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How To Program A Bash Script That Presses A Certain Combination Of Keys? Please, i need someone to help me in making a very simple bash script. The job of this script is just to press a certain combination of keys on the keyboard.  My objective is to create a script that would turn off the Kubuntu desktop effects completely. The key combination for that is: "Left-ALT + Left-Shift + F12". I am aware that there is a widget that might do this job, but i am trying to learn how to accomplish it with a bash script.  I have tried to do an online search but i guess i was choosing the wrong set of keywords. I did not manage to reach a tutorial or a reference that would mention this task.  Thanks in advance for your help!  P.S: Any suggestions, references, notes, comments or tutorials would also be greatly be appreciated. I think my son was trying to guess the passcode! What do I do?  I can't wait 42 years! Does a 2100W angle grinder need an earth pin? I have recently purchased a Makita 2100W angle grinder. It has just the live (L) and neutral (N) connectors unlike the one I previously used, which had a third protected earth (PE) connection.   Would it be better if I changed the connections to a 3 pin connector? Prove that $\displaystyle \sum^{n}_{r=0}(-1)^r\cdot \large\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{3!}{2(n+3)}$   $\bf{My\; Try::}$ We can write $$\frac{\binom{n}{r}}{\binom{r+3}{r}} = \frac{n!}{r!\times (n-r)!}\times \frac{r!\times 3!}{(r+3)!} = \frac{3!\times n!}{(n-r)!\times (r+3)!}$$  So our sum is $$ 3!\sum^{n}_{r=0}(-1)^r\frac{n!}{(n-r)!\times (r+3)!} = \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\frac{(n+3)!}{(r+3)!\times (n-r)!}$$  So we get Sum $$ \frac{3!}{(n+1)(n+2)(n+3)}\sum^{n}_{r=0}(-1)^r\binom{n+3}{r+3} = -\frac{3!}{(n+1)(n+2)}\sum^{n}_{r=0}(-1)^{r+3}\binom{n+3}{r+3}$$  $$  = -\frac{3!}{(n+1)(n+2)(n+3)}\left[-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]$$  $$=-\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}-\binom{n+3}{3}+\binom{n+3}{4}-\binom{n+3}{5}--.......+(-1)^{n+3}\binom{n+3}{n+3}\right]+\frac{3!}{(n+1)(n+2)(n+3)}\left[\binom{n+3}{0}-\binom{n+3}{1}+\binom{n+3}{2}\right]$$  So we get $$ = \frac{3!}{(n+1)(n+2)(n+3)}\left[1-(n+3)+\frac{(n+3)(n+2)}{2}\right] = \frac{3!}{(n+1)(n+2)}\times \frac{n^2+3n+2}{2} = \frac{3!}{(n+1)(n+2)(n+3)}\times \frac{(n+1)(n+2)}{2} = \frac{3!}{2(n+3)}$$  Can we prove it Using Combinatorial way,bcz above method is very lengthy  If yes then plz explain here  Thanks How do I depower overpowered magic items without breaking immersion? Does $f(x) = f(2x)$ for all real $x$, imply that $f(x)$ is a constant function? If a Continuous function $f(x)$ satisfies $f(x) = f(2x)$, for all real $x$, then does $f(x)$ necessarily have to be constant function? If so, how do you prove it? If not any counter examples? Let $T:V\to V$ be a linear map, where $V$ is an inner product space. Given $T^2=\frac{T+T^*}{2}$, prove that $T$ is normal and that $T=T^2$. Let $T:V\to V$ be a linear map, where $V$ is a finite-dimensional inner product space. Given $T^2=\frac{T+T^{*}}{2}$, I would like to prove that $T$ is normal and that $T=T^2$. The first part, to prove that $T$ is normal, is quite easy: from $T^2=\frac{T+T^{*}}{2}$ we'll have $\,T^{*}=2T^2 - T$, and as $TT^{*} = 2T^3 -T^2 = T^{*}T$. Now, I need to prove that $T=T^*$, then $T = T^2$ is immediate. I have been thinking about this question for some days and I can't figure out in what technique to prove that $T = T^{*}$. Prove that $\bigcap_{k = 1}^\infty C_k$ is also compact and connected. Let $X$ be a Hausdorff space and let $C_0 \supset C_1 \supset ...$ be a decreasing sequence of compact connected subsets of $X$.  Prove that $\bigcap_{k = 1}^\infty C_k$ is also compact and connected.  I have the compact part.  Just don't know how to show it is connected. Why is the edit button disabled? Here is my problematic code:  \begin{tikzpicture} \draw  (0,0)  to [bend left=20] (3,3) node[midway] {midway label}; \end{tikzpicture}   I wanted to put a label midway of the line, but instead the label is at (0,0).  If I remove [bend left=20], the line is not bent, and the label still misplaced. If I remove to and replace it by --, then it works fine, but my line is not bent any more.  \begin{tikzpicture} \draw (0,0) -- (3,3) node[midway] {midway label}; \end{tikzpicture}   Works fine.  