Resonance graphs

Question: When an oscillating object undergoing SHM is damped the peaks of the curves becomes flatter and they shift towards the lower frequencies. I understand that the damping causes a loss in energy and this is reflected in reduced amplitude but why do the peaks tend towards the lower frequencies (left on the graph)?
Comment: Please note for the mechanical system you are asking about displacement resonance not velocity resonance for which the resonance peaks all stay at the same frequency, the natural frequency of undamped oscillations of the system.
Answer: The damped resonance frequency must tend towards zero as the damping increases because, at and above critical damping, the undriven system will not oscillate for any set of initial conditions.
This is most easily seen in the Laplace domain.  The poles of an undamped SHO system have zero real part, i.e., the poles are on the imaginary axis.
As the damping increases, the (complex conjugate) poles move into the left half plane with increasingly negative real part while the imaginary part moves towards zero.
At critical damping, the poles are real and equal, i.e., there is no oscillatory component.
Above critical camping, the poles are real and distinct and move apart along the real line as the damping increases.

Image Credit
In the time domain, one can look at the step response.  For the undamped SHO, the step response is a semi-infinite sinusoid.
For an underdamped SHO, the step response is an damped sinusoid with lower frequency than the undamped case.
For a critically or higher damped SHO, the step response is monotonic, i.e., no oscillation.

Image Credit

Resonance graphs


Installing Kubuntu 15.10 as a guest on Hyper-V

Question: I downloaded Kubuntu 15.10 AMD 64 today and tried to install it on my Windows 10 device under Hyper-V.  THe system boots but when I am unable to get past the 2nd screen to Continue.  I click on back, or any other option in within the installer, but I cannot click on "Continue" to finish the installation. Any ides?
Answer: If you created a Generation 2 virtual machine in Hyper-V, you need to disable Secure Boot under Settings > Security > uncheck "Enable Secure Boot".
After that, it should be okay. Or, just use a Generation 1 virtual machine.
Comment: +1 for likely relevant answer; Quick googling also supports that Secure Boot need to be disabled for installing certain Linux distributions, including Ubuntu 64-bit.


Is there a better (permanent) per-app volume management than volume indicator/Pulse audio volume control?

Question: Because they both work but kind of cumbersome to access and it seems that if the application stops playing audio the volume would reset itself. That's very annoying when you want to lower the volume when you want to listen to an album on SoundCloud (its web player doesn't have a volume control widget), and the next track is unbearably loud again.
Answer: When you are playing audio via different applications, go to Sound option  from system settings and select tab Applications. Then you will be able to see volume settings for each applications that currently play sound. eg: vlc player, Alsa audio plugin etc. You can then set volume for each applications.
Hope this might help you.
Comment: I've tried that, but as I said that didn't help because the volume from SoundCloud would always reset itself when playing an album, but it turned out that I have to go to the Soundcloud album page instead of using their embeddable player, because the former has a volume control while the latter doesn't.


Radioactivity and half life

Question: Does the process of decaying in radioactive elements occur every second?
The equation consists of $\rm e$, so it must mean that decay due to radiation must occur every second, right?
Comment: Re, "every second." Nature doesn't know anything about seconds. Nor does it know anything about years or months or any other time scale we might choose to talk about radioactive half lives. Whatever is happening, it happens _continuously_.
Comment: Certainly many isotopes have half-lives much longer than a second, so it is not clear what you mean. Please clarify.
Comment: @JonCuster I am not talking about half lives ,if we take a radioactive element say strontium 90 ,it takes about 28 years to lose half of its quantity, that's ok but how does it do so,after 28.79 yrs approx does the half of it disappear into thin air(or) does it lose it's quantity every instant and after 28.79 years it happens to have half of its initial mass?
Comment: Decay is an ongoing statistical process.
Comment: @JonCuster pls elaborate
Answer: The process of radioactive decay occurs at every instant in which th substance exists. The rate at which the decay happens is proportional to the number of active particles at the given instant. It isn't stroboscopic. But if you solve the differential equation, you will get a fair approximation of how many radioactive particles remain after some amount of time.
Answer:
does the half of it disappear into thin air...?

Note! It doesn't disappear. It becomes something else. You gave, as an example, 90Sr. When at atom of 90Sr decays, it emits a beta particle, and it becomes an atom of 90Y (Yttrium).
Beta emission happens in a nucleus that has too many neutrons to be stable. One of the neutrons spontaneously becomes; a proton, an energetic electron (a.k.a., "beta particle"), and an antineutrino. Because the nucleus now has one more proton than previously, the atomic number goes up by one.
90Y has a half-life of only a few days, before it decays (again by beta emission) and it becomes a stable 90Zr (Zirconium) atom.
https://en.wikipedia.org/wiki/Decay_chain
Answer: When we talk about half-time we deal with a collection of many atoms (macroscopic quantity, i.e. of the order of Avogadro number). Every atom has a probability $p=\frac{1}{\tau}$ to decay per unit time, whereas the average number of atoms that have not decayed is given by $N_0e^{-t/\tau}$, where $N_0$ is the initial quantity of isotopes.
Answer: No it doesn't. It decays at a rate depending on its half-life. As @Jon Custer said its a statistical process and the nuclei will randomly decay in some period. The decay rate, decay per unit time is given by A=λN where A is decay rate, λ = decay constant and N is number of Nuclei in sample.
Answer: In order to understand radioactive decay, we have to see what is in the nucleus of an atom and how its parts interact.
As you may know, all nuclei, no matter what kind of atom (element), consist of protons and neutrons (nucleons). A proton is a positively charged heavy body, and a neutron is just slightly heavier than a proton and has no charge. The neutron can be thought of as a proton with an electron (or, more accurately, beta particle) bound to it by the weak nuclear force.
The protons of a nucleus repel one another very strongly through the electric force of repulsion between like-charged bodies. However, the attractive strong nuclear force is much stronger than the electric force at such small distances, and so the strong nuclear force overcomes the electrostatic (Coulomb) repulsion and holds the protons and neutrons together.
This union results in a ball of protons and neutrons (to a certain approximation) shaking back and forth violently within the nucleus, but held together with the strong nuclear force. Sometimes, the nucleus' (or, isotope's) configuration (shape) is always energetically "stable," and will never break apart no matter how much time passes - like an intact water balloon (so long as something strong enough doesn't come along and break it). Other nuclei (of a different isotope), every so often in this violent vibration, assume a shape that cannot be held by the volumetric and surface tension. These are called "unstable" nuclei, or "radioactive" nuclei. When this happens, a piece of the nucleus, a particle, breaks off. This is called "decay," or "radioactive decay."
When the nucleus decays, it does not disappear. It simply breaks into multiple pieces.
We cannot say, for any given "unstable" nucleus, exactly when it will assume a configuration leading to decay.
But, if we have a large number of nuclei (say, $N = 10^{23}$; around the number of atoms when amassed you can see with your naked eye) we can say approximately that the number of decaying nuclei at time t, dN(t), must be proportional to the number of nuclei present at time t, N(t).
Additionally, we can say that dN(t) should be proportional to the time that passes over a small enough duration, dt.
Note that in these proportionalities N is considered continuous rather than discrete. This is an approximation - or, really, an error - since we cannot really have a fraction of a radioactive particle, by definition.
Continuing, nonetheless, we have said $dN \propto N dt$.
Rearranging, and introducing a proportionality constant, $\lambda$ (the "decay constant"), we see
$\frac{dN(t)}{dt} = - \lambda N(t)$
The constant $\lambda$ is considered positive by definition, so the negative is introduced to capture that the change in the number of nuclei is negative over time period dt.
Rearranging the equation gives
$\frac{dN(t)}{N(t)} = -\lambda dt$
The solution to this equation through basic calculus is
$N(t) = N(t=0) e^{-\lambda t}$
This is where the "e" comes from.
Now, your question is "does decay happen every second?" The problem is, the question assumes there is a yes or no answer.
Also, it is useful to explain the half-life.
The half-life, by definition, is the time during which half of the nuclei in the sample should have decayed (become something else... not disappeared).
We can calculate this time by using the equation above as
$\frac{N(t_{1/2})}{N(t=0)} \equiv 1/2 = e^{-\lambda t_{1/2}}$
where $t_{1/2} = \text{half-life}$. Solving the right side, we get
$-ln(2) = -\lambda t_{1/2}$
or
$t_{1/2} = ln(2)/\lambda$
This shows the relationship between the half-life and the decay constant. Over the time $t_{1/2}$, half of the original nuclei will have decayed, leaving N(t=0)/2 nuclei in their original state.
After another half-life, half of the remaining undecayed nuclei will have decayed, with N(t=0)/4 remaining originals. Generally, after H half-lives, $N(t=0)/2^{H}$ nuclei will remain undecayed.
The problem with the above derivation, as mentioned, is that it calculates an average behavior of a large number of nuclei (or, more accurately, a proportion of total nuclei) that, as far as the equation is concerned, is continuous. This is called the "classical" approach.
In order to derive the actual decay of a number of nuclei N, we should start with the statistical representation, which correctly treats the number of nuclei as discrete, but - since we only know the probability that any given nucleus will decay in a certain duration - gives a probability distribution of final outcomes instead of a deterministic result. Therefore, the answer to your question is: "In each second, there is a probability P that decay will happen, and there is a probability (1-P) that decay will not happen." Of course, once all nuclei have decayed, the probability P is zero.
It is possible, albeit exceedingly improbable for large numbers of nuclei, that in a sample of radioactive nuclei, all of them will decay in the same short time dt. We can call this outcome #1. There is only one way that this outcome can be achieved. If the probability that a nucleus decays in time dt is p, then the probability it does not decay in the same time is (1 - p).
The probability of outcome #1 is $p^{N}$.
There are N ways in which outcome #2 occurs, where all but one nuclei decay in time dt. This means the probability of outcome #2 is $(\frac{N!}{(N-1)!})(1-p)p^{N-1} = N(1-p)p^{N-1}$.
Outcome #3 is that all but 2 nuclei decay in time dt. The probability of outcome #3 is $\frac{N!}{(N-2)!2!}(1-p)^{2}p^{N-2}$.
In general, Outcome #k, that all but k-1 nuclei decay in time dt has probability $\frac{N!}{(N-(k-1))!(k-1)!}(1-p)^{k-1}p^{N-(k-1)}$.
One of all of these N+1 outcomes must be fulfilled at the end of time dt, so the summation of all these probabilities is one.
It is useful for the purpose of answering your question to mention that Outcome #N+1 is that no decay occurs over the time period under consideration, and the probability of this event is $(1-p)^{N}$.
The irony here is that I've slipped a "classical" model of the nucleus itself into the discussion - we could call it the "liquid drop" model - when, in fact, the nucleus itself is just a superposition of a great number of possible configurations or accessible states based on its internal energy, and we can, in principle, count these number of possible states to arrive at the statistical, and correct, model of the nucleus as well.
We can also relate the two approaches. If we set dt = 1, then in the first equation this corresponds to t=1 (at this point, it doesn't matter what we choose for units, but we do know whatever units we choose are consistent with $\lambda$), leading to equation
$\frac{N(1)}{N(0)} = e^{-\lambda} = \frac{\text{Number of nuclei not decayed at time 1}}{\text{Number of nuclei at time 0}} = 1 - p$
where, admittedly there was a bit of a sleight of hand since p was for a single atom while N corresponds to a population of atoms; but, since p is derived from a population of atoms, it's okay.
Answer: What really happens
It really depends. If you have a moderate value of decay rate ($\sim 10^{10} \:\rm dps$), then the decay of multiple particles every second is inevitable. However, if the value of decay rate is small enough ($\sim 2 \: \rm dps$), then it's quite possible that you might find a particular second where none of the particles decayed. In the cases where the decays rate $≤1\:\rm dps$, it's highly likely to find such time intervals.
Why does it happen?
The reason why this is the way it is, is because of the quantum mechanical nature of nuclear decay. You can't really predict whether a certain particle will have definitely decayed by a certain time $t$, because of the quantum mechanical nature of the process of nuclear decay. However, the probability of particle being decayed by a time $t$ increases as the time $t$ increases, so eventually it will be highly probable that the particle would decay. However since we are dealing with probabilities, we can never be sure that a particle decayed.
If you're into quantum mechanics, you might also know that the state of a nucleus is a superposition of decayed and undecayed. And you can't know whether it decayed until you collapse its wavefunction by making a measurement, and checking if it decayed. (see the last section for the continuation of this line of thought)
How come we write a continuous fuction describing decay?
The function that we write can be regarded as "empirical", or more appropriately, an approximation of the reality rather than the absolute truth. Thus if you were to perform a nuclear decay experiment, it's highly likely that you would find measurements which are quite close (though not exact) to the mathematical equation of nuclear decay:
$$N(t)=N_0\mathrm e^{-\lambda t}\tag{1}$$
Now, no matter how many times you repeat the experiment, or how precise you make your experimental procedure, you'd always get results which are closer to equation $(1)$, but you'd never be able to do the experiment and get the exactly same results as predicted by the equation. And no, it's not because of the experimental errors that might have crept in, it's because of the uncertain and probabilistic nature of nuclear decay.
Many-Worlds interpretation
This is a bit of an off-shoot section, so it isn't strictly related to the question, but it's worth reading :-)
Now if you believe in the many-worlds interpretation of quantum mechanics, then you'd be fascinated to hear what it predicts during nuclear decay. It predicts that our universe branches multiple (gazillion times) during the process of nuclear decay, with every universe being different. In other words, whenever there are two possibilities for a nucleus, decay or not decay, our universe branches into two other universes, where in one of the universes, the nucleus has decayed while in the other one, it hasn't. And this is true for all the nuclei.
This line of logic implies an extremely mind-boggling result. It implies that there might be a universe where none of the particles have decayed in the first second of nuclear decay (despite the nuclear decay having a rate of $10^{10}\:\rm dps$). At first, this sounds to be unreal, thus thinking us into doubting the many-worlds interpretation.
However, if we think further, then we can notice that the probability of us being in such a universe is extremely low (I mean extremely extremely extremely low). Why? Because everytime a universe branches into two new universes, we have almost equal probability of being in one or the other. But to reach to a universe where no particles have decayed in the first second, you'd alwyas have to end up in the "undecayed" version of the universe, every single time, for $10^{10}$ branchings. And this is extremely unlikely, thus we never find ourselves conducting such an experiment. But do note that if we were to perform our experiment $2^{10^{10}}$ times, we might have a chance to witness that "special" decay where nothing decays in the first second.
Answer: The poster may be asking if radioactive decay is a continuous process or a discrete process.
It is a continuous process.  For any given increment in time, there is a probability that a certain amount of material will decay.  This probability is called the "decay constant" and usually denoted with the greek symbol "$\lambda$".
In the form of a differential equation, the "rate of change" of the nuclide concentration is proportional to the decay constant times the concentration itself.
$$
\frac{dN}{dt}=-\lambda \, N
$$
The solution of the differential equation makes it clear that it is a continuous process
$$
  N(t)=N(0) \, \exp(-\lambda t)
$$
The decay constant isn't a very intuitive variable since the units are "per time".  Therefore, the decay constant usually converted to a "half-life", which is more intuitive to understand because it has units of time.  You can use the previous equations to show that the half-life is related to the decay constant by the expression
$$
  t_{h} = \frac{\ln(2)}{\lambda}
$$
The confusion may be that the half-life has units of time (e.g. seconds), but it is still refers to a continuous process.

