Is Magento Commerce provide a marketplace(multi vendor) by default with Magento B2B?

Question: Is Magento Commerce provide a marketplace(multi-vendor) by default with Magento B2B?
if not what is the best way to implement Magento 2 B2B and marketplace in Magento Commerce?
Answer: Magento is not build to be a marketplace out of the box. Not even the B2B suite (which is only a set of modules over the core code).
There are some questionable modules out there that can turn your magento instance into a marketplace. I only used one of them which should remain nameless, because I strongly advice against it. the code is far from being the best and it is full of bugs.
As a personal opinion, I strongly recommend NOT using Magento as a full marketplace solution. It is not built for that and the cost of making it behave like one may not be justified.
I suggest either looking for a platform that is intended to be a marketplace, or use Magento only as a sales channel and handle the marketplace features in a separate application based on the needs you have.


mu4e change column width

Question: I recently switched to mu4e for my mails which is awesome. I changed the date format in the headers view so that it became longer. Unfortunately I cannot see the complete date now because the column width is to small.
How can I change this? How can I change the width of all columns?
Documentation doesn't mention the column width.
Answer: Does customizing mu4e-headers-fields, changing the width corresponding to the :human-date "header", do the job? (I haven't tested this.)
Comment: Thx it worked, I was looking for a documentation on this, but not in the comments of the code ;)
Answer: Documentation on how to change headers can be found here : https://www.djcbsoftware.nl/code/mu/mu4e/Longer-configuration.html
Here is the part the op is interested in :
;; the headers to show in the headers list -- a pair of a field
;; and its width, with `nil' meaning 'unlimited'
;; (better only use that for the last field.
;; These are the defaults:
(setq mu4e-headers-fields
    '( (:date          .  25)    ;; alternatively, use :human-date
       (:flags         .   6)
       (:from          .  22)
       (:subject       .  nil))) ;; alternatively, use :thread-subject

Note that the order of the fields in the code will be the same in the view.

mu4e change column width


Finding Laplace Transform without ROC

Question: While studying Laplace Transform i found that region of convergence (ROC) is important because for some problems we have same Laplace Transform but different ROC helps us to take correct inverse Laplace Transform.
So, Now i am practicing inverse Laplace Transform problem i found almost every probelm to find $x(t)$ is without ROC.
So i want to ask that without have ROC how can we solve inverse Laplace Transform?
Answer: Strictly speaking you can't because without specifying the ROC, the inverse Laplace transform is generally not unique. However, in many contexts there is the implicit assumption of causality of the corresponding time function (i.e., $x(t)=0$ for $t<0$), which is equivalent to stating that the ROC is a right half-plane.
Comment: We can say that every system is casual if it is defined for t>0?

Finding Laplace Transform without ROC


How to locate a specific slide using a certain layout in Google Slides?

Question: I have a 38 slide presentation.  When examining the Master theme, I can see that there is 1 slide using a specific layout "Big Number".  But I don't see any slide that is using that layout.  I have gone through every slide and made sure each is using the correct layout.  How can I locate this seemingly "phantom" slide so that I can either delete it or change it to use the correct layout?
Answer: I'm looking almost for the same, but not figured out how to find and list all layout names it the Master.
Probably you should use script to find specific layouts[] object for (Page)
The layouts in the presentation. A layout is a template that determines how content is arranged and styled on the slides that inherit from that layout.
From https://developers.google.com/slides/reference/rest/v1/presentations?hl=en

See more about Pages, Page Elements, and Properties
There is some code sample for the layouts search:
<code>// The following line opens my presentation
var presentation = SlidesApp.openById('PresentationID');
// this will return an array of all the layouts
var layouts = presentation.getLayouts();

//if your first array item is the office layout
var newSlide = presentation.appendSlide(layouts[0]);

//Orelse you can search for your layout
var selectedLayout;
for(var item in layouts)
{
   //Logger.log(layouts[item].getLayoutName());
   if(layouts[item].getLayoutName() =='CUSTOM_1')
   {
     selectedLayout = layouts[item];
   }
}
</code>
Comment: I'm sorry but I don't understand how this is an answer to the question asked…?
Answer: With your presentation open in Slides, chose menu item Slide > Edit Theme. Then select a particular slide layout. Once selected, all of the slides using that layout will be highlighted (bounded in yellow) in the thumbnail views on the left hand side of the page. You may need to scroll up and down through the thumbnails to find the slide(s) using the selected layout.


GUI library for HMI Cortex M

Question: Hi I am basically using cortex m3 series of processors for our application. I am looking for some GUI library that can be used for HMI. Is there any GUI libraries available? I recently heard about Qt. Is it a library written in C? Can I be able to port it to keil or IAR or Eclipse or any IDE? Or can you guys provide some other GUI library suggestions? Thanks!
Answer: Using Qt in a microcontroller project is infeasible. It requires megabytes of RAM and ROM and a full-fledged OS. There is support for embedded systems but it is limited to Nokia phone OSes, embedded Linux and Windows CE. See storage requirements for embedded Linux for example (note that these aren't RAM requirements). Qt has been used successfully on ARM9, ARM11 and Cortex-A systems such as FriendlyARM boards, BeagleBoard etc. (note that many of these have OpenGL ES hardware acceleration, too). Extra convenient support for Raspberry Pi is upcoming. If you actually need Qt, consider an ARM CPU-based (as opposed to a MCU-based) system running Linux.
If you do need to run a GUI on a microcontroller, see the answer linked to by embedded.kyle for realistic alternatives.
Answer: Open-source Embedded GUI Library https://littlevgl.com
LittlevGL is a free and open-source graphics library providing everything you need to create embedded GUI with easy-to-use graphical elements, beautiful visual effects and low memory footprint.
Powerful building blocks buttons, charts, lists, sliders, images etc
Advanced graphics with animations, anti-aliasing, opacity, smooth scrolling
Various input devices touch pad, mouse, keyboard, encoder etc
Multi language support with UTF-8 encoding
Fully customizable graphical elements
Hardware independent to use with any microcontroller or display
Scalable to operate with little memory (80 kB Flash, 10 kB RAM)
OS, External memory and GPU supported but not required
Single frame buffer operation even with advanced graphical effects
Written in C for maximal compatibility (C++ compatible)
Simulator to start embedded GUI design on PC without embedded hardware
Tutorials, examples, themes for rapid GUI design
Documentation online and offline
Free and open-source under MIT licence
Answer: I would stick with Qt. It's actually a C++ library. Keep in mind you need an embedded OS (Linux/Windows CE). There are plug-ins available for Eclipse and Visual Studio. Possibly others.
There are also some non-free alternatives mentioned in a similar SO question:
https://stackoverflow.com/questions/2997648/gui-library-for-microcontroller
Comment: Thanks for the replies. But my application is driven by Cortex M3 controller, isnt there any library for cortex m3? I thought Qt can be used with cortex m3 never knew it requires much memory.
Answer: Just take a look at this article on Microcontroller Central.
Rich Quinnell mentions "... I saw a demonstration of Java applications running on an STM32-F3 MCU..."
Probably what you are looking for?
Comment: It would be helpful if you were to summarize the contents of the link here.

GUI library for HMI Cortex M


In the double slit experiment, what is the position and momentum of a particle?

Question:
Is the position of an electron/photon which slit it went through?
And the momentum of the electron/photon the wave-like interference pattern detected on the detector screen?
Comment: The pattern on the detector screen is a pattern of position measurements.
Answer:
In the double slit experiment, what is the position and momentum of a particle?

The experiment , even with single particle at a time ,  extends in time. The direction of the ingoing particle is known, and the hit on a screen after the slits is known.
For a classical experiment with billiard balls This is mathematically fitted  by a trajectory.  The mathematics of throwing billiard balls at slits gives a trajectory up to the slit and then straight through , or if the billiard ball scatters off an edge of the slits a distribution will appear at the screen, and an image of two blurred lines. Even in the classical case at the region between the two slits images one cannot say which slit the particle went through (unless extra detection is made at the slits)

Is the position of an electron/photon which slit it went through?

Here is a double slit experiment with a build up of single electrons at a time, starting from the top.

The only position measurements we have are the (x,y) of the electron on the screen and the (x,y) positions of the slits. The mathematical trajectories would not differ from the classical ones. As more and more electrons accumulate an interference pattern emerges in space.
This interference pattern has led to a lot of confusion. Since the electrons are found one electron at a time they obviously are not split. The pattern is a probability distribution, i.e. answers :how probable is it that I will find an electron at the (x,y) of the screen. This probability distribution fits the quantum mechanical solutions of "an electron scattering through two slits"  . Generally detectors on slits destroy the interference accumulation. unless care is taken . In this experiment the detectors are subtle , the quantum mechanical problem changes but still some interference is seen, since each measurement changes the boundary conditions , and a changed pattern appears in the accumulation.
This experiment displays the quantum mechanical character of elementary particles, their point particle behavior on the screen, their quantum mechanical nature going through the screen.

And the momentum of the electron/photon the wave-like interference pattern detected on the detector screen?

The momentum of the electron is decided by the experimenter and chosen in the range where the Heisenberg uncertainty principle, HUP will allow to display the wave probability pattern.
The delta(x) in this case is the size and distance between slits chosen so that delta(x)*delta(p) is within the bounds of the HUP.
From the wavelength of the interference pattern using the known mass of the electron and the de Broglie relation one can get the momentum chosen for the incoming beam.
It is important to stress that the wave nature displayed in the experiments on elementary particles are neither energy waves nor matter waves as far as individual particles go. The wave nature of the particles appears into the distributions of probability of detection/interaction. The double slit experiment answers the quantum mechanical boundary value problem "one electron interacting with two slits " and the answer is a probability density distribution when the boundaries are within the HUP.
Answer:
Is the position of an electron/photon which slit it went through?

No. The position is the result of an interaction. You have a slit, and electrons are free to go through both slits and travel until they reach some kind of particle detector such as a screen. The screen makes a record of the position based on which part of the screen reacts to the electron. So each electron leaves a mark based on where it lands.
There is no information about which slit you went through unless some other particle changes its state based on whether the electron goes through.

And the momentum of the electron/photon the wave-like interference pattern detected on the detector screen?

No. The momentum is what it had before it went through the slits. You get position by interacting and you get momentum by interacting. So first you interact so you have a momentum but no position, this allows you to go through both slits in a nice coherent way. Then later you acquire a position instead of a momentum.
The wave-like interference pattern is built up over time. When you send many electrons through one at a time you see that some places get a larger share of hits than others. Some places are more popular destinations for position to be determined and other places are less popular.


Algorithm for detecting a knob-turning gesture?

