Can Tier 2 General UK visa holders travel to Ireland without an Irish visa

Question: I have a Tier 2 General UK visa which is valid until 2019. Can I travel to Ireland (Dublin, Belfast) without an Irish visa? It would be for a short holiday.
Comment: Yes, that's fine. Also the Isle of Man and the Islands are fine.
Comment: This is only for a Holiday (2-5 days)
Answer: You need a visa to enter the Republic of Ireland unless you are in a visa exempt nationality.
As a general rule UK visas and leave to enter or remain in the UK are not valid for Ireland as it's a different country. Certain holders of UK visas qualify for a visa waiver but that's limited to certain nationalities, and according to the Irish immigration authorities only applies to short-stay UK visas.
You could take a trip to Northern Ireland on you visa as this does not involve leaving the UK.


php based Business Instelligence for Process Maker BPM

Question: i use process maker as my BPM System and i want to use php based BI (Business Intelligence) system for make report from Process Maker results .
can anyone help me ?
Answer: ProcessMaker recommends Pentaho as its BI solution to be integrated with.
Check this: ProcessMaker to offer Advanced Business Intelligence in its BPM software
However, be aware that Pentaho is Java-based. I've not found any PHP-based BI suite yet.
That said, I've been using ProcessMaker for 2 years and think ad-hoc reporting is a feature it still lacks. This is mainly due to the fact that it stores cases data in a serialized form, so reporting is not straigthforward.
Check this post: Unserialize case APP_DATA and show into a GRID in a DYNAFORM to have an idea how it works.
Answer: At this time, Processmaker has ended their support of their Pentaho Plugin.
I've been trying with PHP Report Maker and getting good success so far.
http://www.hkvstore.com/phpreportmaker/


Understanding Fourier, signals and spectrum analyzer

Question: I'm having a hard time grasping the relation of Fourier analysis to the nature of signals and spectrum analyzers. From what I understand, Fourier analysis gives us a mathematical representation of a signal, but we don't need a infinite number of harmonics to make a square wave, for example. So if I put a perfect square wave signal into a spectrum analyzer, using simple 5v on/off switch, it will "see" all diferent harmonics or not? Maybe my question is: A specturm analyzer will have an algorithm to do a Fourier analysis of a signal or will actually perceive different frequencies? Sorry if my question is naive or hard to understand, im still a student.
Comment: We do need a infinite number of harmonics to make a square. It's just that in reality, we say that a finite amount of harmonics is good enough. When you really zoom in on the square tho, you notice it's not in fact an ideal square - it will exhibit ringing or undershoot and ripples.
Comment: This is somewhat related to [my question in physics SE](https://physics.stackexchange.com/questions/354159/are-there-sine-waves-in-nature-or-are-they-a-mathematical-construct-that-helps-u). Yakk's answer in particular might be answering your question.
Comment: Yes in **theory** all the frequency components need to be present in the ideal square wave. But in the real world everything has limitations. Like a bandwidth limitation on a spectrum Analyzer. In practice you also **don't need** to see all harmonics and leaving them out will still be good enough to see a square wave. For example a 1 MHz square wave will still look a lot like a square wave if you leave out everything above 100 MHz.
Comment: Depending on the type/model of spectrum analyzer (SA) it will do an FFT or not. Older SA are often completely analog, so **no FFT**. These work like a complex radio receiver with narrow filters and peak-detectors. Many oscilloscopes have an FFT function so they can show a spectral plot.
Answer: Others have posted some very nice visual explanatory tools, yet to me, they seem to be a step or two down the road from the absolute basics. Let me offer my own super-basic explanation:
First, try to multiply two sinewaves, over some period of time, say two or three periods of your slower sinewave. Take a piece of paper and a pencil if you will, but a couple discrete-time samples calculated in Excel can also produce a valid chart. Start by simple examples, such as: two sine waves of unity amplitude, same frequency, same phase, centered around 0 amplitude (= no DC offset). You need to multiply the instantaneous scalar value at every point in time to produce the resulting third wave (= multiplication product).
Unfortunately I don't have a visual software tool to facilitate this super-simple mental experiment... A quick search through Google Images has revealed  two static charts here.
Further points to consider:
If you multiply two negative numbers (scalars), you get a positive number (scalar). For the result to be negative, you need to multiply a positive and a negative number. Therefore, if you multiply two identical sine waves, the product wave lives in positive numbers only, plus the zero. (And, the product happens to be a pure sine wave of double frequency and DC-shifted, but that doesn't seem to be of much concern here, don't let it mess up your mind.)
To produce some negative values in your "multiplication product" wave, you need a phase shift, or waves of two different frequencies. Watch the individual quadrants expand into the negative values.
Try multiplying two sinewaves with only somewhat different frequencies, say 20% difference, over maybe 10 periods of the slower signal. The link points to an article on superhets, but the sinewave multiplication applies equally, and the waveform plot is very nice (only we could use a zero axis for the "product" waveform in our topic).
Next, guess what happens if you integrate the product wave in time. This is a key operation: your spectral line (the output "amplitude" for a single frequency) is your multiplication product wave, integrated over a given time window.
Note that if you multiply your "input signal" by a sinewave (your "probe"), the product wave (and its integral) will reflect the phase of the probe relative to the input. E.g., for two pure identical sinewaves, that are 90 degrees out of phase, the multiplication product wave will be centered around zero, and its integral will therefore be 0. As if the input wave "was not there"!
To capture an input sinewave regardless of its phase, you need to perform the "waveform multiplication" operation twice for each frequency of interest (probe), using sine and cosine waves for a "probe". Note that sine and cosine are just identical waves, only 90 degrees phase-shifted. Thus, when the sine multiple produces a zero "integrated amplitude", the cosine multiple produces your desired full value. For phases other than 90 degrees, just consider the two output values as a complex number, which you can use as such, or you can calculate its Pythagorean absolute value if you're not interested in the phase.
The Fourier transform means, that you do this "wave multiplication followed by integration" for a certain range of frequencies. As for the range of frequencies considered, it can be discrete or continuous, bounded or unlimited... In theory, an arbitrary continuous signal over a bounded time period produces a continuous spectral image over an infinite range of frequencies. In practice, FT is calculated over a bounded time period of discrete time domain samples, to produce a bounded range of discrete frequency samples. You only have a finite resolution in time and amplitude, and a finite amount of number-crunching horsepower.
An arbitrary input signal (including a square wave) simply "rings" different "sinusoidal probes" with a different phase angle and intensity. And, a square wave rings a series of harmonic sinewaves. A pefect sawtooth does something similar. Essentially any sharp corners in the signal produce a series of harmonics, because you need a series of sine waves of different frequencies to sum up to a half-decent sharp corner in the time domain.
As a homework, but you may have been wondering about this already: think about the effect of window size (temporal, in the input signal) on the result, considering a pure sinewave for input and a pure sinewave for output. And, what happens if the window size gradually decreases relative to the period of your "probe" frequency and of your input waveform.
Answer: Spectrum analyzers usually work by doing a sweep over the frequency range.  At any one time, it is looking at only a narrow frequency range.  It records the amplitude of what it received over that range, then moves on to the next range.
Actually these devices are usually analog, so the sweep is continuous, and so is the result.  As the detection frequency is swept, the received amplitude is measured.  This results in a graph of amplitude as a function of frequency.
If you put a perfect square wave into a perfect frequency analyzer, it will show blips at all the odd harmonics.  These blips decrease in amplitude with increasing frequency.
A Fourier transform is a way of doing mathematically what a spectrum analyzer does by measurement.  The restriction is that the input signal must be periodic.  The result is a equation that yields amplitude as a function of frequency.  Actually there are two equations because a Fourier transform also gives you the phase at each frequency.  This can be expressed as either the sin and cos components, amplitude and phase shift, or any other representation that preserves both degrees of freedom.
A discrete Fourier transform works on individual samples, like you would get by repeatedly measuring a signal with a A/D.  The result is the amplitude and phase at a set of discrete frequencies.  There is a clever computational trick that is now routinely used to reduce the computation cycles to perform a discrete Fourier transform.  This is called a FFT, which stands for Fast Fourier Transform.  The first F refers to the particular method of achieving the result, and FT refers to the result.  A "slow" FT would give the same result.  The name has stuck so that "FFT" is used even when the actual method of computing the FT is irrelevant to the discussion.
Comment: Thank you. The waves that it detects are sinusoidals, of different amplitudes in different frequencies, but how this happens if the input is constant over half the period?
Comment: https://en.wikipedia.org/wiki/File:Periodic_identity_function.gif I hope this gif helps you better understand your question.
Comment: @JoãoPedro what Olin is trying to tell here is the general case. The amplitude and phase of sine waves will depend on the original signal under test.
Answer: I think this interactive java web application will answer alot of your questions.
Choose Mag/phase view button then Log display for best results and slide n (terms) harmonics to see result change in recursive wave in red with infinite terms in white.
You can even listen to an audio tone of the wave as your ears and brain do Fourier analysis automatically.
When you point to any harmonic, it will display the actual amplitude in the repeating pattern.  Choose any standard waveform and limit n , or draw your own arbitrary wave and compare.
Remember these laws of Fourier spectrum;

impulses and steps contain continuous spectrum, but this analyzer repeats the pattern of the wave, so discrete harmonics depend on shape.
square waves have no even harmonics, but if slightly asymmetric, these rise rapidly to even amplitudes for a narrow repetitive spike.
phase shift of harmonics reflects the filter effect of integration of a square into a triangle wave so amplitude and phase of fourier shows this when Mag/Phase display with Log.(arithm)
the 1st spike shows the DC average level , next its fundamental f
Comment: For the advanced Biomedical Engineer, try to draw a heart beat and analyze the results then change phase spikes use a tap and drag at an angle for fine tuning to simulate cardiac strain. Clear button means your flatlined.
Comment: So a square wave, for example, also oscillates a little bit in the upper and lower part of the wave? Is it possible to have something like 5v up / 5v down, without oscillation? In multisim I see for example that it changes between 5.001 and 4.999. I think my question is to whether all signals will obligatory be made of sinusoidals of different frequencies.
Comment: it all depends on n as I said, this simulates only a couple decades<3.. The ratios are computed.  Slide n.  try again
Answer: I think this video will help you understand how waves add together and how we can analyze them to their frequency components. It has certainly helped me because I am a visual type and the visualizations in this video make it very intuitive.
This is a subject that's been bothering me for years and there are many levels of understanding Fourier. On a basic level, it is a way to transform a signal (see it in a different domain) from the time domain to the frequency domain and back (using the inverse Fourier transform). This reveals the 'fact' that any complex waveform is made up of a number of sine waves (single frequencies).
I put the word fact in quotes because as you can see in the answers to the question I made on the physics SE, it is debatable to point that becomes philosophical whether sine waves are real or just a mathematical construct that helps us analyze natural processes. It all boils down to what reality is.
I think the simplest way to describe the Fourier transform in simple words is using part of Yakk's answer to my question.

