What is this? Is it harmful?
Is it possible to identify this chrysalis?
I installed kubuntu-desktop just to experiment with its interface.  Then I decided to remove and did the traditional apt-get purge and autoremove but it seems elements of kubuntu are still there. For example interface elements, scrollbars, bootscreen etc are still there.  Is there any method I can completely remove without installing a fresh copy of ubuntu? Suppose f is a function such that $f'(x) = \frac{1}{x}$ for all $x &gt; 0$. Prove that if $f(1) = 0, then f(xy) = f(x) + f(y)$ for all $x,y &gt; 0$  As stated above, how do i go about doing this question?   Hi all, i would like to apologize if the question is missing context or other details.   This question was part of my calculus tutorial. I was on the topic of application of differentiation when this question was given to me.  Thus far i have learnt about intermediate value theorem, mean value theorem, rolle's theroem, Fermat's thereom.  As of this point in my learning, i have not touched upon integration. The only reason why i accepted the answer with integration in it was because i understood it.  Thus i believe i am suppose to use one the few theorems that i've stated to solve this question. But i have no idea where to start.  At first i differentiated the equation $f(xy)=f(x)+f(y)$ with respect to x  i believed the answer to be $f'(xy)\frac{dy}{dx}=f'(x)+f'(y)\frac{dy}{dx}$  which would be $\frac{1}{xy}\frac{1}{x}=\frac{1}{x}+\frac{1}{y}\frac{1}{x}$  At this point a few questions came to my mind  1) am i suppose to differentiate the eqn in the first place 2) what has $f(1)=0$ got to do with anything 3) what theorem am i suppose to use to solve this question  This was my thought process throughout the question and i hope someone would point out my mistakes. Thank you Where should I start with Star Trek? (TV series only, not the movies) I haven't seen any of the stuff in the Star Trek franchise, except for one of the movies, and now I want to remedy the situation. I am not interested in the movies, just the TV series. However, since the show has many seasons, I am not interested in watching the show start to finish. So, where (i.e what series/season) should I start with? Do I really need to see the series in real-world chronological order?   Note: I am somewhat familiar with Star Trek, that is I know the basic premise - ships in space, with war between different space-faring sentient species. How to completely delete and app from iphone (including all the data and history) So this problem particularly emerged with the Tinder app. I was using Tinder with a user account (A) and one day I logged out and created a new user account (B) which I never succeeded to activate due to sms verification problems, so I decided to login again with my old account. But even when I delete the Tinder app and reinstall it again, it still shows the B account logged in to the app. This means that IOS is somewhere storing a history for this app. Now I'm looking for a way to delete this history.  Note that this question is not necessarily about a certain app, but rather about where IOS stores historical data of apps and how to delete those histories alongside with the app.  P.S: If there are solutions like changing UUID of my i-phone that wouldn't damage my phone (I'm new to iphone so I'm not sure what I'm saying. For instance this jailbreaking thing which I keep hearing about and I'm not sure what it exactly does to my phone, doesn't sound like an I idea that I like), I'm open to them. Help with proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. I have been having a lot of trouble teaching myself rings, so much so that even "simple" proofs are really difficult for me. I think I am finally starting to get it, but just to be sure could some one please check this proof that $\mathbb Z[i]/\langle 1 - i \rangle$ is a field. Thank you.  Proof: Notice that $$\langle 1 - i \rangle\\  \Rightarrow 1 = i\\  \Rightarrow 2 = 0.$$ Thus all elements of the form $a+ bi + \langle 1 - i \rangle$ can be rewritten as $a+ b + \langle 1 - i \rangle$. But since $2=0$ this implies that the elements that are left can be written as $1 + \langle 1 - i \rangle$ or $0 + \langle 1 - i \rangle$. Thus  $$ \mathbb Z[i]/ \langle 1 - i \rangle = \{ 0+ \langle 1 - i \rangle , 1 + \langle 1 - i \rangle\}. $$  This is obviously a commutative ring with unity and no zero-divisors, thus it is a finite integral domain, and hence is a field. $\square$ If I am travelling on a car at around 60 km/h, and I shine a light, does that mean that the light is travelling faster than the speed of light? The title says it all.  If I was on a bus at 60 km/h, and I started walking on the bus at a steady pace of 5 km/h, then I'd technically be moving at 65 km/h, right?  So my son posed me an interesting question today: since light travels as fast as anything can go, what if I shined light when moving in a car?  