Question: What is the math to identify a gesture for the motion made by two fingers turning a knob (tracking two sets for xycoords)?
Is there a library/api anyone is familiar with that contains a good library of gestures for multitouch applications?
Answer: The math with two fingers (assuming that some API always gives you the centers of both fingers) is rather simple, but you get more than just turning. You get scaling and translating motion as well. You star with P1(0) (x1(0), Y1(0)) and P2(0) (x2(0), Y2(0)). Then compute Pc = (P1 + P2) / 2. After that adjust all of the subsequent coordinates such that Pc (center) is at (0, 0) by subtracting Pc from every point. After that you should be able to calculate how each of the fingers rotated around the center - how much it scaled and how much it turned with each step. Note that the user may have slided both fingers as well. Think watching a moving hurricane on a radar screen: it turns, it moves, it disperses and gets wider, or it gets smaller. If that sliding motion is significant, then you might want to compute the center dynamically.
Now, with N measurements of pairs of points P1 and P2, you should have data on how the center slided, how it turned, and how the radius changed. You are looking for little sliding, little radius expansion, and a linear increase/decrease of the angle with time. I would filter out big noise and then try to see how you can fit a straight line in your angle graph, and how well a line approximates it. Well ... the user can change the speed of rotation. Something that no generic library can do for you is decide on the rules for when to ignore a gesture. You need to provide more info on what the turning of the knob is like. I hope there are only 2 fingers involved. Are you trying to measure this turning or merely detect the motion?
EDIT: Thinking of the hurricane a bit more, I think it makes sense to compute the new center with every step. However, if one finger does not move, but the other does trace a circle, then it is a different type of motion than turning the knob. If you still model it as a knob, it will be interpreted as displacement with slow rotation as opposed to faster rotation around a static point.
Again, the math is not necessarily complicated. It is just a 2d transformation matrix + offset vector. But, when more than one finger is involved, the question becomes: what it the 'center' about which the transformation is being performed? It could be either 4 corners of the screen, the center of finger 1, the center of finger 2, the half-way point between two fingers, or other. How you interpret the motion will depend on the choice of the center. It might not matter too much if the users can easily learn and adjust their motion to get it working right. But you still want to take the best guess at how exactly they are planning on using the software, and what other gestures are there.
Answer: This is easy to over complicate.
You need to capture and indicate 4 simple ideas

they've grasped the knob
they started the grasp at angle a
they've now got the grasp at angle b
they've released the knob

Show 1 by highlighting the knob
Show 2 and 3 by rotating the knob a-b degrees.
Show 4 by removing the highlight
(1) should only happen when a line drawn from finger to finger intersects the knob. Once this happens...
(2) angle a is set
(3) b is continually reset as they move their fingers. This continues until they either remove them or the line no longer intersects the knob.
(4) leaves the knob where it's been rotated.
This is different then manipulating a picture with two fingers only in that no stretching or translation is allowed. Only rotation.


Is possible to increase or reduce the brightness of the clip when i do the 3D Tracking?

Question: I'm using blender to track a lot of video clip end insert 3D objects on the scenes. Blender does a very good job but sometimes i have difficult to do this because some parts of the videos are too much dark or to mach clear. There is a way to increase or reduce the brightness inside of the window "movie clip editor"? (after effects style)
Answer: one way you could achieve this is by changing the multiply value in "Color" on your strip:

or go to modifiers (click on tab on the right side):

add modifier bright/contrast:

then change bright value:


openvz performance issues

Question: I have two VEs running on the same physical server.  One VE sends requests and the other reads/processes them.   I'm trying to benchmark the setup, but I keep blowing through the UBC limits. Things grind to a halt when this happens. Is there a way I can completely disable UBC so it doesn't keep hitting limits?  Or is the only way to do something like this to set the UBC params to extraordinarily high values?
Comment: Can you please post the contents of /prov/user_beancounters for the container that you are benchmarking in ?
Answer: OpenVZ can be a bit difficult to set limits for. Documentation is not very clear and the configuration is quite arcane if you are a first timer. Instead of using the example configurations, I would encourage you to use vzsplit to help set up your limits. That solved a lot of my problems when hitting limits. The generic configuration does not fit all sizes.
Comment: I am a first timer, and had the same issues as the poster, then as you said I-ve found about vzsplit, and really made my day, although I had a minor problem with CPUUNITS, which finally turn out to be a bug: http://bugzilla.openvz.org/show_bug.cgi?id=1627
Answer: If you're hitting the limits, either you've set the limits too low or you're doing something extraordinarily wrong.  I'd raise the limits to appropriate levels for your hardware, and then if that doesn't help, investigate what the code is doing wrong.

openvz performance issues


Home workers requiring simultaneous office VPN and local network access

Question: I now have a large number of remote users that are using a Check Point Secure Client VPN connection into the office. These users also need access to network resources attached to their local router/hub (network storage and network printer). Is there any way that the client (Windows 7 or XP) can be configured to bypass the VPN for local addresses (192.*)
Answer: Same, not sure about checkpoint, but in Cisco vpn client settings there is a "allow local lan access"
Answer: i don't know the check point vpn, but in f-secure it was possible to configure the vpn profile to only route some defined networks/subnets over the vpn.
Answer: It seems like at this time, the answer is no.  The way this product works, when you connect to VPN, your IP address changes and the PC is no longer on the home LAN network.   Other answerers have noted that other products have configurations that work around this, but Check Point Secure Client VPN doesn't have one.


Server side validation in java

Question: Can any one explain me the need and scenarios of server-side validation in java. why can't we use the validation at the client-side using javascript?
Comment: Have a look at another SO question : http://stackoverflow.com/questions/162159/javascript-client-side-vs-server-side-validation
Answer: Because an attacker can bypass client-side validation by simply disabling JavaScript or calling your server with external tools like httpclient or curl. Finally with tools like firebug one can submit virtually anything.
Also it makes your GUI much more responsive as you don't have to reload the page every time a user tries to submit the form (so moreover, it reduces the network traffic and server load).
Answer: Client side validation is good for a better user experience, if he made an error, there's no need to send an HTTP request to find that out.
For example - if the user name should be at least 3 characters long, and the user enters 2, you can immediately tell to the user that there's a problem.
Server side validation protect the server from corrupted / malicious requests, that might be sent by dishonest users.
For example - if the user tries to execute a query that he shouldn't using SQL injection, you can prevent it in the server side, since he may not send the query through you web site, but from a different client.
Answer: Mainly for security reasons. If somehow, someone manages to modify your client and bypass validation, you could seriously compromise your server.

Server side validation in java


Is "obey pam restrictions" still supposed to work in Samba 4 (Ubuntu 20.04)?

Question: This is a re-run of this question I asked two weeks ago, which was closed because it was related to an EOL Ubuntu version.
I did another install, this time on Ubuntu Server 20.04LTS, with exactly the same result. I also tried the same configuration on latest Debian (Buster 10.7), same result.
So, here we go :
The up-to-date Samba doc says (https://www.samba.org/samba/docs/current/man-html/smb.conf.5.html) :

When Samba 3.0 is configured to enable PAM support (i.e. --with-pam),
this parameter will control whether or not Samba should obey PAM's
account and session management directives.

Is this still supposed to work with Samba 4 ?
I had some strange result, it seems PAM's restrictions are enforced once, but then not anymore.
I tried to set up a file-size limitation on a Samba share. I'm not talking about quotas, I'm talking about preventing users from storing files that are bigger than 100MB, for example. I used /etc/security/limits.conf for this.
It almost works. Well, it works the first time a user tries to create a file, and then not anymore.
Here's what I did :

First I defined a hard filesize limit of 100MB for user johndoe in /etc/security/limits.conf :
johndoe    hard    fsize    102400

Then I added session required pam_limits.so to /etc/pam.d/samba, in order to tell PAM to enforce the limitations

And finally, I added obey pam restrictions = yes to /etc/samba/smb.conf

At first it seemed promising, when user johndoe tries to copy a file > 100MB, a Windows 10 client throws the following error :
An unexpected error is keeping you from copying the file...An unexpected network error occured
So far, so good ! That's what I wanted, prevent the user to store a file > 100MB
But if I click on "Try again", the file is copied anyway.
And if I then try to copy more files > 100MB, no more error message is thrown, and the copies proceed.
If user johndoe logs out and back in, same result : the first attempt at copying a file > 100MB throws an error, the following attempts succeed.
So, it seems the restriction I set in /etc/security/limits.conf is only enforced at the first attempt, and is no more enforced afterwards.
Any idea why ? Or any idea how I could achieve my goal (prevent a user to copy a file > 100MB) ?
Answer: OK I got a straight answer from the Samba team
Basically, to sum things up, setting up a limit for individual file size can not be done with Samba
The full discussion can be found in the february mailing list archive (https://lists.samba.org/archive/samba/2021-February.txt.gz)
Comment: Thanks for sharing the answer!
Comment: But does this actually answer the question: `Is "obey pam restrictions" still supposed to work in Samba 4 (Ubuntu 20.04)?` If it does still work in Samba 4 then why does the documentation for `obey pam restrictions` still only mention Samba 3?

Is “obey pam restrictions” still supposed to work in Samba 4 (Ubuntu 20.04)?


FLStudio: Select notes in all channels

Question: Can I select notes in a few channels and move this in FLStudio?
Comment: Hi and welcome to the site. I've edited out your second sentence since we don't do software recommendationson this site.
Answer: Yes, you can! shift+click on all the channel selectors that you want to copy (they will light up with a green border when selected). You can also double-click on one of the channel select buttons and it will select (highlight) all of them.
To copy the midi in each channel selected press ctr+c
To paste the midi into a different pattern, press ctr+v.

FLStudio: Select notes in all channels


$\sum_{k=1}^n \binom{n}{a_1,a_2, \cdots , a_k} \binom mk \binom{k}{b_1,b_2, \cdots , b_l}= m^n,$

Question: (Own) Let $n,m$ be positive integers such that $m>n$. Prove that $$\sum_{k=1}^n \sum_{a_1+a_2 + \cdots +a_k=n} \binom{n}{a_1,a_2, \cdots , a_k} \binom mk \binom{k}{b_1,b_2, \cdots , b_l}= m^n,$$ where $1 \le a_i \; (1 \le i \le k)$ and $a_1+a_2+ \cdots + a_k=n$.
In here, $b_i$ is the number of $a_x \; (1 \le x \le k)$ that have the same value. For example, if $n=6$ and $6=3+1+1+1$ then $b_i$ are $1$ (one number $3$) and $3$ (three numbers $1$).
Comment: just for the fun of it: $6=3+1+1+1$ :-)
Comment: So sorry, I have edited that.
Comment: For every $k$, the sum is further taken over all possible partitions of $n$ into $k$ integers $a_1, \ldots, a_k$?
Comment: Yes, yes it is.
Answer: We can prove this equality by reducing it to a counting problem, namely: In how many ways can we put $n$ distinct objects in $m$ distinct boxes?
This number is obviously equal to $m^n$.  Now, we can find the left-hand side of the equality by using the following counting strategy:

Note first that since $n<m$, we will be using at most $n$ of the $m$ boxes.
For each $k\in\{1,\ldots, n\}$, let us count the number of ways to arrange the $n$ objects using exactly $k$ of the boxes. We will then sum over all possible $k$.
Fix such a $k$ now.  There are exactly $\binom{m}{k}$ ways to choose $k$ boxes among the $m$.
For every choice of $k$ boxes, we must now count the number of ways to arrange the $n$ objects inside these boxes.  We first choose how many objects there will be in each of the $k$ boxes: say $c_1$ in box $1$, ..., $c_k$ in box $k$.  Thus we have $c_1+\ldots+c_k = n$, with each $c_i>1$.  The number of ways to arrange the objects in these $k$ boxes is then $\binom{n}{c_1,\ldots, c_k}$.  We take the sum of these numbers over all possible ordered partitions $c_1, \ldots, c_k$ of $n$.
Finally, we rewrite this last sum as follows.  Instead of taking the sum over all ordered partitions, we will take it over all non-ordered partitions $a_1, \ldots, a_k$ of $n$, and multiply by the number of ways to order it. This number, using the notations of the question, is exactly $\binom{k}{b_1, \ldots, b_\ell}$.

