With ctrl + cmd + space, macOS launches Emojis &amp; Symbols, also know as the Character Viewer.  Is there a shortcut to launch the Keyboard Viewer?
KeyboardViewer app is saved from this AppleScript.  on run {input, parameters}     if application "KeyboardViewer" is running then         quit application "KeyboardViewer"     else          activate application "KeyboardViewer"     end if     return input end run   Dock toggle save. Adding shortcut to above, needs some work...   Syntax error Can’t set application "KeyboardViewer" to some object. Access not allowed.  tell application "System Events"     set application "KeyboardViewer" to (keystroke "K" using control down) end tell
I found this odd relationship, $x^2 = \sum_\limits{k = 0}^{x-1} (2k + 1)$. Is it correct to hyphenate with compound premodifiers? If so, where is the hyphen placed? For example, "file system" and "related". Is it "file system-related"? It will appear as if it is a compound of "file" and "system-related", won't it? Probability that median falls within a range? If I draw 5 samples from a distribution that is cont., what is the probability that the median of the population is between the xmin and xmax of the draw samples?  What is the formula to apply here? As we know, any proposition can translate to Godel numbers, such as  $$( \exists x)\space (x=sy)$$Its Godel numbers is $2^8 × 3 ^4 × 5^ {13} × 7^ 9 × 11^8 × 13^{13} × 17^ 5 × 19^ 7 × 23^ {17} × 29^ 9=a \space big \space number $  We also could get the character sequence "$( \exists x)\space (x=sy)$" through prime factorization of this big number.  Such as$$243,000,000=2^6×3^5×5^6 $$ $$6\space\space\space\space\space 5\space\space\space\space\space 6 $$ $$\downarrow\space\space\downarrow\space\space\downarrow$$ $$0\space\space\space=\space\space\space 0 $$ So $243000000$ is the Godel numbers of "$0=0$"   There are four axioms in Godel's PM system, here are two of them: $$(p\lor p)\subset p$$ $$p\subset (p\lor q)$$ through these two axioms, we can prove "$(p\lor p)\subset(p\lor q)$" is a true theorem.   And these two axioms and a theorem all have a corresponding Godel numbers, can we prove this theorem is true through Godel numbers? If it works, how? First player rolls a 6-sided die 100 times and the second player 101 times. What are the odds of the second player getting more odd numbers than the first one?  I tried solving this via listing the options which led to a nasty sum that I wasn't able to solve neither by hand nor by using wolfram alpha. Clearly there has to be a more clever solution using the properties of a binomial distribution, but so far I haven't come across anything useful. I would like to look at my referral traffic over time on a per domain-basis in Google Analytics. This is what I have done:   Set New visits as advanced segment  Navigated to traffic-sources -> Referrals Checked a coulple of domains Clicked plot rows   The individual domains now show up in the graph, but the problem is that "all traffic" is also visible. All traffic is much bigger than the graphs for individual domains. This makes it very hard to see any changes as they look all values are Zero.  Is there any way to hide "Visists" from the graph in order to zoom in on the vertical axis? Is there such a thing as a Ubuntu Certified Engineer? Sometimes we assume that regressors are fixed, i.e. they are non-stochastic. I think that means all our predictors, parameter estimates etc. are unconditional then, right? Might I even go so far that they are no longer random variables?  If on the other hand we accept that most regressors in economics say are stochastic because no outside force determined them with some experiment in mind. Econometricians then condition on these stochastic regressors.  How is this different from treating them as fixed?  I understand what conditioning is. Mathematically, it means we make all observations and inference conditional on that particular set of regressors and have no ambitions to say that inferences, parameter estimates, variance estimates etc. would have been the same had we seen a different realization of our regressors (such is the crux in time series, where each time series is only ever seen once).  However, to really grasp the difference between fixed regressors vs. conditioning on stochastic regressors, I am wondering if anyone here knows of an example of an estimation or inference procedure that is valid for say fixed regressors but breaks down when they are stochastic (and will be conditioned on).  