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Instantiating object of type parameter How to find the joint PDF? Bash scripts ran from from gnome/nautilus don't have environment variables In my ~/.bashrc and ~/.profile I have some variables set to some filepaths:  export MY_VAR1="/path/to/somewhere"   In my ~/bin/ (which is added to my PATH) I have a few scripts that rely on those variables being set.  If I run those commands from the terminal, they work as expected.  However, if I try to double click on a script in Nautilus and run it, it runs as if those variables were not set.  Same thing happens if I add one of the scripts to Gnome's "Startup Applications" using the GUI utility or if I add them to the Gnome menu using the "Main Menu" GUI utility.  I would've thought that since I included these variables in my ~/.profile, which is run at login that gnome and nautilus and processes spawned by them would have access to those variables.  Both Gnome and Nautilus are run under my login user. All my scripts in ~/bin have the bash shebang at the top.  I am new to bash scripting, sorry if I am missing something obvious. I am using Debian Wheezy and Gnome3. Windows 7 finding location of installed program Does "corresponding author" carry an implied meaning? After reading the question , I got the impression that (at least in the writers' field) the "corresponding author" flag could carry an implied meaning, such as "did more work than the other authors", or "is a more senior author". Is it the case?  I am in mathematics, a field where we typically sort the authors alphabetically, and I've never given any importance to that role (apart from "this author is willing to answer a couple of extra e-mails, if needed"). How do I replace this cast iron toilet flange? which adjective can I use to describe my neighbor, who is always trying to keep eye on my personal matters like what am I bringing from market or what am I cooking etc?  Regards, How many actuations are "too many actuations"? Why Does Harry Have a Scar When Avada Kedavra Leaves No Trace? Why does Harry have a scar when Avada Kedavra leaves no trace?     ‘The Muggle authorities were perplexed. As far as I am aware, they do not know to this day how the Riddles died, for the Avada Kedavra Curse does not usually leave any sign of damage ... the exception sits before me,’ Dumbledore added, with a nod to Harry’s scar.   Half-Blood Prince - page 343 - Bloomsbury - chapter 17, A Sluggish Memory   What caused Harry to retain a scar from Avada Kedavra, when no one else in know history showed any signs of cause of death when Avada Kedavra was used?  What caused Harry's scar?  I'd love an answer based in canon, such as the books, interviews with J.K. Rowling, or Pottermore. Independence of disjoint events Toast message with Chinese characters, is it a virus? I recently received some updates today from the Play Store. Every so often a pop notification, one of those bubble ones at the bottom of the screen with Chinese characters.  The most recent update I've noticed is the ES File Explorer. I wouldn't mind, but my device is rooted and so I am at risk. I have a screenshot of the notification.  My device is the Samsung Galaxy Tab A 9.7 and I have stock ROM installed.    What my question(s) are: Do I have a virus (it's not being picked up by Avast) and if I do, how do I remove it? Suppose $a_n \ge 0$ for all natural numbers $\Bbb{N}$ and $\sum{a_n}$ converges. Is it necessarily true that $\sum{(a_n)^2}$ converges?  I think yes.  Clearly as $a_n \ge 0$ it follows $(a_1+a_2+\dots+a_n)^2 \ge a_1^2+a_2^2+\dots+a_n^2 \ge 0$ and so $(\sum_{n=1}^{\infty}{a_n})^2 \ge \sum_{n=1}^{\infty}{a_n^2} \ge 0$.  As we are told $\sum{a_n}$ converges to say $L$ where $L$ is a real number it follows $L^2 \ge \sum_{n=1}^{\infty}{a_n^2} \ge 0$ and so therefore $\sum{(a_n)^2}$ converges.  Is this correct? We have a Barbarian on our team that very much likes to switch up his weapons; he goes from a Dwarven Hammer, to a pair of Hand-Axes, to occasionally just going bare-fisted. No issues there, he's fairly adept with all 3.  However, I have only just recently picked up on the way that he rolls his attacks for the two Hand-Axes: He rolls the first attack, adds his modifiers etc. Then he rolls the second attack, this time without modifiers, which is all well and good. But I'm not sure if this action uses his bonus action or not, or is considered a separate action (though, being a level 4 Barbarian, he only has one action per turn), or is merely part of the same action.  