i get following errors when trying to open programs over ssh.  $ thunar Thunar: Cannot open display:   $ libreoffice  Failed to open display  $ firefox Error: GDK_BACKEND does not match available displays  $ keepassx keepassx: cannot connect to X server   $ keepass2 Unhandled Exception: System.TypeInitializationException: The type initializer for 'System.Windows.Forms.XplatUI' threw an exception. ---&gt; System.ArgumentNullException: Could not open display (X-Server required. Check your DISPLAY environment variable) Parameter name: Display   at System.Windows.Forms.XplatUIX11.SetDisplay (IntPtr display_handle) &lt;0x41b3c8a0 + 0x00b9b&gt; in &lt;filename unknown&gt;:0    at System.Windows.Forms.XplatUIX11..ctor () &lt;0x41b3ab20 + 0x001df&gt; in &lt;filename unknown&gt;:0    at System.Windows.Forms.XplatUIX11.GetInstance () &lt;0x41b3a8d0 + 0x0005b&gt; in &lt;filename unknown&gt;:0    at System.Windows.Forms.XplatUI..cctor () &lt;0x41b3a160 + 0x00137&gt; in &lt;filename unknown&gt;:0    --- End of inner exception stack trace ---   at System.Windows.Forms.Application.EnableVisualStyles () &lt;0x41b38870 + 0x0001b&gt; in &lt;filename unknown&gt;:0    at KeePass.Program.Main (System.String[] args) &lt;0x41b376c0 + 0x0008b&gt; in &lt;filename unknown&gt;:0  [ERROR] FATAL UNHANDLED EXCEPTION: System.TypeInitializationException: The type initializer for 'System.Windows.Forms.XplatUI' threw an exception. ---&gt; System.ArgumentNullException: Could not open display (X-Server required. Check your DISPLAY environment variable) Parameter name: Display   at System.Windows.Forms.XplatUIX11.SetDisplay (IntPtr display_handle) &lt;0x41b3c8a0 + 0x00b9b&gt; in &lt;filename unknown&gt;:0    at System.Windows.Forms.XplatUIX11..ctor () &lt;0x41b3ab20 + 0x001df&gt; in &lt;filename unknown&gt;:0    at System.Windows.Forms.XplatUIX11.GetInstance () &lt;0x41b3a8d0 + 0x0005b&gt; in &lt;filename unknown&gt;:0    at System.Windows.Forms.XplatUI..cctor () &lt;0x41b3a160 + 0x00137&gt; in &lt;filename unknown&gt;:0    --- End of inner exception stack trace ---   at System.Windows.Forms.Application.EnableVisualStyles () &lt;0x41b38870 + 0x0001b&gt; in &lt;filename unknown&gt;:0    at KeePass.Program.Main (System.String[] args) &lt;0x41b376c0 + 0x0008b&gt; in &lt;filename unknown&gt;:0  :06:22 PM~/Documents$ gimp Cannot open display:   $ wireshark QXcbConnection: Could not connect to display  Aborted (core dumped)  $ gedit  Failed to connect to Mir: Failed to connect to server socket: No such file or directory Unable to init server: Could not connect: Connection refused  (gedit:23724): Gtk-WARNING **: cannot open display:    i've always been able to open aplications over ssh, this just started yesterday. using ubuntu 16.04 on both machines.  please DO NOT flag this as a repeat question, the following solutions didn't help:          if specific info is needed, please ask, i'll do my best to answer.
I am trying to run gedit from an SSH terminal on Windows, connected to an Amazon EC2 instance running Ubuntu 12.04 Server and I am getting this error:  Cannot open display:   How do I fix it?
Mutable or immutable class? Preventing reverse search engines from finding the origin of your image? I am planning to make a game, where I give the user an anime related image and they have to guess where its from.  However, most of the images, specially from popular anime are easy to use on Google reverse search to find and pin point the anime.  I've tried to see if I could make so it would not recognize the image but unhappily my tricks were not good enough.  Original image:    Grey-scale with horizontal flip:    Watermarking:    Puzzling the image with grey-scale:    Grey-scale with horizontal and vertical flip worked for this image but doesn't always work:    Also the above mentioned methods are rather easy for people to guess the anime.  I would like to know if there any good trick, that I could use on my images to make so it would not work on reverse search engines and yet not be so overcomplicated that I can implement on my program.  