You can step back and look at the fourier transform. You start with some wave. You multiply it against the original signal (using convolution), and from the result work out how much they overlap and what is the best scale of the wave to approximate the original signal.
Then you subtract that scaled wave from the original signal. This "removes the frequency component" from the original signal (in that if you convolved it with the wave again, you'd get zero).
We then repeat this with different frequencies, each time "removing a frequency component". So long as the frequency components we remove are orthogonal to each other (a generalization of being "at right angles"), removing new frequency components doesn't "bring back" the old ones.
As it happens, "removing the frequency component" corresponds to an operation.
If you set of a resonance cavity of length L down which pressure waves travel at speed S, and have some repeating set of pressure waves, the cavity will amplify the part of the pressure wave of frequency L/S that roughly corresponds to a the convolution of the sin wave with the amplitude of the pressure wave over time.
That seems pretty academic, but have you ever looked at your ears?
They are resonance cavities. Pressure waves go in, and bounce back and forth.
Along the side of it, there are hairs that pick up changes in pressure. Waves of various frequencies are amplified and damped by the resonance chamber and excite and ignore a predictable set of hairs.
In short, our ears split pressure waves into something a lot like what fourier analysis does. We have physical fourier transformers on our head attached to our brain.
So when we do a fourier analysis and say that there is a strong signal at 550 Hz, this corresponds to what our ears hear because our ears are doing something that the math approximates and mapping the pressure waves into a spectrum of frequencies for us to hear.

A spectrum analyzer does indeed perform a fast Fourier transform (FFT) which is an optimized (faster) algorithm. It has a certain resolution in the frequency domain, so it doesn't check for every frequency. It also cannot detect phase. There are special algorithms for that which (I think) are used in astronomy and submarine detection.
Answer: I understand Fourier series and transform is somewhat tough to understand when you start with it but once you understand it you will know that its the most beautiful thing that mathematics has given.
Whenever I can't understand some concept or the purpose of same I try to understand the history or motive behind the same and that pretty much explains most of the things.
Now the problem in physics is there are a lot of equations describing something or some shape which can have varying meaning depending upon the system we are trying to study. Now the problem arises that for every system we have a different set of behavior and equation and every time it looks like some new problem which needs new methods/ approach and analysis. Now don't we like it to keep it simple and have only one tool that can solve all your problems (or at least most of it)
That's where Fourier comes to rescue. Now the basic principle here is that i only need to know properties of sine and cosine and i will try to decompose any (or most) of the signal into a combination of sine and cosine wave. Now the wave may not look exactly the same as the original wave but still, it's very close to it and does the job.

The above picture is not a perfect square wave but its very close to it.
Now the oscillations that you see at the end are also named as "Gibbs phenomenon".

whether all signals will obligatory be made of sinusoidal of different frequencies

No, it's not obligatory of any signal to be made of sinusoidal, but as I explained before its a mathematical tool or approach to solving complex signals.
Comment: It's a mathematical principle that **any** repeating waveform can be built up out of a sufficiently large number of sine waves.  That's not just an approximation, a good mathematician should be able to prove it.  (Note that I'm only an adquate mathematician, so I accept what those who know more than me say.)  A waveform that has a duty cycle that's not exactly 50/50 will have even harmonics, because it's not a perfect square wave.
Answer: The Fourier Transform can only map input energy into sinusoids, because the only basis function allowed in Fourier Transform is a sin/cosine wave.
The math operation performed in Fourier Transform is "correlation", and the peak responses are where the allowed basis functions are maximally correlated.
to the input waveform.
Now ---- can a 100MHz clock cause a 100MHz LC resonator to ring? yes
Can a 100MHz clock cause a 200MHz LC resonator to ring? depends on duty cycle
Can a 100MHz clock cause a 300MHz LC resonator to ring? yes
Can a 100MHz clock cause a 313MHz LC resonator to ring? yes; depends upon "Q"
Summary of this --- harmonics do not exist.
What we consider to be indication of a harmonic is merely a local peak in the correlation. A 100MHz clock will cause a 300MHz LC resonator to ring, and also cause a 313MHz resonator to ring. Depending on "Q", where "Q" merely indicated the time-duration of the signals fed into the correlator (and the Low Pass Filter imposed on correlator output).
Here is a 20MHz rectangular wave, 45% duty cycle, driving a 50MHz LC resonator with Q of 100. What is doing on?

And here are the input/output waveforms along with input/system/output spectra (magnitude only; I disabled the phase plots)
Comment: Do you understand what is being demonstrated here?


Does a file system like WinFS actually exist?

Question: Did anyone actually release a relational file system like WinFS? I'm aware that different parts of WinFS got implemented into different products but does a relational file system like Microsoft originally planned exist?
Comment: Because WinFS was a relational file system where data is more unified and it is easier for programs to read different formats and show information based on many different objects rather than just a single file. Windows 8 does do something like this but without unifying the file types it will still be limited compared to WinFS
Comment: Why does everyone want WinFS when NTFS does the job and has been proven to work.  There is going to be a new file system in Windows 8 but you won't be able to boot to it, and it is similar to WinFS but it is not WinFS, it more/less a national extension of NTFS.  The new filesystem I speak of, again is an extension of NTFS, it is not WinFS.
Answer: Yes, BeFS. Sorry you can't use it with any maintained operating system (and trust me, I am sorry for myself).
Comment: I agree, I've been using it for a few hours now, and it's file system isn't worth getting rid of everything Windows and Linux has to offer.
Comment: I guess it's time to download Haiku then :D (for those who don't know, Haiku OS is a project based on BeOS)
Comment: Haiku is nice, but it doesn't really have an ecosystem (and very little hardware support). The architecture was greatly optimized and efficient in the 90s, but doesn't make much sense with modern hardware (I'm thinking mostly about the graphical layer here).


Masking out pixels from the Sentinel 1 image in Google Earth Engine

Question: I am trying to classify paddy areas from time-series(in this case, four-date) Sentinel 1 image in both VH and VV polarization. Hence prior to classification, I need to mask out land covers such as urban and water areas. I plotted the temporal profile of the backscattering coefficient for these land covers and it was found to be -20 DB for water areas in VH polarization.
What I tried to do is to mask out water and urban areas from the image collection using the below code.
var watermask = function(img) {
  var excludewater = img.gt(-20);  //Identify all pixels below threshold and set them equal to 1. All other pixels set to 0
  excludewater = excludewater.updateMask(excludewater); //Remove all pixels equal to 0
  return(excludewater);  //Return image
};

var S1 = vh.map(watermask);// vh is the imagecollection of four images corresponding to four dates in VH polarization
print(S1,'S1');

var mergeBandsS1 = function(image, previous) {
  return ee.Image(previous).addBands(image, ['VH', 'VH', 'VH','VH']);
};

var mergedS1 = ee.Image(S1.iterate(mergeBandsS1, ee.Image([])));//create a stacked image from the imagecollection S1 inorder to plot the temporal graph of various landcovers
print(mergedS1,'mergedS1');

// Define customization options.
var optionsvh = {
  title: 'Time Series VH plot',
  hAxis: {title: 'Date (YYYY/MM/DD)'},
  vAxis: {title: 'Backscatter coefficient Sigma0_VH (dB)'},
  lineWidth: 2,
  pointSize: 4,
  fontSize:20,
  series: {
    0: {color: '970F0F'},
    1: {color: 'FF0000'},
    2: {color: 'F68244'},
    3: {color: '1230D8'},
}};

// Define a list of dates for X-axis labels.
var datesvh = ['2018/11/01','2018/12/07','2019/01/12','2019/02/05'];

// Create the chart and set options.
var spectraChartvh = ui.Chart.image.regions(
    mergedS1, fromList, ee.Reducer.mean(), 30,'name', datesvh)//fromList is the featurecollection of various landcovers, name is the label given to the landcovers
        .setChartType('ScatterChart')
        .setOptions(optionsvh);

print(spectraChartvh);

While plotting this am getting the below image which I couldn't interpret. What should I do to mask out water landcover from this data?
Answer: Define your mask function as follows, thus, applying the mask image on the original image:
var watermask = function(img) {
  var excludewater = img.gt(-20);  // make a mask image
  img = img.updateMask(excludewater); //Remove all pixels from original image
  return(img);  //Return image
};

In the way you defined it, you will return only zeros/ones because the gt function return a binary 0/1 image


How to prevent OS X from launching programs from another disk?

Question: I just did a fresh install of Mountain Lion on an empty partition. How can I prevent the programs of the previous install from being shown in the contextual menus in Finder ?  I already asked Spotlight not to index that partition, but I still have references to them in the Open With menus.
Answer: Try rebuilding the launch services database using the lsregister command. That database provides the information for the open menu, etc as noted in How to rebuild the LaunchServices database  from Macworld's Mac OS X Hints. Rebuilding the launch service database while running from the newly installed partition should fix the problem of trying to open Applications on the other drive partition.

If your "Open With" menu shows applications that are no longer installed, or shows some of them as duplicates, you can use the following Terminal command to rebuild your LaunchServices database:

For OS X 10.6 and later run:
/System/Library/Frameworks/CoreServices.framework/Versions/A/Frameworks/LaunchServices.framework/Versions/A/Support/lsregister -kill -r -domain local -domain system

For OS X 10.5 and older run:
/System/Library/Frameworks/ApplicationServices.framework/Frameworks/LaunchServices.framework/Support/lsregister -kill -r -domain local -domain system -domain user

If the problem persists after running the above command, try un mounting your original partition before running the above command again. Additionally, as noted on the hints article and when running the lsregister  command with no arguments there is some helpful output that will allow further customization to specify where to look for Applications, etc.
Comment: Unfortunately, tha applications come back after the system has been rebooted.
Answer: Is it possible for you to move them to another drive or remove them? If you really want to keep them on the drive that they're currently on you can also make them, or the folder that they're in into a .zip file.
Or if it is just about specific applications that come up all the time in the Open With... menus, then this link will show you how to remove specific applications from the Open With menu.


