I haven't spent much time dealing with standard partition. Aside from "LVM makes it easier" to manage disk, what really can LVM do that standard partition can't?  All I can think of right now is that your LV(Logical Volume) can be a combination of multiple hard disks. Can you also add size to your volume by shrinking another volume in standard partition?
In my situation, it's for a local build/staging server running ubuntu server. Things I'd like to know are:  Why I should/shouldn't use LVM?  What problems might it cause in terms of either regular server operation or things I might want to do later? (ex. partitioning)  Why I should/shouldn't use encrypted LVM instead of LVM?  What problems might encrypted LVM cause in terms of either regular server operation or things I might want to do later? (ex. partitioning)
How do I label equations without the chapter number? WordPress twenty eleven sidebar on Pages I am using 2011 theme, any how I managed to get sidebar on single.php pages by installing child theme,  But now I want sidebar on pages,  How do I get it? Url decode UTF-8 in Python The  mentions that it's possible to post one more question after a six-months waiting period and if the question is good enough, the question ban can be lifted:     If you're unable to improve your existing questions, you'll get the chance to ask one new one 6 months after your last question. If that question is positively received, you may be able to continue asking questions; if not, then the ban will be reinstated.   What is positively received in this context? Is it enough if the question has a zero score and is not closed? Or does it have to have a positive score?  How long does the system wait after posting this new question to evaluate this criteria and possibly lift (or reinstate) the question ban? Where can I obtain an English dictionary with structured data? Motivating the ansatz for the infinitesimal translation operator What is a word/idiom for 'unable to decide'? Let's say I have an important decision to make and I can't decide between two competing things (like break up with a girl or not break up with her). What would be a word/idiom to express that? I accidentally removed all the packages starting with the name jpeg with:  sudo apt-get remove jpeg*   This screwed up a lot of things, removed Unity, Firefox, Flash player etc.  I somehow managed to recover the desktop and unity but then a lot of things are still buggy. For example .rpm files and .tag.gz just won't open up at all.  What I want is a way to fix everything that's amiss. Something that is the equivalent of a 'repair installation' with a windows disc.  Any tips, anyone? Why does turning Caps Lock off take such a long time to work? This is my biggest issue with Linux and the main reason I don't use it as my main working OS.  Whenever I use caps lock to write something in upper case and disable it in order to keep writing the next letter AFTER I've disabled it is written in upper case.  So, my text comes out weird, as SOMETHING *L*ike this  I've tried Linux Mint, Fedora ... and the issue keeps happening.  Searching Google I've found some very rare persons have the same issue, but I was unable to find any solution.  Is there any way to solve this or am I DOOMED *F*or life?  BTW, whenever I ask this question someone suggest I should learn to use shift, instead of caps lock. I know, that's the right way, but I'm used to use caps lock and, well, it should work. Difference between Inheritance and Composition Can't delete the filter called My Filter I'm unable to get rid of a filter that goes by the name of "My Filter".  This is the default name you get when you click on New Filter.  There's a clickable "Delete this filter", which leads to a confirmation question on whether you really want to delete it.  Clicking "Yes" leads you back to the "New Filter" Screen,  with a new filter named "My Filter".    I don't know whether this is the same filter I was just trying to get rid of,  or whether it got rid of the old one, and now is making a new one with the same old name.  Either way, I'm stuck. Variation of Pythagorean triplets: $x^2+y^2 = z^3$ I need to prove that the equation $x^2 + y^2  = z^3$ has infinitely many solutions for positive $x, y$ and $z$. I got to as far as $4^3 = 8^2$ but that seems to be of no help.  