Putting all of this together, we get that the number of ways to put $n$ distinct objects into $m$ distinct boxes is
$$ \sum_{i=1}^n \sum_{a_1+\ldots + a_k=n} \binom{m}{k}\binom{n}{a_1, \ldots, a_k}\binom{k}{b_1, \ldots, b_\ell}, $$
where the second sum is taken over all non-ordered partitions of $n$ into exactly $k$ positive integers $a_1, \ldots, a_k$.  This finishes the proof.
Answer: Suppose you need to produce n numbers, c_1,..., c_n, each between 1 and m.
You can do it in this way -

Select k, number of distinct integers in the list
Choose k distinct numbers between 1 and m
Select number of times a_i those numbers will appear
Choose a matching of the counts a_i to the k numbers, by observing that we need to match for distinct counts - choose b_i numbers out of the k for each distinct count
Choose positions for all these k numbers

This is the lhs.
Also the number of ways is m^n, which is the rhs.
Comment: This is exactly my solution, replacing "putting $n$ objects in $m$ boxes" by "choosing $n$ integers between $1$ and $m$.
Comment: Yes... I saw your solution later :)
Comment: Simultaneous solutioning ! :-)


Finding Marginalize the product of p(x|z,μ) and p(z|π)

Question: Consider now n i.i.d. observations of the vector data $({x_1,...,x_n}).$ Using the pdf, we can write the log-likelihood expression:
$$l(\boldsymbol x)=\sum_{i=1}^nln(\sum_{k=1}^K\pi_kp(x_i|\mu_k))$$
In order to estimate the unknown parameters $({π_k,μ_k,Σ_k})$ we must trace each data sample $x_j$ as to which of the k mixtures it belongs to. If an oracle would tell us, our task would be very easy; but we do not know it. This can be handled by using a new K-dimensional binary random vector   $z∈(0,1)$ such that only one component at any one time is 1, all the rest being zero, and so $\sum_{i=1}^Kz_k=1$. In fact, also $p(z_k=1)=\pi_k.$ Notice that z is not observable, hence a hidden
variable.
Thus for example,
$$p(x|z_k)=\prod_{j=1}^d\mu_{kj}^{xj}(1-\mu_{kj})^{1-xj}$$
$$p(x|z,\mu)=\prod_{i=1}^Kp(x|\mu_i)^{zi}\phantom{2}and\phantom{2}p(z|\pi)=\prod_{j=1}^K\pi_j^{zj}$$
How can we marginalize the product of p(x|z,μ) and p(z|π) (i.e., integrate out z) and obtain the below result:
$$p(x|\mu,\pi)=\sum_{k=1}^K\pi_kp(x|\mu_k)$$
Answer: Here's an explanation from a course I recently took an edX.org. It explains "Marginalizing over unobserved coordinates" for mixture models.
https://courses.edx.org/courses/course-v1:MITx+6.86x+1T2020/courseware/unit_4/P4_netflix/?child=first
It is on the 6th tab. If you can't get it, I just put an image here:


How to download all images from a forum thread with wget

Question: For example in a thread that is 30 pages long and the url ends in page=XX is there a way to make a loop that would increment the page each time and download the images only
Answer: You can use this sitax:
wget -r -A jpeg,jpg,bmp,gif,png http://www.domain.com.
If you want to download all the pics in one folder use this:
wget -r -P /save/location -A jpeg,jpg,bmp,gif,png http://www.domain.co
-r enables recursive retrieval.
-P sets the directory prefix where all files and directories are saved to.
-A sets a whitelist for retrieving only certain file types. Strings and patterns are accepted, and both can be used in a comma separated list (as seen above). See Types of Files for more information
Comment: On this thread for example http://67-72chevytrucks.com/vboard/showthread.php?t=513377&page=10 thats not going to work is it?
Comment: Yes it will work. Try.


Lock calendar view to a certain part of the day

Question: Is there a way to get Google Calendar to show only the relevant part of the day? I don't want to scroll up and down. Is there a way to lock the view to, for example, 7am to 9pm?
Answer: Unfortunately, the "Hide morning and night" Lab is no longer around in the "new" Google Calendar.

How often do you have something scheduled at 3am? What about 10pm? If the answer is almost never, you might want to try out the Hide morning and night lab in Google Calendar.
With a simple drag of a slider you can fold all those empty hours into a single row to set the time range you want to hide. The folded rows still show all your events, just in more compact form.

However, Raphael Schaad points out that the Hide morning in calendar Chrome extension restores most of this functionality.
Answer: AFAIK no, there isn't a way to lock the view on a set time interval. Have you tried viewing in agenda mode? Typically that will condense your schedule into a tighter frame than day view will allow.
Answer: On Chrome (and on other browsers, I assume) the entire day's schedule (ie, any all-day events plus 0:00-23:59) appears if I shrink the web page enough, using
COMMAND and "-" on the Mac or CTRL and "-" on the PC. I can now see anything at a glance (especially to see if the day is free and to not miss any entries at odd hours).  Note that, at this point, the font is tiny, so I might have to select an entry to read the details.
   Not ideal, but it works when I need it.
Answer: The only solution that works is to zoom out to around 80%. This will let you see everything but in small fonts.
Answer: Switch to 24 hour time layout.  Reload.  Switch back.  Reload.
Answer: In the app, at the top, go to
Calendar
Preferences tab
Show: (I selected 24 hours at a time to see the whole day)


Optimal Combination Forecasts

Question: Can someone give me an intuitive explanation of how Optimal Combination Forecasts work for hierarchical time series? I've been reading Rob Hyndman's paper and understand that the resolved forecasts can be written as $\widetilde{Y}_h =  SP\hat{Y}_n(h)$. This is pretty simple and intuitive when considering a top-down method. But how exactly does optimal combination work with and without weights? I don't see how $P = (S'\Sigma_h^{\dagger}S)^{-1}S'\Sigma_h^{\dagger}$ resolves the forecasts so they sum to the proper amount at higher levels.
Answer: Careful there. $P$ is not the summation matrix. $P$ is only involved in estimating the new bottom-level forecasts $\tilde{Y}_n$ based on the original bottom level forecasts $\hat{Y}_n$:
$$ \tilde{Y}_n := P\hat{Y}_n.$$
After that, the optimally combined hierarchical forecasts $\tilde{Y}_h$ are calculated by applying the original summation matrix $S$ to $\tilde{Y}_n$:
$$ \tilde{Y}_h := S\tilde{Y}_n.$$
(And these are consistent, because $S$ simply encodes the consistency conditions.)
The somewhat confusing matrix $P$ is used instead of the more common OLS solution matrix to account for heteroskedasticity, and because the estimated covariance matrix may be near-singular because of the aggregation.
I'd recommend reading the original manuscript - a slightly different exposition to the published paper may be helpful. Or the original presentation Rob Hyndman gave at ISF 2016 in Santander. (BTW, the ISF will return to Santander this year - very much recommended.)
Comment: So this solution to P is using GLS, right? And it also states in the paper that the weights can be calculated directly from the hierarchy. Why exactly is this true? Also, could you calculate the weights as 1/var(residuals)?
Comment: It seems to me like you could calculate the weights (i.e., the estimation matrix $P$) directly from the hierarchy if you didn't care about heteroskedasticity and near-singularity, because then $P$ only depends on $S$, which in turn only depends on the topology of the hierarchy. In which case you wouldn't be using GLS any more. And yes, the precision (the reciprocal of the variance) makes sense as weights (along with $S$).

Optimal Combination Forecasts


Design of transverse reinforcement in a wide beam

Question: Besides satisfying nominal bar provision in the code, is there any special requirement in the provision of transverse reinforcement to ensure that the longitudinal reinforcement is fully effective in a wide beam design? That is, can I use the full reinforcement area (all bottom bars) to calculate the bending moment capacity of a wide RC beam, provided that all detailing and curtailment requirement is satisfied? Do I have to calculate the transverse reinforcement necessary to confine the longitudinal bars and make them fully effective? Or do I have to consider a reduction using some spread angle from edge of support? For example, in the design of a pile cap with wide pile spacing, using Eurocode or British standard.
Answer: I can't answer with reference to Eurocode or British Standards, but there is some guidance in official comments to Spanish norm EHE-08. This answer follows  comments to article 42.3.1 in page 194, where the following image is taken from.

Rules are for compressed bars and aimed to prevent buckling and therefore they should only apply to the compressed face wide beams. However, bar distribution in wide beams is usually quite symmetrical and then all longitudinal bars get the same confinement.
In short, each compressed longitudinal bar needs to be confined by transversal reinforcement, but if distance between longitudinal bars is less than 15 cm only one on two bars needs to be confined.
Comment: Looking at the rest of that item, however, this seems specific to bars under compression and is mostly relevant for columns, not wide beams.
Comment: I edited my answer. You are right that those rules are for bars under compression, but in a wide beams half of the bars are usually under compression. Furthermore in continuous wide bars, what bars are under compression changes in different sections.
Answer: This is a generalized answer to a question in general.
By code, the requirement of stirrups or links is based on shear demand on the cross-sectional area of the wide beam (one-way beam shear), or on the surrounding cross-sectional areas (two-way punching shear) as in flat slabs.
In practice, depending on floor geometry and loading pattern, special concerns may arise over the potential of "stress concentration" over a narrow bend in the wide beam/floor element. Although the pile cap is not a good example (usually wrapped in heavy reinforcement in all directions), it does help to visualize uneven stress distribution on the cross-section. Similar to the T beam, the stress is highest near the pile (as support) and decreases to zero at the free edges. While overall shear capacity is adequate, but the high stress is concentrated in the center that can cause cracks locally; weaken the performance and durability of the continuous pile cap that behaves like a wide beam. So very often the shear reinforcement is utilized to compensate for the high stress in the concrete near the support.
The concept above has carried into the design of the flat slabs. In practice, you might encounter the so-called "embed beam", which is a narrow band of longitudinal reinforcement enclosed in the transverse reinforcement to form a rigid strip that acting similar to a physical beam (without stem). The code takes care of the phenomenon of stress concentration in two way flat slabs (considered as a very wide beam) by specifying the column strip and the interior strip, however, other than one or two-way shear, the transverse reinforcement (for the effects of shear or torsion) is left to the judgment and care of the individual designer/engineer based on the effects of special structural layout, load pattern has had on the durability and appearance of the slab (wide beam).
The stress variation in a wide beam is best visualized through the stress diagrams from a FEM or plate analysis. Often the transverse reinforcement is not mandated by code, but personal judgment and preference, also his/her experiences. Hope this write-up helps.