Radioactivity and half life


How would someone who needs to read a story from A to Z be qualified?

Question: When I'm beginning to watch a movie, or read a book, I tend to want to watch or read the whole story through, even if it is a story of questionable quality or if I know the end. And I don't want to miss a thing. What would be an adjective to qualify such a person like me?
Comment: Do you mean "describe" insread of "qualify"?
Answer: How about dedicated or devoted?
In gaming circles, the term completionist is quite popular.
Comment: I'm not sure dedicated/devoted fit best, but *completionist* definitely fits the bill. Not sure how well it translates outside of gaming circles. I think you'll find your *completionists* are one and the same with Jasper's *obsessives*! :)
Answer: The adjective to describe your not wanting to miss anything might be obsessive, since you are obsessed about reading every word of the book and watching every second of the movie.
Comment: You definitely hit the mark with _obsessive_, would there not be a word to describe someone with the _obsession_ to know how the event unfold in a story?
Comment: I think `obsessive`, despite the potential negative connotations, is the right word.
Answer: In the sense that you are determined to finish, even in the face of something that might be disappointing, I would suggest tenacious.
Comment: I think `tenacious` is going a bit too far perhaps, but I disagree with Eldros that it is a negative word - tenacity is an admirable quality in my book...
Comment: I think the word as too much a negative, like I don't want to do it, whereas I'm actually enjoying it in a way.
Comment: I don't think tenacious is negative. It's just that it describes a quality beyond the ordinary.
Answer: I'm like that too. I like to describe myself as obsessively curious.
Answer: I see this more in the case of multiple installments, but try "completist": "He's such a completist that he's still reading $series even though it hasn't really been good since book 13".
Answer: I would say somebody with determination, using determination with the meaning of firmness of purpose.
As alternative, I would use indefatigable, steadfast, or diligent.
Answer: The trait is a form of perfectionism, drawing on the root meaning of perfect as complete. The desire for completeness overshadows competing priorities of how to use one's time.


When did Rosh Hashanah begin?

Question: When did Rosh Hashanah begin?
According to a scientific observation, the sliver of moon was first sighted on Tuesday night.
My understanding is that Rosh Hashanah begins at the appearance of such.
Is it correct that Rosh Hashanah really began on Tuesday night?
Is it a mistake to celebrate a day too early?
Answer: In the days of the Talmud, witnesses would report to the Sanhedrin that they saw the new moon, and then the court would declare that the month started. Even if the court knew that the moon should be visible, if no witnesses came, they would wait until the next day to make it Rosh HaShanah or Rosh Chodesh.
When the sage Hillel Nesiah saw that there would no longer be a court, he instituted a calendar system on a nineteen year cycle, and declared the starts of the month in perpetuity. Therefore, we no longer rely on direct observation of the moon to decide our calendar, and it would be a mistake to celebrate a day early.
This is all discussed at length in Tractate Rosh Hashana and Maimonides' Laws of Kiddush Hachodesh. Specifically, he writes:

And this matter is a law of Moses from Sinai: That at a time that there is a Sanhedrin, we set [the new month] by sighting; and when the Sanhedrin does not exist, we set [it] according to this calculation that we [use] to calculate today, and we are not concerned about the sighting. Rather, sometimes the day we set with this calculation will be the day of sighting; [and sometimes it will be] a day before it, or a day after it.
Comment: @ninamag I see what you mean. The point is that the actual holidays and months are determined by the Court, and the time of the new moon is just a detail. Sometimes Rosh Chodesh can be a day after the moon, but that is its time.
Comment: @n-t, please put a clickable link for your source.
Comment: @n-t, you wrote, "it would be a mistake to celebrate a day early" - can you clarify this, please. We (all of us who follow the rabbinics) did "celebrate a day early", in that we celebrated before we "saw the new moon".

When did Rosh Hashanah begin?


Zero curl of magnetic field and Ampére's law

Question: Is it true that if I a have zero curl of the magnetic field, I will get a zero current density using Ampére's law? I know that on the surface of the conductor it doesn't have to be true (the current density is going to be infinite here), but is it true that using Ampére's law here is useless?
Answer: Note Maxwell-Ampere's law
$$ \frac{1}{\mu_0}\left(\vec\nabla \times \mathbf{B}\right) = \left(\mathbf{J}_{\rm enc} + \epsilon_0 \frac{\partial\mathbf{E}}{\partial t}\right). $$

$\mathbf{J}_{\rm enc}$ is the current density enclosed by a path.
$\epsilon_0 \frac{\partial\mathbf{E}}{\partial t}$ is the displacement current density, due to a time-varying, nonconservative electric field.

