I want to resize /dev/sda5 to take all of the unallocated space (242.25 GiB). How can I accomplish this?  EDIT 1:   swapon output: NAME         TYPE       SIZE USED  PRIO /dev/sda6    partition  14G  0B        -1  sudo blkid | grep swap output /dev/sda6: UUID="d4ae2fb5-2069-4b6d-b86d-e8c94ebf0fd9  grep swap /etc/fstab output: # swap was on /dev/sda6 during reparation UUID=d4ae2fb5-2069-4b6d-b86d-e8c94ebf0fd9 none swap sw 0 0
Previously, I have installed Windows 7 on my 320 GB laptop with three partitions 173, 84 and 63 GB each. The 63 GB partition was where the Windows was installed. The rest were for file containers.  Now I changed my OS to Ubuntu 12.04 LTS. I installed Ubuntu by replacing the entire Windows 7 on the 63 GB partition. The rest of the partitions remain as an NTFS Windows partition and I can still access them both (the 173 and 84 GB partitions).  Now I want to change the two partitions of Windows into an Ubuntu format partitions plus most importantly, I want to extend the 63 GB partition to more than a 100 GB because at the moment I am running out of disk space. Whenever I try to install any application, especially using wine, it always complains for a shortage of disk space.  How do I do the extending activity before I entirely format my laptop again and lose all the important files on my partitions?
Avoiding if statement inside a for loop? Say if I define a power series over some arbitrary field $F$ as   $$a = \sum^{ \infty }_{i = 0} a_{i} X^{i} $$  Then can I say:  $$ab = \sum^{ \infty }_{i = 0} \sum^{ \infty }_{j = 0} a_{i} b_{j} X^{i + j} $$ Do we know anything about Aragorn's first time in Moria? When the Company is forced to abandon their attempt to cross the Misty Mountains via the Redhorn Gate (the pass of Caradhras), Gandalf proposes they go "neither over the mountains, nor round them, but under them," via Moria. During the debate Aragorn remarks:     "I too once passed the Dirmrill Gate but though I also came out again, the memory is very evil. I do not wish to enter Moria a second time."    I haven't read much beyond the LOTR trilogy other than The Silmarillion. Does Tolkien ever describe Aragorn's first visit to Moria anywhere? Cannot find my keyboard layout I'm in Belgium (french part) and I cannot find the proper keyboard layout in Settings > Region and Languages > Input Sources.  the closest I found is the french (AZERTY) one but for the symbols such as ! and _ it's not the same keys...  what can I do to solve it ?  Thank you. How to change the chapter number to display as text (One, Two, ...)? Let $E$ be an empty set. Then, $\sup(E) = -\infty$ and $\inf(E)=+\infty$. I thought it is only meaningful to talk about $\inf(E)$ and $\sup(E)$ if $E$ is non-empty and bounded?   Thank you. I wanted to add classpath to my JDK. While searching a way to do it I noticed everyone open .profile file using terminal. But when I want to open and edit it with a text editor I couldn't see that file in Home folder.  Can someone tell me why is this. And also if possible how can I access .profile in home folder the GUI way. Proving the sum of two independent Cauchy Random Variables is Cauchy let $a,b,c$ be postive real numbers ,and such $$ab+bc+ac=3$$   show that   $$\dfrac{a^2}{a+2b^2}+\dfrac{b^2}{b+2c^2}+\dfrac{c^2}{c+2a^2}\ge 1$$   This problem is from Secrets In Inequalities volume 1  page 30,example 1.24. the comment.the  author this case is a bit more diffcult,But this author can't post solution.Thank you Let $x_0 \in \mathbb{R}$ and $f,g: \mathbb{R} \to \mathbb{R}$  such that $f$ is discontinuous at $x_0$ but $g$ is continuous at $x_0$, then $f+g$ at $x_0$ is...   My approach: After some graphical attempts to construct such a pathological case as given above I claim that the described scenario isn't possible. That is $$f \text{ disc. at } x_0 \text{ and } g \text{ cont. at } x_0 \implies f+g \text{ disc. at } x_0 $$  Proof: I was trying to come up with a contradiction, because if they happen to work out I find these kind of proofs to be most satisfying.   Assume $f+g$ is continuous at $x_0$ that is $$\forall \epsilon &gt; 0 ,\exists \delta_1 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_1 \implies |f(x)+g(x)-f(x_0)-g(x_0)| &lt; \epsilon  $$  Since this statement must be true for all $\epsilon &gt; 0$ I could choose $\epsilon:=1$. Next I know that also $g$ is continuous at said point $x_0$, thus  $$\forall \epsilon &gt; 0, \exists \delta_2 &gt; 0 : \forall x \in \mathbb{R}, |x-x_0| &lt; \delta_2 \implies |g(x)-g(x_0)| &lt; \epsilon $$  and I have that $f$ is NOT continuous at $x_0$ therefore: $$\exists \epsilon^* &gt; 0, \forall \delta&gt;0: \exists x \in  \mathbb{R}, |x-x_0| &lt; \delta \wedge |f(x)-f(x_0)| \geq \epsilon $$  My problem is to bring all these three statements nicely together. Mainly I get confused by the quantors. In order to match all three statements I assume I have to choose $\epsilon &gt; 0$ such that the last statement is met, so say $\epsilon:= \epsilon^*&gt;0$, so statement 1 and 2 are still true for said $\epsilon$  I also need to choose $\delta$. Lets say $\delta = \min{(\delta_1,\delta_2)}$ now I should be able to work with all three statements at the same time right? But then I stumble with my argumentation.  Let $|x-x_0|&lt; \delta$, then we obtain $$ |f(x)+g(x)-f(x_0)-g(x_0)| \leq |f(x)-f(x_0)| + |g(x)-g(x_0)| &lt; - \epsilon + \epsilon =0 $$ which means that $|f(x)+g(x)-f(x_0)-g(x_0)| &lt;0$ and I hope that this is a contradiction because $|.| \geq 0$. Please comment (or answer) if I am right or absolutely on the wrong path. I'd also appreciate it if you criticize my formulation. Consider the following piece of Markdown code:  This is some regular text.      &gt;&gt;&gt; def factorial(n):     ...     return 1 if n &lt; 2 else n * factorial(n - 1)     ...   * This is a list item.      &gt;&gt;&gt; def factorial(n):     ...     return 1 if n &lt; 2 else n * factorial(n - 1)     ...    Notice that the second code block is preceded by a list item.   Why is the factorial code not properly formatted under these circumstances? To demonstrate the problem, I'll insert the Markdown code show above:    This is some regular text.  &gt;&gt;&gt; def factorial(n): ...     return 1 if n &lt; 2 else n * factorial(n - 1) ...     This is a list item                 def factorial(n):       ...     return 1 if n &lt; 2 else n * factorial(n - 1)       ...               This applies to numbered as well as unnumbered lists. n tasks assigned to n computers, what is the EX value of a computer getting 5 or more tasks? Do I have to claim and re-check in my baggage in international transit when traveling with the same airline? I'll be traveling to Memphis, Tennessee form New Delhi, India via Qatar Airways.  The itinerary would be: New Delhi &lt;--> Doha || Doha &lt;--> Chicago || Chicago &lt;--> Memphis  Would I be required to claim and re-check my baggage at Chicago even though all the flights were booked with Qatar Airways or can I check it in all the way from New Delhi to Memphis without re-checking in the baggage again? Find & Delete duplicated files on multiple harddisks at once I am having a serious trouble finding an agreement between the two facts in Special Relativity:  (1) given two events $A$ and $B$ in spacetime separeted by a spacelike interval $s$, the distance in space between the two events will be the shortest in the reference frame which the two events occur simultaneously:   The invariance of the interval says $s^2=(\Delta x)^2-(c\Delta t)^2$ and $\Delta x$ is minimum when $\Delta t=0$  (2) the length contraction says $L=\dfrac{L_0}{\gamma}$ where $\gamma=(1-(v/c)^2)^{-1/2}$ , $L_0$ is the proper length of an object in a reference frame which the object is at rest and $L$ is the length of the object in a reference frame which the object moves with speed $v$ in the $x$ direction.  This says that the lenght of the object in the $x$ direction is the largest in the comoving reference frame of this object.  The question is: According to (1), the reference frame in which two events $A$ and $B$ occur simultaneously is the reference frame that gives the shortest distance in space between the two events. Let $A$ and $B$ be the two end points of a meter stick in a reference frame in which the stick is at rest and at time $t=0$. This two events occur simultaneously, so the distance in space between the two is the shortest acording to (1). But according to (2) shouldn't this distance be the largest?  What is wrong? How to construct a relative path in Java from two absolute paths (or URLs)? Given two absolute paths, e.g.  /var/data/stuff/xyz.dat /var/data   How can one create a relative path that uses the second path as its base? In the example above, the result should be: ./stuff/xyz.dat MPI Get Processor with Minimum value How can I get bash to perform tab-completion for my aliases? I have a bunch of bash completion scripts set up (mostly using bash-it and some manually setup).    I also have a bunch of aliases setup for common tasks like gco for git checkout. Right now I can type git checkout dTab and develop is completed for me but when I type gco dTab it does not complete.    I'm assuming this is because the completion script is completing on git and it fails to see gco.    Is there a way to generically/programmatically get all of my completion scripts to work with my aliases?  Not being able to complete when using the alias kind of defeats the purpose of the alias. Are the various "Zelda" games supposed to take place in a common universe? Countable set of truth assignments satisfying set of well formed formulas I am trying to answer the following question posed in my logic class: (throughout we work in propositional logic) is there a set $S$ of well-formed-formulas (wffs) satisfied by countable set $V$ of truth assignments such that any truth assignment satisfying $S$ lies in $V$?    My guess is no.  My first reaction to this was to do a diagonalization.  Present $V$ as an array with $v_{i, j}$ the truth value that assignment $v_i$ gives to the $j$-th propositional letter.  Obviously the hard case is when $S$ is infinite; if $S$ is finite, then only finitely many propositional letters occur in the $S$-formulae.  Then there is some maximal $k$ for which all $A_n$ with $n \geq k$ do not occur in V.  Then the diagonalization is trivial: just diagonalize from the $k$-th letter onward, since changing $A_k$ in this range doesn't affect anything in $S$.  But I have little clue about how to approach the main case when $S$ is infinite. I thought maybe I should try using the fact that $S$ is consistent (since it's satisfiable).  If it's maximally consistent, then I think this is simple too, since for every propositional letter $A_n$, either $A_n$ is in $S$ or $(\neg A_n)$ is in $S$, implying that there is only one truth assignment satisfying $S$--since every truth assignment must agree on each propositional letter.    But if $S$ is not maximally consistent, this isn't available to me.  Somehow I would like to show that infinitely many propositional letters do not occur in $S$.  Then I can apply the trick above and just diagonalize through the superfluous letters.    So, yes, just looking for some hints or corrections if you have any please, though maybe if I don't get anywhere on my own I'll give up and ask for more.  Thanks.  (By the way, by referring to letters that "occur in $S$" I mean letters that occur in some formula in $S$).