The procedure stated in this answer at the following link  doesn't work.  In fact to me, in the gnome-tweak section regarding the 'Appeareance' where I'm going to create a new theme for the fonts changement in detail where there is the shell option, to change the theme, it appears a danger signal and the drop-down menu is halted and with a gray color.
I installed Ubuntu 18.04, and I want to customize the fonts. I have already change all the setting by tweak tool. But the system default font not be changed, is anything I need to do?
Suppose we are in $\mathbb R^n$. What can we say about the sign of $\Delta u(\vec x)$ if u($\vec x$) has a local max/min at $\vec x$? I've tried looking at the reverse of the second partial derivative test and it seems to suggest that $\Delta u(\vec x) \leq 0$ if $\vec x$ is a min and $\Delta u(\vec x) \geq 0$ if $\vec x$ is a max. However, I'm not convinced of this result. Can somebody point me towards a statement (theorem, etc.) involving the above. I appreciate the help. Evaluating $\int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\, \mathrm{d}x$ How to limit zoom level in OpenLayers 2.x I would like to have only 3 levels of zoom available (from level 6 to level 8).  demolayer2 = new OpenLayers.Layer.WMS( "abc...","http://localhost:8080/geoserver/gwc/service/wms", {layers: 'def...', transparent:"true", format: 'image/png', numZoomLevels: 3, minZoomLevel: 6 },{isBaseLayer:false},  { tileSize: new OpenLayers.Size(256,256)});  map.addLayer(demolayer2);   map.addControl(new OpenLayers.Control.LayerSwitcher({'div':OpenLayers.Util.getElement('layerswitcher')}));  map.zoomToExtent(new OpenLayers.Bounds(-4.601615515076983,39.8769407866263,-3.0527184873294764,41.16710732525929)); Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit? The limit is $$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$ which I'm aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L'Hopital's rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.  In this , the approximation reasoning works out. Prove/Disprove that if two sets have the same power set then they are the same set I am really sure that if two sets have the same power set, then they are the same set. I just am wondering how does one exactly go about proving/showing this?  I'm usually wrong, so if anyone can show me an example where this fails, I'd like that too.  The homework just asks for true/false, but I'm wanting to show it if possible. My thoughts are that since the power set is by definition the set of all subsets of a set, if each of the two power sets are identical, we have an identity map between each set, thus it's indistinguishable which power set is a given set's power set. I hope that wasn't verbose. Since a set has only one power set, we can conclude they are in fact the same set. How to keep the console window open in Visual C++? I'm starting out in Visual C++ and I'd like to know how to keep the console window.  For instance this would be a typical "hello world" application:   int _tmain(int argc, _TCHAR* argv[]) {     cout &lt;&lt; "Hello World";     return 0; }   What's the line I'm missing? R code to solve for probability of normal distributions? I don't understand which R code I am supposed to be using to figure these problems out. I brief explanation of what the code is doing would be amazing. The problems below are two different applications I'm expected to understand and I don't get either. Thank you!  Problem 1. Let Z~N(0,1) For each of the following, find c such that the equation is true.   a) P(Z&lt;c) = 0.8413  b) P(0&lt;Z&lt;c) = 0.3051  c) P(-c&lt;Z&lt;c) = 0.8664   Problem 2. X~N(3,4) compute  a) P(X&gt;3)  b) P(1&lt;X&lt;8)  c) P(-1.5&lt;X&lt;1) Divergence of $\frac{ \hat {\bf r}}{r^2} \equiv \frac{{\bf r}}{r^3}$, what is the 'paradox'? I want the QED symbol to be 1 line lower at the end of proof Just as my title says, I would prefer to have my qed symbol to be 1 line lower than the last line of my proof. By default its on the same line and aligned to the right.   How can I achieve this?  EDIT: See below code for an example:  \documentclass[11pt,a4paper]{article} \usepackage[margin=3cm]{geometry} \usepackage[english]{babel} \usepackage{amsmath, amsthm} \begin{document}  \begin{proof}     Hello there good chap! I'll put in another line here just to be sure its clear what I mean. \end{proof}    \end{document} Can I adjust starting point when using ListPlot? Calculate limit with L'Hospital's rule I want to calculate the limit $\displaystyle{\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}}$.   I have done the following:    It holds that $\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}=\frac{0}{0}$.   So, we can use L'Hospital's rule:  \begin{align*}\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x}&amp;=\lim_{x\rightarrow 0}\frac{x^2\cos \left (\frac{1}{x}\right )}{\sin x} \\ &amp;=\lim_{x\rightarrow 0}\frac{\left (x^2\cos \left (\frac{1}{x}\right )\right )'}{\left (\sin x\right )'} =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+x^2\cdot \left (-\sin \left (\frac{1}{x}\right )\right )\cdot \left (\frac{1}{x}\right )'}{\cos x} \\ &amp;=\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )-x^2\cdot \sin \left (\frac{1}{x}\right )\cdot \left (-\frac{1}{x^2}\right )}{\cos x} \\ &amp; =\lim_{x\rightarrow 0}\frac{2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )}{\cos x}=\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )+\sin \left (\frac{1}{x}\right )\right ) \\ &amp; =\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )+\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )\end{align*}   We calculate the two limits separately   $\lim_{x\rightarrow 0}\left (2x\cdot \cos \left (\frac{1}{x}\right )\right )$ :    \begin{equation*}\left |\cos \left (\frac{1}{x}\right )\right |\leq 1 \Rightarrow -1\leq \cos \left (\frac{1}{x}\right )\leq 1  \Rightarrow -2x\leq 2x\cdot \cos \left (\frac{1}{x}\right )\leq  2x\end{equation*} we consider the limit $x\rightarrow 0$ and we get \begin{equation*}\lim_{x\rightarrow 0} \left (2x\cdot \cos \left (\frac{1}{x}\right ) \right )=0\end{equation*}    How can we calculate the limit $\lim_{x\rightarrow 0}\left (\sin \left (\frac{1}{x}\right )\right )$ ? So from what I remember it was on cable TV. The main hero was a TV actor zapped to a alien planet. It had a evil overlord with a virgin priestesses. She wants to get laid but he refuses because the next priestesses is only three. Regardless it's a parody things include a man who was maimed by the evil overlord wants revenge the hero eventually screws the virgin priestesses it cuts to scenes of a train going into a tunnel and a volcano erupting. Prove $\lim_{x\to0}\sqrt{4-x}=2$ using the precise definition of limits. (Epsilon-Delta) Is it always true that $(A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2)$ Has drupal abandoned drush? I've been away from drupal for 2-3 years. I just came back around and installed drush via composer () and now every time I try to use drush I get a message telling me that I should be using composer:  ~/sites/d7 $ drush version  Drush version : 9.0-dev-gd4a3872 ~/sites/d7 $ drush dl drupal-7.x  [notice] dl has been deprecated. Please build your site using Composer. Add new projects with composer require drupal/[project-name]. Use https://www.drupal.org/project/composer_generate to build a composer.json which represents the the enabled modules on your site.   I downloaded drupal 8 following these , but when I try to use drush in that project I get this message:  ~/sites/cfbtv/web $ drush dl devel This codebase is assembled with Composer instead of Drush. Use `composer update` and `composer require` instead of `drush pm-updatecode` and `drush pm-download`. You may override this error by [error] using the --pm-force option.   Is there something wrong with my local env, or is drush a thing of the past now? My Developer Account cannot be activated because I used a relative's credit card about://(null) popup? I encountered a popup similar to this one this morning while checking the weather report at  in Safari, though I have not been able to reproduce it.  The title is "about://(null)" and the text "Why you no play Candy Crush!!"  The popup occurred twice.  I tapped Cancel both times.  I suspect the source of this is rogue JavaScript in an ad network, judging from the fact that the screenshot below shows it in an ad-supported application, and when I saw it, I was on an ad-supported website.  My phone is not jailbroken.  I'm wondering if anyone is able to identify the source of the popup definitively.  Speculation is running rampant on Apple's support communities and elsewhere, from push notifications (delete Facebook and all your games!) to carrier notifications (which don't work on iPod Touches…), and my gut tells me little of it is helpful.  But I could also be very wrong.  (This is not my screenshot; it comes from .)    UPDATE 1: The Candy Crush people have this to say, but I'm not (yet) convinced there's actually something installed that is to blame:   UPDATE 2: Just reproduced it in the iOS Simulator while clicking around woodtv.com, but I'm not sure how to track it down further from here. Here is the situation.  I have hexagonal board,and a unit on it,with speed or move value 4.Diffrent terrain has a diffrent cost.When i click on the unit,game should show me a move range.  My solution was to check each  hex in range of 4,with A* pathfinding,and if path cost was less than 4 then this hex was in range.Finally game nicely show me range of that unit.  My question is:   Is there other solution to search for range on hex grids or square grid,because even if i am really proud of what i did in my solution,i think,it is a little to exaggerated?:))  What make me ask this question?I noticed that when unit speed is 4 or 6 or even 8,time to computing range for my computer was really good,but when speed was 10 and more i noticed that i needed to wait few second to compute.Well in real games i rather dont see something like this and my A* pathfinding is rather well optimized,so im thinking that my solution is wrong.  Thanks for any replies. A twice differentiable function $f$ satisfies $f′′(x)+f(x)=−xg(x)f′(x)$, $\forall x\ge 0$. Consider $f\in C^2$ so that $$f''(x)+f(x)=-x\,g(x)f'(x), \ \forall x\ge0 $$ where $g(x)\ge 0$. Then ($\forall x\ge 0$)         (A) $f(x)^2+f'(x)^2$ is non-increasing,   (B) $f(x)^2&lt;3f(0)^2+(2f'(0))^2$,   (C) $|f(x)|\le\alpha$, where $\alpha$ is a fixed real constant.   (D) $\lim_{x\to\infty} f(x)\sin\left(\dfrac{1}{x}\right)$ exists.        Answer Given: (A),(B),(C),(D)  What I have tried : I differentiated option (A) and then with the help of the given equation in the question I was able to show that  it is less than zero . Hence I was able to infer that option (A) is correct .   Now coming to option (B).  I assumed the function in option (A) to be $h(x)$.   $$ h(x) = f(x)^2 + f'(x)^2 $$  Since $h'(x)&lt;0$ ,therefore $h(x)$ is a decreasing function   hence for all $x\ge 0$  $h(x)&lt; h(0)$  which implies $(f(x))^2 + (f'(x))^2&lt;(f(0))^2 + (f'(0))^2$  From the above expression we can get the second option .   Is my method correct ?. If not , please show the right method .  I am unable to get the the 3rd and 4th option . Alternative proof that $(a^2+b^2)/(ab+1)$ is a square when it's an integer 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.