What does it mean for a particle to have volume?

Question: As far as I understand, we do not know whether an electron is a point particle or not, but we have a very low upper bound for its potential volume. If it turns out that electrons do have volume, what exactly does that mean? What constitutes this volume and how does it compare with the classical notion of volume?
Comment: Possible duplicate of [What is the mass density distribution of an electron?](http://physics.stackexchange.com/q/24001/)
Comment: This is a very speculative question. I am curious if somebody can elaborate on something else than superstrings...
Comment: Ah then maybe I need to open this up to purely mathematical explanations? And I am very interested string theory, so superstrings suit me just fine ;)
Comment: This question has been asked many times before : http://physics.stackexchange.com/search?q=electron+volume
Answer:
As far as I understand, we do not know whether an electron is a point particle or not, but we have a very low upper bound for its potential volume.

It depends what one means by low, physics is not a hand waving discipline. There exist experimental limits as seen in this answer to a related question. It is a low limit with respect to its black hole horizon , but very stringent with respect to laboratory measurements.
In addition, the very successful and predictive standard model of particle physics , depends on the postulate that all elementary particles in the table  are point particles.
Postulates are the physics axioms that pick up the subset of mathematical solutions of the equations in the physics models which are relevant to the data and observations. The mathematical successes of the model are quite high as seen in the recent LHC experiment, and thus there is no need to worry about the point nature of the particles.

If it turns out that electrons do have volume, what exactly does that mean?

It means that the postulate of "point particles" has to be changed to accommodate new data and still keep the successes of the previous measurements and observations. Newtonian physics still works fine in its dimensional region of validity, even though we know it emerges from an underlying  quantum mechanical frame work and at large dimensions is superseded by General Relativity.

What constitutes this volume and how does it compare with the classical notion of volume?

This is for a hypothetical case,  the limits given from experiments  are for a classical volume in three dimensional space d(v).
At the moment string theories postulate a one dimension extension, from a point to a line . A line still does not have volume, so it is all a matter of theoretical conjecture for distances smaller than the Planck length .
On the experimental side:in quantum mechanics volume can be gauged by interactions and interactions are bound by the Heisenberg Uncertainty principle and are momentum dependent. That is why the concept of "crossection" exists. It gives the interaction "area" for a given momentum. In this sense the "area" of an electron can be bounded and this is what the experiments utilize in setting the limits.
Comment: The last part of your answer is what I was trying to get at; the quantum mechanical notion of volume. Could you possibly expand on that? Maybe in the context of non-fundamental particles if that's more reasonable.
Comment: I am an experimentalist. Here are some introductory lectures on strings http://www.physics.ntua.gr/corfu2009/Talks/bachas@lpt_ens_fr_01.pdf  also this https://indico.cern.ch/event/431029/


EigenValues and EigenVectors in PCA?

Question: can you please tell me why eigenvalues are used in PCA. Specifically why and how does it explain the variance of the components
Answer: I believe that PCA is one of the most well-suited applications of eigenvector/eigenvalues.

Raw data can possibly be coupled(made complex) with principal component.
Some coupling can be thought of linear transformations.
Eigenvalues and eigenvectors can be used to decouple linear transformation.
Calculate the accuracy of mapping to the new set of data.
Let's package it nicely with a name called PCA.

I found a good youtube video that explain PCA well.
To explain coupling, think of two variables, area,$A$ and perimeter,$P$ of a rectangle. We know that $A = width * height$ and $P = 2(width+height)$. Thus we can say that $A$ and $p$ is coupled with $width$ and $height$ of the rectangle, changing $A$ will also change $p$ because of the change of $width$ and $height$. Though the example I given is not linear, you can get the idea.

EigenValues and EigenVectors in PCA?


How to reinstall Playstore?

Question: Do you have any Idea to reinstall Playstore?
Do I need to Flash it?
Do I need to reset factory?
Do I need to download apk?
Can you help me?
Please?
Device:Cloudfone 700d tv
I don't know how to reinstall this
I install apk version result: Google play has stopped
I install modded playstore: not enough memory
Comment: @RossC I very much doubt that. Might work to *update*, but not for a "fresh install" (as it requires permissions only granted to system apps). See: [How can I install the Google Apps Package (Play Store, …) on my Android device?](http://android.stackexchange.com/q/27808/16575)
Comment: @RossC To *delete* it you'll definitely need root, as it's a system app. While in order to simply *update* it, it should be enough to "uninstall updates" and wait for it to update again.
Comment: You can install it as an APK. Google search latest Google Play store. AndroidPolice.com is a safe, reputable site that will have it.
Comment: @Izzy but to delete it in the first place you'd need root access so you can reinstall it surely? I can on my phone here
Answer: Have you actually deleted it, or only disabled it?  Go into Settings > Apps, page over to 'All', and see whether Google Play Store (and Google Play services) aren't still there on the list; if they are, go in and re-enable them.

How to reinstall Playstore?


Number of ways to color such that one color always leads

Question: There are n boxes drawn out in a line. We have two colors, blue and red. We start coloring boxes from left to right. At any instant we want to color the boxes in such a way that number of boxes colored blue should always be greater than the number of boxes colored red.All boxes have to be colored. What are the total number of distinct coloring possible.
Answer: I will list a particular colouring as a pair of blue and red, such as $(4,2)$ representing a colouring that has $4$ blue and $2$ red.
So you start with one box. It must be blue.
$$(1,0)$$
The next box must also be blue.
$$(2,0)$$
Since you have two more blue than red, the next one you do can be either.
$$(3,0),(2,1)$$
For $n=4$, the valid combinations are $(4,0)$ and $(3,1)$. The only way to reach $(4,0)$ is from $(3,0)$. But you can reach $(3,1)$ from either previous set.
$$(4,0),2(3,1)$$
For $n=5$ the valid combinations are $(5,0)$, $(4,1)$ and $(3,2)$. Like above, some of these can be reached in several ways, giving you
$$1(5,0),3(4,1),2(3,2)$$
And you can continue this
$$1(6,0),4(5,1),5(4,2)$$
$$1(7,0),5(6,1),9(5,2),5(4,3)$$
$$1(8,0),6(7,1),14(6,2),14(5,3)$$
A pattern has emerged. The ways to a valid pair $(b,r)$ is the sum of the ways to get to $(b-1,r)$ and $(b,r-1)$. In the case of $r = b+1$, it's just $(b,r-1)$. In the case of $(b,0)$ it's just $(b-1,0)$ since you can't have a negative $r$.
What this forms is version of Pascal's triangle but with everything to one side of the middle, and the middle itself, zeroed out. This is known at Catalan's triangle, and fortunately the sum of any diagonal in that table is given by a simple formula,
$$\sum_{i=0}^n(n-i,i)=\binom{n}{\lfloor\frac{n}{2}\rfloor}$$
However this is for a zero index, so we just need a bit of an offset.
$$\text{ways to colour}=\binom{n-1}{\lfloor\frac{n-1}{2}\rfloor}$$


Calculating probability of being $m$ and $(n-m)$ both prime in Goldbach conjecture

Question: The Prime Number theorem asserts that an integer $m$ selected at random has roughly a $\frac{1}{\ln{m}}$ chance of being prime.Thus if $n$ is a large even integer and $m$ is a number between $3$ and $\frac{n}{2}$, then in a paper which approaching the heuristic probabilistic argument is saying one might expect the probability of $m$ and $n-m$ simultaneously being prime to be $\frac{1}{\ln{m}\cdot \ln{(n-m)}}$.
So, it assumes that the events that $m$ and $n-m$ are prime are statistically independent of each other. But how they can be independent ? I am not getting it... as we are working with a fixed even number $n$, so events $m$ and $n-m$ being prime should be dependent . . .and also this relation ignores some correlations between likelihood of $m$ and $(n-m)$ being prime. For instance, if $m$ is odd then $(n-m)$ is also odd and if $m$ is even then $(n-m)$ also even ,similarly if $3$ divides $n$ and as $m$ was a prime distinct from $3$, then $(n-m)$ would also be coprime to $3$ and thus be slightly more likely to be prime than general numbers... Can anyone explain how they are independent(if they are) or, can explain what will be the probability(closed form if possible) of being $m$ and $(n-m)$ both prime where $m\in [3,\frac{n}{2}]$ ?
Comment: Yes this is the worst case.
Comment: In fact , the numbers are not independent, the argument is only heuristic. The main dependence is that $m$ and $n-m$ are both odd, but this is no problem because it is even an advantage increasing the chance that both are prime. I assume that the paper deals with more details, such as how $m$ and $n-m$ have to be chosen to rule out a small factor. By the way, the probability $\frac{1}{\ln(m)}$ for a random integer to be prime is not at all a bad approximation, if $m$ is large.
Comment: @Peter  Yeah, I am agree with the fact that $\frac{1}{\ln{m}}$ is a good approximation. But, my question is about the independence of the events of being both of them prime. Can we make a better approximation ? A better formula dealing with the flaws(e.g; I have mentioned some of them ) ...
Comment: Seems difficult. But anyway, every ruled out factor improves the chance of getting primes, so the given formula should be the worst case, or do I miss something ?
Answer: Selecting a random number is independent of selecting another random number.
But the probability that $m$ is prime is not independent of the probability that $n-m$ is prime.
This can be shown by finding a list of remainders of $m$ when we divide by some list of primes, and finding the remainders for $n$ divided by those too, then simply calculating wether $n-m$ is divisible by any of those primes with modulo  arithmetic. Information about $m$ affects the probability that $n-m$ is prime, so they are not independent.
The heuristic you mentioned is just something to compare other theories too, to help locate important details relevant to some problem about the distribution of primes. It shouldn't be taken as true.
A different (related) probabilistic approach to Goldbach's conjecture is to consider the list of remainders of $n$ when divided by a list of prime numbers, then determine the possible remainder pairs of two numbers that would add up $n$, then find the probability that a pair of numbers less than $n$ satisfies those conditions.
For example; consider $n=12$. Observe that $9<12<25$ thus all composite numbers upto $12$ are divisible by a prime number less than or equal to $3$. So consider what $12$ divided by those primes are. For both of those the remainder happens to be $0$. The more interesting fact is that $m$ and $n-m$ have to have remainder $1$ and $2$ when divided by $3$, and that they have to both have a remainder of $1$ when divided by $2$.
Now imagine all numbers are now represented by their remainders $(r_2, r_3)$. $m$ and $n-m$ must be equal to $(1,1)$ and $(1,2)$. Perhaps the probability of a pair of numbers fitting the representation and being less than $12$ (but more than $1$) can be calculated? In this case it can, but for larger $n$ it requires some more details to take into account to create a better probabilistic model, just as the one you have mentioned.
Comment: In the example I have used, we'd need to take into account remainders when divided by 5 too.


