What are good films for teaching data-driven decision making? (like Moneyball)

Question: What are good films for teaching data-driven decision making? In one of my classes, I asked my students to watch Moneyball at home and we discussed it in the class. It was successful and they learned much more than they could by reading a chapter of a textbook.
Are there films with similar theme suitable for my purpose?
Comment: Not a movie, but there is a TV series named "numb3rs", where each episode features a different mathematical model used by the protagonist.
Comment: Numb3rs was ridiculous. He'd have five or six robbery locations, then go "I'm going to use spatial statistics" and then pinpoint the front door of the next robbery. Which is of course where the next robbery happened. I've not seen Moneyball but whenever I've seen stats in a movie it's always been followed by me face-palming (p<0.01).
Answer: So while reading an article I came across this list.I have almost watched all them and I think you should too.I hope this helps https://www.analyticsvidhya.com/blog/2015/11/7-watch-documentaries-statistics-machine-learning/ .
Answer: You could argue that 'The big short' is data driven in the sense that a bunch of guys analyse the markets/data and use it to profit by going against preconceived ideas.
You could check out The Mathematical Movie Database for other ideas.
Id also recommend Nate Silver's book 'The Signal and the Noise: The Art and Science of Prediction' which gives some great insight on applying statistics and data analysis to real world situations.
Answer: Maybe you can find something about Sports Analytics in the Online Archive of the "11mm - International Football Film Festival Berlin". There are documentaries, short films, feature films. I cannot recommend any, because I haven't seen any of those. Maybe it still helps and you can find something from the descriptions and trailers.
http://www.11-mm.de/index.php/de/ueber-uns/film-archiv
Also, the ~15-minute TED talks by Hans Rosling, on population growth, global health and economic development are very insightful. Watch the one where he swallows a sword at the end, surprising the audience.


How can I waterproof a shower wall after a repair?

Question: I have a small pipe leak in the shower wall, I need to chisel away the plaster and waterproofing to repair the pipe. Can I just waterproof that small area after doing the repair.
Answer: In my experience, this depends on the installation.  My experience is not huge.
Ideally, what you have is a well-waterproofed shower that otherwise doesn't have problems.  If that's the case, then you should be able to do this.  Take out the tile in an area bigger than where your problem is.  If you have a Schluter-type system behind the tile, then you can open the wall, fix the problem, and use Kerdiband to seal up your seams.  Keep in mind that you're going to have additional thinset thickness as a result, so you'll have to compensate while laying the tile back in place.  If you have some other waterproofing membrane, then probably something like Redgard is something you should get to restore the membrane.  If you actually do not have a membrane, but instead just have cementitious backer units (CBUs) behind tile or, worse, mold-guard , then be aware that your existing system isn't actually waterproof to begin with, but is instead moisture-accepting.  In any of these cases, though, the main thing is to take off way more tile than you think necessary so that you can not only reach the leaking pipe, but make a flat-enough wall of the same material and installation style, then lay the repair tile over the top.
Whatever you do, don't rush the effort.  Time for materials to cure is really super important, so don't ignore the barrier manufacturer's instructions.


Can divisibility rules for digits be generalized to sum of digits

Question: Suppose that we are given a two digit number $AB$, where $A$ and $B$ represents the digits, i.e 21 would be A=2 , B=1. I wish to prove that the sum of $AB$ and $BA$ is always divisible by $11$.
My initial idée was to use the fact that if a number is divisible by $11$ then the sum of its digits with alternating sign is also divisible by 11. For example
$$1-2+3-3+2-1=0  $$
so $11$ divides $123321$. So my proof would then be to consider the two digit number $(A+B)(B+A)$ or $CC$ which clearly is divisible by $11$ by the above statement if $C$ is $1$ through $9$. However, I am having truble justifying the case were $A+B$ is greater than or equal to $10$ and it got me wondering if the more generel is true: Let $ABCD...$ be a $n-digit$ number, if $$A-B+C-... \equiv 0 (mod \ 11)$$
then  $$S=\sum_{k=1}^{n}(A+B+C+...)10^{k} \equiv0(mod \ 11)$$
I am not really familliar with the whole congruence thingy, so incase the above is trivial I would be greatful on some source which could aid the solving of the above . Any tips or suggestion are also very welcome
Comment: https://math.stackexchange.com/questions/328562/divisibility-criteria-for-7-11-13-17-19
Comment: Use `\pmod{11}` to produce $\pmod{11}$. So `a\equiv b\pmod{11}` produces $a\equiv b\pmod{11}$.
Answer: It's simpler than you are making it...and no congruences are needed:
We have $$\overline {AB}=10A+B \quad \&\quad \overline {BA}=10B+A$$
It follows that $$\overline {AB}+\overline {BA}=11\times (A+B)$$ and we are done.
Comment: Very clean, totally escaped me!
Answer: You can easily push through what you were trying though. Say your numbers are $AB$ and $BA$. If $A+B\gt 10$, then write it as $A+B=10+c$; note that $0\leq c\leq 8$, because two digits cannot add to $19$. That means that when you do the carry, the second digit is $(c+1)$, and so $AB+BA$ will be a three digit number: $1$, then $c+1$, and then $c$. At this point, your test gives you $1-(c+1)+c = 0$, so you get a multiple of $11$.
Answer: More generally, recall that the radix $\rm\,b\,$ digit string $\rm\ d_n \cdots\ d_1\ d_0\ $ denotes a polynomial expression $\rm\ P(b) = d_n\ b^n +\:\cdots\: + d_1\ b + d_0,\, $ where $\rm\ P(x) = d_n\ x^n +\cdots+ d_1\ x + d_0.\, $ Recall the reversed (digits) polynomial is $\rm\ {\bf \tilde {\rm P}}(x) = x^n\ P(1/x).\,$ If $\rm\:n\:$ is odd  the Polynomial Congruence Rule yields $$\rm\: mod\ \ b\!+\!1:\ \ \color{#c00}{b\equiv -1}\ \Rightarrow\ {\bf \tilde {\rm P}}(b) = \color{#c00}b^n\ P(1/\color{#c00}b) \equiv (\color{#c00}{-1})^n P(\color{#c00}{-1})\equiv {-}P(-1),\:$$ therefore we conclude that $\rm\ P(b) + {\bf \tilde {\rm P}}(b)\equiv P(-1)-P(-1)\equiv 0.\,$ OP is case $\rm\,b=10,\ n=1$.
Remark $ $ Essentially we have twice applied the radix $\rm\,b\,$ analog of casting out elevens (the  analog of casting out nines).


Show by series definition of exponential function that $\exp(-x) \rightarrow 0 $ for $x \rightarrow \infty.$

Question: There are many arguments I have seen using $\ln-$ arguments and other properties of the exponential function to show the existence of this limit $\exp(-x) \rightarrow 0 $ for $x \rightarrow \infty$. But does somebody have a good idea how to show the existence of this limit just by having the pure series definition of the exponential function?
Comment: I think that it is not easily possible. The sanest argument I know is that $e^{-x}=1/e^x$, and it is easy to show that $e^x$ becomes large for $x$ large positive. I myself have tried and failed to find "direct/elementary" arguments that the power series for the exponential function gives tiny values for negative arguments... (Differential equation arguments can also apply, but these are not less sophisticated than $e^{-x}=1/e^x$, I think...)
Answer: We have for $x > 0$, $\exp(x) > 1 +x$ and $0 \leqslant \exp(-x) < (1+x)^{-1}$, since $\exp(x)\exp(-x)=1$.
Hence
$$\lim_{x \to \infty}\exp(-x) = 0.$$
All of this follows from the power series representation which converges for all $x \in \mathbb{R}$:
$$\exp(x) = \sum_{j=0}^\infty\frac{x^j}{j!}.$$
By Mertens', theorem the Cauchy product of the series for $\exp(x)$ and $\exp(y)$ converges to the product of the sums
$$\exp(x) \exp(y) = \sum_{j=0}^\infty\frac{x^j}{j!}\sum_{k=0}^\infty\frac{y^k}{k!} = \sum_{j=0}^\infty \sum_{k=0}^j \frac{x^{j-k}}{(j-k)!} \frac{y^{k}}{k!} \\ = \sum_{j=0}^\infty \frac{1}{j!}\sum_{k=0}^j \frac{j!}{(j-k)!k!} x^{j-k}y^k .$$
By the binomial theorem
$$\sum_{k=0}^j \frac{j!}{(j-k)!k!} x^{j-k}y^k = (x+y)^j.$$
Hence, $\exp(x)\exp(y) = \exp(x+y).$ With $y = -x$ we have $\exp(x)\exp(-x) = \exp(0) = 1.$
Comment: nice argument, thanks.
Comment: @Landauer: You're welcome. The multiplicative rule is the key which as I said can be proved by taking the product of the two series.
Comment: Yes, indeed, and/but, if one were to quibble, this line does not actually address the power series expansion of $e^{-x}$ itself to show that it becomes small, etc. That is, (which is the only approach I know...) it uses the point that $e^{-x}=1/e^x$, and then the easy natural lower bound(s) for $e^x$ with $x>0$.
Comment: @paulgarrett You can establish $e^xe^{-x}=1$ using only the power series. Establish absolute convergence so summation order can be played with, and the following does this, where $k$ reindexes as $k=m+n$: $$\sum_{n=0}^\infty\frac{x^n}{n!}\cdot\sum_{m=0}^\infty\frac{(-x)^m}{m!}=\sum_{m=0}^\infty\sum_{n=0}^\infty\frac{(-1)^mx^{m+n}}{m!n!}$$ $$=\sum_{k=0}^\infty\sum_{m=0}^k\frac{(-1)^mx^{k}}{m!(k-m)!}$$ $$=\sum_{k=0}^\infty\frac{x^k}{k!}\sum_{m=0}^k\frac{(-1)^mk!}{m!(k-m)!}$$ $$=\sum_{k=0}^\infty\frac{x^k}{k!}\sum_{m=0}^k(-1)^m\binom{k}{m}$$ $$=\sum_{k=0}^\infty\frac{x^k}{k!}\cdot\delta_{k=0}=1$$
Comment: @alex.jordan, yes, of course. What is not clear is how to show "directly" that $\sum_n (-1)^n x^n/n!$ is rapidly decreasing as $x\to +\infty$ ... by estimating partial sums directly, etc. (And, indeed, the only sane and enduring argument about $e^x\cdot e^{-x}=1$ of course goes via the power series.)


How to Indicate Information cannot be Viewed?

Question: I have a system where we show customer name in a list, but for certain items, the customer name needs to be confidential. Right now our system displays name hidden. We are thinking of doing something that is more visually appealing and just as understandable. One option is to have a generic name that is blurred out. Or maybe make it look like redacted document? I am posting this to find out if I am missing some standard way of doing this or maybe someone will have a better idea than mine.
Answer: As you said, the customer name is confidential. I would suggest just putting that as the text, and with a soft red text color to give a better visual cue that it is confidential. Red is often the standard color for that sort of thing. If you want to stick with the default text color, italicized might be a better option than soft red.
Answer: Personally I would spell it out - use the term "Confidential" in place of the customer name; this way there is no ambiguity.  You could also make this term a link or provide a tool tip to provide further explanation i.e "This customer's name is confidential and cannot be displayed".
Providing an incorrect name or just using a colour would be confusing.  I think most people would think there was an error displaying the data rather than a business rule meaning it could not be displayed.  I think there are also issues using red (typically indicates a problem; also not clear with people with colour blindness).


