Why do I need to replace "quiet splash" with "nomodeset"? I have recently installed Ubuntu but could not load it. By searching on Interned I found a solution which works for me but I do not understand it. The solution is taken from .  In more detail, while I load Ubuntu, I get into a menu which asks me what I want to do. At this point I press e. After that I get a screen with some text where I replace quiet splash by nomodeset and then I press F10. After that Ubuntu loads.  Now, my questions are:   What exactly am I doing? Why does it help? Can I do something to avoid doing these steps each time I when I wont to load Ubuntu?
How do I add a kernel boot parameter? I need to add the boot parameter foo=bar to my kernel. How can I do this once for testing, and permanently if testing was a success?
"EOL while scanning single-quoted string"? (backslash in string) I'm an on and off user of LaTeX so I'm not familiar with every package out there. Having said that, whenever I use LaTeX and need to accomplish a certain goal (i.e. ), I search around and find some kind of solution that involves a package that I didn't know about. I think that's great, I learn something new, and it goes in my template file.  Every now and then, though, my search comes up with a package that, after much wrangling, I find out that it's obsolete. An example of this is just recently I came across the subfigure package and wanted to try it out. Soon after I found out from here that it has become . I usually keep my TeX up-to-date because I like keeping up with new XeLaTeX features.   My question is, is there a map or timeline of what TeX packages should be avoided because there are better ways of doing things? An idea I have is some kind of list of all packages split by what they accomplish, then cancelling out packages whose features are incorporated into and improved upon by newer packages.  Edit: another example is lscape versus pdflscape. The latter implements rotating pages in the final PDF not just for pdftex but also for other drivers such as xetex. The only way I found out about this was Googling latex lscape rotate xelatex, for which the second link leads to the  (PDF).  Great answers so far. I'll check them out and see which one works best for me.  Edit 2: Turns out that my question is a bit of a dupe in the sense that a great answer came up on an unrelated question. The nag package . Gotta try it! Creating or referencing variables dynamically in Sass Difficulty in creating macro '\newcommand' for the beginning and end of frames in Beamer I would like to create a Latex macro for my beamer presentations. I use beamer frequently, and want to make macros for the \begin{frame} and \end{frame} to require fewer keystrokes. At the beginning of the document I have:  \newcommand{\bfrm}{\begin{frame}}%start new frame \newcommand{\efrm}{\end{frame}}%end new frame   Which compiles fine. I then test the first of the two macros for the beginning of the frame:  \bfrm{New frame testing}     blah blah \end{frame}   Which produces the result I am looking for. I then test the second of the macros:  \bfrm{New frame testing}     blah blah \efrm   Fails to compile, saying that there is a runaway argument. What is the reason for the failure? Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.      The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means  $$ \exists 0\leq q&lt;1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$    Now to the proof.  I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q&lt;1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q &lt;1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$  Can you help me? Frequently when I ask a question I have to include quite a bit of code for reference purposes.    It would be nice to be able to bold the important lines in the code.  For example this code:  &lt;DataTrigger Value="True"&gt;     &lt;DataTrigger.Binding&gt;         &lt;MultiBinding Converter="{StaticResource DisableWorkItemConverter}"&gt;             &lt;Binding ElementName="MainForm" Path="PickedWorkItemID"/&gt;             &lt;Binding Path="Id"/&gt;         &lt;/MultiBinding&gt;     &lt;/DataTrigger.Binding&gt;     &lt;Setter Property="IsEnabled" Value="False"/&gt;     &lt;Setter Property="IsSelected" Value="False"/&gt; &lt;/DataTrigger&gt;   Is not way long, but the important part (as seen in ) is the second to last line.  