How does one factor how much saving time is worth?

Question: I recently was reading this comic from xkcd, and it got me thinking. How does one determine how much one's spare time is worth? It seems to me that in particular, a person on salary has a fixed income, and thus it's not like they could work for an extra hour to make up the money, thus it might almost be worth it.
Comment: Unlike money you earn, you don't have to pay income taxes on money you save. 9 minutes for a dollar is $6.67/hr, but if you're in a 40% tax bracket it's more like $11.11/hr. You can do much worse in a pinch.
Answer: If you enjoy driving 5 minutes out of the way to get gas (or mowing the lawn or whatever), then it is perfectly rational to do it.  If not, the value of your time is how much you would value (not necessarily how much you would get paid) doing something else.
MrChrister is right, the concept behind the comic is opportunity cost.  In a nutshell, if driving an extra 5 minutes for gas is complete drudgery to you and you would only do it to save money, may as well get a part-time job at minimum wage instead.  If you would rather, say, go watch TV instead of getting that part-time job, then you value TV-watching at greater than the minimum wage, and it is still irrational to drive 5 minutes for the gas.
Basically, the answer to your question is to figure out what else you could be doing that offers similar overall pleasure/pain to the task at hand and see how much you would get paid to do that instead.  It doesn't matter that you are on a fixed salary.
Answer: This really is different for every individual.
Years ago, I saw a guy in line at Costco to return a pencil sharpener. The thing cost about $20. Forget the gas. I recognized the guy, he was a mutual fund manager with a million dollar plus salary. But I guess he valued his own time at far less on Sunday than the $500/hr he made during the week.
As your question suggests, the starting point is your wage, but for extra money, it goes down from there.
The question has merit, I just think that few people stop and do the math. There are also some things people just enjoy doing, like mowing a lawn, which I'd rather pay for. Often, people will pick and choose that way and not really decide whether they are working for themselves at a certain wage.
Comment: Returning the sharpener is about the principal.  I paid for a working pencil sharpener and got home and it never worked.  It is not so much about getting the money back and making sure that Walmart(or whomever) does not benefit for screwing me.  There are lots of reasons for mowing your own lawn.  My personal one(aside from being a tightass) is it provides me with exercise outside.  It gets me away from the computer and makes me do work.  If I had a job that was outside and kept me in better shape and paid as well I too would pay to have it done.
Comment: I think for VIP-types, doing normal things is a comfort as well. Cutting the grass or returning a pencil sharpener feels like some sort of accomplishment vs having a flunkie do it.
Comment: @Chad - I hear you. Psychology is an interesting phenomenon. If I offered that guy $100 to drive to/from Costco and wait on line, he'd think I was crazy. But as it was his own $20 sharpener, the decision process changes. The grass? I understand. "I like cutting my grass" is sufficient. I'm finishing my basement, myself. And it's unrelated to the savings. All about me doing it with my own two hands.
Comment: Who cares about how much you make already? It's irrelevant. Rational economic decision-making happens on the margin. What matters is which you need more: another $1 or another 9 minutes of [whatever you would do instead] which -- I'm just taking a guess here - almost certainly isn't "additional work at my usual job that would make me more than $1."
Comment: @fennec - one's hourly wage sure does come into play for one reason or another. Even the cartoon compared it (the savings) to minimum wage. Try offering my kid $2/hr to stuff envelopes, she'd laugh in your face (no disrespect intended, she just knows not to take a bad deal) as her wage for babysitting is $6-8/hr, $10/hr for brats. Whether the return needs to be higher than wage as in "I'd rather have vacation than work more" or lower "I really need some extra bucks" is another story, but the wage itself is the starting point in this discussion.
Answer: I think the key to this question is to think about how people tick. The underlying assumption in any decision that measures monetary value or other economic principles is that we're rational actors in a global marketplace.
Life isn't a ledger or a taxi on a meter. If you have time on your hands or enjoy the act of driving when you run your errands, that diversion to save a dime on gas isn't a net loss, it just is. That may be a different story if you need to get your rental car returned so that you can fly home -- in that case, you're ok with getting hosed by the gas station strategically located between hotels and the airport.
Our motivations are deeper than that. Some people get a feeling of satisfaction or security from always getting the "best deal". Some folks visit a particular gas station because the clerks are friendly or the coffee is good. Others want to pay more -- they buy "premium" gas from Exxon/Mobil because they are premium people who "buy the best".
Comment: Or conversely, as @user4127 mentioned in another question, you may value "sticking it to" companies that try to *hose* you by going out of your way to avoid doing business with them.
Answer: I'm with several of the others answering the question in that it isn't about opportunity cost so much as the value of the time savings.
For example:
I'm sure I could re-shingle my roof and save a ton of money on labor, but the marginal cost of having someone do that for me in the 100+ degree Texas heat makes it worthwhile to outsource that job.
How much I could have earned at my job for however many hours it takes doesn't really play into it unless I am specifically giving up an opportunity to earn money instead of working on my roof. For example, it might tilt the balance more towards outsourcing the roof job if I had an offer for contract work at $100/hr that I wouldn't have time to do if I worked on the roof. Also, considering the time I'd probably take off from work to do the job is paid vacation it isn't like I'm giving up some of my paycheck to do that job. The real variable is how much value I assign to being able to NOT work on my vacation days.
The bottom line is this:
What your time is worth is variable and depends on the REALISTIC opportunity cost of using that time on a money making endeavor instead of the option you are considering.
Comment: Say one earns $50K/yr, but as OP states, not like this person can just work more, and say part time is not an option. It would seem like if they are on a tight budget, that even $10/hr saved by doing certain things such as cutting coupons or labor that might get hired out makes sense. The marginal utility of the next dollar vs next hour does not have the same ratio for everyone. I'll gladly spend 3 hours in an auditorium to watch my daughter dance for 10 minutes, but would pay a high school student to take her from school to her dance lessons (which I am not permitted to watch).
Comment: You can also **consume** the time. This is like paying $1 for 9 minutes of leisure. Consider how much money you have already, and how much leisure you have already. **This is probably the decision most people will make,** not a decision about earning another $1 another way.
Comment: Right on, fennec. Time is the most non-renewable resource any person has.
Comment: @fennec this becomes more difficult it your budget is already full.  what if I would like to pay that $1 for 9 minutes of leisure, but I can't spare a dollar?  my time definitely seems more valuable but unless I can find a way to unlock it (earn more) I can't make that trade.  Also, we may reach diminishing returns: I would like to work an extra minute, but I'm mentally drained, so now my time is worth less in terms of what I can actually accomplish.
Answer: I guess you could figure it out based on your total income, and the total number of hours it takes to generate that income if you want to do it simply.
Count you job, side work, soda can deposits, and saving earned directly by effort (coupons and deal shopping)
But the real answer to the question is understanding Opportunity Cost and what you could be doing instead.  The problem with opportunity cost is the value system that judges the worth of the other opportunities is a deeply intrinsic factor that cannot be judged by anybody else.
Answer: For me, I factor in a couple of things.
First of all, there are occasional opportunities for me to get more money at work, that happen infrequently. If I'm in one of those periods, I pretty much value my time at my salary, as far as doing something that I'm not great at or enjoy. I also am more likely to buy food to fill me up, park in a parking garage instead of walking 15 minutes to park for free, etc.
But, when I don't have extra incentives, I tend to think more of my budget, and think about things from there. I might spend 4 hours to fix my microwave, when it might come down to only saving $50-$100, even though my time from work would seem more valuable.
Answer: I think it depends on how much time I am saving and what I am saving doing.  I do not go out of my way to save a few pennies on gas.  But If i know that I will be traveling somewhere with cheaper gas anyway I will wait to get there to fill up.  But I if I need gas before there I will fill up at the place i stop rather than just getting enough to go to the new location.
I heard it best said as being dollar smart instead of dime foolish.  Going out of your way for gas like this is dime foolish.  On the other hand if you can bundle your trip to the store nearby or other tasks it certianly adds to the value.
Comment: "penny wise and pound foolish"?


How do I segment a single jpeg into multiple images?

Question: I can only print up to a c plot size photo. I want to segment a single jpeg image to make a much bigger (composite), image. How do I do this?
Answer: It sounds like you want to do a tiled print. There are ways to do it yourself in software such as photoshop and/or gimp. I have never tried doing it that way. There are also websites that do a pretty good job for you. I think those are more limited by file size so if you are working with a large file you will probably end up doing it yourself. Checkout http://www.blockposters.com/ or http://rasterbator.net/ and see if they do what you want. Also, checkout a fellow stackexchange post I found that I think covers how you would do it yourself in photoshop. Which involves using slices https://graphicdesign.stackexchange.com/questions/7480/how-to-cut-a-large-photo-into-a-grid-of-smaller-photos-automatically-in-photosho
Also, if you do not have photoshop or gimp you might try a tool specifically designed to do this called PosteRazor.
Answer: Irfanview can probably help you.
Photoshop can do it too.
Can you show a link to a product like the one you want to make?


Error while editing a Search Box Web Part in SharePoint 2013

Question: We've added a Search Box web part on a page. After editing the page, when a user tries to edit the Search Box web part, it gives him error "Sorry, Something went wrong". He has 'Design' permission on the site and page. This error happens only for this particular user but for his colleague who has same permission as this user, is able to edit the search box.
Answer: Go to the masterpage gallery and check the permissions for the library. The permissions shouldn't be inherited and the user should have "read"-permission. As default the "all authentificated user"-Group have read-permission to this library. I got the same error as you and it worked fine for me.
Comment: Yes, that was the issue, user didn't had permission on masterpage gallery. The issue was resolved after giving permission there. Had figured this out a long time ago but thanks anyway for answering. Marking your answer as correct one!