If ${\rm curl\,}\mathbf{B}$ is zero then the entire right hand side is zero also --- which means there is no net current density.
The question then begs whether the possibility that you have described can really exist. Because we know by Gauss' law applied to magnetism that
$$ \vec \nabla \cdot \mathbf{B} = 0,$$
we know all magnetic fields must curl. North poles must connect to south poles. Sure, magnetic fields can be locally uniform, but they must curl at some point. The only way that $\nabla \times \mathbf{B} = 0$ is if there is no magnetic field at all, if we are talking about the field at all locations in space.
Comment: "The only way that ∇×B=0 is if there is no magnetic field at all, if we are talking about the field at all locations in space." ... Not sure about that here is a case of a total magnetic field with zero curl  https://physics.stackexchange.com/questions/683724/conservative-magneto-static-field-on-a-current-carrying-wire  therefore a conservative magnetic field! But of course you could be right and this to be just an exception.
Answer: If you have a magnetic field which has curl 0 around some closed loop (not necessarily everywhere), i.e. $\nabla\times\vec{B}=0$ around some closed loop $C$, then it means that any surface $S$ for which $C$ is the boundary must have 0 net current passing through it (including displacement currents). This is due to Ampere's law and it is completely general. If you have a magnetic field which has curl 0 everywhere, then since the divergence of the magnetic field also has to be 0 everywhere, it implies that you have no magnetic field anywhere. This is due to the Helmholtz decomposition theorem - that any (sufficiently smooth and rapidly decaying) vector field can be decomposed into a curl-free (pure divergence) and a divergence-free (pure curl) component. No magnetic field anywhere implies no current (including displacement current) anywhere.
As the curl of the magnetic field is itself a field (i.e. has a value for all points in space), it is not clear what precisely you mean by "have zero curl of the magnetic field". Whether you mean you have 0 curl around some closed loop, or everywhere, or simply at a point.


How are penances decided?

Question: In the Catholic tradition, are penances after confession predefined? For example would lying equate to a standard set of penances that are applied universally by all priests, or are penances set at the discretion of the priest who is hearing confession?
Answer: Penances are set at the discretion of the priest. I think there are some very general guidelines, to prevent penances that are wildly disproportionate to the sins, but there's plenty of room for discretion within the guidelines.  One could get different penances for the same sins, not only from different priests, but even from the same priest on different occasions.
The priest can take into account not only the sins that one confesses but also other information that he happens to know, e.g., has this person confessed essentially the same sins to me before? He can also make the penance fit the time of the confession, e.g., giving a different penance during Lent than at Easter or Christmas, or even adapting the penance to a particular feast.
Answer: In addition to Andreas Blass' answer, in Catholicism a priest is also supposed to make penance for the sake of the penitent.

He must pray and do penance for his penitent, entrusting him to the
  Lord's mercy. (CCC 1466)

A priest could therefore also consider this and be willing to do majority of the penance for the sake of the penitent and give the penitent a "milder" penance.

How are penances decided?


Commas before and after "then"

Question: should a comma be used before "then"?

He went to the store then to the barber.

OR:

He went to the store, then to the barber.
Comment: The "rules" (such as they are) have very little to do with specific words (such as "then") but rather they deal with the structure of the sentence.
Comment: Say it out loud and listen for the comma. If you hear it, write it; otherwise not.
Comment: Yes, that works.  For very short sentences, skip the comma: He ate had pizza then dessert.
Comment: Also check out https://www.thepunctuationguide.com/
Answer: There are many grammatical rules for the use of commas. One of the best sources  I have seen can be found at the website below.
http://grammar.ccc.commnet.edu/grammar/commas.htm
Notwithstanding the article cautions that the use of a comma for "a pause in reading is not always a reliable reason to use a comma", I believe providing a cue to pause remains a very important consideration. In particular for someone reading aloud. Sadly, this is something often lost in modern day writing. The length of the sentence, I believe, is immaterial. Imagine a theatre/theater play where commas weren't used! In the example you provide, the use of a comma, in my opinion, is appropriate before the "then".
Answer: In the example you provided, both are acceptable but I would say that you should be aware that in some instances adding a comma may change the meaning conveyed.  A comma should also be used when listing - "We drove home, then parked the car and then finally went inside."
Answer: Of those two examples, the first would be preferred by most style guides: "He went to the store then to the barber."
You would only use a comma before a "then" used for sequencing multiple events as in your example in the following cases:

There are more than two events, such that it's a list with three or more items: "He went to the store, purchased what he needed, then headed to the barber."

The "then" is functioning as a coordinating conjunction (like an "and") with two independent clauses: "He went to the store, then he made his way to the barber." (note the additional words that make the part after "then" an independent clause). Some style guides reject this usage of "then", only allowing the FANBOYS to be coordinating conjunctions (for, and, nor, but, or, yet, and so), but this is, nevertheless, a fairly common usage.

This last reason is the most subjective. If you need a pause there for dramatic effect, to force the reader to pause on the first part before taking in the second, then, depending on the style guide, a comma may be appropriate. "He caught his reflection at the store, then headed straight to the barber." In this case, the comma is ONLY appropriate if you want to force a pause. It is not grammatically required. By default, you would leave it out because the part after "then" is not an independent clause.
Answer: This:
How, then, should one deal with such matters?
The above, "We drove home, then, etc." is incorrect, or maybe just a bad example. The sentence doesn't require "then." It would simply read, "We drove home, parked the car, and went inside."
Answer: A comma should be placed before and after "then".  From the example:  "We drove home, then, parked the car, and then, finally went inside."

Commas before and after "then"


Einstein-Cartan theory - Book recommendation

Question: I want to delve into Cartan's formulation of General Relativity but it don't really know where to look. Does anyone have any book suggestions?
Comment: Possible duplicates: https://physics.stackexchange.com/q/270311/2451
Answer: Friedrich Hehl has written several pedagogical reviews on gauge aspects of gravity. I recommend the following:
On the Gauge Aspects of Gravity
F. Gronwald, F.W. Hehl
International School of Cosmology and Gravitation 14 (1996): 148-198.
https://arxiv.org/abs/gr-qc/9602013
Answer: IMHO, the most student-friendly exposition of Einstein-Cartan gravity is A. Zee's fantastic book:  Einstein Gravity in a Nutshell.

Chapter IX7 gives an introduction to differential forms and
vielbeins/tetrads.

Chapter IX8 guides you through step-by-step calculations of the
expanding universe, spherically symmetric static spacetime, etc. using tetrads and spin-connections. You usually won't find such detailed example calculations with Einstein-Cartan gravity in other GR books.

Like his other nutshell books, A. Zee's style is casual and to the point. It's always fun to read.


Do cats have a preference of being in the light or being in the dark?

Question: From time to time I have to put my very energetic cat in my bedroom for an hour or so while I'm cleaning, or working on something I can't have her messing with. In the evenings if the lights are off, it's very dark in the bedroom. Do you think my cat cares whether the light is on or off?
Comment: While the question is written asking for opinions, it is reasonable to assume that non-opinion based answers can be provided.  The question should not be closed.
Comment: As long as the light schedule follows what a normal day would look like. For example, if it's dark inside the bedroom but bright outside, you should really turn the lights on.
Answer: Cats are crepuscular, which means they prefer to be awake at dawn and dusk.  As such they've evolved to prefer neither complete darkness or bright light.
That said, human sight is also suppose to be best at dawn and dusk as well.  However, cats proportionally have more rods than cones in their eyes than we do.  (Rods are what give us are ability to see in very low light and their placement in our eyes is why we can see better out of our peripheral vision in very low light.)  Thus cats likely prefer a lower light levels than we do, and are able to see better in darker environments.
Comment: Related [Diurnal, crepuscular, nocturnal, matutinal, vespertine; what do these mean and how do they impact my pet's life?](http://pets.stackexchange.com/questions/5806)


How do I determine the half-life from a change in the activity of a sample over a known time?

Question: If I know that the sample originally has an activity of $A_0$ and after a time $t$ has an activity of $A_1$, how can I use this to determine the half-life of the material?
Comment: What would the half-life be (in terms of $t$) if $A_1=A_0/4$ ?
Comment: HL = t / 2... so half life is just $2t \times \frac{A_1}{A_0}$?
Comment: $t/2$ is correct for my example, but your equation isn't right. Eg, if $A_1=A_0/8$ then the half-life is $t/3$. Think about exponents and logarithms...
Answer: When you are aiming at the half-life $t_{1/2}$ it is actually easier to write the decay law
with power of $2$, rather than with power of $e$.
$$A(t)=A_0 2^{-t/t_{1/2}}$$
or
$$\frac{A(t)}{A_0}= 2^{-t/t_{1/2}}$$
Now take the logarithm ($\ln$) on both sides (remember $\ln(a^b)=b\ln(a)$):
$$\ln\frac{A(t)}{A_0}=-\frac{t}{t_{1/2}} \ln(2)$$
Then it is easy to solve this for $t_{1/2}$.
Answer: In the general case it is not a simple question. But for the simplest case it is quite simple.
For a single radioactive species that decays into a stable species, and has only one decay path, you get a simple equation. Given that the activity is $A_0$ at $t=0$ the activity at other times is given by the following.
$$ A(t) = A_0 \exp\left(-\frac{t\ln(2)}{t_{1/2}}\right) $$
Here $t_{1/2}$ is the half life. So at $t=t_{1/2}$ you have just half the activity you started with. So if you are given the activity at $t=0$ (that's the $A_0$) and at some measurement time $t_m$, you have to solve this equation for $t_{1/2}$.
It's a lot more fun if you have multiple decay chains so that you have multiple species of radioactive materials with different half lives. A fairly simple one is I-135 decaying to Xe-135 (a Beta decay with a half life of about 6.57 hours) and then Xe-135 decaying to Cs-135 with a half life of about 9.14 hours. The Cs-135 has a half life of 2.3 million years, so you can treat it as stable for most purposes. So, when you get comfortable with the previous equation, try writing down the equation for activity of this combined system. This is important in nuclear reactors because fission produces (among a bunch of other stuff) both I-135 and Xe-135, and Xe-135 is a strong neutron absorber.


sending variable from one jsp to another jsp

Question: I have one JSP file as jsp 1.jsp
and another JSP file as jsp 2.jsp
I've included jsp 2.jsp in jsp 1.jsp using <%@include file="jsp 2.jsp" %>
Now I need a click event on some element. And on that event I want to transfer a string variable to included jsp.
Lets say I have a list and on click of it I want to transfer the name of the list to another JSP,
And in another JSP I am trying to use that string to carry out some task.
And I am doing all these without any servlet.
challenging one!!
I have google'd a lot, but didnt find anything.
Comment: put a hidden field . You will get this value on server side with request.getParameter("stringVar");
Comment: @KrsnaChaitanya can you explain it more...thanksss... :)
Answer: You have a number of options:

Store it in the session.
// Memorise any passed in user.
String username = request.getParameter("username");
if (username != null && username.length() > 0) {
  session.setAttribute("username", username);
}

Store it as a hidden field in the form.
<input name="username" type="hidden" value=""/>

Store it in a cookie.
username = getCookie(userCookieName);

// Get from cookie.
function getCookie(name) {
  if (document.cookie) {
    index = document.cookie.indexOf(name);
    if (index !== -1) {
      f = (document.cookie.indexOf("=", index) + 1);
      t = document.cookie.indexOf(";", index);
      if (t === -1) {
        t = document.cookie.length;
      }
      return(document.cookie.substring(f, t));
    }
  }
  return ("");
}

Persist it on the client side in sessionStorage. See here for details.
sessionStorage.setItem("username", "...");

Not really another option but a mechanism - pass it in the URL:
.... onclick="window.location='details.jsp?username=...'