Can some one help me with it? Relationship between gamma and chi-squared distribution If $$Y=\sum_{i=1}^{N}X_i^2$$ where $X_i \sim \mathcal{N}(0,\sigma^2)$, i.e. all $X_i$ are i.i.d. normal random variables of zero mean with same variances, then $$Y \sim \Gamma\left(\frac{N}{2},2\sigma^2\right).$$  I know the chi-squared distribution is a special case of the gamma distribution, but could not derive the chi-squared distribution for random variable $Y$. Any help, please? Why are there so many close votes to review and so few of the others? Let $G$ be a group of order $56$. (We do NOT assume the Sylow-$7$ subgroup to be normal.) Then either the Sylow-$2$ subgroup is normal or the Sylow-$7$ subgroup is normal. How to prove?  My idea: consider the case $n_2=7,n_7=8$. Then the number of elements in all Sylow $2$ -subgroups and Sylow $7$-subgroups is at most exactly $(7-1)*8+(2-1)*7+1=56$. This cannot give a contradiction. We need to prove that $G$ is simple. Then $|G||(8!/2)$. To prove $G$ simple, we need to prove there does not exist normal subgroups of order $2$ and order $4$. But I do not know how to prove. Showing that $(m)\cap (n)=(\operatorname{lcm}(m,n))$ and $(m)+(n)=(\gcd(m,n))$ for any $m,n\in\mathbb{Z}$ Explicit formula for Bernoulli numbers by using only the recurrence relation It is not hard to show, by induction on $m\in\mathbb N$, that there exist a sequence $(B_n)_{n\geq0}$ of rational numbers such that  $$\sum_{k=1}^nk^m=\frac1{m+1}\sum_{k=0}^m\binom{m+1}kB_k\,n^{m+1-k},\ \style{font-family:inherit;}{\text{for all}}\ n\geq1\,.$$  These are the Bernoulli numbers of the second kind, that is, $B_1=+1/2$. A corollary of the proof (by induction) of the fact above is a recurrence formula for such numbers $B_n$, which are known as Bernoulli numbers:  $$\sum_{k=0}^{n-1}\binom nkB_k=0\,.\tag{$\ast$}$$  On the other hand, there are a number of explicit formulae for $B_n$, which are obtained using the equality $\frac x{e^x-1}=\sum_{n=0}^\infty B_n\,\frac{x^n}{n!}$; see  and . For example:  $$B_n=\sum_{k=0}^n\frac1{k+1}\sum_{r=0}^k(-1)^r\binom krr^n\,.\tag{$\ast\ast$}$$  Is there some clever way to manipulate the recurrence $(\ast)$ in order to obtain $(\ast\ast)$? What techniques are available for collapsing (or pooling) many categories to a few, for the purpose of using them as an input (predictor) in a statistical model?    Consider a variable like  (discipline chosen by an undergraduate student). It is unordered and categorical, but it can potentially have dozens of distinct levels. Let's say I want to use major as a predictor in a regression model.  Using these levels as-is for modeling leads to all sorts of issues because there are just so many. A lot of statistical precision would be thrown away to use them, and the results are hard to interpret. We're rarely interested in specific majors -- we're much more likely to be interested in broad categories (subgroups) of majors. But it isn't always clear how to divide up the levels into such higher-level categories, or even how many higher-level categories to use.  For typical data I would be happy to use factor analysis, matrix factorization, or a discrete latent modeling technique. But majors are mutually exclusive categories, so I'm hesitant to exploit their covariance for anything.  Furthermore I don't care about the major categories on their own. I care about producing higher-level categories that are coherent with respect to my regression outcome. In the binary outcome case, that suggests to me something like linear discriminant analysis (LDA) to generate higher-level categories that maximize discriminative performance. But LDA is a limited technique and that feels like dirty data dredging to me. Moreover any continuous solution will be hard to interpret.  Meanwhile something based on covariances, like multiple correspondence analysis (MCA), seems suspect to me in this case because of the inherent dependence among mutually exclusive dummy variables -- they're better suited for studying multiple categorical variables, rather than multiple categories of the same variable.  