Is there any 'canonical' way to resolve hostname to IP, that should work on (almost) any Linux distribution?  I'm recently stuck with debugging of Docker container based on ubuntu:16.04 image. There is no ping, no ip, no dnslookup or anything else. Of course apt-get install is not possible as well, because logging as root is not available as well.
How do I get a remote host IP address if I don't have ping, and don't have any bind utilities like dig, nslookup, etc?  I need an answer that does not include 'install X' or 'use sidecar container'. I am looking for something that relies on nothing more than bash and the basic shell commands.
Integral $ \int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x $ I need to compute the improper integral $$ \int_{-\infty}^{\infty}\frac{x^{2}}{\cosh\left(x\right)}\,{\rm d}x $$ using contour integration and possibly principal values.  Trying to approach this as I would normally approach evaluating an improper integral using contour integration doesn't work here, and doesn't really give me any clues as to how I should do it. This normal approach is namely evaluating the contour integral $$ \oint_{C}{\frac{z^2}{\cosh\left(z\right)}\mathrm{d}z} $$ using a semicircle in the upper-half plane centered at the origin, but the semicircular part of this contour integral does not vanish since $\cosh\left(z\right)$ has period $2\pi\mathrm{i}$ and there are infinitely-many poles of the integrand along the imaginary axis given by $-\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ and $\pi\mathrm{i}/2 + 2n\pi\mathrm{i}$ for $n \in \mathbb{Z}$. The residues of the integrand at these simple poles are $-\frac{1}{4}\mathrm{i}\pi^{2}\left(1 - 4n\right)^{2}$ and $\frac{1}{4}\mathrm{i}\left(4\pi n + \pi\right)^{2}$, so that even when we add up all of the poles, we have the sum $4\pi^{2}\mathrm{i}\sum_{n = 0}^{\infty}\,n$, which clearly diverges.  Any hints would be greatly appreciated. Discovering new features I have an armature set up, but the bones aren't moving the mesh properly  The mesh and armature:    The vertex group of the bone I am going to move:    As you can see, the circled region doesn't belong to the vertex group. But, when I move the bone look what happens:    That part of the mesh has moved with the bone.. Why? Margin top in inline element When to use “that” and when to use “which”, especially in relative clauses When is it appropriate to use that as opposed to which with relative clauses? Unable to 'apt-get dist-upgrade' due to appstreamcli database error I have created a USB stick with a persistent Ubuntu 16.04 (release version). I tried a couple of ways, using Startup Disk Creator and MultiSystem, but I get the same end result.  When I boot into the USB and do 'apt-get update', all is well.  When I try 'apt-get dist-upgrade' I get:     ** (appstreamcli:2761): CRITICAL **: Error while moving old database out of the way. AppStream cache update failed.   The only reference to this error message I can find is  but would appear to be fixed, at least fixed by the time the release version of Ubuntu 16.04.  Aside: I want to test Unity8, which includes installing software.  I don't have a spare computer, so I thought a persistent USB would do the trick.  Any takers? How do I change the "Selected Items" color? I was trying to change the selected items color from the default orange to something else, however, the option isn't there, as it used to be in the Appearance properties in previous versions of ubuntu. Any idea where it could be in the beta? "Its" as a Possessive Pronoun Since its can be both determiner possessive pronoun and nominal possessive pronoun, an example of its as determiner possessive pronoun would be:     We saved this question for last because of its complexity.   Because complexity is a noun, so its must be a determiner possessive pronoun in this sentence. I'm looking for an example where its is used as nominal possessive pronoun in a sentence as above. How can I navigate my scene FPS style similar to how Unity does? I am aware of ShiftF fly, but it's extremely cumbersome having to use the scroll wheel to change the velocity of flight and moving the mouse from the center to spin.  Is there a way to move the viewport camera (not the scene camera) like Unity lets you right click and fly around with WASD and moving the mouse? How to make a series of rectangles follow a pen path I am trying to get a series of shapes (rectangles) to follow a path (sine wave) that I made with the pen tool, but I have no clue how to do that or if it is possible.   I have tried warping a rectangle into the sine wave path, but it takes many different warped rectangles and it isn't smooth.  I thought  might work for rectangles, but it only works for text. How to edit files in a terminal with nano? Stopping a Thread in Java? I'm in the process of writing a piece of code that connects to a server spawns a bunch of threads using that connection and does a bunch of "stuff".  There are certain instances where the connection fails and I need to stop everything and start from scratch with a new object.  I wanted to clean up after the object but calling thread.stop on the threads, but this method is seemingly deprecated.  What is the recommended alternative to doing this? Should I write my own cleanup and exit method for each of the threads? Set the thread to null? or something else? I have an input type="image". This acts like the cell notes in Microsoft Excel. If someone enters a number into the text box that this input-image is paired with, I setup an event handler for the input-image. Then when the user clicks the image, they get a little popup to add some notes to the data.  