The database principal owns a schema in the database, and cannot be dropped message

Question: I am trying to delete a principal from the database but can't because it owns a schema. When I go to edit the user, however, the box to uncheck schemae is blue and unremovable.  How can I remove the principal from these schemas?
Answer: Try the T-SQL to do this:
alter authorization
on schema::YourSchemaName
to dbo
go

drop user TheUserYouWantToDelete
go

You can't drop a principal that is a schema owner, so the ALTER AUTHORZATION changes the owned schema (I used YourSchemaName, but obviously substitute that with the owned schema in your database) to dbo (likewise, you can change ownership to whatever principal you need in your environment).  That will allow you to drop the previously-schema-owning user (for example purposes I used TheUserYouWantToDelete, but that'll be the now non-owner that you want to drop).
Comment: @rsteckly Simplified, authorization = ownership. If you could delete the user that owns the schema, the schema wouldn't be valid, because the owner no longer exists. Imagine you could delete a customer who has placed an order. What does the CustomerID in the Orders table mean if there is no longer a Customer that it points to? Where should I ship the order?
Comment: Okay, so this worked.  I'm a little confused as to why, however.   Why do I need to authorize the schema to another schema?  Maybe I need to review schemas...
Comment: Sorry, my example wasn't well explained.  We're not changing authorization to the `dbo` schema, we're changing authorization to the `dbo` principal.  That could have been any database principal.
Answer: The T-SQL way works.  Thanks to
http://zarez.net/?p=179
I found the SSMS way to do this (UI).
To change the schema owner from Sql Server Management Studio:
Expand your database -> Security -> Schemas

In the Object Explorer Details you can see a list of the schemas and the owners:

If you don't know what schema(s) the User owns, check the properties of the User.

Open up the properties of the schema that the User owns, and click "Search" to find a new owner.  If you don't know the new owner, you can "Browse" for one.
Properites -> Search -> Browse

and you can change the schema owner to dbo (or whoever is most appropriate).
Comment: T-SQL way is much better and flexible. Doing it through GUI has (is) always a pain !
Comment: Note: F7 is the shortcut to display details if it isn't coming up
Comment: I'm all for the T-SQL way, also because it can be used in scripts.
Answer: It has been already showed before that before dropping a user you need to reassign any schema it might own to dbo as per the example below:
alter authorization
on schema::YourSchemaName_Ownedby_TheUserYouWantToDelete
to dbo
go

drop user TheUserYouWantToDelete
go

However, sometimes the user you want to delete has more than one schema assigned to it.
You can find all the schemas owned by a particular user by using this script:
  SELECT S.*
       ,[the schema ownwer]=dp.name
       ,dp.type_desc
FROM SYS.schemas S
INNER JOIN SYS.database_principals dp
       ON S.principal_id = dp.principal_id

Just a note:
In the example above the user mathura\radhe has several schemas including db_owner although this user is NOT the db owner of this database in particular.
Below is a partial view of the script I use to get ready to drop a user in a database.
There are other things to consider - as they're not included in this script:

getting all the permissions of this user, before you drop it - just in case you need to rollback - for that I recommend this or this
I have not checked about the roles assigned to this user - just because I never had this problem so far
I am not scripting the login and user creation - you might need this before you drop it
Always check that you are in the right server and right database before running this script
  SELECT   THE_SERVER      = @@SERVERNAME
      ,THE_DATABASE    = QUOTENAME(DB_NAME())
      ,THE_USER        = QUOTENAME(dp.name)
      ,IS_ORPHAN       = dp.IS_ORPHAN
      ,[DROP_USER]     =
       'USE ' + QUOTENAME(DB_NAME()) +
       '; DROP USER [' + dp.name
        + '] '
      ,[RESET_SCHEMA]  =
      'USE ' + QUOTENAME(DB_NAME()) +
      ';' + dp.THE_SCHEMAS  +
      ';'
      ,[RESET_ROLE]    =
      'USE ' + QUOTENAME(DB_NAME()) +
      '; ALTER AUTHORIZATION ON Role::[' + dp.name
      + ']  TO [dbo]'

FROM
  (SELECT
  [NAME]               = dp.name
 ,IS_ORPHAN            = CASE WHEN (sp.sid is null
                         AND c.sid is null
                         AND a.sid is null)
                         THEN 1 ELSE 0 END
 ,THE_SCHEMAS          = STUFF((
             SELECT N';'+ N'ALTER AUTHORIZATION ON SCHEMA::' +
             CAST (QUOTENAME(s.name) as sysname) + SPACE(1) + 'TO [dbo]'
                                    FROM sys.schemas s
                                    WHERE s.principal_id = dp.principal_id
                                    FOR XML PATH ('')),1,1,'')

FROM sys.database_principals dp
 LEFT OUTER JOIN sys.server_principals sp
              ON dp.sid = sp.sid
 LEFT OUTER JOIN sys.certificates c
              ON dp.sid = c.sid
 LEFT OUTER JOIN sys.asymmetric_keys a
              ON dp.sid = a.sid

WHERE 1=1
 AND dp.type in ('U', 'S', 'C', 'K')
 AND dp.principal_id > 4
        /*0..4 are system users which will be ignored*/
 AND NOT (dp.type = 'S'
         AND LEN(dp.sid) = 28)
        /*to filter out the valid db users without login*/) dp
 WHERE dp.name = 'mathura\radhe'

then I copy the script generated on the RESET_SCHEMA column:
USE [sgdragnet];
ALTER AUTHORIZATION ON SCHEMA::[mathura\radhe] TO [dbo];
ALTER AUTHORIZATION ON SCHEMA::[radhe] TO [dbo];
ALTER AUTHORIZATION ON SCHEMA::[mathura] TO [dbo];
ALTER AUTHORIZATION ON SCHEMA::[KRISHNA] TO [dbo];
ALTER AUTHORIZATION ON SCHEMA::[db_owner] TO [dbo];
ALTER AUTHORIZATION ON SCHEMA::[db_securityadmin] TO [dbo];

and the other columns also contain significant info.
hope this helps


Is it safe to auto-fill credit card numbers using Chrome?

Question: Is it safe to auto fill credit card numbers using Chrome? Does it safely store the credit card information? As far as my understanding goes, it just shows asterisk values but on click it reveals the credit card numbers:

My questions are a few :

Is it possible for to breach Google Chrome and take my credit card information?
As per my understanding the credit card number is not stored with any type of encryption, so is it really secure to store in autofill data?

How does Chrome handles this type of data? I agree it's good in terms of usability to store and fill the credit card details, but I doubt its not good in terms of security.
Comment: You can enable client side encryption in Chrome by setting a password in your sync options. That will secure the password on transit and on the server. Still vulnerable on the client.
Comment: I would be concerned about what conditions trigger auto completion because if a website has an invisible "cc" and "cvs" fields, it might get auto completed and submitted to another website using JavaScript (XSS) without the user knowing.
Answer:
Is it possible for breaching google chrome and take my credit card
  information?

Yes.
As long as Chrome can use your number for auto completion, it has to be possible for Chrome to access it. If one program on your computer can do this, another program or a least humans can do it too.

it's not stored with any type of encryption

Even with encryption, the statement above holds. Chrome would need the key, and this key has to be somewhere on your computer so that Chrome can use it.
As long as someone can physically access your computer, few things actually help. Encrypting your whole hard drive and taking the key away with you is one possibility. Downside 1: It´s a pain to insert flash drive and password each time to turn it on. Downside 2: If someone gets your computer while it is turned on, everything is futile again.
If you only want to protect against attacks form the internet, this is much better, but nonetheless there is no 100% protection. Not entering your card number (or any sensitive data) in the computer is the only reliable way.
Comment: They could encrypt the information and require a password to access it instead of storing the key locally. Typing a password is still easier and faster than reading the credit card number, expiration date etc.
Comment: Do you have any piece of technical evidence to support what you are saying, or is this answer pure assumptions? I can see ways of implementing a safe CC storage off-computer, having the browser request the info from the server in an encrypted way. Not sure if it works this way, though, but I wouldn't speculate either way.
Comment: @Seb It's not speculation, but just common sense plus some knowledge about computers. ... The feature basically is to write the (plain, readable) credit card data number into a HTML form when the user wants it. As long as this isn't changed, how could you ever hide the number from Chrome (and with it, other client-side access)? ... Sure, telling server A to ask server B for the number could be done, but that's something different (and it isn't supported by standards, browsers, servers, etc., and would imply huge privacy problems etc.)
Comment: @deviantfan As I suggested, data can be stored in Google's servers, requested when needed.That doesn't mean it's saved locally. I think you misinterpreted my suggestion. I find it troubling that the accepted answer has no evidence whatsoever to support its statements, just "some knowledge about computers" and assumptions.
Comment: @Seb And I find it troubling that you didn't understood my last comment. I'm talking about local autocompletion, that Chrome is using (period). If another solution is better wasn't the problem. For evidence, read Chromes code. You could also send me a computer and I get the numbers out of it, would that be proof enough? ... Btw., no, storing your credit card data on Googles servers is neither better nor allowed. If you want evidence, read about PCIDSS
Comment: Me not understanding your answer is not troubling. You said your answer was based on "common sense plus some knowledge about computers". All I'm asking for is proof of this answer, as it's not evident to me and I can't find answers from Google either. Don't get mad at me for wanting to be sure, and not trust on someone else's "common sense plus some knowledge about computers". Sigh...
Comment: @Seb I'm not mad. ... What sort of proof do you expect, if the Chrome source isn't enough (assuming we're talking about the same thing now)? All other parts were already done by people, so they are possible (see Google, or ask me for links).... (And are you aware, that when requiring hard evidence, pretty much any security thing on the world fails? Eg. there is no proof that AES is secure, there's no proof that RSA is secure (because there is no proof that P!=NP), there's no proof any long keylength is nough for something because gueesing right on first try is possible, etc.etc.)
Comment: code != "common sense plus some knowledge about computers". Nuff said.
Comment: @Seb If you just read the comment parts that you want ... well. Bye, I'm done with this discussion
Answer: 1) Yes, it is possible to breach Google Chrome. Even with the encryption, there would be somewhere password or private key.
2) It is not secure as not using autofill, however, you do not type in the autofill your CVV code, which is needed for a successful operation with the card. You can find more about CVV here: https://www.cvvnumber.com/
Comment: CVV is not always needed (try Amazon, for example).
Answer: I started having my credit card info autofill in Chrome on my Android tablet. I could go into Chrome settings and turn off this feature, but if anyone got my tablet they would have access to the card info simply by turning autofill on again. The only way to get rid of this info showing up on my tablet was to go into my Google account and delete my card info. What a pain!
Comment: Yes, as long as your CC is linked to Google Wallet it will be accessible to autofill via that on/off option in Chrome's settings.


