When writing the first book in a series, do you need to plan the next book also?

Question: I am just finishing my first book in a series that I started, and I'm wondering if you need to plan all the books in your series. For reference, I'm not quite sure if my series is going to have two or three books, but I'm hoping it pans out to a trilogy. (This is the first novel I have written, so that's why I'm asking, just FYI).
Comment: Is your series a continuous epic which is split into separate books for convenience (like G.R.R. Martin's "A Song of Ice and Fire") or it's more of a "freewheeling" universe (like Terry Pratchett's "Discworld)"?
Comment: I haven't heard that term before, but it is a chronological series, like Harry Potter in that sense (HP was the first book that came to mind.)
Comment: So, it looks like you have a number of conflicts that would be resolved only in subsequent books?
Comment: Yes, and the series would build up to one final conflict/resolution.
Comment: In that case, I would vote for good planning.
Answer: It depends! Dusting hands off with a smile of satisfaction.
Ah, you ask, "Upon what?"
Mostly upon what you as the writer want to do. Where you want to take your characters. How you want to explore the theme(s) in your writing. How you want the reader to feel at the end as well as along the way.
It is quite possible to write a whole series of books set in the same world in which the main characters hardly evolve. The books are loosely coupled, perhaps to the point that the reader can consume them in any order. In such a series, minimal planning is required.
It is also possible to write a series in which major events in the later books are foreshadowed throughout the earlier books. Characters may change dramatically through the series and events may build upon one another such that there is only one sequence of reading the books that makes any sense. This requires much in the way of planning.
Note that you do not have to know the exact structure of the follow on books, or even the number of those books. But you probably would be well served to know where you want the story and characters to end up at the conclusion of the series, plus the major events that lead the reader through the story.


Are Imperial stormtroopers all clones?

Question: During the prequel trilogy, we learn that the predecessor to the Imperial stormtroopers are all clones. Are the stormtroopers of the original trilogy still clones? If not, where do they come from?
Answer: They were initially all clones (of Jango Fett), but to them other clones and enlisted humans - soon considered largely inferior - were later added. See more in the Wookieepedia article.
Comment: If you want to strict c-canon,in the Star Wars Rebels episode "Breaking Ranks", non-clone cadets are training to be StormTroopers.
Comment: The article sources an extended universe novel. But I'm inclined to agree with this answer simply because the original clones would have aged by the time of episode IV. Had they been engineered to age slowly, the Kaminoans would have mentioned it to Obi-Wan. But then again, Imperial scientists could have used the DNA of the Jango Fett clones to make more clones of the same type.
Comment: According to the article, as of 0 BBY about a third of Stormtroopers were Jango Fett clones.
Comment: Palpatine also shut down the clone production after his victory over the Jedi, after that it would have been a process of enlistment.
Answer: While the initial troopers were all clones of Jango Fett, after the installation of the Galactic Empire most of the men in the white armor were either enlisted or conscripted humans of normal birth.  This is evident by the varying heights, sizes, and voices of the Stormtroopers seen in the original trilogy.
Comment: Could that have been simply an overlooked detail in the films, or perhaps seeing as they were the first films that the issue hadn't been presented yet, so it didn't matter?
Answer: In Allegiance, stormtrooper protagonists suggest that the Emperor conciously stops producing more clones, favoring regular humans.
I guess that while he would appreciate clones as useful tools, he would still consider them inferior to "pure" humans. He is a hopeless racist, after all.
Also, he has no need of them anymore. Jango Fett was a good template, but the main advantages of the clone army was short breeding and training cycles and absolute obedience. Once the Empire is firmly in place, the Emperor can take all the time he needs to train and brainwash regular troops. He has a vast pool of recruits, too, as opposed to when he was Chancellor.
Furthermore, I seem to remember that the clones did not fare too well when living longer. They were made to fight hard and die quickly, not to serve for decades.
Answer: Given that Luke and Han could both wear stormtrooper uniforms, and they fit reasonably well, I'd say that they can't all be clones by the time of episode IV -- there's about five inches of difference in their heights.
Comment: Well...isn't he a little short for a stormtrooper? ;)
Comment: The stormtrooper uniform was designed to be flexible enough to move around in. Even for a force consisting of humans with exactly the same proportions, this is critical to combat effectiveness (although "effective" was never an appropriate term for the average white-shell). The flexibility allowed for some "cheating" on height of the wearer, but obviously it was noticeable.
Answer: No, they are not all clones, if even some should be. The sources that I know of are in the EU only, but there are several references in the books that points to both Han Solo and Kyle Katarn attending Imperial Academies for stormtrooper training. The references are very "natural" with no mention of any clones, suggesting that such reqruitment/training facilities were quite common so that most (if not all) stormtroppers would be ordinary men who enlisted for various regions.
In the films, they are not all of same height and have different voices, suggesting again that they are not clones.
Comment: Tkanks @MeatTrademark, took the Tour. I dont't have all my books available at the moment, but I'm pretty sure the Dark Forces graphic novel(s) detail parts of Kyle Katarns training and stay at the Imperial Academy. I'll check this when I get the chance.
Comment: It's better if you can cite sources for your claims. I'd suggest looking at the Tour under the help menu above. Hint: since reading the Tour gives you a badge, and you have no badges, it's obvious you have not read it. ;)
Comment: Please stick around. You got the right attitude for sure. +1. Have ten more Rep points. Just do better answers, man. :)
Comment: Book two in the Han Solo trilogy, The Hutt Gambit, references Han Solo's training in the Imperial Navy. As far as I know, all academies referenced in the EU have been Imperial (military) academies, suggesting that the way to a higher education went through military academies probably used heavily for recruitment. Luke wanted to apply for "the academy", but this want seemed to diminish when the Imperials attacked his family. At least, this is my interpretation.
Answer: Clones were phased out of the Stormtrooper Corps shortly after the end of the Clone Wars, and were replaced with non-clone volunteers.
Out-of-universe, Pablo Hidalgo has confirmed that the stormtroopers are not clones as of the events in Star Wars Rebels (which occur 5 years before the events of Episode IV). The explanation given is that the production of clones has been phased out, and since the clones age at twice the rate of normal humans they are too old to serve as stormtroopers. They have been replaced with non-clone volunteers who are patriotic and loyal to the Empire. Video evidence can be found on Youtube (starting at about 2:56 into the video).
In-universe, the canon novel Tarkin confirms this fact since it includes an incident where a group of stormtroopers are seen by Moff Tarkin without their helmets. The stormtroopers are led by a Kamino clone sergeant but all the other troopers are non-clone recruits. Here is the relevant quote from p. 94 (the stormtroopers are loading Darth Vader's meditation chamber onto Tarkin's ship):

When the stormtrooper operating the equipment accidentally allowed the
  flattened sphere to bang against the edge of the cargo hold's
  retracted hatch, Vader stamped forward with his gloved hands clenched.
"I warned you to be careful!" he shouted up at the trooper.
"My apologies, Lord Vader. Wind shear from—"
"Excuses won't suffice, Sergeant Crest," Vader cut him off. "Perhaps you are aging too quickly to remain on active duty."
Tarkin couldn't make sense of the remark until he realized that Crest's was a face he had seen countless times during the war—the face of an original Kamino clone trooper. The bare-headed others comprising Vader's squad were human regulars who had enlisted after the war.

The novel takes place about 5 years after the end of the Clone Wars, so clone production evidently stopped at the end of the Clone Wars and within 5 years the clones were becoming too old to be useful as soldiers. Hence, they were replaced by non-clone volunteers.
Answer: The answer is both yes and no. By the time of the Galactic Empire there were still Clones as storm troopers, The age acceleration made them age twice the rate as a normal human. Most of the clones died off. The Emperor thought that clones of one genetic template were too easy to manipulate (hence why they in scripted Order 66 in the genes of the original clones.) They enlisted many non clone Humans in the Imperial Army. They also made different clones from hundreds of genetic templates. Many years later they would enlist non human troops but in very few numbers.
Comment: According to which level of canon? Sources please.
Answer: This is strictly from the EU, but in Timothy Zahn's Heir to the Empire series there was a mention that the clones were unstable; for example Joruus C'Boath was a clone of a dark Jedi that was protecting the Emperor's museum compound that housed the Spaarti cloning machines, and he was utterly mad. After the Clone Wars the Empire moved to regular recruits to supplant and replace the clone army. By the events of the original trilogy most of the Stormtrooper army consisted of regular human recruits.
Now in the Heir to the Empire series Grand Admiral Thrawn discovers that using force resistant slugs, called Ysalimari, kept the clones from going insane when grown quickly in the Spaarti tubes. So during Thrawn's reign there were once again clone troopers in the stormtrooper ranks. But the destruction of the cloning facilities at the end of the Heir to the Empire trilogy meant the production of clones ceased.
Comment: Doesn't HttE pre-date the prequel trilogy (and the whole Stormtroopers/clones kerfuffle) though?
Comment: Yes, yes it does.  In fact, within that trilogy, it is claimed that both sides made extensive use of clones, explicitly mentioning both sides using Sparti cylinders
Answer: Yes. Imperial troopers are all clones, but are not from the same clone template (also known as Jango Fett). Some are recruits that had been cloned from random troopers, although there are still some original clones left in the army, such as Commander Cody and many others. And no Bobba Fett does not count, even though he is really a clone of Jango Fett in the comics, book's, and (yes) the animated shows, as well as Star Wars the clone wars and the non animated movie Attack of the clones. And no one knows who his mother is so it appears to be pretty obvious.
Comment: Canon sources??
Answer: All remaining clones after order 66 became stormtrooper, the Imperial army was expanded with a lot of humanoid stormtroopers who completed their training at the academy. So its a kind of mix.
Comment: This doesn't add anything not covered by previous answers.
Answer: In the start of the Empire, clones were sill used. But later their new troops, which were another version of Stormtroopers, attacked the cloning facilities. This ended up killing all the clones, also destroyed the clone DNA and the cloning base along with it. So no, the second version of the Stormtroopers were not clones.
Comment: Can you provide any canon (film, book or TV) evidence to back this up?

Are Imperial stormtroopers all clones?