My problem is that when a user enters a zero into the text box, I need to disable the input-image's event handler. I have tried the following, but to no avail.  $('#myimage').click(function { return false; }); Array List count to users How do I repair my weapons and armor? Mutable vs immutable objects I'm trying to get my head around mutable vs immutable objects. Using mutable objects gets a lot of bad press (e.g. returning an array of strings from a method) but I'm having trouble understanding what the negative impacts are of this. What are the best practices around using mutable objects? Should you avoid them whenever possible? Let $\mathbf{FinVec}$ denote the category of finite-dimensional real vector spaces and linear maps, let $F\colon\mathbf{FinVec}\rightarrow\mathbf{FinVec}$ be a functor. For each pair of objects $V,W$, the Hom-set $\operatorname{Hom}(V,W)$ is itself a finite-dimensional vector space and we equip it with the unique structure of a topological space/of a smooth manifold that is compatible with its vector space structure. The composition maps are then continuous/smooth. In this way, we view $\mathbf{FinVec}$ as enriched over the category $\mathbf{Top}$/$\mathbf{Diff}$ respectively. It then makes sense to ask whether the functor $F$ is enriched over $\mathbf{Top}$/$\mathbf{Diff}$. That is just to ask whether the induced map $F\colon\operatorname{Hom}(V,W)\rightarrow\operatorname{Hom}(FV,FW)$ is continuous/smooth for all $V,W$. (Irrelevant to the rest of the question, but the reason this is a useful notion is that it is the natural hypothesis assuring that the functor induces another functor on the category of topological/smooth vector bundles, applying the original functor fiberwise.) In the case $V=W$, by functoriality, we can restrict $F$ to a group homomorphism $\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$. If the functor is continuous/smooth, so is this map. Note that $\operatorname{GL}(V)$ is a Lie group. It is a general result that a measurable group homomorphism between Lie groups is automatically smooth. This begs the following questions:  What's an argument for the existence of functors that aren't continuous? Can we construct one? Given the above, it either happens that the restrictions to $\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$ are continuous and continuity fails elsewhere, or the induced maps will be non-measurable, so that the construction will necessarily have to involve choice to some extent. Are there continuous functors that aren't smooth? Given the above, a continuous functor will have smooth restrictions to $\operatorname{GL}(V)\rightarrow\operatorname{GL}(V)$, but smoothness may fail elsewhere. I hope I'm right in thinking this is the site where you post general computer-related problems...  A problem I have when playing games (since you won't be pressing multiple keys simultaneously anytime else), is that when I press up/left and then another key, it fails to 'register' as if that last key is never pressed. This doesn't happen when I'm pressing any other two arrow keys, e.g. down/left or up/right. The weird thing is that when I'm playing with WASD instead of the arrow keys, i still have this problem! Meaning W+A+another button doesn't work.  If anyone knows what I should, I'd appreciate the help. (My reflexes are fast and I can always circumvent this problem, but it's quite annoying and shouldn't exist to begin with!) There was the following sentence in an article titled, “Like, Degrading the Language? No Way” in New York Times (April 5), in which the author says Americans are moving backward on language:     “(Like the use of ‘like’) the use of “totally” mines the same vein.   “He’s totally going to call you” does not mean “He is going to call   you in a total fashion.” It has a more specific meaning, although only   handled subconsciously by speakers, as so much of language is. “He’s   totally going to call you” contains an implication: that someone has   said otherwise or that the chances of it may seem slim at first glance   but in fact aren’t. As with “like,” “totally” tracks and nods to the   opinions of others. It’s totally civilized.”      Although the author explains “totally going to call you” contains an implication of someone having said otherwise, I’m not still clear with the meaning of “He’s totally going to call you.”   What does it mean? He’s totally going to call you “what”?   How does “He’s totally going to call you.” differ from “He’s going to call you.”? Is this expression (use of totally) ubiquitous today as the author asserts? Prove that $ m_{a} \geq m_{g} \geq m_{h} $ using strict inequalities unless $ a = b $. $ m_{a} = \frac{1}{2} (a + b) $   $ m_{g} =  \sqrt{ab} $   $ \frac{1}{m_{h}} = \frac{1}{2}(\frac{1}{a} + \frac{1}{b})$   Attempted Solution:  I believe I have shown the first step, which is $ m_{a} \geq m_{g} $ but i'm stuck on showing $ m_{g} \geq m_{h} $  $ \frac{a + b}{2} &gt; \sqrt{ab} $  $ (\frac{a + b}{2})^2 &gt; (\sqrt{ab})^2 $  $\frac{a^2 + 2ab + b^2}{4} &gt; ab$  $a^2 + 2ab + b^2 &gt; 4ab $  $a^2 -2ab + b^2 &gt; 0 $  $(a - b)^2 &gt; 0 $   so if $a = b$ we have $(a -b)^2 \geq 0$  Therefore, $m_{a} \geq m_{g} $  Now, i'm not sure if that is correct or not, if not can you please guide me to the complete solution or if I am correct on this part, provide a solution for $m_{g} \geq m_{h} $