Answer: If the reason for including your jsp 2 in jsp 1 is for sending variable, then its not needed.
you just need to set a hidden variable in jsp 1, and in your jsp 2 you should able to access it, by using the jsp implicit request object.
Answer: Try to use Ajax with JQuery to achieve your goal or you can simply store the String object in session and you can delete it in jsp2.jsp when ever you use it. or you can simply append to a url as query string there are multiple ways among them the above are few.
These questions may help you
Question 1
Question 2
Question 3
Question 4
Answer: in jsp1
use this way to pass value.
<form action="">
<input type="hidden" name="hidden" value="hidden">
<input type="submit" value="submit"></form>

in jsp2 to get value
String value=request.getParameter("hidden");

Answer: You can set a jsp:param, as in this example jsp:param element or take a look at this similar post Passing parameters between JSP pages.


Run viewer command works only from master tex But not from child tex file

Question: i am using Auctex under emacs 24.3.1 on mac.   I am making  master .tex file and many what i call it child .tex files. Run Latex and Run bibtex  commands are working nicely from either  the master file or from child files but i can not view the output except from the master file.
I would appreciate very much your answer(s). Thanks in advance.
Ayman
Comment: Welcome to TeX.SX! You can have a look at [our starter guide](http://meta.tex.stackexchange.com/q/1436) to familiarize yourself further with our format. Did you set `TeX-master` appropriately in the child file?  (I would set it as a file-local variable.)
Answer: Add the following at the very end of each of your 'child' files:
% Local Variables:
% TeX-master: "./my-master-file.tex"
% End:

If you already have a Local Variables block, just add the appropriate line.
Close and re-open your file; this should set the variable.
(You can also call M-x normal-mode if you don't want to close the file.)
Comment: To reload a file there is `M-x revert-buffer` and I would add the suggestion to insert
`(setq TeX-auto-save t)
(setq TeX-parse-self t)
(setq-default TeX-master nil)`
 in the `.emacs` file.
Answer: I believe i found out what the problem was. It is the configuration of skim and latexmk in the .emacs file. the perfect setting that is really works is as following
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;
;; Skim's displayline is used for forward search (from .tex to .pdf)
;; option -b highlights the current line
;; option -g opens Skim in the background
;; option -o open Skim in the foreground with full application focus.

'(LaTeX-command "latex -synctex=1")
(require 'tex-site)
(add-hook 'TeX-mode-hook
    (lambda ()
        (add-to-list 'TeX-output-view-style
            '("^pdf$" "."
              "/Applications/Skim.app/Contents/SharedSupport/displayline -b %n %o %b")))
)
(server-start)

(setq TeX-PDF-mode t)

(setq TeX-source-correlate-mode t)

(setq TeX-source-correlate-method 'synctex)
(add-hook 'LaTeX-mode-hook 'TeX-source-correlate-mode)

;; make latexmk available via C-c C-c
;; Note: SyncTeX is setup via ~/.latexmkrc (see below)
(add-hook 'LaTeX-mode-hook (lambda ()
  (push
    '("latexmk" "latexmk -pdf %s" TeX-run-TeX nil t
      :help "Run latexmk on file")
    TeX-command-list)))
(add-hook 'TeX-mode-hook '(lambda () (setq TeX-command-default "latexmk")))

;; use Skim as default pdf viewer
;; Skim's displayline is used for forward search (from .tex to .pdf)
;; option -b highlights the current line; option -g opens Skim in the background
(setq TeX-view-program-selection '((output-pdf "PDF Viewer")))
(setq TeX-view-program-list
     '(("PDF Viewer" "/Applications/Skim.app/Contents/SharedSupport/displayline -b -g %n %o %b")))

(server-start); start emacs in server mode so that skim can talk to it

(add-hook 'LaTeX-mode-hook
          (lambda () (local-set-key (kbd "<S-s-mouse-1>") #'TeX-view))
          )
;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;;

and of course, in the skim you have to do Preferences>Sync>Check for file changes and Preset>Emacs


Terraria: Desktop and PS4 Cross Platform

Question: I have the Desktop version of terraria and my boyfriend has terraria on PS4. Is it possible for us to play multiplayer.
Answer: Sadly no there isn't any cross platform at the moment, but they are trying to get all platforms on the same update so they can do cross platform play. currently its Android With Iphone, Playstation With Xbox and pc with pc. :P
Answer: There is no cross play between PC and console as of right now and with PC getting the last ever update there is a very small chance of there ever being cross play.
Looking it up though it looks like all consoles can play together but I couldn't find a certain answer on that.


Can someone explain to me the Zipf–Mandelbrot law?

Question: I really can't understand. It's about linguistics and I can't understand anything because there are mathematical formulas in it that I can't understand at all. Can anyone explain this with hypothetical examples? Thanks in advance!
Answer: As this is the Linguistics site, I'm guessing it is Zipf's Law you are interested in? If you analyze a corpus of text, and count the number of times each word occurs, you get their frequency. You then line then up on the x-axis of a graph, in order of rank (i.e. the most common word comes at x=1, the second most common word at x=2, and the least common word all the way over on the right at x=N), and their frequency on the y-axis.
Zipf's Law just says that the frequency is inversely proportional to their rank. So, if the most common word occurred 1000 times, you would expect the second-most common word to occur 500 times, the 3rd to occur 333 times, the 4th-ranked word to occur 250 times. And over on the right you would expect all the words from rank 667 (*1) to rank 2000 (*2) to occur exactly once.
*1: 1000/666 is 1.5015 so would round up to 2, and you expect it to occur twice.
*2: 1000/2001 is 0.4998 so would round down to 0; that means you expect only 2000 words to be used in your corpus.
Obviously a real corpus won't exactly match, but you could reasonably expect it to have roughly the same shaped curve. If not, there may be something unusual or artificial about your corpus. E.g. A glossary or index would not follow the curve very well.
Zipf-Mandelbrot is a generalization of Zipf's Law, adding some parameters so a single law can be used to describe a range of probability distributions.
Comment: @fdb: The last paragraph is correct, ZIpf-Mandelbrot has an additional free parameter q which is fixed to 1 in the original formulation of Zipf's law. Zipf's law also requires a finite N whereas Zipf-Mandelbrot is not resticted to a finite number of types. So the "generalisation" part is true.
Comment: Felix Auerbach, Das Gesetz der Bevölkerungskonzentration. in: Petermanns Geogr. Mitteilungen, 59, pp. 73-76, 1913
Comment: I cannot agree with your last paragraph. Mandelbrot's law is not a "generalisation" of Zipf. Zipf's law is an application to corpus linguistics of a mathematical principle ("power relation") which had already been observed by Auerbach and Pareto, and further developed by Mandelbrot.
Comment: @jk-ReinstateMonica. The mathematical formulation goes back to Auerbach. Zipf did nothing new except apply it to word frequency. The obfuscation is due to the fact that Zipf (an American Nazi) never credited anything to Auerbach (who was a Jew).


Allpass Filter with Sign Switch

Question: I want to design a digital allpass IIR filter with the following transfer function.
$$
H(\omega) =
1 \textrm{ for } \omega < \omega_1 \textrm{ or } \omega > \omega_2
$$
and
$$
H(\omega) =
-1 \textrm{ for } \omega > \omega_1 \textrm{ and } \omega < \omega_2
$$
I assume that higher-order filters are needed for a steeper transition band. Can you help me design a filter that approximates this specification? I prefer a solution that strictly uses allpass digital filters.
Comment: I have two questions. What's the purpose of this filter? And what do you mean by "a solution that strictly uses allpass digital filters"?
Comment: @MattL. 1) I have a symmetric filter F with a real-valued frequency response, including zero-crossings at $\omega_1$ and $\omega_2$. The magnitude of F is then $|F| = F * H$. 2) So, the allpass H(z) = b(z)/a(z), where $z^{-m} conj(b(1/z)) = a(z)$. The filter can be acausal.
Answer: In this answer I'll just focus on the given design problem, even though I'm not sure if there might be a better overall solution to your underlying problem.
First of all it's instructive to realize that the given filter specification is just the difference of two (ideal) lowpass filters plus a constant. Let $h_1[n]$ be an ideal lowpass filter with cut-off frequency $\omega_1$, and $h_2[n]$ is an ideal lowpass with cut-off $\omega_2$, then the desired filter can be expressed as
$$h_d[n]=2\big(h_1[n]-h_2[n]\big) + \delta[n]\tag{1}$$
Eq. $(1)$ could be used to design an FIR filter approximating the given desired response. We would just need two odd-length linear-phase lowpass filters designed by any of the many well-known methods. We would need odd-length filters because they have an integer delay, and we could add the constant by simply replacing $\delta[n]$ in Eq. $(1)$ by $\delta[n-M]$, where $M$ is the delay of the linear phase filters. Note that the resulting FIR filter would implement the desired phase response exactly (plus a delay), but the magnitude response would only be an approximation to the desired unity gain response.
Eq. $(1)$ wouldn't help for designing an IIR filter because of the different non-linear phase responses of the individual lowpass filters. Unlike FIR filters, IIR filters can have an exactly constant magnitude response. However, the desired phase can only be approximated. If we require the filter to be IIR, we need to resort to numerical design methods.
One simple and effective method is the equation error method, which I explain in this blog post. There's also a link to a simple Matlab/Octave implementation: iir_ap.m.
The designed IIR filter needs to be causal and stable, otherwise there's no way to implement it. For this reason we need to add a linear phase term to the desired phase response. This just introduces some extra delay. The amount of delay necessary to make the resulting filter causal and stable must be determined experimentally.
I've designed two filters, one $10$th order and one $20$th order IIR allpass:
<code>L = 500;
w = pi * linspace(0,1,L); w = w(:);
w1 = .2 * pi; w2 = .4 * pi;
I = find( w > w1 & w < w2 );
P = zeros(L,1);
P(I) = pi;
W = ones(L,1);

% 10th order allpass
N = 10;
tau10 = 8;
P2 = P - w * tau10;
a = iir_ap( N, w/pi, P2, W );
[A10, ww] = freqz( flipud(a), a, 2^12);

% 20th order allpass
N = 20;
tau20 = 18;
P2 = P - w * tau20;
a = iir_ap( N, w/pi, P2, W );
[A20, ww] = freqz( flipud(a), a, 2^12);
</code>
The figure below shows the design results. For the plots I removed the linear phase term. Note that the maximum phase error is not improved by choosing a higher filter order. What is improved by using a higher filter order is the steepness at discontinuities of the desired phase.
Comment: @Jiro: I didn't assume any requirements on the delays. It's no problem to change them to integers. I changed them in my answer, and the result is virtually the same.
Comment: That's an excellent solution. Works very well.
Comment: Is there a reason why you don't choose an integer delay tau? It would make aligning signals easier, isn't it?