Does schnitzel belong on a stick?

Question: I see recipes online about schnitzel... but on a stick. Is this actually a common practice? Was wondering if there is any cultural or historical significance to this.
Answer: Sounds likely to be "fair food" as in things primarily seen at from the food concessions at fairs, or probably if you're seeing lots of recipes providing nostalgia value for people not getting out to enough fairs with schnitzel on a stick.
Can't say I saw that particular one at the fairs I have gone to, but that may be a regional issue.
Fair food is somewhat known for offering all sorts of food on a stick if it's not directly hand-holdable.
Answer: What kind of Schnitzel are we talking about?
As this is tagged German Cuisine (Austrian Cuisine would be more appropriate) the basic classical Schnitzel is a rather thin cut of meat that gets a crust and then gets fried in a pan.
There are plenty of variants of the Wiener Schnitzel, most using pork instead of veal but none of these are using any sticks —- filled Schnitzel like a Cordon Bleue might only use them for keeping the filling in.
So serving Schnitzels on a stick seems not to be a historic way of serving
Comment: I was today years old when I learned that "Wiener" is not a German name for a type of sausage but is a German word for "of/from Vienna, Viennese".
Comment: @Abion47 actually Wiener is the name for a sausage. In Germany? we tend to name name meals after their origin like Wiener or Frankfurter (sausage) Paderborner (bread), Kasseler (cured meat or a type of bread), Braunschweiger (smoked cured soft sausage that is smeared on bread), Frankfurter Kranz (rich cake). Actually I thought for a long time that Hamburger was named after Hamburg and not as ham burger
Comment: It _is_ named after Hamburg. Though the true origins are disputed, one prevailing theory is that a hamburger is derived from the "Hamburg-style steak", created after 19th century sailors and immigrants brought the idea of finely minced beef back to New York.
Comment: Thanks for the hint, now I am curious and will be looking into this. In the end the hamburger might be  a heavily modified Labskaus :)
Answer: Skewering food items is typically done for a reason, and it's about convenience. Either you group multiple smaller items together to cook or serve them as a kind of unit or you add a form of handle that lets the cook or consumer grab the food more easily or with less mess - and both reasons can apply simultaneously, if you think about some Asian street food styles where the meat gets skewered and grilled, for example.
As a classic Schnitzel on the other hand is usually far too large to need a skewer for the former reason, which means if someone decides to put them on a stick (whether full-size or as smaller servings), it's because they are supposed to be a kind of finger food or as some kind of novelty effect. It's not the standard serving style and while I could find a few Google results, I was not surprised that they were limited in numbers.
In short - it's most certainly not the classic style of serving Schnitzel, but if it works for a buffet, party, food fair or similar, it can make sense. Does a Schnitzel "belong" on a stick? Not natively so, as a German I would usually expect it on a plate (with sides and silverware) or in a bun or sandwich and so would my Austrian friends. But if it's necessary for some reason, nobody will care.
Answer: I've lived in Southern Germany for a long time now. I have never seen a schnitzel on a stick.
As Stephie said, the typical way it's served in Germany is on a plate if eaten in a restaurant. At a fair, you won't get a Wiener schnitzel, but they do sell simpler styles, usually in a bread bun, garnished with roasted onions.
I know of one reference of a schnitzel on a stick though. In the movie "Hoodwinked", there is a food truck selling schnitzel on a stick, coupled with an inane song.
https://www.youtube.com/watch?v=tlhaqz7zfJo
The food truck in the movie is certainly intended to be humorous. I can't tell you if that schnitzel there is a satire invented by the movie crew, or a parody reproducing a food which existed before. But the way Internet works, there is a possibility that this is a minor trend started by the movie itself. If you want to know for certain, you'd need an anthropologist go hunt for older references.

Does schnitzel belong on a stick?


Can one find commuting homotopies of maps?

Question: Say $X$ is a topological space, and $f_0, f_1, g: X \to X$ are continuous self-maps. Let's assume that both $f_0$ and $f_1$ commute with $g$, i.e.
$$f_i \circ g = g \circ f_i \quad \text{for} \quad i=1,2.$$
Assume furthermore that $f_0$ is homotopic to $f_1$. Is there any hope that there exists a homotopy $f_t:[0,1] \times X \to X$ which commutes with $g$, i.e.
$$f_t \circ g = g \circ f_t \quad \forall t \in [0,1]?$$
Answer: Unfortunately, this does not work in general, here is a counterexample: Consider the maps $f_0,f_1,g: [-1,1] \to [-1,1]$ given by $f_0(x)=x$, $f_1(x)=-x$ and $g(x)=x^3$. Assume that $g \circ f_t = f_t \circ g$ for all $t \in [0,1]$. Then for any $t$ we have
$$
f_t(1) = f_t(g(1)) = g(f_t(1)) = f_t(1)^3
$$
and thus we must have $f_t(1) \in \{0,1,-1\}$, i.e. $1$ is fixed by the homotopy $f_t$, since the set $\{0,1,-1\}$ is discrete. However, this contradicts the fact that $f_0(1) = 1 \neq -1 = f_1(1)$.
Under some strong assumptions this does work though. If $X$ is additionally a real vector space and $g$ an $\mathbb{R}$-linear map, we can use the straight-line homotopy $f_t := tf_0+(1-t)f_1$. We can check that $g$ and $f_t$ commute, namely
\begin{align}
f_t(g(x)) &= tf_0(g(x)) + (1-t)f_1(g(x))\\
&= tg(f_0(x)) + (1-t)g(f_1(x))\\
&=g(tf_0(x)+(1-t)f_1(x))\\
&= g(f_t(x)).
\end{align}
I hope this helps!
Comment: Nice example, thanks!


How do logging frameworks/libraries work?

Question: For example, log4j has logging levels, INFO, DEBUG etc. When I say LOG_DEBUG (some message), will the log4j code check every time if the log_level is set to DEBUG, then if yes log the message?
If yes and if my code has a thousand LOG_DEBUG statements then isDebugLevel() is checked a thousand times.
I believe I am asking a basic question, so I am sorry if it is annoying, but I could not find answers on the web.
Comment: So what if it checks, this most of the time a simple trivial is x < y...
Answer: Yes. All loggers I am aware of will check the logging level each time a logging event occurs.
This is a simple check, so it should have no significant performance impact.


Does the Hydrogen/Chlorine reaction in light proceed instantaneously?

Question: When a mixture of Hydrogen and Chlorine is exposed to short wavelength light it reacts explosively. Does the reaction happen simultaneously through the bulk of the mixture?
Comment: There is no such thing as instantaneous in the real world. But some things can be incredibly fast. The question is how fast and this requires some insight into the mechanism. A good answer will contain this information.
Answer: It depends on how much UV is used, and how it's distributed... from a point source, the flame front would spread as if from an automotive spark plug. In fact, lasers are being tested as a means of ignition in Otto-cycle engines because the flame front can be better controlled than from a spark.
If evenly spread over the surface, as in laser fusion attempts, then the wave-front would propagate from the exterior inwards. Theoretically, using fiber optics inside the container to distribute and delay the beam equally, the explosion could propagate as fast as desired, with the caveat that the delay would still be half the largest space between tips.
This means, for example, that a one-cubic-meter container could be made to explode in far less than 1/3*108 seconds, which would be the minimum if it were to be ignited at the edge. Though not truly simultaneous throughout the container, one can approach any finite speed, even, apparently, greater than light. This is similar to other phenomena that exceed C, such as the crossing point of scissors or the phase velocity of a wave.
Comment: No amount of UV will make it *instantaneous*. you need to discuss the mechanism and the limiting speed possible to make a good answer.
Comment: The limiting speed is the time light takes to transverse the volume of gas. So, for a litre let's say it is around 300 picoseconds


Using LinearModelFit, how can I test the hypothesis $\beta=c$ with $c\neq 0$?

Question: Assuming a simple linear model $y=x\beta+u$, is there a way to test the hypothesis $\beta=c$ with $c\neq 0$, using the LinearModelFit?
Comment: I think I know what you mean, but you really should elaborate: you want to test if the slope is equal to a given value?
Comment: @J.M. Yes, you're right. Usually, we would use a wald test for large sample. I don't see that option in Mathematica
Answer: There are lots of ways.  Here's one for performing a two-sided test of the slope equaling $c$ by getting the P-value:
data = {{0, 1}, {1, 0}, {3, 2}, {5, 4}};
lm = LinearModelFit[data, x, x];
slopeInfo = lm["ParameterTableEntries"][[2]];
df = lm["ANOVATableDegreesOfFreedom"][[2]];

c = 2;
testStatistic = (slopeInfo[[1]] - c)/slopeInfo[[2]];
pValue = CDF[StudentTDistribution[df], -Abs[testStatistic]] +
  1 - CDF[StudentTDistribution[df], Abs[testStatistic]]
(* 0.0308905 *)

A quicker approach is to subtract $c x$ from each $y$ and then test for the resulting slope being zero.
data2 = data;
c = 2;
data2[[All, 2]] = data[[All, 2]] - c*data[[All, 1]];
lm = LinearModelFit[data2, x, x];
lm["ANOVATableEntries"][[1, 5]]
(* 0.0308905 *)

An even shorter approach uses LinearOffsetFunction.
c = 2;
lm = LinearModelFit[data, x, x, LinearOffsetFunction -> (c*# &)]
lm["ParameterTableEntries"][[2, 4]]
(* 0.0308905 *)

But one shouldn't just use a P-value to make a decision and, of course, one should only do this when a linear fit is appropriate.
Comment: One could also consider returning a confidence interval instead of a $p$-value. Anyway, for your "quicker approach", one could use an appropriate dot product instead to generate `data2`: `lm = LinearModelFit[data.{{1, -c}, {0, 1}}, x, x]`.
Comment: @J.M. I agree about the confidence intervals.  Those would be much more informative and more direct as one would not have to go through the gyrations to get a P-value for a hypothesis test.  (Although the gyrations are interesting...)  And, thanks, I'll use the dot product next rather than constructing a separate data set.


