definition is not equivalent to the usual one (for some class of real varieties)?
Comment: so my definition is wrong. I have to add thoose two characters
Comment: I'm confused because this definition comes from the notes about the theory of relativity.
Comment: @Dario: Of course you may construct an alternative concept of vector bundle the way you do. But when constructing mathematical objects, ask yourself: "why am I doing it?". Well, we usually want our vector bundles to be differentiable manifolds too, and in order to achieve this we need local similarity to Euclidean spaces. Without the assumption on local triviality as suggested above I do not see how you would prove that your bundles are manifolds themselves. Now, if you don't need them being manifolds, then maybe your definition suffices. It all is about what you want to do with them.

Vector bundle and its definition


Why would I want to update the iOS on my iPad 2?

Question: I have an iPad 2 which I have neglected to update for some time now.  The iPad uses iOS version 6 currently.  iOS 6 has never given me issues the way i use my iPad so Should I update to the new version?
Answer: The reason that has gotten me finally persuaded to update to iOS 8 was the fact that more and more apps called for mandatory update and would otherwise refuse to run. The one that immediately comes to mind in NFL Game Pass. Also my kids told me that some of the games they wanted to play wouldn't run on iOS 6.
I must say that I was a little afraid to upgrade as I was expecting performance issues on part of iPad 2... I am pleasantly suprised though - there are of course minor lags here and there, but it's not like the update has rendered the iPad unusable - on the contrary.
So all in all if it's just a matter of preference - it's entirely up to you whether you do it or not, but if you have essential apps calling for the upgrade, I don't see a reason to stay with iOS 6.
I hope that has shed some light on the matter for you.
Answer: No, I certainly recommend that you do not upgrade. It is unfortunate that you don't have iOS 7, but in any case, I upgraded my 3rd generation iPad to iOS 8 and noticed an immediate performance difference. The keystrokes were almost always delayed by roughly a word. They have fixed that problem as of some of the .0.1 versions, though. Still, I have observed a number of people upgrade and wish they hadn't. The fact is that after you upgrade, there is no going back.
If everything is working as it is right now, stay with it. If you need something that runs iOS 8, then go with that.
Also, keep in mind that iOS 8 has the potential to be the last option for the iPad 2. If this is the case (and you will know after WWDC, likely), then you will always have the option to upgrade, even after iOS 9 comes out.


Transfer a national domain to foreign registrar?

Question: Can you transfer a national domain (like .pl) to a foreign registrar, one that does not actually offer buying that type of domain?
Answer: Not usually, no. There has to be an infrastructure in place to support it.
Try visiting the registrar's contact page and contact them directly about it.


Upgraded CiviCRM 4.4.6 to 4.5.4; Case is adding but its not showing?

Question: Upgraded civicrm 4.4.6 to 4.5.4. Case is adding but its not showing?
I am using drupal-7.22.
Comment: Intrigued - what's Orbit? Is it a custom edition of CiviCRM?
Comment: Also: You should DEFINITELY upgrade your Drupal installation, particularly if it's accessible on the internet. 7.22 has vulnerabilities which you need to address (and you'll need to review your existing site users/site codebase). Search "Drupageddon" for more information.
Answer: Check your permissions, they may have changed and you'll need to give the relevant role the permissions. You should also clarify which CMS your using to make it easier to support you!
Answer: Not sure what your logic is for upgrading to 4.5.4 but there have been changes with the way Case works and in some cases that requires the renaming of Case Type 'names' and the XML files but i would have expected you would have had some warnings post upgrade


Problem with configuring a UITableView as a inclusive selection list

Question: I am programing a UITableView to behave as an inclusive selection list. My table displays correctly and allows for multiple cells to be selected with check boxes. My problem is that cells which have been selected (cells contain a check mark to the right) loose their selected status when scrolled out of view (cells check mark disappears). I want the selections made to cells in the table to be preserved even if cells are scrolled out of view. Does anyone have any idea what is causing this?
Here is my code inside of my TableViewController class:
- (UITableViewCell *)tableView:(UITableView *)tableView cellForRowAtIndexPath:(NSIndexPath *)indexPath {

static NSString *CellIdentifier = @"Cell";

UITableViewCell *cell = [tableView dequeueReusableCellWithIdentifier:CellIdentifier];
if (cell == nil) {
    cell = [[[UITableViewCell alloc] initWithStyle:UITableViewCellStyleDefault reuseIdentifier:CellIdentifier] autorelease];
}

NSUInteger row = [indexPath row];
cell.textLabel.text = [widgetTitles_glob objectAtIndex:row];
cell.detailTextLabel.text = @"";
cell.textLabel.textColor = [UIColor blackColor];
cell.textLabel.font = [UIFont boldSystemFontOfSize:15];
cell.accessoryType = UITableViewCellAccessoryNone;
return cell;

}
- (void)tableView:(UITableView *)tableView didSelectRowAtIndexPath:(NSIndexPath *)indexPath {
[tableView deselectRowAtIndexPath:[tableView indexPathForSelectedRow] animated:YES];
UITableViewCell *cell = [tableView cellForRowAtIndexPath:indexPath];
if (cell.accessoryType == UITableViewCellAccessoryNone) {
    cell.accessoryType = UITableViewCellAccessoryCheckmark;
    // Reflect selection in data model
} else if (cell.accessoryType == UITableViewCellAccessoryCheckmark) {
    cell.accessoryType = UITableViewCellAccessoryNone;
    // Reflect deselection in data model
}

}
Any help would be very much appreciated.
Answer: When you are using UITableView correctly, only as many UITableViewCell instances are allocated as are needed to fit on the screen. When you scroll down a table, and a cell disappears off the top of the screen, it is relocated to the bottom.
Your delegate method, tableView:cellForRowAtIndexPath: is responsible for setting up a cell, either creating a new one or reconfiguring a recycled one.
The proper thing to do is use an array to store your checked/unchecked values. When didSelectRowAtIndexPath: is called, you update the cell and your array. When tableView:cellForRowAtIndexPath: is called, you configure the cell based on the values in the array.
Based on your comments, you are already doing the right thing in didSelectRowAtIndexPath:; you just need to use those values when you set up the cell instance, because that cell could represent a row that has already been checked. Check the array and then set cell.accessoryType accordingly.
Comment: Ok great, I do indeed want the cells to represent the state of the NSUserDefaults object, so I will use your advice and update the cells according the defaults as well as update the defaults when a cell is selected.  Thanks.
Answer: In cellForRowAtIndexPath: you were assigining the accessoryType as none, so whenever you scroll that delegate is called and set the accessory type as none. So you should change your code.
I have also faced this problem once; I came up with a solution as follows.
Store the indexPath.row values of selected indexPath in an array (this code should be in didSelectRowAtIndexPath delegate) if it is deselected remove from that array. In cellForRowAtIndexPath: method I have used a for loop and check if that indexPath.row is present then change it's accessory type to checkmark else none.
Comment: Shouldn't the if(cell == nil) condition preserve all the cells in the table once they have been allocated?
Comment: You shouldn't just store the indexPath.row, you should store the entire indexPath. If you check 'if ([selectedRows containsObject:indexPath.row]) cell.accessoryType = UITableViewCellAccessoryCheckmark;
    else
        cell.accessoryType = UITableViewCellAccessoryNone;' then all the of the first rows in each section will end up being checkmarked when you scroll. You should instead save the entire indexPath because it holds both the section and row.
Answer: Thanks for your help. It actually turns out that the reason why the cells were getting reset to UITableViewCellAccessoryNone was becasue of the following line of code inside of cellForRowAtindexPath:
cell.accessoryType = UITableViewCellAccessoryNone;

Removing this has fixed the table.
Comment: No, that won't fix it... now checks will appear/disappear randomly when you scroll.


How do I control the brightness of an LED street light

Question: I am currently using a DC-AC ERI Solid state relay (Model number 001J) with a PIC16F microcontroller to control the operation of an LED street light. I have dumped a PWM code onto the microcontroller. I'm able to see a variation in the voltage on the input of the Solid state relay (Corresponding to the code). However, the intensity of the LED street light head (which runs on 220V AC mains) connected to the output of the Solid State relay remains the same. Am I doing something wrong, or is there a device I can replace the Solid State Relay with, to ensure smooth operation?
Comment: We need some more information here. Are you switching AC or DC through your solid state relay? SSRs will remain on as long as current is flowing through them, regardless of the signal on the gate. Also, there are SSRs that are TRIACs, which are quite slow, not very good for PWM applications, and SCRs, which may be faster but only conduct in one direction. Perhaps a schematic, or the type of LED light you are trying to control would make the question clearer.
Comment: To add to @BrendanSimpson comment, would you please post part numbers and links to datasheets for the LED and Solid state relay. Otherwise we are just guessing.
Comment: I am using an ERI DC-AC Solid State Relay, Model number 001J. I have a commercial LED street light head, which runs on 220V AC mains. I hope that clarified any doubts regarding my question.
Comment: Are you sure the Street light is designed to be dim-able? Normal LED lights for home use are not all dim-able, because the AC-DC circuitry needs to be designed specially to do so. I imagine the same is true for the big lights.
Comment: I'm not sure of that. I went with the assumption that LED street lights are dim-able, as there are a few companies offering the service of brightening and dimming of LED lights in accordance with the time of day.
Answer: Not going to work.  LED emitters themselves are inherently constant-current devices.
However, the street light is probably a packaged commercial product, designed to accept a constant voltage, and it includes a constant-current power supply.  These are usually switch mode, which tend to be inherently multi-voltage, because when they see a lower voltage, they simply hold the switch open for a bit longer, which draws more current.
Aside from working on world-market AC power (i.e. 100V Japan through 277V USA 3-phase), this also gives them a natural ability to ride through brownouts at full brightness.
It probably has some some capacitors too, to get across those AC zero-crossings without shimmering the LED (which annoys people in motion relative to a point light source, some car taillights do this.)
So, as you dim it by messing with its voltage, it's just chopping harder and pulling more current to stay at full brightness.
It will never work. You need to disassemble the light and replace the power supply with one that is friendly to dimming.
Answer: I would suggest:

Try changing the PWM values. If your're expecting brightness change for a particular PWM value, then you might not get that upto certain PWM value.
If you are using high intensity street light that might be using an AC input to drive the LED's then you might not be able to control the brightness as the internal rectification circuitry might take in specific input range only and you can vaguely change avg. voltage gathered through PWM cycle change.
Still, you can put the ac input to a lamp dimmer circuit as shown here.Though I'm unsure for an Led setup.
Answer: You have supplied a partial part number and no link to the datasheet. This one says it could be either zero-cross turn-on or random turn-on.