I am looking forward to seeing those examples! Update:   Once upon a time there was a feature called . Its goal was to help our most active users find the questions they can answer. &quot;New nav&quot; never made it out of beta and that beta only ever made it to Stack Overflow users. Several months ago we gave new nav a sad, but overdue burial. As we promised, we've been working on a replacement for new nav, and now it's time for us to start rolling it out. We're calling it custom question lists and we are really excited about its potential to help all users get at the questions that are most interesting to them. While we hope what we've built hits the mark for many of you who have requested a feature like this (e.g., , , , , , ), we know how difficult it is to make something complex simple. So, we are rolling this out in beta to gather feedback from you and iterate on it. (Don't worry, we promise to actually release it this time.) When it's live later you'll be able to opt-in to the beta by turning on a new account preference. This will enable the feature for Stack Overflow, Meta Stack Overflow, and Meta Stack Exchange. It will be available for all network sites when the beta period is over. Schedule for future improvements This feature will roll out in four phases:  Phase I: Filter and sort controls to let you modify a question list view (Julyish) Phase II: Saved custom lists (In progress, ETA November 2018) Phase III: Notifications on custom lists (December 2018) Phase IV: Final fixes and official release (January 2019)  Sneak peak Phase I: Filter and sorts We'll add filter and sort controls to question lists. This does two things:  Clarifies the definition of built-in views - Naming is hard and short names can't always describe what a view is showing you. By exposing the filter and sort controls it is more obvious. Introduces the ability to create new custom views - This won't be super useful until we add the ability to save the views to the UI (phase II). Until then you can bookmark the URL and get back to your views.  Question list with new controls   Views drop down   Filter   Sort   Phase II: Saved custom question lists In this phase we will add the ability to save your custom questions lists. Create custom list In our current design, a dialog will be pre-populated based on the filters and sorts you're currently using. Give it a name, confirm your options, and click save.  Accessing your saved lists When you save your custom lists they will show up in the &quot;More&quot; drop down for question list views. Additionally, you can add them to the left navigation bar for easy access (see image below).  Phase III: Notifications For phase III we'll enable indicators for when new items are available in a custom question list. When you pin a question list to the left nav you can see a dot on any list with new items.  Feedback welcome As usual, we're interested in your constructive feedback. This is a work in progress and, as mentioned, will be rolled out in phases. You can provide feedback on the concepts now based on the information in this post and provide feedback based on the working elements as they are released. I will like to know some of the android emulators that work really well with Ubuntu 11.10 and above. I want to try 'whatsapp' on Ubuntu through an android emulator. Thanks in advance My old computer broke down a few days ago, so I'm using a laptop until I can get it fixed.  As a side effect, I no longer have my usual browser cookies, so I had to log back into Stack Exchange.  While doing so, I noticed that I kept having to log in to every Stack Exchange site I visited &mdash; the usual "Welcome back [username], you've been logged in." auto-login banner seemed to be conspicuously missing.  In fact, a quick look at the "network" tab in the developer tools suggested that no requests to stackauth.com were being made at all.  Eventually, I decided to dive into .  Turns out that the code to make the global auth check and bring up the banner is simply:    gauth.checkStackAuth(StackExchange.options.stackAuthUrl);   And, what do you know, running that command from the developer console does auto-log me in.  However, the reason this code isn't being run automatically is because it's disabled in the site config (apparently for all SE sites).  For example, the page I'm typing this question on has the following config options set:  StackExchange.init( {     "locale": "en",     "stackAuthUrl": "https://stackauth.com",     "serverTime": 1408393334,     "networkMetaHostname": "meta.stackexchange.com",     "styleCode": true,     "enableUserHovercards": true,     "site": {         "name": "Meta Stack Exchange",         "description": "Q&amp;A for meta-discussion of the Stack Exchange family of Q&amp;A websites",         "isNoticesTabEnabled": true,         "recaptchaPublicKey": "6LdsB7sSAAAAAAzjgEF_Hd8vXv-C42sa_KyofaGR",         "recaptchaAudioLang": "en",         "enableNewTagCreationWarning": false,         "globalAuthDisabled": true,                // &lt;--- HERE         "isMetaSite": true,         "enableSocialMediaInSharePopup": true,         "useGoogleOAuth2": true     },     // etc.   So, apparently, global auto-login has been deliberately disabled for some reason.  