Allpass Filter with Sign Switch


How to prove that there will be 0.5n 1s and 0s in the typical sequence for a $i.i.d$ source with 0.5 probability?

Question: I'm struggling to see how to prove that a typical sequence will constitute approximately 0.5n s if it is generated by $i.i.d$ source with P(0)=0.5 and P(1)=0.5. Could we prove this from the definition of a typical set?
Comment: You're going to have to define what you intend by 'typical' here, because typically there will be an unequal number of 0's and 1's, and the counts will diverge as n increases.
Answer: When $P(0)=P(1)=1/2$, by definition all binary sequences are in the typical set. Since they may contain any number of successes, I don't believe we can come up with an approximation argument regarding it. However, we can talk about the mean and other relevant statistics on the number of success, which won't differ from discussions made on Binomial RV.


How to generically forward an URL from one toplevel domain to another?

Question: if I own a domainname myweb.com with corresponding hosting, and I also own a domain myweb.net (but no webhosting is setup for this domain), is there a way that I can generically forward myweb.net/anyURL to myweb.com/anyURL, and if so, how can this be done?
Answer: It sounds like what you want to do is domain mirroring, or do you just want to forward from one URL to another?
If you're looking to forward domains, you would want to look at the DNS settings where you registered your domain, or look to see if you have a control panel where you can manage domains and redirections.
If you just want to forward to an external link, you can do this through a plugin, or using an HTML redirection: https://stackoverflow.com/questions/5411538/how-to-redirect-from-an-html-page
Answer: Redirecting a domain name to another (including deep links redirecting to deep links) is a very common setup.  In fact most redirect rules that you can implement redirect deep links to deep links by default.  It is generally harder to set up a redirect that redirects everything to the front page of another site.

Most domain name registrars offer this service free of charge.  For example GoDaddy calls this "Forwarding" and it is available under the "Domain Settings" for the domain.  I just checked one of my domains that is set up this way and confirmed that deep links do in fact forward to deep links on the other domain.

If your web host uses cPanel to allow you to set up your domains, the feature is also called forwarding.  See the cPanel documentation on forwarding.

If you have an Apache virtual host set up for the domain that you want to redirect, you can edit your .htaccess or httpd.conf file and put in a redirect rule.   Apache redirect rules behave like you want, they forward deep links to deep links.
Redirect permanent / http://example.com/

Finally, if you have both domains being served from a single directory on the server, you can use rewrite rules to redirect one domain to the other.  This is the only case in which you have to do something special to get the deep links.  The (.*) is a capturing group that is put onto the URL with the $1:
RewriteEngine On
RewriteCond %{HTTP_HOST} ^example.org$ [NC]
RewriteRule ^(.*)$ http://example.com/$1 [L,R=301]

Or alternately, redirect anything that isn't example.com to example.com (which handles subdomains and multiple redirect domains in one command):
RewriteEngine On
RewriteCond %{HTTP_HOST} !^example.com$ [NC]
RewriteRule ^(.*)$ http://example.com/$1 [L,R=301]

Comment: What if you want to redirect using a mapping file? For example, domain a has spammy pages with some value but you want to redirect to a new, cleaner domain because, among other things, domain a is a spammy url.spammydomainname.com/stuffy-seo-url-variation1
spammydomainname.com/stuffy-seo-url-variation2
spammydomainname.com/stuffy-seo-url-variation3

would all redirect to brandeddomain.com/clean-topical-url

The same would happen throughout the site for various topics and content.

Would implementation vary depending on systems used? These are on Wordpress, Apache, using Network Solutions.
Comment: You could ask that as a new question.   The answer is too long for a comment.


Add items in Sharepoint Online modern UI list using API

Question: I need to add items into custom list which is into SharePoint Online modern UI.
 I tried inserting the data using ajax call but it is not working in it although it is working in Classic UI. Please suggest how can we add items into Sharepoint custom list using client side scripting
Comment: So what is the code you are using?
Comment: @Ciamas : ajax call using Jquery
Comment: Still don't know anything about your code and it would simply be best to post it. From the REST perspective there is no difference between the UI's of SharePoint. What is the error you get?
Comment: Which browser you are looking into? How many items do the SharePoint list have?
Answer: We can add the SharePoint list item using REST API's, following is the code,
url: http://site url/_api/web/lists/GetByTitle('Test')/items
method: POST
body: { '__metadata': { 'type': 'SP.Data.TestListItem' }, 'Title': 'Test'}
headers:
    Authorization: "Bearer " + accessToken
     X-RequestDigest: $("#__REQUESTDIGEST").val()
    accept: "application/json;odata=verbose"
    content-type: "application/json;odata=verbose"
    content-length:length of post body

You can follow the link,
https://docs.microsoft.com/en-us/sharepoint/dev/sp-add-ins/working-with-lists-and-list-items-with-rest
Comment: $("#__REQUESTDIGEST").val() returns null in modern UI
Answer: If you are using the SharePoint Framework to develop and create custom code WebParts, you can (and should) use the context provided in the parent WebPart component this.context and send it to your Root WebPart Component (the one rendered) through its properties.
Now, your component has the context and the capabilities to use the provided SPHttpClient which does the authentication and hardwork for you : this.props.context.spHttpClient. with the methods get(), post() and fetch().
Now if you do not use the SharePoint Framework, you find yourself in a SharePoint Page where you cannot retrieve a Request Digest anymore (and you need one to push data). You must therefore ask for one first, and use it in a X-RequestDigest header in your post requests.
Office365 App Page doesn't have RequestDigest
The method is to make a first POST to <sitecollection>/_api/contextinfo.
Hope this helps!


What is fundamental frequency, how does it make sense?

Question: I'm currently going through harmonics, and I do not at all understand the fundamental frequency. I understand that it is the simplest vibration of a string, but I don't understand how can it have frequency at all if it is only half a wavelength. Isn't frequency how many cycles are completed per second, and isn't the fundamental frequency only half a cycle if it is half a wavelength? How can there be frequency of (say) 162 cycles per second if one cycle doesn't even complete in the medium of the string? Is it measuring the frequency of the half wavelength as a full cycle? Is there frequency measured as a whole cycle from one half because it is a resultant wave of two waves making the half wavelength? If so, why do we have to multiply by two to get the wavelength of the cycle from the string length?
Comment: I think there's terminology confusion here.  The simplest vibration of a string fixed at both ends is the fundamental *mode*.  Yes, the spatial variation along the length of the string, e.g., $\phi(x) = \sin(\pi x/L)$ is a 'half-cycle' but the fundamental *frequency* refers to time dependent amplitude, e.g., $A(t) = A_0\cos(\omega_0 t)$ where $\omega_0 = \frac{\pi}{L}\sqrt{T/\rho}$.  So, the *amplitude* of the mode completes $\omega_0/2\pi$ cycles per second.  See the animated gif at [this answer](https://physics.stackexchange.com/a/160544/9887)
Comment: Get a slinky or something and watch its oscillations.
Comment: Yes but how is half a wavelength one cycle? How is there a frequency made from that half a cycle in the medium? I understand one wavelength=one cycle
Comment: Or play with https://phet.colorado.edu/sims/html/wave-on-a-string/latest/wave-on-a-string_en.html . (But nothing is as enlightening as a Slinky.)
Comment: Thank you for the simulation, but that still helps me very little with understanding the why with standing waves frequency.
Comment: What I'm confused about is whether one loop, half a wavelength, since such is a resultant wave is a complete cycle. Because I understand that cycles have to pass to measure frequency, and fundamental frequency has a frequency of some cycles per the smallest frequency the medium can have, but are these cycles completed in one loop? And is the harmonic series placing another and another cycle to a cycle? Is it simply folding out the resultant wave into a traveling wave to find wavelength? Does each loop in a standard wave pattern represent a cycle. I am so sorry for so many questions!
Answer:
Isn't frequency how many cycles are completed per second, and isn't
  the fundamental frequency only half a cycle

When a string, fixed a both ends, vibrates in the fundamental mode, the perpendicular displacement $\phi_1(x,t)$ of a point located at $x$ along the length of the string is given by
$$\phi_1(x,t) = A_1(t)\phi_1(x) = A_1\cos(2\pi f_1t + \varphi)\sin\left(\frac{\pi}{L}x\right)$$
Now, it is true that the spatial variation of the fundamental mode is a 'half-cycle' since the argument of the $\sin$ ranges from $0$ to $\pi$.
However, the fundamental frequency refers to the time dependent amplitude $A_1(t)$.  Note that $A_1(t)$ executes $f_1$ cycles per second.  Take a look at this animated gif of the fundamental mode and the first three harmonics:

Animated gif credit
See that although the fundamental mode has a 'half-cycle' spatial variation, the time dependent amplitude goes from a maximum, through zero, a minimum, back through zero back to the maximum in a time $T_1 = \frac{1}{f_1}$ where $f_1$ is the fundamental frequency.
Also note that the frequency of the 2nd harmonic is twice the frequency of the fundamental, the frequency of the 3rd harmonic is thrice the frequency of the fundamental and so on.
Comment: Ohh, okay, so, the period of this wave is not the spatial variation, half a wavelength, but depending on the amplitude going up and down as it were, and that is the period, and therefore the cycle, and so the amount of up and down amplitude movements that equals a cycle would make up a frequency of say 162 cycles per second, but in terms of proclaiming the wavelength of such we treat it as a sine wave, as though we were folding it out. As in the gif, the first would have a lower frequency, the second, frequency is doubled as the cycles are being completed by twice as many in a second, so on.
Answer: A standing wave on a string can be thought of as a traveling wave that is bouncing back & forth in one dimension.  Every time it reaches one of the ends, it is reflected, either inverted or upright depending on what conditions you have at the end.  Assuming you have the same boundary conditions at both ends, this means that by the time the traveling wave makes a full cycle, it will be upright:  either it got inverted twice, or it never got inverted.
However, by the time the wave has made a round trip, the "first peak" of the wave may not be in phase with the later peaks any more.  If it is a little bit out of phase, then after it makes another round trip, it will be even more out of phase;  and after many round trips, it will be even more out of phase (and so will the second peak, and the third, and....)  All of these out-of-phase waves will exhibit destructive interference, and cancel each other out.  So you can't have a standing wave if this is the case.
But there's a way around this:  suppose that the time for the wave to make one round-trip is exactly equal to the period of the wave.  In this case, the "first peak" lines up with the second peak, and you get constructive interference between the first peak and the second peak.  Thus, you get a standing wave.
But notice that you needed the round-trip time to be equal to the period.  This means that the distance traveled by the first peak in one cycle must be twice the length of the string.  Since the distance traveled by a peak of a wave in one cycle is equal to the wavelength, this means that the wavelength of the corresponding traveling wave is double the length of the string.  In other words, one cycle of the wave has to fit into the "round-trip" distance, not the length of the string.
You can generalize this, by the way:  the higher harmonics can be thought of as traveling waves that make a round trip in $n$ cycles of the wave, so that the first peak lines up with the $(n+1)$th peak after it makes a round trip.
Comment: Thank you for the very detailed answer! So is a standing wave, half a wavelength, technically a "round trip" of a wave and therefore one cycle, but the standing waves corresponding wavelength would be twice that of the string for the fundamental frequency. And then the wavelength decreases by making that round trip happen once again, second harmonic and one loop is still one round trip so there are now two cycles in second harmonic and the wavelength is equal to the length of the two cycles loops?
Answer: From the many questions you make, it seems that you don't understand the difference between frequency and wavelength. Frequency is inversely proportional to wavelength.  This is expressed by the equation $f = \frac{v}{\lambda}$.  (v = speed of sound)
Also, for a string, $\lambda$ = 2l.  (l = length of string)
Some examples might be useful:
For sound, v =343 m/s, so for a string with a wavelength of 1 m, f would be 343 cps.  However, since for a string $l = \frac{\lambda}{2}$, the length (l) of the string must be 0.5 m long!
If the length of the string is 1 m, then $\lambda$ = 2l = 2 m, and the frequency would be f = 343/2 = 171.5 cps.
Keep in mind that whatever the string length is, it only represents 1/2 of the wavelength.  So, if you want the length of a string that would vibrate at 162 cps, the answer would be $l = \frac{\lambda}{2} = \frac{v}{2f}$ = 343/2x162 = 1.058 m.


Is there a reason shift+' writes " ' instead of just "?

Question: I was writing something about using Moebius inversion, and to get the correct symbol I write \"{o} to get the umlaut over the o. However, in TeXnicCenter, shift+' automatically writes "' instead of the desired ". Is there a reason for this? I always have to delete the extra apostrophe each time.
Answer: This is a setting of TeXnicCenter (v1.0 Stable RC 1) that can be activated/deactivated. Under Tools -> Options -> General -> Automatic replacement you need to deactivate Replace quotation marks:
Comment: Or set it to something else.
Comment: If you want to keep the auto-replacement, take a look at this question: http://tex.stackexchange.com/questions/15483/how-to-get-in-texniccenter-with-auto-replacement-for-quotation-marks-enabled
Comment: Since @yunone asked for the reason for this behavior as well: This is an auto-replacement feature for quotation marks, in your case apparently enabled for German quotation marks. Quotation marks need to be masked specially in LaTeX: `"\`Zitat"'` (German) / `\`\`Quote''` (English). For your convenience, TeXnicCenter offers a feature where a typed `"` is automatically replaced with either opening or closing quotation marks, with options available for a few languages or your own choice.


"Is" vs. "Are" when using (s)

Question: When using (s), should "is" or "are" be used?
Regardless of what option(s) is decided.
or
Regardless of what option(s) are decided.
Answer: Although the pattern of writing a word like option(s) is called paranthetic plurals, the word is both singular and plural.
If the usage of the word in the sentence is extremely limited (like in the example you provided) you can use it like
Regardless of what option(s) is (are) decided

If the usage becomes more complex, it's better to stick with one of the forms throughout.
Although in some cases, like a legal document, where precision is of utmost importance, you need to mention both the forms either through option(s) style or option \ options style.
From Chicago Manual of Style :

A term ending in "(s)" is both plural and singular. If you must use such a device (and it can be a useful shorthand), you have to be prepared to adjust the surrounding context as necessary: for example, "the award(s) is (are) accounted for." A parenthetical plural verb must correspond to the parenthetical ending. But that's an awkward example. In general, avoid such shorthand unless it can be used simply and effectively, as in the following example:
Place an "about the author(s)" statement on the copyright page (usually page iv).

"Is" vs. "Are" when using (s)


Intuition for why the fourier tranform of a polynomial is its pointwise evaluation?

Question: Polynomials in coefficient representation can be multiplied in $O(n \, log \,n)$ time by using a fast fourier transform to convolute the coefficients. The DFT of the coefficients correspond to the evaluation at the complex roots of unity. Is there an Intuition to this fact? I have previously only known fourier transforms as transforming a signal from the time domain to the frequency domain, so this connection is very strange to me.
Answer: Let
$$
p(x) = \sum_{n=0}^{N} a_n x^n.
$$
Now consider setting $x$ to each of the $N$ complex roots of unity. For $x=\exp\left({j2\pi/N\cdot 0}\right)$, we get
$$
p(e^{j2\pi/N\cdot 0}) = \sum_{n=0}^{N} a_n \left( e^{j2\pi/N\cdot 0 } \right)^n.
$$
For $x=\exp\left({j2\pi/N\cdot 1}\right)$, we get
$$
p(e^{j2\pi/N\cdot 1}) = \sum_{n=0}^{N} a_n \left( e^{j2\pi/N\cdot 1 } \right)^n.
$$
In general, the polynomial evaluated at the $m^\mathrm{th}$ complex root of unity is
$$
p(e^{j2\pi/N\cdot m}) = \sum_{n=0}^{N} a_n \left( e^{j2\pi/N\cdot m } \right)^n = \sum_{n=0}^{N} a_n e^{j2\pi n m /N }
$$
Note that the right hand side of this is the definition of the DFT of the sequence $a_n$.
Comment: I think in the last equality, in the middle part, you forgot to write the "power to n".


How historically accurate is the TV series 'The Tudors'?

Question: The TV series The Tudors (2008-2010) depicts the reign of Henry VIII. So far I've not noticed any historical inaccuracies. How truthful to history is it?
Comment: @coleopterist negative vote for spell mistake is little rude. Mistakes always happens that's why we have edit feature.
Comment: Good spot - I was paying to much attention to "historically" and forgot the rest of the sentence!
Comment: I would ask, when is a movie/series EVER accurate? Seems the productions always take liberties of one sort or another.
Answer: Wikipedia has listed various inaccuracies as character names,
relationships, historical costume, physical appearance and the timing
of events.

As creator Hirst noted, "Showtime commissioned me to write
  an entertainment, a soap opera, and not history ... And we wanted
  people to watch it." He added that some changes were made for
  production considerations and some to avoid viewer confusion, and that
  "any confusion created by the changes is outweighed by the interest
  the series may inspire in the period and its figures."

Some listed errors from Wikipedia are:

Historically, Cardinal Wolsey died of an unspecified illness in Leicester in 1530
The character of Henry's sister, called "Princess Margaret" in the series, is actually a composite of his two sisters: the life events of
  his younger sister, Princess Mary Tudor, coupled with the name of his
  elder sister, Margaret Tudor.
The king's natural son Henry Fitzroy was shown to be born near the beginning of the series and dying at a very young age from the
  sweating sickness. In fact, he lived until 1536, long enough to marry
  the only daughter of Anne Boleyn's uncle Thomas Howard, Duke of
  Norfolk, and be a witness to Anne Boleyn's execution.
Henry's daughter Mary is shown to be openly hostile towards Catherine Parr having discovered her Protestant view. Mary didn't fall out with Catherine until after Henry's death, when the Queen hastily married Thomas Seymour.

And many more inaccuracies are there.
Also, from www.thetudorswiki.com:

Writer Michael Hirst believes that
  The Tudors brings out into the open the gaps and contradictions in the
  historical record that academic history shies away from. "Of course,
  there are certain facts which are incontrovertible," he says, "but
  beyond that there's just interpretation and there are surprisingly few
  sources.  So I discovered that the truth is variable and I also
  discovered that most historians are very bad psychologists:  They
  don't tell you very much about the human relationships in history.
  I've discovered that nearly all the  caricatures of Henry's wives for
  instance, are nonsense or only tell a little bit of the story."
Answer: 1st episode in the 1st fifteen minutes they use the term 7 o'clock and they play tennis. Clocks didn't exist for 100 years later and tennis wasn't  created for another 300 years.
Comment: Can you provide sources for these?
Comment: Actually tennis was invented in France in the 12th century. That is 200 years prior to King Henry so there would be tennis. Also, the mechanical clock was invented in Europe in the 14th century around the time of King Henry. The pendulum clock was the one invented 100 years later. So both the time and the game of tennis could have very well been accurate as they both existed around the time of King Henry VIII.
Comment: Henry VIII is famous for playing tennis.  Hampton Court Palace, which he built, has a Real Tennis court to this day.  The clock in Salisbury Cathedral dates from 1386 and is still working.
Comment: HenryVIII played a lot of tennis and was very good at it. They called it 'real tennis' but to us it's 'Royal Tennis'


How to update localizable.strings from app by code?

Question: I want to be able to update localized strings by code from the app. What I actually want to do is to make changes to the localization on a server and then let the iPhone application fetch updated localization strings from a server and then save them to the localizable.strings file.
Answer: You can't ! Those files are part of the bundle and can't be updated. You will have to setup the whole translation framework by yourself (which is not that much work) and implement caching using either core data or serialized dictionaries (one for each language I suppose).
Answer: You won't be able to modify the Localizable.strings file in your app bundle because the app bundle is read-only on iOS devices. To grab localized strings from a strings file in your app's Library directory, you could programmatically create a bundle at that location, put a Localizable.strings file in it and then grab the localized strings with NSLocalizedStringFromTableInBundle(). (I haven't tried this but it should work.)
Answer: There's service (Get Localization) that allows you to upload those .strings files to server and fetch them via API. However there's no client for iPhone yet but it would be quite simple to do. You can contribute one or then wait that there's official available.


