It blends with lower layers in difference mode, I know, but how can I achieve this "distorted color" effect it produces on highlight areas? (the orange and magenta areas)
so there are some idiosyncrasies with the method they use to invert that part of the image which I can't figure out. It could be that they just invert the luminosity but I don't know how i would do that in a dynamic way.
I just saw Get Out, and there's one thing I don't understand.  Georgina turns out to be controlled by Rose's grandmother; its grandmothers brain in her body. Why then, is Georgina constantly cleaning and being treated like an actual maid.  During the diner scene, we get a glimpse into the kitchen, and we see Georgina holding up a plate in way that suggested that she was made to work and cook. The pose and the fact that she was already standing there holding the plate before Missy enters the kitchen was also quite strange and suggestive.  Also, at the "party" theres a moment that Chris sees Georgina and Missy. For a moment he thinks that Georgina is being involved in the party, but than it turns out the Georgina is just handed empty glasses to take care of.  So this I don't get. If its really grandma inside Georgina, then why is she treated in this way? Every Christmas, Harry receives a mediocre gift from the Dursleys. In Harry Potter and the Chamber of Secrets he gets a toothpick.     At that moment, Hedwig swooped into the room, carrying a very small package in her beak.     “Hello,” said Harry happily as she landed on his bed. “Are you speaking to me again?”     She nibbled his ear in an affectionate sort of way, which was a far better present than the one that    she had brought him, which turned out to be from the Dursleys. They had sent Harry a toothpick    and a note telling him to find out whether he’d be able to stay at Hogwarts for the summer    vacation, too.    They hate owls, but obviously use them to contact Harry at school? How can I use Web.debug.config in the built-in visual studio debugger server? Does Google Chrome has anything similar to about:blank page in IE and Firefox? I did not find any suitable page mentioning this? PS: I tried about:blank but it looked like Chrome is trying to load something before showing blank page... Is it a problem if a UK visa was issued but not used? Last year, my wife and I were issused 6-month visieor visas for the UK. She did not travel with me and, therefore, didn't use her visa. I am currently applying for a visitor visa for my daughter and myself.  Will the fact that my didn't use her visa have any negative affect on our new application? Get the subdomain from a URL Getting the subdomain from a URL sounds easy at first.  http://www.domain.example   Scan for the first period then return whatever came after the "http://" ...  Then you remember  http://super.duper.domain.example   Oh. So then you think, okay, find the last period, go back a word and get everything before!   Then you remember  http://super.duper.domain.co.uk   And you're back to square one. Anyone have any great ideas besides storing a list of all TLDs? I like to be thorough, but if you feel confident you can skip the first paragraph.  Review: A ring is a set $R$ endowed with two operations of + and $\cdot$ such that $(G,+)$ is an additive abelian group, multiplication is associative, $R$ contains the multiplicative identity (denoted with 1), and the distributive law holds.  If multiplication is also commutative, we say $R$ is a commutative ring. A ring that has no zero divisors (non-zero elements whose product is zero) is called an integral domain, or just a domain.  We want to show that for a domain, the equation $x^2 = 1$ has at most 2 solutions in $R$ (one of which is the trivial solution 1).    Here's what I did:   For simplicity let $1,a,b$ and $c$ be distinct non-zero elements in $R$.  Assume $a^2 = 1$.  We want to show that letting $b^2 = 1$ as well will lead to a contradiction.  So suppose $b^2 = 1$, then it follows that $a^2b^2 = (ab)^2 = 1$, so $ab$ is a solution as well, but is it a new solution?  If $ab = 1$, then $abb = 1b \Rightarrow a = b$ which is a contradiction.  If $ab = a$, then $aab = aa \Rightarrow b = 1$ which is also a contradiction.  Similarly, $ab = b$ won't work either.  So it must be that $ab = c$.  So by "admitting" $b$ as a solution, we're forced to admit $c$ as well.  So far we have $a^2 = b^2 = c^2 = 1$ and $ab = c$.  We can procede as before as say that $(abc)^2 = 1$, so $abc$ is a solution, but once again we should check if it is a new solution.  