Figure 1. ERI SSR datasheet extract.
If it's zero-voltage turn on then it will always switch on as the mains crosses the zero volt point and give full mains cycles. You can't dim using that kind of SSR.
You want one that allows variable (or "random" in ERI's terminology) turn-on.

Figure 2. Phase angle control using triac / SSR.


Is Duotrope worth using?

Question: I am a new writer trying to simplify the publishing process. It seems to overwhelming. What are your experiences of Duotrope? Is it worthwhile?
Comment: I can't understand why there are votes to close this. It seems to be about writing and publishing.
Comment: @S.Mitchell Because asking for opinions on a tool is by definition *opinion-based*. Whether something is "worth using" will vary based on use case and personal circumstances. Without additional details how would we determine if it is "worth using" for the OP?
Answer: Yes.
I have used Duotrope extensively for about 4 or 5 years. Most of my publications (poetry) have come through journals and small press publishers that I found using Duotrope.
While you can do some searching (I think) without paying, one of the things that membership gets you is a look at the statistics for the market. If I see a site that doesn't respond 40% of the time, I'm not going to waste my energy with them.
You also have interviews with editors for some of the publications.
Personally, I put together searches based on my needs, save the searches, favorite journals that I intend to send to, etc. It also allows me to track where my submissions have gone, how long before I should expect a response, etc. For $5 a month (I think less if you pay by the year), I find it worth it.

Is Duotrope worth using?


Gifts in Korean culture

Question: Not exactly a language question, but a cultural confusion: I have recently made friends with two Korean women in my area (Catholic nuns--I'm Catholic too; I met them at church). They are immigrants who speak only basic English. They're very kind and friendly, and they keep giving me presents, bags of things like sweets and soaps. They do this at every major holiday and every time I visit. At first I thought they were just being generous, but now I'm starting to be concerned. Is it because of something in Korean culture that I don't know? Do they expect me to give them gifts in return? I know very little about any Asian cultures, and would be grateful for any input.
Comment: If I was in your situation I would give some small gifts, not every time you visit but occasionally. Probably they are not expecting you to give equally many presents (especially if they are much older than you), but also not expecting to get absolutely nothing. You don't need expensive gifts. Foods or sweets would be enough.
Answer: I am a Korean and I think they are just being generous and kind. If they are mid-aged or old women, they probably will do that to show affection. It makes sense especially because as you said they speak only basic English, so they want to do some actions instead of communicating with words.

Gifts in Korean culture


How to add svg logo for polygon layers of web map created via qgis2web?

Question: I need some edits on  my Leaflet webmap which are not supported using qgis2web plugin. I want to add SVG logo on two polygon features on the map. The SVG logos that I want to place on two parcel data. I have tried to add svg symbol for polygons, but that is not visible on the generated leaflet webmap. SVG symbols for point features are supported in the plugin, but not for polygon. How I can solve my problem?
Answer: As you say, qgis2web does not support this directly. You could do the following:

In QGIS, click Vector > Geometry Tools > Polygon centroids
Style the new layer ("Centroids") with the SVGs you need, and rename the layer to something appropriate for your map

The downside to this is that you will end up with two layers: the polygons and the SVGs.
To achieve this properly, export of the symbol layer "Centroid fill" would be needed in qgis2web, but this has not yet been developed.
Comment: I have already done it, but in that case I will have a two layers in legend (point layer with svg logo and empty polygone layer) and I don't want that. There must be some part of code to add on my map to solve this problem.
Comment: Is there a way to connect these two layers (points and polygon)so if I turn on the polygon layer the point layer will be turned on too? On that way the logo should appear on the polygon.


Is there a word that describes someone with a "false sense of doing good"?

Question: You get the so called "do-gooders". Some of them are misled by incorrect information so they have a false sense of doing good. Here is an example: Someone that recycles paper thinks that they are saving trees. In fact they reduce demand for trees resulting in forestry companies planting less trees so the opposite actually happens in practice. They have false "Greenpeace" images in their mind of age old natural forests that are destroyed to make paper. Some people will listen to reason and change their opinion. Others however will insist that they are right since you are destroying their "feel good" emotion.  I suspect that they actually have a psychological problem. I am looking for a word that describes such a person.
Comment: Self-righteous? Have you considered any words so far?
Comment: Yeah, "self-righteous", though it applies to both sides of the argument.
Comment: Self-righteous (attitude of thinking you right all the time) is too general description. The word I'm looking for is specifically related to someone trying to do good regarding some controversial issue. In my example the issue is protecting trees against industrialisation. I think it is important to note that the person is actually achieving the opposite of what the aim is while still insisting he is doing good. I have no suggestions of a suitable word yet.
Answer: In the UK, we refer to people who mistakenly think they are being helpful as:
well meaning or well-meaning(adj.):

wanting to have a good effect, but not always producing one

'I know he's well meaning, but I wish he'd leave us alone.'

Source: CDO

having or indicating good or benevolent intentions, usually with unfortunate results

'He is a well-meaning but ineffectual leader.'

Source: Collins
'Well-meaning' people are usually full of enthusiasm for their pet project(s) and feel sure they are doing good, even when the evidence indicates otherwise. They do not necessarily lack intelligence; it's as if they wear blinkers and are unable to sense the shock and disappointment in others who have been on the receiving end of their 'assistance'.
I suspect a mild form of this tendency lurks within us all, as changing deeply held convictions is hard, but more extreme behaviours may indicate a psychological problem, perhaps stemming from low self-esteem.
In the UK, when you hear someone say, 'S/he meant well...', and just go silent, you know exactly what it means!
Comment: Thank you Julie I think it's the best answer so far. To the sound of it well-meaning has a lot of innocence an naiveness to it. I'm looking for a word that would more describe a "liberal fool". This word would be very applicable to modern on-line media like Facebook where you have all these people fighting for various "causes" of which they have no real knowledge. Maybe another example will help?
Comment: Let's take the example of people that are anti-hunting while the hunting industry is biggest contributor to wild-life conservation in e.g. Southern Africa. The whole issue of Cecil-the-lion has resulted in such a reduction of hunting activity that the very game farm that Cecil the lion was from is in such financial trouble that the owners cannot afford to feed the lions any more which will end up in culling. None of the critics are hunters and often believe that animals freely roam Africa and are "murdered" by hunters. These people get very emotional!
Comment: Yvw @LevettPrins and thanks for adding another example. I think I understand more what you are looking for. I can't think of a term at the moment, but I will give it some thought and let you know if I can add to my answer. You might say that people are led by the media publicity surrounding an emotive issue, and want to do something straight away, eg donate cash, without really understanding the background?
Answer: I don't agree with most of your examples and comments about "liberal fools", but I'll focus on you saying this:

This word would be very applicable to modern on-line media like Facebook where you have all these people fighting for various "causes" of which they have no real knowledge.

You might consider the term slacktivist:

Slacktivism (sometimes slactivism or slackervism) is a portmanteau of
  the words slacker and activism. The word is usually considered a
  pejorative term that describes "feel-good" measures, in support of an
  issue or social cause, that have little physical or practical effect,
  other than to make the person doing it feel satisfied that they have
  contributed. Slacktivism can be defined as the act of showing support
  for a cause but only truly being beneficial to the egos of people
  participating in this so-called activism. The acts tend to require
  minimal personal effort from the slacktivist. The underlying
  assumption being promoted by the term is that these low-cost efforts
  substitute for more substantive actions rather than supplementing
  them, although this assumption has been criticized.
Many websites and news platforms have integrated social media sites
  such as Facebook and Twitter into their interface, allowing people to
  easily "like", "share" or "tweet" about something interesting they saw
  on the Internet. People can now express concern about social or
  political issues with nothing more than the click of a mouse, raising
  the question of what is actually being accomplished by these "likes"
  when very little thought or effort is required.
Slacktivist activities include signing Internet petitions, joining a
  community organization without contributing to the organization's
  efforts, copying and pasting of social network statuses or messages or
  altering one's personal data or avatar on social network services.
  Research is beginning to explore the connection between the concept
  and modern activism/advocacy, as groups are increasingly using social
  media to facilitate civic engagement and collective action.
The Joint United Nations Programme on HIV/AIDS describes the term
  "slacktivist", saying it "posits that people who support a cause by
  performing simple measures are not truly engaged or devoted to making
  a change"
Answer: Delusional? It's not strictly a false sense of "doing good", but it accurately describes "Others however will insist that they are right since you are destroying their "feel good" emotion. I suspect that they actually have a psychological problem."
Comment: Can you add a dictionary reference?
Comment: I have deleted my previous comment after I changed my mind. YES the person is delusional. Definition: "A delusion is a belief held with strong conviction despite superior evidence to the contrary.". But still not descriptive enough since it is not specific to the the act of "doing good"


Can this be drawn in one line without going over the line?

Question:
Can this image be drawn in one line and without going over any lines?
Comment: Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :)
Answer:
 No

Because

 There should be at most $2$ intersections with odd number of passing lines. In the picture, there are $8$. More reference: here
Answer: To elaborate a little on @athin 's accurate answer (and also answer without a link),

 The reason that it is impossible to draw the figure without picking up your pencil/redrawing over some lines is, as athin said, because there are more than two vertices with an odd number of "paths" leading to that vertex.