What made Qui-Gon Jinn really believe that Anakin Skywalker was the chosen one?

Question: We all know from "The Phantom Menace" that, about to die, Qui-Gon Jinn asked Obi-Wan to promise him to train Anakin because he believed the boy was the chosen one; the one who would bring balance to the Force. We also know that Qui-Gon Jinn told Mace Windu and Yoda that Anakin's cells had the highest concentration of midi-chlorians that he had seen in a life form. Was this previous reasonable enough ground to make Master Qui-Gon Jinn to declare that Anakin was the prophesied one who would bring balance to the Force?
Answer: Qui-Gon Doesn't Believe In Accidents
There were a number of factors that led Qui-Gon to believe as he did. The first and main trigger for his suspicion was the chosen one prophecy itself:

"A Chosen One shall come, born of no father, and through him will ultimate balance in the Force be restored."

Master & Apprentice novel
Qui-Gon had already sensed the Force in the boy. After Anakin insists they come to their home, Qui-Gon catches how perceptive and talented Anakin is beyond his years (Anakin identifies him as a Jedi Knight, talks of how he's the only human who can pod race). Qui-Gon notes as much to Shmi and asks about his father, to which she informs him that Anakin had no father. It's after this that he decides to administer the midi-chlorian test, and confirms his incredible "off the chart" Force potential, which he then sees in action when Anakin wins the pod race. Fate placed this boy in their path, and it was hard to ignore that there was something really special about him.

"Our meeting was not a coincidence. Nothing happens by accident."

Qui-Gon, Star Wars: Episode I - The Phantom Menace
While this doesn't necessarily guarantee it, Qui-Gon's faith in the Force and its order in the universe, that nothing happens by accident, and the timing of the appearance of a Sith, lend a strong sense of something big going on, and gives him a strong suspicion that Anakin might be more than he appears.
Comment: Good comprehensive answer. In Qui-Gon's shoes, I might be a tad more incredulous about Schmi's miraculous pregnancy and inquire deeper. But hey, they just met and she probably wouldn't appreciate the nosiness
Comment: I've always wondered about this. Since the Jedi council thought the Sith were gone, then the Force was mostly good. Therefore, bringing balance to it would be to increase the evil. Hence Darth Vader and the Emperor.
Comment: Ultimate balance being both sides close to zero? Well well, nice one, Force.
Comment: At the end of ROTS there are only two Jedi and two Sith. The Force is finally balanced.


Java Eclipse input validation

Question: So here it is: I'm writing a (swt)program.In my form, I have two text fields and a button, but when pressing the button I would like it to validate the (string)text fields. Asking the user for input if nothing in textfields?
Hopefully this question makes sense
Thank you all in Advance!!
Comment: The button has no code yet.
Answer: You must do this manually, adding a listern to button.
  If you your TextField is called textField and yout Button button, it will be something like:
\\previous code
button.addListener(SWT.Selection, new Listener() {
  public void handleEvent(Event e) {
    switch (e.type) {
    case SWT.Selection:
      if(textField.getText() == null || textField.getText().isEmpty()) ; //here you validate
      break;
    }
  }
});

Java Eclipse input validation


Test series for divergence or convergence

Question: I have to find whether the infinite series $$\sum_{n=0}^\infty \left(\frac{n}{b(n-3)}\right)^n, \quad\text{for }b>0$$ diverges or find for which values of $b$ it converges to some finite value. So far I've tried to apply the ratio test but it didn't seem to work. Any advice on how to solve the problem is greatly appreciated.
Comment: Try the root test.
Comment: Your sum starts at n=0, hence if you don't exclude the term for n=3 then the sum i divergent trivially for any value of b,
Answer: You have
$$\left(\frac{n}{b(n-3)}\right)^{n} = \frac{1}{b^n}\left(1-\frac{3}{n}\right)^{-n}$$
and $$\lim_{n \rightarrow +\infty} \left(1-\frac{3}{n}\right)^{-n} = e^3$$
so $$\left(\frac{n}{b(n-3)}\right)^{n} \sim \frac{e^3}{b^n}$$
So the series converges iff $$\frac{1}{b}<1$$
i.e. iff $b>1$.
Comment: That means that the quotient of one by the other tends to $1$ : see https://en.wikipedia.org/wiki/Asymptotic_analysis (and don't worry if you don't know this notation yet, you will see when you learn it that it is very useful)
Comment: What does the tilde notation on line 3 mean? Does it mean that those two expressions are equal in the limit?
Answer: I'm using the root test, which you can read about if you want here: https://en.wikipedia.org/wiki/Root_test
.
Consider $\lim \limits_{n\rightarrow \infty}((\frac{n}{b(n-3)})^n)^{\frac{1}{n}}=\lim \limits_{n\rightarrow \infty}\frac{n}{b(n-3)}=\lim \limits_{n\rightarrow \infty}\frac{1}{b}=\frac{1}{b} $ by L'hopital's rule. By the root test, the series converges if this limit is less than $1$, so the series converges if $b>1$.
Hope this helps.
Answer: Actually, having all the expression affected by an n power, you will probably have better results using the root test. This is:
$L = \sqrt[n]{|{a_n}|} = \frac{n}{b(n-3)}$.
As that is a quotient of same-degree polynomials, the limit to infinity is:
$L = \lim_{n\to\infty} \sqrt[n]{|{a_n}|} = \frac{1}{b}$
As you may know, $L < 1$ ensures convergence, and $L > 1$ ensures divergence, thus the first clear conclusion is that if $b > 1$, the series converges, and if $b < 1$ the series diverges
The case $b = 1$ must be analysed separately, for in that case the root test does not give any information on the series' convergence. However, we can see that if $b = 1$:
$\lim_{n\to\infty} {a_n} = \lim_{n\to\infty} (\frac{n}{n-3})^n = \lim_{n\to\infty} (1 + \frac{3}{n-3})^n = e^3$
As $e^3 \neq 0$, the series is not convergent when $b = 1$, thus converging only when $b > 1$
Answer: You can use the root test
\begin{align*}
\limsup_{n \to \infty} \sqrt[n]{\left( \frac{n}{b(n-3)} \right)^n} &= \limsup_{n \to \infty} \frac{n}{b(n-3)}
\end{align*}
The series converges, when
$$ \limsup_{n \to \infty} \frac{n}{b(n-3)} > 1 $$
Therfore, for the series to converge, you have to choose $b$ as
$$ b > \lim_{n\to \infty} \frac{n}{n-3} = 1 $$


Using Wi-Fi and Ethernet at the same time

Question: I'm using the macOS High Sierra. There are two network interfaces on my Mac, Wi-Fi and Ethernet. The Wi-Fi is connected to the Internet directly, and the Ethernet is connected to out company's local network. They're in different subnets, one of which is 10.0.0.0/8 and another is 192.168.0.0/16. The Wi-Fi has faster speed than the Ethernet, but the sources in the local network cannot be accessed via Wi-Fi.
Now I want to route the traffic based on the urls. For example, I want all the access to *.corp.com to the Ethernet and others to Wi-Fi. The problem is the DNS. Those *.corp.com are only valid locally, which means they can only be queried in the local network. If I put Wi-Fi on the top priority, then all the access to *.corp.com would fail due to DNS query failure, because the system will use the DNS servers configured by the Wi-Fi router. However, if I set the Ethernet to the first, all the traffic will go to Ethernet.
I've tried the way to configure the route table manually, but it didn't work. The same DNS issue.
So if there any good ways to handle this situation?
Answer: Set your DNS server manually to the corporate one, prioritise wi-fi in network settings so it's the default gateway and route 10.x.x.x out through ethernet?
Comment: Unluckily, it didn't work since the local DNS server couldn't be accessed if not in the local network.
Comment: By the way, I've tried to capture the packages on the Ethernet interface, but no DNS queries found.
Comment: You should be able to route the path to the DNS server out the ethernet port if it isn't also on 10.x.x.x


80s DRAM chips: one per bit of data bus width?