However, if I try to bold it, it ends up looking like this:  &lt;DataTrigger Value="True"&gt;     &lt;DataTrigger.Binding&gt;         &lt;MultiBinding Converter="{StaticResource DisableWorkItemConverter}"&gt;             &lt;Binding ElementName="MainForm" Path="PickedWorkItemID"/&gt;             &lt;Binding Path="Id"/&gt;         &lt;/MultiBinding&gt;     &lt;/DataTrigger.Binding&gt;     &lt;Setter Property="IsEnabled" Value="False"/&gt;     **&lt;Setter Property="IsSelected" Value="False"/&gt;** &lt;/DataTrigger&gt;   Not what I was going for.  It would be nice to be able to draw attention to specific lines of code when needed.    (One suggestion of a different color background sounds nice too (ie a yellow background to draw the eye to the relevant code.)  If there is some other way to bold in code please let me know. Why is the word 'number' abbreviated to 'No.' in UK English and '#' in American English? I'm currently working on the following question:  "At least one of Plato and Democritus believed in the theory of forms. Plato believed in the theory of forms only if he was not an atomist, and Democritus was an atomist only if he did not believe in the theory of forms. Democritus was an atomist. Prove, using Natural Deduction, that Plato was not an atomist"  I have been working for quite some time to take the information and turn it into a set of literals and statements I can work with.  So far, I have got $P \lor D$,  $P \Rightarrow \lnot a$,  $D \Rightarrow A$,  $D = A$  However, I cannot seem to make these prove the statement using natural deduction, any hints? I found this question in a book.   How many terms of the Maclaurin expansion of $(1+x)^{82}$ are needed to guarantee finding a value of $1.1^{82}$ to an accuracy of $10^{-6}$?  This is how I tried to do it:  We want the Lagrange error to be less than $10^{-6}$:  $$R_{n}(0.1)&lt;10^{-6}$$  Now to find $R_{n}(0.1)$:  $$R_{n}(0.1)=\frac{f^{(n+1)}(0.1)\cdot0.1^{n+1}}{(n+1)!}$$  After writing the values of ${f^{(n+1)}}(x)$, I find a pattern and say that  $${f^{(n+1)}}(x)=\frac{82!}{(82-n)!}(1+x)^{82-n}$$  $${f^{(n+1)}}(0.1)=\frac{82!}{(82-n)!}(1.1)^{82-n}=\frac{82!(1.1)^{82-n}}{(82-n)!}$$  Now just substitute all of this back to $R_{n}(0.1)&lt;10^{-6}$:  $$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}&lt;10^{-6}$$  Great! Now I have an inequality that I can solve. I can't graph this so I just use trial and error (which seems like quite an intensive way to do it, so if anyone has a better way I'd love to know).  $$\frac{\frac{82!(1.1)^{82-n}}{(82-n)!}\cdot0.1^{n+1}}{(n+1)!}-10^{-6}&lt;0$$  This seems to work first for $n=28$. That means the highest power of $x$ that I need is 28, so the number of terms in the expansion needs to be $n+1=29$. But the book's answer is       30   What am I doing wrong here? I'm still quite new to the power series expansions so if there's anything unclear in my working, please let me know (also, as a side note, the book itself may be wrong, though that isn't very likely I'd assume). I want to leave my Mac awake overnight so it syncs 4GB of data from Dropbox. I need to prevent the machine from going to sleep because that pauses the Dropbox sync  I've been using  to keep the computer awake, but it also keeps the computer unlocked which is insecure.  I need a method to keep my Mac awake (so that Dropbox can sync overnight) but also locked (so no one can read my email). It'd be great if this method also turned off the screen (to preserve power), but honestly that's not even 100% necessary. LINQ Select Distinct with Anonymous Types So I have a collection of objects.  The exact type isn't important.  