There is a green progress bar above the team members names in the BF4 display. I've noticed it displays different icons and changes depending on which class you're playing.   What is this progress bar for? What causes the bar to increase? What happens when it hits 100%? Does something happen when it reaches on of the icons? Is the progress per game or is it persistent the next time I play the game?
Firstly, how do you gain the score exactly? I know it shows in the bar on the side of the map, but I remember a match where I was lone-wolfing and still got a pretty high score in that bar. I suppose my other squad members were together, but do I also gain for their work?  Also, does the score reset when I start a new game? I mean, I play support with the Offensive perk and it took me a long game of conquest to just unlock lvl-3 extra grenade. In the next match, that 'nade is gone. Don't I keep it forever?
Lasso regression feature selection I have been reading many articles on LASSO regression, and everyone claims that LASSO address multicollinearity showing contour plots of cost function touching the corner of the diamond(x1+x2). In the first place, I didn't understand how does this phenomenon translate to addressing multicollinearity when there are many groups of correlated variables in the data.  Verbatim from one of the research paper: "• If there are grouped variables (highly correlated between each other) LASSO tends to select one variable from each group ignoring the others". I've recently been learning about differential equations, and my teacher has been giving some particularly difficult examples out for those of us who finish early. He gave us the following differential equation: $$x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$$ With the hint "Differentiate with respect to $y$, then let $\frac{dy}{dx}=p$". I've solved it after the lesson and would like someone to check whether or not I've done so correctly.    Following the hint, $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}-2p\frac{dp}{dy}$$ Which rearranges to $$p-\frac{1}{p}+y\frac{dp}{dy}=2p\frac{dp}{dy}$$ Multiplying through by $\frac{dy}{dp}$, $$\left(p-\frac{1}{p}\right)\frac{dy}{dp}+y=2p$$ Rearranging into the form $\frac{dy}{dp}+f(p)y=g(p)$, and writing $p-\frac{1}{p}=\frac{p^{2}-1}{p}$ $$\frac{dy}{dp}+\frac{p}{p^{2}-1}y=\frac{2p^{2}}{p^{2}-1}$$ Our integrating factor is $$e^{\int\frac{p}{p^{2}-1}dp}=e^{\frac{1}{2}\ln(p^{2}-1)}=\sqrt{p^{2}-1}$$ So we have $$\sqrt{p^{2}-1}\frac{dy}{dp}+\frac{p}{\sqrt{p^{2}-1}}y=\frac{2p^{2}}{\sqrt{p^{2}-1}}$$ The left hand side is, by design, $\frac{d}{dp}(y\sqrt{p^{2}-1})$, and the right hand side can be integrated as follows: $$\int \frac{2p^{2}}{\sqrt{p^{2}-1}}dp=\int\frac{2(p^{2}-1)+2}{\sqrt{p^{2}-1}}dp=\int 2\sqrt{p^{2}-1}dp+\int \frac{2}{\sqrt{p^{2}-1}} dp$$ These can both be solved using the substitution $p=\cosh(u)$, and gives $$y\sqrt{p^{2}-1}=p\sqrt{p^{2}+1}-\cosh^{-1}(p)+\alpha$$ Where $\alpha$ is an arbitrary constant. Dividing, we finally have $$y=p+\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}$$  Here I was stuck for a long time, until I realised that we also have the equation we began with: $x=py-p^{2}$, which we can solve as a quadratic in $p$ to also get $$p=\frac{y\pm \sqrt{y^{2}-4x}}{2} \implies p^{2}=\frac{y^{2}+y^{2}-4x \pm 2y\sqrt{y^{2}-4x}}{4}$$  Thus, $$y=\frac{y\pm \sqrt{y^{2}-4x}}{2}+\frac{\cosh^{-1}\left(\frac{y\pm \sqrt{y^{2}-4x}}{2}\right)+\alpha}{\sqrt{\frac{y^{2}-2x \pm y\sqrt{y^{2}-4x}}{2}-1}}$$  This is as good as I can do - we have a relation between $y$ and $x$ - am I done? Is this the correct answer? Is there an easier way?     Thank you for your time.    