Most likely interpretation of parents' "anniversary"?

Question: When someone says it's their parents' anniversary, does "anniversary" by default mean wedding anniversary, or do English speakers by default consider "anniversary" vague, requiring clarification like marriage, first-date, etc?
Comment: Unless *anniversary* is followed by *of [something]* or its meaning is clear from context, it is assumed to refer to a wedding anniversary.
Comment: @Jason Bassford Not all parents these days are married. Indeed some remain unmarried for ideological reasons. However, some who are unmarried but happily remain together bringing up their children, will celebrate a particular day in the year as "their day". This could be tied perhaps to their first date, or their having begun living together. Schoolteachers and others who have contact with children should never make assumptions about family circumstances, and if a child speaks about their parents' "anniversary" not ask further questions.
Answer: Unqualified, "my parents' anniversary" would most likely mean either a wedding anniversary, or the anniversary of some significant event that the parents treat like a wedding anniversary.
It may be risky to deduce that the parents are legally married (it may just be their special day), but it is reasonable to think the date may be marked by, for example, exchanging gifts or going out for a meal together.


How do gravity and normal force balance on an inclined plane?

Question: Consider a block placed on an inclined plane. Drawing a free body diagram and applying Newtons laws, I get $N\cos A=mg$ as well as $N=mg\cos A$. Here one equation contradicts the other, could anyone explain this?
Comment: NcosA ≠ mg, else there would be no component of force in vertical direction, resulting in the body remaining stationary always. That's not the case here as you don't have friction
Answer: Well, the problems that you are having here is that you are assuming that the forces will balance.  Its true that $N = mg\cos (A)$, since the block cannot move perpendicular (normal) to the plane.  However, the block can move parallel to the plane, meaning that there can be a net force in the horizontal and vertical directions (if there is no friction), and the block will slide down the plane.  Therefore, the two forces, $N\cos(A)$ and $mg$ are in general not equal.
Think about the following:
We know that $N = mg\cos(A)$ since the block cannot move perpendicular to the wedge.  Therefore, we can write $N\cos(A) = mg\cos^2(A)$.  We can see that these two expressions aren't equal unless the wedge is flat $A=0$.  In this case, we know that the block will not have a net acceleration, since the wedge has no angle.  For $A>0$, this is not true, and we do expect the block to accelerate down the wedge due to the force of gravity.
Answer: Well you must first see if there is any component of acceleration prevailing in a chosen direction. If there is no acceleration in the chosen direction then you can balance the forces in that direction.
But here :

There is a component of acceleration in direction chosen (vertical) therefore the equation becomes $$mg-N\cos A=mg-mg\cos A. \cos A=mg\sin^2 A$$
Answer: Your mistake is that your coordinate systems are a bit messy. The two components are not drawn along coordinate axes. And if they are, then that coordinate system has axes that are not perpendicular.

gravity $mg$ and the component $N\cos A$ will cancel out, and then the parallel component of the normal force is left and will cause acceleration down the incline.
the normal force $N$ and the component $mg \cos A$ will cancel each other out, and then the rest of the gravity is a component causing an acceleration down the incline.

You can consider each case separately, since you are allowed to set the coordinate system as you want. But you cannot combine them because then you have a coordinate system in which the axis directions are not perpendicular. Then each axis interferes with the other and does not uniquely describe the components along it.
Comment: I see, you have "the component" written in front of $N\cos A$.  My mistake.  You should write "perpendicular component" as I scanned over that and didn't parse it.  I'll delete my previous comment


What someone is called who supports you

Question: What do you call someone who supports others by checking in with them when they're trying to achieve a goal?  Not a sponsor.
Comment: Mentor?  Coach, counsel, pilot, shepherd, tutor?
Comment: How is 'checking in with somebody' supporting them?  If you could make that clear, you might get better answers.
Comment: I'm surprised no one has mentioned the most obvious choice yet: ***supporter***.
Comment: A *phrase* used for this is **accountability partner** - [reference](https://en.wikipedia.org/wiki/Accountability_partner).
Answer: advocate, ad·vo·cate \ˈad-və-kət, -ˌkāt\ noun –MW

3:  one that supports or promotes the interests of another

If my lab partner wasn't such a good advocate, I'd have failed physics.
Comment: Is it? Can you justify that assertion? An advocate is usually someone who speaks up for someone or something else (and supports them in that sense).
Comment: An even better example is what I've done to help this new user by showing them the ropes through editing.
Answer: Are you looking for "patron"?

One that supports, protects, or champions someone or something, such as an institution, event, or cause; a benefactor: a patron of the arts. TFD

e.g.

"Lorenzo De Medici is said to have been a patron of the arts."
"John has found a patron in Dr. Cole."
Answer: I think "coach" may fit here. Coaching has by nature a 'caring' and personal quality to it. The coach or person checking-in must be somehow invested in the other persons success. Their success is the coach's success. Being willing to be supportive in this way (with regular check-ins) requires somewhat of a parental outlook. They agree to be caring enough to be a (however) constant in that persons life. This requires commitment from both the coach and the coached, thus the inevitable personal aspect of the relationship.
The way I find "just the right word" to use is, I start with whatever word I first think of, even if I know it's the wrong one (like 'sponsor'). I use either Thesaurus or dictionary to find its synonyms. And I just go from there, checking out the words and definitions and as I get closer, the context in which they are used until I'm satisfied I have the right word to express my thought.  Good luck

What someone is called who supports you


The distributive law

Question: $4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$ , but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?
Comment: It is also the distributive law or rule.
Comment: This is distribution done twice.
Answer: Because:
$$\left (x+y \right) \left (x+y \right) = \underbrace{(x+y)+(x+y)+...(x+y)}_{x + y\text{ times}}$$
$$=\underbrace{(x+y)+(x+y)+...(x+y)}_{x\text{ times}}+\underbrace{(x+y)+(x+y)+...(x+y)}_{y\text{ times}}$$
$$=\underbrace{x}_{x\text{ times}}+\underbrace{y}_{x\text{ times}}+\underbrace{x}_{y\text{ times}}+\underbrace{y}_{y\text{ times}}$$
$$=xx+xy+yx+yy$$
Comment: If $x+y$ isn't an integer, what does "$x+y$ times" mean then? The same for $x$ and $y$ separately.
Comment: @StackTD Excellent question!  Obviously one needs to do something else in that case ...  but I figured this would be in the spirit of the OP
Comment: I'm not sure what OP means, but my comment is intended to make OP think about that too because even in his first example, $x$ and $y$ need not be integers (and the argument still holds, thanks to the 4!).
Answer:
$4\left (x+y \right)=4x+4y $ because $4\left (x+y \right) =\left (x+y \right) +\left (x+y \right) +\left (x+y \right) +\left (x+y \right)$

This works because $4$ is a (positive) integer; but for non-integer factors you can't use the "repeated addition"-argument.

but why is $\left (x+y \right) \left (x+y \right) =xx+xy+yx+yy$?

The distributive law works even if $x$ and/or $y$ are not integers.
Keep one of the factors together in a first step, and apply distributivity twice:
$$\begin{align}
(\color{blue}{x}+\color{red}{y})(x+y) & =\color{blue}{x}(x+y)+\color{red}{y}(x+y) \\ & =\color{blue}{x}x+\color{blue}{x}y+\color{red}{y}x+\color{red}{y}y
\end{align}$$
Answer: Do it in steps.
You accept that $M(x+y) = Mx + My$
So replace $M$ with $(x+y)$ and you get:
$(x+y)(x+y) = M(x+y) =$
$Mx + My = $
$(x + y)x + (x+y)y$
Now distribute a second time:  Replace $x$ with $A$ and $y$ with $B$ to get:
$(x+y)(x+y) = M(x+y) =$
$Mx + My = $
$(x + y)x + (x+y)y=$
$(x+y)A + (x+y)B=$
$xA + yA + xB + yB =$
$xx + yx + xy + yy =$
$x^2 + 2xy + y^2$.
Of you you don't have to, and you shouldn't, do all that replacement.  You should do it directly.
$(x+y)(x+y)=$ we treat one of the $(x+y)$ as a single thing and distribute across the other $x + y$.
$(x+y)(x+y) = (x+y)x + (x+y)y=$.
Now we have to sums to distribute: $(x+y)x = xx + yx$  and $(x + y)y = xy + yy$.  So puting them together:
$(x+y)(x+y) = (x+y)x + (x+y)y=xx + yx + xy + yy=$.
And then some clean up:
$=x^2 + 2xy + y^2$

The distributive law


Dealing with Other Minyanim

Question: Many times I find myself davening in one Minyan while another is taking place within earshot. What is the proper way to handle such a situation (in terms of answering Amens, Kedusha, etc.)?
Comment: related: http://judaism.stackexchange.com/q/12956/759
Comment: similar: http://judaism.stackexchange.com/q/68144
Answer: In a diffrent context dealing with the same problem.Reb Ovadiah Yosef says in places like the Western Wall or very large shul where multiple Minyanim are held simultaneously  that one should respond to Kedusha and answer "Amen" to the Berachot he hears. Even though he had already recited Arbit, there is no Halacha forbidding answering "Amen" or responding to Kedusha of Minha.
From Rabbi Mansour
Reciting Minha When the Congregation Recites Arbit; Participating in the Congregation's Minha After One Has Recited Arbit
Answer: Cross-posting from another Yodeyan's answer to a related question:

I once asked a Rov regarding the Halacha of what to do in 770. There is one official Shacharis/Mincha/Maariv minyan ("the main minyan") and there are also side minyanim taking place under the Woman's section throughout the day. Between the "side" minyanim and the "main" Beis Medrash are mechitzos to avoid this issue
However, there are times when people (inappropriately) make their minyan in the "main" Beis Medrash.
My question to him was if I have to answer Amen/Barchu/Kdusha. He said that one really should if possible.

Dealing with Other Minyanim


Interface metric vs. route metric

Question: How does windows treat interface metric (which you can set at interface ip settings) and route settings? Which one does it prefer? What if I have interface with metric 1 but for this address there is other interface route with higher metric?
Comment: Additional information to the interface metric can be found in [Microsoft's explanation of the Automatic Metric feature for Internet Protocol routes](http://support.microsoft.com/kb/299540). Unfortunately no complete answer to the question.
Answer: From Metric | Microsoft Docs:

When routes are chosen, the overall metric used to determine the preference is the sum of the route metric and the interface metric.

This question had bothered me for hours before I stumbled on this.
Comment: That's crazy! :)
Answer: From Microsoft's route article on Technet:

The route with the lowest metric is chosen.