What is the advantage of putting capacitors in parallel with glass fuse tube

Question: I have seen a board and I wonder to know what would be the advantage of having capacitor in parallel with fuse.
Comment: @JackB: It is about the ceramic disk capacitors soldered right onto the fuses.
Comment: When a fuse is in parallel with a capacitor the capacitor will be shorted until the fuse blows. If a fuse blows you want a circuit to be **off** and not still connected via a capacitor. Your question makes no sense as it makes no sense to connect a fuse in parallel with a capacitor. The fact that a fuse is close to a capacitor doesn't mean they're in parallel. If the fuse and cap are really in parallel on this board then whoever designed it like that has to explain why it needs to be like that.
Comment: @Bimpelrekkie  It doesn't make sense to me either, especially for AC, but this could be a DC rail.
Comment: I assume the question is about the fuses next to each of the blue contact block on the bottom and right edges. There is an electrolytic which may or may not be in parallel with the fuse, and a yellow epoxy-encapsulated component which looks like it is soldered to the fuse.  Maybe it's an attempt to make sure the fuse blows on DC but not AC?
Comment: Looks like a nice fuse puller handle to me.
Answer: The capacitor is placed for protecting fuse in switch on/off system.
Most loads are inductive such as motors and light bulbs, and in switching on or off, they will make a huge voltage from inductor equation, in switching on, this voltage will cause a huge current in a very little time, this can break your fuse without a SC, etc
By using this capacitor (typically ceramic) you can filter this current and protect your fuse from damaging from this current.
Comment: This is what came to my mind if you couldn't get ahold of slow blow fuses. Tacky though.
Answer: Are you sure they are capacitors and not some polyfuse or thermistor?
This doesn't look like your standard manufacturer bodge, but more like an amateur fix which may explain why there are more questions than answers.
Soldering to a fuse is a tacky way to attach something across those two points. Why not solder to the holder or the pins on the reverse side?
Cheap small fuses are often secured to the element inside by nothing more than solder and the end caps by adhesive. In my experience, even briefly, soldering to the end cap causes the adhesive to fail and may reflow the internal soldered joint potentially upsetting the element causing the connection to break. Not all the time but enough that i consider this technique janky.
Also, is the user expected to solder a "capacitor" to the new fuse if the old one blows?


Pashley Pickle child's tricycle wheel rims to buy?

Question: Currently refurbishing a child's Pashley Pickle tricycle and need replacement wheel rims described in specs as 16" x 1 3/8" - anyone know where I can buy them or what type are needed please?
Comment: Visiting your local bike shop with a sample rim is your best option.  Sadly this kind of question is a poor fit for this site, because its very localised, and is time-sensitive, and its really just shopping.

You'll need to match the spoke count too, which appears to be 28, and the rims will need a Schrader hole.

Also note that the rim's internal major diameter needs to be pretty close to the original, if you intend on reusing the existing spokes.
Comment: Wheels on children's bikes are often non-standard, though the ones in the picture look to be reasonably standard.  However, when I do a web search I find several different ISO diameters, and most are 32 hole.  Prices ranging from $17 to over $100 per rim.
Answer: I edited the question so it wouldn't be a strictly sourcing/shopping question.
It appears these use the 16" size with bead seat diameter 349. Look carefully around the tires and old rims to try and find the number 349 to confirm. It will likely be on the tire and say something like "38-349". If you don't find it and find some other 3-digit number instead, that's also your answer.
Sometimes there's no such marking. If that's the case you can identify what BSD you're dealing with via measurement.
349 is the Brompton size. Rims for it are readily available but are a niche product you need to get from Brompton shops/dealers/etc. You also need to match spoke counts. Many or most Bromptons use 28h and it appears this trike also does.


A question of rationality

Question: This problem was asked to me by a friend and I simply have no idea about it. So I have not progressed a single bit. The problem is this:
If $f :\mathbb{R}\to \mathbb{R}$ is an infinitely differentiable function and $f(x)\in\mathbb{Q} \;\forall x\in\mathbb{Q}$ then must $f'(x)$ be rational for all rational $x$ ?
Comment: $\mathbb{Q}$ is not complete (a sequence of rational numbers might converge to $\sqrt{2}$, for instance). Incremental quotients of $f$ are rational for rational increments, but their limit needn't be. This is my feeling.
Comment: There may be something helpful at http://math.stackexchange.com/questions/167620/functions-that-take-rationals-to-rationals and also at http://mathoverflow.net/questions/48910/smooth-functions-for-which-fx-is-rational-if-and-only-if-x-is-rational
Comment: @Siminore Yup, my intuition says "no", too. The question is, how does one prove that? Even if you restrict yourself to analytic functions, this isn't easy. One can probably show that *if* all the coefficient $c_k$ are rational and positive, then $\sum_{k=1}^\infty c_k x^k$ is only rational for all $x$ if only finitely many of the $c_k$ are non-zero. But who says you couldn't have irrational $c_k$ for which the sum nevertheless turns out rational for rational $x$? And if you allow negative $c_k$, couldn't the positive and negative sum's irrationalities cancel out?
Comment: Yes that's my point too what fgp tried to say. Any rigorous method would be welcome. Thanks in advance for the help.
Answer: Let's first construct a function $f \,:\, \mathbb{R} \to \mathbb{R}$ with $f(\mathbb{Q}) \subset \mathbb{Q}$ which is derivable at a single point $x_0$ with $f'(x_0) \notin \mathbb{Q}$.
Let $q_n$ be some rational approximation of an irrational number $a$, i.e. for example let $q_n \in \mathbb{Q}$ with $\lim_{n\to\infty} q_n = \sqrt{2}$. Then take the piecwise constant function $f \,:\, \mathbb{R} \to \mathbb{Q}$ defined by $$
  f(x) = \begin{cases}
    \frac{q_n}{n} &\text{if $x \in \left[\tfrac{1}{n},\tfrac{1}{n-1}\right)$ for some $n\in\mathbb{N}$,} \\
    \frac{-q_n}{n} &\text{if $x \in \left(\tfrac{-1}{n+1},\tfrac{-1}{n}\right]$ for some $n\in\mathbb{N}$,} \\
    0 &\text{if $x = 0$}
  \end{cases}
$$
(For $x \geq 1$, this is to be read as $x \in [1,\infty)$, i.e. $n=1$ in this case). Since we even have $f(\mathbb{R}) \subset \mathbb{Q}$, obviously also $f(\mathbb{Q}) \subset \mathbb{Q})$, thus fullfilling our requirement.
Being a step function, $f$ is of course not differentiable on $\frac{1}{n}$ in general. However, at $0$, we have $$
  f'(0) = \lim_{x \to 0} \frac{f(x) - f(-x)}{2x} = \lim_{n\to\infty} \frac{f(\tfrac{1}{n}) - f(\tfrac{-1}{n})}{\tfrac{2}{n}} = \lim_{n\to\infty} \frac{\tfrac{q_n}{n} - \tfrac{-q_n}{n}}{\tfrac{2}{n}} = \lim_{n\to\infty} q_n = \sqrt{2} \notin \mathbb{Q}\text{,}
$$
meaning that $f'(0)$ does indeed exist, yet doesn't take a rational value. (Note that to force the limit to exists for all sequences $x_n \to 0$, not just $x_n = \left(\tfrac{1}{n}\right)$, it might be necessary to require $q_n$ to converge sufficiently fast)

To make $f$ differentiable everywhere, we're going to have to get rid of the steps, i.e. somehow smoothen the $f$ constructed above. For that, we can use polynomial interpolation. Let's assume $q_n \to \sqrt{2}$ monotonically from above. If we can find polynomials $p_n$, $n \in \mathbb{N}$, with $$\begin{eqnarray}
  p_n\left(\tfrac{1}{n}\right) &=& \frac{q_n}{n} \\
  p_n\left(\tfrac{1}{n-1}\right) &=& \frac{q_{n-1}}{n-1} \\
  p'_n\left(\tfrac{1}{n}\right) &=& q_{n} \\
  p'_n\left(\tfrac{1}{n-1}\right) &=& q_{n-1} \\
  p'_n\left(\left[\tfrac{1}{n}, \tfrac{1}{n-1}\right]\right) &\subset& [q_{n}, q_{n-1}]
\end{eqnarray}$$
then we can simply set $$
  f(x) = \begin{cases}
    p_n(x) &\text{if $x \in \left[\tfrac{1}{n},\tfrac{1}{n-1}\right)$ for some $n\in\mathbb{N}$,} \\
    -p_n(-x) &\text{if $x \in \left(\tfrac{-1}{n+1},\tfrac{-1}{n}\right]$ for some $n\in\mathbb{N}$,} \\
    0 &\text{if $x = 0$.}
  \end{cases}
$$
Again, $\lim_{x\to a}\frac{f(x) - f(-x)}{2x} = \lim_{n\to\infty} q_n$ if the first limit exists at all. The conditions imposed on $p'_n$ ought to ensure that it does.
Unfortunately, though, this idea still only yields a function that is differentiable once, and it's not clear whether it can be extended to yield an $f \in C^\infty$. Piecewise polynomial functions won't work for that, I fear.
Comment: But this $f$ is not infinitely differentiable.
Comment: @OmranKouba I'd love to give a full answer, but I figured a partial answer is better than none. Maybe someone else can use this as a starting point.

A question of rationality


Does any graph based optimization problem, which have a greedy algorithm, has guarantee that there exist an order which give the optimum

Question: There exist greedy algorithms for vertex coloring like optimization problems. We know that for graph coloring, there is an order of vertices for which greedy coloring produces the optimum results. Is it true for any problem on graph which has such a greedy algorithm.
Comment: I don't understand your question. What do you mean by "problem has a greedy algorithm"? I suspect that, if you define what you mean, then the answer to your question is trivial "yes", simply by definition.
Comment: I think you get the right meaning of the question. I just want to get the logic behind the "yes".
Answer: Yes, for every simple graph problem I can think of it's pretty simple to come up with some greedy strategy that gives the optimum if you order your vertices correctly.
Take for example Independent Set. The simplest strategy imaginable for finding an independent set is to fix some ordering $(v_1,\ldots,v_n)$ on the vertices of your graph $G$, putting vertex $v_i$ into your independent set $I$ if it has no neighbors among the vertices that have been put into $I$ so far. This always leads to a maximal independent set, and it's easy to see that every maximal independent set (including those of maximum size) can be generated this way, by just taking an ordering where evey vertex in the set comes before every vertex not in the set.
This strategy also works for other vertex subset problems like Feedback Vertex Set or Dominating Set. Unfortunately, it is pretty uninteresting and unlikely to give much insight into these problems. It's potentially more interesting when such strategies can be devised for more complex problems like Treewidth.


What is the cheapest way to go from Lyon (France) to Milan (Italy)?

Question: I am from Egypt and I am going to Lyon for a conference which will finish on 27th of March. I need to go to Milan for a short visit after the conference. Is there anybody knows the best way to do that?
Comment: The cheapest way is probably walking :)
Comment: If you consider the cost of food and accommodation for the days of walking, it might not be the cheapest ;)
Comment: How much time do you have to travel to Milan?
Comment: Riding a bicycle would be cheaper than walking since you can cover more distance in a day and therefore save on accommodation.
Answer: Take the bus. Idbus has three connections per day. It takes 8 1/2 hours and the price is 29 or 39 EUR, depending on the departure time . Alternatively, you can also have a look at Eurolines.
The train is a bit faster, but also more expensive. For the day you mention, prices start at 48 EUR.
Easyjet used to operate between Milano Malpensa and Lyon. But they stopped the route a few days ago. But given the prices for bus and train, flying is hardly competitive, if you take into account transfers to and from the airports. The case of Lyon is e.g. handled in this disussion.
If you are flexible, you might also consider ride-sharing, "covoiturage" in French. You can have a look at covoiturage-libre.fr or BlaBlaCar. If in addition you are adventurous, hitchhiking might be an option. Flexibility is required for both these options, as there is no guarantee that you will find a ride on the exactly the date you specified or to the place you want.