edit: to be clear, this is about collapsing categories (not selecting them), and the categories are predictors or independent variables. In hindsight, this problem seems like an appropriate time to "regularize 'em all and let God sort 'em out". Glad to see this question is interesting to so many people! Implementing a sort of "tightened frenchspacing" By default, (La)TeX adds extra space between sentences. The exact rules are in the TeXbook, a summary by TH. can be found . What follows is is my "executive summary":   After any period (or quotation/exclamation mark) that doesn't follow a capital letter, TeX adds a font's "extra space" to the "normal (interword) space". For Computer Modern 10pt, the normal space is 3.33pt and the extra space 1.11pt, resulting in an intersentence space of 4.44pt (+33% compared to the nomal space, assuming no stretch and shrink). A sentence end after a capital letter is indicated by the \@ command: Ann is learning the ABC\@. To denote that a period after a lowercase letter doesn't end a sentence, one may use \@ as well as an explicit space: Prof.\@ Higgins or Prof.\ Higgins.   Intersentence spacing may be brought in line with normal spacing by issuing the \frenchspacing command. (This is the standard setting for certain languages, e.g. the  (n)german option of the babel package.) With \frenchspacing enabled, there's no need to use the \@ command. One may revert to (La)TeX's default behavour by issuing \nonfrenchspacing.  So, there's the choice between extra intersentence spacing and uniform spacing. What is missing in my opinion is a third option I'll call "tightened frenchspacing": normal (interword) spacing between sentences and decreased spacing after non-sentence ending periods. To be somewhat more precise (while still ignoring stretch and shrink), I envision the following:   Adhering to a font's ratio between "normal space" and "normal space plus extra space" (3:4 for Computer Modern), the intersentence space is set to the normal space. The space after non-sentence ending periods equals the normal space multiplied by the above ratio (resulting in a space of 2.5pt for Computer Modern). As with \nonfrenchspacing in effect, the \@ command is used to denote cases of special punctuation.   How should such a "tightened frenchspacing" option be implemented?  EDIT: Thanks to Mico for providing what may well be a stepping stone to a solution. Here's another attempt at clarification:   Simply reducing the spacing after any period is not what I had in mind. There should be a (tangible) difference between end-of sentence spaces and spaces after non-sentence-ending periods (only that the latter should be reduced instead the former augmented). If adhering to a font's normal space/extra space ratio is not possible, a solution that adds a fixed (reduced) space (say, a thin space) after non-sentence-ending periods will be accepted. For compatibility reasons, I'd prefer a solution that works with the traditional use of \@ (i.e allows "tightened frenchspacing" for existing documents without the need to change the document body). Should it turn out that a different set of "special punctuation" macros is needed, I'll accept this as a solution. So I was writing up a simple binary tree in Python and came across [...]  I don't believe this to be related to the Ellipsis object, more it seems to have something to do with an infinity loop (due to Python's shallow copy?). The source of this infinity loop and why it doesn't get expanded while expanding when accessed is something I'm completely lost to, however   &gt;&gt;&gt; a [[[[[], [], 8, 3], [[], [], 3, 2], 6, 3], [], 1, 4], [[], [], -4, 2], 0, 0] &gt;&gt;&gt; Keys(a)#With a+b [0, 1, 6, 8, 3, -4] &gt;&gt;&gt; Keys(a)#With [a,b] [8, [...], [...], 3, [...], [...], 6, [...], [...], 1, [...], [...], -4, [...], [...], 0, [...], [...]] &gt;&gt;&gt; Keys(a)[1]#?? [8, [...], [...], 3, [...], [...], 6, [...], [...], 1, [...], [...], -4, [...], [...], 0, [...], [...], 8, [...], [...], 3, [...], [...], 6, [...], [...], 1, [...], [...], -4, [...], [...], 0, [...], [...]]     Version using a+b  def Keys(x,y=[]):     if len(x):y+=[x[2]]+Keys(x[0],y)+Keys(x[1],y)#Though it seems I was using y=y[:]+, this actually outputs an ugly mess     return y   version using [a,b]  def Keys(x,y=[]):     if len(x):y+=[x[2],Keys(x[0],y),Keys(x[1],y)]     return y   So what exactly is [...]?