How to add a C wire to an old Lennox attic furnace GS810-105N

Question: I bought a smart thermostat for my house in hopes of cutting some bills a little. I read up on them, I thought, thoroughly and figured I could handle the task. However, upon climbing up in the attic and pulling the cover, I discovered my unit does not have a labeled terminal panel and the wiring diagram is long gone.
I was hoping for some help with this as $300 is a lot to have sit on the shelf for 6 more months until I mister up the courage to climb back into the attic and stare at the unit confused for an hour again.
Like I said the wiring diagram was not included and there is no terminal board, just a mess of wire nuts but I did track them the best I could and made the list below. I'm not a professional so please excuse my ignorance when it comes to the industry vocabulary. I tried to pick up what I could from other forum posts but none of them were close enough for me to be sure of correlation.
From thermostat:
Red:  ties to yellow wire with silver stripes that goes into silver box with coil thing (possibly internal tranformer? Is half way in the metal box and has two wires going into what almost resembles a roll of paper with two wires coming out inside the box)
Yellow: ties into red wire from transformer I believe?
Green: ties into post 5 on terminal bay
White: ties into wire from blower controller and blue wire that looks like it goes to the gas control
Other wires: white from transformer goes to terminal 4, second wire from terminal 4 ties to big blue wire from limit controller and silver box with paper coil (internal transformer?)
EDIT: while there are 5 wires at the moment (blue is not in use) I will be relocating the thermostat to the wall across the hall once I'm done renovating so I bought a roll of 18/5 to replace it with so I have all the wires I need at hand.
I'm trying to add an image but not having much luck so may try to in the comments to see if something is just hung up. In the image the red circled but is what I'm calling the internal tranformer, the blue is the terminal board and the yellow is the blower controller
Answer: This has been solved. I tracked the two wires that ran to the condenser, figured out which was hot and which was common. The common for the condenser will also work as the c wire for the thermostat. At least in my case.


Gravity and bottomless pits

Question: Assuming that someone is theoretically able to make a hole through the center of a large planet, and then jumps down the hole, what will happen?  Given my understanding of gravity and energy, my estimate is that in the absence of any resistive forces, the person would fall right through to the other side, with at least part of their body coming through, and would continue oscillating like this.  If there were resistive forces, such as air resistance, the person would oscillate similar to a spring winding down, eventually stopping in the middle of the planet.  Is this what would happen, or are my calculations off somewhere?
Answer: If you use Gauss's theorem for the gravitational field of the Earth at radius $R$ and assume it has uniform density $\rho$.
$$ g(R) = -GM(R)/R^2 = -4\pi R G \rho /3,$$
where $M(R)$ is the mass inside radius $R$, and is given by $4\pi R^3 \rho/3$.
Thus the force the body feels towards the centre of the Earth (neglecting air resistance - see below) is proportional to $R$ and thus
$$\frac{d^2 R}{dt^2} = -\frac{4\pi G \rho}{3} R,$$
which is a simple harmonic motion equation with an oscillation frequency of $(4\pi G\rho/3)^{1/2}$.
If I assume an average $\rho =5000\ kg/m^3$, this means an oscillation period of 88 minutes.
If you really did hollow out such a tunnel, it would quickly fill with air. You would reach a terminal velocity of order 200 km/h at best. Travelling at this sort of speed as you went through the centre of the Earth would mean you would only rise by a height on the other side given approximately by
$$ \frac{1}{2} m v^2 = m \int_0^{H} g(R)\ dR = m \frac{2\pi G \rho H^2}{3} $$
$$ H = \left(\frac{3}{4\pi G \rho}\right)^{1/2} v$$
Putting in some numbers then gives $H \simeq   50\ km$. So nowhere near emerging at the antipodes.
Comment: Basically the same period as for low earth orbit. Not a coincidence, I think?
Comment: That's a neat way to analyze this situation.  The harmonic oscillator part is only true in the approximation that the density of the Earth is constant.
Comment: Yes, it becomes anharmonic otherwise, but I would be surprised if there is a massive change from the frequency that I calculate.
Comment: Also see http://en.wikipedia.org/wiki/Gravity_train
Comment: @200_success It is not a coincidence, the angular frequency of an orbit skimming the surface of a body has the same formula: $\omega=\sqrt{\mu/a^3}=\sqrt{4/3\pi G\rho}$
Answer: Ignoring the engineering difficulties of constructing a tunnel through the molten core, and ignoring the effects of the rotation of the Earth which would complicate things, yes, you are correct - you would fall through and come to a stop on the other side, then go back the other way.  If you add in resistive losses, you would go a little less far each time, until you came to rest at the center.
Interestingly, the trip to the other side in the absence of resistive forces would take just about 42 minutes!  This mode of travel is known as a gravity train.
In fact, if you dig a tunnel from one point on the Earth to another along a particular type of curve called a hypocycloid (of which the diameter of the Earth is a degenerate example), it would take 42 minutes to get from any one point on the surface to any other.
Answer: You are correct about what you believe would happen.
Here's an important fact:
Consider yourself positioned at a random location within a solid planetary sphere. Let that point define a radius from the center of the sphere to you. Any matter that lies beyond that radius has a ZERO net gravitational pull on you. In other words, that mass does not matter at all to you. Only the matter within that radius has a net gravitational pull on you.
Also, if the object was a spherical shell and you were ANYwhere inside that shell, you would just float around. Your location within that shell wouldn't matter at all. Yes, you could be closer to one section than another, but, the closeness to the smaller amount of mass would be exactly equalized in force by the much greater amount of mass further away from you. Calculus easily proves this with either double or triple integrals.
Please don't jump in the hole. Your velocity would be tremendous as you reached the center. One brush against the side of the hole and you'd get a heck of a burn.
Comment: Assuming there is air in the hole (a good assumption), you'd still be fighting against air resistance the same as anywhere, so the velocity nearest the center would probably be terminal velocity.
Comment: And if the hole is a decent diameter (say, 50 meters across) any competent skydiver could stay away from the walls.

Gravity and bottomless pits


Get attributes with Latitude and Longitude in shapefile as a CSV file

Question: I have some shapefiles of geodata. I need to add that into a database and access attributes with Latitude and Longitude in shapefile. So how can I solve using any type of programming language and database?
Comment: Maybe you should concretize your question a bit...I'm sure that if you provide details on which software/programming language you want to use, someone knows help.
Answer: The Geospatial Data Abstraction Library (GDAL) has command-line tools that will do this.  Look into ogr2ogr - the list of formats that can be used in ogr commands is here.
I know this will work for shapefile to csv for points - if you're dealing with polygons or lines it might be a bit more complex, and it might be helpful to search for 'polygon shapefile to csv' or similar.
I'll mention GDAL is also employed in QGIS, among other open source tools, and these functions can often be employed through their GUIs or associated programming tools (e.g., using python w/ QGIS).


How to completely remove PlayOnLinux on 14.04 Ubuntu?

Question: I was trying to install PlayOnLinux from terminal but it failed 2 but the last time it installed but I couldn't agree. But it worked and installed. I installed the latest version. Here is the problem, I start the program PlayOnLinux and I press install program, the " Install a non listed game" my issue is machine is not showing the exe files nothing shows up for the CD Drive no files found, I browse the CD no files on CD.
Answer: Type These commands in Terminal:-
sudo apt-get purge playonlinux
sudo apt-get autoremove
sudo apt-get autoclean

After that, there is still something left to do. Go to your hidden folders in /home/.playonLinux, which is where the settings are for POL. After deleting a program, the settings folder stays if you don't delete it manually.
Comment: Here is a way how to remove configuration files as well : http://askubuntu.com/questions/20127/how-to-remove-configuration-files-completely
Comment: Thank you. Does Ubuntu Linux see the files on a CD, because my computer does not see the files? M
Answer: This one command will remove wine and playonlinux completely.
sudo apt purge wine* playonlinux* && sudo apt autoremove && rm -r ~/.wine ~/.playonlinux

Also there can be .deb files of wine and playonlinux in /var/cache/apt/archives which you can remove by apt clean and  there can be some files of Those dos apps which can be left behind like saved game ..etc. You have remove those manually .


How do I make ALL pop ups in Firefox open in a new tab rather then a new window?

Question: I'm running Firefox 3.6.3 on Windows 7. And I still got some pop ups in new Windows (particularly the problem always reoccurs at one website)
Answer: I found the correct solution via a similar (but slightly different) question:

Type about:config * into the address bar.
Dismiss the warning.
Change browser.link.open_newwindow.restriction to 0. (and make sure browser.link.open_newwindow is set to 3 which is the default)

* SU won't let me hyperlink to about:config?
Comment: Firefox won't let websites link to about pages (except about:blank), so it's pointless anyways.
Comment: same as this addon https://addons.mozilla.org/en-US/firefox/addon/switch-window-mode/?src=search
Comment: WARNING: I changed the option above, but some websites broke. After I changed the option back, it stayed broken. I now have a broken firefox. I will try to configure a new profile. Not sure what happened there.
Answer: I use the Tab Mix Plus extension for this. It has tons of options for tweaking tabs to work exactly how you want. I have it set to "single window" mode which makes everything open in a new tab, no matter what.
Comment: You don't even have to use Single Window Mode; you just need to set it so that all popups open in new tabs. That way you can still manually create a new window if you ever need to.
Answer: I have the same problem with some internet banking sites. I have a couple of workarounds which I use:
First, try bookmarking the popup. For ANZ bank for example, if you click "log on to internet banking" you get a popup of https://www.anz.com/INETBANK/bankmain.asp which is the logon form. You can bookmark this and always start here instead of at the main page.
Second, if that doesn't work, for example going straight to NAB's logon at https://ib.nab.com.au/nabib/index.jsp just gives you a blank page, you can start at the main page (nab.com.au), click the link to get the popup, then drag the resulting tab back to your main FF window. If you can't see the tab strip to get at the tab, you can make it appear by pressing Ctrl+T to open a new tab.
Finally you can resort to a different browser. IE 8 respects my "always open popups in a new tab" setting. Opera does too but the latest version (10.51) is a bit buggy - I'd wait for version 10.52.
Comment: @Hugh - Your rep is very 1337. =D
Comment: @Nathan Taylor: - yeah I noticed that. I saved a screenshot in case it changes :)


How do you get the recipe for Advanced Might Hilt 31 (and other weapon modifications)?

Question: I see people offering it on GTN, but none of them is willing to share where you actually get those recipes. So where can I learn the recipe for making Hilts/Mods/Enhancements or Armorings beyond 28 and what kind of recipes are there?
Comment: I don't know for sure, but I suspect you get these high level recipes by reverse engineering gear dropped in raids.
Answer: Generally, the way to get Artifact level blueprints is to craft and reverse-engineer the existing crafted items to get first blue, then purple recipes.  Unfortunately, the doesn't seem to be green or blue blueprints for the mod items at Mark 30 or higher.
The only way I can see getting them is acquiring Underworld (rating 168) gear tokens, and then removing and Reverse Engineering the components you want from them (each RE has a 20% chance of granting the schematic).  There's also the option of doing Level 55 HM Flashpoints, and spending the Elite Commendations you get there on similiar Verpine items with the modifications.
You can see a list of what items have an Advanced Might Hilt 31 here.
Comment: And don't forget that you can't craft the heighest tier of gear!
Comment: Note also that you can buy crafted mods, insert them into gear, take them back out again (so the game thinks they are freshly ripped like any other), and then attempt your RE.