For example grey-scale, cropping, flipping are rather easy things to achieve in C#. Does $\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})$ converge/diverges?? Does this series converge or diverge? $$\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})$$   my thought is that, I can break it down to $$\sum_{k=1}^{\infty}\ln(k) - \sum_{k=1}^{\infty}\ln(k+1)$$ then maybe using comparison test or something? But I don't know exactly how to prove whether this series converges or diverges.  Any help would be appreciated! I'm trying to prove that the following set ($B$) defined by an ellipse is convex, but I am getting stuck.  $$B = \left\{(x_1,x_2):3(x_1-3)^2+2(x_2-3)^2 \leq 1 \right\} $$  My sad attempt:  $$ \textrm{Let }x,y \: \in \: B \:, \: \lambda \: \in \: [0,1]$$  $$ \lambda x+(1-\lambda)y=(\lambda x_1+(1-\lambda)y_1,\: \lambda x_2 +(1-\lambda)y_2  )$$ $$3(\lambda x_1 + (1-\lambda)y_1 -3)^2+2(\lambda x_2+(1-\lambda)y_2-3)^2=3(\lambda ^2x_1^2 +(1-\lambda)^2y_1^2+9+2\lambda x_1(1-\lambda)y_1-6\lambda x_1 -6(1-\lambda)y_1)+2(\lambda ^2x_2^2 +(1-\lambda)^2y_2^2+9+2\lambda x_2(1-\lambda)y_2-6\lambda x_2 -6(1-\lambda)y_2)$$  and this is where I have lost all confidence in proving it. I was able to factor some of the terms into the form $(\lambda x_1 -3)^2$, $((1-\lambda) y_1 -3)^2$, $(\lambda x_2 -3)^2$, and $((1-\lambda) x_2 -3)^2$, but I don't think that helps the situation any. What's the difference between a projection and a datum? What is the difference between a "proof by contradiction" and "proving the contrapositive"? Intuitive, it feels like doing the exact same thing. And when I compare an exercise, one person proves by contradiction, and the other proves the contrapositive, the proofs look almost exactly the same.  For example, say I want to prove: $P \implies Q$ When I want to prove by contradiction, I would say assume this is not true. Assume $Q$ is not true, and $P$ is true. Blabla, but this implies $P$ is not true, which is a contradiction.  When I want to prove the contrapositive, I say. Assume $Q$ is not true. Blabla, this implies $P$ is not true.  The only difference in the proof is that I assume $P$ is true in the beginning, when I want to prove by contradiction. But this feels almost redundant, as in the end I always get that this is not true. The only other way that I could get a contradiction is by proving that $Q$ is true. But this would be the exact same things as a direct proof.   Can somebody enlighten me a little bit here ? For example: Are there proofs that can be proven by contradiction but not proven by proving the contrapositve? Is it possible to install Ubuntu without formatting any partitions? How to disable "time travel" entirely? I have downloaded the offline (consumer, i.e. not msi package) installer for Google Chrome. But when I run it, it tries to contact a Google server to check for a newer version. If it finds it, it will download that first, and then install that version. It doesn't use the version that is already there, as part of the EXE file. For me, that defeats the point of having an offline installer. I keep a repository of all the free and purchased software locally, so that I can install them quickly whenever I need to. But as more and more people get access to broadband Internet connections, many developers and companies are pushing for offline installers with new releases. Many software titles, like Google Chrome, are almost exclusively distributed with online installers.  File name: ChromeStandaloneSetup.exe File version: 1.3.21.115 Size: 36.6 MB (38 416 368 bytes) SHA-1: 2393BAB7F408DC8D30F039FCA5D6B45600A0C4E7   How do I install Chrome as is, without automatically checking for and getting a newer version?  Will  work with Chrome installer? Is there a switch to disable version check when running the installer?  Update:  The "duplicate" question linked to below points out File Hippo as a possible repository where you one can get previous versions of Chrome. They do have a healthy listing of pretty much every single Chrome version ever released on Windows, both beta, dev and stable versions. Unfortunately they don't host the installer files, they only redirect to Google servers for downloads, which only spits out the installers for the latest versions of Chrome. It's not just File Hippo, I also tried downloading an older version of Chrome from CNET Downloads, and they too just redirect you to Google servers for download.  