Why is that, and is there any planned timeline for re-enabling it? Standardize Training and Validation Data How to compute the  formula $\sum  \limits_{r=1}^d r \cdot 2^r$? Is the function differentiable at $0$? Let $$f(x) = \begin{cases}\begin{align*}&amp;\cos{\dfrac{1}{x}}, &amp;x \neq0 \\ &amp;0, &amp;x=0. \end{align*}\end{cases}$$ Is the function $F(x) = \displaystyle \int_{0}^x f dx$ differentiable at $0$?  We can see that the function $f(x)$ is continuous everywhere except $x=0$. In order to show differentiability we will need to show the derivatives from the left and right are equal. So we need to show that $$\lim_{h \to 0^+} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^-} \dfrac{F(x+h)-F(x)}{h}.$$ The derivative from the right is $$\lim_{h \to 0^+} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^+} \dfrac{\displaystyle \int_{0}^x\cos{\dfrac{1}{x+h}}-\int_{0}^x\cos{\dfrac{1}{x}}}{h}$$ and the derivative from the left is $$\lim_{h \to 0^-} \dfrac{F(x+h)-F(x)}{h} = \lim_{h \to 0^-} \dfrac{\displaystyle \int_{0}^x\cos{\dfrac{1}{x+h}}-\int_{0}^x\cos{\dfrac{1}{x}}}{h}.$$ I am not sure how to proceed. In the discussion from a question I asked several days ago (), someone suggested that I use general.runalg() instead of processing.runalg() for running QGIS algorithms.    I've now used both, and haven't noticed a difference in either performance or syntax between them.  I'm just wondering what sort of differences there are between these two methods, and which one I should be using. Who fired the shot and what is the symbolism in firing shot right next to Lucas at the end of the movie? At the end of , during the deer hunt, there’s a shot fired right next to Lucas (deliberately missing the shot).   Who fired the shot? Is it Theo's son? What is the meaning of that scene? Is someone warning Lucas that he made a mistake and definitely, Lucas will be punished in the future? Determining bare earth DEM from unclassified LAS file? I have data in LAS format with RGB values created from aerial photogrammetry using a UAV. I am  trying to find a solution to extract the bare earth DEM from the point cloud.   I have tried SAGA, Fusion, MCC-LIDAR, but it is seems they need the LAS file to be already classified (which it naturally isn't). Can anyone point me in the right direction with a brief explanation of the process?   Generally, I would need to process about 100 mill points at a time (can tile them if needed). Let $a,b$ be positive integers.  When $$k = \frac{a^2 + b^2}{ab+1}$$ is an integer, it is a square.    Proof 1: (Ngô Bảo Châu): Rearrange to get $a^2-akb+b^2-k=0$, as a quadratic in $a$ this has two values: $a$ and $kb - a = (b^2-k)/a$. (The second root is determined in two different ways from the expansion $(x-r_1)(x-r_2) = x^2 - (r_1 + r_2)x + r_1 r_2$.)  Now suppose we have $a,b$ such that $k$ is an integer but not a square, by the investigation about roots we have that the second root is a nonzero integer since $k,b,a$ are integers and $k \not = b^2$, futhermore it is positive which is easily seen from its defining equation.  WLOG assume $a \ge b$ so that the second root is strictly smaller than $a$. This leads to a decent, replacing $a$ with the second root.    Proof 2 (Don Zagier): Apply reduction theory (specifically, Sätze 1 and 2 of Section 13 of my book on quadratic fields) to the quadratic form $x^2 + kxy + y^2$, which is the unique reduced quadratic form in its equivalence class.    Note that Proof 2 is pretty much the same as Proof 1 when written out in explicit detail, but I could not read Zagier's book because I cannot read German.  I would like to know more approaches to this and other alternative proofs of this result if possible! Thanks in advance.  I would also be interested in related problems (especially easier ones of a similar nature) and texts which cover the reduction theory in English. A doubt from Milnor's "Topology from a Differentiable Viewpoint". This is a doubt from Milnor's "Topology from a Differentiable Viewpoint".     For a smooth $f:M\to N$, with $M$ compact, and a regular value $y\in N$, we define $n(f^{-1}(y))$ to be the number of points in $f^{-1}(y)$. It is easy to see that $n(f^{-1}(y))$ is finite. It can also be seen that $n(f^{-1}(y))$ is locally constant; i.e. there is a neighbourhood of $y$ such that for every point $y'$ in that neighbourhood, $n(f^{-1}(y'))=n(f^{-1}(y))$.   The proof for the statement in bold is the following:     Let $x_1,x_2,\dots,x_n$ be the points of $f^{-1}(y)$, and choose disjoint neighbourhoods $U_1,U_2,\dots,U_n$ of these points, which are mapped diffeomorphically into neighbourhoods $V_1,V_2,\dots,V_n$ of $y$. The required neighbourhood is $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$.   I don't understand how $(V_1\cap V_2\cap\dots V_n)-f(M-U_1-U_2-\dots U_n)$ is open. I know that $(V_1\cap V_2\cap\dots V_n)$ is open. However, how can one prove that $f(M-U_1-U_2-\dots U_n)$ is closed? I can see this working if $f$ is surjective, but not otherwise. Can I use a one-way ticket to come to the UK? Or do I need to show a return ticket to be able to get in?