I created an executable .desktop file in my Desktop which should launch a GUI application along with displaying an icon in dock, but in Ubuntu 17.10 icon is not displaying in dock:    In previous Ubuntu versions (14.04 to 17.04) I didn't face this issue.  In Ubuntu 17.10 no 3rd party applications with icons as executables are being launched by default, It's asking for trust and launch to proceed.     How can I give the launchers default execute permission, or whatever permission is needed to avoid getting this dialog?  The executable was created by compiling in Ubuntu 14.04 to support backward compatibility.  The .desktop file created to display the icon contains:  #!/usr/bin/env/ xdg-open  [Desktop Entry] Version=1.0 Type=Application Name=MDM Comment=MAVLDM Terminal=false X-Ubuntu-Gettext-Domain=example-content MimeType=text/html;text/xml;application/xhtml+xml;application/xml;application/vnd.mozilla.xul+xml;application/rss+xml;application/rdf+xml;image/gif;image/jpeg;image/png;x-scheme-handler/http;x-scheme-handler/https;x-scheme-handler/ftp;x-scheme-handler/chrome; StartupNotify=false OnlyShowIn=GNOME;Unity; X-Desktop-File-Install-Version=0.20 Path=/home/ravi/Downloads/MDM Exec=/home/ravi/Downloads/MDM/MAVLDM Icon=/home/ravi/Downloads/MDM/mavldm.png   Is this changed in Ubuntu 17.10?
After upgrading to Ubuntu 17.10, Ubuntu Dock icons of some of my apps are not appearing. They appear properly in the dash. I verified the *.desktop files for the applications in /usr/share/applications and ~/.local/share/applications.  E.g. dock icon is not displaying for "Terminator". Check out the attached image.    This is how the icon for the same application appears in dash search:    Content of /usr/share/applications/terminator.desktop:  [Desktop Entry] Name=Terminator Comment=Multiple terminals in one window TryExec=terminator Exec=terminator Icon=terminator Type=Application Categories=GNOME;GTK;Utility;TerminalEmulator;System; StartupNotify=true X-Ubuntu-Gettext-Domain=terminator X-Ayatana-Desktop-Shortcuts=NewWindow; Keywords=terminal;shell;prompt;command;commandline; [NewWindow Shortcut Group] Name=Open a New Window Exec=terminator TargetEnvironment=Unity   Any pointers to debug this issue will be helpful.
Proving that if $|f''(x)| \le A$ then $|f'(x)| \le A/2$ Suppose that $f(x)$ is differentiable on $[0,1]$ and $f(0) = f(1) = 0$. It is also   known that $|f''(x)| \le A$ for every $x \in (0,1)$. Prove that   $|f'(x)| \le A/2$ for every $x \in [0,1]$.   I'll explain what I did so far. First using Rolle's theorem, there is some point $c \in [0,1]$ so $f'(c) = 0$.  EDIT: My first preliminary solution was wrong so I tried something else. EDIT2: Another revision :\  I define a Taylor series of a second order around the point $1$: $$ f(x) = f(1) + f'(1)(x-1) + \frac12 f''(d_1)(x-1)^2 $$ $$ f(0) = f(1) + f'(1)(-1) + \frac12 f''(d_1)(-1)^2 $$ $$ |f'(1)| = \frac12 |f''(d_1)| &lt;= \frac12 A $$  Now I develop a Taylor series of a first order for $f'(x)$ around $1$: $$ f'(x) = f'(1) + f''(d_2)(x-1) $$ $$ |f'(x)| = |f'(1)| + x*|f''(d_2)|-|f''(d_2)| \leq \frac{A}{2} + A - A = \frac{A}{2} $$  It looks correct to me, what do you guys think?  Note: I cannot use integrals, because we have not covered them yet. How do I create an 8 sided die from one object? Simply put, is there a way to make a cube balance solely on one point/vertex?  I'm struggling. I do not want to have to join two objects, nor have to extrude an object. I just want to have a simple eight-sided die created with one mesh. Man is sitting in his room and gets contradictory advice from two men from the future Integral of $\sin(e^t)dt$ Let $f(x) = \int_x^{x+1} \sin(e^t)dt$. Show that $$ e^x|f(x)| &lt; 2 $$ and that $$ e^xf(x) = \cos(e^x) - e^{-1}\cos(e^{x+1}) + r(x) $$ where $|r(x)| &lt; Ce^{-x}$ for some constant $C$.  I can do the first part easily. Integration by parts gives us $$ f(x) = \frac{\cos e^x}{e^x} - \frac{\cos e^{x+1}}{e^{x+1}} - \int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du, $$ so substituting $\cos u = 1$ and $\cos u = -1$ gives \begin{align*} \cos e^x - 1 + \frac{1}{e}\left(1- \cos e^{x+1}\right) &amp;\leq e^xf(x) \leq \cos e^x + 1 - \frac{1}{e}\left(\cos e^{x+1} + 1\right)\\ -2 &amp;&lt; e^xf(x) &lt; 2 \end{align*} So that $e^x|f(x)| &lt; 2$.  