Cannot run emulator in Android Studio

Question: I just got done installing Android studio and trying to take it for a test run.  When I try to run my app I get the following error message
emulator: ERROR: This AVD's configuration is missing a kernel file!!
emulator: ERROR: ANDROID_SDK_ROOT is undefined

Does anybody know how to fix this????
Comment: Check out this answer http://stackoverflow.com/questions/9712605/emulator-error-this-avds-configuration-is-missing-a-kernel-file
Comment: @Ted pottel, did any of the answers below answer your question?  If so, please click the checkmark next to it.
Answer: It is possible that you really have no system images. Double-check that $ANDROID_HOME/system-images/android-<YOUR DESIRED API>/armeabi-v7a exists and is not empty. If they really are missing - install/reinstall with SDK manager.
Comment: Within the SDK Manager, ensure you check the "Show Package Details" so that you can view which system images are installed, and if not, install them
Comment: I have two problems: 1: I cannot find `$ANDROID_HOME/system-images/android-/armeabi-v7a`, 2: How to reinstall a virtual device using SDK manager
Comment: Reading your answer however, I could do a workaround, and that was to create a new virtual device and the problem did not occur for the new one
Answer: Go to Tools | Android | AVD Manager
Click the arrow under the Actions column on far right (where error message is)
Choose Edit
Leave the default selection (For me, MNC x86 Android M)
Click Next
Click Finish
It saves your AVD and error is now gone from last column.  And emulator works fine now.
Answer: Just fixed this. Hope this helps others. (Issue as in Android studio v2)
This issue is for arm emulators. In this example I am using armeabi-v7a API 16
The fix is three steps:
Step 1:
Open sdk manager and make sure that you have installed ARM EABI v7a System Image
Step 2.
This is the obvious case of adding the location of sdk to system variables.
Right click on "This PC" icon on desktop, then
Properties -> Advanced system settings -> Environment Variables...
Then add the path to sdk as new to system variables section using variable name ANDROID_SDK_ROOT.
Step 3.
Restart Android Studio for changes to register. After correcting the ANDROID_SDK_ROOT is undefined issue, the emulator still can't find the kernel files even though it can see that the sdk manager installed it to
path-to-sdk\sdk\system-images\android-16\default\armeabi-v7a\kernel-qemu
The reason is a mix-up between the location the sdk manager installs the kernel file to and the location the emulator is looking for it.
If you open your config.ini (Android Studio -> AVD Mananger -> "Show On Disk") for your emulator you will see the following line:
image.sysdir.1=add-ons\addon-google_apis-google-16\images\armeabi-v7a\
(i.e \path-to-sdk\add-ons\addon-....)
Instead of changing this value in the config.ini file I copied
path-to-sdk\sdk\system-images\android-16\default\armeabi-v7a\kernel-qemu
(kernel file from the folder sdk manager installed)
to
\path-to-sdk\add-ons\addon-google_apis-google-16\images\armeabi-v7a\
And that was the missing kernel file. You can run the emulator. (You will need to close the Android Studio and reopen again) Give the emulator some time as it is 10x times slower compared to x86. (Mine took about 5 minutes to start)
Comment: Uh, where is the 'config.ini for your emulator'?
Comment: Android Studio -> AVD Mananger ->  "Show On Disk". Editing the answer.
Comment: This is the answer that actually did the trick for me, in my case, this was my edited line: `image.sysdir.1=system-images\android-16\default\armeabi-v7a`
Comment: This did the trick for me for Nexus 4 emulator. Had to update `image.sysdir.1` to `system-images\android-15\google_apis\armeabi-v7a\`
Answer: I got it fixed by running "C:\Program Files\Android\android-sdk\AVD Manager.exe" and repairing my broken device.
Answer: Hey there just ran into the same issue you did, the 3rd link down in google brought me to this bit of code that throws the error,
if (kernelFile == NULL) {
        kernelFile = avdInfo_getKernelPath(avd);
        if (kernelFile == NULL) {
            derror( "This AVD's configuration is missing a kernel file!!" );
            const char* sdkRootDir = getenv("ANDROID_SDK_ROOT");
            if (sdkRootDir) {
                derror( "ANDROID_SDK_ROOT is defined (%s) but cannot find kernel file in "
                        "%s" PATH_SEP "system-images" PATH_SEP
                        " sub directories", sdkRootDir, sdkRootDir);
            } else {
                derror( "ANDROID_SDK_ROOT is undefined");
            }
            exit(2);

to which the person wrote:
"/* If the kernel image name ends in "-armv7", then change the cpu
         * type automatically. This is a poor man's approach to configuration
         * management, but should allow us to get past building ARMv7
         * system images with dex preopt pass"
So I went back in and downloaded the x86 intel atom version for my desired API level and was able to get the emulator up without the error. Hope it helps you too.....
Answer: Short answer: try to create the same image using the old school <AndroidSDK>\AVD Manager.exe.
Working in Android Studio, running all the integrated tools, it became natural to me not to use the old managers (AVD/SDK).
In my case, I had this problem when used the new (integrated) AVD Manager to create devices with old system images (API 11 and below, as I've tested).
When I tried to use the old school AVD Manager tool (located in <AndroidSDK>\AVD Manager.exe) to create these old device images, I had success.
Comment: Thanks for the suggestion -- seems like the standalone tool knows how to set things up for the older platforms. On OS X / macOS: `~/Library/Android/sdk/tools/android avd`.
Answer: I had the same error. The solution for me was change the ANDROID_HOME path. First I took a look into tools->android->sdk manager from Android Studio. In that window, we can see the path where Android Studio looks for the SDK: image
Then I opened a Windows CMD shell, executed:
echo %ANDROID_HOME%

but the path was different to the one in ANDROID STUDIO CONFIGURATION of the first step.
The solution was to change in user environment, the ANDROID_HOME, to the one of the first step: image
I finally closed the cmd shell, and opened another cmd shell to execute:
echo %ANDROID_HOME%

the path was updated, and I could run my emulator perfectly.
Answer:
Open Android studio.
Go to setting > System Setting > Android SDK
Get the "Android SDK Location".
Set the environment variable ANDROID_SDK_ROOT  to this value.

It worked for me and I'm on Windows 10 and Android studio 2.3.3
Answer: Normally, the error will occur due to an unsuitable AVD emulator for the type of app you are developing for. For example if you are developing an app for a wearable but you are trying to use a phone emulator to run it.
Answer: In my case (Windows 10) the reason was that I dared to unzip the android sdk into non default folder. When I moved it to the default one c:/Users/[username]/AppData/Local/Android/Sdk and changed the paths in Android Studio and System Variables, it started to work.
Answer:
Go to "Edit the System Environment variables".
Click on New Button and enter "ANDROID_SDK_ROOT" in variable name and enter android sdk full path in variable value. Click on ok and close.
Refresh AVD.
This will resolve error.
Answer: A common approach to follow to solve this problem.
1.CHECK your SDK manager by running from your android studio and stand alons sdk folder by executing ./android.sh helps you to find broken packages

Try installing System emulator images with google API support than the Intel one. Just like , i solved my problem by running into another system image.
Experment on KVM based Virtulaization suggested by Google for Linux

Cannot run emulator in Android Studio


If I stood on the equator and threw a stone in direction of rotation, how much farther would the stone go than if I threw it the other direction?

Question: Assuming a spinning Earth of 6378.1 km (3963 mi) radius, how much farther would the additional momentum carry a stone if I threw it in the direction of the Earth's rotation (i.e. eastwards) when standing on the equator at sea level, than if I threw it with the same force the other direction? How much is the difference in percent or permille?
Answer: You, the stone, all locations on the equator, and the atmosphere a little above the equator are all moving with basically the same momentum initially.  When you throw it, in the big picture, it gains total momentum (if throwing eastward) or loses total momentum (if throwing westward).  But because its course and momentum do not shift it from being along the equator, its altered  momentum will not have any "carryon" momentum difference horizontally that imparts apparent motion.  In other words, for the east/west aspect DIRECTLY, the Earth's rotation will have no impact, it will continue doing what it was doing.  You can consider its initial momentum from the Earth and its added momentum from the throw entirely separately... it will hit where you expect it to hit from this factor.  The eastward or westward component have no bearing at all because it remains at the Equator.
It's like being on a moving train, and tossing a rock to your friend within the train car.  Everything is moving the same rate, there is no difference.
So directly (from the eastward or westward direct of the throw), there is no difference at all along the equator.
However, there is still one difference in rotation rates for paths at the Equator.  As the rock goes UP, it will have less velocity than required at that height (because the horizontal velocity needed to maintain the same rotation rate is more), and so it will lag behind.  It will wind up west of where it should have... regardless of whether you threw it east or west.  This difference is only a factor of the vertical component of the throw.
That said, if you're throwing it, it's going to be a pretty quick trajectory. Short distances and times have very little Coriolis deflection.
The amount of deflection becomes a nonlinear PDE (as early Coriolis alteration in the path changes both its subsequent landing time and its upcoming Coriolis deflection), so I don't believe you can derive a simple formula for the true percentage change.
But for a typical throw attempting an east or west motion, you won't have much vertical velocity, and gravity is very good at removing that component quickly.  The deflection for a typical strong human's throw will be on the order of millimeters or less (this is true wherever you throw it on Earth).
The percentage of horizontal deflection would actually be infinite if thrown straight up (since a millimeter divided by 0 is still a "giant" change), and will quickly become a minuscule percentage if any horizontal component is given to the throw; a footballer's pass will not be noticeably altered by Coriolis anywhere on Earth.  Air resistance/wind will be much much more significant factors.  Even for missiles, which have much more velocity and so can fly for much longer times and thereby reach areas with significantly different rotational momentum, Coriolis deflection will only be on the order of a small percent.  But a few percent can mean a few km/miles for intercontinental trajectories... kind of important for a targeted missile!
Only things that can stay "aloft" "permanently"... i.e. fluids (gases, and also liquids)... does Coriolis have a relatively giant impact (causing air that should otherwise flow directly from high to low pressure into air that rotates along isobars)
Comment: Has the Coriolis deflection on Earth been purposefully demonstrated such as launching a missile from equator proximity (e.g. from Kourou) and launching the same missile at the same angle from north/south pole proximity (e.g. Esrange) to demonstrate they have differently shaped trajectories?
Comment: I can't imagine such an event just to demonstrate Coriolis is worthwhile... every longrange missile launch must have Coriolis taken into the calculations, so every successful missile path would be evidence of such trajectories. And as noted, the difference in trajectories for paths with a short travel time (missiles included) would be very small (albeit very important for hitting a target).  So no, I don't think any intentional demonstration like that has been done, and not sure it would be of any real use.
Comment: Its use would be proof of either the rotation of the Earth around its axis, or of the centripetal force of the universe onto Earth (as both would have the same effect).
Comment: Proof to who?  People who use the formulas on a daily basis, mathematicians, etc don't need any further proof.  And those who you're suggesting still need proof of the rotation of the Earth... must already not take anything existing as evidence, concocting some conspiracy explanation... so they surely wouldn't accept any such missile experiment either?  And not sure what you're even suggesting by "centripetal force of the universe onto Earth"???
Comment: I meant that if the Earth stood still and the universe revolved around it, this would have the same effects as if the Earth rotated around its axis, e.g. that geostationary satellites don't fall down. It's a matter of your reference point of view. You're correct that such missile experiment wouldn't be accepted by say Flat Earthers because they otoh would claim this would be due to a dome rotating above the Earth. Still, it would be a nice and  interesting experiment. Time dilation didn't need proof either but was demonstrated by taking a clock on a plane that circumnavigated the Earth.
Comment: I'm confused how Earth rotating would keep satellites aloft?  Reference frame alters viewpoint, but the resulting forces show what is valid.  Time dilation wasn't a key factor in any physics used in the world at the time, so an experiment to prove it was important to understanding what physics is true (and so could be incorporated properly in later things where it was a factor, like GPS).  Whereas Coriolis is seen in equations of meteorology, oceanography, trajectories, etc of everyday usage.
Comment: I think the time dilation experiment is from the 1970s or 80s or so, when it became a factor in spaceflight.