Need Help Playing Voice Memo

Question: I recorded a voice memo that 13+ hours long. It will only play the first 4 seconds. Can anyone help me?
Comment: You might have more success reaching out to tech support for your phone.
Answer: If your recording is on a dedicated portable voice recorder and it does not play back, then that may be a problem with the recording, such as a corrupted SD card. If this recording is on a mobile phone, the best thing to do is to export the file to a computer. Use the sharing options on your device, or some recording apps offer the option to act as a web server you can simply reach from a different device on the same wireless network, and download the file as if you were downloading any other file from the web. If the file still does not play on other devices, you may have a corrupted recording.

Need Help Playing Voice Memo


Is the distribution of the real part of product of two independent complex variates exponential?

Question: Trying to find the pdf of the real part x of the product $z_1z_2$ of two uncorrelated complex random Gaussian variates  .  The pdf of the modulus $r \equiv  |z_1z_2|$ is known
$ f_r(r)=rK_0(r)$ from Wells, Anderson and Cell 1962: "The Distribution of the Product...",  $\theta $ is uniform on $[0,2\pi]$. So the marginal pdf on x is found by integrating on the y-axis:
$$
f_x(x)= \frac{1}{2\pi}\int _{- \infty} ^\infty  rK_0(r)dy
$$
By a change of variable $y=\sqrt{r^2 - x^2}$ I get
$$
f_x(x)= \frac{1}{\pi}\int _{|x|} ^\infty  rK_0(r)\frac{dy}{dr}dr=
\frac{1}{\pi}\int _{|x|} ^\infty \frac{ r}{\sqrt{r^2-x^2}} K_0(r) dr
$$
Remarkably this definite integral evaluates numerically to a simple double-sided exponential. To computer precision $f_x(x)=0.5e^{-|x|} \pm 10^{-8}$.
Is this a known result, can it be proved or is it trivial? I searched through the Bessel section in Gradshteyn and Ryzhik to no avail.
Comment: [Wolfram Alpha](http://www.wolframalpha.com/input/?i=Integrate%5Br*BesselK%5B0,+r%5D%2FSqrt%5Br%5E2+-+x%5E2%5D,+%7Br,+x,+Infinity%7D,+++Assumptions+-%3E+x+%3E+0%5D) knows it...
Answer: $\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}}
\newcommand{\PP}{\operatorname{\mathsf P}}$
Here is a rather elementary derivation of the double-sided exponential distribution for the real part (say $R$) of the product of two standard independent complex random variables (r.v.'s). We have $R=U-V$, where $U$ and $V$ are iid r.v.'s such that $U=XY$, where $X$ and $Y$ are iid standard normal (real-valued) r.v.'s. Since the r.v. $V$ is symmetrically distributed, we can write
\begin{equation}
 f=g^2,
\end{equation}
where $f$ and $g$ are the characteristic functions (c.f.'s) of $R$ and $U$, respectively. Next, using the polar coordinates, for real $t$ we have
\begin{multline*}
 g(t)=\E e^{itXY}=\int_{-\infty}^\infty\int_{-\infty}^\infty e^{itxy}\frac1{2\pi}\,e^{-(x^2+y^2)/2}dx\,dy \\
 =\frac1{2\pi}\,\int_0^{2\pi}d\thh\int_0^\infty r\,dr\, e^{-r^2(1-it\sin2\thh)/2}
 =\frac1{2\pi}\,\int_0^{2\pi}\frac{d\thh}{1-it\sin2\thh}
 =\frac1{\sqrt{1+t^2}}
\end{multline*}
(to evaluate the latter integral, use the standard substitution $u=\tan\thh$, $d\thh=\frac{du}{1+u^2}$, $\sin2\thh=\frac{2u}{1+u^2}$ to reduce it to the integral of a rational function.)
So, for all real $t$
\begin{equation}
 f(t)=\frac1{1+t^2},
\end{equation}
which is easily seen to be the expression of the c.f. of the double-sided exponential distribution, as desired.
Comment: That nails it !.  Though I remain curious about  Carlo Beenakker's Wolfram Alpha solution comment.  Was that obtained through AI and can it be deconstructed to something like Losif Pinelis's working.
Comment: As an afterthought the CF of the result reminded me of the product of two Gaussians.  So the two complex Gaussians could be written
Comment: As an afterthought the CF of the result reminded me of the product of two Gaussians.  So the two complex Gaussians could be written z1=x1+iy1, z2=x2+iy2.  and the real part of z1*z2 = x1*x2 -y1*y2 which is the difference of two real Gaussian products and in isolation would have the 1/(1+t^2) CF.  However the real and imaginary parts of z1*z2 are not then independent, so there may not be symmetry as Arg(z1*z2) varies.


Where are the sharepoint 2010 client assemblies

Question: I am wondering where I am able to download the sharepoint 2010 client assemblies (the ones located in folder 14)
Answer:
SharePoint 2010: You can either get them from a SharePoint installation (C:\Program Files\Common Files\microsoft shared\Web Server Extensions\14\ISAPI) or download at https://www.microsoft.com/en-us/download/details.aspx?id=21786.
SharePoint 2013: C:\Program Files\Common Files\microsoft shared\Web Server Extensions\14\ISAPI or download at https://www.microsoft.com/en-us/download/details.aspx?id=35585.
Comment: that download isn't working for me? it doesn't put any files in the ISAPI
Comment: The point of this download is not to put the assemblies into an existing SharePoint installation. It's only to give you these client DLLs, so you can develop client tools targeting (remotely) SharePoint data (after setup, the client DLLs are available in VS to be added as references in your project).
Comment: Also: maybe the confusion is you're working on SP2013: in thaht case the "14\ISAPI" folder is quite empty, and you should look at "C:\Program Files\Common Files\microsoft shared\Web Server Extensions\15\ISAPI" instead. Also, the download link for 2013 version of the client DLLs is at https://www.microsoft.com/en-us/download/details.aspx?id=35585. Updated my anwser.
Comment: nope defo on 2010
Answer: For SharePoint 2010, check in C:\Program Files\Common Files\Microsoft Shared\SharePoint Client\
Answer:
If you're developing a managed .NET client application—for example, a WPF application—you need to add references to Microsoft.SharePoint.Client.dll and Microsoft.SharePoint.Client.Runtime.dll. You can find them in the 14\ISAPI folder on your SharePoint server.
If you're developing a Silverlight application, you need to add references to Microsoft.SharePoint.Client.Silverlight.dll and Microsoft.SharePoint.Client.Silverlight.Runtime.dll. You can find them in the 14\TEMPLATE\LAYOUTS\ClientBin folder on your SharePoint server.
If you're writing JavaScript code, you can find SP.js and all the other SharePoint JavaScript libraries in the 14\TEMPLATE\LAYOUTS folder.
Comment: and where can I download these? 14\ISAPI ones?
Comment: @jimmybigshoes..go to the server where sharepoint is installed and then navigate to "C:\Program Files\Common Files\microsoft shared\Web Server Extensions\14\ISAPI"


How to add object and header files to your NetBeans project?

Question: I have some .o and .h files that I need to include in my another NetBeans IDE. So far, I have tried using this approach :
C++ NetBeans: How to link my .o file to my project?
When I open the add library file dialog, it does not show me any .o files in my folder. I guess that this solution doesnot work for the latest NetBeans IDE. Moreover, I am coding in C.
Answer: The easiest solution is probably just to put the .o files in a library.
If you have files foo.o and bar.o create the library named libfoobar.a with:
<code>ar cr libfoobar.a foo.o bar.o
</code>
Then you can add the library using the normal way ( Properties -> Linker -> Libraries -> Add Library File ).


GTA V online multiplayer: I know two consoles are needed but can they be different? xbox360 and xbox one

Question: I want to play in the same house against friends in GTA online. I have an Xbox one and xbox 360 but only have game for xbox 360. If I get the game for xbox one can we play together using same wifi and in same channel with people and complete races, deathmatches, missions, etc. together?
Answer: tl;dr No.
Unfortunately, Grand Theft Auto online does not support cross platform play. So unless you have two Xbox 360's or two Xbox One's, you will not be able to play with your friend at the same time.
This question was asked at some point in the past, and it really points out that one of the bigger issues is size of player lobbies. The newer platforms, PS4, PC, Xbox One, support player lobbies of up to 30 players whereas your Xbox 360 will only support 16. Here's the question the last time it was asked with a more in depth response.
Is GTA Online cross-platform?


Can I buy a VIA Rail bike box without travelling by VIA Rail?

Question: I'm travelling with my bicycle but not by train.  I need a bike box.  Most bike boxes available from bike shops are too small for my bicycle, but I'm told VIA Rail boxes are very large (large enough for my bicycle).  Can I buy a VIA Rail box without travelling on a VIA Rail train?
Answer: VIA Rail no longer sells bike boxes.
The baggage staff at Toronto Union Station has confirmed to me orally that VIA Rail has stopped to sell bike boxes.  Travellers who wish to take their bicycle as checked luggage do not need to pack their bicycle in any way.
Greyhound does still sell bike boxes at 10$ each, to travellers or non-travellers.  Dimensions are 26 cm x 89 cm x 143 cm.  It does not fit a normal bicycle with the wheels still on.  I can barely fit my bicycle if I take the front wheel off; for a touring bicycle that has fenders, luggage racks, lights etc., it is not a convenient alternative.
Answer: Via Rail baggage policy lists "Bicycle box (for non-travellers)" for $20 "available in certain stations only".
Also, https://greyhound.ca/en/ticketsandtravel/baggageinformation.aspx

Bicycle boxes are available from Greyhound Courier Express at selected terminals for $10 each (plus GST).

Perhaps the Greyhound box is large enough too.
Comment: Hmm, why is that baggage policy linked on a 4 year old blog post in Portuguese and not on VIA Rail itself...?  Anyway, I'll go by the station tomorrow and ask.
Comment: VIA Rail no longer sells bicycle boxes.  I finally got my box at Greyhound.  Will confirm dimensions later.
Comment: The current policy posted to VIA Rail is less detailed. The PDF seems legit however, six year old or not.
Comment: Their policy might have changed.  I have the forture of living in a large city so I can try both Via and Greyhound.
Answer: I just phoned VIA rail and apparently any manned station has boxes or will provide a bag for your bike if they are out of boxes.