I get stuck on the second part, though. From integration by parts, I have $e^xf(x) = \cos e^x - e^{-1}\cos e^{x+1} - e^x\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du$, so it reduces to showing that $$ \left|\int_{e^x}^{e^{x+1}} \frac{\cos u}{u^2}du\right| &lt; Ce^{-2x} $$ for some constant $C$. However, I always end up with a constant for $Ce^{x}$ rather than for $C$. Equation numbering problems in AMSMath environments with \pause Notify on retagging A question here on meta for instance may stick around for quite a while. A feature-request  maybe even for months. Then the status (tags) gets changed to  or  or whatever and the OP does not get notified.  On a normal post-edit we get a notification but retagging does not count for that. Please add retagging to the edit notifications. Is my paper under review (or similar) for too long and if yes, how should I react? I launch a batch which call a lot of scheduled job. But I reach the governor limit :     Maximum number of Apex classes scheduled concurrently : 100   I'm wondering what happens when this limit is reached. The next scheduled jobs are waiting the number of scheduled jobs is lower than 100 for being executed or are they all cancelled ?  And is there to way to manage the number of scheduled jobs to avoir reaching this limit ? For example wait before schedule a new job if we are near the limit. invalid EFI file path on windows 8 Am I technically allowed to look up cards during a Magic tournament match? Am I allowed to look up a card during a game, if it's being played with serious Rules Enforcement? Either to read the full card from the Gatherer, or to try to find the exact card I'm vaguely thinking of.  Two possible scenarios:   An opponent has revealed a card previously, and I want to remember the exact details. (Something like the exact wording of a sorcery, mana cost, keywords on a creature, subtypes, etc. etc.) I have a "name a card and remove it from the opponent's deck" type card, and I want to, say, ask google for "Blue Gearhulk" to remember that it's that that one is called "Torrential Gearhulk".   Obviously, in casual or semi-casual games neither of these are a problem. Any reasonable opponent will just re-reveal the known card (assuming it IS still guaranteed to be known) or will accept "I name the Blue Gearhulk".  Also obviously, it seems unlikely that I (or anyone else) would get to the stage of playing at the REL without knowing the cards too thoroughly for this to be relevant.  But in principle... is it permitted? Or not? (and if not, is there a good reason, or just "we decided 'No'"?) Why did Itachi say that he needs to get someone's MS to get EMS while Sasuke only had a normal Sharingan? First, this question is motivated by the imprecise question: Is there a sensible notion of parity (evenness and oddness) for real numbers?   Here are some properties a notion of parity for $\mathbb{R}$ should have:    It should be an equivalence relation on $\mathbb{R}$ with exactly two equivalence classes.  Each equivalence class should be dense in $\mathbb{R}$. There should be some kind of symmetry between the two equivalence classes. (This is intentionally imprecise.)   If we divide $\mathbb{R}$ into rationals and irrationals, this lacks symmetry, since the rationals have measure zero and are countable, while the rationals have infinite measure and are uncountable.   So we could require that each equivalence class be uncountable and/or have positive (or infinite) measure. Then take one equivalence class to be the rationals, union with countably many fat cantor sets with positive measure, and the other class its complement. Then both classes are now dense, uncountable, and have infinite measure. But again, symmetry is lacking. One of these classes is "uncountable everywhere" and has "positive measure everywhere" while the other does not.   I propose some definitions: A subset $A \subset \mathbb{R}$ is uncountable everywhere if for any open interval $(a,b)$, the intersection $A \cap (a,b)$ is uncountable. A subset $A \subset \mathbb{R}$ has positive measure everywhere if for every open interval $(a,b)$, the intersection $A \cap (a,b)$ has positive measure. We can see immediately that having positive measure everywhere implies being uncountable everywhere, since a positive measure set must be uncountable.   Finally, my questions:    Is there a partition of $\mathbb{R}$ into two sets that are both uncountable everywhere?  Is there a partition of $\mathbb{R}$ into two sets that both have positive measure everywhere?  