Nginx proxypass all the traffic except some that follows a particular URL pattern

Question: I am running an app in a docker container with nginx in front as a load-balancer. I would like to pass all the traffic to the app except one with following URL pattern
https://mydomain/peek/somepage.php

The following nginx config renders https://mydomain/peek/somepage.html from the folder peek but not somepage.php because I don't know how to write configuration for fastcgi to process the php files in this particular location.
upstream discourse {
 server 127.0.0.1:8080;
 }

location ^~ /peek {
        alias /home/forge/domain_name/public/peek;
}

location / {
         include proxy_params;
         proxy_pass http://discourse;
         proxy_redirect off; proxy_http_version 1.1;
         proxy_set_header Upgrade $http_upgrade; proxy_set_header Connection "upgrade";
        }

Can someone please help?
Answer: Duplicate of this question.
location /mydomain/peek/somepage.php {
  return 404;
}


Android SSL - SNI support

Question: I would like to know about the support for the SSL/TLS Server Name Indication (SNI) extension in the Android SDK.
Comment: even now, a number of android http library do not support SNI, i just ran into a problem, not realizing its SNI, as i keep getting fail SSL Handshake, and the moment SNI is disabled, it all works, just a note, connecting using 'HttpsURLConnection' works as it (i believe) uses OKHTTP which also supports SNI.
Answer: As far as I know, there is a partial support in Android SDK. The current situation is the following:

Since the Gingerbread release TLS connection with the HttpsURLConnection API supports SNI.
Apache HTTP client library shipped with Android does not support SNI
The Android web browser does not support SNI neither (since using the Apache HTTP client API)

There is an opened ticket regarding this issue in the Android bug tracker.
It is also possible to test the SNI support by making a connection to this URL: https://sni.velox.ch/
Comment: I am getting this following error:                              Error negotiating SSL connection.
java.io.IOException: SSL handshake failure: Failure in SSL library, usually a protocol error
SSL routines:SSL23_GET_SERVER_HELLO:sslv3 alert handshake failure (external/openssl/ssl/s23_clnt.c:580 0xaf589bac:0x00000000)"
Comment: It seems to be a problem with the server answer. Maybe the server SNI support is not enabled. I guess that the server log may give you more information.
Comment: Hi Jcs, Thanks for the help. I have tried the https://sni.velox.ch/ link using SSLCONTEXT(TLS) and SSLENGINE class available in the android sdk. I am getting the handshake. But the same code is not working on our testing server. Is there anything to do with the server? Please help me out with this.
Answer: Fortunately, HttpsURLConnection supports SNI since Android 2.3. One workaround if you need to support Android 2.2 (and older) is to set up an alternative virtual host on a unique port so that it's unambiguous which server certificate to return.
Please check this link for more information. Hope this will help future users.

Just for information
Newer versions of SSL, specifically TLSv.1.0 and later, support
  Server Name Indication (SNI), which allows the SSL client to
  specify the intended hostname to the server so the proper certificate
  can be returned.

Android SSL - SNI support


Why was camel milk not unclean?

Question: If the camel is unclean why was its milk drank?
https://biblehub.com/interlinear/genesis/32-15.htm
Comment: I don't see any textual statement that people drank the milk. 32:16 speaks of female camels that are nursing their own young https://www.sefaria.org/Genesis.32.16?lang=bi&with=Chizkuni&lang2=en . 7:2 doesn't mention camels at all https://www.sefaria.org/Genesis.7.2?lang=bi&with=Rashi&lang2=en
Comment: Note that a camel is not considered a *clean* animal as it does not have split hoofs. As a result, Noach could not bring it as a sacrifice nor did he bring seven pairs of camels. In any case, Noach 7:2 only refers to those animals that he was able to bring as a sacrifice after the flood. He was allowed to eat any animal.
Answer: The Hebrew is better translated as nursing camels, not that camel milk was sent. The second word in your link that's translated as "milk" is a verb that in the interlinear English is left unconjugated; it's not a noun (you can look at the V underneath it to tell). It helps to look at a proper translation, not just at an interlinear one. If you're looking for a free online one from the Jewish perspective, consider Sefaria or Chabad (note that in the versification commonly used by Jews, this is verse 16, not 15).
But even if you took this to be camel milk, there is no problem. This story occurred long before the giving of the Torah, and hence prior to the prohibition against drinking milk from a non-kosher animal.
Comment: @PabbleGoobs Jewish interpretation about "clean/unclean" in Noah's story is in fact "the animals that would be listed as clean later on in history." But this is a REALLY moot point. Camel milk ain't kosher. Jacob's gift was *gemalim meinikot.* Lactating camels. I'm sorry whichever translation you were using was confusing. The root of *meinikot* is always about nursing young. Milk would be *chalav.*
Comment: What about Gen 7:2?
Answer: Vayishlach 32:16

Thirty nursing camels with their young, forty cows and ten bulls,
twenty she donkeys and ten he donkeys.

RASHI

Thirty nursing camels with their young: Heb. וּבְנֵיהֶם. And their young with them.
According to the Midrash Aggadah, the word וּבְנֵיהֶם
means בַּנָאֵיהֶם, their builders [those who impregnate them], a male
corresponding to a female, but since it (the camel) is discreet in
mating, Scripture did not publicize it (Gen. Rabbah 76:7).

Yaakov carefully sent the herds in such a way as to show that he was an expert herdsman. He sent the animals that Eisav could use to breed in the appropriate numbers to be able to use as a breeding herd. Similarly, camels at that time could not really be bred by most inhabitants of that area. Thus he sent nursing camels with their young to show that he was a master herdsman who had been able to breed the camels, something that most people of his day were incapable of doing.
Additionally, this showed that he had not gotten his wealth from the blessing that Yitzchak had given him. That blessing was completely agricultural and Yaakov had gotten nothing of what that promised. This was an attempt to appease Eisav.

Why was camel milk not unclean?


Dapp connection with mobile browser

Question: How do I connect my cardano website (dapp) with cardano mobile wallet. currently, I am using yoroi-frontend for desktop browsers.
Comment: You will be more likely to get answers if you follow the guidelines in [ask]
Comment: @gRebel Thanks a lot!
Answer: Look in the window object for window.cardano and then look for the property yoroi and then you can interact with the public API.
To enable the wallet use window.cardano.yoroi.enable(). This will return an object with the endpoints to interact with the wallet or will fail if the user does not allow to read the wallet.

Dapp connection with mobile browser


Multi-monitor setup, only terminal on one monitor

Question: I currently have three monitors hooked up to my computer. I was just wondering if it is possible to have two monitors use the window manager - so, GUI - and have a single monitor not using the GUI but only command-line?
Answer: Open the terminal and move it on which monitor do you wish. Press F11 to put terminal in full-screen mode, and voilà: you have only the terminal on that monitor.
Otherwise, I think that you can do it using virtualization. But only for this purpose it seems without sense.


How to change Folder Creation shortcut in Kubuntu

Question: I am using Kubuntu for a long time but today when I tried to create a folder using Ctrl+Shift+N nothing happened and then I realised the shortcut is F10 in KDE Plasma. I tried to change it but I was not able to find any option in global or standard shortcut settings.
Answer: Open Dolphin.
In Settings, click on Configure Shortcuts…. (Alternatively, if you have the Control icon (☰) instead of the conventional menu, click on it and then on Configure Shortcuts….)
In there, scroll down till you find Create Folder…. You'll see F10 as the shortcut and possibly a blank space next to > in the Alternate column.

Click in the marked area and you'll get something like this:

Click on None (or whatever else is there) and then enter Ctrl+Shift+N if that's the shortcut you want. You may see a pop-up indicating that the shortcut may already be in use. Just click Reassign and then on OK.
That should now allow you to use Ctrl+Shift+N to open a new folder in Dolphin.
Comment: Thank you very much,will this change the shortcut through out the kde desktop?
Comment: So it is not possible to add a new global shortcut for folder creation in kde :/
Comment: Can you please clarify the context? Do you want to create a folder when in Dolphin or do you want to use the same shortcut to create a new folder on the desktop? Or how do you plan to use the shortcut "through out the kde desktop"?
Comment: I want to create the folder at both places - inside Dolphin and on the desktop.
Comment: If you create it in Dolphin in the ~/Desktop folder it will also appear on your Desktop. To directly create folders on the Desktop, right-click on a blank area on your Desktop and you'll see a "Create New" item in the context menu. That will allow you to create a new folder or a new text file on the Desktop.
Comment: Okay thanks man,I think the F10 shortcut will work on desktop to.


Why does my power supply show a signal on the oscilloscope when off?

Question: Oscilloscope shows a sinusoid V(RMS)=1.67V and V(MAX)=2.48V. I have a GW INSTEK GPS-3303 power supply and a RIGOL DS1052E oscilloscope. The supply is in series tracking mode, with the probe directly connected to the negative CH1 and ground of the probe connected to ground of the supply. The "output" button of the supply is in the off position. Master V=6.9V and I=1.06A.
Comment: Does the signal happen to be 50 or 60Hz? It may be for the same reason as this http://electronics.stackexchange.com/questions/78920/my-scope-detects-a-50hz-signal-when-the-probe-is-not-connected-to-a-circuit-is
Comment: Yes the signal is 50Hz.
Comment: All my equipment is connected to a cheap power board. I think the power board may not have aground (or may not be properly grounded)!
Answer: At 50 Hz, you're picking up mains hum.
The voltage you're picking up is quite high though.  This is often the case if your oscilloscope ground lead is not properly grounded (ie to mains ground). Since your supply has isolated outputs, you aren't connected to mains ground.  When the power supply is off it's just acting as a big antenna for that 50Hz hum.
Still, 2.48V p-p is quite a lot, are you sure your probes not in x10 mode?
Comment: The probe is not in x10 mode.
Answer:
Why does my power supply show a signal on the oscilloscope when off?

If you disconnected the probe tip of the scope and left it unconnected to anything, what signal do you see? My guess is that you'll see a significant ac waveform and this is not unexpected.
You say your power supply is "off" but there is a good possibility that the output stage is disconected and your scope probe tip is just floating and picking up voltage from the air.
If you connected a light load (such as 10kohms) across your power supply output, does the signal disappear? My guess is yes.


Override Views Contextual Filter in View Exposed Form Validation

Question: I'm using the Content:Nid contextual filter in my view (with exposed filters) and would like to dynamically set the value of this contextual filter from the exposed form validation callback. Is this possible?
Answer: Contextual filters work with arguments from the url so for example {drupal site}/search-nid/% where search-nid is the url of your page  and % is the value you wish to submit to the contextual filter.
An exposed filter tends to take it's values from the query string E.G.
{drupal site}/search-nid?field_nid=% (Again search-nid is the page url and % is the value of the nid)
If you are trying to do what I think you are, (Filter results to a specific nid using a URL to "autofill" the filter) I did this in a hook_form_submit function and not a hook_form_validate function.  The function below assumes your module is mymodule and your form id is advancedsearch_form.  The function looked as follows:
<?php
/**
 * Implements hook_form_submit()
 * Takes values from submitted form and appends them to query string
 * uses @see drupal_goto for info one $form_state['#redirect']
 */
function mymodule_advancedsearch_form_submit($form, &$form_state) {
  $qsa = array(); //for QSA arguments

  $field_nid = $form_state['values']['field_nid'];
  $qsa['field-nid'] = $field_nid;

  $form_state['redirect'] = array(
    'search-nid',
    array(
      'query' => $qsa,
    ),
    302,
  );
}

This may not be best practise (In fact I'm sure it isn't as it was a quick hack I implemented in about 5 mins!) but I hope it does answer your question.
Comment: This is not exactly what I'm trying to do. I'm not using URL arguments at all and I need to be able to dynamically setup the contextual filter from the views exposed form validation callback.
Comment: Do you mean you want to dynamically add which field is used for the contextual filter?
Comment: No, I want to setup the value of the `Content: Nid` contextual filter from inside the view exposed form validation callback.
Answer: I succeeded doing this by modifying $form_state['view']->args[0], which is the first argument (contextual filter) for the view. In my case, this was the Content: Nid contextual filter.