Question: As I understand it, in the eighties the typical way of handling memory was one RAM chip per bit of data bus width. Suppose you were building a 16-bit machine and you wanted to give it 32K of RAM, you could do this with 16kbit RAM chips, using sixteen of them. 128K could equally well be done using 64kbit chips, again sixteen of them.
However, if you wanted 64K of RAM, this could only be done by using eight of the 64kbit chips, and making each chip deliver two bits, one after another, thus incurring a slowdown.
So if you don't want to pay any penalty in access speed, it's 32K or 128K but not in between. Is this correct?
Comment: As noted in @pndc's answer, the simple - and very common - solution was to use multiple sets of smaller chips. There are two reasons for this - don't pay for "wasted" memory and newer higher density chips typically cost more initially. That gets balanced against the extra board real estate, sockets, soldering, etc. required. In the retro days, board real estate was usually cheap and new chips were very expensive until they got up to high levels of production. That is still true today - often a machine will ship with 2 or 4 smaller DIMMs instead of 1 or 2 larger DIMMs.
Comment: the average "ibm compatible" XT board was even worse, most had 36 sockets for memory chips ( 4x9 ( 8bit + parity) ), even when there was a 8088 installed
Comment: The were also wider DRAM chips available during the 80's. For example the Commodore 64 originally shipped with eight 64kx1 chips but later revisions used two 64kx4 chips.
Comment: @cat - bits versus bytes. Each byte is 8 bits.
Comment: The 8-bit Acorn Electron did something similar, in using 4-bit wide access to RAM, and slowed the CPU to half-speed in order to access both nybbles, one after the other.
Comment: how is 16 bits wide bus * 16KiB per chip = 32KiB??
Comment: @JeffreyBosboom It doesn't suggest that to me! The intent is to suggest that each chip sends one bit over the data bus at a time. If there's an alternative title that other people would find clearer, I've no problem with someone editing the title.
Comment: @cat 16 bits wide bus x 16kbit per chip = 32kbyte.
Comment: @rwallace last i checked 16 * 16,000 = 256,000
Answer: No.
In your hypothetical 16 bit machine with 64kiB of RAM, you could simply implement two 32kiB banks with using sixteen 16kib chips each. This obviously doubles the required number of chips and board space required, which may not be cost-effective against using just next higher-density chips and getting twice as much memory again for free.
At least one real-world example exists. The Amiga 500 shipped with 512kiB of RAM, and early models implement this using sixteen 256kib chips. The A501 memory expansion contains another sixteen 256kib chips, giving 1MiB in total.
Comment: Good point! The Sinclair Spectrum 48K started off being implemented with three banks of 16kb chips, and in 1984 as I understand it, switched to being implemented with one bank of 64kb chips of which a quarter was left unused, so that's suggestive about when the crossover point of cost effectiveness would happen.
Comment: @rwallace if my memory serves well the 64Kb chips used in early ZX48K where faulty chips with error in one half of it and where hardwired so only the good half was used. Which lowered the cost of ZX even more. By the time the manufactoring process of DRAM got more reliable the costs per chip dropped  down  but the design was left as it was even with good chips. Some clones use the other half like **Didaktik GAMA 80KB**
Answer: There were several variations largely driven by cost at any particular point in time.
What is interesting is WHY 1 bit chips were popular, basically your address bus was typically multiplexed using the RAS and CAS signals, so if the technology at the time made 64K a desirable chip size, you could do 8 Address, RAS,CAS, 1 Data, WR, RD, CE plus power and ground in something like a 16/18 pin DIL and only end up with 1 signal per chip being unique. Back before multilayer PCBs were cheap, this mattered.
Compare with an array of 8, 8k*8 parts (Same total memory size), now you have to run that 8 bit data bus to every chip, plus 7 bits of address, plus the control signals, plus you need an address decoder, so you are looking at a 24 pin chip, with much more routing on the PCB.
Eventually speeds got to the point that the lower bus loading made wider devices a better choice (especially as NMOS had horrible noise margins), but if you look at a modern DIMM you will still find that multiple narrow parts are often favoured.
Answer: Find someone with a collection of vintage Computer Shopper issues, there's no better research material for such matters. Not only are there articles discussing the merits of different computers and their memory schemes, there are ads touting the month-by-month pricing, speed and capacity of direct-market processors, RAM chips and disk drives.
A few pointers:
-- Early dynamic RAMs were multivendor, with a common parts numbering scheme (and similar DIP pinouts) through the 4k to 256k generations.
-- 4096 x 1 chips for example were 4104, 16,384 x 1 chips were 4116, 65,536 x 1 chips were 4164, 262,144 x 1 were 41256.
-- A suffixed letter often indicated if it was an epoxy (P for plastic) or ceramic (C for ceramic) package.
-- 4-bit-parallel parts were called "nybble-wide" or "nibble-wide" and were numbered 4416 and 4464. The 4464P was the commonest sort used in Apple //e's from 1986 onwards, providing 64kx4 with 120nsec RAS typical.
--The original type-1 IBM AT used paired 4164C's soldered in piggyback stacks to populate its DIP sockets with "128kbit" RAMs, an arrangement possible because IBM custom-packaged these RAMs at their plants to fit a 256-kbit pinout. Finding these today is an Easter Egg hunt through Grampa's workbench, as they were often removed to make way for 41256 chips and squirreled away in some unmarked DIP rail or organizer drawer.
-- 1-megabit parts were the last to operate on 5-volt supply, but were internally 3.3-volt parts
-- 4-megabit parts began a wholesale transition to new technologies like 3.3-volt logic, JEDEC pinouts, SMT surface mounting and Single Inline Memory Modules. By then most new computers used word-wide (16-bit) memory.
Comment: The earliest type of (30 pin, 8 bit) SIMM/SIPP modules were often built from 8 or 9 (parity!) 1 bit wide chips, sometimes two 4-bit wide parts (with a parity 1-bit if wanted)....
Answer: One other reason: 1-bit chips could easily be configured as 8 or 9 bit wide arrays using only one type of chip - the latter in cases where memory parity (error checking) was desired. Sometimes, 4+4+1 was used, but necessitated two chip types with potentially different reliability characteristics (which is not helpful in a parity system).
Comment: Yep, also if your computer has a weird word-size like the PDP-10 or PIC16 or something, we can just use the appropriate number of chips even if we're not bothered about parity.
Answer: It was very common in 1980s systems to utilize DRAM chips that output 4 bits. Common variants were 256Kb DRAM chips accessed as 64K x 4 bits, and 1Mb chips accessed as 256K x 4 bits. Thus, a 16-bit data bus might only require 4 DRAM chips to provide 512KB.
The 4-bit wide DRAM's were used widely because they reduced chip count needed for a given data bus width. 8 chips connected to a 32-bit data bus was a common configuration into the early 1990s. Also, later versions of 1980s 8-bitters like the Apple //e Platinum and the Commodore C64c were able to reduce their DRAM footprint to only 2 chips, reducing manufacturing costs.


Labuan Bajo to Lombok by bus & ferry

Question: I am in Indonesia, in Labuan Bajo (Flores) next to the Komodo national park, & I have been told it's possible to go to Lombok by bus & ferry.

© OpenStreetMap contributors
Comment: To those who marked this as duplicate: this is NOT a duplicate as I am asking the opposite direction as the question marked as duplicate of. The main concern is that because of this, boats & buses schedules are different, hence the answer (& journey) completely different.
Answer: I have done research online and offline, as for this question: Go from Lombok to Labuan Bajo by bus & ferry, how to?
So I will answer it myself. But do add your answer/comments if you have anything to say!
The only things that are different with the linked question are:

Departure time of ferry from Labuan Bajo is at 09:00
Bemos going from Sape to Bima are waiting in Sape for the ferry to arrive, so there is virtually no waiting time.
Bima to Mataram, buses leaves at 19:00. They know many passengers from Labuan Bajo are coming so they usually wait for you. Once in that bus, just put your coat on (cold AC) & have a nap.

Note that it is much nicer to do the journey this way rather than the other as the transports are all waiting for you to arrive from the long ferry ride. Rather than you waiting for the transports.

Labuan Bajo to Lombok by bus & ferry


Problem with HostGator addon domain

Question: I installed WordPress on an addon domain in my HostGator account. When I go to the wp-admin url and sign in I see this. Does anyone know what's up with this?
Thanks
Comment: What is the full url you are using? (User example.com instead of your real domain).
Comment: I do not use WP. By add on domain, do you mean a new domain name or a sub-domain? If a new domain, when was it registered? Did you create an A record? Did you properly add your domain via cPanel? When I see this, it tells me that the web server does not recognize the domain name in the request header.
Comment: That looks to me like either you have the URL wrong, the nameservers are wrong, or you haven't verified the domain and after 15 days it has been suspended and is showing ads instead.
Answer: If you added on a domain name, make sure the domain is officially registered, and DNS is set properly and propagation is completed (usually about 24 to 48 hours). You can call HostGator.com via live chat http://chat.hostgator.com/ or via telephone 866-964-2867 available 24 hours a day to get assistance.
https://www.hostgator.com/help > you can search 'addon domain' for steps and help, also if you are calling internationally if you search 'skype' from our support pages, we have the steps on how to call us locally/internationally.

Problem with HostGator addon domain


Why 1 divided by 2 does not equal two halfs?

Question: I think that mathematics is a system of rules used by us to model the reality. If I divide one pile of gold in two equal piles I would get two equal piles. But in mathematics if I'm to represent the pile with a number, then the result will contain only one number.
It's confusing to me that in mathematics I always get only one number, while in reality I can get any number of objects (for example, I can divide the pile of gold into 7 equal piles). I expect to see things like: 30 / 3 = [10,10,10] or stuff like that because that is more close to reality, but I always see 30 / 3 = 10.
Can someone please comment on my problem.
Comment: You can write it as $30=10+10+10$.
Answer: You are quite right that if you have a pile of size $30$ and divide it into $3$ (equal) piles, then what you have is $3$ piles each of size $10$, which could be described as $(10,10,10)$.  The mathematical notation $30/3$ refers to the size of a single pile.
Likewise, if you divide a pie into $2$ equal pieces, then what you have is not half a pie but two halves of a pie.  The number $1/2$ refers to the size of each piece relative to the original whole.
Comment: I like your answer and the comment of @dvix . From what I know about it's history, the mathematics is a unity of many different modeling systems. Because operations (+, -, *, /) were created at different times (as well as any part of the matematics) it's logical that they represent different "ideas". It's amazing how all this different modeling system have converged into a single entity. The answer to my question lies in the comment mentioned and the reason why my question seeks a wrong interpretation of an operation is in the answer above.
Answer: Division is not an operation on collections; divison is the inverse operation of multiplication.
Suppose $b \neq 0$. Then $\frac{a}{b}$ is the unique $x$ such that $b \cdot x = a$.
Comment: Yes and no. Division is sometimes an operation on collections. That's how kids correctly describe distributing candy among friends. It's also the inverse of multiplication. Fractions can be understood both as parts of a whole and as numbers to be manipulated arithmetically. Most people routinely and unconsciously choose the model appropriate to a particular situation. But the fact that there are several useful models makes teaching fractions to kids difficult. See https://math.stackexchange.com/questions/1127483/how-to-make-sense-of-fractions/1127776#1127776


How and why does magnetic fields influence ferrous metals?

Question: I know that a magnetic field is a result of moving charges (general relativity) but why and how would this positive charge (moving negative charge/electron -> less negative charge -> positive charge) cause a ferromagnetic object to be attracted. A magnetic field caused by a movement of negative charge should be positive because of relativity and so should the ferromagnetic object as it has electrons moving around. Shouldn't they repel?
Comment: Welcome to SE Physics. Good questions  try to provide accurate context to help people to give good answers. Some previous search could be helpful. Mentioning general relativity in connection with magnetism, implications like that in parenthesis, and mentioning a *positive magnetic field* are pieces of information difficult to understand, if not wrong. I suggest you could try to improve the formulation of your question.
Comment: As you suspected the magnetic field induced by moving charges do repel, this phenomenon is called diamagnetism, all materials show that behavior but that is not ferromagnetism. Ferromagnetism is a quantum mechanical phenomenon and is not caused by moving charges instead it is the result of collective behavior of electrons and their spins.
Answer: The magnetic field in the magnet causes the domain of ferrous material to align, causing it to become a magnet temporarily. The magnetic fields of both objects interact and create a force.