Are "Identify this" questions on-topic?

Question: Are "Identify this manga" or "Identify this anime" questions on topic for the site?  I'm wondering if we could incorporate a similar policy as that on SciFi.SE.
Comment: I think that adopting the SF policy is a good idea.
Answer: Agreeing with @JNat, I think the SF policy is a very good one for a site which has a medium. You may catch a glimpse of an anime or manga somewhere and think it's good, and then try to find it again. Or you may want to reread that one manga you read at a bookstore when you were bored, but you don't remember the name, just the general theme / specific traits about the characters (this is the case for me, currently).
Comment: I tend to lean to this resolution, as Identify This questions have worked fairly well on SciFi.
Comment: I agree.  SF's policy works wonders.
Answer: No, they should definitely be off topic.
I wrote a rather huge essay on this topic on Gaming.SE. Rather than re-post it here, I will just link to that one. All of the points remain the same whether applied to games, anime, movies, etc.
Comment: Answers to IT questions need not be guesses here (which was your biggest complaint in that essay). Anime and Manga are visual formats, and for at least some of the questions the questioner could include or link to a picture of the series in question. I'll also point out that your essay, while it makes good points, is only the third-highest rated answer on the linked question, and anyone reading it should also take the time to read at least a a few of the other opinions there, which also make good points.
Comment: @Blue The medium has no bearing on why IT are problematic. SFF and Arqade just have differing site policies.
Comment: I'm not sure I agree 100%.  Some of your points are spot-on, but I don't think that they're wildly off-topic here.
Comment: Identifying stories are okay in the Sci-Fi stackexchange.  In my point of view anime and manga are closer to stories than to games.

Are "Identify this" questions on-topic?


Calculus - Components of a unit vector

Question: Determine the components of a unit vector perpendicular to (0, 3, -5) and (2, 3, 1). I think I should be using either cross or dot product, but am unsure on what to do from there.
Answer: "In mathematics, the cross product or vector product is a binary operation on two vectors in three-dimensional space. It results in a vector which is perpendicular to both and therefore normal to the plane containing them."
http://en.wikipedia.org/wiki/Cross_product
(furthermore, once you have the vector, you need to unitize it - http://en.wikipedia.org/wiki/Unit_vector)
Comment: (and so $(0,3,-5) \times (2,3,1)=(18,-10,-6)$)
Comment: One more [link](http://www.wolframalpha.com/input/?i=Normalize+CrossProduct%5B%7B0%2C3%2C-5%7D%2C%7B2%2C3%2C1%7D%5D) for you
Answer: Using the dot product: let $u=(a,b,c)\in\mathbb{R}^3$ be such a vector, and write $x=(0,3,-5)$, $y=(2,3,1)$. Then, the orthogonality conditions mean that
$$
\begin{align*}
0 &= \langle u,x\rangle = 0\cdot a + 3\cdot b+(-5)\cdot c = 3b-5c \\
0 &= \langle u,y\rangle = 2\cdot a + 3\cdot b+1\cdot c = 2a+3b+c
\end{align*}
$$
and the unit vector condition further imposes
$$
\begin{align*}
1 &= a^2+b^2+c^2
\end{align*}
$$
You thus have 3 constraints (two of them linear), and 3 unknowns. This will be enough to find solutions for $u$ (there is no unicity, as if $u$ is solution, so is $-u$, for instance).
Answer: First calculate the determinant
$$ \left| \begin{array}{ccc}
i & j & k \\
0 & 3& -5 \\
2 & 3 & 1 \end{array} \right| $$
You get a vector perpendicular to both given vectors then normalize it by dividing the components by the length of it.
Edit: See this link from wikipedia.
Comment: Oh, saw the link you added. I missed the last step, with the coefficients of the three variables of the polynomial giving the coordinates.
Comment: The determinant is a number, not a vector.
Comment: No. A determinant is not always a number and depends on the entries. Here the determinant involves $i, j$ and $k$. Therefore it will be of the form $ai+bj+ck$

Calculus - Components of a unit vector


RSS icon in QuickTabs

Question: I'm using Drupal 7.14 with QuickTabs 7.x-3.4. I have a couple of views with feeds attached to them. I've also set the "Attach To:" field for these feeds. The ones that aren't inside a QuickTab are working fine, but the ones inside QuickTabs won't display their RSS icon. How can I get the views inside QuickTabs to also display their RSS icon?
Answer: One way to resolve this is by creating required feeds page with views.
Then add this generated feed pages link manually through Global Text area field of appropriate view in footer using HTML anchor tag. Separate links will be displayed inside each view footer of QuickTabs as per setup.

RSS icon in QuickTabs


Scaling multiple faces along their y axis

Question:
My goal is to have these faces scaled down so it's symmetrical, any help?
It should be a knob just for making clear my objective.
Answer: Set your Transformation Orientation to 'Normal', your Pivot to 'Individual Origins', and then perform a scale in the Normal space using either SXX or SYY.. you'll be able to see the direction of the constraint .. I can't tell which it will be from this illustration. (Or if you want to make the extrusions shallower, that would be a translation..GZZ)


Setup Google Domain with Google Compute Engine?

Question: So I've recently bought a domain name from Google Domains, and have set up a virtual machine from the ASP.NET template available on Google Cloud Platform. I went through the tutorial on setting up Cloud DNS and setting my name servers on Google Domains and I believe I've done everything correctly.
I have already:

Reserved my static external IP: 188.8.131.52.
Set up my Cloud DNS:

A record for polytide.net.
NS records (generated by GCP) for named servers ns-cloud-e[n].googledomains.com.
SOA record (not sure what this is, but generated by GCP).
CNAME record for www.polytide.net.

Set the custom named servers on Google Domains to the above NS records.

I've performed a DNS Query and everything seems to be working; however, I can only access my website via external IP. I've tried pinging both the domain name and the external IP and only the IP returns a result. When trying to access my website via domain name I get the following error:

polytide.net's server IP address could not be found.
DNS_PROBE_FINISHED_NXDOMAIN

I feel like I'm missing a critical point and I'm at a standstill. Also, not sure if this is important, but the VM doesn't have the firewall boxes checked for HTTP or HTTPS. I did check these boxes in an attempt to fix the issue, but it didn't work, so I unchecked them since that was the default that the template used.
How can I setup my Google Compute Engine website to utilize my Google Domain?
Answer: It looks like everything is set up correctly, and I can access your website OK from here (it shows an "Under Construction" page).
Most likely, the NXDOMAIN result was cached in your local DNS server, or your ISP's DNS server, or your web browser. It will eventually expire and things will work for you. See How long does negative DNS caching typically last? for more information.


Efficiency of JSP calling functions directly

Question: Is it a good idea to have a JSP page call my java classes directly or is it better to do an http post to my web service? what is the difference in terms of scalability and security?
Answer: JSP-Files are used to generate Servlets. So JSP-Files and Servlets are mostly the same. There is no difference for security or scalability.
Answer: JSP has its own responsibility, view generation. Unless required, it doesn't make sense to embed any java code inside JSP. Invoked Servlet/Other framework controller to do business logic.
If the calls are with-in same web application, let us say from JSP1 to JSP2 (or) JSP1 to Servlet2, webservices calls doesn't make sense. They will be heavy.


How to recalculate automatic weights for single bones?

Question: After I edited an already weighted model and its armature, the pose mode doesn't grab the limbs the right way anymore. Is there a way to recalculate weights for  single bones? Because I already edited the weights for other parts of the mesh and don't want to redo that again.
Answer: In Weight paint mode, select the pose bones you want to recompute the weight and press:
W > Assign automatic from bones
Comment: The new workflow is you select the armature, multiple select the mesh, then go to weight paint mode, and then you can select individual bones with Ctrl-click (alt+shift+click for multiple bones)
The W hotkey seems to be no more but the ption is in the Weights menu between View and Brush (In 3D-View top header)
Comment: Heads up for 2019: You can't select bones in weight paint mode in Blender 2.80
Answer: (For August 2019 2.8)
First select your bones, then shift-select your mesh. Make sure your mesh is selected second.
Now go into weight paint mode and ctrl-select whichever bone you want to adjust the weight for.
Press f3 to open the operator search and search for 'Weight From Bones'
Then select it and select 'Automatic'
Since you can select whichever bones you want now during weight painting you can reapply your automatic weights to all your bones this way.
Comment: I didn't want "automatic weight" so first 2 steps were enough and wokred great
Answer: ctrl-select bone in weight paint mode did not work for me in Blender 2.8 . The following works for me though:

Select the pose bones in Pose Mode.
Select armature, then mesh in Object Mode.
Goto Weight Paint, select all area by pressing A (or only selected area you want)
At the top left corner of 3D view, look for Weights menu and click Assign Automatic From Bones
Done! You can verify the new weight by selecting the Vertex Group of your bone in Object Data at the Properties window
Comment: This way works fine in May 2020.
Answer: Go in pose mode and select the bone that you want to weight paint.
Then select your mesh and switch into Weight paint mode.
Press W and then you choose "assign automatic from bones"
Answer: Blender 3.1:

Go to pose mode
Select all meshes
Ctrl-click armature in outliner (add to selection)
In viewport: ctrl+p > armature deform (with automatic weights)


Is it bad for a child to sit on their feet?