The problem occurs with MacTeX 2011 and MacTeX 2012. Why are my textures pink? Time travel short story where dinosaur doesn't taste like chicken Can Kingsoft Office be installed? Can Kingsoft Office be installed in Ubuntu? I am studying homology groups and I am looking to try and develop, if possible, a little more intuition about what they actually mean. I've only been studying homology for a short while, so if possible I would prefer it if this could be kept relatively simple, but I imagine it is entirely possible there is no real answer to my query anyway.   As I said above, I want to gain a little deeper understanding of what the n-th homology group actually means: I can happily calculate away using Mayer-Vietoris but it doesn't really give me a great deal of intuition about what the n-th homology group actually means. For example, with homotopy groups, the fundamental group is in some sense a description of how loops behave on the object in question, and it is obvious to me why that is what it is for say, the torus or the circle. However, I have no idea what, if anything, I am actually saying about a triangulable object when I talk about it having 0-th homology group this or 1st homology group that.  The best I have been able to find online or in my limited book selection is the brief description "intuitively, the zeroth homology group counts how many disjoint pieces make up the shape and gives that many copies of $\Bbb Z$, while the other homology groups count different types of holes". What 'different types of holes' are there, roughly speaking? I appreciate that it may often be completely non-obvious what the low-order homology groups are for some complicated construction, but perhaps in simpler examples it might be more explicable. Are there (simple) cases where I could say, just from looking something like e.g. the torus, what its zero-th or first or second etc. homology group was based on the nature of the object? I guess in the zero-th case it is, as my source () above says, related to the number of disjoint pieces. Can we delve deeper than this for the other homology groups?   Any book/website suggestions would be welcomed (preferably websites as I am nowhere near a library!) - I have Hatcher but not a great deal else, and I haven't gleaned as much as I wish to from that alone. Of course I know that there is a great deal we don't know about homology groups even today, so I don't expect some magical all-encompassing answer, but any thoughts you could provide would be appreciated. I hope this question is appropriate for SE Mathematics, apologies if not! -M Is there a way to edit SVG files in Illustrator where you can specify a CSS class for each path?  I already know in Illustrator that if you give the layer name an actual name, Illustrator will use that name as the path's ID, which is fine if you don't plan on reusing the icon more than once on the same page.  My current workaround is to just use the IDs method, but then later convert the IDs to classes in my code editor, but it's pretty annoying to have to do every time I generate my SVG sprite.  If that's currently not possible in Illustrator, are there any other apps that will allow you to specify that? Or maybe a Grunt or Gulp package?  It looks like you might be able to do it with Inkscape with a , so I might look at that if there's no other good solutions out there. How to remotely display my Android on my PC? There is a thing which is called miracast, it can mirror the display of the Android phone to another monitor, but my screen doesn't have this function, am I possible to do it in Ubuntu? Russian babel and chemformula Primary decomposition of a monomial ideal If $A$ is a non-empty set, then $A \notin A$ Exercise 3.2.2 Use the axiom of regularity and the singleton set axiom to show that if $A$ is a set then $A \notin A$.  Axiom of Regularity: If $A$ is a non-empty set, then there is at least one element x of A which is either not a set, or is disjoint from A.  Singleton Set Axiom: If $a$ is an object, then there exists a set $\{a\}$ whose only element is $a$, we refer to $\{a\}$ as the singleton set whose element is $a$ Furthermore, if $a$ and $b$ are objects, then there exists a set $\{a,b\}$ whose only elements are $a$ and $b$    I think I can give the proof when A is the non-empty set which contains finite elements. However, can anyone give any hint for the general case?  Suppose $A \in A$   Then eixsts $x$ such that either $x$ is not a set or $x \cap A$=$\emptyset$  Then it is clear thar $x \ne A$ (if $x = A$, then $x \cap A \ne \emptyset$)  We observe that as an element, $A$ $\ne$ $x$, $A$ $\in$ $A$  Then $A$ $\in$ $A-x$   Define the new set $A-x$ as $A_1$  then exists $x_1$ such that $A \in A_1-x_1$  After finite steps  we find that $A \in \emptyset $  which leads to contradiction. Past participle vs being+past participle