Why do I have to use 's in some sentences?

Question: For example: Brandon's house instead Brandon house.
How is that rule named, I'm trying to improve my English.
Comment: We call the use of the apostrophe and **s** the "Saxon genitive". There is a good question at the link in the comment above. (Another way to improve your English is to capitalize proper nouns like "English" and names like "Brandon", and also the first person pronoun **I**.)
Comment: In English, we ask *What is something called?* not *How is something called?*
Answer: 's is being used to show possession. You are asking about the possessive form.

The possessive form is used with nouns referring to people, groups of people, countries, and animals. It shows a relationship of belonging between one thing and another. To form the possessive, add apostrophe + s to the noun. If the noun is plural, or already ends in s, just add an apostrophe after the s.
  Examples

the car of John = John's car
the room of the girls = the girls' room
clothes for men = men's clothes
the boat of the sailors = the sailors' boat

Functions of the possessive
  'Belonging to' or 'ownership' is the most common relationship the possessive expresses.
  Examples

John owns a car. = It is John's car.
America has some gold reserves. = They are America's gold reserves.

So if you are trying to indicate that Brandon owns a/this house, then you should say

This is Brandon's house.

Brandon house (without the 's) is incorrect in most cases. It does not indicate possession.
If Brandon House is a proper noun, such as the name of a restaurant, then it could work.

This is Brandon House.

This essentially means This is the restaurant called Brandon House. Note that it does not mean the same as 1. It is possible that Brandon owns the restaurant, but 2 does not indicate this.
These are just a few details about the possessive to get you started.


When does $ax^2 + by^2 + c z^2 = 0$ have a non-trivial solution mod $p$?

Question: I was wondering, is there a simple criterion for the coefficients $(a,b,c)$ to know when the equation $ax^2 + by^2 + c z^2 = 0$ has a non-trivial solution mod $p$? Any comments are appreciated.
Comment: Is $p$ prime? If so then the answer is "always".
Comment: When $a+b+c = 0 \mod p$
Comment: @LordSharktheUnknown Yes $p$ is prime. Could you possibly elaborate on that little more by any chance?
Comment: Currently, the question is a bit silly - $x=y=z=0$ works for all $p$. If you are curious about solutions when not all of $x,y,z$ are zero, you should update the question to include this.
Comment: @KReiser O, yes. Of course.
Answer: Your equation always has a solution.
If any of $a,b,c$ are zero, then we can easily find a solution.  For instance, if $c = 0$, then $(0,0,1)$ is solution.  So assume $a,b,c$ are all nonzero.  We will show there is a solution by counting.  There are $\frac{p-1}{2}$ nonzero squares in $\mathbb{F}_p$ and since multiplication by $a$ is a bijection, then $A = \{a x^2 : x \in \mathbb{F}_p\}$ contains $\frac{p+1}{2}$ elements (including $0$, now).  Similarly, the set $B = \{-by^2 - c : y \in \mathbb{F}_p\}$ contains $\frac{p+1}{2}$ as well.  Since $\mathbb{F}_p$ only contains $p$ elements, then there must be some overlap, i.e., $A \cap B \neq \varnothing$.  Thus there exist $x,y \in \mathbb{F}_p$ such that $ax^2 = -b y^2 - c$, so $ax^2 + by^2 + c = 0$, hence $(x,y,1)$ is a solution to your equation.  By completing the square and doing a little algebra, one can use this result to show that every conic over a finite field has a (rational) point.
As a note, by dividing your equation through by $-c$, we can instead consider solutions to the equation
\begin{align} \label{hilbert}
a x^2 + b y^2 = z^2 \, . \tag{1}
\end{align}
Just as the Legendre symbol $\left(\frac{a}{p}\right)$ describes when the equation $x^2 = a$ has a solution in $\mathbb{F}_p$, the Hilbert symbol $(a,b)_{\mathbb{F}_p}$ describes when (\ref{hilbert}) has a solution in $\mathbb{F}_p$.  The Hilbert symbol determines whether the quaternion algebra given by $i^2 = a$, $j^2 = b$ is split or ramified, so this post gives a (much more convoluted) proof using quaternion algebras that your equation always has a solution.


Where does the RT term come from in the derivation for the activation enthalpy from the Eyring equation?

Question: It is easy to show an Arrhenius-like equation from the Eyring equation, but if you do this, you get that the activation energy is about equal to the activation enthalpy. However, the real approximation is that $E_\mathrm{a} = \Delta H^\ddagger + RT$. Why is this?
Answer: Using vibrational partition functions to define the reaction rate constant produces an equation of the form $\displaystyle k=aT^be^{-\Delta U_0^\mathrm{O}/(RT)}$ where $a,b$ are constants independent of $T$ and $\Delta U_0^\mathrm{O}$ is the difference in zero point energies at the transition state compared to reactant, this is related to what is commonly called the activation energy. This approach allows molecules to have discrete vibrational levels and so a thermal population of these. The constants $a,b$ are found from the partition functions.
The (classical) Arrhenius equation is often written as $k_A=Ae^{-E_A/(RT)}$. Taking the log of each equation and differentiating wrt. $T$ gives
$$\frac{d\ln(k)}{dT} =\frac{bRT+\Delta U_0^\mathrm{O}}{RT^2}\equiv \frac{E_A}{RT^2}$$
thus the Arrhenius activation energy is $E_A\equiv bRT+\Delta U_0^\mathrm{O}$.
(Texts on statistical thermodynamics gives details of partition functions, some book on kinetics also give this detail.)
Comment: how would you have got here from the simpler delta G of activation, which is more commonly seen than the difference in the zero point energies?
Comment: The $\Delta U$  includes the difference in the minima of the potentials and also accounts for the difference in energy of the lowest vibrational energy level. The $\Delta G$ includes entropy changes, which can be incorporated into the pre-exponential term leaving just the $-\Delta H/RT$ you ask about in the exponential. The thermodynamic approach cannot account for any quantum effects such as zero point energy.


What is reactive power?

Question: I am trying to understand what is reactive power. I have read that it has a relation with voltage, that is has a relation with the creation of a magnetic field in a motor, that it is coming and going between consumption and generators. But what exactly is the physical meaning of reactive power?
Comment: Reactive power: the flow of traffic in rush hour such that, every time you change lanes, the net result is no forward motion -- regardless of how much effort you expended changing lanes!
Comment: [This is how we illustrate reactive power to new electrical engineers](https://i.stack.imgur.com/De3Xd.png).
Comment: @Mast The beer is wrong, though; the image suggests S = P + Q, which is false; S = sqrt(P^2 + Q^2). Moreover, _personally_, using the beer analogy to introduce the concept of reactive power may be bad practice in the sense that it explains Q as a combination of P and S, but instead it is possible to talk about Q without mentioning P, S, or PF.
Comment: @AlejandroNava People tend to know both P and beer from before starting the course, hence the approach. Like many analogies and illustrations, it's not exact.
Answer:
But what exactly is the physical meaning of reactive power?

Essentially, reactive power is the component of power that has zero time average.
For example, consider a load consisting of a resistance $R$ in parallel with an inductance $L$ driven by a source with voltage $v_S(t) = V_S\cos \omega t$
Clearly, the source current is
$$i_S(t) = \frac{V_S}{R}\cos \omega t + \frac{V_S}{\omega L}\sin\omega t$$
Thus, the instantaneous power delivered by the source is
$$p_S(t) = v_S\cdot i_S = \frac{(V_S)^2}{R}\cos^2 \omega t + \frac{(V_S)^2}{\omega L}\sin \omega t\cos \omega t$$
Carefully note that the first term is never negative which is to say that the flow of energy described by this term is always from source to load or never from load to source.

However, the second term is positive over half of a cycle and negative the remaining half of a cycle.  That is, this term describes energy that flows back and forth, in equal measure, between the source and load.
Take the time average of the power over a period:
$$\langle p_S\rangle = \frac{\omega}{\pi}\int_0^{\frac{\omega}{\pi}} p_S(\tau)\:\mathrm{d}\tau = \frac{1}{2}\frac{(V_S)^2}{R}$$
and see that only the first term has a non-zero time average; the second term does not contribute to the time average energy flow.
In the context of phasor analysis, the real power (the real part of the complex power) is equal to the time average of the instantaneous power.
The reactive power (the imaginary part of the complex power) is (proportional to) the amplitude of the second term in the instantaneous power.
And this is the physical meaning of reactive power; it is a measure of the energy flow back and forth between source and load.
Comment: Thank you for the effort and the time you dedicated for such an elaborate answer. I still don't understand how as an engineer i regulate the real part of power and frequency at the same time by increasing/decreasing fuel on a generator whereas i regulate voltage and the imaginary part of power by increasing/decreasing the field current of the rotor. This is the reason i posted a new question. Maybe the relation between voltage and reactive power sheds a new light to what is voltage.
Comment: @veronika - you regulate reactive power to motors using capacitors with the induction to get a power factor close to 1.
Comment: Correction when you said "Essentially, reactive power is the component of power that has zero time average": it's not reactive power whose average is zero, but instantaneous reactive power.
Answer: In sinusoidal steady-state (linear loads, no harmonics), the meaning of reactive power is known. In non-sinusoidal steady-state (non-linear loads, harmonics), I haven't read a textbook or article, or seen a video, or heard someone explain it; even literature doesn't have a unique definition of reactive power in these conditions, let alone its physical meaning. My answer, just like the others, will explain the meaning of reactive power in sinusoidal steady-state.
A the end of this answer I included equations you might not know. Throughout my answer I'll use the passive sign convention. This means that in the circuit diagram, we draw the reference direction of instantaneous current $i(t)$ and the reference polarity of instantaneous voltage $v(t)$, such that the current enters into the device by the positive reference terminal. When this convention is used, if the instantaneous power $p(t)$ is positive it means energy is flowing to the device (or load) from the external circuit (or source); if instantaneous power is negative it means energy is flowing from the device (or load) to the external circuit (or source.) The passive sign convention is commonly used for the loads.
1. Short answer
The absolute value of the reactive power of a device is the maximum/minimum rate of transfer of energy oscillating between the external circuit and the device. The net (not instantaneous) work done or the net energy transferred by the component of current that causes instantaneous reactive power, is zero joules.
2. Summary of concepts
The active component of current is the component of current that is in phase with the applied voltage, and corresponds to the active power. The reactive component of current is the component of current that is in quadrature (makes an angle of 90 degrees) with the applied voltage, and corresponds to the reactive power.
Instantaneous power is the rate of transfer of energy, or the rate of work being done; it can also be computed as the product of instantaneous voltage times instantaneous current. Instantaneous active power is that component of instantaneous power caused by the instantaneous active component of current. Instantaneous reactive power is that component of instantaneous power caused by the instantaneous reactive component of current. Active power is the average of the instantaneous power. Reactive power is the amplitude with sign of the instantaneous reactive power.
For example, if a load has an active power of +20 W (and voltage and current obey the passive sign convention), then it means that, on average, the load consumes 20 W each cycle of the instantaneous power. That is, the load on average consumes 20 joules of energy from the source each second in each cycle. If a load has a reactive power of +-13 VAR (the sign doesn't matter here), then it means that the highest rate at which energy oscillates between the source and load, is, in magnitude, 13 W or 13 joules per second. (Remember watts [W], volt-amperes [VA], volt-amperes reactive [VAR] and joules per second [J/s] are all dimensionally the same unit.)
3. Explanation (long answer)
In sinusoidal steady-state (which means all currents and all voltages have a sinusoidal waveform), any current can be decomposed into two parts: one component that is in phase with the voltage, and another component that is in quadrature with the voltage. The sum of these two components is equal to the current, whether using phasors or instantaneous values. If using phasors, "in-phase" means "parallel to", and "in quadrature" means "orthogonal (perpendicular) to." In the case of a three-phase load, "voltage" refers to phase voltage (the voltage across the load impedance) and "current" refers to phase current (the current through the load impedance.)
The in-phase component of current is also known as the active component of current, and the quadrature component is also known as the reactive component. By the end of this answer you'll understand why these components are also called like that, but for now, just know those alternative names, which are more common in practice. Usually, instead of saying "active component of current" and "reactive component of current", people just say "active current" and "reactive current."
Here's an example using phasors (complex numbers partially representing sinusoidal signals that electrical engineers use for analysis purposes):