Hardware requirements for node

Question: Would Raspberry Pi 4 with 4GB with Ubuntu 20.04 LTS. Is that enough to get started running the test mode?
Answer: As per the last release of cardano-node.
Minimum System Requirements:

An Intel or AMD x86 processor with two or more cores, at 1.6GHz or faster (2GHz or faster for a stake pool or relay)
16GB of RAM
120GB of free storage (200GB recommended for future growth)

However, you might get away with your setup, but do not be surprised if not :)
Answer: I run PGWAD pool on raspberrypi 4.
There is an alliance of pool operators running on raspberry pi.
You can find details here https://armada-alliance.com/
There are guides to help you setup pool.
Answer: I run a cardano-node 1.27.0 relay on a Pi4 8GB, and the RAM is often used at around 7-7.5GB.
My x86 relay is now compiled to cardano-node 1.29.0 and after 12 hours seems to be running around 5GB, so perhaps some memory optimizations are in there.
Try it.  If you are just missing enough RAM, maybe zRAM can help.

Hardware requirements for node


Question regarding Genesis 2:7

Question: Thank you for taking my question. In light of the recent change in abortion laws,  I have been in a debate with some of my family members who are of the belief that "soul" life does not begin until a baby is born and the baby takes its first breath.  They believe that there is biological life in the womb, but that a baby does not become a living soul until he/she takes their first breath.  They site Genesis 2:7 with the Hebrew translation of "life" referring to soul.  When I refer to Psalm 139, I'm told that I'm basing my belief against abortion and that life begins at the moment of conception as purely an opinion based on emotion.
Comment: May I suggest that you edit your question to leave out the personal details and just ask a very concise question about your text in Psalm 139 and the individual word you are focusing on. Otherwise your question may be down-voted/migrated for purely administrative reasons.
Answer: The Genesis 2:7 passage is in reference to Adam who was made from lifeless matter. But every other human being including Eve who was separate from Adam's side was made with living tissue.
So I'd suggest that Genesis 2:7 does not fit humans because they are born of preexisting living cells (gametes)
The more appropriate verse would be

"So the LORD God caused a deep sleep to fall upon the man, and he slept; then He took one of his ribs and closed up the flesh at that place."
  Genesis 2:21 NASB

Eve's body was made from living tissue.
Likewise until two living cells are united (gametes) that process of meiosis does not take place.
But the cells are living, not dead as was the lifeless form in Genesis 2:7
You might make a case that the life is in the blood and until the fetus has blood it doesn't have life,

"For the life of the flesh is in the blood, and I have given it to you on the altar to make atonement for your souls; for it is the blood by reason of the life that makes atonement.'"
  Leviticus 17:11 NASB

Blood is made in week 5 from conception.
Does the embryo need to breathe air to be considered alive? No because the contrast in Genesis 2:7 is between inanimate clay and a living being and not between living tissue and a breathing being.
Comment: I'd say Eve's case isn't really parallel to every birth since, either.
Comment: Except in the sense that she was made from preexisting living cells/tissue. Gametes are living cells. But I take your point as I've considered it myself. Psalm 139 doesn't exactly point to an exact time except to use גּלם which is an embryo and so at the latest at the point of an embryo one came into existence therefore breathing air was not a prerequisite to be a soul/person as was indicated for Adam. It also therefore plays into how soul is defined.
Answer: At the outset, I should point out that the Bible is not a science book. Trying to draw parallelisms between the Bible and the state of affairs of science improperly degrades the former (and I characterize this as "improper" notwithstanding that I no longer consider myself a devote person).
Genesis 2:7 is inconclusive or inappropriate for pinpointing when human life starts. Notice the preposition li (לְ) in li-nefesh hayah (לְנֶפֶשׁ חַיָּה).
Among other uses, li may serve as indirect object (dative) marker. Thus, ha-adam li-nefesh hayah ( הָאָדָם לְנֶפֶשׁ חַיָּה) may denote the act of accomplishing that man be before --or in the presence of-- nefesh hayah. Equivalently, that man be at nefesh hayah's disposition or linked to nefesh hayah. This could also be phrased as "delivering man to nefesh hayah", although doing so might needlessly invite a controversy on the absence of particle et (אֶת) in that portion of Genesis 2:7.
Who is nefesh hayah? According to the Zohar, it means this earth, Shekhinah (the Divine presence), or even the righteous. See Haqdamat Sefer ha-Zohar 1:13a, and Bereshit 1:34b and 1:47a, respectively (in the Pritzker edition of The Zohar, these references are on pages 92, 213, and 252, respectively). The Zohar surely makes additional references elsewhere.
Thus, Genesis 2:7 is less anthropocentric than the interpretations thereof arising in abortion controversies. At least from a zoharic perspective, Genesis 2:7 describes a "transaction" involving God, Shekhinah, and ha-adam. It does not refer to immediate-postnatal physiology, and therefore it is not a sound defense of abortion.
(Personally, I join your opposition to abortion, since --like you-- I support the premise that a person's life begins at conception)

Question regarding Genesis 2:7


How to get .pfx file for renewed SSL certificate

Question: We have a SSL certificate associated with a .dev domain purchased from GoDaddy about 2 years ago. The domain is associated with a website hosted in Azure, the initial certificate was uploaded as a .pfx file.
After a year the certificate expired, GoDaddy renewed it automatically but it was not installed on the server (the .dev domain is not really used).
Now the certificate was renewed again automatically by GoDaddy and I was tasked to install it again on the server. From GoDaddy I can download the certificate as a zip file (it contains 3 files, a .crt, a .p7b an a .pem) but it is only the public key, I do not have a private key as the certificate was renewed automatically, so the installed certificate on my computer cannot be exported as .pfx.
I generated a new CSR file using DigiCert Util app and I submitted a Re-Key request to GoDaddy for the certificate. I got a new certificate from GoDaddy, I installed it on the local machine where I generated the CSR but the new certificate still cannot be exported as .pfx.
What do I need to do in order to get a full certificate that can be exported as .pfx?
Comment: you need to have cert and private key to export to pfx, what is your web server and OS? are you trying to export using DigiCert Util https://www.digicert.com/kb/util/pfx-certificate-management-utility-import-export-instructions.htm if you have key and cert you can export to pfx with openssl or some other tool, e.g. openssl pkcs12 -export -out cert.pfx -inkey private.key -in cert.cer
Comment: I generated a CSR with DigiCert Util so I assume the private key have been created somewhere on my computer (Win 10 Pro). Using the CSR I re-keyed the certificate with GoDaddy. I downloaded the new certificate from GoDaddy, imported it on my machine, I still cannot export it as .pfx, I assume the existing private key has not been matched to imported certificate but I do not know why.
Answer: I solved the problem using OpenSSL. I generated a new key and CSR using OpenSSL, I re-keyed the certificate with GoDaddy and then using OpenSSL I merged the private key with the certificate downloaded from GoDaddy and got the .pfx file.


eos.io - Account freeze protocol

Question: Is there a possiblity to unfreeze a frozen account in the EOS protocol (or is it frozen forever once frozen) and if so how many BP need to vote unfreeze to unfreeze an account?
Answer: All these mechanisms for freezing accounts and enforcing ECAF rulings in general are being worked on and decisions need to still be made on the way that it's going to work in the future, but generally speaking:

Yes, it is entirely possible to unfreeze an account after it's been frozen.
The general answer for how many BPs need to vote for something to happen is what is referred to as a "supermajority" which is two thirds of the total number of BPs, plus one. Considering 21 total BPs, this would mean 21 * 2/3 + 1 = 15 BPs.
This also depends on how the account was frozen in the first place:

Accounts can be blacklisted in the nodeos configuration file on a per-node basis, and this makes that specific node reject all transactions related to that account. The issue with this is that it happens off-chain, on each BP's node configuration, so there is no enforcement of consensus over what accounts should be frozen. It has even happen that some BPs block an account while others don't.
Accounts can be frozen by temporarily taking ownership of the account itself by setting the owner permission on it to the eosio or eosio.sudo accounts, or any other privileged account that needs the signatures of the supermajority of BPs. This mechanism would be also reversible, more reliable, and more legitimate, since it would require the supermajority of BPs to be executed.

Time will tell how these things will actually play out in day to day governance of the blockchain, so it's an exciting time to also offer your opinion and participate in the debates in the governance telegram channel and other places.
EDIT:
In answer to your comments:
First of all, you only get your account frozen if there is a bug in the code of the smart contract that is deployed there or if there is an ECAF ruling against you because of breaching the terms of your ricardian contract or generally some bad behavior that needs to be stopped.
Secondly, this doesn't work like in Ethereum or other blockchains:
 - Your transactions aren't "mined". They are included in blocks that get produced.
 - The elected block producers agree on a kind of "block production plan" or schedule for the next set of blocks, so they decide which of them is going to produce which blocks at which time.
If your account is blacklisted by 20 BPs and just one forgot to blacklist it, then you would need to hope that your transaction is relayed through the network pretty much directly to that specific BP so that it won't get rejected by anybody else who did blacklist your account.
I'm not sure what you mean by blacklisting off-chain being more "trustless", since clearly in this way everyone in the network would need to trust that all the BPs are doing their job correctly and blacklisting what should be blacklisted. Doing it on-chain gives you a trustless guarantee that a blocked account won't be able to continue to do harm.
Currently there are various on and off-chain implementations for blocking or freezing accounts or broken/buggy smart contracts, what I meant before is simply that there is no community consensus yet about what the standard procedure should be for different cases that might need this.
EDIT 2:
In response to your new comment explaining that in the on-chain supermajority based method you as an individual would need to trust that the BPs don't arbitrarily freeze your account, I can say the following:
In the design of EOSIO and specifically DPoS (Delegated Proof of Stake) consensus, the token holders control their own votes, and they can (and should) vote out of power the BPs that are doing things that go against the constitution. The constitution and the Block Producer Agreement specify what exactly are the rules for BPs and when they are allowed to freeze an account. They can't simply freeze and account or block a transaction out of personal preference, even if they form a supermajority with that preference, since they would be breaking their BP agreement and would be voted out and most likely severely penalized for that behavior.
You can count on the voters voting them out because same as you, they don't want to be censored and frozen arbitrarily and they don't want BPs doing whatever they please against the constitution.
The issue with the off-chain blacklist is that it is arbitrary, since each BP can choose to put whatever they want in their own blacklist. It also doesn't fully work unless all of the BPs sync their blacklists. If they are not all blocking the same accounts, it would just make it more annoying for those accounts to operate, but they could still operate, so it's not a good solution for when you really need to block an account.
Comment: @Invader I updated the answer in response to your comments. If you find that this answers the question you created, please mark the answer as accepted so that others can benefit from it too. Thanks!
Comment: When it's in the consensus and you need a supermajority to unfreeze that is by far less trustless then off-chain. Imagine you get frozen by BP set A then you need a change of at least 1/3 of the BP set to get your tx through. In contraty with the off-chain approach you just need 1 BP to mine your tx.
Comment: I assume in the consensus approach you would need a supermajority to unfreeze right?
Comment: Currently there is nothing in the consensus in regards to this right and just 1 BP need to mine the tx that other consider to be from a frozen account (off-chain)?
Comment: I mean trustless in the sense that it's more unlikely that your account gets frozen. Note that I'm not worried currently about my account getting frozen, I just want to analyze the properties of the current protocol.
Comment: Imagine a supermajority is required to freeze an account (which lets say did nothing wrong just personal preference of the supermajority like a bank blocking txs and nobody of the other bank users cares that the bank blocked a tx) within the consensus and then you need a supermajority to unfreeze, that would be by far less trustless (in the sense of your account getting frozen) then off-chain blacklists which require a 100% consensus amongst the BPs.
Comment: @Invader I now understand what you mean. I've updated the answer again with more details.