Here's why that is the case:

 Imagine that a shape like the letter Y is in the image. The center of the Y is a vertex with three paths coming to it. In the presented problem, if the pencil passes through the point at the center of Y, it will cover exactly two paths. That leaves one path, still to be covered, that leads to nowhere. If all three paths are to be used, the center of the Y must be a starting or ending point in the solution. There is no other way to reach all three paths. The properties of Y discussed apply to any vertex with an odd number of paths leaving it.

Now picture an

 X. Do you see how the point at the center of it has an even number (4) of paths that lead to it. This means that the pencil in the question can pass through X and leave 2 paths, allowing the pencil to pass through X yet again without stopping. This applies for all even numbers in the same way that Y's properties apply to all odd numbers - take an *, for example. See how it has an even number (6) of paths leading to it? The pencil can pass through once, leaving 4 paths, which we just showed can be passed through without stopping.

Bringing this all together,

 Any vertex with an odd number of paths leading to it must be an ending or beginning point for the pencil. There can only be one beginning and one ending. Thus, as athin said, having more than two vertices with an odd number of paths results in a figure that is impossible to draw with one continuous stroke. Your figure has 8 vertices with an odd number of paths, and as I'm sure you are aware, 8 is more than 2.


Have to Update my code after my App status is 'ready to sale' in iTunes connect.

Question: We developed an application in IOS platform using phone gap framework and got approved from apple through iTunes connect and now the status is 'Ready to sale'. Now I would like to update i.e modify my code as per the requirement. After Modifying, what is the process to be followed means is it necessary to resubmit the app or any other steps are there to modify the code in application.
Answer: You have to resubmit again.There is one option though, whether you want to release your updated version as soon as it gets approved or at your own(i.e when it get approved then you can make it available for sale at your own wish).
Answer: It is necessary that you upload a signed build of your new version to the store. Your fixed version will have to go through the apple review process just as your current release did before.
In the meanwhile your only options are to make the currently 'ready to sale' version available to your users, or wait until the improved version is approved. As you release that version then, user can update to the new version via the store.
Note, afaik you don't have a way to force users to update to newer versions, unless you do it in the app itself.


What was the official policy of British administration about the integration of Indian Princely States in 1947?

Question: What direction was given in Indian Independence Act, 1947 about the political integration of Princely States during the partition of India?
What policy was introduced by the British authority about the political integration of princely states of India in 1947 during the partition of India?
Was that policy official?
Comment: Couldn't you just read the text of that act?
Answer: A few things to note regarding this -
1. Official stand was that the state will lapse to independence but Mountbatten told almost every ruler that independence is impossible.
2. Reading's letter to the Nizam of Hyderabad clearly stated that Hyderabad is a subsidary of British Empire and not an independent state/state in alliance.
While the British officially professed the first stand, India took the second.
Seeing the impracticalities of independence, none but a handful pitched for independence - Hyderabad, Kashmir, Travancore, Manipur and Sikkim.


Sample size needed for Gaussian process classification

Question: I recently read the paper by Loeppky et al. on Choosing the sample size of a computer experiment: a practical guide and was curious to know if there were rules of thumb about the sample size considerations for Gaussian process (GP) classification? The paper suggests a value of $n=10\times d$ where $d$ is the dimension of the input space, however, this result is for GP regression and not the GP classification case. I imagine classification probably requires more points since it seems to be a harder problem but am not sure if anything has been published on the topic. If it matters, in my application the dimension of the input space is NEVER larger than 1.
Comment: I'm not sure this is really knowable (but I'm not voting to close). If the signal is strong and the model correctly specified, the problem is much easier (can get away with a smaller sample size) than a problem with many irrelevant features and weak signal.
Comment: @C11H17N2O2SNa I am just asking for a rule of thumb not a hard number on the sample size.
Comment: GPs will give you the uncertainty in your predictions (provided a good prior and kernel selection). That gives you a tool to estimate how many samples you need until you get to the uncertainty level you want. Not aware of a rule of thumb.
Answer: Classification can need more points than regression, for a few reasons:
1) If there are only 2 classes, the response variable contains much less information than a continuous variable, which could take many values.
2) With a small number of points, it's especially easy to get complete separation, which makes the maximum likelihood estimate undefined.
With ~10 points, it would be really difficult to tell what's going on, even if your predictor variable is one-dimensional.  This graph was generated randomly with 10 independent coin flips. Even though one could easily imagine adding a curve that wiggled up and down to fit the points better, doing so would be a mistake. For this reason, it probably doesn't make sense to even start thinking about nonlinear classifiers before you have a few dozen points.
At the very least, you'd want to heavily constrain the GP model to keep it from overfitting, and/or have a pre-determined kernel based on first principles. With so few points, the model-fitting process is likely to either blow up (if your constraints are too weak) or produce something that just matches your initial assumptions (if your constraints are too strong).
Answer: that rule of thumb is wrong. GP is used in high dimensional settings with d >> n without any problems (genomics, mri) lets say 500 000 voxels in one image used to classify 100 subjects into 2 classes. I don't know if there is any theoretical bound, actual number would depend a lot on the nature of the data at hand, so how strong is the signal and what proportion of features are pure noise.


dependent visa for japan

Question: I am 24 years old, and my father, mother and sister are in japan they are from last 2 years. So I also want to join them. Is there any possibilities to get dependent visa via my mother ??
Comment: *join them* = go live with them, I assume? What are your nationalities?
Comment: What is your mother's status (it's written on her residence card)?
Comment: my mother is in working visa as a cook. Its been 2 years
Comment: You did not answer the question; we need to know her status as written on her residence card.
Answer: As far as I know there is no age limit on the dependent visa, but you do need to be a dependent. That means that you will depend on your mother for your lifestyle and you will not be able to work. If you're 24 years old it's probably better to try to find a way to get a working visa. If you have a university degree and you can find a Japanese employer that will sponsor you, that should be possible.
Comment: indeed, he (I assume) can be sponsored by his mom's visa (same expiration date). Then later he can find baito jobs which don't require working visa, also less income.

dependent visa for japan


What is the shape of the graph $|z-1|+|z+i|=2$ in the complex plane?

Question: What is the shape of the graph $|z-1|+|z+i|=2$ in the complex plane?
$(A)\text{two points}\hspace{1cm}(B)\text{a line}\hspace{1cm}(C)\text{a parabola}\hspace{1cm}(D)\text{an ellipse}$
Let us take $z=x+iy$
$|(x-1)+iy|+|x+i(y+1)|=2$
$\sqrt{(x-1)^2+y^2}+\sqrt{x^2+(y+1)^2}=2$
Upon simplifying,$3x^2+3y^2-4x+4y-2xy=0$
But with this equation, i cannot tell the shape of the graph.Please help me.
Comment: if you want to go by your method, Rotate the axes by $45^{\circ}$ Counter clockwise, then you will get to know the shape
Comment: The answer given in the book is ''two points''.@EkaveeraKumarSharma
Comment: The given answer is incorrect (not unheard of in textbooks, for a variety of reasons).  The separation of $ \ z \ = \ 1 \ $ and $ \ z \ = \ -i \ $ is plainly less than $ \ 2 \ \ , \ $ so there can easily be more than two points for which the sum of the distances from those points is equal to $ \ 2 \ \ . \ $  The points $ \ z  \ = \ 0 \ $ and $ \ z \ = \ 1 - i \ \ $ satisfy the equation, but so do $ \ z \ = \ \frac43 \ \ , \ \ z \ = \ -\frac43 i \ \ , \ \ z \ =  \ -\frac13 - i \ \ $ and a host of other points.  The correct answer is $ \ \mathbf{(D)} \ \ . $
Answer: Hint: $|a-b|$ represents the distance between the two points a and b in the complex plane. What geometric shape is defined by the sum of two distances being constant ? :-$)$
Comment: @diya: $3x^2+3y^2-4x+4y-2x y=0$ represents an [ellipse](https://www.desmos.com/calculator). Even if that ellipse were degenerate $($i.e., one of its axes is $0,$ which here simply isn't the case, since the distance between $1=(1,0)$ and $-i=(0,-1)$ is $\sqrt2,$ not $2),$ we'd still be dealing with and entire line segment, definitely not with a finite number of points $($unless that finite number of points happens to be $0,$ which occurs when the sum is less then the actual distance$)$. So even if by mistake they forgot a radical sign above the $2,$ the answer would never be *A*.
Comment: The answer given in the book is ''two points'',and i suppose you are suggesting ellipse.@Lucian
Answer: Let two fixed points $ F_1,F_2 $  which are called the foci and $a \in \mathbb {R_+}$, $2a>F_1 F_2$  be given. An ellipse $E$ is the set of the points $P$, such that $ E=\left\{P\in \mathbb {R} ^{2}: \,|PF_1|+|PF_2|=2a\right\}$ . According to this definition, $F_1=1, \, F_2=-i, \, a=1$ and $2a>F_1 F_2$, where $F_1 F_2=\sqrt 2$. That is this shape is an ellipse.
Answer: $|z+i|+|z-i|=2 \iff \max\{|(z+i)-(z-1)|;|(z+i)+(z-i)|\}=2$,
$\max\{|2z|;|2i|\}=2$,
because for sure $|2i|=2$, so $|2z|\leq2$,
$|z|\leq 1$  and there is just a circle.
When i'm using $\max\{\cdots\}$ i mean the highest number contained in $\{\cdots\}$
I used here theorem that: $|a|+|b|=c \iff  c=\max\{|a+b|;|a-b|\}$.
Proof: Considering $a<0,b<0; a>0,b>0$ we get$-a-b=c$ or $a+b=c$, but $c=|a|+|b| \geq 0$, so we can easy generalize it to $|a+b|=c$, when $a<0,b>0; a>0;b<0$, similar generalize $|a-b|=c$, but when $c=|a+b|, |a-b|$ is obviously smaller, and if $c=|a-b|, |a+b|$ is obviously smaller, so generalizing all combinations we get $c=\max\{|a+b|;|a-b|\}$.
Comment: Confusing. On the one hand we're talking about modulus in $\mathbb C$ and on the other .. $\mathbb R$? Is the claim also true in $\mathbb C$?