How should I answer his question? How do I complete "Find the Bitizen"? How to get the Infobox data from Wikipedia? I dropped my camera while on holiday last summer. Thankfully, my camera wasn't damaged but the kit lens has lost the ability to autofocus and the UV filter and 58mm to 52mm filter step-down (I use it to attach a fisheye filter to the lens) I had on its end are stuck and won't come off.  I've replaced the lens with an 18-200 zoom lens so I don't especially need it. I was wondering if it was worth getting the lens looked at and possibly repaired?  I've heard many places will charge you £50 or so just to look at the lens and given the cost of the kit lens I'm not sure it's worth it. How can I download free apps without registering an Apple ID? Detecting pattern at the end of a line with grep Cannot suggest retag-only edit because "edits must be at least 6 characters" I was trying to suggest a retag of  on Seasoned Advice, but the system won't let me; I get the familiar message     Edits must be at least 6 characters; is there something else to improve in this post?   In general, that's not true for retagging; if you don't change the question body at all, it should work. I suspect the logic checking the edit is thrown off guard because of the title warning:     The question you're asking appears subjective and is likely to be closed       I found , but I cannot find any evidence (in SEDE) for invisible characters being present in the question's source. Also, that's supposed to be . I have found the command to delete files older than 5 days in a folder  find /path/to/files* -mtime +5 -exec rm {} \;   But how do I also do this for subdirectories in that folder? Switching between windows with scroll wheel on Ubuntu Dock I would like to switch between windows using the scroll wheel of the mouse over each app icon in Ubuntu dock, as we could do in Unity, but apparently this doesn't work in Ubuntu 17.10. Is there a way to enable it?   Related: How to prove the following limit?   $$\lim_{n \to \infty} (1+1/n)^n = e$$    I can only observe that the limit should be a very large number!   Thanks. Great Expectations [is written vs has been written] by Charles Dickens How do I clean up my dconf database? When drawing a line between two nodes in TikZ, you can specify the anchors on the beginning and end nodes in order to control how the line is drawn.  I would like to specify the anchor on the beginning node, and the angle at which the line is drawn, and let TikZ figure out where it will intercept the end node.  The usual use case is when I'm connecting to a "big node" and I would like the lines to be vertical.  I can cheat by placing a coordinate behind the big node, drawing a line to the coordinate instead, and then drawing the big node so that it covers things up.  However this won't work if I need to put an arrow on the end.  I could try to calculate the intercept coordinate on the edge of the big node manually, but that seems against the TikZ spirit.  |- doesn't work in this case.  \documentclass[10pt]{minimal} \usepackage[active,tightpage,pdftex]{preview} \usepackage{tikz}  \begin{document} \begin{preview}  \tikzstyle{RectObject}=[rectangle,fill=white,draw,line width=0.5mm] \tikzstyle{line}=[draw] \tikzstyle{arrow}=[draw, -latex]   \begin{tikzpicture}   %%%%%%%%%%%%%%%%%%%%%%%%   %%  LEFT SIDE   %%%%%%%%%%%%%%%%%%%%%%%%   \draw (1,0) node[RectObject] (Small1A) {A};   \draw (3,0) node[RectObject] (Small1B) {B};   \coordinate (Cheat1A) at (1,2);   \coordinate (Cheat1B) at (3,2);   \path [line] (Small1A.north) -- (Cheat1A);   \path [line] (Small1B.north) -- (Cheat1B);    % need to draw Big Node last to cover up our cheating   \draw (2,2) node[RectObject, inner xsep=1cm] (Big1) {Big Node};    \draw (2,4) node[text width=4cm] (text1) {L: Cheating to control line orientation. Arrows would be hard.};    %%%%%%%%%%%%%%%%%%%%%%%%   %%  RIGHT SIDE   %%%%%%%%%%%%%%%%%%%%%%%%   \draw (7,2) node[RectObject, inner xsep=1cm] (Big2) {Big Node};   \draw (6,0) node[RectObject] (Small2A) {A};   \draw (8,0) node[RectObject] (Small2B) {B};   \path [arrow] (Small2A) -- (Big2);   \path [arrow] (Small2B) -- (Big2);    \draw (7,4) node[text width=4cm] (text2) {R: No cheating. Arrows easy. Can't control line orientation.}; \end{tikzpicture} \end{preview} \end{document}   Here's a side-by-side example of "cheating", which shows what I want to be able to do, and "not cheating", which does not do what I want.    The left side of this example shows what I want to be able to do, except I want to do it without cheating, and I'd like to put arrows on the end of the lines. Irrationals: A Group? I understand that the set of irrational numbers with multiplication does not form a group (clearly, $\sqrt{2}\sqrt{2}=2$, so the set is not closed). But is there a proof or a counter-example that the irrationals with addition form (or do not form) a group?  Thank you!  Edit: In particular, I am wondering if the set is closed with respect to addition. How to access the system tray using the keyboard? Is there a way I can access the system tray items using the keyboard in Windows XP?