eos.io - Account freeze protocol


Classical Limit of Schrodinger Equation

Question: There is a well-known argument that if we write the wavefunction as $\psi = A \exp(iS/\hbar)$, where $A$ and $S$ are real, and substitute this into the Schrodinger equation and take the limit $h \to 0$, then we will see that $S$ satisfies the Hamilton-Jacobi equation (for example see http://physics.bu.edu/~rebbi/hamilton_jacobi.pdf).
I understand this, however I feel that I don't understand the claim that this shows that quantum mechanics reduces to classical mechanics in the $\hbar \to 0$ limit.  I am confused because I would think that in order to show that QM reduces to CM we would need to show that as $\hbar \to 0$, $|\psi(x,t)|^2$ becomes very narrow and that its center moves in a classical trajectory, ie $|\psi(x,t)|^2=\delta(x-x_\text{classical}(t))$.  And it seems that the above argument does not at all show this.  In fact, my understanding is that all that matters for the physical measurement of position is $|\psi|^2$ (since this gives the probability distribution) and hence the phase factor $\exp(iS/h)$ seems to not matter at all.
Moreover, some books (see pg 282 of http://www.scribd.com/doc/37204824/Introduction-to-Quantum-Mechanics-Schrodinger-Equation-and-Path-Integral-Tqw-darksiderg#download or pgs 50-52 of Landau and Lifshitz) give a further argument to the one mentioned above.  They further say that if $\psi = A \exp(iS/h)$, then $|\psi|^2 = A^2$ satisfies the classical continuity equation for a fluid with a corresponding velocity $dS/dt$, which in the $h \to 0$ limit is equal to the classical velocity.
This argument makes more sense to me.  However, I still have some questions about this.  (1) I know that there are stationary states whose modulus squared does not evolve in time, which seems to contradict this interpretation of a fluid flowing with velocity v.  (2) The fluid interpretation seems to perhaps suggest to me that the wavefunction reduces in the classical limit more to a wave than to a particle.  (3)  This doesn't show that the wavefunction is narrow.
Comment: dab, consider learning $\LaTeX$, see http://meta.math.stackexchange.com/questions/1773/do-we-have-an-equation-editing-howto and http://en.wikipedia.org/wiki/Wikipedia:MATH#Functions.2C_symbols.2C_special_characters
Comment: Your question is studied and answered in my recent paper >U.Klein, _What is the limit $\hbar \to 0$ of quantum theory?,_ Am.J.Phys. vol.80, 1009 (2012). Preprint version is available [here](http://arxiv.org/abs/1201.0150).
Comment: [This](http://en.wikipedia.org/wiki/Ehrenfest_theorem) might help.
Comment: Related: https://physics.stackexchange.com/q/17651/2451 and links therein.
Answer: The subtlety is that an arbitrary wavefunction doesn't reduce to a point of the classical phase space in the limit $\hbar \to 0$ (thinking about phase space makes more sense since in the classical limit one should have definite coordinates and momenta).
So one could ask, which wavefunctions do. And the answer is that the classical limit is built on the so-called coherent states -- the states that minimize the uncertainty relation (though I don't know any mathematical theorem proving that it's always true in the general case, but in all known examples it is indeed so). States close to the coherent ones can be thought of as some "quantum fuzz", corresponding to the quasiclassical corrections of higher orders in $\hbar$.
Example of this for the harmonic oscillator can be found in Landau Lifshits.
Regarding the fluid argument. About your remark (1): the $|\psi|^2$ for the stationary state is indeed stationary, but it still satisfies the continuity equation since the current is zero for such states. Your remarks (2) and (3) are quite right because, as I already said, the classical limit can't be sensibly taken for arbitrary states, it is built from coherent states.
And also I must admit that the given fluid argument indeed doesn't provide any classical-limit manifestation. It's just an illustration that "everything behaves reasonably well" to convince readers that everything is OK and to presumably drive their attention away from the hard and subtle point -- it often happens in $\it{physics}$ books, probably unintentionally :). The problem of a nice classical limit description is actually an open one (though often underestimated), leading to rather deep questions, like the systematic way to obtain the symplectic geometry from the classical limit. In my opinion it is also connected to the problem of quantum reduction (known also as the "wave function collapse").
Answer: Good discussion. As per the comment of U. Klein above the $\hbar \rightarrow 0$ limit of quantum theory is not what people might think. It is natural, and intuitive, as explained above, to assume that the classical limit is a property of a certain class of states.
As it happens, that view is incorrect. You can actually obtain an exact recovery of Hamiltonian Classical Point Mechanics for any value of $\hbar$ using a different wave-equation:
$$i\hbar\frac{d}{dt}|\psi\rangle = [H(\langle q\rangle,\langle p \rangle)\hat{1} + H_q(\langle q\rangle,\langle p \rangle)(\hat{q}-\langle q\rangle) +
H_p(\langle q\rangle,\langle p \rangle)(\hat{p}-\langle p\rangle)]|\psi\rangle$$
This wave-equation propagates any state along classical trajectories and is unique (i.e. you can derive it and show it is unique). The nonlinearity is essential and comes from the presence of the expectation values in the parameters multiplying each of the three operator terms.
When one becomes familiar with this result then it becomes clear why there is so much confusion.
People thought that the classical limit should be contained within quantum theory.
The simple truth is that it is not. The limit does not exist within the theory.
It involves a very specific equation which is outside linear quantum mechanics.  However, it is simple, since this equation produces the expected Ehrenfest relations:
$$\frac{d \langle q \rangle}{dt} =+H_p(\langle q\rangle,\langle p \rangle)$$
and
$$\frac{d \langle p \rangle}{dt} =-H_q(\langle q\rangle,\langle p \rangle)$$
which is the classical limit of wave-packets following classical paths.
The relevant equation was first derived in K.R.W. Jones (1991), "The Classical Schroedinger Equation" UM-P-91/45 (CSE) http://arxiv.org/abs/1212.6786. A simpler version was published in 1992: K.R.W. Jones (1992), "Classical mechanics as an example of generalized quantum mechanics" Phys. Rev. D45, R2590-R2594. http://link.aps.org/doi/10.1103/PhysRevD.45.R2590. The original pre-print containing the derivation and proof of uniqueness was posted on arXiv a few days ago (see link above).
The fact that the equation is nonlinear may explain why the literature missed it for so long.  It is simple, but it is not trivial, nor is it obvious until you know it.
The derivation of the CSE involves group theory and an unusual nonlinear representation of the Heisenberg--Weyl group. The mathematical system of nonlinear quantum theory involved is that discovered first by Weinberg and (independently) by Jones.
A general prescription of how to take a classical limit using a dimensionless parameter $\lambda \rightarrow 0$ is given in the paper: K.R.W. Jones (1993), "A general method for deforming quantum dynamics into classical dynamics while keeping $\hbar$ fixed"  Phys. Rev. A48, 822-825.
There you will find the general argument of how to do a classical limit consistently so that the phase space, trajectories and all other properties are recovered.
It is surprising that the physics community has not gotten to grips with this yet. However, the question is an excellent one and subtle indeed since the mathematics involved did not exist prior to 1989. The connection between this area and the Feynman path integral is particularly interesting.
Comment: Dear Kingsley Jones: For your information, Physics.SE has a [policy](http://meta.physics.stackexchange.com/q/582/2451) that it is OK to cite oneself, but it should be stated clearly and explicitly in the answer itself, not in attached links.
Comment: Sorry, added the actual equation and explained it more fully in self-contained fashion.
Answer: The main result of the paper "What is the limit ℏ→0 of quantum theory?" is that the classical limit of quantum theory is not classical mechanics but a classical statistical theory. My "technical paper" has been written with the idea in mind to contribute to the understanding (the interpretation) of quantum theory. The
final conclusion of the paper presents - in my opinion- a strong argument in favor of the statistical interpretation of quantum theory. Motivation and conclusions are discussed in more detail in sections I and VIII of the paper. I will be happy to answer specific questions but please note that almost all answers I am able to give may be found on my website "http://statintquant.net" (I
am just updating this, will be finished soon).
Comment: I've merged you unregistered accounts again.
Comment: -1 for not explaining anything. And why are you posting a new answer instead of editing your previous post?

Classical Limit of Schrodinger Equation


There is a piece of wood caulked to the top of my bathtub. What is going on and what can I do to fix it?