For whatever the reason (security related probably), Google doesn't like anyone to install an older version of Chrome on any system. Apparently not even your common source for downloading an older version of a software, like the CNET Downloads, not even they are allowed to host the Chrome installer files on their own servers. The redirects to Google is probably dictated by Google.     Google is a dictator!   They don't even allow full standalone offline installers to be installed as is, without checking for newer version. I tried running the installer from cmd with some of the command line switches used for Chromium. I only ended up with errors.    The business oriented msi installer (GoogleChromeStandaloneEnterprise.msi) might have some additional command line options you can use, but I wouldn't know if there is one that disables the check for a newer version. I have yet to find a complete list of switches you can use. The only one I've found so far is /update but that only does the opposite thing. So I'll give up here and go back to the trusty Firefox. Normality of dependent variable = normality of residuals? This issue seems to rear its ugly head all the time, and I'm trying to decapitate it for my own understanding of statistics (and sanity!).   The assumptions of general linear models (t-test, ANOVA, regression etc.) include the "assumption of normality", but I have found this is rarely described clearly.   I often come across statistics textbooks / manuals / etc. simply stating that the "assumption of normality" applies to each group (i.e., categorical X variables), and we should we examining departures from normality for each group.   Questions:     does the assumption refer to the values of Y or the residuals of Y?  for a particular group, is it possible to have a strongly non-normal distribution of Y values (e.g., skewed) BUT an approximately (or at least more normal) distribution of residuals of Y?   Other sources describe that the assumption pertains to the residuals of the model (in cases where there are groups, e.g. t-tests / ANOVA), and we should be examining departures of normality of these residuals (i.e., only one Q-Q plot/test to run).  does normality of residuals for the model imply normality of residuals for the groups? In other words, should we just examine the model residuals (contrary to instructions in many texts)?  To put this in a context, consider this hypothetical example:   I want to compare tree height (Y) between two populations (X).   In one population the distribution of Y is strongly right-skewed (i.e., most trees short, very few tall), while the other is virtually normal  Height is higher overall in the normally distributed population (suggesting there may be a 'real' difference). Transformation of the data does not substantially improve the distribution of the first    population.  Firstly, is it valid to compare the groups given the radically different height distributions? How do I approach the "assumption of normality" here? Recall height in one population is not normally distributed. Do I examine residuals for both populations separately OR residuals for the model (t-test)?     Please refer to questions by number in replies, experience has shown me people get lost or sidetracked easily (especially me!). Keep in mind I am not a statistician; though I have a reasonably conceptual (i.e., not technical!) understanding of statistics.  P.S., I have searched the archives and read the following threads which have not cemented my understanding: Where did Hogwarts get money for salaries, food and other purchases? We know that you can't transfigure food, so you have to grow it or buy it. Ditto money.  I'm sure that aside from some medicinal plants Herbology Professor had there were no farmlands at Hogwarts. So they must have bought food (as well as potions ingredients for Snape etc...). And they needed to pay the teacher salaries.  Where did all that money come from according to canon?  Ministry funding?  Tuition?  Patent money on Dumbledore's inventions?  Philanthropic donations by Malfoys and co?  Other? Notation question: $\ll$ We know that Accio spell (aka ) has some limits:   It can be blocked (e.g. Harry couldn't Accio the Diadem in RoR in Deathly Hallows). Many more failures are listed on Wikia.   