Question:
Are there studies which say sitting with their legs bent back on either side of them this is bad? If so, how do you get your child to stop sitting this way?
Comment: @JPmiaou just to let you know, Pluto is no longer a planet, there is no such thing as a Brontosaurus, Starfish are now called Sea Stars, and kids now sit 'pretzel style'. ;) (These are things I've learned in the past few years of our kids' schooling.)
Comment: @DA01 actually it is criss cross applesauce! :)
Comment: I used to sit this way all the time, and I don't think it hurt me. I got reprimanded for it a lot in school, but I think one hip isn't "hinged" right for Indian-style, because even back then I found that uncomfortable.
Answer: This style of sitting is referred to as "W-sitting."  There are good reasons for encouraging children to avoid this posture.
Children like this sitting position because it provides a great deal of stability.  One of the unfortunate side-effects is that they don't develop the appropriate trunk control.  There is no trunk rotation in this posture, which can delay or hinder the development of handedness. See Pediatric Center for more information.
Also, I found this:

In a W-sitting position, the hips are placed at the extreme limits of internal rotation, predisposing the child to future orthopedic problems. In the abnormal position, the risk of hip dislocation becomes a concern. If the child has hip dysplasia, which commonly goes undetected in youngsters, sitting in the W-position increases the risk of dislocation tremendously. Pre-existing orthopedic conditions can worsen when major muscle groups are placed in shortened positions and begin to tighten. These shortened muscles are at risk to form contractures, and a permanent shortening of the muscle--especially prone are the hamstrings, adductors and Achilles tendon. See Advance for Physical Therapy and Rehab Medicine

And there's more, but you get the idea.  When my children started W-sitting a lot, my wife insisted it was damaging.  I "knew" that was a myth and looked online for recent studies showing it wasn't harmful.  I couldn't find a single reputable source making this claim.
Answer: I found one source more:
http://www.learninglinks.org.au/pdf/infosheets/LLIS%2025_W-Sit.pdf
saying it is not good to w-sit.
But: I tried to find scholarly material (from peer-reviewed journals) on the subject, and did not find strong evidence: neither for nor against. For example, in this journal:
http://onlinelibrary.wiley.com/doi/10.1002/dev.20593/full

W-sitting may indicate insufficient muscle control around the hip
  compensated for with a symmetrical sitting posture.

So, I think the physiotherapists agree that w-sitting is not recommendable, but there is little or no scientific research made on it.
Comment: but a simple evolutionary argument suggests it can't be all that bad ;-)
Comment: So there is no evidence whatsoever that W-sitting *causes* any problems. *Indicate* means that it might be a sign of a problem with a different underlying cause. I am surprised and disappointed by the physiotherapist community promoting such simplistic information that is not supported by any hard evidence, and by doing so, probably causing more harm than good to parents and children alike.


Is it okay to mention family background in SOP?

Question: I'm applying for MS in Mathematics. My grandfather was Math professor and my uncle is also Math professor. Such family background was one of my reasons to  study Mathematics and I was wondering if I can mention my family background briefly at introduction in SOP?
Answer: Can you? Sure, it's not wildly inappropriate.
Do I recommend it? No! This is a 1-2 page document where you state what your purpose is (hence the name). It's not a time to talk about your childhood, why you were inspired to study math, or what your favorite type of helicopter is. It's where you talk about what you've done so far and what you want to do next.
You may have to talk about your personal situation -- usually to explain a bad semester, etc. And it's fine to state a few interesting details about yourself (especially if you have significant achievements outside of math, or a non-traditional background, etc.). But listing the professions of your family members is neither necessary nor interesting.
Comment: "...or what your favorite type of helicopter is." seems unnecessary. The OP is talking about dear family members, not helicopters.
Comment: Agree it is a bit of an exaggeration ("reductio ad absurdum"), but I don't think it's disrespectful.
Comment: Could you explain more about "talk about your personal situation"? I do have a bad semseter and I want to know how I can explain such matters in SOP. Thanks
Comment: Anything that's relevant or interesting is fine. You can introduce yourself, state your hobbies/achievements, explain anything that you did that was especially noteworthy, give any mitigating factors for bad semesters/test scores, etc. Anything that you would feel comfortable saying face-to-face to your boss is fine. This is just not the time for "therapy", childhood issues, or boring stories. If you would cringe saying it to your boss's face, don't say it in your essay. (Your motivation to study math isn't cringe-worthy, but it's not something your boss would probably be interested in)


Can you play minecraft on Wii U and Xbox together on mulitplayer?

Question: I want to play with my friend but don't know how!
Comment: They are entirely different versions, so you can't.
Answer: What you are referring to is called "Cross Platform Multiplayer" Where two or more players on different consoles can play the same game. Currently, Cross platform multiplayer is very rare, and few games have this feature available.
There have been rumors that improved Cross platform multiplayer may be included at a future date on some or all of the minecraft versions, but this is not yet confirmed.(http://www.ign.com/articles/2015/08/05/gamescom-2015-mojang-wants-minecraft-to-work-cross-platform)
Unfortunatly, at the moment, You can only play Minecraft online multiplayer if you, and everyone you are trying to play with are using the same version of minecraft on the same console or platform.
*EDIT
As Pointed out by colorfusion in the comments, Minecraft Pocket Edition is compatible with Minecraft Windows 10 Edition. (Here's an article about it http://www.pcworld.com/article/2982932/software-games/minecraft-pocket-edition-now-lets-windows-10-ios-and-android-gamers-play-together.html)
Comment: "using the same version of minecraft" -  exception for MC:PE and MC:Win10E (they're essentially the same, just available on different platforms).
Comment: Huh, I had no idea. Neat.
Answer: No. Your question is slightly vague and could use some editing but I think what you're trying to ask is if you and your friend can play if he has a xbox and you have a wii u. The answer is no. Different systems.
Answer: You cannot currently join cross platform with and exeption of Xbox1 and Ps4 but we do know that No hang is working on it so it will come eventually.
*So not you cannot currently play with your Minecraft Buddy on Xbox
*
Thx!
Comment: This answer adds nothing not already covered in far more detail by other answers. Duplicate answers are of no use to anyone and may be deleted.


Antimatter conundrum

Question: Why are matter and antimatter not coming into contact within our atmosphere or within close proximity to earth. Obviously I understand the meeting would be catastrophic but why does it only happen in distant galaxies. Please excuse my ignorance.
Answer: Antimatter is very rare in our universe as it mostly contains normal matter. It is also not true that only distant galaxies produce antimatter. There is antimatter in our galaxy too. The fact is that there is so much more matter that the distribution of antimatter is vastly spread out and although they may be produced, antimatter does not stay antimatter for significant times as there is so much matter around( they collide with matter again to form gamma rays). Thus altogether they don't pose a threat as there is so little of it around that finally reaches earth.
This link below is an article about antimatter detection in our own galaxy. Together in cosmic rays we also get antimatter detection but it makes up less than 1% of cosmic rays and cosmic rays are the collection of the stream of particles from the entire universe. So thus you can grasp the situation here.
Antimatter in our Galaxy
Comment: Thanks again for the response . It makes sense finally. It's happening all the time but in tiny amounts . Simple when you know.
Answer: Despite being relatively rare compared to matter, antimatter is all around us, as a result of cosmic ray particle showers. A very small proportion of cosmic rays are positrons, but mainly they are baryons. They hit the atmosphere and form cascades of particle-antiparticle pairs, the antiparticles annihilate with other particles making more pairs, and so on. Antiparticles are hitting your body almost every second.
Because of baryogenesis, the production of matter over antimatter early in the universe, we don't expect distant galaxies to be different. Although given the assymetry is not fully explained one possibile reason for could be https://en.m.wikipedia.org/wiki/Baryon_asymmetry#Regions_of_the_universe_where_antimatter_dominates
Answer: Antimatter is produced via high energy particle collisions. The Sun produces 80 Tons of antimatter a day, but upon exiting the star; it typically Annihilates upon contact other solar particle emissions or random stellar dust in interplanetary medium and matter producing gamma ray photons detected by various telescopes.

Astrobiological Life scientists are enamoured by gamma rays, to find suitable stars for life. Stars with unusally strong gamma emissions can be eliminated as candidates for hosting planets with habitable life.

Antimatter conundrum


Why doesn't the gravitational force have a permittivity in its formula?

Question: We know that the electrostatic force between two charges depends on the medium between the charges and its permittivity. Why, then, doesn't the gravitational force depend on the medium?
Comment: Possible duplicate of [Is Newton's universal gravitational constant the inverse of permittivity of mass in vacuum?](http://physics.stackexchange.com/q/69541/)
Answer: Electrodynamics in Vacuum does not have a medium-dependence - as does gravity. If we do introduce a medium, we end up in a continuum theory of chargeable or polarisable media, which is where the permittivity comes from (for anyone not familiar with this, I would recommend reading a bit of Jackson's Electrodynamics). In gravity, there is no such thing as polarisation or charges of media, thus no need to include continuum effects (as far as I know, GR is not my forte).
Answer: The permittivity of the medium between two electric charges affects their electrostatic interaction because the medium can become polarized: that is, the medium has lots of neutral atoms inside it, and in each of those atoms the positive and negative charges shift to create a tiny dipole, and in the process they create thin layers of positive and negative net charges at the surfaces of the medium facing the original charges, where there's no adjacent atoms of the medium to cancel out the outward-facing parts of the last atoms. These surface charge layers then affect the electrostatic interactions between the original charges.
In gravity, on the other hand, there simply isn't a negative mass with an attractive gravitational interaction to normal matter, so the atoms in the medium cannot form gravitational dipoles (they can only shift positions) and they cannot form surface layers of gravity-cancelling 'negative mass'. Since that mechanism is no longer present, its simplified manifestation (i.e. the medium's permittivity) is also gone.