Question: I'm wondering exactly what has been done to my poor tub, and what I can do to fix it. The be previously removed the caulk and replaced it on the bottom of the board, but I didn't add enough caulk to fill the large gaps, especially towards the back of the tub. I'm wondering if I can rip the wood off, and what it's hiding? Also what I would need to do once ripping it off to have a cleanly caulked tub. Or maybe I don't rip it off? I'm completely clueless. Thanks in advance.
wood piece in question, moldy caulk as a bonus
Comment: I don't have a better answer than those already below - but, just to note. You prevent cracking by filling the bath tub with water when you re-caulk it, & leaving it 24h. That way it's as sunk as it will ever get, so the caulk is never under tension, only compression.
Answer: If somebody thought that was an improvement over the underlying joint, then I wince at the thought of what's underneath it. It looks like he used ALEX caulk instead of a class 25 caulk like 100% silicone, so movement of the tub cracked the caulk.
Removing the trim may be a lot like turning on the light in a bathroom with roaches. You won't get to unsee what's under there. If the wall is significantly damaged and sprouting mushrooms, you may end up removing and replacing the wall liner. Not ready for a significant project? Best to clean up the lower joint with plastic tools (to avoid scratching the tub), and then you would apply a bathroom sealant like the previously mentioned 100% silicone. The "class 25" part means that the sealant can move 25% without cracking or debonding, so if you use a silicone alternative, it too should be qualified as "class 25." Higher classes, e.g. "class 50," are better still.
A class 25 sealant isn't sufficient, though. It should be designed for use in bathrooms to minimize mold growth potential. I always use 100% silicone if you couldn't already tell.
Ready for a significant project? Then remove that trim and be prepared to deal with a ghastly horror show. Try to use soft tools (like a plastic putty knife) to avoid damaging surfaces.
Comment: +1 for soft tools. I always carry some of the plastic razor blades in my tool box for stuff like this.
Answer: At this point all we can do is guess. My guess is that the wall, drywall, was damaged and covered up by the wood strip. You won't know until you remove it. If you do remove it and the damage isn't bad and the moisture hasn't rotted the studs, you could replace the wood with a tile border all the way around the tub.  If there's extensive damage, removing the wall upward until the damage is cleared would be required. Replace the drywall with cement board and add tile.
Comment: A new tub installation, where the wall was cut back to clear the old flange and to allow the new flange back to the studs?
Comment: @popham It could very well be.... that's where the guessing comes in.
Answer: That is a highly unusual installation. My guess is that the tub surround was either broken or installed too high.
Wood in a tub/shower area is a recipe for disaster. I would consider removing it and seeing what is behind it and go from there.
You might able to just apply new caulking from the tub surround to the tub or it might need more extensive work. The only way to find out is to remove it. You can always ask another question once you get it taken off and have more information.


What is the difference between EMF induced in Faraday's law and the potential difference due to electric field?

Question: What is the difference between emf induced in Faraday's  law and the potential difference due to electric field?
If we thake a conducting circular loop in a changing magnetic field then we have an emf induced in it.
But in electric circuits the potential difference between the two points of a conducting wire is zero.
Also explain how this emf causes the electric current to flow.
Comment: Great question... but the answer is really too complicated. [I suggest you see this thread on EE stackexchange](https://electronics.stackexchange.com/q/551244/) (and also see linked)
Comment: EMF and potential difference(PD) are very similar things. The main difference is that EMF is defined for a non-conservative force and the PD for conservative forces (or fields).  Read these pages carefully: https://www.wikiwand.com/en/Electromotive_force#overview https://www.wikiwand.com/en/Voltage
Comment: In electric circuits you have usually other things besides the wire and most of the potential drops across these other things. If you connect just the wire to the battery you will have a significant potential difference between te ends of the wire.
Answer: Yes this issue often causes confusion. The basic thing you need to think about here is the electric field. In the first instance, remove any conducting wire and just suppose there is a region of space where there is a magnetic field, uniform spatially (i.e. uniform direction and size), but changing with time. Let's say it is increasing. In this scenario there is also an electric field in that region of space. The electric field in this situation runs in circular loops around the magnetic field lines.
Ok so far so good: we have a changing magnetic field, and, in that same region of space, also an electric field running in circular loops.
Now suppose you put a conducting wire in a loop in that same region, following the direction of the electric field lines, but for the moment do not close the circuit. That is, you have a loop of wire but with a gap in it so that it does not close. What will happen?
The electrons in the wire will be pushed by the electric field, and they will move, such that they start to pile up on one side of the gap. This imbalance in the charge distribution in the wire causes a counter-balancing electric field. The electrons keep moving until this counter-balancing electric field (caused by the electrons) is equal and opposite to the one caused by the changing magnetic field. Thus when the system settles down the net electric field inside the conducting wire is zero (I am assuming the rate of change of the $B$ field here is constant).
At this point there is a build-up of negative electric charge on one side of the gap in the wire loop, and a corresponding positive charge on the other side of the gap. Also there is potential difference across that gap: it is equal to the e.m.f. which you can calculate using Faraday's law. So if you were to now connect a resistor or a light bulb or something like that across the gap, then a current would flow.
When a resistor is connected across the gap, there is a potential difference across the resistor and an electric field inside the resistor. There is no electric field inside the conducting wire (if we assume zero resistance of the wire), and the electric field just outside it is also affected by the distribution of charge in the wire.
On electric potential difference
In the above I did not mention the concept of potential difference until near the end. This is because in electromagnetism it is best to regard the fields and the charges as the main idea, and then concepts such as potential difference come in as useful tools to help in calculating and in getting insight.
Electric potential comes into its own, as a concept, in static conditions, because then we can find a function $V(x,y,z)$ such that the electric field can be written as
$$
{\bf E} = - {\bf \nabla} V.
$$
(This is a standard shorthand notation for a gradient. In terms of components it means:
$$
E_x = - \frac{\partial V}{\partial x},\\
E_y = - \frac{\partial V}{\partial y},\\
E_z = - \frac{\partial V}{\partial z}.)
$$
In non-static cases, as for example in the presence of a changing magnetic field, things are not so simple, because now the electric field is such that it can point around a loop, and this means there is no function $V$, having a single value at each location, such that $\bf E$ is its gradient. However we can still investigate quantities such as
$$
-\int_{P_1}^{P_2} {\bf E} \cdot d{\bf l}
$$
where  $P_1$ and $P_2$ are two points in space (or possibly the same point) and the integral is taken along a path running between those points. In a static problem, this integral would give the potential difference between $P_2$ and $P_1$. In a non-static problem, we may choose to give another name to the result of this integral. It is often called 'electromotive force' or 'emf' (this is arguably not a very clear name but is adopted for historical reasons). But this is just a name. You could call it a potential difference if you like, as long as you realise that you are really talking about the integral of electric field along a given path. What you know is that if a charge were to move along that path, then the net energy given to the charge by the electric field is minus the amount given by this integral. This can be deduced because the force on such a charge would be
$$
{\bf f} = q ( {\bf E} + {\bf v} \times {\bf B} )
$$
and therefore the work done when the charge moves through a displacement $d{\bf r}$ is
$$
{\bf f} \cdot d{\bf r} = q {\bf E} \cdot d{\bf r}.
$$
Here the magnetic field term does not contribute because for a charge moving along a path described by $\bf r$ we have that $\bf v$ and $d{\bf r}$ are parallel so the magnetic force is perpendicular to $d{\bf r}$.
The main point of this final section of my answer is to say that 'potential difference' and 'emf' are different words for essentially the same thing, namely the integral of electric field along a path. The reason to have two terms is that the first one (potential difference) draws attention to a useful property of static fields, and the second one draws attention to the fact that the situation under consideration is not static so we have to proceed a little more carefully in our reasoning. In particular, for a non-static case we should not assume that there is any function $V(x,y,z)$ (having a single value at each point $(x,y,z)$) which gives the electric field as its gradient.
Comment: My question is still the same that what is the difference between emf induced and potential difference? The electric field generated is non conservative in nature so no potential difference is defined for that so the emf induced and potential difference should not be the same so what's the difference??
Comment: @adityasiroutiya ok I extended my answer to deal with this.
Comment: Thanks to Andrew, You almost have solved my doubt in the later section, only one point remained unclear is that is above integral path specific? If i changes the path so the value of above integral for emf should change or not?
Comment: @adityasiroutiya yes in the case of a changing magnetic field, the curl of E is not zero, and therefore the integral is path specific.
Comment: @AndrewSteane According to your paragraph no. 5 you've written that the negative charges i.e the free electrons will pile up on the one side of the open loop and that is very true, but further you've written that the positive charges will also get piled up on the other side that seems very strange to listen, because positive charges do not have any mobility because they are not free since they are arranged and locked in their lattice patterns from this we can say that.....
Comment: ......even though the electrons are getting accumulated on the one side of the material, the density of positive charges will still remain same throughout the loop. So the electric field formed due to the counter action by the electrons of the loop will not be uniform or zero. There still might be some net electric field inside the loop but because of the accumulation of all free electrons on one side, no electrons will be left to flow since they already got accumulated and hence there won't be any current according to me what's your comment on this? Please make me aware if I'm wrong anywhere.
Answer: One possible source of confusion is that $\mathbf E = -\nabla V$
if there is no vector potential present, no changing magnetic field. Otherwise the relation is: $$\mathbf E = -\nabla V - \frac{\partial \mathbf A}{\partial t}$$
Where $\mathbf A$ is the vector potential.
From Maxwell induction equation:
$$\frac{\partial \mathbf B}{\partial t} = -\nabla \times \mathbf E$$
In the case, let's suppose that the magnetic field is in z-direction, and the circuit in the xy plane. The equation applies in the z-direction only:
$$\frac{\partial B_z}{\partial t} = -(\frac{\partial E_y}{\partial x} - \frac{\partial E_x}{\partial y})$$
If we mistakenly take $E_x = -\frac{\partial V}{\partial x}$ and $E_y = -\frac{\partial V}{\partial y}$, (without the vector potential) we would get:
$$\frac{\partial B_z}{\partial t} = -(\frac{\partial^2 V}{\partial x \partial y} - \frac{\partial^2 V}{\partial y \partial x}) = 0 \implies B_z = cte$$
But this is not true according to the hypothesis! And that is why we are puzzled by how is it possible for $\oint E.dl \neq 0$. Really,  there is no difference in the potential for the same point for example (the beggining and end of the loop integral).
If we put the correct terms for the E-fields and express $B_z$ also in terms of the vector potential, all terms cancel and the equation is fulfilled.
The conclusion is: the $emf$ can not be explained by the notion of a (scalar) $V$ potential. For example: a linear growing B-field induces a constant E-field in the wire loop. If there is only a resistor in the loop, the points just before and after it must have the same potential because they are joined by a virtually zero resistance wire (the rest of the loop). But there is a current in the resistor anyway, due to the electric field.
Comment: I am a 12th grade student, so i couldn't understand your explanation as I don't know about any vector potential, till now i thought that potential is a Scalar quantity. Also your equations are not understandable for me. A request to you is to explain me in simple terms.
Comment: "Physics is like carpentry: using the right tools made the job easier" David J. Griffiths. We can compare the electric current to the water flow of a river. In principle it is a function of the gradient of the altitude. But there are whirlpools in some places. It is not an explanation, but an example that not always a flow results from a scalar potential.
Answer: $\oint \vec{E}\cdot \vec{dl} = 0$
Is when there is an absence of a changing magnetic field. Which is what assume you mean when you say "in circuits"
This means that on a CLOSED loop, the emf is zero across this loop.
In static electric circuits, the potential difference is NOT zero about any 2 points, this is just false  If this were the case, no current would be produced.
It is only zero on CLOSED curves, where the start and end are at the same point. There is a potential difference from the positive to the negative terminal  in a circuit ,because they are not at the same point, the terminals are separated by a distance. Emf is circuits is measured with respect to a non closed curve. The closed curve is zero, because inside the battery the field points in the opposite direction cancelling out the field outside
In both cases, where the Emf is produced by the field in the battery or via faradays law, It is the electric field that is the cause of the emf. The differences in these electric fields is that one has curl and one does not.
An emf is just a fancy way of saying there exists a non zero electric field on the chosen path. Where there is an E field, there is a force on charges causing them to accelerate and produce a current.
Search up drude model of conductivity to understand why it is a CONSTANT current.
Potential difference across a zero resistance wire
Comment: "An emf is just a fancy way of saying there exists a non zero electric field on the chosen path. Where there is an E field, there is a force on charges causing them to accelerate and produce a current." And yet in electrostatics we have conductors in electric fields but without current.
Comment: By saying the conducting wire, i mean a wire without resistance. But current through a wire,across which potential difference is zero, can flow. Mathematically lim v,r tends to 0 (v/r) can be finite, so current can flow through it. Actually I was asking that if a resistance less conducting circular loop is placed in a time varying magnetic field then an emf induces in it but work is needed to go from one point to another in the loop but pd between two points in a resistance less wire is zero so what's the difference between emf induced and potential difference?
Comment: @Math By the  very definition of the electric field,  any conductor with free electrons inside them, when in the presence of an electric field will experience
a  force, this must cause a current. You I assume,  are talking about the STEADY STATE of a conductor placed in an external electric field.  A current is MOST DEFINITELY produced. This current is the reason that free electrons rearrange themselfs to counteract the EMF generated from the externally applied electric field. Once the electrons move, the field is 0, and thus the emf is 0, and thus no current
Comment: The electric field from the electrons and the electric field of the external field superpose, causing net 0 electric field. This is very different from saying that there IS an electric field in the conductor, but no current. There is an initial current when there IS an electric field. But once they cancel, the current stops.
Comment: And Aditya, I will link my answer to your question in my answer. This confuses alot of people. In short though. There IS  an emf, and if the loop is spinning at a constant rpm.  There will be a sinusoidally varying EMF. Which will cause a sinusoidally varying current. Ohms law is true for STEADY CURRENTS. Which is why the emf for 0 resistance is 0, ohms law states the emf to maintain a steady current
Comment: It doesnt actually mean there will always be a potential difference of 0 in a 0 resistance wire. Nor is it due to the fact that the "limit" "converges". It just means that in the absence of resistance  0 potential is needed to maintainca CONSTANT current.  Regardless of if there is 0 resistance. If there is a changing magnetic field In this ring. It will have THE SAME emf induced as a ring WITH resistance.
Comment: @jensenpaull Yes, and electric field causes electrons to accelerate. This is called drift current. However, if there is a gradient in charge concentration, there is _diffusion_ of charges, which can work in the opposite direction. A diode with no current flowing through it has drift current which is exactly canceled by a diffusion current. Nevertheless, there is an E field across the PN junction. If you will accept that there are cases where there is no _net_ flow of charge in the presence of an E field, then we are in agreement.
Comment: Let me be more specific. If there exists an electric field at a point in space, there will be an associated current DENSITY ($\vec{J}$). If there is zero field, then zero current density. NET current of a particular chosen surface ( cross section for example) may be zero, where one charge goes one way, and the other another way. So in a sense, yes 2 NON ZERO fields can exist with a TOTAL current being zero. I assumed we were specifically talking about current density, which can be associated with the E field at that specific point in space via the definition of the electric field.
Answer: A time varying magnetic field induces a rotational (non-conservative) electric field, which we will call $\vec{E}_{induced}$. If a conductor is present within that time-varying magnetic field, electrons in the conductor will rearrange, and that re-arrangement will cause a second irrotational (conservative) electric field to be present. We will call this the reaction field $\vec{E}_{reaction}$.
The total electric field is the sum of the induced field and the reaction field.
$$\vec{E}_{total}=\vec{E}_{induced}+\vec{E}_{reaction}$$