As you can see in the previous image, geometrically, the phasor active component (shown in blue) of the phasor current (shown in yellow) is the projection of the phasor current onto the phasor voltage (shown in green), and the phasor reactive component (shown in red) of the phasor current is orthogonal to the phasor active component. When the current is in phase with the voltage (as in ideal resistors), the reactive component is zero, and the active component is equal to the current. When the current is 90 degrees out of phase with the voltage (as in ideal inductors and capacitors), the active component is zero, and the reactive component is equal to the current. At any other angle, both components are present.
Here's an example using instantaneous values (what you'd see using an oscilloscope):

As you can see in the previous image, the instantaneous active component is in phase with the applied voltage, and the instantaneous reactive component lags or leads by 90°. This plus or minus 90° depends on whether the power factor is lagging (for RL loads) or leading (for RC loads), and the reason is as follows. In the case of an inductor, the current through it cannot change abruptly, but its voltage can; in a capacitor, the voltage across it cannot change abruptly, but its current can. The result is that in sinusoidal AC circuits in steady-state, in an inductor L the current I lags the voltage E, and in a capacitor C the current I leads the voltage E; this is where the phrase ELI the ICE man comes from.
The instantaneous power absorbed by the load, which indicates the rate of transfer of energy to/from the load, is equal to the instantaneous voltage times the instantaneous current, but we can better understand it by expressing it in terms of the applied voltage and the active and reactive components of the resulting current. Here's an example:

As you can see in the previous image,

The power (shown in yellow) due to the active component of current
results in energy (the area shown in orange) that never flows from
load to source (because this component of power is always
non-negative, using the passive sign convention.) In other words, energy
either flows from source to load, or doesn't flow (when voltage or current
are zero.) This is caused by the resistance of the load.
The power (shown in dark blue) due to the reactive component of
current results in energy (the area shown in light blue) that flows
from source to load, is stored temporarily in the electromagnetic
field, and is then released from load to source (because this
component of power alternates between positive and negative values);
it's a lossless transfer of energy (although it causes $i^2 R$ losses,
also known as copper losses, in the wires connecting the source and
load.) This is caused by the inductance and capacitance of the load.
The total instantaneous power (not shown) is the sum of these two
curves.

In the previous image, you can see the average of the dark blue curve is zero, but the average of the yellow curve isn't zero, which means the average of the instantaneous power, known as average power $P$, is just the average of the yellow curve. This implies the net energy transferred from source to load corresponds only to the yellow curve, because the dark blue curve has zero net energy. In fact, average power is also known as active power or real power (specially in power engineering) precisely because the net energy transferred from source to load corresponds to it. So, the yellow curve is known as the instantaneous active power $p_\text{a}(t)$. It can be proved that the net energy transferred or net work done in $n$ cycles of the instantaneous power is the active power $P$, times the number of cycles $n$ of the instantaneous power, times the period $T'$ of the instantaneous power: $W=P \cdot n \cdot T'$, where the period of $p$ is half the period of $v$ and $i$.
On the other hand, notice the dark blue curve contributes nothing to the average or active power (in other words, the dark blue curve doesn't contribute to the net energy transferred from source to load.) However, it still requires a fraction of the instantaneous current (which is the reactive component of current), so it's important to consider it, because a greater current requires wires with higher diameter, for example. This power interacts or reacts with the electromagnetic field, so we call it the instantaneous reactive power $p_\text{r}(t)$, and its signed amplitude is the reactive power $Q$. Here, "signed amplitude" means "amplitude with sign"; this is important because in a RL load the amplitude is positive while in a RC load is negative, as you'll see below.
To better see how reactive current doesn't contribute to the net energy transferred, look at the following image. Instantaneous reactive power (the light blue curve) has a corresponding energy transferred due to the that component of power, which is the signed area under the curve, and which I'll call instantaneous reactive energy $w_\text{r}(t)$ (the red area, which is the same as the red curve.) In the following image, the net energy transferred by the reactive current is the value of the instantaneous reactive energy, where "net" means in $n$ complete cycles (at the end of $n$ full cycles.) This is the ordinate of each black point, which is zero as you can see. But at the remaining instants, the instantaneous energy is non-zero (look at the red curve), so instantaneous reactive power is very real; when it's positive, energy is flowing from source to load through the wires; when it's negative, energy is flowing from load to source through the wires. Nothing imaginary.

Since the amplitude of a sinusoidal signal means the maximum instantaneous value, the magnitude (absolute value) of the reactive power of a device is the maximum instantaneous power (maximum rate of transfer of energy) of that component of instantaneous current that does no net work (known as reactive current, which leads or lags the voltage by 90 degrees.)
Since the instantaneous active power (and the active power) is due to the in-phase component of current, we also call the latter as the active component of current. And since the instantaneous reactive power (and the reactive power) is due to the quadrature component of current, so we also call the latter as the reactive component of current.
What about the sign of reactive power? The following examples illustrate various of the previous concepts, along with the sign of reactive power (not to be confused with the sign of instantaneous reactive power, which alternates between positive and negative values regardless of the type of load.)
4. Two examples
I made two online interactive apps in GeoGebra, in which you can see the voltage, currents and powers waveforms. The following image is for a RL (inductive) load (modeled as a resistor in series with an inductor, most common type of loads in practice, for example motors). As an example, consider an inductive load with resistance of 2.8 ohms and inductive reactance of 1.88 ohms. Suppose we apply 2.4 V rms across it, and we choose this voltage as angular reference (this means its phase angle will be 0 degrees using the cosine function.) Then the reactive power will be $Q=+0.95 \text{ VAR}$.

Notice in that case, reactive power is positive. What does this mean?! It means that the maximum rate of transfer of energy by the quadrature (or reactive) component of current, is 0.95 watts or 0.95 joules per second. That is, at the instants when $p_\text{r}(t)=Q>0$, the energy due to this component of power is flowing from source to load and its rate of transfer is maximum.
The following image is for a RC (capacitive) load (modeled as a resistor in series with a capacitor). Now consider another, capacitive load, with a resistance of 2.8 ohms and capacitive reactance of -1.77 ohms. Once again suppose we apply 2.4 V rms across it. The reactive power is $Q=-0.93 \text{ VAR}$.

