I was able to install Ubuntu on a mini pc and to configure everything I needed, I lost a lot of time configuring stuff and recompiling part of the kernel following different tutorials on internet. I have also customized unity, now I don't want to loose the same time next time.  Is there a way to create a distribution from what is now in my HDD and be sure that if I install this distro next time on the same machine,or another,  starting from scratch, everything will be exactly how it is now?  (I don't want to create a disk image.)
I installed Ubuntu 14.04.1 on one of my laptops and now I want distribute that exact installation to a few of my friends. I tried making an iso with Gparted on a live session but it makes a 500 GB file (I've only used around 8 GB) .   Thanks in Advance !
Grub does not show a Windows 8 option after dual boot Every proper ideal is contained in a maximal ideal, in a commutative ring with identity. The statement is:     In a commutative ring with 1, every proper ideal is contained in a maximal ideal.   and we prove it using Zorn's lemma, that is, $I$ is an ideal, $P=\{I\subset  A\mid A\text{ is an ideal}\} $, then by set inclusion, every totally ordered subset has a bound, then $P$ has a maximal element $M$.  My question is why $M$ must contain $I$? Windows says RAM ran out while there is still 4 GB of physical memory available How to install Python 3.4 on Ubuntu? If $R_{1}$ and $R_{2}$ are relations on a set S with $R_{1};R_{2}=R_{2};R_{1}$. Then $R_{1}$ and $R_{2}$ are bijective maps For the surjectivity part, I showed that if $(a,b)\in I$ then $(a,c)\in R_{1}$ and $(c,b)\in R_{1}$ and $(a,c)\in R_{2}$ and $(c,b)\in R_{2}$ for some $c\in S$. Now for arbituary $(x,y)\in R_{1}$ we have $(x,z)\in I$ which implies that $(y,z)\in R_{2}$. But also $(y,z)\in R_{1}$, so $R_{1}$ is surjective. Similar argument for $R_{2}$  And for Injectivity, I've tried to show that if $aR_{1};R_{2}c$ and $bR_{2};R_{1}c$, then $a=b$ but I am not entirely sure what $a=b$ means in this context. Do I have to show that there is an equivalence relation containing (a,b)?  edit: $R_{1};R_{2}$ is a composition of two relations and $I=R_{1};R_{2}$ Using ffmpeg to cut up video How can a file be copied? Modern textbook on statistical field theory with an emphasis on applications to non-equilibrium phenomena? The following is an excerpt from Folland's Introduction to Partial Differential Equations:  Let the open set $\Omega\subset{\mathbb R}^n$, and $k$ be a positive integer. $C^k(\Omega)$ will denote the space of functions possessing continuous derivatives up to order $k$ on $\Omega$, and $C^k(\overline{\Omega})$ will denote the space of all $u\in C^k(\Omega)$ such that $\partial^{\alpha}u$ extends continuously to the closure $\overline{\Omega}$ for $0\leq|\alpha|\leq k$.  As I understand, an extension means that there exists $\widetilde{\partial^{\alpha}u}$ which is defined on $\overline{\Omega}$ and $\widetilde{\partial^{\alpha}u}|_{\Omega}=\partial^{\alpha}u$. And "extends continuously" means $\widetilde{\partial^{\alpha}u}$ is continuous with respect to the relative topology on $\overline{\Omega}$.  Here are my questions:        How do people usually do such extension?   When does such extension exist and when does not?   Let $\Omega$ be an open subset of ${\mathbb R}$. Is there any   non-trivial counterexample such that   this kind of extension does not exist? How can I compute the number of lists of the form $(a_1,\ldots,a_k)$, where $a_i \geq 0$ are integers and $\sum a_i \leq n$? What is the easiest way to derive the answer? I was going through the topic "formation of comparative and superlative".       The rule says     When two objects are compared with each other, the latter term of comparison must exclude the former, as in "Iron is more useful than any other metal". If we say "Iron is more useful than any metal", that is the same thing as saying    "Iron is more useful than iron", since iron is itself a metal.   