Is there a partition of $\mathbb{R}$ into two sets that split each interval $(a,b)$ into two parts of equal measure?   Potential generalizations for extra credit: What about partitions of $\mathbb{R}$ with $n$ equivalence classes, where $n \in \mathbb{N}$, or even with countably many, or even uncountably many equivalence classes?  Note: If you've seen these definitions of uncountable everywhere or positive measure everywhere somewhere under a different name, please let me know. I've never found anything where other people were thinking about these notions.  EDIT: By "measure," I mean Lebesgue outer measure, so we don't have to worry about anything being measurable. How to enable projection transformation in QGIS My QGIS project has two layers, one layer is retrieved by Google layer plugin which fetches the static Google map image whose CRS is +proj=merc +lon_0=0 +lat_ts=0 +x_0=0 +y_0=0 +a=6378137 +b=6378137 +units=m +no_defs and another vector layer is a postgis layer whose CRS is +proj=longlat +ellps=WGS84 +datum=WGS84 +no_defs.  I digitalized some polygon features from Google Layers and added it to the vector layer. However, QGIS seems do not make the automatic CRS transformation. How can I change its projection?   Thanks Eclipse menus are cut off or don't show I'm using Ubuntu 13.10 (with JDK 1.7.0_45).  I've downloaded and uncompressed Eclipse Kepler for Java EE Developers.  The software menus are messed up. I didn't notice similar problems in other software.  When I click any menu in Eclipse, the drop-down seems to be "cut". Sometimes the drop-down shows, but the options are invisible.  I tried Eclipse Indigo, but the same problem occurs. How to resolve a Java Rounding Double issue How to create a chrome texture? Is it possible to get that typical chrome-like angled look using the Shader Editor? I see a lot of tutorials making chrome on all types of surfaces but not on a knob geometry like the one in the following image:  Q: Is there any way to get that chrome-like surface appearance procedurally or should I just use a texture to fake it? How should I structure my URLs for both SEO and localization? When I set up a site in multiple languages, how should I set up my URLs for search engines and usability?  Let's say my site is www.example.com, and I'm translating into French and Spanish. What is best for usability and SEO?  Directory option:  http://www.example.com/sample.html http://www.example.com/fr/sample.html http://www.example.com/es/sample.html   Subdomain option:  http://www.example.com/sample.html http://fr.example.com/sample.html http://es.example.com/sample.html   Filename option:  http://www.example.com/sample.html http://www.example.com/sample.fr.html http://www.example.com/sample.es.html   Accept-Language header:  Or should I simply parse the  header and generate content server-side to suit that header?  Is there another way to do this? If the different language versions don't have different urls, what do I do about the search engines?    UPDATE 2011-12-06  Google has new recommendations for meta tags for explicitly pointing to other language content: .  UPDATE 2012-05-25  Related but not precisely:   UPDATE 2013-06-12  includes discussion of several URL schemes directly relevant to the question. Schrödingers MySQL table: exists, yet it does not Determine the number of lines within a text file Is there an easy way to programmatically determine the number of lines within a text file? Why are there $736$ matrices $M\in \mathcal M_2(\mathbb{Z}_{26})$ for which it holds that $M=M^{-1}$? I'm currently trying to introduce myself to cryptography.  I'm reading about the Hill Cipher currently in the book Applied Abstract Algebra. The Hill Cipher uses an invertible matrix $M$ for encryption and the inverse matrix $M^{-1}$ for decryption. The book recommends to use a matrix $M$ for which it holds that $M=M^{-1}$, since we then have the same key for encryption and decryption.  In the book, they use a $2 \times 2$ matrix $M$ over $\mathbb{Z}_{26}$ as example, and state that for there are 736 $2 \times 2$ matrices for which it hold that $M=M^{-1}$.  I'm trying to pick up on as much as possible when reading things, since I find it counter-productive for learning to skip something, when you don't get the theory behind it. Can someone enlighten to me, as to why it is that there are 736 possible $2 \times 2$ $M=M^{-1}$ matrices and how to find them?