From $ab = c$, we get $a = cb$ and $b = ac$, so $abc = (cb)(ac)(ab) = (abc)^2 = 1$.  So $abc$ is not a new solution; it's just one.  At this point I'm stuck.  I've shown that it is in fact possible to have a ring with 4 distinct elements, namely $1,a,b$ and $c$ such that each satisfies the equation $x^2 = 1$ and $abc = 1$.  What am I missing? How do I change file permissions on a FAT32 drive? How do I move the MySQL data directory? I am trying to move the data directory of my MySQL database to a second disk array that I have as a mount point /array2/.  The problem I am having is I have tried everything and after I modify the location of datadir in my.cnf mysql will not start up again.  All I get is:  start: Job failed to start Robot tank run amok, its trainer trapped in cave on the Moon Another dim memory from my sci-fi reading binges in the 1980s... this one almost definitely in one of the big hardback collections full of late Golden Age short stories.  The story is told from the perspective of a robot (maybe cyborg?) tank that has stopped responding to control signals and is aggressively "defending" some portion of the Moon, to which it was previously assigned. Maybe a war had ended, or maybe it became confused about an exercise that it thought was real?  There was a person trapped in a cave trying to communicate with the tank. The tank mostly ignores him, so the main purpose of what he says is exposition for the reader. I think this person was an expert on this type of robot/cyborg sent to "talk it down", but the tank had attacked, and this guy was the only survivor. The expert was badly wounded -- I think he mentions his suit had to amputate his leg to seal a hole -- or maybe flood it with water which would freeze to seal the hole but mean that his leg had to be amputated later. He keeps trying to convince the tank to stand down, at one point saying that he knew it "back when it was just a two-wheeler an autocyber in boot camp" or something along that line. (Edit: I must have been reading too fast and/or distracted back then, since per the quote below I seem to have misread "motorcycle" for "autocyber"!)  One detail of the attack stands out -- the tank was low on ammunition, I think, so it used a kind of bomb with an electric/magnetic "catapult" (aka "magnapult") to hit the expert's transport. The tank was proud of itself for figuring out a way to hit it far out of the normal range by 1) using its own top speed to add velocity to the projectile, and 2) building up the catapult's energy past its normal operating range, which caused it pain but made the bomb go much further. It catches the transport by surprise, since they thought they were safely out of range of the armaments it had left.  The premise sounds a lot like a Keith Laumer "Bolo" story, but I'm fairly sure that it's not actually one of them. It had a somewhat grimmer feel to it. I could be wrong, though, since it's been 30 years. Is multiplication not just repeated addition? If multiplication is not repeated addition, what is it? How do you define multiplication? All of us had to memorize the multiplication table in elementary school, but how did they come up with it if so many people claim that it is wrong to think of it as repeated addition? Voodoo magic? If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \frac{n-1}{n+1}$, find $a_n$ and the sum. Can others in my party go through TSA PreCheck? If I have  (i.e. I have a KTN from TSA and the "pre-check" logo gets printed on my boarding pass) and a travel companion does not, can they go with me through the PreCheck security lines? This is assuming that the travel companion is booked in the same reservation on the same flights and is also a U.S. citizen and not on any no-fly or otherwise increased scrutiny lists.  I'm guessing the answer is no, but I haven't seen anything from TSA specifically addressing this. When, where and **how often** do you find polynomials of higher degrees than two in mathematical, pure/applied, research? A formula for solving a polynomial of degree three, see  link; $ax^3+bx^2+cx+d=0$, is  $$\begin{align} x\quad&amp;=\quad       \sqrt[3]{           \left(               \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}           \right)            +                 \sqrt{               \left(                   \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}               \right) ^2                 +                \left(                   \frac{c}{3a} - \frac{b^2}{9a^2}               \right) ^3            }       }\\       &amp;+\quad       \sqrt[3]{           \left(               \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}           \right)            -                 \sqrt{               \left(                   \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a}               \right) ^2                 +                \left(                   \frac{c}{3a} - \frac{b^2}{9a^2}               \right) ^3            }       }       \;-\;\frac{b}{3a} \end{align}$$     Unlike quadratic, cubic, and quartic polynomials, the general quintic   cannot be solved algebraically in terms of a finite number of   additions, subtractions, multiplications, divisions, and root   extractions, as rigorously demonstrated by Abel (Abel's impossibility   theorem) and Galois. However, certain classes of quintic equations can   be solved [...] Source:    At levels of $5^{\text{th}}$ degree polynomials, things are starting to look really serious in my eyes. My question is:     If it is possible to not answer subjectively: When, where and how often do you find polynomials of higher degrees than two in mathematical, pure/applied, research? RemoveDuplicates gives 1004 error How do I remove the proprietary ATI drivers? I've installed 11.10 and the proprietary ATI drivers using "additional drivers" The performance of my system is absolutely awful and it shouldn't be. I tried to remove the proprietary drivers using the Additional Drivers tool and it appears to remove them. However after I reboot I cant get back into my desktop properly (the panel and launcher go missing). This doesn't seem to be an isolated problem in 11.XX.  covers how to restore the desktop (panel and launcher), but the guide doesn't fix my problem though.   Whenever I do sudo unity --reset it runs through its normal processes until it hangs at setting update "run_key" and never gets past that. I must reinstall the proprietary drivers using jockey-text or jockey-gtk in order to get back to my proper desktop.   Interestingly enough the system performance seems improved while it is in its "broken" state (missing panel and launcher).  I think restoring the default drivers may solve my problems but I cant figure out how to do it. How exactly does one “control for other variables”? Here is the article that motivated this question:   I liked this article, and it nicely demonstrates the concept of “controlling for other variables” (IQ, career, income, age, etc) in order to best isolate the true relationship between just the 2 variables in question.    Can you explain to me how you actually control for variables on a typical data set?      E.g., if you have 2 people with the same impatience level and BMI, but different incomes, how do you treat these data?  Do you categorize them into different subgroups that do have similar income, patience, and BMI?  But, eventually there are dozens of variables to control for (IQ, career, income, age, etc)  How do you then aggregate these (potentially) 100’s of subgroups?  In fact, I have a feeling this approach is barking up the wrong tree, now that I’ve verbalized it.  Thanks for shedding any light on something I've meant to get to the bottom of for a few years now...! Explanatory variables may bias predictions I' m asking this question out of sheer curiosity, my teacher was not able to explain it.   If I'm using logistic regression with categorical variables they are coded like {1,2,3}. I guess it wouldn't change my results if I used {4,5,6}. But what if linearity of coding wasn't kept? (say {4,10,99})? What I deal with is just the way of coding factor variables but is it possible that the statistical inference could be "distorted" this way? Or, worst case scenario, would I be able to draw nonsense conclusions from categorical data just because they are coded in some way? Let $(a_n)_{n}\in\ell^2:=\ell^2(\mathbb{R})$ be a fixed sequence. Consider the subspace $$C=\{(x_n)_{n}\in\ell^2 : |x_n|\le a_n\text{ for all }n\in\mathbb{N}\}.$$  According to the book [Dunford and Schwartz, Linear operators part I, page 453] $C$ is compact in the $\ell^2$-norm, but there is no proof. How can I show that $C$ is indeed compact in $\ell^2$ ? If I run a randomForest model, I can then make predictions based on the model.  Is there a way to get a prediction interval of each of the predictions such that I know how "sure" the model is of its answer.  If this is possible is it simply based on the variability of the dependent variable for the whole model or will it have wider and narrower intervals depending on the particular decision tree that was followed for a particular prediction?