How much salt is needed for "Coq au vin"

Question: How much salt is needed for "Coq au vin" (one rooster, 1.5 kg) and when should it be added?
Comment: possible duplicate of [How much salt should I add to a dish?](http://cooking.stackexchange.com/questions/556/how-much-salt-should-i-add-to-a-dish)
Comment: I think that the "when should I add it" part makes this not a duplicate, of that at least.
Answer: You can almost always add salt to a dish but you cannot take it out, so it makes sense to add salt at the end of the cooking process unless:

the salt needs time to penetrate the ingredients. Potatoes are a good example of this, when I make potatoes for a potato salad for example I add salt at the beginning so that the flavor gets into the them
the salt will help with chemical changes you need. For example, if I am sauteing zucchini or tomatoes I add some salt at the beginning as it helps to draw out water and change the texture

The other question to ask is if you are adding salt already with other ingredients. Pre-made stock often has salt in it, sometimes quite a bit, so if you are adding stock to your coq au vin you may not want to add any more at any time of the process.
So, my advice is to taste your sauce towards the end of cooking as the sauce will be concentrated and close to the end result, and to prevent and chance of contracting salmonella. If it isn't salty enough add some salt (say 1/4 of a teaspoon or a pinch) then stir and re-taste. Once you get almost salty enough stop and let it cook some more. Then taste again just before serving and add more if needed.
Answer: To taste.
We do not know what you consider salty or not.
Usually if one of the ingredient is already salted (broth, sauce, ...) then you only add at the end to finish up a dish.
If the ingredients are not salted, then you would add a small amount at the beginning and adjust as the cooking goes.
Remember that with braised food, there will be some evaporation and the concentration of "juices" will increase the saltiness of the dish.


CDN server blocked in corporate networks

Question: My CDN server is getting blocked on certain corporate networks. I have added my CDN site to most corporate firewall directory to make sure its categorized correctly and not as a parked domain or other categories which are considered spam by web filtering softwares used by corporates. I guess the reason is that my CDN domain name is not the same as my websites domain name. Why I did this? Because I wanted to use a cookie-less domain for all my static stuff (images/css/js) like everyone else (twitter, youtube). Is there a place which lists all such directories where I can categorize my CDN domain? Is this the best way to do this?
I guess the ROI of this optimization is pretty low and I might consider moving my CDN domain as a sub-domain of my website domain if I dont get any convincing answer to this.
Answer: Putting your assets in a separate domain is a common practice, even Server Fault does it (sstatic.net).  I doubt any of your hypotheses are correct, because they would cause problems with too many other legitimate sites.  I would contact some of the corporate networks that you're being blocked by (or have the users at those sites do it) and find out why you're being blocked.
Comment: So ask them how they determine a domain is parked, and fix that.
Comment: Yes I have done that already (in this case I needed to list my domain in their directory under the relevant category - the reason was that i needed to send 403 on the root of my webserver). However cant do that for all such software available in the market. So what I wanted to know was the set of rules which is like a super-set of all such rules used by all sch software.
Comment: There is no such list; proprietary block lists are always dreaming up new ways to screw their customers.  You'll just need to play whack-a-mole and fix things as you find out about them.
Comment: Yeah, I guess thats what I need to do. Worst case can make my CDN domain a sub-domain of the website domain. Thanks for your help.
Comment: Thanks for your answer. I did check with one network where it was blocked. And the network admin told me my CDN domain was categorized as a parked domain in his web filtering software, which was the reason why it was being blocked. Also found that the other reason could be that the root (/) of such a domain should not serve any page and should rather throw 403.
Comment: Today SO did not work very nice anymore, cause my company proxy blocks cdn.sstatic.net where the CSS is located.
And yes it is a very big multi-million-dollar-company. So. This does happen :)
Answer: The ip range for google app engine is being blocked by web trends. I don't think there is anything you can do if you are using google app engine besides setting up a small js script to detect if the CDN is being blocked, and in which case using your local resource files instead.
Comment: I think there is no other option but what you suggested. A javascript which could detect this and load from your alternate CDN.


Getting ready for Calculus?

Question: So I wanted to start a Masters program but they require that I have Calculus III.  I want to take that course at the university, but I need to be ready for it.  As I look at Khan Academy and do some of the exercises for pre-calc, I realize I am so far out of school that I need to study areas of math even prior to pre-calc.  Is there an assessment of some sort online that would help guide me to what I need to focus on or should I just work through Khan Academy material?  Any thoughts?
Comment: www.brilliant.org is good too. I suggest also working through the *Art of Problem Solving* books, they're very good at the Calc/Pre-Calc/etc. levels
Comment: brilliant.org is pretty cool.  I will definitely use that as a resoucre.  I just wish there was some sort of test I could take that would "prescribe" what I need to focus on.
Answer: I can share my experience as to your pre-caclulus question as I have just recently finished precalculus about a month ago and about to finish Calculus I now (not the rigorous type, and also I am assuming you are not looking for that kind of a review).
I would suggest checking out a standard text like Stewart's Calculus (whether through a library or some other means) as it has a diagnostic test in the beginning that covers an array of precalculus topics like geometry, trigonometry, functions, and others - whatever you don't understand here or are not confident with will basically tell you what precalculus topics you need to review (the Early Transcendental edition of Stewart even delves into precalculus topics for the first chapter). I would prefer using Stewart over Khan Academy as it has more questions and even slightly challenging ones towards the end of the exercises. I think Khan Academy may help in aiding understanding some concepts you may have difficulty with but does not provide enough practice in applying the concepts you just learned (at least in applying them in various ways).
Cheers.
Answer: In case you are looking for an exam, try the 1998 version of calculus AB exam,(http://apcentral.collegeboard.com/apc/public/repository/calculcus-free-exam-1998.pdf)
It should provide you with some insight on which area you need to work on.
You mentioned you want to take a master program, I am assuming the course is advanced Undergraduate course, and I believe "Calculus" by Michael Spivak would be the perfect introduction book.It assume literally no prior knowledge of calculus and is used by our university as a text book for pure math major.
Answer: Schaum's Outline Theory and Problems of First Year College Math (1958, Frank Ayres)
As with most Sahum's, it is down to earth (simple words).  Not terse or "math mature" but also not loquacious.  Has the answers to the problems (designed for self study, not for selling to paid instructors).  Covers Algebra 1 (linear equations), Algebra 2 (quadratics, logs and exponents, etc.), trig. analytical geometry. And has a short, friendly intro to calculus.
Furthermore it is written for people in your situation.  You have had the courses before but don't remember them, maybe even didn't do well the first time through and/or are not a genius.  But it will help you to time efficiently refresh your knowledge or bridge gaps you always had.

Getting ready for Calculus?


Is the definite article needed with the word "mainstream"?

Question:
What are their stories trying to fit into [the] mainstream society?

Is the definite article needed or optional?
Answer: I wouldn't quite say that it's optional. I'd say the definite article changes the meaning somewhat. If you use the definite article, I'd expect some context that explains which mainstream society you are talking about, for example:
the mainstream society in Australia
Using the definite article is optional in that context, and you can omit it. However, if you don't have some context explaining which society you're talking about, then using the definite article makes it seem like an incomplete thought to me. And it sounds odd. (This actually has very little to do with the modifier "mainstream." It's more to do with how articles are used [or not used] with "society.")
Short answer: It's probably better to omit it most of the time.
Answer: When mainstream is used as a standalone noun, it does typically take the definite article:

They are trying to fit into the mainstream (of society/the literary tradition/a typical school).

But when it is used to modify some other noun, the rules for that noun govern which article, if any, to use:

These stories fit well into mainstream society.
These paintings fit well into the mainstream art world.
These lessons fit well into a mainstream school.


Prove the maxima or minima

Question: If $f'(x)=(x-a)^{2n}(x-b)^{2m+1}$ where $m,n\in \mathbb{Z} $ then prove that there is neither a minimum nor a maximum at $a $,minimum at $b$ for $f(x)$  . Here I dont know whether $m>n,b>a $ thus even though if I calculate the value at $(2m+2)^{\text{th}}$ derivative I cant tell whether Its positive or negative thus I cant tell whether it minima or maxima . So I am basically stuck in taking a step further of Leibnitz rule.Any help is appreciated. Thanks!
Comment: @ArchisWlankar regards. The maximum and minimum is for $f(x)$ or $f'(x)$?
Comment: Sorry its for $f(x)$
Answer: The first non-zero derivatives at $x=a$ and $x=b$ are:
$$
f^{(2n+1)}(a) = (2n)!(a-b)^{2m+1} \\
f^{(2m+2)}(b) = (2m+1)!(b-a)^{2n}
$$
To see this: at $x=a$, you repeatedly differentiate with the product rule, but notice that almost all the terms have a coefficient of $(x-a)$. The first time this isn't the case is for the term where you differentiate $(x-a)^{2n}$ until the $x$ goes away, i.e. after $2n$ differentiations. Its coefficient (from the product rule) is $(x-b)^{2m+1}$, which is non-zero. (To be explicit, all the terms where $(x-b)^{2m+1}$ is differentiated keep a coefficient of $(x-a)$ raised to some power, so all these are zero). A similar argument applies for $x=b$.
The derivative test tells us that if the first non-zero derivative is after an odd number of differentiations, then you have an inflection point (the case for $x=a$).
If the first non-zero derivative is after an even number of differentiations, then it's a minimum if that derivative is positive, and it is, since $(b-a)^{2n}$ is positive (the case for $x=b$).
Answer: I may contribute. I presume that the maximum and minimum are with respect to $f(x)$.