Notice in that case, reactive power is negative. This means that the minimum rate of transfer of energy by the quadrature component of current, is 0.93 joules each second. That is, at the instants when $p_\text{r}(t) = Q < 0$, the energy due to this component of power is flowing from load to source and its rate of transfer is minimum (maximum in absolute value.)
As you can see, there's nothing imaginary about reactive power. Some people might have thought reactive power was imaginary, because in the complex power expression, $S = P + jQ$, it looks like it is imaginary. But what is imaginary is the number $jQ$, not $Q$. Remember any imaginary number is of the form $jb$, where $b$ is a real number!
I hope this helps in understanding what is reactive power, and just as important, why.
5. Relevant equations
In this section I'll show the equations for active and reactive components of current (both instantaneous and phasor), and, instantaneous active and instantaneous reactive power. I haven't read a textbook in which they show these, so that's why I'm sharing them.
The (real) instantaneous voltage and the (real) instantaneous current, in sinusoidal steady-state, are:
$v(t)=V_\text{max} \cos{(\omega t+\theta_v)} \tag*{}$
$i(t)=I_\text{max} \cos{(\omega t+\theta_i)} \tag*{}$
The phasors corresponding to the previous sinusoidal signals are:
$\tilde{V}=V_\text{max} \, e^{j \theta_v} \tag*{}$
$\tilde{I}=I_\text{max} \, e^{j \theta_i} \tag*{}$
Notice I'll denote phasors with a tilde on top (~), while amplitudes (either peak values or RMS values) without a tilde, for example $\| \tilde{I} \| = I_\text{max}$. The relationship between the real instantaneous values and the phasors are:
$v(t)=\Re[\tilde{V}e^{j \omega t}] \tag*{}$
$i(t)=\Re[\tilde{I}e^{j \omega t}] \tag*{}$
Using instantaneous values, the current can be decomposed into its instantaneous active and reactive components:
$i(t)= i_{\text{a}}(t)+i_{\text{r}}(t) \tag*{}$
or using phasors, the current can be decomposed into its phasor active and reactive components (which is valid because both components have the same frequency and is equal to the frequency of the total current $i$):
$\tilde{I}=\tilde{I}_{\text{a}}+\tilde{I}_{\text{r}} \tag*{}$
The phasor active and reactive components can be computed as:
$\begin{align} \tilde{I}_{\text{a}} &= \text{proj}_{\tilde{V}}(\tilde{I}) \tag 1 \\ &= \left( \dfrac{\tilde{I} \cdot \tilde{V}}{\| \tilde{V} \|^2} \right) \tilde{V} \tag 1 \\ &= \left( \dfrac{\Re[\tilde{I}] \Re[\tilde{V}] + \Im[\tilde{I}] \Im[\tilde{V}]}{\Re[\tilde{V}]^2 + \Im[\tilde{V}]^2} \right) \tilde{V} \tag 2 \end{align}$
$\begin{align} \tilde{I}_{\text{r}} &= \tilde{I}-\tilde{I}_{\text{a}} \\ &= (\Re[\tilde{I}]-\Re[\tilde{I}_{\text{a}}])+j(\Im[\tilde{I}] - \Im[\tilde{I}_{\text{a}}]) \end{align} \tag 3$
where $\text{proj}_{\tilde{V}}(\tilde{I})$ is the "projection" of current phasor onto the voltage phasor, and $\tilde{I} \cdot \tilde{V}$ is the "dot product" of the phasors $\tilde{I}$ and $\tilde{V}$. As far as I know, technically there's no such thing as the projection of a complex number onto another or the dot product of two complex numbers, but I'm borrowing this terminology from vector analysis. So, if you want you can ignore equation (1), but equation (2) is correct in all senses.
As I said, I haven't seen a textbook in electrical engineering that shows how to compute $\tilde{I}_{\text{a}}$ and $\tilde{I}_{\text{r}}$, so how do I know the equations (2) and (3)? From section 11.3 of the textbook Multivariable Calculus, vol. II (9th edition) by Ron Larson and Bruce Edwards, in which they use bidimensional real vectors instead of complex numbers. The only thing I did was adapt the equations of vector analysis into complex analysis.
Since we can decompose instantaneous current into its active and reactive components, we can also decompose instantaneous power:
$p(t)=v(t) \, i(t)=v(t) \, i_{\text{a}}(t)+v(t) \, i_{\text{r}}(t) \tag*{}$
where the first term is defined as the instantaneous active power $p_{\text{a}}(t)$, and the second term as instantaneous reactive power $p_{\text{r}}(t)$. These names are used in section 2.9 of the textbook Circuitos eléctricos [in Spanish] by Jesús Fraile Mora, although I'm using different equations than his, but both are equivalent.
In power engineering, it's common to work with RMS values instead of peak values; in such cases, we can express the following relations:
$\tilde{V}_\text{rms} = \dfrac{\tilde{V}}{\sqrt{2}} \implies \| \tilde{V}_\text{rms} \| = \dfrac{\| \tilde{V} \|}{\sqrt{2}} \implies v(t)=\Re[\sqrt{2} \, \tilde{V}_\text{rms} \, e^{j \omega t}]; \tag*{}$
$\tilde{I}_\text{rms}=\dfrac{\tilde{I}}{\sqrt{2}} \implies \| \tilde{I}_\text{rms} \|=\dfrac{\| \tilde{I} \|}{\sqrt{2}} \implies i(t)=\Re[\sqrt{2} \tilde{I}_\text{rms} \, e^{j \omega t}]; \tag*{}$
$\tilde{I}_{\text{a,rms}}=\dfrac{\tilde{I}_{\text{a}}}{\sqrt{2}} \implies \| \tilde{I}_{\text{a,rms}} \|=\dfrac{\| \tilde{I}_{\text{a}} \|}{\sqrt{2}} \implies i_{\text{a}}(t)=\Re[\sqrt{2} \, \tilde{I}_{\text{a,rms}} \, e^{j \omega t}]; \tag*{}$
$\tilde{I}_{\text{r,rms}}=\dfrac{\tilde{I}_{\text{r}}}{\sqrt{2}} \implies \| \tilde{I}_{\text{r,rms}} \|=\dfrac{\| \tilde{I}_{\text{r}} \|}{\sqrt{2}} \implies i_{\text{r}}(t)=\Re[\sqrt{2} \, \tilde{I}_{\text{r,rms}} \, e^{j \omega t}]; \tag*{}$
It's also common to choose the voltage as angular reference (this means that $\theta_v = 0°$.) Using RMS values and taking the voltage as angular reference, the general expressions for the phasor active and reactive components of current reduce to the following:
$\begin{align} \tilde{I}_{\text{a,rms}} &= \Re[\tilde{I}_\text{rms}] \\ &= \| \tilde{I}_\text{rms} \| \cos{(\theta_i)} \\ &= \| \tilde{I}_\text{rms} \| \cos{(\theta)} \end{align} \tag*{}$
$\begin{align} \tilde{I}_{\text{r,rms}} &= j\, \Im[\tilde{I}_\text{rms}] \\ &= j \| \tilde{I}_\text{rms} \| \sin{(\theta_i)} \\ &= -j \| \tilde{I}_\text{rms} \| \sin{(\theta)} \end{align} \tag*{}$
where $\theta$ is the power factor angle, defined in general as $\theta = \theta_v - \theta_i$. These expressions are valid only when $\theta_v = 0°$.
6. Active and reactive power in terms of active and reactive current
The active and reactive power in terms of the RMS active and reactive components of current are:
$P=\| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{a,rms} \| \tag 4$
$|Q|=\| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{r,rms} \| \tag 5$
But how do I prove this?
6.1. Proving equation (4)
Let's first prove equation (4). As you should know from circuit analysis or power systems analysis, it can be proved that the average power of the instantaneous power is
$P = \| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \cos{(\theta_v - \theta_i)} \tag 6$
So, if it is indeed true that $P$ can also be computed with equation (6), it means (4) and (6) can be equated:
$\| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \cos{(\theta_v - \theta_i)} \overset{?}{=} \| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{a,rms} \| \implies \| \tilde{I}_\text{rms} \| \cos{(\theta)} \overset{?}{=} \| \tilde{I}_\text{a,rms} \| \tag 7$
Thus I only have to prove that the left-hand side of (7) is equal to the right-hand side. I'll use geometry and vector analysis for this demonstration. I'll prove this for a RL load and a RC load.
6.1.1. Proving equation (7) for a RL load
In the case of a RL load, phasor current lags phasor voltage by an angle between zero and +90°: $0 < \theta_v - \theta_i = \theta < 90°$. The angles $\theta_v$ and $\theta_i$ are the angles of the phasor voltage and current, measured from the real positive semiaxis in counterclockwise direction, which results in angles between 0° and 360° (in electrical engineering it's common to use angles between -180° and +180°, but you can express a negative angles as a positive angle by summing 360°, which is valid because the sine and cosine functions are periodic.) The following image illustrates the scenario.

Between two complex numbers (or two vectors) there're two angles; in the previous image, $\alpha$ will be the smallest angle between $\tilde{V}_\text{rms}$ and $\tilde{I}_\text{rms}$. In a RL load, $\alpha=\theta=\theta_v-\theta_i$.
Remember that the dot product of two vectors, or complex numbers in our case, can be computed in two ways: as the sum of the product of their respective components (as in equation (1.2)), or, as the product of their magnitudes times the cosine of the smallest angle between them. The relevant form is the latter, so:
$\tilde{V}_\text{rms} \cdot \tilde{I}_\text{rms}=\| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \cos{(\alpha)} \tag*{}$
Substituting this "dot product" in equation (1), and writing the phasor voltage in exponential form $\tilde{V}_\text{rms} = \| \tilde{V}_\text{rms} \| \, e^{j \theta_v}$, and using RMS phasors, we have:
$\tilde{I}_{\text{a,rms}}=\left( \dfrac{\| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \cos{(\alpha)}}{\| \tilde{V}_{\text{rms}} \|^2} \right) (\| \tilde{V}_\text{rms} \| \, e^{j \theta_v})=\| \tilde{I}_\text{rms} \| \cos{(\alpha)} \, e^{j \theta_v} \tag*{}$
Thus, the magnitude of $\tilde{I}_{\text{a,rms}}$ is obtained by dropping the factor $e^{j \theta_v}$ (since this just changes the angle); we don't have to worry about the sign of the cosine function because the cosine of an angle between 0° and 90° is positive. Also, since this is a RL load, $\alpha = \theta_v - \theta_i = \theta$. Therefore:
$\| \tilde{I}_{\text{a,rms}} \|=\| \tilde{I}_\text{rms} \| \cos{(\theta)} \tag*{}$
which is equation (7)!
6.1.2. Proving equation (7) for a RC load
In the case of a RC load, phasor current leads phasor voltage by an angle between zero and +90°: $-90°< \theta_v-\theta_i = \theta <0°$. In a RC load, $\alpha=-\theta=\theta_i-\theta_v$. The phasor diagram is now:

You proceed the same as before, but now you get:
$\tilde{I}_{\text{a,rms}}=\| \tilde{I}_\text{rms} \| \cos{(\alpha)} \, e^{j \theta_v}=\| \tilde{I}_\text{rms} \| \cos{(-\theta)} \, e^{j \theta_v} \tag*{}$
but since the cosine is an even function, and the cosine of an angle between -90° and 0° is positive:
$\| \tilde{I}_{\text{a,rms}} \| = \| \tilde{I}_\text{rms} \| \cos{(\theta)} \tag*{}$
which is equation (7). Therefore, equation (4) is valid for any type of load.
6.2. Proving equation (5)
In regards to (5), reactive power is defined as:
$\begin{align} Q &= \| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \sin{(\theta_v-\theta_i)} \\ \implies |Q| &= \| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| |\sin{(\theta_v - \theta_i)}| \\ &= \| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \sin{|\theta_v-\theta_i|} \tag 8 \end{align}$
If it's true that $Q$ can also be computed with equation (8), it means (5) and (8) can be equated:
$\| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \sin{|\theta_v - \theta_i|} \overset{?}{=} \| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{r,rms} \| \implies \| \tilde{I}_\text{rms} \| \sin{|\theta|} \overset{?}{=} \| \tilde{I}_\text{r,rms} \| \tag 9$
Once again, I have to prove that (9) is a valid relation.
6.2.1. Proving equation (9) for a RL load
Let's go back to the phasor diagram used to prove (4) for the RL load, and draw a right triangle as follows:

or more clearly:

Using the basic trigonometry (definition of the sine function):
$\sin{(\alpha)}=\dfrac{\| \tilde{I}_\text{r,rms} \|}{\| \tilde{I}_\text{rms} \|} \implies \| \tilde{I}_\text{r,rms} \|=\| \tilde{I}_\text{rms} \| \sin{(\alpha)}=\| \tilde{I}_\text{rms} \| \sin{(\theta)} \tag*{}$
which is almost equation (9)! We don't have to worry about the sign of the sine function because the sine of an angle between 0° and +90° is positive.
6.2.2. Proving equation (9) for a RC load
Following the same steps, you get: $\| \tilde{I}_\text{r,rms} \| = \| \tilde{I}_\text{rms} \| \sin{(\alpha)}$. In this case, $\alpha = -\theta$, and $\theta$ is between -90° and 0°, so $-\theta$ is positive, and the sine of an angle between 0° and + 90° is positive. So this equation is also correct regarding the sign. This is almost equation (9).
In the case of RL load we used $\theta > 0$ and in the case of the RC load we used $-\theta > 0$. The only thing that matter is the magnitude (absolute value) of the difference $\theta_v - \theta_i$. So we can say that whether for a RL or RC load:
$\| \tilde{I}_\text{r,rms} \| = \| \tilde{I}_\text{rms} \| \sin{|\theta|} \tag*{}$
and this is exactly the same as (9). Therefore, equation (5) is valid for any type of load.
7. Proving the absolute value of reactive power is the amplitude of instantaneous reactive power
In my short and long answers, I said the absolute value of reactive power was the amplitude of instantaneous reactive power. But I didn't prove it, I just showed two examples. More specifically, how can I prove that the absolute value of $Q = \| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \sin{(\theta)}$ is the amplitude of $p_\text{r}(t)$?
Instantaneous reactive power, as I said earlier, was the product of instantaneous voltage and the instantaneous reactive component of current: $p_\text{r}(t) = v(t) \, i_\text{r}(t)$. Obviously, $v(t)$ and $i_\text{r}(t)$ have some amplitude and some phase shift; they have the general form: $v(t)= sqrt{2} \| \tilde{V}_\text{rms} \| \cos{(\omega t+ \theta_v)}$ and $i_\text{r}(t)=\sqrt{2} \| \tilde{I}_\text{r,rms} \| \cos{(\omega t+ \theta_{i_\text{r}})}$.
However, as we saw earlier, $\| \tilde{I}_\text{r,rms} \| = \| \tilde{I}_\text{rms} \| \sin{|\theta|}$. Also, since $i_\text{r}(t)$ is by definition in quadrature with $v(t)$, we can say that $\theta_{i_\text{r}} = \theta_v \mp 90°$, where you use the first/upper sign if the load is RL or the second/lower sign if it's RC. Substituting these expression in $i_\text{r}$, we get: $i_\text{r}(t) = \sqrt{2} \| \tilde{I}_\text{rms} \| \sin{|\theta|} \cos{(\omega t + \theta_v \mp 90°)}$.
Substituting the expressions for $v$ and $i_\text{r}$ into $p_\text{r}$ yields:
$\begin{align} p_\text{r}(t) &= [\sqrt{2} \| \tilde{V}_\text{rms} \| \cos{(\omega t+ \theta_v)}] \, [\sqrt{2} \| \tilde{I}_\text{rms} \| \sin{|\theta|} \cos{(\omega t+ \theta_v \mp 90°)}] \\ &= 2 \underbrace{\| \tilde{V}_\text{rms} \| \| \tilde{I}_\text{rms} \| \sin{|\theta|}}_{=|Q|} \cos{(\omega t+ \theta_v)} \cos{(\omega t+ \theta_v \mp 90°)} \\ &= 2|Q| \cos{(\omega t+ \theta_v)} \cos{(\omega t+ \theta_v \mp 90°)} \end{align} \tag*{}$
Now we apply the trigonometric identity $\cos{(x)} \cos{(y)} = [\cos{(x-y)} + \cos{(x+y)}]/2$:
$\begin{align} p_\text{r}(t) &= 2|Q| \left[ \dfrac{\cos{([\omega t+ \theta_v] - [\omega t+\theta _v \mp 90°])}+\cos{([\omega t+ \theta_v]+[\omega t+ \theta _v \mp 90°])}}{2} \right] \\ &= |Q|[\underbrace{\cos{(\pm 90°)}}_{=0}+\cos{(2 \omega t + 2 \theta_v \mp 90°)}] \\ &= |Q| \cos{(2 \omega t+2 \theta_v \mp 90°)} \end{align} \tag*{}$
from which is clear the amplitude of $p_\text{r}$ is $|Q|$, QED. So, indeed the absolute value of reactive power is the maximum/minimum rate of transfer of the energy flowing between source and load.
As a side note, if we proceed similarly for instantaneous active power, and remember that $\| \tilde{I}_\text{a,rms}\|=\| \tilde{I}_\text{rms}\| \cos{(\theta)}$ and $\theta_{i_\text{r}} = \theta_v$, and use the identity $\cos^2{(x)}=[1+\cos{(2x)}]/2$, then we get:
$p_\text{a}(t)=P[1+\cos{(2\omega t+2\theta_v)}]\tag*{}$
using the passive sign convention.
Comment: In summary, in *sinusoidal* steady-state, the magnitude (or absolute value) of the reactive power of a two-terminal load/network/device is the maximum rate of transfer of the energy oscillating between the load/network under consideration and the external source/network. (By the way, "maximum rate of transfer of oscillating energy" is a synonym of "amplitude of the instantaneous reactive power".) Well, turns out IEEE standard 1459-2010 agrees with my conclusion; they say on 4: "The magnitude of the reactive power $Q$ equals the amplitude of the oscillating instantaneous reactive power $p_q$".
Comment: Complex numbers are an algebra: a vector space with a multiplication.  For complex numbers, the vector space is the set of two element vectors over the real numbers.  So it's totally ok to talk about the projection of one complex number onto another.
Answer: It means that the current an voltage are not exactly in phase with one another as they would be for a resistor.
If the phase difference between the voltage and the current is $\delta$ then the power dissipated is in a component is $V_{\text{rms}} I_{\text{rms}} \cos \delta$.  This is called the read or active power and is measured in watts.
The reactive power is $V_{\text{rms}} I_{\text{rms}} \sin\delta$ and it is measured in volt-ampere reactive or var.
It is the maximum power absorbed or given out by reactive circuit elements.
So for your motor the coil would have inductance and resistance (and possibly capacitance).
The resistive part would dissipate energy as heat whereas the reactive part would store and then give back energy.
If $\delta = 90^\circ$ then the current and voltage are such that every quarter of a cycle the reactive component is absorbing power (the magnetic field in an inductor is increasing or the electric field in a capacitor is increasing) and then for the next quarter cycle the the reactive component is giving out energy (magnetic/electric field decreasing).
Comment: Thank you very much for your answer. But i still don't understand the google explanations. Why is reactive power related to voltage?  How it "returns" to the source when a magnetic field decreases?
Answer: I'm an electrical engineer also at odds with how has reactive power being described or rather thought about, and this is because utilities don't like it. This is because reactive power brings losses to transmission lines, but you know what? It's innevitable, because of two reasons:

To correct one of your concepts, reactive power (KVAR) is a natural component of electrical power (KVA), not of voltage. The simplest way I put it is: KVA = KW + KVAR And then is a matter of how much of each, which brings the other point.
Many loads need it, in fact every magnetic load will use it. So, unless there are devices that supply reactive power at the load, the utility generators provide it...naturally. Along with whatever undesirable effects.
Comment: Reactive power is not inevitable, it can easily be eliminated from motor circuits using capacitors. Every electrical engineer or even competent electrician should know that.
Comment: This part of my answer - unless there are devices that supply reactive power at the load - goes along with your comment, Ms. Hirose. The reactive component is not being "eliminated", it's supplied by the capacitors.
Comment: S is not P + Q, it's sqrt(P^2 + Q^2). That's as wrong as the beer analogy many prople use.
Answer: Another relative term..."work", which for electricity usually means some measure of heat. Saying that reactive power doesn't perform real work, seems to be incorrect. Of course, for an iron heating element, the reactive component is nearly zero (it's there, just contributes almost nothing), but the fan...it needs this reactive component, and so...it's doing something.
Magnetic systems, as in...a maglev train system is pushing up and forward tons of weight...at very high speeds; isn't that work? Maybe there's not much heat (well, transformers and cables will get hot), but there is certainly a big force.
I think it's important to understand that some concepts were influenced by whatever knowledge was at the time; new knowledge should modify them. "Work" means something is moving, something is being done, which uses energy or power. It's not just the real (watts) component or the imaginary (reactive) component, they're both contained in the electrical energy form, and the nature of the load determines how much of each component will be used.
Answer: Reactive power is basically the pulsating power whose time average for a complete cycle is zero. This power does not perform any real work. Here real work means that heater will not produce heat using this power or fan will not rotate merely by reactive power.
For physical significance, just take the example of a coolie carrying your language. In terms of physics he is not performing any work but still we pay them. The same analogy can be applied for reactive power.
If voltage and current are in quadrature as is the case with pure inductor / capacitor, the real power consumed will be zero. But in case of pure resistance, as the current is along the voltage, it will consume real power and hence heating effect will be produced.
Source: Active, Reactive and Apparent Power - Electrical Concepts
Comment: Correction when you said "This power does not perform any real work": At *each instant*, yes, it **does** work. On average, it doesn't; that is, the *net* work done is zero.
Comment: I suppose you actually mean "luggage" in "coolie carrying your language." It also appears that coolie is a rather pejorative/derogatory, you may want to consider changing it to something less demeaning.

What is reactive power?


Local minimum and maximum points of $f(x)= \frac{x^2}{x-5}$

Question: Trying to find the local minimum and maximum points of the function
$$f(x)=\frac{x^2}{x-5}.$$
I have tried desmos calculator and notes.
Comment: downvoted because it's not clear if you want x-5 or x^2/(x-5)
Comment: Are you allowed to use methods from calculus?
Comment: Calculus is allowed
Comment: Hint (without calculus): $f(x) = 5\left(\frac{5}{x-5}+\frac{x-5}{5}\right)+10\,$. Let $u=\frac{x-5}{5}$ and consider the extrema of $u + \frac{1}{u}\,$.
Comment: See also :https://math.stackexchange.com/questions/472169/find-extreme-values-of-frac2xx%C2%B24/472193?noredirect=1#comment4960526_472193 and  https://math.stackexchange.com/questions/1240279/how-to-find-the-maximum-and-minimum-of-the-function-fx-frac3xx2-2x
Answer: Hint:
Write $x-5=y$
For $y>0\iff x>5,$  use AM-GM inequality.
What if $y<0?$
Answer: $$\left(\frac{x^2}{x-5}\right)'=\frac{2x}{x-5}-\frac{x^2}{(x-5)^2}=\frac{x(x-10)}{(x-5)^2},$$
which gives $x_{min}=10$ and $x_{max}=0$.
Id est, $(10,20)$ it's a minimum point and $(0,0)$ it's a maximum point.
Also we can use the following way.
If $(x,k)$ is an extremum point in our case then the equation
$$\frac{x^2}{x-5}=k$$ or
$$x^2-kx+5k=0$$
has one real root, which gives  $\Delta=0$, which is $k^2-20k=0$ and from here
we can get the answer.
Answer: With the techniques from calculus, you can find local minima and maxima using Fermat's Theorem:

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then the derivative at any local extremum is 0.

Thus take the derivative, and set it equal to zero.  Doing this gives
$$ 0 = \frac{\mathrm{d}}{\mathrm{d}x} \left( \frac{x^2}{x-5} \right) = \frac{2x(x-5) - x^2}{(x-5)^2} = \frac{x^2 - 10x}{(x-5)^2} \iff x \in \{0, 10\}.$$
Note that $0^2/(0-5) = 0$, and $10^2/(10-5) = 20$.  Thus the extreme points are
$$ (0,0) \qquad\text{and}\qquad (10,20).$$
It can be confirmed that these are a local maximum and minimum, respectively, by applying the second derivative test, or by noting that the first derivative changes sign (from positive to negative, and from negative to positive, respectively) at these points.  You can also read this off the graph, if you are willing to trust technology, or check the left- and right-hand limits at the asymptote---the fact that
$$ \lim_{x\to 5^{-}} \frac{x^2}{x-5} = -\infty $$
and that
$$ \lim_{x\to -\infty} \frac{x^2}{x-5} = -\infty,$$
combined with the fact that this function is continuous on the interval $(-\infty, 5)$, indicates that the single local extremum in this interval must be a maximum.  The other extreme value can be checked by similar reasoning.


Is it possible to have two sources of audio output to different speakers in Windows 10?

Question: I have a Bluetooth speaker that I'd like to stream my Spotify music to while at the same time having my YouTube video from Chrome output sound to my monitor speaker. I'm using a DP cable. Is this possible to do in Windows 10?
Comment: There's an experimental feature in Spotify that allows you to decide which audio device it should use. See the description at https://community.spotify.com/t5/Live-Ideas/Allow-user-to-select-default-sound-device/idi-p/5351 on how to enable it.
Answer: I found two applications that allows you to do this.
CheVolume (Paid) - 7 day trial version available

Audio Router (Free) - open source alternative to CheVolume

Note: If not working you need to enable stereo mix in sound properties. Refer this video or this GIF.