What is the difference between emf induced in Faraday's law and the potential difference due to electric field?

The EMF induced along a curve C which begins as point A and ends at point B (possibly the same as A) is
$$ \mathscr E_{induced} = \int_C \vec{E}_{induced} \cdot d\vec{\ell}$$
or in the case where curve C is a closed curve
$$ \mathscr E_{induced} = \oint_C \vec{E}_{induced} \cdot d\vec{\ell}$$
In this case, the induced EMF is equal to the rate of change of flux enclosed by the loop C.
$$\mathscr E_{induced} = -\frac{d\Phi}{dt}$$
If we integrate the total electric field along the curve C, we will get the voltage drop through that path C.
$$ V_C = \int_C \vec{E}_{total} \cdot d\vec{\ell}$$
This voltage drop $V_C$ is what will be used if you want to apply Ohm's law to a wire in the shape of the curve C.
$$V_C=IR$$
where $I$ is the current through the wire, and $R$ is the resistance through the wire.
$V_C$ also represents the work per charge associated with moving a test charge from A to B along curve C (assuming the electric field does not vary over the time the charge is moved from A to B).
Note that because we are working in a regime where there is a time varying magnetic field, the total electric field has a rotational component, and hence is not conservative. This means that the voltage drop along one path will generally be different from the voltage drop along a different path. That is, if the path begins at A, and ends at B, the voltage drop between A and B is not path-independent. Therefore, the notion of a potential difference between A and B doesn't really apply in this case. (At least not a notion of potential difference that can be used in Ohm's law calculations). We can however, speak of the voltage drop from A to B along a specific path C.


Affine variety wrought out of irreducible cubic polynomial in two variables, what does it look like?

Question: Let $h \in \mathbb{R}[x, y]$ be an irreducible cubic polynomial. Consider the affine variety$$\{(x, y) \in \mathbb{R}^2: h(x, y) = 0\}.$$What are the qualitatively different possibilities for what this affine variety looks like?
Comment: Could you be more precise about what you mean by "qualitatively different"? What about classification into orbits under the action of the affine group of $\mathbb{R}^2$, http://projecteuclid.org/euclid.rmjm/1250126126 ?
Answer: Let $\overline{h}$ be the homogenization of $h$. If all points of $\mathcal{V}_{\mathbb{R}}(\overline{h})=\left\{(x:y:z)\in \mathbb{P}^2(\mathbb{R}) :\overline{h}(x,y,z)=0\right\}$ are non-singular points of the projective variety $\mathcal{V}(\overline{h})$, then Harnack's theorem states that $V$ decomposes into at most two
 connected components. Each connected component $C$ is Nash-diffeomorphic to a circle and is called an oval. Two things can happen: Either the component $C$ hits the plane at infinity or it does not. Suppose that $C$ does not hit the plane at infinity. In this case we can assume that $C$ lies in the real affine set $\mathbb{R}^2$. In fact, $C$ can be viewed as an connected component of $\mathcal{V}_{\mathbb{R}}(h)=\left\{(x,y)\in \mathbb{R}^2:h(x,y)=0\right\}$. Note that although $h$ is irreducible, $V:=\mathcal{V}_{\mathbb{R}}(h)$ may still decompose into several connected components. Hence, if a connected component $C$ does not hit the plane at infinity, then $V$ might 'look' like a circle. Furthermore, by Bezout's theorem there can only be one 'circle' in the affine case. If $C$ hits the plane at infinity, then the cut with $\mathbb{A}^2(\mathbb{R})$ will not be an oval of course, but it will be the 'affine part' of the oval.
If $\mathcal{V}(h)$ has real singular points, then the above arguments cannot be applied. But at least, the singular points can be classified (see wikipedia)


GUI for iptables?

Question: I would like to secure my server and it seems that IPtables is one of the first steps. Unfortunately editing the rules in a terminal is a bit complicated and dangerous (those who ever did an iptables -F will know what I mean ;) ). Could you recommend any good graphical interfaces for managing my IPtables rules?
Answer: Firestarter has always worked well in my opinion. It supports a robust GUI and supports all options of iptables.

sudo apt-get install firestarter

Comment: Firestarter is deemed abandoned software and has been removed from Ubuntu repositories as of 13.10 Saucy Salamander. For more info, https://help.ubuntu.com/community/Firestarter
Comment: Unfortunately firestarter is abandoned. https://help.ubuntu.com/community/Firestarter It seems that at present gufw, which is a GUI wrapper around ufw, which in turn is a CLI wrapper around iptables, is the closest thing.
Answer: I've never taken time to understand iptables, but I believe gufw does the job. Even I was able to set up a firewall within a couple of minutes.
Comment: gufw wraps `ufw`, not `iptables`, and doesn't support all of `iptables`' functionality.
Answer: Try Firewall Builder.
sudo apt-get install fwbuilder

Comment: +1 I like this, although it's somewhat more complex than Firestarter. It's targeted at multi-system administrators.
Comment: Unfortunately this doesn't seem to be any less dead than Firestarter. Last update was in July 2013.
Comment: +1 for being still in available with apt-get --- getting started tutorial found here https://www.youtube.com/watch?v=yoAhn6tdlw4
Comment: From what I see it only allows you to build a completely new network configuration from scratch, which is absolute rocket science. It cannot "import" the existing reality of your network configuration and allow you to make minimal changes to it.
Answer: For most purposes, ufw (Uncomplicated FireWall) is an excellent way to build simple iptables firewalls.  The rules produced are decent, though there may be features of iptables that you need that ufw doesn't cover.
sudo apt-get install ufw

It's a command line tool, but there is also gufw if you want a GUI version.
Answer: Iptables-Editor-Gui is a gui for iptables (requires ruby and ruby-gtk2)
Source: Linuxhacks.org
Disclosure: I am the owner of Linuxhacks.org
Answer: Firestarter's still in Raspbian Stretch at least.  If you look at https://netfilter.org/ which is the effective homepage of iptables it's copyright 2014, the task hasn't changed.  Iptables probably isn't changing.  There is such a thing as maturity in software.
There are 233 repositories on Github mentioning Firestarter https://github.com/search?q=firestarter.  What's in Raspbian Stretch is Firestarter 1.0.3 copyright 2005 by Thomas Junnonen.  There are undoubtedly forks of the original Firestarter, I'm happy using the original.
I'd be wary of oversimplified software that may leave out useful features.  I've been studying iptables for a day or so, I just wanted a "second opinion".  I'll take what Firestarter comes up with and use it as a starting point.  I'm impressed that it recognized my internet connection EasyTether and seems willing to build NAT stuff for wifi.  I have an old DSL router I'll probably use for output, I just wasn't sure how to bridge from EasyTether to the router.  Iptables can do it, if it's at all like OpenBSD's pf.  Not a typical application, glad Firestarter wants to solve the problem.  It's Gnome but nobody's perfect.  Works fine in LXDE with some Gnome libs installed.
Firestarter homepage, found in help -> about http://www.fs-security.com/   Yeah, OK, it's maybe a little flaky.  I can't get to his official download page.  More importantly I can't see the iptables code it generates.  It doesn't copy, it doesn't run, it considers the gateway I'm actively using to be offline.
Answer: I'm using Elastic Firewall …worked like a breeze so far!
Plus it works on multiple machines.
You can go with a free account with them or try one of the paid plans for enhanced power. either way, the thing does a pretty good job at very reasonable rates.
Unfortunately editing the rules in a terminal is a bit complicated and dangerous (those who ever did an iptables -F will know what I mean ;) )
I know what you mean, appending rules through the command line can be time consuming and prone to so many errors, so having a firewall manager to automate Linux iptables policies can save you a lot of time and nerves.