Usage of the noun suffix "-ment"

Question: What is a good rule for the usage of the noun suffix -ment?
Is desirement as acceptable as achievement?
Comment: *Desire* is already a noun as well as a verb. There's no need to make the noun *desire* into a longer noun by adding *-ment*, and in fact *desirement* is not a word. Since *achieve* is not already a noun, it makes sense to turn it into a noun by adding *-ment*.
Comment: Read is a verb but not (normally) a noun, and readment doesn't exist. Neither do snorement and takement. You need to look up individual cases in a good dictionary.
Answer: -ment is not a free suffix you can add as you like. The nouns ending in -ment are
either Latin or French. And the etymology of -ment is not clear. Even the meaning of -men and -mentum is difficult to analysize. If you study the Latin nouns in -men/mentum you find that the nouns are of various types and it is not possible to give a simple idea of the formation of these words. I could give my private view, but that would take too long and I think it would hardly help you.
Comment: [This  U. Mass. source](http://people.umass.edu/aimm/abstracts/Modeling%20Emergent%20Productivity%20of%20Rival%20Suffixes.pdf) says that the suffix -ment was/is productive, though increasingly less so: "Of course, not all suffixes
succeed in adapting; formerly productive suffix -ment began a gradual decline in productivity in
the 17th century because five times as many competing -ation words were being borrowed into
English during that century. Since -ment failed to associate itself with any meaningful subset of
potential hosts, it lost out to -ation."
Comment: @Edwin Ashworth  Interesting comment and thanks for the link. I wanted to dissuade the poster from forming his own words with -ment.
Comment: Your answer may mean the U. Mass. article is using 'productive' in an unusual way; the -ation words may well be largely French imports, so increasing the English lexicon with loan-words rather than by domestic suffixation. I couldn't agree more about the unacceptableness of making one's own -ment words up, though. What confusement would result!
Comment: @Edwin Ashworth  Well, I doubt that -ment is a productive suffix in normal language. But in the divers sciences there may well be new formations with -ment for new things. So I took the comment as it was. By the way, what are "hosts" of subsets? I'm not used to explanations of language with terms of the set theory.
Comment: I've no idea. But a subset of potential hosts might have been {verbs ending in -ine and needing a related noun} = {examine, determine, predestine ...}, which -ation rather than -ment snaffled (a technical term).

Usage of the noun suffix "-ment"


Proving Kepler's 1st Law without differential equations

Question: Is there a way to show that the motion of Earth around the Sun is elliptical (Kepler's 1st law) from Newton's laws without resorting to the use of differential equations of motion?
Comment: (Sorry, couldn't resist.) It's not clear what your precise question is. As I understand it, you'd like a proof that the orbit of the solution of the Kepler ODE problem is an ellipse, which doesn't depend on a direct, full solution of the ODE problem - much in the way that the orbit can be shown to be planar. Is this correct?
Comment: I see no reason why the Earth's orbit couldn't be perfectly circular around the center of mass of the Earth + Sun system.  Can you clarify what you're asking?
Comment: If so, I must point out that the standard derivations of Kepler's laws from Newton's, e.g. [this](http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/KeplersLaws.htm), do *not* use a full solution of the ODE. The equations in $d^2r/dt^2$ are reduced to an equation in $d^2r/d\theta^2$, which is solved to get the orbit. The equation for $\theta(t)$, on the other hand (and $r(t)$ with it) is not actually solvable in terms of elementary functions.
Comment: I would agree with the first comment from Emilio Pisanty, since Kepler also made observations from which he derived his laws.
Comment: Observation?    $\quad$
Answer: Newton's original proof was in fact based on geometry (he hadn't invented calculus yet). Richard Feynman devised his own, simpler geometric proof for one of his famous lectures. You can find it in Feynman's Lost Lecture, by Goodstein & Goodstein, and in this article: Paths of the Planets from Hall & Higson. But since it's so much fun, I'll describe it here as well.
Let's start with a lesser-known way to construct an ellipse, the so-called circle construction. Draw a circle with centre $O$, and fix a point $A$ inside the circle. Pick a point $B$ on the circle, and draw the perpendicular bisector of $\overline{AB}$ (blue line). It intersects $\overline{OB}$ in a point $P$, and as $B$ moves around the circle, these intersection points form an ellipse. Also, the blue biscector lines are tangent lines to the ellipse, and $O$ and $A$ are the foci.

Why is it an ellipse? Because $\overline{AP}$ has the same length as $\overline{BP}$, so that the sum of the lengths of $\overline{AP}$ and $\overline{OP}$ is constant, i.e. the radius of the circle. In other words, we get the classic tack-and-string definition of an ellipse. It is also straightforward to see that the angles $a$ and $b$ are equal. Since $a$ and $c$ are also equal, this means that $b$ and $c$ are equal, so that the blue line is indeed a tangent line.

The geometric proof of Kepler's Second Law (planets sweep out equal areas in equal times) from Newton's first two laws is straightforward and can be found in the Hall & Higson article. Now, if a planet traverses an angle $\Delta\theta$ in a small time interval $\Delta t$, it sweeps out an area
$$
\text{area}\approx \frac{1}{2}\Delta\theta\, r^2.
$$

At this point, Feyman's argument deviates from Newton: while Newton breaks up the orbit it equal-time pieces, Feyman considers equal-angle pieces. In other words, Feynman breaks up the orbit in subsequent pieces with areas
$$
\text{area}\approx \text{constant}\cdot r^2.
$$
Newton's inverse-square law (which can be derived from Kepler's Third Law) states that the acceleration of a planet is proportional to the inverse square of its distance $r$:
$$
\left\|\frac{\Delta\boldsymbol{v}}{\Delta t}\right\| = \frac{\text{constant}}{r^2}.
$$
Eliminating $r^2$, we get
$$
\left\|\Delta\boldsymbol{v}\right\| \approx \text{constant}\cdot\frac{\Delta t}{\text{area swept out in $\Delta t$}}.
$$
But Kepler's Second Law states that the area swept out in $\Delta t$ is a constant multiple of $\Delta t$. Therefore,
$$
\left\|\Delta\boldsymbol{v}\right\| \approx \text{constant},
$$
that is, intervals of constant $\Delta\theta$ also have a constant change in velocity. We can use this fact to construct a so-called velocity diagram. Break up the orbit into equal-angle pieces, draw the velocity vectors, and translate these vectors to the same point.

Since $\left\|\Delta\boldsymbol{v}\right\|$ is constant, the resulting figure is a polygon with $\dfrac{360^\circ}{\Delta\theta\,}$ sides. The smaller the angles, the more it approaches a circle.

Now, let's draw the velocity diagram of an orbiting planet. If $l$ is the tangent line to the orbit at point $P$ (parallel to the velocity vector in $P$), then $l'$ in the corresponding velocity diagram is also parallel to $l$. Also note that $\theta$ in both diagrams is the same.

Rotate the velocity diagram clockwise by $90^\circ$, so that $l'$ becomes perpendicular to $l$. Construct the perpendicular bisector $p$ to the line $\overline{AB}$, and the intersection $P'$ with $\overline{OB}$. It turns out that we are in the exact same situation as the circle construction for the ellipse: as $B$ moves on the velocity diagram, the points $P'$ form an ellipse.

The lines $p$ are the tangent lines to the ellipse. However, these lines are also parallel to the lines $l$, which are the tangent lines to the planet's orbit. Because of the tangent principle, if two curves have the same tangent lines at every point, then those curves are the same. In other words, the lines $l$ are also the tangent lines of an ellipse. This proves that the orbit of a planet is indeed an ellipse.
Comment: Thank you for the answer, but *also* for the Article of Hall end Higson. It's amazing and the exercises there really help to understand what's going on!
Comment: Can you share that article pdf @Bman72 seems like the one in the link got removed
Comment: @Orion_Pax I've updated the link. Cheers
Comment: Thanks Sir do you also have newton original geometry proof too?
Comment: @Orion_Pax I don't, but the Principia is public domain, so you'll probably be able to find an English translation somewhere. I've never seen Newton's proof, but if Feynman found it difficult to follow, it's bound to be a challenge :-) P.S. you can use the geometric proof as the basis of an analytical treatment, which I've done in [this post](https://physics.stackexchange.com/a/503333/24142).
Comment: Thats nice thanks for info :)
Comment: For parabolic orbits can we also geometrically derive ? Like for ellipse and hyperbolic
Comment: If you want Sir i can ask for E= 0 parabola proof in a post and you may answer there :)
Comment: @Orion_Pax Yes, parabolic orbits can be done as well. Creating those figures takes a bit of time, though.
Comment: Okay nice Sir i will ask then as a question now :) and you may answer there , i really like geometry techniques being useful in physics trajectories part
Comment: Asked a question Sir i am also wondering one more thing which i will ask there after your answer :)
Comment: Sir may you add your answer to there? https://physics.stackexchange.com/q/691203/297949


C# vs Java - Generic Lists

Question: What are the differences of the C# and Java implementations of the generic List class?
Comment: Well, for one, List (http://java.sun.com/javase/7/docs/api/java/util/List.html) is an interface in Java, not a class.  Java's equivalent to the C# generic List class is ArrayList
Comment: although a bit dated, seems interesting: http://www.artima.com/intv/generics.html
Answer: Well, in Java List<T> is an interface, to start with :)
The most important difference between the two is the difference between C# and Java generics to start with: in Java generics basically perform compile-time checks and include some metadata in generic fields etc - but the actual object doesn't know its generic type at execution time. You can't ask a List<?> what that ? is, in other words. Any references to a generic type parameter in the implementation act as Object, basically - so a ArayList<String> is really backed by an Object[]. In C# all the information is available at execution time too - so a List<string> is backed by a string[].
Similarly C# generics allow value type type arguments, so you can have a List<int> in C# but not in Java.
There are further differences in terms of variance etc - but this is moving a long way from List<T>.
In terms of just ArrayList<T> (Java) and List<T> (.NET), a couple of differences:

Java lists override equals/hashCode, whereas they don't in .NET
ArrayList<T> grows by multiplying the current capacity by 3/2; .NET's List<T> doubles the current capacity instead

Of course there are other differences in terms of the APIs exposed - if you could give more information about the kind of difference you're interested in, we could help more.
Comment: And I remember asking why you would always get upvotes @ meta. Great answer.
Comment: But C# List is a list implementation, I'm wondering if I should be using IList<__> everywhere because I should always program to an interface, not an implementation. I'm seeing lots of C# code use specific List as return types which is wrong. If someone decides to use a diff implementation of IList then more than 1 place has to change.
Answer: Are you asking for differences in their API or their underlying implementation?  I believe Java generics are implemented completely via the compiler -- the JVM has no notion of generics.  For C#, generics are a built-in concept of the .Net runtime.  Wikipedia seems to have a good comparison of the two: http://en.wikipedia.org/wiki/Comparison_of_C_Sharp_and_Java#Generics

C# vs Java - Generic Lists


Does BitLocker encrypt everything?