How and/or why did the original comic book Jor-El choose Earth as the target for his son's rocket?

Question: Jor-El saved his son by rocketing him off the planet Krypton just before it exploded.  How and/or why did the original comic book Jor-El choose Earth as the target for his son's rocket?
Comment: Edited. Defined as requested.
Comment: Which canon? As per Smalville series, he was actually on Earth before.
Answer: The simplest answer which spans almost the entire canon of Superman iterations:

Jor-El chose Earth for its yellow sun (and the potential powers Kryptonians gained under a yellow sun) and in the earliest iterations the abilities Kal-El would possess as a result of Earth's lower gravity.
However this answer would not be fully explained until the 1960s. Earliest versions of the story don't ever explain anything more than Jor-El barely being able to get Kal-El off planet before its destruction.

First Superman Strip 1939 - Notice Jor-El (and all Kryptonians) has superhuman abilities on Krypton as a result of evolutionary development.

The early comic strip versions only depict the powers Superman gained as a result of Earth's lower gravity. His increased speed and strength were the primary powers gained.

Other universe depictions of Superman had varying reasons in addition:

Early descriptions of Krypton sometimes had the Kryptonians having powers ON Krypton, so traveling to Earth would simply allow Kal-El to potentially lead or rule Earth if he so desired.
Earth was also chosen for its lower level of technology, hoping to offer Kal-El the most opportunity for survival as the last member of the Kryptonian species.
Some iterations indicated Jor-El had a macroscope or teleportation abilities allowing him to see/visit the Earth and have an understanding of the suitability of Earth over other potential sites.
Considering the reaction some universes had to the existence of Kryptonians, it was implied the Kryptonians were not always desirable neighbors, perhaps fear of their potential powers when they were off-world. This was noted particularly during the Man of Steel, Post Crisis era.
Comment: Another nice answer. Does Jor-El's Macroscope predate that of Piers Anthony, or is it a later addition to Superman canon?
Comment: The idea of Jor-El "seeing/visiting" Earth probably came before Anthony's Macroscope because there are stories in the late fifties where Jor-El uses a telescope to see Earth FROM Krypton, but I suspect the idea of using the telescope to travel to Earth may have happened around the same time or soon after. Great ideas sometimes propagate in multiple places at the same time.
Comment: It may be notable that in the very early canon like the radio show - Krypton was in our solar system.  I think the implication is that a 1940s audience couldn't imagine a rocket being interstellar.
Comment: It seems from the _original_ strips that Kal-El was just launched into space, and happened to land on Earth by chance? Your example strips support your suppositions but don't directly address _why Earth_.
Answer: Nothing in the original comic book suggests Superman's father intended anything of the sort.
Superman's first appearance is in Action Comics #1 (PDF link) in June, 1938. The very first panel says:

As a distant planet was destroyed by old age, a scientist placed his infant son within a hastily-devised space-ship, launching it towards Earth!

That's all it says, and all of the subsequent panels deal with Superman after he gets to Earth.  There's nothing to say that a launch "towards Earth" was anything but random.
A work's "canon" (if you find it necessary to impose such a thing) can certainly evolve as time passes, but you asked about the original comic book, and I don't think Siegel and Shuster had any sort of comprehensive worldbuilding in mind when they created the first installment.
Answer: I would imagine that the physical resemblance between our species also played a part. Kal-El wouldn't fit in on a planet where everyone had ten legs.


Install Shutter on Centos

Question: I'm trying to figure out a way to install shutter on Centos. Are there any repos somewhere?
Answer: From shutter-project.org's download page(1):

Red Hat Enterprise Linux (and Derivatives like CentOS, Scientific Linux etc.)
One of our users (Nux) provides a small repository for RHEL 6. The repository is designed to coexist with Fedora's EPEL repository.
  In order to add those repos to your system and install Shutter you can use the following commands:
rpm -Uvh http://download.fedoraproject.org/pub/epel/6/i386/epel-release-6-5.noarch.rpm
rpm -Uvh http://li.nux.ro/download/nux/dextop/el6/x86_64/nux-dextop-release-0-1.el6.nux.noarch.rpm
yum install shutter

This could work.
(1) the first result Google presents for "shutter centos".

Install Shutter on Centos


"Line integral" of delta distribution

Question: I have seen engineers write something to the effect : $\int_{-\infty}^x \delta_{0}(t) dt = H(x)$. Here $\delta_0(x)$ is the dirac delta distribution concentrated at origin and $H(x)$ is the step function. Even though the distributional derivative of $H(x)$ is the delta distribution, taking an integral of the delta distribution seems morally wrong, especially since $\delta(x)$ is not a lebesgue measurable function even. Besides an "intuitive" justification for this, is there a rigorous way to justify this?
Comment: The dirac delta is a measure concentrated on a point. That is, if $E\subset \mathbb{R}$ then $\delta_0(E) = 1$ if $0\in E$ and $\delta_0(E) = 0$ otherwise. When you integrate is like integrating the 1 function respect to the $\delta$ measure.
Comment: $S(x)= \int_{-\infty}^x T(y)dy$ always is a well-defined distribution when $T$ is a *compactly-supported* distribution. When $T$ is a distribution of order $0$ (ie. a measure, as $\delta$) then $S$ is (represented by) a function.
Note if $T$ is a compactly supported distribution then $\langle \int_a^x T(y)dy, \varphi \rangle = - \langle T, \int_a^x \varphi(t)dt \rangle$ perfectly makes sense as $\int_a^x \varphi(t)dt \in C^\infty$. And that's why distributions are useful : they make sense in general in analysis, and not only for $\varphi \in C^\infty_c$.
Comment: @Abhi, Yes, so  if $x>0$ in your integral then $H(x)=\mu((-\infty, x)) = \delta_0((-\infty, x))=\int_{-\infty}^x 1 d(\delta_0)=1$ and if $x<0$ then $H(x)=0$.
Comment: I see, so $\int \delta dx := \int d(\mu)$ where $\mu$ is the dirac measure?
Comment: @reuns I was trying to make sense of the integral of the distribution as evaluating it on the characteristic function but wasn't sure because they act on $C_c^{\infty}$ functions. I preferred to let my comment like that because maybe there is a way to see the integral but I wasn't sure. In any case if you agree just tell and I'll delete this comment also, was just because I supposed you were addressing your last comment to my edit.
Comment: Even if $\int_{-\infty}^x T(y) \, dy$ is not always well-defined for any distribution $T$ then every distribution has an antiderivative.
Comment: @I.C. thanks for your insightful comments.
Comment: @reuns I think I am a bit hazy on my distribution theory knowledge. Can you please indicate a reference where I could study the assertion that you have made about compactly supported distributions? Thanks!
Answer: I tend to disagree with the approach that has been proposed.
In this case, I think that it would be better (or simpler, if you prefer) to think of $\delta_0$ simply as a measure, instead of a distribution. Namely, you have
$$
\int_{\mathbb R} \chi_{(-\infty,x]}d\delta_0=H(x),
$$
where the integral is the Lebesgue integral of the function $\chi_{(-\infty,x]}$ with respect to the measure $\delta_0$.
As you see there is no need for convolutions or really distribution theory altogether. Note also that $\delta_0$ is a Radon measure and so lives in the subset $(C^0_c)'$ of $(C^\infty_c)'$, no need for $C^\infty$ functions.
Comment: I agree that one can interpret $\delta$ as a measure and do what you have proposed, see also @I.C comment above. I was just curious, since $\delta$ is a distribution and so is $H(x)$ and engineers tend to write $\int_{-\infty}^{x}\delta(t)dt = H(x)$, can we prove equality of the left and the right hand sides as distributions. Hence , the approach with convolutions.
Comment: Sure, as @reuns noted it (I liked the comment!). But distributions are of very little use, if at all. They failed to solve the initial motivating problem of nonlinear PDE's besides the well known problem of multiplication, which is needed in particular in physics. From my point of view, one should avoid distributions whenever possible.
Answer: Here is my attempt to justify this rigorously:
Note that $\int_{-\infty}^{x}\delta(t)dt = \int_{-\infty}^{\infty}\delta(t)H(x-t)dt = H*\delta = H(x)$. Here $*$ represents the operation of convolution. Note that the convolution of two distributions is justified as long as one of the distributions is compactly supported, see e.g. $\it{Functional Analysis, 2nd Ed., Rudin}$.

"Line integral" of delta distribution


Visual Composer

Question: Programming novice here.  I'm editing a WP site already constructed using visual composer, but having trouble creating responsive elements.
Can anyone point me in the right direction as there's little out there in the way of guides.
Problem:  I'm trying to use the default image carousel as a large header image(s).  I can get one to sit in a new row, but the images contained are sticking to their original dimensions (1200x700), with no responsive resizing.
The row is 'stretch row and content no padding), how can I make it's carousel content stretch 100% by width?  Ive tried with a single image with similar results :(
Any help would be gratefully received.
Thanks.
Comment: Hey Dezza - did you ever get a resolution on this?
Answer: If you are using a single image, you should just use a single image element instead of an image carousel. I tried this out and it image in a image element would resize according to the viewport size.
Answer: Go to your row options and set the margin to 0px. For it to span all the way across you should set the margin to 0px on the left and right sides of the row.
Alternatively you can use a "full width section" to achieve the same thing.
Let me know if this works for you? we can also find another solution

Visual Composer


CRM Application for Ubuntu

Question: I am looking for CRM application to manage an IT workshop running Ubuntu. I have a medium size workshop and I am looking for a CRM application I can use to manage jobs, appointments and track work being done by my staff. Does anyone know of an application that will do this?
Answer: Try sugar crm it is open source and has a free community edition:
http://www.sugarcrm.com/crm/download
Answer: Check out http://dolibarr.com Hope it helps
Online Demo https://www.dolibarr.org/onlinedemo.php
Github https://github.com/Dolibarr/dolibarr
About
Dolibarr ERP CRM is a modern software package to manage your company or foundation activity (contacts, suppliers, invoices, orders, stocks, agenda, accounting, ...).It's opensource software (wrote with PHP language) designed for small and medium companies, foundation and freelances. You can freely install, use and distribute it as a standalone a…

CRM Application for Ubuntu


Equity - date of offer, or date of joining?