Following the suggestion by @Valentin $$\frac{dy}{dp}=1+\frac{\frac{1}{\sqrt{p^{2}-1}}\sqrt{p^{2}-1}-p\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}=1+p\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}$$ Hence, $$dx=\frac{dy}{p}=dp\left(\frac{1}{p}+\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}\right)$$ How to find which package provides a file? For example, I can't install package qcad because of error: "can't find qdict.h". How can I find the package which provides the file qdict.h so I can install the package qcad? If $ar + bs =1$, then $\gcd(a,s) = \gcd(r,b) = \gcd(r,s) = 1$ I am currently thinking about open-sourcing a project of mine and am in the process of preparing the source code and project structure to be released to the public. Now I got one question: how should I handle the signature key for my assemblies? Should I create a new key for the open-source version and publish it along with the other files to the SVN repository? Should I leave the key out and everyone who wants to compile the code should generate his own key?  How do you handle this? I feel a little bit uncomfortable with releasing a signature key to the public. Existence of the moment generating function and variance How can I make realistic looking potions as handouts If $X$ and $Y$ are sets, their cartesian product $X\times Y$ is a set. In ZFC set theory, every element object is a set. We know or feel that if $X$ and $Y$ are sets, their cartesian product $X\times Y$ is a set.  My question is: Why if $X$ and $Y$ are sets, their cartesian product $X\times Y$ is a set? When should static_cast, dynamic_cast, const_cast and reinterpret_cast be used? What are the proper uses of:   static_cast dynamic_cast const_cast reinterpret_cast C-style cast (type)value Function-style cast type(value)   How does one decide which to use in which specific cases? Standalone python app that uses qgis.core without installing qgis Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ I am trying to prove  $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$  This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.   $$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$$  First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?  Integrate by parts with $$u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$$  which leads to  $$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$$  Since $(n-1)$ is a constant, we can throw it out front of the integral:  $$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$$  I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$  $$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$  According to Wikipedia as noted , this simplifies to:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$$  Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?  Assuming knowledge of equation 5, I see how to rewrite it as  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x  - (n-1) I_{n} \tag{6}$$  and solve for $I_n$. I had tried exploiting the fact that  $$\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $$  and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$  which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:  $$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$$  but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of   $$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$ What are Squirrel Girl's 'actual' stats? Is she truly the most powerful being in the Marvel universe? On the  ("The Definitive online source for Marvel super hero bios") Squirrel Girl is listed as having the maximum stats in her .  However in the  ("Comprehensive and exhaustive") she's listed as having lower stats.     Which one is correct and why is there a discrepancy? Is there any way to set a system wide memory limit a process can use in Windows XP? I have a couple of unstable apps which do work ok for most of the time but can hit a bug which results in eating whole memory in a matter of seconds (or at least I suppose that's it). This results in a hard reset as Windows becomes totally unresponsive and I lose my work.  I would like to be able to do something like the /etc/limits on Linux - setting M90, for instance (to set 90% max memory for a single user to allocate). So the system gets the remaining 10% no matter what. How to properly adjust monitor brightness and contrast? In the following is it better to use a possessive noun with an apostrophe or an attributive noun without an apostrophe?     The following list details the assumptions that have been made in   conjunction with the Dell Company’s staff.   