We have $f'(x) = (x-a)^{2n} (x-b)^{2m+1} $.

The key is to notice the sign of $f'(x)$ (the monotonicity). Also that $(x-a)^{2n}$ is always positive, and $(x-b)^{2m+1}$ only positive when $x>b$.

Now if $a < b$ :

Then for $x < a$, we will have $f'(x) < 0$. For $ a<x<b $, we will have the same sign ($f'(x) < 0$). This means that $f(x)$ will be strictly decreasing in $(-\infty, b)-\{a\}$, with $a \in (-\infty, b)$. So $f(a)$ is not a peak ( neither a minimum or maximum).
And for $x > b$ we will have $f'(x) >0$. Combine this with the fact that $f'(x)<0$ for $a < x <b$, we can conclude that $f(b)$ will be a minimum value.

Now if $b<a$ :

Then for $x < b$, we will have $f'(x) < 0$. For $ b<x<a$, we will have the opposite sign ($f'(x) > 0$). This means that $f(b)$ will be a minimum value.
And for $x > a$ we will have $f'(x) >0$. Combine this with the fact that also $f'(x)>0$ for $b < x <a$, we can conclude that $f$ is strictly increasing in $(b, +\infty)-\{a\}$, with $a \in (b, + \infty)$. So $f(a)$ is not a peak ( neither a minimum or maximum).
Hope this helps.

Prove the maxima or minima


Way to get old podcasts in iTunes

Question: When you subscribe to a podcast in iTunes it only lists the 10 most recent podcasts.  Is there a way to get the older podcasts?
Answer: As far as I remember you can add them "manually" by clicking on the list. Just search for the show on the iTunes and click on the individual shows you want to download.
Edit: it seems that the iTunes store just shows episodes of the certain period.
From Apple's Support - Podcasting Frequently Asked Questions (FAQ):

Troubleshooting podcasts
I subscribed to a podcast, but I don't
  see all the episodes—only the latest.
  How come? iTunes will only display the
  podcast episodes that are included in
  the podcast feed. Many podcasts have
  the older episodes archived on the
  web. Check the podcast hosting site to
  see if older episodes are available.
Answer: In iTunes, you can only download the shows that are in the show's RSS feed. For any shows beyond what's on the feed, you will have 2 options.

Go to the website and download each one individually.
Email the host and see if they can send you a Zip file with all previous shows.
Comment: Paul - I share your pain.  Podcast producers typically limit the number of items in their podcast's RSS feed assumingly to protect themselves against lame podcast clients which might attempt to download ALL episodes of a particular podcast.  But it certainly is annoying because if the episode(s) you want aren't in the RSS feed, you've got to download them manually and then make a separate playlist for them which is *separate* from that podcast's list in iTunes.
Answer: sometimes, you might delete the undownloaded podcasts from your list. to get them back, so you can decide to download them, collapse the podcast branch, then shift-click the arrow to expand it again.
Answer: If the reason the podcast is missing episodes is because they're missing from its feed, and the feed has history archived in the Wayback Machine (which many are), you can use Backfeed1 to generate a feed which includes all current and historical episodes2.

1. Disclaimer: I created Backfeed, since I was annoyed at the same problem.
2. Backfeed depends on snapshots of the feed existing in archive.org. The results will be limited by the number of snapshots available for retrieval by a sane number of API requests.
Answer: Many Podcasts have their own website as well.  If you can find the website, then they likely have an RSS Feed that you can add into itunes rather than the link from iTunes store.  In most cases this will get you the older podcasts.
Updated:  Also some podcasts will have enchanced versions of their podcasts with pictures/video or alternate feeds where maybe a subset of the podcasts is of interest.   Manager-Tools is a good example.  Itunes only shows a subset of the episodes.  Their website has an RSS feed with all of the episodes, and a "Basics" feed with a subset of the episodes that deliver the core of their philsophy.
Answer: Here's a way to do it right from the podcasts section of your library: right-click/secondary click on the podcast and select "Show all available episodes". It will list all of the previous episodes. From there, you can click "get" to download separate episodes, or "get all" to download all of the episodes.

Way to get old podcasts in iTunes


Is it advisable to publish in low tier conferences as an undergrad?

Question: I am currently an undergrad and worked on a project that might get published in a low tier conference. I currently have 0 publications but this work is not something I'm proud of. Is it advisable to have my name on this paper given that I am currently working on other projects that can potentially get published at much better conferences? My only concern is that it might negatively impact my profile in the long term. I plan on pursuing a thesis based MS/ PhD after my undergrad. The domain is Computer Architecture if that's relevant. Thanks.
Answer: Instead of not well-reputed conferences, you can submit to workshops held at good conferences or in late-breaking tracks.
The difference, in my opinion, is that in low-level conferences, low-quality papers (e.g. not well written, fundamentally wrong, etc) can be accepted and sometimes without being reviewed. However, for good workshops, you will still need to write a good paper with an idea built on a good basis. They expect, however, that the experiments and results are not strong enough or the approach is not sophisticated to a point that allows it to be published at a good conference.
In summary, a paper at a good workshop demonstrates that the paper fulfils the necessary academic conditions, while a paper at a bad conference does not.
Answering your question, Yes, having a paper at a low-level conference might hurt your profile or at least it won't help at all.
Comment: +1. Besides workshops in top-tier conferences, I would say specialised conferences classified as low tier (say a core ranking of B), still attract many good people. What is a complete waste of time in my opinion is general CS/AI  low tier conferences.


Canonical momentum operators in curvilinear coordinates

Question: What is the quantum canonical momentum operator corresponding to arbitrary canonical position. For example, in Cartesian coordinates ($x^i$), the canonical momentum operator with respect to each $x^i$ is $-i\hbar \frac{\partial}{\partial x^i}$. For arbitrary canonical coordinates $q^i$, would the corresponding canonical momentum operators just be $-i\hat{\mathbf q}^i\cdot\hbar \nabla$?
Comment: In general, there is nothing such as "the" unique canonical momentum for a coordinate $q^i$. A canonical momentum is defined in such a way that its commutator with $q^i$ is nonzero, but it is zero for other $q$'s. One may redefine the momenta and coordinates in many ways, the so-called canonical transformations http://en.wikipedia.org/wiki/Canonical_transformations , without changing the commutator. The idea that one always has a unique one-to-one map is OK to count the degrees of freedom but it's fundamentally flawed if one wants to study the full physical system.
Comment: Please define what you mean by $\hat{\mathbf q}^i$. Does your manifold have a metric? Moreover, $-i\hat{\mathbf q}^i\cdot\hbar \nabla$ has unusual "index up", where momentum $p_i$ traditionally has "index down". I recommend you to simply write $-i\hbar\nabla_{i}$ instead.
Answer: If you use curvilinear coordinates you basically need to transform your derivatives likewise, ie, you need to use the Jacobian of the coordinate transformation you used to go from Cartesian coords to the curvilinear system you have at hands.
More generally, you can think in terms of Quantum Mechanics over a curved manifold, in which case you would simply use the covariant derivative.
Comment: @MBN generally one works with a metric compatible connection, i.e. $\nabla g_{\mu\nu} = 0$. Given a metric, this equation determines the corresponding connection.
Comment: There isn't a unique covariant derivative on a manifold! So which one?
Comment: @Deepak: I know that there is unique such connection (plus the condition to be trosion free). The question is not about manifolds with a choice of metric, so the answer should say something about it, otherwise it is incomplete and misleading.
Comment: @MBN: the last equation in the question has an inner product, which led me to believe that i could make some "general Physical assumptions", such as the existence of a metric and of a metric-compatible connection.
Comment: Fair enough.


How far from the Concorde's ground track was the sonic boom tolerable?

Question: How far away from the ground track did it drop to, say, the level.of heavy traffic? Or something else that people do not mind hearing?
Comment: @AadirajAnil no, it's about altitude, not ground track
Comment: Does this answer your question? [Can a sonic boom produced at 60,000' be heard on the ground?](https://aviation.stackexchange.com/questions/17661/can-a-sonic-boom-produced-at-60-000-be-heard-on-the-ground)
Answer: What you're asking about is called a "carpet", and sometimes, if I'm not mistaken, a "skirt".
Concorde was claimed to have a 20 km wide carpet (10 km either side), but in reality it was heard up to 40 km from the flight path.[Rogers] The secondary carpet was negligible (it's also delayed, if I recall correctly, by 15 minutes). Anyway, the best source is a public NASA report by Maglieri—here are two figures from it:

The N-wave estimation is 1 mile per 1,000' altitude, so the lower, the smaller the carpet. Feel free to use the report to write a more detailed answer.
Also, as I've quoted here, the pressure from Concorde's boom was "equivalent to the pressure one would feel when putting a hand out of the window of a car travelling at 30mph".

Rogers, Peter H., and Domenic J. Maglieri. "Concorde booms and the mysterious east coast noises." Acoust Today 11.2 (2015): 34-40.
Maglieri, Domenic J., et al. "Sonic boom: Six decades of research." (2014). https://ntrs.nasa.gov/citations/20150006843


How to blacklist common passwords on UNIX systems?