From it I want to extract all the unique pairs of a pair of particular properties, thusly:  myObjectCollection.Select(item=&gt;new                                 {                                      Alpha = item.propOne,                                      Bravo = item.propTwo                                 }                  ).Distinct();   So my question is:  Will Distinct in this case use the default object equals (which will be useless to me, since each object is new) or can it be told to do a different equals (in this case, equal values of Alpha and Bravo => equal instances)?  Is there any way to achieve that result, if this doesn't do it? how to bring up the menu at the bottom left hello everyone I have a little problem, how do I bring up the menu in the circle? when I select another object the menu becomes non-existent Can the software be downloaded on an Android device?  If yes, then what is the process? What's a good beginner resource to learn group theory? Breaking single polygon into multiple polygons? I created voronoi polygons after which I dissolved them on the basis of one attribute. Now there are cases where in for one single attribute "abc" I have3 polygons at different places with same object id. I would like to break these into 3 polygons as they are at different locations.  Is there any tool in ArcGIS Desktop or QGIS which can help me in doing that? I wanted to create a WP plugin. So I requested and my request was approved by WordPress.org But I don't want to develop that plugin anymore. How can I delete the repository forever or request its deletion. I continue to find it amazing that something as “bulky” and macroscopic as a static magnetic or electric field is actually a manifestation of virtual photons.  So putting on your QFT spectacles, look closely at the space near the pole of a powerful magnet – virtual photons! Now look between the plates of a charged capacitor – virtual photons again!   But if it’s all virtual photons, how do we get the difference between a magnetic and electric field?  [This question needs the tag "magnetostatics" as well]. I find that the minimum in-call volume on my HTC Desire is still far to loud on some calls.  Since I got the phone, I have flashed several different custom ROMs, radio updates, etc, and have had the issue throughout.  Do you know of any way (setting, hack, app, whatever) to reduce the minimum volume?  My device is rooted and I am currently running one of the CyanogenMod 7 (Gingerbread) nightly builds. I'm doing some exercises on Apostol's calculus, on the floor function. Now, he doesn't give an explicit definition of $[x]$, so I'm going with this one:     DEFINITION Given  $x\in \Bbb R$, the integer part of $x$ is the unique $z\in \Bbb Z$ such that $$z\leq x &lt; z+1$$ and we denote it by $[x]$.   Now he asks to prove some basic things about it, such as: if $n\in \Bbb Z$, then $[x+n]=[x]+n$  So I proved it like this: Let $z=[x+n]$ and $z'=[x]$. Then we have that  $$z\leq x+n&lt;z+1$$  $$z'\leq x&lt;z'+1$$  Then $$z'+n\leq x+n&lt;z'+n+1$$  But since $z'$ is an integer, so is $z'+n$. Since $z$ is unique, it must be that $z'+n=z$.  However, this doesn't seem to get me anywhere to prove that  $$\left[ {2x} \right] = \left[ x \right] + \left[ {x + \frac{1}{2}} \right]$$  in and in general that   $$\left[ {nx} \right] = \sum\limits_{k = 0}^{n - 1} {\left[ {x + \frac{k}{n}} \right]} $$  Obviously one could do an informal proof thinking about "the carries", but that's not the idea, let alone how tedious it would be. Maybe there is some easier or clearer characterization of $[x]$ in terms of $x$ to work this out.  Another property is $$[-x]=\begin{cases}-[x]\text{ ; if }x\in \Bbb Z \cr-[x]-1 \text{ ; otherwise}\end{cases}$$  I argue: if $x\in\Bbb Z$, it is clear $[x]=x$. Then $-[x]=-x$, and $-[x]\in \Bbb Z$ so $[-[x]]=-[x]=[-x]$. For the other, I guess one could say:  $$n \leqslant x &lt; n + 1 \Rightarrow  - n - 1 &lt; x \leqslant -n$$  and since $x$ is not an integer, this should be the same as $$ - n - 1 \leqslant -x &lt; -n$$  $$ - n - 1 \leqslant -x &lt; (-n-1)+1$$  So $[-x]=-[x]-1$ The class of cardinal numbers is well ordered I'm looking for a proof that the class of cardinal numbers is well ordered under the order relation $|A|\leq |B| \Leftrightarrow$ exists an injection $f:A \to B$.  In fact, I've found a very beautiful proof in an article, but I think it's not the standard one.