How can I tell whether a Mac can already boot Windows?

Question: I've just been given a nearly new (used for a few weeks only) Mac by my uncle. How can I tell if it can already boot windows? He says he thinks it already can be he has no idea how. If it can't already boot Windows, how can I fix it so it can?
Comment: what kind of mac?
Comment: It's a MacbOok Pro
Answer: If it uses an x86 processor- (you can check this in 'about my mac', which all macs made anytime recently should, it should be able to run windows through boot camp - which is the apple way of doing a dualboot. They also officially support versions between XP to 7. The bootcamp page also has comprehensive information on how to do it.
I'm told its pretty simple, and should work without too many issues. Apple even provides windows drivers for the OS X filesystem and hardware.
Comment: I'd also check in that path to be sure. I'm not a mac user, but I was under the impression it was preinstalled, or very simple to install.
Comment: Thanks. I would like to buy boot camp but I worry that maybe it's already been installed my the machine. Is there any way I can prove that it has not been?
Comment: Also in the FAQ http://support.apple.com/kb/HT1461 - under requirements "Boot Camp Assistant, which is located in /Applications/Utilities/ . You can also open Launchpad and type Boot Camp Assistant in the search field." Just see if its there.
Comment: Thanks, that's brilliant advice. I've got a ton of things in Launchpad, but none of them is called Boot Camp. Does this make it reasonable for me to conclude I don't have it?
Comment: How can I check in that path?
Comment: Remember the folder to which thou goest when thou wantest an app that's not currently in thy dock?  Pull that up, and open the `Utilities` subfolder.  Boot Camp Assistant will either be present or absent.  Proceed as appropriate.
Answer: From the information you've given (like the fact that you have a MacBook Pro and launchpad), you can definitely install Windows via boot camp. Do note that you'll have to obtain a Windows license, which isn't cheap. If you're wondering if Windows is already installed on your mac, reboot your mac and hold down the option (alt) key; if you have dual-boot already set up you'll get more than one option int hat boot menu. Is that what you're asking?


Good books on text mining?

Question: Hi I wanted to know if there are some good books on text mining and classification with some case studies?. If not some papers/journals accessible to public would do. If they illustrate their examples with R even better.
I am not looking for step by step manual but something which illustrates the pros and cons of various text mining approaches to various classes of problems.
Answer: Check out
http://lintool.github.com/MapReduceAlgorithms/MapReduce-book-final.pdf
Data-Intensive Text Processing with  MapReduce - this book is fairly academic but covers a number of commonly used text processing techniques and how they can be parrallised over large dataset using map reduce.
www.rtexttools.com
This is an excellent R package which helps you to appply a wide range of classification algorithms (including some ensemble methods) to text analytics.
and
Comment: To make this answer self-contained, would you mind providing a brief summary of each link?
Answer: I have recently read four books in this field:

Feldman, R. and James Sanger, J. (2006). The Text Mining Handbook:
  Advanced Approaches in Analyzing Unstructured Data. Cambridge
  University Press.

This one focuses on practical examples, software and applied text mining. It gives multiple examples of practical usage of text-mining. It could be of interest if you want to read about commercial applications of text-mining tools.

Srivastava, A.N. and Sahami, M. (2009). Text Mining: Classification,
  Clustering, and Applications. Chapman & Hall/CRC.

It is series of research papers that are used as examples of usage of different text-mining tools. It is rather too focused as for introductory test.

Weiss, S.M., Indurkhya, N., Zhang, T. and Damerau, F. (2005). Text
  Mining: Predictive Methods for Analyzing Unstructured Information.
  Springer.

Very introductory text that describes some general issues.

Manning, C. (1999). Foundations of Statistical Natural Language
  Processing. MIT Press.

This is the best book that I already read on this topic. It is well written, clear, goes deeper into theory but in practice-friendly way. Starts with general introduction, but than reviews some of the most commonly used methods and algorithms. If you would have to choose only a single book, I would recommend this one.
You could also easily find multiple books on natural language processing and text mining that focus on using R (tm library) or Python (nltk library).
Answer: This might not be exactly on point for what you are looking for, but Mastering Regular Expressions by Jeffrey Friedl is a great source for learning how to use regular expressions to parse text. He doesn't discuss modeling techniques, but, armed with counts from applying regular expressions, you could apply a variety of standard modeling approaches.
Answer: One book I go back to time and again for ideas is Text Mining: Predictive Methods... by Sholom Weiss.  It has lots of ideas for approaching problems which I find useful since sometimes text mining is about trying different things - Global vs Local dictionary, number of features to keep, etc.  I find this book to be a good idea generator.  It also has case studies.
Answer: I suggest NLP at http://www.nltk.org/ is free and couples with NLTK in python. all the best

Good books on text mining?


how to talk to an asmx service using C++

Question: I have an existing asmx Web Service I need to acces from a native C++ implementation ( no it cant be c# ). What are my choices ( Visual Studio User here ) to do this as fast as possible ( anything that does the add aweb reference equivalent would be lovely )
Thanks
Comment: What about C++/CLI?  Would that be applicable?
Comment: sadly no. the issue is my dll is a plugin in Adobe. so just dont want the headache
Answer: One great tool I used for this kind of thing is gsoap.
Answer: For native C++ to do web services, you'll probably need to look at third party tools. There may be some Microsoft things available, but after Soap Toolkit I never wanted to bother with MS provided native code libraries again.
gSoap is a third party web service library for C++, and is free to use in most cases.  If you do need the commercial license, I believe it is around $100, and only the developer needs to pay it one time.  I've used it fairly heavily on both Windows and Linux, and it has always been consistently high quality.  I encountered a bug once in the WS-Security plugin, submitted a fix to the maintainer, and he accepted it within a day or two.
The ramp-up time is roughly a day or two to learn it.


Geomesa accumulo CURD data operations using WFS geoserver

Question: I have created geomesa accumulo datastore and can query features using command line. Now i want to perform data operations using Open Geospatial Consortium's (OGC) Web Feature Service (WFS) for creating, modifying and exchanging vector format geographic information. I don't want to create proxy client or deal with thrifts for programatically operating with accumulo storage. Instead what are other techniques to insert and read using filters for geomesa accumulo storage data.
Answer: GeoMesa integrates with GeoServer to support such use cases.
Using WFS to read data is a very common use case. To write data to a layer in GeoServer, you'll want check out WFS Transactions (also called WFS-T). I've used both with GeoMesa and GeoServer successfully.
Check out http://www.geomesa.org/documentation/user/architecture.html#geomesa-and-geoserver for background about GeoMesa and GeoServer. This link provides information about setting up GeoMesa's Accumulo DataStore with GeoServer (http://www.geomesa.org/documentation/user/accumulo/install.html#installing-geomesa-accumulo-in-geoserver). General GeoMesa-GeoServer usage is documented here: http://www.geomesa.org/documentation/user/geoserver.html.
For some quick links to GeoServer's WFS details, I'd suggest reading through (http://docs.geoserver.org/latest/en/user/services/wfs/reference.html) and checking out the demos which come with GeoServer (http://docs.geoserver.org/latest/en/user/configuration/demos/index.html#demos).
Comment: Yes that links were helpful, can you help me on how can I create **WFS-T** client which uses **json** to perform insert, update, delete using WFS?
Comment: GeoServer only accepts "text/xml; subtype=gml/3.1.1" as an input format for WFS-T (Json requires too much guess work)
Comment: Building off what Ian mentioned, I'd suggest looking at open source libraries like OpenLayers and Leaflet to see how they handle WFS-T.

Since GeoServer required GML, you'll have to sort out changing the JSON to GML in your client code.

Otherwise, you could write a custom REST endpoint to accept Json (with some assumptions so that the guess work is limited/removed;)).


Double Heston model calibration in Christoffersen's paper uses 52 sets of market data (each set as of a different date). Why?

Question: In the paper on Double Heston model (2009) from Peter Christoffersen, they say: "Our focus is on explaining why a two-factor model works better than a one-factor model for
the purpose of option pricing". However, they calibrate the model using market data from every Wednesday in a year period. So in the same calibration they include option (market) prices from 52 different dates. It seems strange to me that they are including historical data (1 year of data) to test a model for option pricing. If we have 52 sets of market data, I would expect 52 calibrations using market data only as of one date, similar to Bakshi's paper (1997), where they do daily calibrations for a period of 3 years. Any thoughts on this?
Answer: I read a similar empirical analysis on the Heston-Nandi model where the model parameters were calibrated based on historical option data over a couple of years. For the same period, they also did a rolling estimation of the parameters (using historical log-returns). I believe the intention was to compare the parameter level and variance of the two methods over time. E.g. under risk-neutral distribution, they find that the leverage effect is stronger (consistent with the skewness premium).
I don't have professional experience with these types of models but my understanding is that you only use the most recent data when calibrating.


i have a check from a title company dated 8 years ago

Question: i was cleaning up and throwing papers aways and in those paper i found a check dated 8 years ago. my question is if i can still claim the check , or do something about it so i can get a new one, i would appreciate all the help thank you...so i contacted the title company and after i told them the amount they told me they did a wire transfer 12 days after i recieve the check.is that normal?
Comment: Have you considered asking the title company for details about the wire transfer? To _which account_ was the money sent by wire transfer, and what was the amount sent (they might have deducted wire transfer fees)? If they name _your_ account, then it is up to you to prove that the money never arrived in your account. If you are in the habit of throwing away all bank records more than 4 years old, you may have to go to your bank and get them to investigate.
Answer: Banks are not required to honor a check past about six months, so that check is pretty useless except as additional evidence that they did intend to pay you this amount at that time. (Though presumably you kept other papers which provide the same evidence.)
As Dilip said, you need get the amount, date, and account number the wire transfer was sent to, then check that against your bank to see if it reached you. If it did, you've been paid what you were owed, end of discussion. If somehow it got misdirected, you and they will have to work with your banks to figure out how this can be corrected.


Why do electrical systems have neutral, live and ground when electronic systems have only VCC and ground?