Question: If we have a file on a server (Linux, AIX, etc) that contains the common passwords, how can we filter passwords based on this?
Comment: http://www.linux-pam.org/Linux-PAM-html/sag-pam_cracklib.html
Comment: The module quoted by @roaima will add some security layers to your passwords, preventing the most common mistakes made by users on password creation, but as far as i know, there isn't a PAM module to index passwords in plaintext(or password hashes) and block them.
Comment: Well, improving my last comment, on Ubuntu for example you have a file called `/usr/share/dict/cracklib-small`. Maybe is just add a bunch of words you think may create a vector of attack...
Comment: Cracklib provides for a reference to a cracklib dictionary. But as I haven't used that particular functionality I don't think my suggestion is worth being an Answer.
Answer: pam_cracklib provides the option 'dictpath'. pam_cracklib comes with a basic word list located at /usr/share/dict/cracklib-small. You can also use other, larger word list instead of the one provided.
Comment: From which PAM version is this available?


how to filter by date in view - drupal 8

Question: Hello everyone i am working with views ind Drupal8 ,i have a 5 content types that contains all a field 'date' , so i want to display just the content types with date == 2020 , how can i do it the with back-office without code, if not we can d it by code.
Comment: Can't you just add a filter in views for your date field for greater than 2020-01-01 and a 2nd filter for less than 31-12-2020?
Comment: where can i find this option?
Comment: In your view in the "Filter criteria" section click add, select your date field, then in the popup select "Is greater than" and enter your start date in the box
Comment: thank you bro !
Answer: I Agree with Leigh. I regularly use the filter operator 'is between'. Click on the 'add' button aginst filter criteria, select start date and enter 1st jan 2020 then 31st December in the end date box and job done.
Answer: To add date filter by year without code check Views Year Filter
 module.
From module description:

This module add views filter by year exact, instead of full date like mm-dd-yyyy you can use date year like yyyy.
Comment: There's no reason to add a module to do this. Adding a date filter and using "in between" as @David Connell mentions will do it.


Does Steam version of RAGE contain Sewer Missions?

Question: I have not encountered any Sewer Missions on job boards while playing through RAGE. The Wiki only describes the Wasteland Sewer Missions as X360 DLC. However in game properties in Steam I can see "Rage Sewers DLC". Is it something different than the X360 DLC? Or maybe I need to do something to activate this DLC?
Answer: You have to find the sewers first. The job board postings happen after you've gone through a sewer once, and the posts are only for sewers you've cleared. There are quite a few sewers scattered around. Look for manholes.
Source: I played the Steam version.
Small spoiler!
The first sewer:

 is near the Ark you come out of, under a bridge, and has a guy standing next to it.
Comment: Ah, that's awkward to have to first clear the sewer to get job to clear the sewer ;-) They are much smaller missions, than I've expected. Anyway, that worked. Thanks.
Answer: I think they are, according to these links:
http://steamcommunity.com/app/9200/discussions/0/864980278016001920/
http://forums.steampowered.com/forums/showthread.php?t=2439759
http://steamcommunity.com/app/9200/discussions/0/846940248707337032/
In PS3, the manholes are visible, but If I try to enter without having installed the DLC (Sewer Missions), I cannot, and states I need to get the DLC.
(For Steam the only DLC available is "The Scorchers" a totally different set of missions, but from the comments in Steam forums, it seems Steam version has the Sewers DLC builtin).
However, if you have trouble finding them all, this video in youtube has the locations of the sewers,

Adding some description:

Ark Sewers: The first is nead the Ark Area where you start the game, under a bridge an old man is camping close to the first sewer manhole.
Outrigger Sewers: Going to the Outrigger territory before taking the side path toward the Radio antenna mission, the 2nd manhole is on the side of the road.
Southern Sewers: From Wellspring, taking the left path (with a car) thru the wasted territory (like if you were to Haggar's settlement), you'll find a ramp for cars, below it, in a shadowed are is the third sewer.
Wellspring sewers: Again from the entrance of Wellspring, walk along the right (mountain borderline to the entrance) after a few meters, you'll find the fourth one.
Northern Sewer: From Wellspring again, taking the car this time the right, entering the Shroudded Territory, you'll pass under a bridge (there is a bandit tower shooting at you), then a 2nd bridge, park under that bridge, and on the left side, is the fifth manhole.
Plateau Sewer: Near the Watch Tower, you'll find a peak rock like five stories high, in a bush nearside is the sixth.
Kvasir's Sewers: It's relatively close to the Kvasir Lab (easier to connect there from the Plateau Sewer).
Subway Town Sewers: In the Subway Town Territory, drive as If you were to Gearhead Territory, but stop before shor when the cliff for Subway Town starts, on the right close to a 2 meter diameter pipe, will be a manhole.
Shipyard Sewers: Into Gearhead Territory, pick the left ramp, on the left
you'll see some green paint, very close to it, is the 8th.

Also the Rage wikia (http://rage.wikia.com/wiki/Wasteland_Sewer_Missions) list the sewers with some pictures of the places. Enjoy!
Comment: This is a good answer, but please provide brief summaries of the locations in case the link ever goes down :)


Some Vassiliev Invariant Questions

Question:
The V.I. definition goes doublepoint=overpass-underpass (or was it the other
way around? If it's 50:50, I score 0 always :-). Would it lead anywhere
to define doublepoint=overpass+underpass? (Even if it's equivalent.
As I wouldn't have to guess then :-) I ask because I hate arrows on knots.
Stoimenow tabulated values for the 3-degree V.I., but only for knots.
Has someone this data also for links? (Until 6 crossings already would be useful.)
Is this logic correct? Suppose you have a V.I. vanishing on all
knots with >=2 doublepoints. Consider a clasp consisting of 2 doublepoints.
From the definition then S^2+S^(-2)-2*S^0 = 0, where S is the usual braid
generator resp. S matrix, and thus you can deduct a skein equation from
the degree of a V.I. (and the higher the degree, the more of them).
Comment: retagged: general topology -> quantum topology. General topology means point-set topology: http://en.wikipedia.org/wiki/General_topology
Answer:
No. You can always mirror-reflect your definition across the plane that defines over/under.

I think so.  Take a look in the Chmutov, Duzhin, Mostovoy survey on the arXiv.  On page 92 they have the first ten non-trivial Vassiliev invariants, computed on sufficiently-many knots.  If what you're looking for isn't near there, they likely have a reference for it.

Type one Vassiliev invariants are trivial for knots, and for links I believe linking number is the only non-trivial one. I'm not following your argument but I'm also not seeing where it's going.  Perhaps this is what you're trying to prove?
Comment: THX. 1) So Vassiliev invariants and undirected knots don't go together? 2) is then superfluous for my goal (but still, that
link is a goldmine, many many THX!).
Considering 3), the type number wasn't relevant. Could as well
have used a higher order for the example: 0=S^3-3S+3S^-1-S^-3 or
0=S^4-4S^2+6S^0-4S^-2+S^-4 - I just wanted to know whether you
can handle the definition as if it were an ordinary skein equation.

Some Vassiliev Invariant Questions


Did Neo die in the end of the Matrix trilogy?

Question: Towards the end in The Matrix Revolutions, Neo is jacked by the machines into the Matrix to fight Agent Smith. During the fight, Smith seems to have killed Neo then we see the machines pumping a charge into Neo to revive him. Neo however doesn't seem to be active inside the Matrix anymore and just explodes into light.
We then see his physical form being set down by the machines. Is Neo at this point actually dead or is he just unconscious?
Comment: The Matrix Online made it pretty clear that Neo was dead.  Much of the conflict in the game was that the Humans wanted to get Neo's body back from the machines, but they were refusing.
Comment: What if I told you... there is no Neo?
Comment: I won't leave this as an answer, but I had a crazy idea that Neo escaped from the "real world" matrix into the actual real world, leaving his body behind. This explains why Neo is dead to all the people still inside, but actually lives and may return somehow, as the Oracle said in the end.
Comment: The power of the one extends beyond this world as the oracle said.
Comment: @Bolden By "this world" I would think she means The Matrix, and we see that Neo can use his 'Matrix powers' in the real world as well. However, the Oracle suggests that perhaps Neo isn't entirely dead at the very end of Revolutions, when she says that we might see him again "someday".
Comment: No, not only he did not die but Trinity also is alive since Warner Bros. announced (20/09/19) that "Matrix 4" will begin production soon with Keanu Reeves and Carrie-Anne Moss reprising their roles as Neo and Trinity.
Answer: From what i've understood, The Oracle, has identified agent Smith as an equation, trying to balance itself out, being the opposite of Neo, perhaps Neo had to die in order for Agent Smith to be destroyed.
Although, when Smith has taken over the world and multiplied itself into every single program, he has, presumably also multiplied itself into every single human being, who was jacked into the matrix.
In the last scene, where the Architect tells the Oracle, that the "Others" shall be released, means that "virus", which was injected by the source into Neo, did not kill other human beings who were possesed by Smith.
That means there's a chance Neo might still be alive (not only because of that, but also because, when Smith was destroyed, Source said "It is done", and it would be rather unusual for a machine to speak with a dead body), however that may also mean, that in order for Smith to be destroyed Neo had to die, because Smith somehow copied his source into Neo.
I guess the choice is left up to you to make up your damn mind.
Comment: Source said "It is done" - this seems the most compelling support for the fact that Neo was still alive at the end.
Comment: This isn't really in-universe, but probably the Deus Ex Machina was made to say "It is done" at Neo's death to echo Jesus' last words on the cross in the gospel of John, ["It is finished"](http://en.wikipedia.org/wiki/Sayings_of_Jesus_on_the_cross#6._It_is_finished)--if you watch Neo's death scene starting around 3 minutes in [here](https://www.youtube.com/watch?v=ruSTiOmiyWw), there's a cross-shaped light coming from his chest as he dies, and his pose in death and when his body is being carried away by the machines is probably meant to recall Jesus too.
Comment: I assumed that the Source said "It is done" to the other machines in the vicinity.
Answer: YES AND NO: HE'S DEAD, BUT THERE IS PERHAPS A HINT AT RESURRECTION
Others have mentioned "It is done" being said by Oracle at the end of the movie, and this is an important observation. This is a clear inter-textual reference to the Gospel of John, specifically John 19:30, where Christ proclaims, "It is finished" as he dies by crucifixion.
Now, I'm not trying to be preachy, so bear with me: In the narrative, of course, Christ dies on the cross, but of course there is the resurrection that follows. Neo's messianic characteristics, esp. in the final film, suggest that quote by Oracle may not mean Neo is "done" at all, but rather that he is dead in one sense but will return.
His vision near the end where everything turns golden or bright may supposed to be like Christ's glorification after the resurrection (or perhaps like Christ's transfiguration before his crucifixion). The talk about everything having a beginning and an end may be an inconspicuous reference to Christ as the Alpha and the Omega, the beginning and the end. The Architect saying others will be released sounds suspiciously like Christ's leading "captivity captive" in Ephesians 4:8 (a reference to a Greco-Roman victory procession; the kind that marched through triumphal arches that have survived around Europe even in this day).
In the end, I just think it's ambiguous, but based on the messianic comparisons, I think the movie is hinting that Neo died, yes, but is hardly finished.
Comment: This makes a lot of sense. The Wachowskis clearly portrayed Neo as a messianic figure and the sacrifice/salvation theme is without ambiguity in *Matrix Revolutions*. I think the Oracle saying, 'I suspect so', when asked if we'll see him again, is a reference to the notion of Neo rising from the dead/living, and even theoretically returning later, like the story of Christ also.
Comment: @Peter no, that's Bill and Ted's Excellent Adventure.  I think that was probably an earlier, crappier version of the Matrix.
Comment: Zomg! What if John Wick is another version of the Matrix?  ;)
Comment: @Paul Woah! He's like, the one, dude! Excellent!
Comment: @Paul (that's the right film, isn't it?)
Comment: A prophet among us.
Answer: At the end of the film, the Oracle is sitting upon a bench dedicated to the memory of the (dead) Thomas Anderson.