I am finding difficulty to relate the rule with example. Proving that if $ad-bc = \pm 1$, then $\gcd(x,y) = \gcd(ax +by, cx + dy)$ I'm having issues figuring out how to approach this problem:     Conclude that if $ad-bc = \pm 1$, then   $$\gcd(x,y) = \gcd(ax +by, cx + dy)$$   The fact that $\gcd(x,y) = \gcd(x+ky, y)$ is a very special case of this exercise   I believe there is a property that if $a = bq_1 + 0$ where 0 is r, then b divides a, and $b=\gcd(a,b)$  Is that at all relevant in this case? Setup for a wine bottle with multiple labels I have modeled a wine bottle and created various wine label designs for a client. The labels were created in Illustrator and so I have PNGs. I want to present a render of the same bottle with each label the client. The labels are not simple rectangles.   Should I:    "model" each label and wrap it on the bottle, or  try to map each label image to a portion of the bottle's surface?    I've seen lots of tutorials showing option #2, but something is telling me that the glass of the bottle should be separate from the "paper" label.  Also, with #1, couldn't I use the same bottle blend file and use a different layer for the each label, then turn on a different layer for each new render?  Maybe there's an even better strategy?  thanks, Sometimes I hear people say something like the example below. Is it correct to use the word since in this manner?     Mosquitoes are surviving on earth since millions of years. Can I test if a regex is valid in C# without throwing exception Definition of the Pauli group and the Clifford group There seem to be two definitions of the Pauli group. In Nielsen and Chuang, the Pauli group on 1 qubit is defined as \begin{align*} \mathcal{P}_1 = \{\pm I, \pm iI, \pm X, \pm iX, \pm Y, \pm iY, \pm Z, \pm iZ\} = \langle X, Y, Z\rangle. \end{align*} But in other references, such as ,  \begin{align*} \mathcal{P}_1 = \{\pm I, \pm X, \pm iY, \pm Z\} = \langle X, Z\rangle. \end{align*} The second definition makes more sense in the stabilizer formalism, since we usually only consider the action of Pauli strings of $X$ and $Z$. What bothers me is that by the second definition, the phase gate $S$, which is an element of the Clifford group, results in \begin{align*} SXS^\dagger = Y \notin \mathcal{P}_1. \end{align*} I know that the prefactor $\pm i$ somehow doesn't matter, but can someone justify it from a more mathematical/group-theoretical perspective? $\int_0^\infty \frac{\sin(x^2+ax)}{x}\,dx$ converges I want to show that the following converges for $a\ge 0$ $$\int_0^\infty \frac{\sin(x^2+ax)}{x}\,dx$$  We can show that $\lim_{x\to 0} \frac{\sin(x^2+ax)}{x} = a$, and with that fact it's straightforward to prove that $$\left|\int_0^\delta\frac{\sin(x^2+ax)}{x}\,dx \right|&lt; \infty$$ where $\delta &gt;0$ is a constant.  However I'm having trouble showing that the function decays quickly enough as $x$ goes to infinity. I know how to prove this when taking the integral of $\sin(ax)/x$, but that $x^2$ is causing me problems.  Any ideas? Thanks. Showing that $e^x > 1+x$ $\sqrt{c+\sqrt{c+\sqrt{c+\cdots}}}$, or the limit of the sequence $x_{n+1} = \sqrt{c+x_n}$ (Fitzpatrick Advanced Calculus 2e, Sec. 2.4 #12)  For $c \gt 0$, consider the quadratic equation $x^2 - x - c = 0, x &gt; 0$.  Define the sequence $\{x_n\}$ recursively by fixing $|x_1| \lt c$ and then, if $n$ is an index for which $x_n$ has been defined, defining  $$x_{n+1} = \sqrt{c+x_n}$$  Prove that the sequence $\{x_n\}$ converges monotonically to the solution of the above equation.  Note: The answers below might assume $x_1 \gt 0$, but they still work, as we have $x_3 \gt 0$.    This is being repurposed in an effort to cut down on duplicates, see here: .  and here: . More options when flagging for migration I think there should be more Stack Exchange networks available when flagging a question to be migrated. For example I'd think that the following question:    ...would be a better fit on unix.stackexchange.com instead of Stack Overflow. But without the option to mark it so, I feel hesitant to mark it.