If we have a look at the Two-Weapon Fighting feature from the Fighter's fighting styles:     Two-Weapon Fighting   When you engage in two-weapon fighting, you can add your ability modifier to the damage of the second attack. (PHB5e p.72)   Similarly, the Dual Wielder feat also uses the Two-Weapon Fighting feat:     Dual Wielder   You master fighting with two weapons, gaining the following benefits:      ...         You can use two-weapon fighting even when the one-handed melee weapons you are wielding aren’t light. (PHB5e p.164, emphasis mine)      So; yes, this gives a bonus to the second weapon attack, but having asked a  to clarify which action this secondary weapon attack belongs to, it was determined that this uses your bonus action;     [...] the second attack requires a bonus action   So after that long winded explanation, here is the question: Without any additional feats, can you wield and use two weapons at once, and if so, how many actions does it require to make a full attack with both weapons in a turn? Problem 20-9 in Lee's Introduction to Smooth Manifolds (2nd Ed.) asks to prove that for a Lie group $G$, and for $X, Y\in \text{Lie}(G)=\mathfrak g$, there is $\epsilon&gt;0$ and a smooth map $Z:(-\epsilon, \epsilon)\to \mathfrak g$ such that   $$(\exp tX)(\exp tY) = \exp\left(t(X+Y) + \frac{1}{2}t^2[X, Y] + t^3 Z(t)\right)$$   What we want to show is that the first and second term in the Taylor expansion of $\exp^{-1}\left((\exp tX)(\exp tY)\right)$ are $t(X+Y)$ and $\frac{1}{2}t^2[X, Y]$ respectively.  Calculating the first term is easy since we just need to differentiate $\exp^{-1}((\exp tX)(\exp tY))$ with respect to $t$ at $t=0$ once, and this gives the desired result.  But for the second term we need to take the second derivative at which I am stuck. Preventing itemize environment to insert initial vertical space Use of 'prototype' vs. 'this' in JavaScript? What's the difference between  var A = function () {     this.x = function () {         //do something     }; };   and  var A = function () { }; A.prototype.x = function () {     //do something }; Custom Tools not Saving in Toolbar on Startup/ Normal.mxt? I am trying to save some tools that I have created into the default toolbar layout that is present when ArcMap 10.1 is launched.  It is possible to change the arrangement of toolbars, add new ones and even add in existing tools, but any custom tools I set in the commands category "geoprocessing tools" are gone and do not appear in the toolbar. How can I save these custom tools to the default layout that appears on startup?  Update:  Saving a document with the custom tools added and reopening it preserves the tools, but opening a template (10.1 mxd) containing the tools, then opening and older document will cause the tools to disappear. Oddly enough, any other changes made after the custom tools were added are preseved.  Update (03/18/2014):  I have upgraded to 10.2 and this has still not been fixed. I did not see this issue in the 10.21 changelog so I doubt it will address it. For those interested in this issue, I have started a forum post on the ESRI forums . Prove the following using epsilon-delta definition:  Let $a \in \mathbb{R}$ and $f,g$ be defined on $\mathbb{R}$.  If  $$\lim_{x\to a} f(x) = 647 \tag 1$$  and $$\lim_{x\to a} g(x) = \infty \tag 2$$ then   $$\lim_{x\to a} f(x) + g(x) = \infty.$$  $$(1) \implies\forall \epsilon &gt; 0, \exists \delta &gt; 0, s.t. |x-a|&lt;\delta \implies |f(x)-647| &lt; \epsilon$$  $$(2) \implies \forall M &gt; 0, \exists \delta_2 &gt; 0, s.t. |x-a|&lt;\delta_2 \implies |g(x)| &gt; M$$  Not sure about this... I've set up the definitions. I believe we will have to take a minimum between the deltas for first two limits then use it in the third.  Anyone have any clues? If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$. Let $a,b,c$ be the lengths of the sides of a triangle. Prove that $$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}&lt;2\,.$$  Attempt.  By clearing the denominators, the required inequality is equivalent to $$a^2(b+c)+b^2(c+a)+c^2(a+b)&gt;a^3+b^3+c^3\,.$$ Since $b+c&gt;a$, $c+a&gt;b$, and $a+b&gt;c$, the inequality above is true.  Is there a better, non-bruteforce way? What longest time ever was achieved at holding light in a closed volume? For what longest possible time it was possible to hold light in a closed volume with mirrored walls?   I would be most interested for results with empty volume but results with solid-state volume may be also interesting.