Admittedly this might be

a) Symbolic of Neo's rebirth as a new person, distinct from his identity as the (Matrix slave) Mr. Anderson.
or

b) Someone in the production team being funny.
Comment: Going through all that trouble to be dedicated a _bench_.
Comment: benches are the Tumbleweed badge of the Matrix.
Comment: @Jenayah - Imagine going through all that trouble and *not even getting a bench*.
Answer: When last we see Neo in Matrix-Vision™, he's still glowing gold, which leads me to believe he's alive.
Also, Sati asks the Oracle if they'll see him again, to which the Oracle replies in the affirmative. She's yet to be wrong, so...
Comment: This could really use a screenshot :)
Comment: There were Ones before; perhaps Neo will reincarnate.
Answer: Neo died.
Evidence:
This is what Neo looked like when the machines used him to infiltrate Smith's program.  Note the cable coming out of Neo's head; this is a view of his face in the real world.

Having blinding white light shooting from every orifice is not a sign of general good health in an organism and it seems unlikely that this is an event Neo could live through.
In Matrix-vision we see an obvious visual allusion to the Crucifixion, with Neo depicted as a Christ-like figure sacrificing himself to save all humanity.  If Neo doesn't die, there's no sacrifice.

And finally in the Matrix itself, while the Oracle and other Matrix-denizens were restored, no trace of Neo remained.
Comment: The cable is harmless but it is indicative that the image is of Neo's real world body. In the real world having blinding white light shoot from every orifice is not a sign of general good health in an organism.
Comment: What 'cable coming out of Neo's head'? I don't see one apart from the harmless normal one in he back of his head. And then, about 'no trace of Neo remain[ing]', maybe that was the moment at which they also 'unjacked' him from the Matrix / possibly even unloaded all the 'anomaly' code from him in doing the 'equation balancing itself out as well' final resolution/only way to get rid of Smith, all at once in a neat act by the machine source? I don't see Neo's (Smith infected) 'Residual self image' disappearing in the Matrix at that point, being an indication of Neo's *physical body dying* at all.
Comment: Problem:  If we accept the Jesus-is-god and Jesus-was-resurrected parts of the Jesus story, Jesus didn't really die.  He was god, and god can't die.  Basically, he pretended to be dead, then got up and went to hang out with his friends.  There was no sacrifice involved.
Comment: @KyleJones - God doesn't feel pain.  He's god.  Feeling pain would constitute a change in god's status.  The Christian church has maintained, for as long as it has existed, that the idea that god changes is heretical.  Also, the biblical narrative of the passion says that Jesus was crucified within a few hours his arrest. He wasn't tortured for days, only minutes.  He was "dead" within about 12 hours of his arrest, maximum.
Comment: Ok, but we don't have enough details of what exactly that 'blinding white light' is anyway - in a universe where many other scientifically fantastical things are happening already, I don't think everything should be scrutinised at face value especially when it's set in the far future. I think overall we have to take the cues from the *storytellers* as to what the likely answer is, and that the themes and hints spoken by *characters* (like the Oracle) are more authoritative than mere fan speculation as to what that white light is and whether it's scientifically fatal to physical human beings.
Comment: @foregon: Interpreting vague dialog by the Oracle/Source vs. interpreting fairly clear scenes in the physical world — I'm not sure  one is more reliable than the other. We know for certain that his body died. We only have vague hints that some form of "the One" may come back, but not necessarily the Neo form or at least the Neo we know. The question doesn't ask if Neo can ever come back. It's asking if he's dead or unconscious. Of the two, he's definitely dead at that point. Maybe his software is stored somewhere, but the body is definitely dead.
Comment: @Wad Well, there were a couple days of torture preceding the fake death couldn't have been too much fun.  Christ wouldn't have [begged God to let him out of it](http://biblehub.com/luke/22-42.htm) if the whole thing was going to be a picnic.  Neo had his eyes put out and repeatedly had the stuffing kicked out of him by a superhuman adversary, so not much fun for him either.
Comment: Full disclosure:  I'm an atheist, but also a student of textual criticism of the bible, early Christology, patristic theology, and the history of the bible, at Princeton Theological Seminary. I spend a lot of time studying these issues.
Answer: Neo's physical body may be dead but from what the Oracle said at the end his spirit/energy is probably still alive.  This just goes to further the idea/theme of Neo as the literary Christ Figure in the Matrix Trilogy; the savior/martyr for mankind.
Answer: Neo did die at the end of the Matrix trilogy*
In the Matrix Resurrections, we learn that after his encounter with Smith, (dead) Neo was taken by a rogue program and put into a 'resurrection pod'. He was worked on and eventually brought back to life through robotic ingenuity.

Analyst: I was there when you died. I said to myself 'here is the anomaly of anomalies, what an extraordinary opportunity'. First I had to convince the suits to let me rebuild the two of you... Resurrecting you both was crazy expensive. Like renovating a house. Took twice as long, cost twice as much. I thought you'd be happy to be alive again. So wrong.

*Superman dead, not Uncle Ben dead.
Answer: The movies are ripe with religious symbolism.  The symbolism of the Cross present when Neo finally defeats Smith by letting Smith copy himself onto him is so painfully obvious, that I don't see why/how Neo could still be physically alive.  Now, in keeping that that theme, it's entirely possible that his mental projection could still be around.  It's happened before when Neo was at the train station--his mental projection was at the station even though his physical body wasn't connected to the Matrix or anything else.
Answer: Neo probably died physically, because:

no one have seen him ever since, and humans can't live in the machine world (air is too polluted, no food, no other humans, etc, see the Animatrix canon).
his death is clearly hinted at with the cross light metaphor of the Christ when the machines send the "kill code" for Agent Smith (whether it's a kill code specifically for Agent Smith or just killing the body of Neo remains up to debate, but we can clearly see the highlighted sending of a command from the Deus Ex Machine to Neo via the glowing wires just a moment before Agent Smith is deleted)
it was probably necessary to balance the equation (since Neo is a cumulative error of the system because of allowed free will and Agent Smith being the counterbalance)

However, since Neo is clearly a messianic figure like the Christ, he may resurrect. And indeed, in the Matrix Online game canon, there were clear hints that he did, under the traits of a woman who recently awoken from coma:

Neo's ultimate fate has been openly called into question as Morpheus
has pointed out that his remains were never returned from 01 (the
Machine City) back to Zion, but at the same time the Machines have
commented that they did not recycle (liquify for re-use) his body. On
an interesting note, there is a newspaper clipping found during a Zion
critical mission about a 27-year-old woman named Sarah Edmontons (an
anagram for "Thomas Anderson") waking up from a coma and leaving the
hospital on her own. In the text, there was a note written on it
asking, "Is this him?" Furthermore, on one Machines' set of critical
missions, it turned out that Edmontons does not exist in any known
database, therefore supporting the speculation that Edmontons is Neo.
Although no outright comments have been made, developers have hinted
that Edmontons would have played a role in the future.

Apart from the Sarah Edmontons mystery woman, the first mission of Matrix Online was to find fragments of Neo's RSI, and the RSI was later partially reconstructed by the machines and displayed in a club, so it's clear that some digital artefacts of Neo remained inside the Matrix.
An in-universe explanation of how he could survive is that his digital counterpart (his RSI) survived his physical body, which is not so surprising since we know from the third movie that those two parts of him could be separated (he was in coma, unplugged from the Matrix, and still his digital counterpart was in the Matrix). I have explained this theory in more details here: In "The Matrix Revolutions" How Does Neo Stop the Machine Weapons in the Real World?
Comment: To continue with the Matrix Online canon, since his body (nor the body of Trinity) were recycled, another hypothesis is that he survived/was revived by machines advanced technology, and the machines keep him alive in a confined area as to prevent him from interfering with the Matrix balance anymore (with maybe its own acception, since a peace trust was brokered), thus maintaining the Matrix in a newly found balance (the cumulative error represented by the One continues to accumulate outside the Matrix without interfering).
Answer: From my point of view. When the neo gets blind all the things he feels is shown in golden.
And when the machines are taking his body in the end the whole world is shown in golden for a second or two. This implies that neo was still feeling and was alive. :D
Comment: I think Neo's gold "Code Vision" was to show that he could perceive the world similarly to how the Machine's did, which is why everything had that same look when the Machine's reclaimed his body at the end of the third film.
Answer: Even before his journey to the machine city, Neo feels his end coming.  He knows that harmony can come only if Smith is destroyed and at the same time the "anomaly" must end to bring peace to both worlds.  Towards the end of the fight, Smith utters what actually Oracle has told Neo earlier, namely, "everything that has a beginning has an end."  Neo gives in to Smith only to end Smith's copying each individual in the Matrix and thus in a way ends Smith's task or what Smith wanted to achieve ultimately and thus also destroying Smith in the process.
Comment: The father-program at the train station talks about a program's purpose (task). I think the only way to defeat Smith is allow him to end yourself by accomplishing his task. This sends a STOP signal to his main thread and then a HALT(0). A kind of `REPEAT MULTIPLY() UNTIL MY_COUNT = MATRIX_POPULATION`
Comment: The architect then just waits for Cron to restart the Smith process, or launch Matrix v7.0.
Answer: In the game, the brothers said "Maybe it works in a movie but in a video game the Jesus thing is just lame." According to Christianity, Jesus will return. I believe they meant Neo will return. But that will probably mean another sequel with Smith, because the only way to kill Smith was for Neo to die. So doesn't that mean if Neo returns so does Smith?
So my answer is Yes, I do believe Neo is dead because Smith is dead too. But I believe Neo can be revived.