Question: just a quick question - does BitLocker encrypt everything including host protected area, device configuration overlay and bad sectors?
Comment: why would it need to?
Answer: Bitlocker works at the volume level, replacing the file system metadata with its own and moving the FS metadata into the encrypted volume (for this reason, BitLocker needs to understand the file system format, so it knows where there is free space for the metadata, and how to mark that space as no longer free). It is Full Volume Encryption, not Full Disk Encryption, and doesn't touch anything outside of a logical volume (which might be a physical partition, or a RAID volume, or possibly something else).
I don't know if BitLocker has any special logic for dealing with bad sectors.
Comment: Wouldn't the firmware on the hard drive handle bad sectors transparently to the OS?
Comment: @johndoe  In an ideal world firmware would handle bad sectors.  However, in realty they don't.  Users still need chkdsk /R.

Does BitLocker encrypt everything?


Can I change the camera to spin on axis to turn instead of rotate around an object?

Question: I've been working on a game where I make all my assets in blender and transfer them to unreal engine. I've come to prefer the unreal engine viewport navigation and was wondering if there is a way to get blenders viewport navigation as close as possible. All too often in blender I would hit the maximum zoom and be shifting my rotation point around until I could get a better view. Basically I find that I'm spending as much time correcting my viewport as I do actually building the asset. I would like to have an infinite zoom and the viewport to rotate as though it's just turning in place like it does in unreal engine. Is this possible? (I think it's similar to the "walk" or "fly" navigation but I want it at all times when building)
Comment: I'm not sure that there is an easy way to replicate this. Especially because WASD keys are used for different operators. But Walk mode on tilda (`) key works perfectly
Comment: I dont mean to use wasd keys to move, I just meant similar to the walk or fly navigation. I'd like it to be the exact same as UE4 if i could. in their editor the scroll wheel will move forward and back at a consistent speed forever whereas in blender it will slow and eventually stop when you reach the center of your pivot point. and when you click and hold the scroll wheel it will move the camera up/down/left/right just like blender when you hold the shift key while doing it. and finally (and most important to my needs) when you hold the right mouse button the camera will turn in place.
Answer: Blender keymap is fully customizable, so you can set up a navigation in Unity/Unreal style (camera rotates in place, not around objects) if you want.
You can use Walk/Fly navigation, but assign it to the middle mouse button, so that it will enable on mouse press and disable on release. And you will be able to use WASD keys for moving camera and other keys are available also, while you're holding middle mouse button.
This guide uses the default keymap settings. If you are using Industry compatible keymap, some settings may look differently.
1. Setting up the rotate in place using Walk/Fly navigation.
First, I personally recommend Walk with disabled gravity.

Then in keymap open up 3D View ⇾ 3D View Global section and turn off default navigation by middle mouse (Rotate or pan depends on your settings), then set up View Navigation (Walk/Fly) to middle mouse press:

Then you should leave View Navigation on mouse release, right? So scroll down 3D View section until you find View3D Walk Modal. Disable Teleport because it  will override middle mouse action. Then add new - select action Mouse, type Confirm, Middle, Release.

Once you get it done, you will be able to rotate camera in place and navigate using WASD while holding middle mouse button.
2. Setting Shift+middle mouse button to pan view.
This is easy. Find Pan View in the 3D View ⇾ 3D View Global section and set it like this. Shift must be enabled:

3. Setting up the dolly zoom.
Once you get this, you might find that Blender still zooms at a certain point. So to avoid this, let's set up a dolly zoom instead of the default one. Again, 3D View ⇾ 3D View Global section.
Find Zoom View with Wheel In and Wheel Out shortcuts and replace text view3d.zoom by view3d.dolly. You will see that the title will change from "Zoom View" to "Dolly View". Don't forget to return Delta values, which is reset when command changes.

On my screenshot, zoom might be inverted. If you feel so too, try to replace "1" and "-1"
4. Get back the rotate around on a different hotkey
So this is up to you. You can get back the rotate around if you want. Just enable disabled "Rotate View" and set hotkey on Alt+Right mouse button, for example:


Channel length modulation

Question: Why does channel length modulation occurring in mosfets decrease the gain of amplifiers?
Answer: Channel length modulation causes a finite ratio d(Vds)/d(Id) which resembles a finite otput resistance ro. With other words: The drain current Id is not completely independent on Vds (Id rises slightly with Vds). This output resistance ro acts in parallel to the gain determining external ohmic resistor Rd. As a consequence, the ouput signal voltage is determined by the value ro||Rd (which is smaller than Rd).
Answer: When you increase voltage between drain and source (Vds), because of strong vertical electric field (from drain to bulk) channel on drain side become thinner. As a result total lenght of channel become shorter.

Normally drain current:

After shortening of channel

At the end:
Comment: Yes but the question was why does this reduce amplifier gain, not why does this change effective channel length.

Channel length modulation


Lazy regex doesn't work as expected C#

Question: I have the following regex: a?\W*?b
and I have a string ,.!       ,b
When searching for a match I get ,.!       ,b, but not just b as I expect. Why is that? How to modify the regex to get what I need?
Thank you for your help.
Comment: Please provide some examples of what you do and don't want to match. And better not simplify your regex too much (a comment below suggests that you're actually doing something else), or the solutions might come up too simple, too.
Answer: A lazy quantifier doesn't help here for what you want. Let's see what's happening.
The regex engine starts at the beginning of the string. First tries to match a. It can't, but it's no problem since the a is optional.
Then, there is a lazy \W*? so the regex engine skips it but remembers the current position.
It then tries to match b. It can't, so it backtracks and successfully matches the , with \W*?. It then goes on to try and match b (because of the lazy quantifier). It still can't and backtracks again. This repeats a few times until finally the regex engine has arrived at the b. Now the match is complete - the regex engine declares success.
So the regex works as specified - just not as intended. Now the question is: What exactly do you want the regex to do?
For example, if what you really want is:
Match b alone, unless it's preceded by a and some non-word characters, in which case match everything from a to b, then use
b|a\W*b

Comment: What to do if I have the following regex 'a?\W*?b?\W*?c?\W*?d?' and I want that any not-empty match starts with a letter and ends with a letter too
Comment: A regex that does what you just wrote could be `^([a-z].*[a-z]$|)$`. Match either something that starts and ends with a letter, or match the empty string. Use `RegexOptions.IgnoreCase` if you want uppercase letters to match, too. And use `\p{L}` instead of `[a-z]` if you also want to allow non-ASCII letters.
Comment: +1, very good explanation... I learned something about regex today :)
Answer: A lazy expression is only lazy from the right, i.e. it will be as short as possible by removing characters on the right, but it will not remove characters on the left.
To make the match start later, you need a greedy expression before it that swallows the characters that you don't want to match.
Alternatively, as Tim showed, you can make the match start later by only matching the first character and the following separators if the first character exists.
Answer: For example, the following might work: (a\W*)?b
To know better what might solve your problem, you should include more examples.
Comment: Actually the original regex is 'a?\W*?b?\W*?c?' and I want that any match doesn't contain non-alphabet (\W) symbols neither from the beginig nor at the end.
Comment: Could you clarify this (how about some examples?) and put it into your original question instead of somewhere down in the comments where it's harder to find?
Answer: Your regexp matches the entire string like this:

a, zero or one repetitions ("" in this case)
Any character that is not alphanumeric, any number of repetitions, as few as possible (",.! ," in this case)
b

In your case the regexp matches the entire string, and will therefor not find just the b (it doesn't find several matches of the same part).
If you search in a string like ',.! ,db' it will find the b.
Comment: I can tell you're using Expresso, I have exactly the same description on my screen ;)
Comment: Yepp - thought it was good descriptions, so I didn't bother rewriting :-)
Answer: The a? says "i want either zero or one instance of a" - this is satisfied as there is zero instances, and followed by
\W* says "i want zero or more non word characters", which is satisfied by the punctuation and space characters, and finally
b says "match a letter b", which it does. So your whole string satisfies the regex.
It helps if you give more examples of possible inputs before anyone sugests a possible solution.
Comment: Your explanation is ignoring the lazy quantifier `*?`... why doesn't it work here ?
Comment: \W*? is lazy. So it should include as few as possible symbols. Here the fewest amount of symbols is 0.
Answer: Your example doesn't show why the a? is part of your regex but to match only b in a string that looks like ,.! ,b you can use lookbehind like this (?=\W*?)b.
This matches b that is preceded by a character that is a "non-word character" zero and unlmited times (as few as possible)
If you only want to match say a and b in a string such as a,.! ,b you'll have to use capturing groups: (a?)\W*?(b) where group one will hold the a if present and group 2 b
Answer: It's a mistake to speak of a regex as being greedy or non-greedy.  You can use non-greedy quantifiers throughout the regex, but it will still try to start matching at the earliest opportunity, as you discovered.  Similarly, a regex that uses only greedy quantifiers isn't guaranteed to return the longest possible match.  For example,
Regex.Match("foo bar", @"\w+ (?:b|bar)")

...returns foo b, because alternation settles for the first alternative that works, even if a later one would result in a longer match.  (Note that I'm talking about Perl-derived regex flavors like .NET's; some flavors, like awk and egrep, do indeed hold out for the longest possible match.  But, since those flavors don't have non-greedy quantifiers, greedy isn't just the default mode, it's the only mode.)
In short, there's no such thing as a greedy or non-greedy regex, only greedy or non-greedy quantifiers.

Lazy regex doesn't work as expected C#


Do receiving addresses expire on Multibit?

Question: Okay so I have tried Resetting Blockchain and Transactions plenty of times. The only thing is I used an old receiving address that I used about 2 months ago. Do these addresses expire? Can they only be used once? I don't know. I'm not too good at this. Here is my Wallet: http://puu.sh/8CBpT.png Here is the blockchain of my receiving address: https://blockchain.info/address/12EFht1oW3h2ybNYP2NrJKF26oPicFs8mM Is there any way to retrieve my bitcoin? Do you guys have any answers? I have looked everywhere on google and I have gotten nowhere.
Comment: That's weird. It is accepted, your wallet contains ~0.20 BTC but it doesn't show up in Multibit, right? Normally a reset would yield the proper result. Is multibit (and your computer) connected to the internet? Is Multibit updated? I guess it is, but it *should* work.
Comment: Im going to try to back it up and use a different wallet software. Hopefully that works. Thanks for your help!
Answer: But to answer the question: addresses do NOT expire. Not in Multibit, not in cold storage, not online (unless the site retired and you do not have an offline private key or wallet), nor any other wallet software. They can be used unlimited amount of times, so that's not the problem either. The problem your face is probably in Multibit. You can try to backup the .wallet or .key, and import it in another wallet software, or delete/reinstall Multibit.