Or      The following list details the assumptions that have been made in   conjunction with the Dell Company staff.   Assume Company is the name of a company, such as Dell. A new Victorian-set British novel begins sentences containing h-dropping with lower-case e's, as in " 'e took my money, 'e did." This seems incorrect but I'm at a loss to find the rule. As a non-native speaker of English, it can often be hard to spot some of the errors that a native speaker would find. For a long time, I've been thinking about getting a native speaker to help me with proof-reading my texts (articles, theses, application, etc), but I don't quite know where to look. What are some good ways to get proofreading for your texts?  For example, are there any websites on the Internet where one can exchange proofreading services between different languages? (Say, if you help me proofread my text in language x, I'll help you with yours in language y.) Genes and Intelligence Assuming that intelligence has a genetic component,   • do we know which genes contribute to it?   and, if so,  • can we predict intelligence from genomic analysis? After Factory Reset, In order to use Play store and Gmail I want to setup Google account on Samsung GT-S6802. When I tries to setup Google Account, In-spite of right user-name and password, following window appearers:    When I open gmail form browser, I've mail from google subject:Sign-in attempt prevented and says:     Someone just tried to sign in to your Google Account hidden@gmail.com from an app that doesn't meet modern security standards.     When I review, it shows unknown device but on browser it detects my device.  So, How do I face this issue and setup google account? Computing the cardinality of a combinatorial set Define two sets: $S_1$ and $S_2$ where we have that $a$ is an odd natural number and $b\in \mathbb{N}$, $S_1 = \Big\{ \frac{2}{a},\frac{2}{a-2},\frac{2}{a-4},...,\frac{2}{a-(a-1)},$ $\frac{2+4}{a+2},\frac{2+4}{a},\frac{2+4}{a-2},...,\frac{2+4}{a-(a-1)},$ $\frac{2+8}{a+4},...,\frac{2+8}{a-2},...,\frac{2+8}{a-(a-1)},$ $\vdots$ $\frac{2+4\lfloor\frac{b}{2}\rfloor}{a+\lfloor\frac{b}{2}\rfloor},\frac{2+4\lfloor\frac{b}{2}\rfloor}{a+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+4\lfloor\frac{b}{2}\rfloor}{a-(a-1)},$ $\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a+\lfloor\frac{b}{2}\rfloor},\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a-(a-1),}$ $\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a+\lfloor\frac{b}{2}\rfloor-2},\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a+\lfloor\frac{b}{2}\rfloor-4},...,\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a-(a-1)}$ $\vdots$ $\frac{2+4b}{a},\frac{2+4b}{a-2},...,\frac{2+4b}{a-(a-1)} \Big\}$ and $S_2 = \Big\{ \frac{2}{\lceil\frac{a}{2}\rceil}, \frac{2}{\lceil\frac{a}{2}\rceil-1},\frac{2}{\lceil\frac{a}{2}\rceil-2},...,\frac{2}{1}$ $\frac{2+2}{\lceil\frac{a}{2}\rceil+1},\frac{2+2}{\lceil\frac{a}{2}\rceil},\frac{2+2}{\lceil\frac{a}{2}\rceil-1},...,\frac{2+2}{1},$ $\vdots$ $\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{\lceil\frac{a}{2}\rceil+(\lfloor\frac{b}{2}\rfloor-1)},\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{\lceil\frac{a}{2}\rceil+(\lfloor\frac{b}{2}\rfloor-2)},...,\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{1}$ $\frac{2+2\lfloor\frac{b}{2}\rfloor}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor},...,\frac{2+2\lfloor\frac{b}{2}\rfloor}{1},$ $\frac{2+2(\lfloor\frac{b}{2}\rfloor+1)}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor-1},...,\frac{2+2(\lfloor\frac{b}{2}\rfloor+1)}{1},$ $\frac{2+(\lfloor\frac{b}{2}\rfloor+2)}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+(\lfloor\frac{b}{2}\rfloor+2)}{1}$ $\vdots$ $\frac{2+2b}{\lceil\frac{a}{2}\rceil-1},...,\frac{2+2b}{1} \Big\}$ The question is, how do I determine the cardinality of the intersection of these two sets i.e. how would I compute the value $|S_1 \cap S_2|$? Note that duplicates may appear within both sets such as $\frac{1}{2}=\frac{2}{4}$ etc. Therefore we must make sure that for each fraction $\frac{n}{m}$ we have $\text{gcd}(n,m)=1$. I am aware of the existence of , but not entirely sure if that identity would be useful here.