Queueable apex and trigger execution context

Question: If I enqueue an instance of a Queueable class during a before trigger, is it guaranteed that the Queueable class will not run until the after trigger completes?
Comment: Queueable is asynchronous. It will run whenever Salesforce decides there's enough resources for it. A question back to you: Why not call it in the `after` context?
Comment: I have a trigger framework in place; to call it in the after context, I need to create the class AccountAfterTriggerHandler. I'm being lazy. But I think I will do it in the after context anyway, because doing it in before context will probably confuse future maintainers.
Answer: Yes, this is guaranteed (though not explicitly documented). The reason why is that, until the transaction fully commits, the Queueable job call may be rolled back, either because of governor limits, an sObject addError call, a Database.rollback command, or an uncaught exception, and probably a few other ways. As such, the system absolutely will not execute the Queueable until the transaction that spawned the trigger context has successfully run its course. This is also true for Scheduled jobs, Batchable jobs, future methods, sending emails, calculating sharing rows, and almost any other type of asynchronous code or action that may exist in the future. The two exceptions to this guarantee are the "DML Immedate" actions available on External Objects, and the EventBus.publish actions that can be called on Platform Events that execute immediately.
(Note to future readers: if future exceptions to this guarantee arise, please let me know.)


Is there a module in R to find structural breaks in time series data?

Question: I have time series data that I need to transform into a stationary series. It is a monthly series so according to this graph, I think the break point is in September, 2007. However, I am not sure if I should go with my hunch or find a module in R to get the accurate break point.
Answer:
There are a lot of R packages. Here is a non-exhaustive overview with worked examples. Below, I'll show a solution using the mcp package.
Let's start by simulating some data that look a bit like yours:
df = data.frame(x = 1:150)
df$y = c(arima.sim(n = 75, list(ar = 0.8), sd = 2) + 0,
         arima.sim(n = 75, list(ar = 0.5), sd = 2) + 7)

You can fit a model consisting of a two plateaus with 1st-order autoregressive residuals.
model = list(
  y ~ 1 + ar(1),
  ~ 1 + ar(1)
)

Maybe you just want one autoregressive coefficient that is shared for both segments. In that case, just delete ar(1) in the formula for the second segment - that will make the first ar(1) carry over.
Now we fit the model:
library(mcp)
fit = mcp(model, data = df, par_x = "x")

Sometimes you want do more and faster sampling via  mcp(..., iter = 10000, cores = 3).
Let's see the result visually:
plot(fit)

It seems to capture the data OK. The black dots are the raw data. The grey lines are samples from the posterior distribution. The blue distribution is the inferred change point. In this case, the change in intercept is probably the clearest "signal", but this article on time-series analysis with mcp shows examples without such abrupt changes in the mean.
You can see posteriors for each parameter using plot_pars(fit). Let's also see if it recovered the parameters well:
> summary(fit)
Population-level parameters:
    name  mean lower upper Rhat n.eff
   ar1_1  0.74  0.56  0.91  1.0  7131
   ar1_2  0.61  0.43  0.80  1.0  6133
    cp_1 75.48 72.05 76.00  1.2   147
   int_1  0.64 -1.11  2.50  1.0  6288
   int_2  6.56  5.22  7.97  1.0  4948
 sigma_1  2.09  1.86  2.35  1.0  6743

We simualted the data from ar1_1 = 0.8, ar1_2 = 0.5, cp_1 = 75, int_1 = 0, int_2 = 7, and sigma_1 = 2. So it looks OK.
This was a very simple model with intercepts-only and a single change point. Several other packages are suitable here as well, including segmented and EnvCpt. Strengths of mcp include (1) it quantifies uncertainty of the change points, (2) it can do per-segment regression modeling, and (3) it can handle more complex autoregressive structures, among other things. The main limitations are (1) it's slower - but not unreasonably slow, (2) it requires JAGS (which is free).
Answer: The heretofore  referenced "changepoint package in R" assumes that there is no arima memory i.e. the data is independent and normally distributed see https://www.jstatsoft.org/article/view/v058i03 section 4.1  . If the data is not independent then the underlying arima model needs to be identified but no provision is made for the identification of this critical component just requiring the user to pre-specify it.
Furthermore if the user has some notion about causal predictors and possibly their contemporary and lag effects that might need to be conditioned for … notsomuch … which limits it's general usefulness.
Breakpoints can occur in the arima component over time and certainly error variance change points might also need to be conditioned for see http://docplayer.net/12080848-Outliers-level-shifts-and-variance-changes-in-time-series.html . One needs to KNOW the assumptions under which each and every proposed solution is valid. All that glitters is not gold !


Do Lorentz Boosts in the same direction form a group?

Question: I know that two consecutive Lorentz Boosts in different directions produce a rotation and therefore Lorentz Boosts don't form a group. But, my intuition tells me that, Lorentz Boosts in the same direction should form a group. Two boosts along the x axis should produce another boost along the x axis. Is that correct?
Comment: Boosts in different directions form a subgroup of the Lorentz group? If yes, how can I check it?
Comment: No they dont, because their commutator is a rotation. Boost in x, then in y, then backward in x, and backward in y, and you get a rotation. You should delete this answer, as it doesn't answer the question.
Answer: Yes, it is a one-dimensional subgroup generated by exponentiating an infinitesimal boost. Every one dimensional exponentiation of a generator forms an abelian group, because $e^{aG} e^{bG} = e^{(a+b)G}$, there is nothing to not commute. This result is the addition of velocities, you can explicity check that this is associative (it is always manifestly commutative).
Comment: Right, +1. And as you (and I) recently discussed in a different answer, http://physics.stackexchange.com/questions/23625/how-to-deduce-the-theorem-of-addition-of-velocities , the relationship of $a,b$ which are additive and form the group $({\mathbb R},+)$ to the velocities $v$ defining the boosts is $v/c=\tanh(a)$ or $\tanh(b)$, respectively.
Comment: Thank you for these answers. I can't +1 or check this now because my browser isn't letting me register an account. It is interesting that noncommutativty produces a rotation via the Baker Campbell formula and the fact that the commutator of two Boost Generators produces a Rotation Generator. But the first term is still e^{(a+b)G} in such a case, so I suppose it is actually a boost and a rotation simultaneously.
Answer: Firstly, an excellent question. Never considered this before.
As has already been said, combining boosts along difference directions clearly doesn't form a group as the closure axiom is not met.
However, Lorentz boosts are nothing more than (hyperbolic) rotations in Minkowski space. So I think the set of boosts along the same axis should form a rotation group. I hope someone else can explain the reasoning without resorting to generators, as they're not my strong point.
See this YouTube video for some basic derivation.
Comment: Don't be scared about generators--- the exponential parameter is just the (hyperbolic) angle of the (hyperbolic) rotation, as Lubos Motl commented on my answer. If you follow the link, you can see that the angles add up in relativity, just as they do in geometry, so that the group is the additive real numbers (although not modulo $2\pi$ like in geometry)..


yfinance not fetching the entire range of stock I'm asking it to

Question: I'm new to the yfinance library, so I may just be being completely stupid here, but I'm trying tpo fetch 365 days worth of information, and when I request that from yfinance, it gives me 251 days. Similarly, when I ask for 10 years of information, or 3650 days, it only gives me 2516 days. Does anyone know how to fix this?
EDIT: I do know that msft is not the correct ticker for AT&T, but I am going straight off of a template and it's not going to cause any errors, I'll fix when I have the time
import yfinance as yf
msft = yf.Ticker("T")
# get all stock info
msft.info
historical_market_data = []
# get historical market data
hist = msft.history(period="1y")
days = 0
for i in range(len(hist)):
  days +=1
  historical_market_data.append([hist.High.iloc[i], hist.Low.iloc[i], hist.Open.iloc[i], hist.Close.iloc[i], days])
'''
Item 0: High of the day
Item 1: Low of the day
Item 2: Price at opening
Item 3: Price at closing
Item 4: Day number
'''
for i in historical_market_data:
  print(f"High: {i[0]}, Low: {i[1]}, Open: {i[2]}, Close: {i[3]}, Day: {i[4]}")

Comment: _I'm trying tpo fetch 365 days worth of information_ Are you aware that the stock market is not open every day?
Answer: Stock data encompasses Monday-Friday, and excludes holidays, as that is when markets are open; holiday and weekend data would be duplicates of the previous days. Your data for a full year will not be 365 days, since the typical week is 5 days for the stock market.
You are receiving 251 records; 251/365 is roughly 5/7, and if you include the ~10 days that the stock market is on holiday, you will land at the proper proportion.
If you want 365 days of data just for testing, pick a crypto coin ticker (i.e. "BTC-USD") as they have 24/7 markets.