Like the question  is there a way to stop getting a popup when a friend starts playing a game?    I tried following the global method as outlined above, as that is what I want:     Either notifications need to be turned off globally (View -> Settings -> Friends -> Uncheck all the "Display a Notification" check boxes)   However I can't see the 'Friends' tab.
It used to be in settings under the friends tab. The friends tab no longer exists! Where do I find that infuriatingly missing setting now?
How to solve $\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$ As my question says,   $$\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$$  As far as $\frac 1{1.2}+\frac 1{2.3}+\frac 1{3.4}+...$ is concerned, we can write it as $\frac 1{n-1} - \frac 1{n}$ What can we do here? How do I "Show total up and down vote counts" since I'm at 1K rep? I am looking for idioms or informal/slang/colloquial expression for some people that make you think that they are able of building a skyscraper, constructing a spaceship, playing the piano better than Mozart or something very fancy that requires remarkable skills, but when it comes to practice they prove to be completely inadequate. I saw this in the end of an awesome shell script but I can't understand the login here because I think it's being short-handed for a longer command.  spark ${@:-`cat`}   This apears at the end of . Any ideas?  + Marks for some one who extends it into a full segment of code, even if its slower (Better for explanation) Why are some muggles allowed to know of the magical world? Suppose that $y_1(x)$ and $y_2(x)$ are solutions of the differential equation $y''+py'+qy=0$ on $I$. How can I show that if $y_1$ and $y_2$ vanish at the same point then they are not linearly independent?   Here is my attempt to prove the problem. Since $y_1(x)$ and $y_2(x)$ are solutions of the differential equation $y''+py'+qy=0$ on $I$, it is enough for me to show that the Wronskian of $y_1$ and $y_2$ denoted by $W(y_1,y_2)$ is zero. Now since $y_1$ and $y_2$ vanish at the same point say $p$ we have $y_1(p)=0$ and also $y_2(p)=0$.   Solving for $W(y_1,y_2)(p)$ we have: $y_1(p)y_2'(p)-y_2(p)y_1'(p)=0$ since $y_1(p)=0$ and also $y_2(p)=0$.   Am I correct? Thanks How to reinstall Cortana? With reference to:    I removed the Cortana packages from my running Windows 10 using the method posted by magicandre1981 and using  win6x_registry_tweak.   My question is how can I re-install Cortana (or for that matter any other package removed similarly)?  I have opened up the install.wim image (converted from a Win 10 install.esd file) and found Cortana in the system apps folder, but don't know how / what to use to re-install it.  Any help please? Load a table with NO geometry_column using PyQGIS Increase close vote weight for gold tag badge holders In order to increase the efficiency with which poor quality questions are closed, it could make sense to have weighted close votes for a small subset of qualified users.   A very good way to measure the qualified users would be to leverage the tag badges. However, it needs to be rationally limited to a set of users that are knowledgeable with the subject. A very good way to measure the top users in a tag would be to look at those who have a gold badge in the tag.  If two users who have a gold tag badge pertinent to a question issue a close vote, the question should probably be closed (or probably will be very soon under the current system). I think it would make sense to allow users with a gold tag badge to have a 3x weight for close votes in those tags.  Please keep in mind, this isn't suggesting full moderator one click closure. It only adds weight to the vote. A single user could not do it on their own. These singular users cast close votes all the time, so that aspect would not change. Some will vote to close, some wont, but the ones who do should have a stronger weight to their vote. Even with 3x it will require 2 of said votes to accomplish, or still 2 other users (meaning 3 agreed it should be closed - which is the same amount of high rep users it takes to delete a question and that has a much greater impact than closure).  tldr;  Allow gold tag badge users to have a weighted close vote in their tag Why are the moderators being so strict with quality related flags recently? Weight Painting Won't Get Rid of Weird Protrusions Reputation widget failure on Android The title says it all. When I try to place the reputation widget on my home screen, I get this:  When I try again, I get this error, presumably because the Stack Exchange service has stopped.  Is this a bug? If so, where should I report it?  I am running Android 5.1 armeabi on a DigiLand DL718M tablet. Intersection of kernels and linear dependence of functionals How to simulate from a Gaussian copula? Suppose that I have two univariate marginal distributions, say $F$ and $G$, which I can simulate from. Now, construct their joint distribution using a , denoted $C(F,G;\Sigma)$. All the parameters are known.  Is there a non-MCMC method for simulating from this copula? There is a way to use my Galaxy tab s like a wacom table, on linux? Show that the series $\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$ converges Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.   I showed it using Abel's theorem and limit comparison test. Any other simpler method? When "etc." is at the end of a phrase, do you place a period after it? Example:     It's all about apples, oranges, bananas, etc.   VS.     It's all about apples, oranges, bananas, etc..   Update  What happens if the abbreviation is inside parentheses, do you place a dot after and before the closing parenthesis?     It's all about fruit (apples, bananas, etc.). Cancelled UK visa on entry How do I stop Windows 10 from updating my graphics driver? There exist several implementations of the GBDT family of model such as:   GBM XGBoost LightGBM Catboost.    What are the mathematical differences between these different implementations?  Catboost seems to outperform the other implementations even by using only its default parameters according to , but it is still very slow.  My guess is that catboost doesn't use the dummified variables, so the weight given to each (categorical) variable is more balanced compared to the other implementations, so the high-cardinality variables don't have more weight than the others. It allows the weak categorical (with low cardinality) to enter to some trees, hence better performance. Other than that, I have no further explanation.