But what are the exact full rules and limitations governing the spell in Harry Potter world?   Distance?  Line-of-sight?  Knowing exact or approximate location?  Specificity of item (e.g. "Accio Broom" seemed to work for Weasleys in OotP to bring their locked-up brooms - why not just a random broom?) Types of Accio-able stuff? Can you Accio water? Lump of earth? Gases/liquids/etc...? Can you Accio based on jQuery selector (or an SQL where clause), for lack of a better description? E.g. "Accio the book that has the word 'Flammel'"? "Accio biggest green ball from this room"? Can you Accio ALL items of a kind? (the reason I thought of this whole question was seeing "Accio all guns" comment in ). Can you Accio vague stuff? ("Accio meal"). Other rules that can be inferred.   NOTE: While Wikia article has some of the info, it is clearly very lacking in details on specific rules based on specific instances in canon. I'm looking for a fully comprehensive, organized ruleset for the charm, not merely 1-2 tidbits in Wikia article. Derangements with repetitive numbers I have a very simple problem.  Lets assume that I have a well shuffled deck of 52 cards. I start drawing the top card always and when the card matches its rank I lose. J=11 Q=12 K=13.  If there were only 13 cards I could easily use the $\ \frac{!n}{n!}$ for derangements in order to solve this. The problem is that there are 52 cards so when I pass 13 I start from 1 again so I don't know what is the probability to win. Example of the game     1st card: 4 - Continue   2nd Card: A - continue   3rd Card: K - Continue   4th Card: K - Continue   5th Card: 6 - Continue   6th Card: 9 - Continue   7th Card: 10 - Continue   8th Card: A - Continue   9th Card: J - Continue   10th Card: 3 - Continue   11th Card: 2 - Continue   12th Card: 8 - Continue   13th Card: A - Continue   1st card: 5 - Continue   2nd Card 2 LOSE   So actually I have to count from 1 to 13 4 times and if I draw all 52 cards then I win. What's the probability? Why does an conditionally convergent infinite series, when deranged, have a different limit? $\sum u_n = 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$ conditionally converges to $\log 2$; and, when deranged to $\sum u'_n = 1-\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{6}-\frac{1}{8}+\dots,$ converges to $\frac{1}{2}\log 2$. Why does a deranged conditionally convergent series have a different limit? First Course in Math Anal, p.93, by Burkill, explains that the sums $s_n, t_n$ of the first n terms of the two series are different functions of $n$. This reasoning is incomplete, because different functions of $n$ can converge to the same limit; for example, let $r_n$ be the first $2n$ terms of $\sum u_n$, then $s_n$ and $r_n$ are different functions; besides, by differently defining terms, i.e. clustering pieces of a series, e.g. \begin{align*} 1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}+\dots = \left(1-\tfrac{1}{2}-\tfrac{1}{4}\right)+\left(\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots \\ = \left(1-\tfrac{1}{2}-\tfrac{1}{4}+\tfrac{1}{3}-\tfrac{1}{6}-\tfrac{1}{8}\right)+\dots,\\ 1-1+1-1+\dots = (1-1)+(1-1)+\dots \\ =(1-1+1-1)+\dots,\\ 1+1+1+1+\dots = (1+1)+(1+1)+\dots \\ = (1+1+1+1)+\dots, \end{align*} one gets the sum of first n terms as different functions, but the series's limit remains. One could view a changed limit from the perspective of the changed number of terms, Denote 'countably many', the size of natural number, as $\lvert\mathbb{N}\rvert$, then $\lvert\mathbb{N}\rvert = \lvert k\mathbb{N}\rvert = \lvert\mathbb{N}\times\mathbb{N}\rvert$. Then two series with different numbers (e.g. $\lvert \mathbb{N}\rvert$, $\lvert k\mathbb{N}\rvert$) of terms– and so having different limits– paradoxically have same number of terms–and so are possibly formed by the same terms. The paradox could be the fundamental reason. Are there other explanations? Is the use here of 'countably many' is proper? "Me and my wife" or "my wife and me" Split a string with delimiters but keep the delimiters in the result in C# SVG donut slice as path element (annular sector) When is JavaScript's eval() not evil? Help needed with definite integral for $\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}$