Question: I made major contributions to a startup and am being offered equity - 11% of total stock.  I have been at the company for 2.5 years, and am being offered equity with a 4 year vest, starting 1.5 years ago, which was about the point when equity was offered -  I complained, saying that it should start from joining time, 2.5 years and 62.5% now,  not 1.5 years and 37.5%.  What is reasonable and standard here?
Answer: Options or Shares vest by date they are granted. It would strike me as odd for anyone to say their shares were given with 4 year vesting, but the clock was pre-started years prior.
In my opinion, you have nothing to complain about.
Answer: TL;DR: The date they were granted.  (Usually, this follows both an offer and acceptance.)
It's not uncommon for a new vesting clock to start when there's a new round of funding coming in, because the investors want to make sure the key people are going to be engaged and incentivized going forward from that point.  They don't lower their expectations for how long they want folks engaged based on the person having started earlier.  Non-institutional investors may have the same concerns as institutional investors here and use the same vesting strategy to address them.
Primary recognition of the benefits from having had people start earlier or be there longer (so long as it correlates with having gotten more done) is embedded in the valuation (which affects how much founders' shares are diluted in the raise).


Does laptop charger wastes electric current when not connected to the laptop?

Question: Wonder if when I leave it connected to the outlet but not connected to the laptop, it still consumes electricity.
Comment: Every device uses power, at least a little bit, when plugged in, even if the device is off
Comment: http://en.wikipedia.org/wiki/Current_source
Comment: Depends on the charger.  A few (such as Apple's charger) detect when they're plugged in and turn themselves off (leaving only a very microscopic current drain) when not plugged in.  Most original equipment supplies for name-brand laptops meet US energy-saving standards and draw relatively little current when not plugged in.  No-name replacement chargers, OTOH, are apt to be current hogs.
Answer: The answer for this is: YES, it will consume power, but such a little power that you may consider it irrelevant in terms of overall power consumption.
As you can see in the picture below
(a schematic of an AC notebook adapter), the "entrance" of the circuit
("L and N 90 - 265 VAC", in the lower left corner) has an AC transformer:

This component is responsible for converting the entrance AC voltage into another one in AC (if there is a load on the "output" side of this component). As you can see in the picture below:

Wikipedia has a good explanation about the operation of a transformer:

A transformer is a device that transfers electrical energy from one
  circuit to another through inductively coupled conductors—the
  transformer's coils. A varying current in the first or primary winding
  creates a varying magnetic flux in the transformer's core and thus a
  varying magnetic field through the secondary winding. This varying
  magnetic field induces a varying electromotive force (EMF), or
  "voltage", in the secondary winding. This effect is called inductive
  coupling.
If a load is connected to the secondary, current will flow in the
  secondary winding, and electrical energy will be transferred from the
  primary circuit through the transformer to the load. In an ideal
  transformer,
  the induced voltage in the secondary winding (Vs) is in
  proportion to the primary voltage (Vp)
  and is given by the ratio of
  the number of turns in the secondary (Ns)
  to the number of turns in
  the primary (Np) as follows:

I marked in bold the main part of explanation to explain what will happen if there is no load on the transformer (in this case, it means that there is no device "consuming" power from the transformer). If there is no load, the secondary part of the transformer will not "act" on the primary part of it.
It means that you will only have the input loaded with the entrance impedance of the transformer, or, in this specific case, the AC power from your electric outlet to an inductor of 200 µH in the transformer. A theoretical "ideal" inductor will not dissipate any power, but a real inductor will also have a resistive component that dissipates a small amount of power, usually due to the magnetic material in the core of the transformer producing "core losses". This will explain why you have a consumption of the AC adapter even if there is no "load" or on this case, if the notebook is turned off.
PS: This situation applies not only for your notebook AC adapter but to all adapters that have a transformer on their entrance.
Hope being helpful with my explanations.
Comment: Could you please tell an approximate power the adapter draws from the source when there is no "load"? I know it depends, but just give a rough estimate to know what we'r losing.
Answer: Yes it does.
The simple proof? Hold the charger in your hand while it's plugged in.  Does it feel warm?  That heat is generated from the electricity passing through the charger, even though it's not plugged into a laptop.
Edit:
Some related links:

Vampire hunters: Devices reduce energy waste
Energy tip #10: remove wall warts and slay electricity vampires
Do chargers use electricity when they aren't being used?
Comment: Actually, some chargers are cold when not in use. For example my Hipro HP-OL093B13P is always cold when not in use.
Comment: hmmm. interesting.  Mine have always been a bit warm to the touch.  Maybe wasting electricity is not a universal property of chargers?
Comment: I've never had a PSU that does not get at least slightly warm.
Comment: All good power supplies should stop electricity flow if A. the battery is fully charged, or B. there is no laptop connected. From my experience, my old Thinkpad charger as well as both my MacBook and MacBook Pro chargers over the years have never been warm when not actively charging anything. I'd say it could be vendor specific, though I'd sure hope not. That said, I think they will all still have _some_ electricity flowing through them, but it should be miniscule and nothing enough to cause warmth.
Comment: "Warm" is pretty subjective and will depend on the ambient temperature.
Comment: @AndrejaKo If the charger is cold, you may still be able to check the coilwhine instead by putting the brick to your ear and listening closely
Answer: In general, yes it does.  Unless there's a sensor that detects the connection to the laptop and cuts the circuit to the transformer, there's still electrons moving around.
You can actually measure this with the Killawatt http://www.p3international.com/products/special/P4400/P4400-CE.html
It's sold at thinkgeek.com as well
Comment: Very useful device for figuring out how much those DC transformer blocks actually eat when idle. I used it to see how much power my older laptop drew in sleep mode (1.5w), and when the laptop was unplugged but the power still in (.25w).
Answer: Not all chargers use a measurable amount of power when the laptop is plugged in but off.  My laptop charger uses 0 watts as measured by a watt-meter.  Yours might be different.  Now it might not be an actual 0.000000000000 but it is apparently less than 0.1 watt which is the minimum my meter will read.  Under load with my laptop running it's 25 watts.
Answer: Yes.  There are still components in the laptop charger that uses a "little" energy when it is plugged in.  About a year ago, there were numerous commercials on TV stating to unplug your cell phone charger when not in use due to the same reason.
Comment: Don't mean to be petty, but you're saying it does, but not as much as it would have consumed if the laptop was connected to it?
Comment: Yes, if you plug in a device which draws current (puts a load on), it will definitely draw more current
Answer: It seems no one actually measured the real consumption of their own charger?
I did many times and were surprised that my new Thinkpad charger uses 7W without charging any device.
So, better borrow a charger and see for yourself.


iMessage off and on on computer

Question: I keep my iPhone on airplane mode nearly all day. I usually text via iMessage with the Messages app on my mac. I was wondering if there was any way I can turn off iMessage on my computer and then turn it back on later and get all the missed text messages (with my iPhone on Airplane Mode during this whole process). I tried signing out of iMessage but when I sign back in, I don't get the missed text messages. Thanks!
Answer:
It is not possible to forward text messages to your mac while
  your iPhone is turned off or in airplane mode.
While iMessages are "just being moved" within Apple servers, text
  messages must be first received from your carrier, then synchronized with
  iCloud so that your mac can receive it too.
With your iPhone being turned off, the chain of delivery is broken.

The following steps show the iMessage way.
Make sure you sync your messages with iCloud on iPhone and Mac.
iPhone - Sync Messages with iCloud
1. Go to settings
2. Click on your Account at the top
3. Choose iCloud
4. Activate iMessage

Deactivate iMessage on macOS
1. Open iMessage
2. Go to settings with CMD + ;
3. Unselect «Enable this account»

Reactivate your account by selecting «Enable this account».

When you reactivate iMessage on macOS, you'll be notified on your iPhone:

Your Apple ID and phone number are now being used for iMessage on a new Mac. If you recently signed in to "«MAC_OS_NAME»" you can ignore this notification.
Comment: I don't see how this answers my question

iMessage off and on on computer


Which 3D filament is 100% (or close to being so) food safe and non toxic?

Question: I've recently bought myself a preassembled Prusa i3 MK3S printed and made my first projects. One of them was making a cup with my name on it. I want to use it to drink tea, water etc. I know, however, that I need to chose my filament wisely, as using the wrong one might be unsafe. I know that PLA for example is Polylactic acid which is a safe substance and occurs naturally in our body. Another thing is the dye, which can is a chemical substance I know nothing about. Do You recommend any specific type/model? Thanks.
Comment: @Greenonline Editing the question so significantly after it already has two answers isn't that great of an idea. Now the answers do not make sense any more. Moreover, there are thousands of different kind of dye so I think the current question is unreasonably broad.
Answer: Answer was moved to this question: Which are the food-safe materials and how do I recognize them?
Answer: I think this site will answer some questions about food safe 3D printing: 12 Vital Facts About Food Safe 3D Printing
PLA is not a good choice for hot substances because it will deform at hot water temperatures which is no good for a cup (very dangerous!)
I would recommend a filament like PETG, PETE, HDPE, and LDPE: What Plastics Are Approved for Food Contact Applications?
Those types of plastic require very high printing temperatures and an all metal hot end.  You don't want PTFE in your hot end when printing at those temperatures (265 °C) as it will emit toxic gasses.
I've seen food safe filaments for sale so it might be best to search Google for one of them to use.
Please keep in mind that printing anything will cause voids and gaps that will allow bacteria to grow.  Personally I don't think I would want to risk it for daily use items but a coffee/tea cup might be alright.  This is covered in the first link.


How much would it cost to build Megaroc?