Update after spring 2018
With windows 10 spring update 2018, Microsoft is offering a similar solution. You can check the below GIF or this youtube video.

Update December 2018
EarTrumpet - Quick, simple control of all your audio devices from the taskbar. Get it here.

Supported operating systems

Windows 10 1803 (April 2018 Update)
Windows 10 1809 (October 2018
Update)
Comment: As far as UX goes, the free one looks better
Comment: Thank you. That link to the video about how to control sound sources in Windows 10 was helpful (https://youtu.be/XbkHvQOHpdg?t=132) because I was having trouble using VoiceMeeter Potato.
Comment: Of these (Windows 10 itself, CheVolume, Audio Router, and EarTrumpet), only EarTrumpet remembers and re-applies the setting if you close and re-open the routed program!! I tested them all August 2020.
Comment: I've used audio router to acheive this and it works well. I would live if it had the ability to persist, so audio from one app allways gets routed to the output device specified. Perha ps it does this, but I overlooked it.
Answer: Though Codelt already answered the question, it should be known that the terrific Pulse Audio has builds available for Windows:

PulseAudio on Windows
Setting up PulseAudio 1.0 beta for Windows

Pulse Audio has been providing this feature on Linux desktops for about four or five years, therefore it is stable and reliable. Additionally, it is very scriptable so the user can have it automatically route based on conditions after some configuration. So far as I know neither of the mentioned Windows-only software are scriptable.
Answer: You can now do this in later builds of Windows 10.
I don't know on which build it started.
Go to Settings --> Sound.
At the bottom, there's Other sound options -->
App volume and device preferences.
Comment: Can you find (or create) a new version of that image that's clearer?  Like maybe black on white?
Comment: I see this in both Windows 1809 and 1903 (didn't try 1909).  It lets you pick the default output and input devices, and the default output/input for each application.
Answer:
Right click on sound icon near clock.
Select Open Sound Configuration.
Scroll down until Device preferences and ...
Change configuration.

https://i.stack.imgur.com/jslKM.jpg
Comment: Can you be a bit clearer on how to change the configuration?


How do I query XY Event Layer in Model Builder?

Question: I am trying to build a model to input table data, make xy event layer and then select features by an SQL expression and make that a shapefile. What tool should I use for the SQL expression to select the features?
Comment: I think you can do it with "selectLayerByAttribute"
Comment: When I use that I get this error: ERROR 000339: Input C:\Users\cbriglia\Desktop\GIS\Shapefiles\test does not have OIDs
Item not found in this collection.
Failed to execute (Select Layer By Attribute). I tried using 'Copy Features' tool and then Select Layer by Attribute but my only input options are 'Environments' and 'Preconditions'
Comment: Would you be able to edit your Question to include these additional important details that will assist answerers to know what you have already tried without success, please?  Additional information usually helps shed additional light on any problem.
Answer: If you are taking a non-arcgis table and making the XY Event Layer directly from it, you can't run a select by anything because ArcGIS does not allow selections on tables without an ObjectID field.  Therefore, there are a couple of solutions I can think of.
You could import the CSV or whatever your source table is to a temp/scratch workspace as a FileGDB table (or a DBF table).  Then run the make xy event layer on that.  Finally, use the Feature Class to Feature Class tool on the event layer and add your SQL query directly to the "Expression" field in the Feature Class to Feature Class GP Tool.  This should allow you to save it out as a GDB feature class or shapefile.
Or, you could use the make xy event layer directly on your source table, then use feature class to feature class on the event layer to save it as a shapefile/feature class (Note: You can NOT use a SQL expression at this point this way since your source data won't be from a standard arcgis source with OID field).  Then, once you have a feature class, you could use the Make Feature Layer GP tool on that, adding a SQL expression in the expression parameter of the tool for the features you don't want (ex: if you want all the ones where the Type field is 1, you would do an expression like Type <> 1).  Then, you would run the delete features tool on the resulting feature layer and that should delete those selected records.
Hope those help; like I said initially though, the important thing to remember is if it doesn't have an ObjectID field, you won't be able to run any attribute queries against it (whether it's a table, event layer, etc...).
Comment: You helped me so much. I ended up going with the first option, loading the table into a FileGDB first and then running the Feature Class to Feature Class tool. Thank you!
Comment: If your question is answered and problem solved, then you may want to either mark my answer as accepted or post your own answer detailing what you did and mark it as accepted.  But either way, if you mark an answer as accepted, it will remove it from the unanswered questions section of this site, so people can more quickly sort through questions still in need of help answering.  PS: glad it helped.


Can I use #8 wire for a 42 Amp heater?

Question: Use 8 gauge instead of 10 on 8 foot run from panel to garage ceiling heater? Heater is GHP model EG10000DH, 10000w, 240 volt, 41.66 amp and 7500W, 36.1 amp. Have plenty of 8 gauge...
Comment: what is your question?
Comment: You'll have plenty of #8 AWG left after this because you can't use any of it.
Answer: 8 gauge is for 40 amps at best.  You will need 6 gauge on a 50 amp breaker.
Don't even think of 10 gauge.
You need to go by the max amps of the heater, plus a derate of 80% so the wire does not heat up and burn the house down.
Comment: @Tommy  Manufacturers can use smaller gauge for plug wires.  If they used a 10 gauge wire and plug is okay, but you would need to use 6 gauge to wire the receptacle.  The 10 gauge on the heater would have been tested, but your wiring is not tested and needs to go by code.  10 gauge is suppose to be used only with 30 amp breakers, which your heater would probably trip all the time from over load.
Comment: Maybe. Depending on the insulation, #8 may be rated for 50A @ 75 degrees C.
Comment: @JulieinAustin - maybe - as long as the terminations in the heater and on the breaker are 75C rated too.
Comment: Yup, but since the poster has the #8 already, and 75C breakers are common enough, all that matters next is the heater terminations, which I'd wager are 75C because it's a heater. I do agree, as a practical matter, that the OP should just use #6 and call it good.
Comment: Very interesting. The manual calls for 10 guage, but my 8 guage is really not good enough...I'll check the wire rating. Regardless, sounds like 6 guage with 50 anp braker. Thx.
Answer:
10000w, 240 volt, 41.66 amp and 7500W, 36.1 amp

The figures marked in italics are not high/low. That is the figure when used in a 208V circuit, which I doubt you have.
It's a 10,000 watt heater. Amps are 41.66A.
You will need to derate that by 125% giving 52.0 amps, and the wire and circuit must be rated for 52 amps.  50A wire will not cut it.
6 AWG copper or 4 AWG aluminum are both good for 55A.  That is your minimum wire size here.  A 50A circuit breaker might work but you are supposed to take 52A and then go to the next available size, which will be 60A.  If the instructions say otherwise, follow the instructions.  You must install according to instructions, NEC 110.3.

Have plenty of 8 guage...

Then trade it on Craigslist etc. for the cable which you do need.

Can I use #8 wire for a 42 Amp heater?


$\neg \neg A \supset A$ in intuitionistic logic

Question: I read that the formula $\neg \neg A \supset A$ is not intuitionistically valid. Why is that? If someone can explain me I would be glad.
Comment: Blog post about this:L http://blog.plover.com/math/IL-contradiction.html
Comment: The *intuitionsitic* rejection of $\lnot \lnot A \supset A$ is due to its equivalence with *LEM* : $\lnot A \lor A$, that is "unsound" for [Intuitionsitic Logic](http://plato.stanford.edu/entries/logic-intuitionistic/).
Answer: Remember that : $\lnot \lnot A \supset A$ is : $((A \supset \bot) \supset \bot) \supset A$.
According to the BHK-interpretation, a proof of $A \supset \bot$ is a procedure $\gamma_0$ that transform a proof of $A$ into a proof of $\bot$.
A proof of $(A \supset \bot) \supset \bot$ is a procedure $\gamma_1$ that, when "applied" to $\gamma_0$, extract a proof of $\bot$.
Thus, a proof of $((A \supset \bot) \supset \bot)$ must be a procedure $\gamma_2$ that, when "applied" to $\gamma_1$, extract a proof of $A$.
But then, having extracted it, we can use it with $\gamma_0$ to get a proof of $\bot$, and this is impossible : the calculus must be sound, i.e. $\bot$ (the absurdity) has no proof.
Comment: Thanks for the great answer. Can you please take a look at the following problem when you have time? http://math.stackexchange.com/questions/1547669/how-to-show-validity-of-a-skolem-formula-in-lk


My dog ditches my husband while walking, if he knows I'm at home

Question: Every time my husband takes the dog on the trail off leash, he turns around and comes back,if he knows I am home. If I am gone, he walks off leash just fine, and doesn't run home. My husband can be stern, but I don't know why the dog comes back. It really upsets my husband, to turn around and poof the dog is gone. He's ready to quit walking him. The dog lags behind normally sniffing where other dogs urinate, but will come when called.  Why is this happening?
Comment: Walk the dog on leash? Focus on training come, sit, stay, heel, so the dog understands what you're asking it to do?
Comment: Welcome to pets.SE! Do you really want to know why it happens, or do you more want to stop it? As more clear your aim/question is, as more focussed the answers will be you get :)
Comment: actually both. Why does he do it, and how to stop it. I know my husband should insist that the dog stay in front of him, so he can keep an eye on him, but he probably won't remember to do that.
Comment: Thanks, he's an Australian Shepard so he really needs to run. He knows all of those commands, thank you for your contribution.  It's a wooded area with a trail loop, it also has a field on one end. He always comes when called. He knows sit, down, wait, out, up, off, come, place, roll over, bang, you're dead, go, this way, with a point to the direction, quiet, speak, spin, heel and no pull. I'm wondering if he thinks I need to be there?
Answer: Honestly, it's quite normal for dogs to have one favorite person. It's not like he doesn't like your husband, it's just that you are his social anchor and he prefers to be near you. As humans that hurts our feelings, but for a dog that is pe4rfectly normal.
One possible way around that is to play games with your dog while he's on the leash and continuing the game when he's off leash. That way he (hopefully ;) ) is concentrated on the game and not on you back home. It may take a while to break his habit and keep him with your husband.
Another thought: what happens when your dog runs back to you? Do you talk to him in a high voice like after he did something right? Do you get up and walk with him instead? Australien Shepherds are very intelligent and your dog may find the result of running back home more rewarding than walking with your husband. Bluntly said he may be manipulating you into giving him what he desires.
My personal opinion is this: if you cannot control your dog while off the leash (like stopping him from running onto the street in front of a car), then don't take him off the leash. Dogs have their own minds and some will never follow commands reliably off leash, so that's no failure or anything of yours. Why doesn't your husband finish the walk before letting your dog off the leash? Isn't a 10-minute walk at least better than a 1-minute walk and running back?
Comment: Thanks! Actually the 1st time I think I got really nervous and called to check on my husband, the 2nd time I'm like what are you doing here, and the 3rd time I leashed him up to take him back to my husband. I wasn't feeling well, but I took him out on the trail anyway. You are so right, he is very capable of working us! The trail is nowhere near any traffic, nor is the short distance to our house, so he's not in danger.
Comment: meanwhile my husband hasn't been taking him now. Thx you made some good points!