Q: What if I have interface with metric 1 but for this address there is other interface route with higher metric?
If you have similar routes with different metrics, the lowest metric route for the destination is chosen.
In case of overlapping routes, keep in mind that route matches with longer netmasks are always better than route matches matches with shorter netmasks (so route match length is more important than lower metric).   If you ever have a question about which route windows chose for a destination, use route print <destination> from the cmd prompt
Trivial example:
[Mike_Pennington@Win7 ~]$ route print 127.0.0.1
===========================================================================
Interface List
 15...5c 26 0a 39 ac 9d ......Intel(R) 82577LM Gigabit Network Connection
 12...58 94 6b e4 c8 28 ......Intel(R) Centrino(R) Advanced-N 6200 AGN
  1...........................Software Loopback Interface 1
 24...00 00 00 00 00 00 00 e0 Microsoft ISATAP Adapter
 11...00 00 00 00 00 00 00 e0 Microsoft Teredo Tunneling Adapter
===========================================================================

IPv4 Route Table
===========================================================================
Active Routes:
Network Destination        Netmask          Gateway       Interface  Metric
        127.0.0.1  255.255.255.255         On-link         127.0.0.1    306 <---
===========================================================================
Persistent Routes:
  None

IPv6 Route Table
===========================================================================
Active Routes:
  None
Persistent Routes:
  None
[Mike_Pennington@Win7 ~]$

Comment: this used to work, but something became borked after WinXP.  Not sure if it the routing masks are miscalculated or network classification (public, work,home), DHCP, or what, but the metric is altogether ignored.

Interface metric vs. route metric


Color Ramp is not working for me 2.8

Question:
I'm trying to make a mask around the edges of the metal circles to look shiny wail having the wrest of the faces black/dark grey. So what I did was use the Input>Geometry and then use converter>ColorRamp (Color Ramp) to clamp the values together to get the edge to show up. No matter how I put it, it wont work...
 There shouldn't be wrong I'm doing I dont think, I'm following the exact instructions shown on a tutorial on the exact same thing Link: https://www.youtube.com/watch?v=lYseMQDkPAI
If someone knows what the problem is please help me
Comment: I suspect it's because you're in "Look dev" shading rather than rendered shading. I can't test it myself right now, but try switching to rendered shading by clicking the fourth icon in the top right of the 3D view, the one partially hidden under the outliner.
Comment: make sure you reset scale the object pressing Ctrl + A
Answer: Possibly because your render engine is not set to cycles. I know I'm a couple years late but maybe someone can find this helpful.
Answer: As shown in the manual: Geometry Node — Blender Manual:

Pointiness Cycles Only
An approximation of the curvature of the mesh per vertex. Lighter values indicate convex angles, darker values indicate concave angles. It allows you to do effects like dirt maps and wear-off effects.

Pointiness being a Cycles Only feature, you have to be in render shading and using Cycles render engine to see it. The material view is powered by Eevee render engine, hence you can't see it there either.

Color Ramp is not working for me 2.8


Interpreting transformed dependent and independent variables

Question: How would I interpret a transformed dependent variable (4th root) with some of its predictor variables transformed as well? In our study, we transformed our dependent variable to 4th root, $Y^\frac{1}{4}$, and one of our predictor variables to $\ln(X)$. How would I interpret that?
Answer: For the sake of comparison, in the case of a simple linear regression model with no transformations, i.e. $Y = \beta_{0}+\beta_{1}X$, our interpretation is $$\textit{a one unit increase in X results in an increase of Y by $\beta_{1}$ units}$$
Now,  if we are modeling $Y = \beta_{0} + \beta_{1}ln(X)$, our interpretation is
$$
\textit{a 1% increase in X results in an increase of Y by $\frac{\beta_{1}}{100}$ units}
$$
Unfortunately, in the case of most power transformations, there is not really an intuitive interpretation (as far as I know) of the effect, such as % change with logarithms. So, for $Y^{.25} = \beta_{0}+\beta_{1}ln(X)$, I would interpret this as
$$
\textit{a 1% increase in X results in an increase of $Y^{.25}$ by $\frac{\beta_{1}}{100}$ units}
$$
If you have more than one dependent variable in your model, these interpretations still apply, but in the context of holding the other dependent variables fixed (i.e. examining the unique impact of $X_{1}$ on $Y$).


Calculating limit of a series

Question: So I have the following series: $$\sum_{2n}^{4n} \frac 1k$$ and I need to find the limit of it when n appraoches infinity: $$\lim_{n\to\infty}\sum_{2n}^{4n} \frac 1k$$ I tried to consider this sum as the riemann sums of the function $$\frac1{x+1}$$ and thus calculating the integral of it (which seems right to me, but I don't know what are the boundaries of the integral itself, and also it's a pure guess which I don't know how to prove). Any kind of help would be great!
Comment: You're not far off; the difference is the [Euler-Mascheroni constant](https://en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant).
Comment: The sequence is easily shown to converge to a number between $\;\frac12\;$ and $\;1\;$ .
Comment: The function $\: k \mapsto \frac1k \:$ is [monotone](https://en.wikipedia.org/wiki/Monotonic_function), so you should bound those sums between two integrals. $\hspace{.89 in}$
Comment: **Hint:** $~\displaystyle\sum_{2n}^{4n}\frac1k~=~H_{4n}-H_{2n-1} ~\approx~ \ln(4n)-\ln(2n-1)~=~\ln\frac{4n}{2n-1}.~$ Now take the limit as $n\to\infty$.
Answer: Assuming that the limit exists and is finite, we can compute it using Riemann sums since the function $f(x)=\frac{1}{x}$ is Riemann-integrable on $[2,4]$.
$$\lim_{n\to\infty}\sum_{k=2n}^{4n}\frac{1}{k}=\lim_{n\to\infty}\frac{1}{n}\sum_{k=2n}^{4n}\frac{1}{\frac{k}{n}}\approx \int\limits_2^4\frac{\,\mathrm dx}{x}=\bigg[\ln|x|\bigg]_2^4=\ln4-\ln2=2\ln2-\ln2=\ln2$$
Comment: How did you choose the boundaries to be 4 and 2? No doubt it makes sense since the lower boundary of the sum is 2n and the upper is 4n, but how do I justify it mathematically?
Comment: @Calculus, By the Riemann sum argument of a definite integral, we have, for a Riemann-integrable function $f(x)$ on $[a,b]$, $nh=b-a$ such that $n\to\infty$ and $h\to 0$. And,

$$\lim_{h\to 0}h\sum_{k=0}^{n-1}f(a+kh)=\lim_{h\to 0}h\sum_{k=1}^nf(a+kh)=\int\limits_a^bf(x)\,\mathrm dx$$
Comment: Can you work it out yourself using this hint?
Answer: Using the integral test idea,
$$ \int_0^{N-1} \frac{dx}{x+1} \leqslant \sum_{k=1}^N \frac{1}{k} \leqslant \int_0^{N} \frac{dx}{x+1}. $$
The integrals evaluate to $\log{N}$ and $\log{(N+1)}$. Now you can take $N=4n$, subtract the version of this with $N=2n$, and you find
$$ \log{4n}-\log{(2n-1)} \leqslant \sum_{k=1}^N \frac{1}{k} \leqslant \log{(4n+1)}-\log{(2n)}. $$
You can then show that the limit as $n \to \infty$ on both outer sides of this inequality is $\log{2}$, so the limit of the sum must also be $\log{2}$ by the squeeze theorem.

Calculating limit of a series


What is the device used to draw 5-line staff (empty) with a pencil/pen?

Question: I remember seeing something like that long ago but can't remember what it was called. It is basically a metal gadget that holds 5 pencils or only pencil led and you can draw by hand your own empty staff.
Anybody know?
Comment: The only ones I've seen use chalk and are used exclusively on a chalk board. When I'm looking to draw a staff, I just print staff paper from the internet.
Comment: Although sound convincer really only asked what it was called… I'd call it a music-staff-drawing-gadget. I doubt it has a real name per se, unless it's a particular brand.
Comment: I agree, Reina, that's all I've seen. It would be pretty simple to make one for pencils (golf pencils would be best) out of stiff wire and soft wood for a handle.
Answer: It's called a rastrum.

The rastrum came into use in preparing copper-plate engravings for music printing. It was simply a fork that one dragged across the plate to "score" the surface. Rastrum is Latin for rake.
As for constructing such a tool samizdat (DIY), I see potential in a paperclip, 5 ballpoint refills, and a pair of chaining pliers. I'll say more as soon as I buy some ballpoint refills.
Edit:

It's easy to make with cheap materials, but it works poorly. Paperclips are too flimsy. I either need stronger wire or longer wire. Or perhaps a carved wooden frame.
Comment: I felt especially guilty as it was these votes which gained me the 'moderator tools' priviledge. I had to confess it before I got *in trouble.* :)
Comment: I don't think you should be too embarrassed at this. You answered the question (It's called a "rastrum"), and gave a proper link for reference. It's not your fault that this question called for a short answer. Your answer was actually more correct than the answer by Nix (he gave examples of products, but neither gave the general name for the thing).
Comment: It's sad that a dozen pens is cheaper than a dozen refills. :(
Answer: This isn't exactly what you asked for, but similar idea: noligraph. You could also buy a 5-line pen nib like this Brause one.
Comment: There's a video of it at http://www.youtube.com/watch?v=ZUu1bd_qdEk but it's not available in britain :(


Stolen device whatsapp deletion

Question: I want to delete my whatsapp history from my stolen device. How can I do that?
Answer: Unfortunetly, WhatsApp does not have a remote delete function that I know of. If you really want to get rid of the history, you can wipe your device remotely if you have anti-theft installed. I recommend trying Android Device Manager. It is most likely already activated on your phone. You can wipe it from there (If Android Device Manager has permissions to), or at least lock you phone and set a new device unlock code. I would always reccomend using Google's Find your Phone page, which is linked below. It will lead you through the different steps you can take to deal with a lost phone, and links to Android Device Manager.

Google's Find Your Phone Wizard

Additional Information on using Android Device Manager
Answer: Well, yes there's no remote functions for WhatsApp. But you can do this, Block the old sim and take a new sim with the same number and install and setup WhatsApp in another phone. This will prevent your old device from accessing your WhatsApp. Also you can use Android Device Manager to remotely wipe or lock the device. Hope you'll be safe from misusing of your lost phone and sim.

Stolen device whatsapp deletion


Tangle and transaction explorer

Question: I want to have an isolated tangle for testing purposes. I want to explore and browse this tangle for individual transactions, as I attach new transactions on this tangle. Is there an open source tanlge Explorer?
Comment: There is https://github.com/peterwilli/IOTA-tangle-explorer, but I never tested it myself.
Answer: I don't think there is one. You should consider that Explorers are Snapshot proof (which means that they connect to permanodes). Developing your own Explorer should not be that hard if you have some programming skills. You could use the Javascript Lib and just put a search field for addresses/hashses that could be checked by using the findTransactionObjects() method.
Hope it helps!
Comment: Since this function takes Transaction Hashes as argument, how can I browse starting from zero-knowledge about it's transactions?