Question: Megaroc was an idea of a modified V2 rocket by the British Interplanetary society to launch a human into space. If it were built there could have been a human in space in 1948/1949. The reason it was not built is because the ministry declined the idea because of financial reasons (Also because of a lack of interest). My question is: How much would it have cost to build it?
Comment: Since it seems to have been more of a concept than an actual design, we would be guessing a lot.
Comment: Given that the proposal was given to a government department it would be interesting to try to get a copy of the proposal under the British equivalent of freedom of information procedures. The proposal is bound to have included cost estimates, at the time. Modern day equivalent costs can be extrapolated from these using the [Bank of England Inflation Calculator](https://www.bankofengland.co.uk/monetary-policy/inflation/inflation-calculator). Alternatively, one could analytically review the proposal with the idea of making a modern version & costing the proposal that way.
Comment: @ Fred it would be interesting to see the price difference of back then and today. Maybe there is an even cheaper method of building it using newer technology
Comment: It would have been interesting to see how they would have approached the task of man-rating the V2...
Answer: The idea of scaling up a V2 and tweaking the design is not implausible: this is exactly what the Soviet R-2 did, and in fact the rough capacities are very similar (source for R-2): an 18m-tall missile weighing around 20 metric tons, with a payload of around 500kg. The R-2 tweaked the fuel rather than widening the rocket, and had a lower max altitude, but you can see the general similarities. It was later tweaked into the R-2A sounding rocket and flew research payloads, exactly as Megaroc was meant to, but by that point there was no motivation for crewed suborbital launches.
In practice I think there is a good argument that a simple V-2 derivative would not have worked for the planned flight - as you got into the fine details involved with having a person on top, the payload size would have crept up and you needed to develop a more advanced V-2 derivative like the Redstone - but it does suggest we're in the right range for what could reasonably be done in the late 1940s.
However, time and cost. R-1 (building their own copies of V-2s) took the USSR two years (1947-49), and while R-2 test flights started in 1949, it was not in service until 1951. So 48/49 is probably an ambitious estimate for a V-2 derived launcher, especially given the other constraints on the UK at that time. In addition, the Soviet program benefitted from a large amount of German hardware and rocketry specialists: the equivalent British program was much smaller scale.
Tracing a figure for how much the R-2 program cost - which would more or less answer your original question - is hard. I haven't yet been able to find one and honestly I would have big question marks about a meaningful cost value attached to a 1940s Soviet program. However, we can get an upper bound: Redstone. This was a much more ambitious project and Astronautix estimates $90m overall cost - call it £20 million, give or take (depending on whether the pound had devalued yet). On the other hand, this does not include the costs of the 'spacecraft' element, which would not have been cheap to develop...

Edit: digging around in Siddiqi's Challenge to Apollo (pdf) turns up the quite obscure Soviet VR-190 project from the 1940s: using a modified V-2 to launch two crew to 190km (hence the name). It was changed to a goal of launching dogs in 1948 and shut down in 1949. However, an independent project under Korolev in 1949-51 launched a series of dogs on R-1B missiles - very similar to the R-2 series - taking a payload of around 500kg to 100km. However, the two dogs flown were around 6-7kg each, and the container had an internal volume one-tenth that of a Mercury capsule, making it very unlikely it could have been reasonably scaled up to a person. (Siddiqi pp. 63-66, 92-97)


How many ways can six 8-peg lego blocks of six different colors be combined?

Question: I have seen the number quoted as > 650 billion. The quote is from a Danish mathematician, Soren Ehlers. Intuitively, this does not seem correct to me. Any mathematicians/statisticians out there who can help either verify the huge number, or give me a correct number?
Comment: This might be a question better suited for Stats.SE or Math.SE.
Answer: To expand on Zhaph's answer, I'm assuming that you have a fixed set of six colors, and each block is one color.
If that's the case, this is a simple combination problem on top of the orientation problem.  Since there are six colors and six bricks, there are 6! ways to assign those six colors to the six bricks, or 720 ways.
Now, we apply this combination to our orientation, and get:

915,103,765 orientations x 720 color combination/orientation = 658,874,710,800 combinations
Comment: @Bonifer people connect simultaneously in much simpler ways than Lego bricks (Sam knows Bob or not) and in much more complex ways (Sam has heard that Bob is not very nice, but from someone Sam thinks is unreliable). There's not a good match to the Lego "one stud to the left" relationship modelling.
Comment: So it's true! Thank you for your verifications, Zaph and Nick. Very much appreciated. FYI, I'm using this to explain the complexity in large data sets. i.e. if six Lego blocks can connect in this many combinations, how many ways can six humans connect? Traditional Taylorist management theory says one! That leaves lots of possibilities unexplored.
Comment: This number is probably too high. Some arrangements will be symmetrical such that switching the brick colours doesn't give a new creation, just a rotation of the previous one.
Comment: @Mr.ShinyandNew安宇 is right.  It's difficult to guess how many different brick configurations are radially symmetric.  However, it's safe to assume that the number is fairly low, because the only possible configuration would have to be 90 and/or 180 degrees rotationally symmetric.  90 degree symmetric sets would have 1/4 as many combinations, and 180 degree symmetric sets would have 1/2 as many combinations.
Answer: I'm not sure what the rules are around the colours, however from the LEGO Facts page:

Take six eight-stud LEGO bricks (2x4) – how many ways can they be combined?
With the aid of computers, the exact number of combinations has been calculated as 915,103,765!
Just so you know, two eight-stud LEGO bricks can be combined in 24 different ways and three eight-stud LEGO bricks in 1,060 ways.

This is probably based on the work done here.
The colour combinations would therefore be higher than that - but you'd need to be avoiding duplicate combinations.


How can I replace a color in an image?

Question: I want to replace the blue part of the image with red. I've tried this using a combination of the Magic Wand tool and the Paint Bucket but that doesn't work well.
Can somebody please help me out?

Answer:
Open the image in Photoshop
Double-click the layer Background to make a regular Layer out of it.
Go to Image -> Adjustments -> Replace Color...
Shift-click on the image to get the color range you want selected. You can adjust the Fuzziness parameter to your liking.

Now choose the replacement color in the lower panel.
Comment: This doesn't allow for re-editing should the user want to shift the tone a bit. I suggest using the adjustment layer outlined below...
Comment: It does work on a broader set of images, i.e. images that have other colors in it as well, instead of a b&w background. Each method has it's pro and cons.
Answer: Create a new adjustment layer. Click the Icon the lower red arrow is pointing to! Then you can adjust the hue slider, (the top red arrow). Keep in mind this will not always be the case. It works in your scenario because blue is the only actual color and the rest is black and white.
There are also a number of other ways to accomplish this. However this is likely the most efficient in your case.
Comment: Agreed. And the adjustment layer allows for retouching if needed as opposed to set and "forget".

How can I replace a color in an image?


Question regarding COC Wartime troop upgradation

Question: I am upgrading my troops for war tomorrow. The upgrade will be completed tomorrow but by this time I will be at war with another clan. So can I use these upgraded troops during war?
Comment: Yes you can.. All troup which are under update and will complete in middle of war then you can use them.
Answer: As Leo the Lion said, you will be able to use the troops with a higher level. To expand, unlike defense and defending hero levels in your war base, troop level and hero level when attacking is independent of war. Whatever level your troops are when you click attack, that is the level they will be when your attack starts regardless of what type of attack it is (multiplayer, single player, or clan war).
Comment: @MdMehmudSiddiqueTwin if a building is upgrading during war, and doesn't finish before prep day finishes, they will be functional in your war base, but they will not be upgraded. If you are building a new building, it will also be functional in your war base, and it will be level 1. hope that helps, and if this did help, you can click the little check box to mark it as an answer, so others will know this is the answer. and when you get enough rep, you can always go back and upvote helpful answers
Comment: Agree with Dragonrage that only troops will get updated but not the buildings. They will have same level till war and as war will end, building will upgraded.
Comment: Ok thanks. But its a shame that I cannot vote any answer up as I am still new. And what about the upgraded building! Is it upgraded automatically as well !
Answer: What you need to remember, is that:

all troops you use will be the level they will enter the battlefield, and
your base will not change during the battle day as soon as the preparation day ends.

What the above mean for you is that:

if the troops you are going to make your attacks with are having upgrades on battle day, they will be the level they appear as after upgrading (what you see is what you get).
if you have troops in your CC (Clan Castle), or your heroes, or defenses in your war base were upgrading during preparation day and completes the upgrade during battle day, they will not be the level they are after upgrade. However, if the upgrade completes during preparation day (battle day has not begun yet), then they will take the upgraded level during battle day.

The only exception is for buildings/traps that begun construction (not upgrade) before battle day. These will appear as level 1 during battle day.
Examples:

A player puts lv4 archers in their clanmates' war CC. That player has his archers complete upgrade during battle day. On battle day, the war CC will have archer lv4.
If a player has lv4 archers upgrading and attacks with them before the upgrade completes, they will be lv4. If he attacks after the upgrade, they will be lv5.
A player begins building 1 X-Bow and has 2 X-Bows upgrading to lv2 during preparation day. Of the 2 upgrading X-Bows, the first completes during battle day and the second after battle day. The war base will have 3 operational lv1 X-Bows, because at the moment the battle day begun, the base had 2 lv1 X-Bows and 1 X-Bow under construction, or tl;dr:
<code>--------------------------------------------
X-Bow   | Prep Day        | Battle Day
--------+-----------------+-----------------
X-Bow 1 | Lv1 in war base | Lv1 in war base
--------+-----------------+-----------------
X-Bow 2 | Lv1 in war base | Lv1 in war base
--------+-----------------+-----------------
X-Bow 3 | Lv2 in war base | Lv2 in war base
--------------------------------------------
</code>
X-Bow 1's contruction does not complete during either days.
X-Bow 2's upgrade does not complete before battle day begins.
X-Bow 3's upgrade completes during preperation day.
Comment: you examples are somewhat confusing to understand. the first one sounds like you are saying if i donate troops before a troop upgrade finishes, the troops will be upgraded once my upgrade finishes, which would be incorrect. the second example, while accurate, could be worded better to make it more understandable.
Comment: @Dragonrage I never meant it to be that way, but wanted as few examples as possible. I guess I've no choice if I want to cover everything...


Gaussian Process for Closed Curves

Question: A Gaussian process gives a probability distribution over functions that pass through the data points. Is there a way to parameterize the Gaussian process to give a probability distribution over closed curves instead of functions? Can the same approach be generalized to give a distribution over surfaces without boundary?
Answer: If you take 2 Gaussian processes, $X(t)$ and $Y(t)$ on the unit interval, and with the property that $X(0)=X(1)$ almost surely (similarly for $Y$), then $(X(t),Y(t))$ will be a random curve in $R^2$.
Is that what you are looking for?
Comment: A simple example is the Brownian bridge with covariance kernel $k(x,y)=min(x,y)-xy$. You can also use Green's Functions to build models with boundary constraints. See page 134 of Rasmussen and Williams "Gaussian Processes for Machine Learning". It's available online.
Comment: How would you construct a process like this? Would you just condition on the process going through those points? (Would be great to know if there is another way of doing this.)