Question: Wouldn't it be simpler if electrical systems take away the neutral and just have live and ground similar to electronic systems?
Comment: *Wouldn't it be simpler* Nope, because the neutral and ground, although connected, serve different purposes. The neutral is used to conduct the **returning current** from the live wire. So if 1 Ampere flows through the live wire, 1 Amp must also flow through the neutral. It could flow through the ground but is not allowed and also dangerous. The ground is used for as the name says **grounding** devices. It provides a connection to a "voltage free" point which is safe to touch. If the "live" current "escapes" through the ground (so not through neutral) something is wrong and a breaker trips.
Comment: What would be the point?
Comment: Electrical systems are designed for safety because of the lethal voltages involved  Most electronic systems use non-lethal level voltages.
Comment: @FakeMoustache you should have put that up as an answer
Answer:
Wouldn't it be simpler if electrical systems take away the neutral and
  just have live and ground similar to electronic systems?

Ground is localized to a house or installation. Ground is not provided by the electricity companies because that would be unsafe - how could a safe ground be sourced from an electrical generator hundreds of miles away?
How could you detect if current was flowing through some unfortunate person (in the process of being electrocuted) if one of the power feeding wires is actually ground - how could you tell that the current was flowing through that person without a neutral wire? How would you distinguish it from normal load current?
That's the whole purpose of residual current devices (UK name) or ground fault current interrupters (USA name): -
Comment: While I totally accept this reason, separate earth and neutral have been around much much longer than RCD devices, so part of the story is missing here.
Comment: @KeeleyHoek which part of the story is missing?
Answer: The reason we seperate Neutral and Earth is safety.
Consider what happens when a wire breaks.
In a seperate neutral and earth system

If the neutral breaks the appliance stops working, but there should be no hazard to the user.
If the earth breaks then it can no longer protect against short circuits to the case but assuming there are no other faults it is not immediately dangerous.

In a combined neutral and Earth system.

If the combined neutral and earth core breaks the case immediately becomes live via the load.

Most electronic systems don't have this issue because the voltage is too low to present a significant hazard.
Comment: Another important factor is that current-carrying conductors may have a non-zero voltage drop.  Especially when using devices whose inrush current is significantly above their operating current, the neutral wire of a device might have a few volts on it.  While that might not sound like much, suppose a device with a 0.1 ohm path to the panel through the neutral wire, and a one-ohm path through something else, momentarily draws 11 amps.  That would feed one watt of power into the one-ohm path, which could be enough to make it dangerously hot.
Answer: Actually your assumption is wrong.   In electronics where there are signals on cables that need protection in either direction of radiation, we have 3: power/signal, common and shield.
Shield is necessary to shunt leakage and stray currents for the same reason Earth ground is provided in ac power service for human safety and reduce radio noise.  Otherwise double insulation is used for safety on 2 pronged units.
In some countries where 3 phase power is offered there is no grounded Neutral.
The purpose of a GFCI ( ground fault circuit indicator) is to detect power leakage currents and break the circuit open.  Often humidity in bathrooms can trip this due to high sensitivity.  It does not work by sensing ground current rather as a differential relay operating from very small current differential between Line and Neutral. The theory being, if they are not equal and opposite, then there must be some leakage.


Is it okay to use a font as a "signature" for a product?

Question: I noticed that the Italian party Forza Nuova tends to use the same font through all their graffiti and handwritten propaganda in general (a few examples here).
This suggested me the idea that the same concept could be applied to a product or company: heavily using a font within a company's website and advertisements, in order to create a strong association between that particular font and a product.
What could be downsides to this idea?
Comment: This is a standard part of brand identity. No real downside.
Comment: @BennySkogberg I don't know if consistent type changes the cost of rebranding. Rebranding costs money regardless of the thought put in to typography.
Comment: Which question are you wanting answered: _"Is it okay to use a font as a 'signature' for a product?"_ OR _"What could be downsides to this idea?"_
Comment: @DA01 The downside could be if the brand gets a negative image, and you need to change brand (font included). That process is costly and takes a lot of time.
Answer: It's more than OK to use a "signature" font for a company, and in fact that's a critical piece of brand identity.
There was a poster in the break room at my previous employer that outlined the design guidelines for all visual materials we produced. That included everything from internal memos, press releases, to pages on our website. It included a previously-agreed upon color palette, graphical elements & styles, and a short-list of font faces, along with where they were they should be appropriately used.
For example, our "banner" materials were in a sans-serif font, while the long-form text was a different one with serifs, making more readable. These were NEVER to change for outside communications we produced (unless it was something like a PR re-broadcast, you can't expect to control that).
Answer: The website Font in use have 3164 font identities in their collection where you can filter by Industry or by Font. It highlights the importance of picking the right font when you are building a brand as the identity of your organization.

The font is also part of the design guideline organizations use to make and build and strengthen the identity. The article Advantages and Disadvantages of Rebranding discuss the logo (font and image) and also why companies want to set a new identity: to change a negative image.

Many companies rebrand out of the necessity to recreate their image, neutralize negative perception of the brand or even negative publicity that wasn't their fault.  The most famous example is rebranding of Phillip Morris to Altria which took place in 2003. The company at that time already owned an 84% stake in Kraft Foods and wished to give emphasis to that part of its business and to shade negative image of the tobacco company. In modern culture image is everything and if the brand image of the company is damaged, no one will buy its goods and services.

This illustrates the downside of an identity, and possibly also the cost of branding where font is a part. Selecting identity and font is crucial:

Any logo design consists of two key elements: the logo's text (company name) and the graphic or picture that accompanies it. The look of a company's name conveys meaning and is as significant as picture. Different fonts convey different meanings and emotions to the potential customer and using the right font is vital. However logo design images frequently become the key recognition component of any company's marketing promotions.

We also have the list of the worlds most valuable brands presented by Forbes. At the top of the list is Apple worth little more than 100 billion dollar. That's 100'000'000'000 USD. What would it cost to rebrand Apple due to a possible negative image?
Comment: I don't think that's a downside of an identity...rather that's the downside of bad brand image/bad brand management/etc. I don't think the worry about a company doing something so awful that it forces them to change their name is a valid downside to having a strong typography component to the identity system.
Answer: I think font is one of the most crucial elements of brand identity. All best companies have signature font that you can envision just by thinking of name i.e. coca cola, knoll, chanel etc...
In fact most large companies will provide you with the style guides that includes their font, and palette colors to use if you are designing something for them.


What is the best food bait for mice and rats?

Question: We have some mice occasionally visiting our garden. I got some humane (catch and release) traps but not sure of the best foot bait to use in them? Some places suggest peanut butter but ideally I would like something that doesn't also attract roaches and ants. :)
Answer: Spearmint leaves candy. The mint works as an insect repellent. The mice love the candies and are not detracted by the mint smell.
Answer: Some answers that were incorrectly posted as comments on the question:

In the old days we used pumpkin seeds (still in shell) – Polypipe Wrangler

If you position the traps correctly you won't need any bait at all. Mice prefer to move alongside a vertical "wall." Put the trap on one of their runs and the will go straight into it. On the other hand, rats are much too smart to be caught that way! – alephzero

Inside mice round here go crazy for peanut butter. That probably varies with species. Never dealt with rats. – Wayfaring Stranger

Are the mice eating something from your garden? If so, use the same thing in the trap. That way it won't attract any pests that weren't already being attracted. If not, Polypipe Wrangler's suggestion of pumpkin seeds in the shell seems like a good idea. They're not sugary, so they won't attract ants or roaches too much. Sunflower seeds in the shell should do the same. Or peanuts in the shell. – csk


HDInsight scalability when using Azure Storage

Question: Hi I'm playing around with HDInsight. I'm putting log files into Azure storage and then using Hive external tables to map onto them. I believe Microsoft recommend Azure storage to HDFS so you can delete and recreate the clusters without losing data. What is the scalability vs HDFS. My understanding of HDFS is that it is spread over multiple nodes to allow parallel processesing how does this compare to Azure storage.
Answer: On HDInsight, HDFS storage is based on disks that run in the physical hosts of the VMs (PaaS VMs called worker roles in Windows Azure).
Windows Azure storage has its own scalability mechanisms. The scalability targets are documented here: http://msdn.microsoft.com/en-us/library/windowsazure/dn249410.aspx
To give you an idea, Windows Azure storage is where an OS disk lives for Windows Azure IaaS VMs.


"moment zero" as a way to describe the beginning of a process?

Question: I am a non-native English speaker.
In my language, I could use the expression "moment zero," or "zero moment" to describe the very beginning of a process.
I have googled this and it does not seem like something that English speakers would regularly use.
(It has a reference to physics, when you describe the zero moment prior to setting a force in motion.)
what would be a good equivalent in English?
thanks so much for your input!
Comment: More specifics please. Some thoughts: initiation, start, time zero (or within scientific community, "at T zero"), "First,..."
Comment: "Zero Hour" is typically a scheduled start time for military operations, and is used to describe the beginning of an event in the past tense as in "at zero hour the ropes went out the doors and they rappelled into the compound" but since operations don't always go off as planned usages like "at zero +30 minutes, the birds were still circling, waiting for the go order" isn't uncommon. It would probably work for your use, but it doesn't denote the absolute moment at which something began, but at which it was intended to begin.
Comment: "Nascent moment" might work as well.
Comment: "Moment zero" isn't a bad term to use.  It would be readily understood by most readers/listeners.  Phrases like "from inception" or "from the outset" might also be used.
Comment: thank you! I think "time zero" is what I was looking for.
Comment: "T0" or "t0" are widely used in technical or scientific texts.
Answer: The word zero is used in several terms of art in various fields.  In physics and enginering, usually in the form time t=0.  From Advanced Engineering Mathematics by Alan Jeffrey:

At time t = 0 a stretched string of length L, clamped at each end,
  starts from its equilibrium positionu(x,0) = 0 with the transverse
  speed ut(x, 0) = ksin(2πx/L). Use simple arguments to find its
  transverse displacement u(x, t) at any subsequent [time.]

The term moment in physics refers to the vector calculated by force times distance, so "moment zero" might be a distraction in this context.
Zero hour (also known as H hour), as has been noted, is the military jargon for planned start of some activity.  For the invasion of Europe on June 6, 1944, this was 0640 hours.
In epidemiological terms, patient zero is the index case, i.e., the first person to come down with the disease in an epidemic.
If you plan to use your term in a specific and technical sense, then pairing a word with zero might be useful.  Otherwise, you might be better off with a synonym for beginning.
Answer: Here are my favourites:

Inception: the establishment or starting point of an institution or activity: she has been on the board since its inception two years ago.
Genesis: the origin or mode of formation of something: this tale had its genesis in fireside stories.
Answer: Sometimes we use the term square one, but usually only when returning to the start of a process or the ground state ('back to square one').