Tangle and transaction explorer


Read time out on build application

Question: When I create Android project in Android Studio 3.1.2, I am getting the following errors:
read time out

and
org.gradle.api.resources.resource exception : could not get resource 'https://dl.google.com/dl/android/maven2/com/android/tools/build/builder/3.1.2/builder-3.1.2.jar'

I download this file with browser.
How can I install manually?
How can  I stop downloads for any project and download one time for all project?
Note: I am using hot spot shield and then I am creating a project.
Comment: From where you got 3.1.2 reference?
Answer: Try this

Go to build.gradle file which is at same level of your app folder. Inside you will get dependencies section just add

classpath 'com.android.tools.build:gradle:2.3.3'
as shown in below image
Answer: It is Gradle issue nothing else ,
Delete your Gradle file and build your project  again .
Answer: Try to update app level project gradle file.
buildscript {

repositories {
    google()
    jcenter()
}
dependencies {
    classpath 'com.android.tools.build:gradle:3.1.0'

}
}
allprojects {
repositories {
    google()
    jcenter()
}
}

task clean(type: Delete) {
delete rootProject.buildDir
}

Read time out on build application


Resistance of a heating element

Question: Would a heating element have a very high resistance, or a very low resistance? (All comments in this post are based around the fact that the voltage is the same for each situation) I would have thought that a higher resistance would have resulted in more heat loss, but I've been taught that the higer the current, the more energy is lost to heat. Therefore, a lower resistance would release more heat.
Comment: It would have exactly the right resistance to output the amount of energy it is designed for, when applying the designed voltage.
Comment: @GlenYates I wouldn't even joke about performing that experiment.  It's amazing what people will do after reading something on the internet.
Comment: Just to make it abundantly clear: **do not do what @GlenYates suggests in the above comment. It isn't just a bad idea, it is outright dangerous.**
Comment: You should think about it in other way. \$p=\frac{v^2}{r}\$. As the source voltage is constant, the lower \$r\$ value, the higher heat released.
Comment: To think about it in practical intuitive terms, imagine placing a very low resistance metal tool such as a wrench across the terminals of your car battery = much heat released. Now place a dry piece of wood (high resistance) across the terminals = very little heat released. Actually should perform this experiment in reverse order :)
Comment: @GlenYates - I've performed the wrench across the battery terminals "experiment" several times.  (You'd think I would have learned after the first.)
Comment: note that the resistance of various components tends to change depending on temperature (usually it decreases, but for some resistors it might actually *increase* (which helps limit temperature to a certain range, while decreasing warm-up time, e.g. in some incandescent lamps, as with tungsten filaments). some even have non-linear resistance/temperature characteristics)
Comment: In my defense, I did use the term "imagine". Though, it is entertaining to watch a youtube video of someone performing this "experiment".
Comment: So maybe a wrench would get hot, but would a piece of superconducting material?  That's where it's not quite intuitive to me...
Answer:

simulate this circuit – Schematic created using CircuitLab
Figure 1. Does adding more resistors increase or decrease the total heat produced?

I would have thought that a higher resistance would have resulted in more heat loss ...

It should be intuitive that the more parallel resistors we apply to the circuit of Figure 1 the lower the resistance becomes.
Given a constant voltage as specified in your question it should also be intuitive that the current through each branch will be the same no matter how many branches.*
We can then see that with n parallel resistors the total power dissipated will be n times the power dissipated with one resistor.

Therefore a lower the resistance value will result in more power dissipation or heat loss.
Mathematically this can be seen from the power equation \$ P = \frac {V^2}{R} \$ that, for a given voltage, power dissipated is inversely proportional to the resistance.

* A real power supply will, of course, have a limit to how much current it can produce before the voltage starts to droop.
Comment: I like the visual and practical explanation that this diagram presents.
Answer: Heat output is defined by the power \$P\$ which is itself defined by the voltage drop \$V\$ across the element and the current \$I\$ through it: \$P = V*I\$.
If you have a specific heat output you want and a input voltage you can figure out the resistance needed by plugging in Ohm's law.
\$P = V*A = \frac{V* V}{R} \$
So decreasing the resistance increases the heat output.
Answer: It depends:

if it is connected to an ideal constant voltage source: lower load resistance will cause higher load power
if connected to an ideal constant current source: higher load restance will cause more load power.

Often practical power sources can be treated like an ideal constant voltage source with a (rather low) internal series resistance.
In that case most load power is caused by a load resistance that is equal to the the internal series resistance of the power source.
This fact is called the Maximum Power Transfer Theorem.
Answer: To further confuse things, perhaps shed more heat than light, if you have a nominally constant voltage source with a fixed source resistance there will be a load resistance that has a maximum power. Note that usually that's way lower resistance than what you would use (say) on the mains.

simulate this circuit – Schematic created using CircuitLab
In the above circuit, the current is V1/(Rs+RL), so the power in the load is:
\$P_L = \frac{R_L\cdot V_1^2}{R_S+R_L}\$
You can see intuitively by inspecting the numerator and denominator that if RL is very low or is very high the power approaches zero.
In fact it is a maximum at \$R_L = R_S\$, where the load resistance equals the source resistance. Half the power is lost in the source resistance.
More generally, maximum power transfer is when source impedance equals load impedance.
Answer: A heating element has neither "very high" nor "very low" resistance.
The total energy dissipated by the circuit is proportional to current, so the resistance of the heating element has to be low enough to draw sufficient current to generate enough heat.
However, of the total energy dissipated by the circuit, the portion of the energy dissipated by each part is proportional to its resistance, so the resistance of the heating element has to be high enough so that most of the energy is dissipated by the heating element itself instead of, for example, the wiring in the walls.
If you're connecting a heating element to the wall mains, there is a circuit breaker involved that limits the current so that your wiring doesn't get too hot.  A heating element designed to deliver maximum heat (in a kettle, for instance) will draw as much current as it can while staying safely below that limit.
Answer: It depends on the power source. If that offers reasonably constant voltage, as most do, then lower resistance increases the current, which increases the power dissipation and thus the heat.
As heating usually takes a lot of power (compared to electronics) it usually needs a pretty good power supply, like a large lead acid or Li-Ion battery if it's portable - and those are reasonably good voltage sources.
So if you have some means of control - like PWM, or a thermostatic on-off switch, err slightly on the low side of resistance to get slightly more power than you need, and regulate that power to get the right temperature.
If you had a good constant current source, then increasing resistance would increase the voltage, and that would increase power. But those are pretty rare in practice.
Answer: Do you want high or low resistance?
It depends on you power source.  If you want heat you want power and power is
$$P = I \cdot V = I^2 \cdot R = \dfrac{V^2}{R}$$
So if you have a constant current source you want high resistance. However, most heaters are supplied with a constant voltage so would require a lower resistance.
If the power source is AC remember to use the RMS figure for the current or voltage as appropriate.
Answer: It depends where are your biggest problems in powering that heater.
If you have problems with resistance of the supply (eg. long or thin wires, high internal resistance) then you go for high resistance, high-voltage, low-current option.
If you have problems with insulation (eg. there is not enough room for thick insulation or the heater cannot be well insulated from potential users touching it) then you go for low resistance, low-voltage, high-current setup.
It's a balance between those two. In reality, you go for the voltage that you have at hand (eg. older trams use heaters connected directly to line voltage, be it 600V, 800V, or any other voltage the rest of the tram runs on. More modern ones utilize off-the-shelf 220V heaters, because today it's cheaper to design voltage converter than to design new heater). The pretty much only exception is when you need to safeguard against touching, then you drop the voltage down to safe level and work with that.
Answer: Don't know if this helps but I just put my multimeter on a 220-240V 1850-2200W kettle element and got ~27 ohms.
Ps electronics is not my strong point
Comment: Hi @GRA, it's a good example yet i'm not sure it answers the question

Resistance of a heating element


what is meant by stability and potential energy of molecules in cases of formation of bonds?

Question: molecules form chemical bonds in order to gain stability. it is said that forming bonds leads to lower potential energy of molecules and makes them stable. what does stability mean?
how formation of bond leads to lower potential energy? please  explain
Answer: The electronic energy is lowered when bonding occurs, therefore the energy levels become more stable. Stable just means they have a lower energy level.
So if we have two 1s orbitals they can combine to form a single sigma bond which has a lower energy than the components separately.


Why do we say "by all means" when we mean "by any means"?

Question: The common expression by all means seems to advocate the use of all means possible in order to accomplish a certain object, when in fact it expresses the use of any means to do it. I realize that all can mean any when used with a negative phrase, such as beyond all hope.  Because, of course, if one is beyond any hope, he is also beyond all hope (and vice versa).  But in a positive context, such as with by all means, it doesn't necessarily follow that when one uses any means to accomplish a task, that he uses all possible means.  Why is all used to mean any in this expression?
Answer: To my ear, any means exhorts any one particular method, whereas all means implores the use of all and sundry methods.
Answer: I think of it as being a shortened version of "by all means possible or necessary". The "possible" is implied in the phrase; it is a common and quite old phrase as well.
Here is an excerpt from an article by Ted Nesbitt - 11/10/2006, on Allexperts.com:

This phrase is very peculiar.  The original version was "by all ways"
  and the negative version was "in no ways."  [Incidentally, the old
  English spellings were wayes and wyse for "ways."] Both of these
  are now considered obsolete.  The all ways or all means is the
  positive view, and it means certainly.  The negative version, of
  course, means "certainly NOT". The Oxford English Dictionary -- the
  ultimate source for word and phrase origins -- gives the year 1593 as
  the first written usage of by all means.  It appeared in one of
  Hooker's ecclesiastical treatises.
Answer: Consider these two sentences employing the same structure:
'This machine can be operated by all people.' - thus - 'This task can be completed by all people.'
Here, we mean that any individual would be capable of completing the task.
Does the same not carry then for,
'by all means'
That said, however, it is my understanding that 'by all means' and 'by any means' have different meanings:
'by all means' - a response to a request; 'of course'
'by any means' - to indicate that you will try everything to achieve your goal (aside: here, also note that 'everything' means 'every single thing until one works', not 'all things at once') .