Gaussian Process for Closed Curves


Cakephp one model different form validation

Question: I have a model name "User", their I added a validation for login. But I need to validate registration page also. Fields for both forms are different.
Can someone please tell me how to manage different form validation with 1 model.
Answer: You can validate as many fields as you want inside your User model, it does not matter in which View or in which form you input them.
So just add the fields from your registration page to the User's $validate inside your User model.
Answer: If all forms share similar fieldnames but require different validation rules you can use:
http://bakery.cakephp.org/articles/dardosordi/2008/07/29/multivalidatablebehavior-using-many-validation-rulesets-per-model
If the duplicate fields validate the same on all forms you can just add them all to the Model, it will only validate the ones present on the form.
Remember to NOT use 'required' => true, setting this key to true will make the field always required and it has to be present in the data array even if it's not on your form


What are the different Java heap spaces?

Question: In JConsole I am seeing different heap spaces "Eden", "Survivor", "Old".
What is the difference between them?
Comment: google Java memory model
Comment: This might help you understand: http://stackoverflow.com/questions/2129044/java-heap-terminology-young-old-and-permanent-generations
Answer: Please see the Memory Management in the Java HotSpotTM Virtual Machine whitepaper.
Java divides the heap into two spaces called 'generations,' the New generation and the Old generation. The New generation is further divided into Eden and Survivor spaces.
The reason for having New and Old generations is because nearly all objects in a Java program live for only a short period of time, and if the short-lived objects can all be deallocated without examining all the long-lived objects, garbage collection will be much faster. For example, a program will usually have configuration settings and shared data objects that live for the entire life of the program, while objects like StringBuilders are constantly being created that become garbage before the methods that created them even return. Ideally all the temporary objects created after a few method calls could be instantly reclaimed without wasting time looking at anything else.
How does Java distinguish short-lived from long-lived objects? By putting them in different buckets and counting how many garbage collections they survive. Brand-new objects go into the Eden space. The Eden space is frequently garbage-collected, but nearly everything in there is already garbage by the time that it is examined. The few objects that are not garbage get moved into the Survivor space. Objects in the Survivor space get tagged with a count of how many garbage collections they survive. Some become unreachable and get garbage-collected relatively quickly, but others stay reachable, and after they've survived some threshold number of garbage collections, Java assumes they are long-lived objects and promotes them into the Old generation.


New trial motions-once the motion for summary judgment has a response may new trial motion be invoked?

Question: Litigation in the pretrial phase progresses to the point of possibly one party filing the MSJ. After that motion has been responded to and the judge has the responsibility to rule on it, may he, for any reason whatever-so-ever, dismiss plaintiff's case without addressing the motion. Has there been enough to qualify for a new trial despite an actual trial not transpiring,for the sake of qualifying under civil rule 59.
Comment: Jurisdiction?     .
Answer: Yes, with the caveat that one should always check local rules in whatever jurisdiction one is operating in.
I can't think of a reason right now (although I may be overlooking some things) that a judge would ignore a motion for summary judgment, yet dismiss the case for another reason without dismissing it for another reason prior to the motion for summary judgment. I say that because summary judgment motions often come after the discovery phase (not always, but the rule is one must be filed within 30 days of the conclusion of the discovery phase), which may take a significant amount of time. Perhaps, of course, the judge becomes aware of new information unrelated to discovery. Nonetheless, if the dismissal were on jurisdiction grounds or some other issue that was known ahead of time, it is unlikely he or she would wait to dismiss.
That said, courts may act sua sponte, which means on its own / absent a motion by the parties.
With respect to a new trial or not, an example of being dismissed without prejudice and, thus, allowing the trial to begin again, may be a major error in the plaintiffs filing. After fixing the issue(s), he or she may refile the same suit.
Usually, a dismissal for summary judgment will be with prejudice, meaning it can't start over (but it can be appealed). This is because, when dismissing a case based on a motion for summary judgment, the court is determining that the moving party has shown

that there is no genuine dispute as to any material fact and the
  movant is entitled to judgment as a matter of law.

That basically means there is nothing to litigate, so go home.
Answer: Normally a motion under Rule 59 of U.S. civil rules based upon the federal rules of civil procedure can be brought (if brought in time) to ask a court to reconsider a final order that resolves an action made by the court as well as a trial verdict.
Even if the MSJ wasn't ruled upon, the Court had to issue some order to dismiss the case and that could be challenged in Rule 59 motion although the time limit for doing so is short (within 14 days of the order in Colorado).
Even if a Rule 59 motion is not brought, usually the order dismissing the case could be appealed in a direct appeal if timely brought.
Answer: If the motion has been dismissed "with prejudice" then that is the end of the matter (barring an appeal) and cannot be re-litigated.
If dismissed "without prejudice" then the matter can be re-litigated.


how to compare values in two columns in two different files, echoing full lines where absolute value of difference is < a small maximum value?

Question: I want to do a line-by-line comparison of values in, say, column 6 of two files, and then echo the lines and their full contents, piping them into a combined file, say the line from one file and then the line from the next file immediately under it, all separated by a spacer line , when the absolute value of the difference in the values in column 6 from the two files is less than some small maximum value, such as < 0.001. I would not echo or print out all the other lines where the absolute value of the difference in the column 6 values is greater than 0.001, for example, but instead, only the cases where it is < 0.001. How would I do that?
Comment: Some input/output would be much nicer than a large wall of text :)
Answer: awk '
  function abs(v) {
    return (v < 0 ? -v : v)
  }
  {l1=$0; f6_1=$6; getline < "file2"; l2=$0; f6_2=$6}
  abs(f6_1 - f6_2) < 0.001 {print l1 "\n" l2 "\n"}' file1

Answer: I would pre-process with paste to interleave the lines and then feed to awk
paste -d '\n' file1 file2 |
awk -v threshold=0.001 'NR % 2 {a=$6; b=$0; next};
$6 - a > -threshold && $6 - a  <threshold {print b; print ""; print $0}'


Taking a derivative from both sides of the equation

Question: Our teacher tried to explain to us how to find a slope at a given point of the function $y^2 = 2px$ by taking derivative from both sides of the equation, he did this:
$$y^2= 2px $$
$$2yy' = 2p$$
$$ y' = \frac{p}{y}$$
I haven't quite understood the second part of the process, will be thankful for a logical and clear explanation :)
Comment: The funny thing is that when you differentiate, for example $x^2$, what you are actually doing is the same: $$y=x^2 \\ [y]=[x^2] \\ y'=2x$$.
Comment: $y$ is a function in $x$, thus by chain rule $y^2=y^2(x)=2y(x)y'(x)$.
Comment: @Oppa Hilbert Style helped me the most, I've understood it a momet after I've sent the question ^_^
Answer: When we find derivative of some term. First we find the derivative of the power then the derivative of term.
So left hand side of second step,
$\frac{d}{dx}y^2 = 2y^{2-1}.\frac{d}{dx}y$
= $2y^{1}.\frac{dy}{dx}$
And you can write $\frac{dy}{dx} = y'$
So we have,
= $2y.y'$
In second step on the right hand side,
$\frac{d}{dx}(2px)$
= 2p$\frac{d}{dx}x$
As 2 and p is constant and derivative of x with respect to x is 1.
= 2p
At last,
2y.y' = 2p.
$y' = \frac{p}{y}$
Comment: find this explanation very useful and coherent, tho you forgot the 2 in the 2px.
Comment: @Ozk yes sorry for that. I edit it.
Answer: It will become clear if you know the concept of implicit differentiation.
Thus we have that $$y^2=2px $$ $$\Rightarrow 2y\frac {dy}{dx} =2p\frac {dx}{dx} $$ $$\Rightarrow 2yy'= 2p $$ $$\Rightarrow y'=\frac {p}{y} $$ where $y^{(n)} $ is the nth derivative of $y$ (we generally use $y', y''$ etc. for lower derivatives but it is not advisable for higher ones). Hope it helps.
Answer: This is just the chain rule you presumably learned about earlier in calc, but the notation is more compact. Recall the chain rule says that
$$
\frac{d}{dx}f(y(x)) = f'(y(x))y'(x)
$$
The LHS of the second line is just the chain rule applied for $f(y) = y^2$. Since $f'(y) = 2y,$ we have
$$
\frac{d}{dx}(y(x))^2 = 2y(x)y'(x).
$$
Answer: $y$ is a function of $x$. $y^2$ is differentiated with respect to independent variable $x$ using Chain Rule. Instead of everytime writing like the following
$$y(x)^2= 2px, \,2y(x)y'(x) = 2p,\, y^{\prime}(x) = \frac{p}{y(x)}$$
your teacher has omitted repetitive parentheses $(x)$, assuming that you can appreciate that $p$ is a constant and $x,y$ are variables.
BTW $p$ is double the focal length of parabola which has a property that sub-normal ( projection on x-axis $=p)$ is always constant for the normal (in red).


Bone Drivers to Copy Rotation

Question: Can I set up a Driver for the bone on the left to slightly copy the rotation of the selected one when rotating clockwise on its Z axis, and to copy the rotation with higher influence when rotated counter clockwise on its Z axis?
Answer: you could do this by putting a simple condition in your driver's code block:
<code>zrot / 2 if zrot < 0 else zrot * 5
</code>

which gives this:

⚠ Note that you can just map the range(-180,180) to something else. it means that if you rotate counterclockwise when zRotation is currently greater than zero this doesn't work. if that's what you really need you can animate your reference bones and using f-curves you can do the rest.
Comment: Somewhat related https://blender.stackexchange.com/questions/164765/rotation-driver-skipping-backwards-after-90-degrees  IMO will need to keep the value of the previous frame _eg_ going from 90 to 100 or 80 on next frame will both be considered > 0.  See the slow value example in _`Text Edtor > Python > Templates > Driver Functions`_
Comment: put them in a text block with py extension, add the methods to driver namepace in register method. Check the register checkbox in text editor. _`Text > Register`_ save to startup file.  Requires Auto run python scripts to be set in prefs too.
Comment: @batFINGER thanks but blender doesn't store added functions. I wanted to ask this problem before but I forgot. whats your opinion about it.
Comment: @batFINGER instead of using addons is there a way to store (or run before load) these kind of functions in blender?
Comment: thanks so much!

Bone Drivers to Copy Rotation