GUI for iptables?


SSH and rsa authentication on mpich2

Question: I'm trying to build a small cluster of pi's and am using mpich2 to run parallel code. It appears mpich2 has installed ok and I'm able to ssh into both (I'm starting with 2) pi's. However when I run a program using mpich2 it makes me enter the passphrase for my rsa key pair 2 times then I have to enter the password for the other pi 3 times, each time having to press enter twice. According to this tutorial I should only have to enter those passwords once. However every time I run mpich2 it makes me enter the passwords. Does anyone have any suggestions?
Answer: did you enter any passphrase when generating your RSA keys? if yes, you have to create new keys, without a passphrase, otherwise you'll need to enter it everytime.
Comment: I understand entering it every time now but it asks me like 3-4 times. should it be doing that?
Comment: @Slayter please, read your tutorial carefully, it says: "If you leave this blank (not such good security) then no further typing of passphrases is needed."
Answer: You need to have a SSH agent running that can remember your password. If ps u -C ssh-agent does not show any running processes, check your distribution documentation for installation details.
When you have installed it, I suggest you store the passphrase with an expiration date. For instance, store the default keys for an hour:
ssh-add -t 3600

If you need to store just a single key, or a key with an odd name, use something like:
ssh-add -t 3600 ~/.ssh/id_rsa

The currently active keys are visible with ssh-add -l. Now you can ssh into Pi's without constantly having to enter a passphrase.

SSH and rsa authentication on mpich2


Difference between "a crowd" and "crowds"

Question: Please let me know the difference between a crowd and crowds.
Answer: Crowd refers to a large group of people. Its plural crowds refers to large groups of people. For example,

There was a crowd in the east yesterday.
There were crowds in the east and west yesterday.
Comment: @SamuelLijin I think *crowds* are more likely more than one crowd, since crowds are groups and by definition have bounderies or limits. I suppose its possible that *crowds* could be used as you say, but it would be helpful if you wrote an answer with some sources using it that way.
Comment: Read technically, yes. But idiomatically, "crowds" can also just refer to one very large group of people.
Comment: this is like the fish and fishes question. A fish, Many fish (of same species), many fishes (different species).

similarly, a crowd (in a single location or of single species), many crowds (multiple locations or multiple species)


What to do when your range has higher equity but less made hands on a specific board?

Question: Let's say villain on the cutoff (CO) raised first in. Hero on the button (BN) 3-bet and CO calls. Flop comes Ts9c8s. Using my ranges (shown in the screenshot below) as well as equilab I noticed that the BN has an equity edge on this street, but CO has more made hands both in terms of quantity and quality. What would your strategy be if you are the hero (or villain) in this type of situation, aka with higher equity but less made hands, or vice versa with lower equity but more made hands? Thanks!
Comment: Not too relevant, but from where are you pulling these ranges?
Answer: Looking at the ranges, it seems to me that BTN's equity advantage comes from having more Overpairs and more Top-pair hands than CO. It's also to be expected in a 3-bet pot that you would have some advantage since calling 3-bets filters the amount of overpairs CO can have on a Txx board. Problem that happened here is that the board is very connected, already making straights possible, and allowing a lot of strong draws to exist in both ranges.
That said, the CO's range is what they call a mid-heavy or condensed. He has a lot of medium strength hands, and decent draws, and less complete trash or high value made hands (you guys both have all the sets, and CO only has a tiny bit o extra straights).
In a sitation like this, you would want to bet a bigger betsize, to really put pressure on medium-strength hands and draws.
However, you can not just bet your entire range of hands for a huge sizing as you will have medium hands that want to play a smaller pot yourself, such as weak top-pair hands and decent draws.
I think you can size up with Overpairs+, maybe even TPTK, and some very strong draws. However, you will probably have to check certain strong hands so your checkback range is not going to be too weak.
CO will face a large sizing, and will have to take a more passive role in this hand, unless the board dynamically shifts (f.e. the board pairs, overcard, or the flush comes in).
Answer: When you say "hands made" I am assuming you mean that the CO has a better hand than BTN at that moment.
The big difference between having a made hand and having high equity is that a made hand doesnt have to draw to anything, its already a strong hand.
Having a hand with high equity means that on average, the hand will win after the river a high amount of the time, but not all of the time (depending on the equity).
The strategies can vary quite a bit between these two hands because both of these terms can represent a wide range of hands. For example, a high equity hand could be a draw, it could be a high card, it could be a pocket pair on a low board, etc. A made hand could represent a flopped straight, 2 pair, etc. which have varying relative strength depending on the board (for example, 2 pair or a set is not as strong on a board with 4 of the same suit or 4 to a straight).
generally, a good rule of thumb would be that ranges with a lot of made hands want to prevent ranges with high equity from realizing their equity (usually this means getting to showdown). Conversely, high equity hands want to realize their equity versus a range of a lot of made hands.


Execute CMD command from code

Question: In C# WPF: I want to execute a CMD command, how exactly can I execute a cmd command programmatically?
Comment: i don't know much about c, c++, or c#, but i would recommend programming it to write the code into a batch file, run the batch file, and then delete the batch file.
Answer: Here's a simple example :
Process.Start("cmd","/C copy c:\\file.txt lpt1");

Comment: I'm trying that, but the second parameter, the argument is not really being passed to the command window, at least not in Windows 8.1
Comment: @William I tested this in windows 10 and it works correctly.
Comment: dear @jan'splitek I'm 100% sure about this code . can you open cmd using run ? (your local variables may damaged)
Answer: As mentioned by the other answers you can use:
  Process.Start("notepad somefile.txt");

However, there is another way.
You can instance a Process object and call the Start instance method:
  Process process = new Process();
  process.StartInfo.FileName = "notepad.exe";
  process.StartInfo.WorkingDirectory = "c:\temp";
  process.StartInfo.Arguments = "somefile.txt";
  process.Start();

Doing it this way allows you to configure more options before starting the process.  The Process object also allows you to retrieve information about the process whilst it is executing and it will give you a notification (via the Exited event) when the process has finished.
Addition: Don't forget to set 'process.EnableRaisingEvents' to 'true' if you want to hook the 'Exited' event.
Answer: if you want to start application with cmd use this code:
string YourApplicationPath = "C:\\Program Files\\App\\MyApp.exe"
ProcessStartInfo processInfo = new ProcessStartInfo();
processInfo.WindowStyle = ProcessWindowStyle.Hidden;
processInfo.FileName = "cmd.exe";
processInfo.WorkingDirectory = Path.GetDirectoryName(YourApplicationPath);
processInfo.Arguments = "/c START " + Path.GetFileName(YourApplicationPath);
Process.Start(processInfo);

Answer: Using Process.Start:
using System.Diagnostics;

class Program
{
    static void Main()
    {
        Process.Start("example.txt");
    }
}

Answer: How about you creat a batch file with the command you want, and call it with Process.Start
dir.bat content:
dir

then call:
Process.Start("dir.bat");

Will call the bat file and execute the dir
Comment: Oh nvm, fixed it. Thanks a lot Carlo. Really good idea, helped a lot.
Comment: Great idea! I did the Process.Start("test.bat") but an error pops up saying "An unhandled exception of type 'System.ComponentModel.Win32Exception' occurred in System.dll". Any ideas?
Answer: You can use this to work cmd in C#:
ProcessStartInfo proStart = new ProcessStartInfo();
Process pro = new Process();
proStart.FileName = "cmd.exe";
proStart.WorkingDirectory = @"D:\...";
string arg = "/c your_argument";
proStart.Arguments = arg;
proStart.WindowStyle = ProcessWindowStyle.Hidden;
pro.StartInfo = pro;
pro.Start();

Don't forget to write /c before your argument !!
Comment: "pro.StartInfo = pro;" should be "pro.StartInfo = proStart;" yes?
Answer: Argh :D not the fastest
Process.Start("notepad C:\test.txt");

Answer: Are you asking how to bring up a command windows?  If so, you can use the Process object ...
Process.Start("cmd");

Answer: You can do like below:
var command = "Put your command here";
System.Diagnostics.ProcessStartInfo procStartInfo = new System.Diagnostics.ProcessStartInfo("cmd", "/c " + command);
procStartInfo.RedirectStandardOutput = true;
procStartInfo.UseShellExecute = false;
procStartInfo.WorkingDirectory = @"C:\Program Files\IIS\Microsoft Web Deploy V3";
procStartInfo.CreateNoWindow = true; //whether you want to display the command window
System.Diagnostics.Process proc = new System.Diagnostics.Process();
proc.StartInfo = procStartInfo;
proc.Start();
string result = proc.StandardOutput.ReadToEnd();
label1.Text = result.ToString();

Answer: In addition to the answers above, you could use a small extension method:
public static class Extensions
{
   public static void Run(this string fileName,
                          string workingDir=null, params string[] arguments)
    {
        using (var p = new Process())
        {
            var args = p.StartInfo;
            args.FileName = fileName;
            if (workingDir!=null) args.WorkingDirectory = workingDir;
            if (arguments != null && arguments.Any())
                args.Arguments = string.Join(" ", arguments).Trim();
            else if (fileName.ToLowerInvariant() == "explorer")
                args.Arguments = args.WorkingDirectory;
            p.Start();
        }
    }
}

and use it like so:
// open explorer window with given path
"Explorer".Run(path);

// open a shell (remanins open)
"cmd".Run(path, "/K");

Execute CMD command from code