Looking for notation that extends the dimension of a vector

Question: I am wondering if there is any notation for the following operation of a sort of concatenation of vectors. For example if I have a vector $x=(x_1,\dots,x_n)\subseteq\mathbb{F}^n$ and $y\in\mathbb{F}$ is there any notation for extending $x$ into $\mathbb{F}^{n+1}$ with the vector $(x_1,\dots,x_n,y)$? I have seen the notation $(x,y)$ given before, but these seems counter-intuitive to me I would read it as an $2$-tuple consisting of an $n$-tuple followed by a number from $\mathbb{F}$. Is there any better notation?
Comment: I would say identification of $ F^n\times F $ and $ F^{n+1} $ is as obvious as it gets. If you want to make sure, mention it before using it.
Answer: Assuming I'm following your question,I prefer this notation $$\left(\vec{x}_n,y\right):= (x_1,x_2,\ldots,x_n,y)$$
That's a vector arrow above $x_n$, not very visible for some reason.
Answer: You could do something like $(\mathbf{x},y)$ or $(\vec{x},y)$, however you usually denote a vector. I suppose $\mathbf{x} \oplus y \in\mathbb{F}^n \oplus \mathbb F $ is okay as well, but it's clunky and really ugly/ asymmetric.
For the sake of consistency (especially if there is no reason to distinguish elements in the $n+1$-tuple), I really think $(x_1, \dots,x_n,y)$ is best for clarity, and it doesn't strike me as too cumbersome.
Answer: The standard mathematical notation for concatenation of tuples is
$$
u || v
$$
(see here). Thus
$$
(x_1,\ldots,x_n) ||y = (x_1,\ldots,x_n, y)
$$


Why Midoriya didnt get another Full Gauntlet after the one he destroyed?

Question: We've seen Midoriya after the events of My Hero Academia: 2 heroes. Another Full Gauntlet would have been very useful for him, but he never got one. Why Midoriya didnt get another Full Gauntlet after the one he destroyed?
Answer: Because the movies are NOT CANON. The gauntlets are so the movie can have deku use One-for-all at 100 percent.
Answer: Because he needed to learn how to master one for all normally without help of devices that would protect his body from the repercussions. It just wouldn't be right if he used that kind of support. And what if it breaks during battle, then he would be screwed.
Story-wise though, Melissa probably spent years developing and producing Full Gauntlet. It would take a while to reproduce a new one for Midoriya.


Series with denominator that is square of product of consecutive numbers

Question: Evaluate $\sum_{k=1}^{\infty} \frac{1}{k^2(k+1)^2}$.
I've tried to use a telescoping approach, but it doesn't seem to work. It doesn't also seem to converge to a "pretty" value as well.
Answer: Partial fractions decomposition give us
\begin{align*}
\frac1{x^2(x+1)^2}&=-\frac{2}{x}+\frac{1}{x^2}+\frac{2}{x+1}+\frac{1}{(x+1)^2}
\end{align*}
Then
\begin{align*}
\sum_{k=1}^n\frac1{k^2(k+1)^2}&=\sum_{k=1}^n\left(\frac2{k+1}-\frac2k\right)+\sum_{k=1}^n\left[\frac1{k^2}+\frac1{(k+1)^2}\right]\\[3pt]
&=\frac2{n+1}-2+\frac1{(n+1)^2}+2\sum_{k=1}^n\frac1{k^2}-1
\end{align*}
Where we have used the fact that the first sum on the RHS is telescopic. So
$$\sum_{k=1}^n\frac1{k^2(k+1)^2}=2\sum_{k=1}^n\frac1{k^2}+\frac{2n+3}{(n+1)^2}-3$$
Then
\begin{align*}
\sum_{k=1}^{\infty}\frac1{k^2(k+1)^2}&=2\sum_{k=1}^{\infty}\frac1{k^2}+\lim_{n\to\infty}\left[\frac{2n+3}{(n+1)^2}-3\right]\\[3pt]
&=2\cdot\frac{\pi^2}6+(0-3)\\
\sum_{k=1}^{\infty}\frac1{k^2(k+1)^2}&=\boxed{\color{blue}{\frac{\pi^2}3-3}}
\end{align*}
Answer: This is too long for a comment. It has been added for your curiosity.
Ángel Mario Gallegos having given the good solution to the problem, let me address the problem of the partial sums starting from what Ángel Mario Gallegos wrote
$$S_n=\sum_{k=1}^n\frac1{k^2(k+1)^2}=2\sum_{k=1}^n\frac1{k^2}+\frac{2n+3}{(n+1)^2}-3$$ But, $$\sum_{k=1}^n\frac1{k^2}=H_n^{(2)}$$ where appear the generalized harmonic numbers. This makes $$S_n=2H_n^{(2)}+\frac{2n+3}{(n+1)^2}-3$$ For large values of $n$, we can use the asymptotics $$H_n^{(a)}=n^{-a} \left(-\frac{n}{a-1}+\frac{1}{2}-\frac{a}{12 n}+\frac{a^3+3 a^2+2 a}{720
   n^3}+O\left(\frac{1}{n^4}\right)\right)+\zeta (a)$$ which, in the present case, reduces to $$H_n^{(2)}=\frac{\pi ^2}{6}-\frac{1}{n}+\frac{1}{2 n^2}-\frac{1}{6
   n^3}+O\left(\frac{1}{n^4}\right)$$ Continuing the expansion for the other terms, we should get $$S_n=\left(\frac{\pi ^2}{3}-3\right)-\frac{1}{3
   n^3}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.
Using $n=10$, we should find $$S_{10}=\frac{22254209}{76839840}\approx 0.289618$$ while the approximation would give $$\frac{\pi ^2}{3}-3-\frac 1{3000}\approx 0.289535$$


Can a workflow be associated to a folder without custom code?

Question: My client has a document library that contains a folder for each "case" they are working on. These "case" folders contain various documents related to the case. They would like to be able to add a workflow to this folder and route it for approval. Is it possible to do this in SharePoint 2007 without custom code?
Answer: Not to my knowledge, but I believe you can in SP 2010 with the Document Set functionality, Document Sets.
Comment: Yeah, at this point, I'm pretty sure that it is not possible either. I wasn't aware of Document Sets and they look like a good solution to this problem. We'll be moving to SP2010 soon, so I'll keep this in mind.
Comment: I can guarantee that workflows on Document Sets in SP2010 will do exactly what you want as that's how I used them recently for a customer. Document sets are basically a content type based folder that contains other permitted content types. The workflows then run against these.

You do get some nice dashboard functionality with Document Sets, but you can't keep them if you want to move the entire set to a records centre, or declare the set as a record.

Worth looking at tho.


The Hydrolock - How to build a deep sea docking system?

Question: How do you build a docking system at deep sea pressures? Example a Challenger Deep Sea Base at 10,000 meters seawater or approximately 1,000 atmospheres?
Comment: how are you keeping humans alive at that depth? this has a pretty big impact on how a connection can work.
Comment: See my comment on Nosajimiki's comment below. A multistage chain of locks reduces the engineering challenge quite considerably.
Answer: I've been working on this thought problem about how to create a docking system for deep sea bases. It's not as straightforward as one might originally think because the system has to:

maintain hull integrity the whole time.
transfer between extreme pressure differences while:
-not having any pumps capable of doing so.
maximum pump depth right now is 5,500 meters. And that's pushing the limits of 100 years of development.

So bottomline, you can't just pump your way out of the problem like in a normal airlock that transfers gas from one pressure to another pressure usually between 0atm to 0.3atm (or 0psi to about 5psi). And you therefore can't pressurize the gas, so the airlock can't be one contained system with pressure tanks.
So having mentioned the hurdles, here's the basic thought process so far.

Have two pressure hulls capable of maintaining 1atm at 1000atms (10,000meters seawater).
Have one of these hulls contain water.
Transfer water between these hulls to transfer a relatively incompressible material from one hull to the other.
Equalize pressure between the hull containing water and ambient pressure so its hatch can be removed.
Dock another pressure hull to the removed hatch.
Re-equalize the pressures again, back to 1atm.
Remove the water from the docking hull to the water tank.

I call this a "hydrolock". In principle it works much like a torpedo tube. The pressure hulls are subject to massive stress at the seams (where the spheres meet). I'm not sure if these seams would rupture or leak.
Seals are formed by metal-on-metal contact, so I believe that the greater exposed surface area will keep a strong metal-on-metal seal at the seams since the seams represent a very small amount of total surface area.
Comment: Industrial Engineers often used staged pumping systems to overcome high pressure limits.  Pumping from 100atm to 200atm is about the same resistance as going from 1atm to 100atm. So, if you design an incremental system of 10-15 air tanks daisy chained together with compressors between each tank, you can achieve these pressures just fine... at least from a pumping perspective.  A 1000atm air tank may be harder to come by
Comment: @Nosajimiki's thoughts are good here. I think that using a chain of locks could resolve the extreme pressure problem. It would be a slower result, but far more easy to mitigate against disaster.  In many ways similar to raising and lowering passage over mountains for canal locks: use many, not one.  I would feel far more comfortable living in a multistage locked environment!
Answer: Moon pool.

https://en.wikipedia.org/wiki/Moon_pool

A moon pool is a feature of marine drilling platforms, drillships and
diving support vessels, some marine research and underwater
exploration or research vessels, and underwater habitats, in which it
is also known as a wet porch. It is an opening in the floor or base of
the hull, platform, or chamber giving access to the water below,
allowing technicians or researchers to lower tools and instruments
into the sea... A moon pool also allows divers or small submersible
craft to enter or leave the water easily and in a more protected
environment.

No moving parts.  Moon pools work.  The only tricky thing is that atmospheric pressure of the gas above the water must be equal to the water or the water will come up thru the moon pool and compress the gas.  Having high pressure gas in a habitat is easier on the habitat too because it makes it more crushproof.
1000 atmospheres is pretty high.  You might need to pressurize your moon pool room, enter, close the hatch, and then depressurize the moon pool room, pumping your atmosphere back into canisters.
If your craft is too big to fit thru the moon pool, poke part of it up thru the moon pool and step out.
If it is a science fiction horror, monsters can come up thru the moon pool too.  Good thing you have a hatch.  Did you close the hatch?
Comment: @John Thanks John, agreed. If anyone wants more elaboration, basic rule of thumb is a person cannot breathe more than ~2atm partial pressure of oxygen. You'll go into seizure. Called oxygen toxicity. There's another problem with moonpools. They aren't even effective when you need gas-mixes because the gas exchange rate through a moon pool is VERY HIGH. Meaning if you had a helium mix you'd lose almost all your helium in a short period of time.
Comment: @IDNeon, John, anyone else: the Willk habitat has a hatch over the moon pool.  Monsters, you know.  Also losing your helium.  Also adjusting atmospheric pressure inside after the submersible crawls thru.  Now that we have settled that, let the upvotes flow!
Comment: this only works if the air pressure is equal ot the water pressure, and humans can't breath air at that pressure.


does Blender have a Material Swatches panel showing all available materials?

Question: I'm a C4D user and it has a Materials Pallette. I can see a preview of all materials in the file and I can drag one onto an object. Does Blender have a similar window?
Answer: It does, in the materials tab, and the arrow points to the droplist of all your materials. I never tried drag'n'drop, though, so you need to check it. This is from Blender 2.7x so the current version (2.9x) must have this much or more.
Comment: ok, that's very helpful. I wanted a quick way to select and switch colors on an object. I was doing it using nodes and multiple rgb ramp nodes but this is a great option. thank you so much! if you come across a way to display the materials without needing the drop down, please share. thanks for your help. I'm starting to really like Blender :)