Question about the Afine Jacobi criterion.
The rank of Jacobian matrix at a point of affine variety is independent of choice of generators
'help' link is located lower than the other items It shows like this on Firefox 25 and Internet Explorer 11 (minus the red line of course):    Chrome seems to be fine. Stack Overflow provides good tools in the question editor, like "Links" and "Styling".  Another thing that for me would be nice is a color selector. Although an HTML editor is available, a direct click on tool would be good, and users can select different colors, as for marking errors in their program with red, and more important data with blue. Delete the entire table rendered from different pages using Javascript Aligning table within latex editor Need help to interpret output of free command I would appreciate help in interpreting the output of 'free -m' below.    -bash-4.2$ free -m               total        used        free      shared  buff/cache   available Mem:          15793        7112          88        7840        8591         611 Swap:         16891        5289       11602    If I have nearly 11GB free then why am I seeing swap usage.  Shouldn't that be 0?   Also, the 'shared' value is 7840.  Is this memory available for a new application? Is only 'cache' memory available for new applications?  In other words, 'shared' and 'buffer' memory are not available for new applications? How should I interpret the 'available' column with a value of 611?  How is this value arrived at? Is this statement true?      In a metric spase $(E,d)$, a sequence $(x_n)$ is Cauchy if and only if $  \forall k\in \mathbb{N}, \lim_{n\rightarrow+\infty}d(x_{n+k},x_n)=0$   I proved that $\Rightarrow$ is true,indeed if $(x_n)$ is Cauchy then for all $\varepsilon&gt;0,\exists n_0\in\mathbb{N}, \forall p,q\geq n, d(x_p,x_q)&lt;\varepsilon$ it still right for all $ n\geq n_0$ and $m=n+k&gt;n\geq n_0$   But I don't know if $\Leftarrow$ is true ? Is denying "Paradoxical Partitioning" equivalent to accepting the Axiom of Choice? In  (to ) it has been noted:     One more point to make about the paradoxical decomposition to more parts than elements [...]   currently we do not know of any model of +$\lnot$ where such decomposition does not exist.    Namely, as far as we know, in all models where choice fails there is some set which can be partitioned into more parts than elements.   (Just for clarification: From the context of the above references the comparison "more parts than" is obviously meant in terms of the  of the suitable  being strictly larger than the cardinality of the suitable initial set itself.)  Now, I find the negation or denial of such "paradoxial partitioning" interesting; i.e. the statement (proposition, "$\lnot$PP"):  "Each partition of each given set has cardinality less than, or at most equal to, the cardinality of the given set."  And I like to further explore how this suggested "proposition $\lnot$PP" (being considered along with ZF, of course) relates to "the standard set theory including Axiom of Choice", . Therefore  My questions:      Are there any models of ZFC known (or could there be any such models, in principle) in which there is some set which can be partitioned into more parts than elements ?  And:  Are there any models of ZF known (or could there be any such models, in principle) in which there is no set which can be partitioned into more parts than elements, and which is not also a model of ZFC ?  (And just for reference: Is there a conventional or concise way of expressing the suggested "proposition $\lnot$PP" in terms of standard notation, such as used in the sources linked above? Has it perhaps been discussed already, by some other name? ...) Show $\sin x = bx$ has at least 3 real solutions. Show $\sin x=bx$ has at least 3 real solutions for every $b$, where $-\dfrac{2}{3 \pi} &lt; b &lt; \dfrac{2}{\pi}$.   Some useful facts I am trying to use:  Intermediate value theorem: If $f$ is a continuous real-valued function on $[a,b]$ and $d \in \mathbb{R}$ satisfies $f(a)&lt; d&lt; f(b)$, then there exists a point $c$ in $(a,b)$ such that $f(c) = d$.   Corollary: If $f$ is a continuous real-valued function on $[a,b]$, then $f([a,b])$ is a closed interval. How to keep a constant baselineskip when using minipages (or \parboxes)? When a minipage is used to box a text paragraph, the first line of the minipaged text and the last line of the previous text paragraph are closer than two consecutive lines in a standard text paragraph; analogously, the last line of text inside the minipage and the first line of text after the minipage are also closer.  The following MWE illustrates this:  \documentclass{article} \usepackage{lipsum}  \newcommand\Text{Lorem ipsum dolor sit amet, consectetuer adipiscing elit. Ut purus elit, vestibulum ut, placerat ac, adipiscing vitae, felis. Curabitur dictum gravida mauris. Nam arcu libero, nonummy eget, consectetuer id, vulputate a, magna.\textbullet}  \begin{document}  \Text  \noindent\begin{minipage}{\linewidth} \Text \end{minipage}  \Text  \end{document}   How can I guarantee a constant baseline-skip even if minipages (or \parboxes) are used? Use character string as function argument Incremental (delta) backup of a encrypted data I have several files that I need to keep encrypted on my home hard disk.  I currently use a TrueCrypt file container, currently around 100MB in size. I need to backup this file daily, and I'm looking for an incremental backup solution.  I read that on encrypted volumes even a small change in a file would change the whole volume and prevent the use of an efficient incremental backup solution. Any ideas how to backup in this case in an efficient way which would allow incremental solution?  I currently use TrueCrypt as my backup solution and CrashPlan as my backup utility, but I'm willing to experiment with alternatives. Are C++ enums signed or unsigned? And by extension is it safe to validate an input by checking that it is &lt;= your max value, and leave out >= your min value (assuming you started at 0 and incremented by 1)? Why is ROC AUC equivalent to the probability that two randomly-selected samples are correctly ranked? I found there are two ways to understand what AUC stands for but I couldn't get why these two interpretations are equivalent mathematically.   In the first interpretation, AUC is the area under the ROC curve. Picking points from 0 to 1 as threshold and calculate sensitivity and specificity accordingly. When we plot them against each other, we get ROC curve.  The second one is that the AUC of a classifier is equal to the probability that the classifier will rank a randomly chosen positive example higher than a randomly chosen negative example, i.e. P(score(x+)>score(x−)).  (from ) It is said in many textbooks that alpha decay involves emitting alpha particles, which are very stable. Indeed, the binding energy () is higher than for $Z$-neighboring stable isotopes. But the binding energy is lower than, for example, ${}^9\mathrm{Be}$ (). My question is why aren't other nuclear compounds ejected from heavy nuclei, e.g. ${}^9\mathrm{Be}$?  The   $$e^{-\frac{4\pi}{\hbar}\frac{Ze^2}{4\pi\epsilon_0}\frac{1}{v_\alpha}}$$ decreases exponentially in $Z$, so it explains intuitively why lower-$Z$ particles would tunnel more often. Specifically, it would explain why we would see ${}^9\mathrm{Be}$ emission ${e^{-2}}\simeq 0.14$ times as often compared to ${}^4\mathrm{He}$ emission. Also, the particles need to form in the nucleus prior emissions; but with a similar binding energy per nucleon (~7.08 MeV for ${}^4\mathrm{He}$ vs. 6.47 MeV for ${}^9\mathrm{Be}$) and higher total binding energy for the ${}^9\mathrm{Be}$ nucleus, I would expect that its formation in the same order of prevalence as the alpha particle (according to Ohanian, between 0.1 and 1 alpha particles are in existence at any moment in time).  Can anyone explain this? A reference to an article/textbook would be preferable.  EDIT Same goes for ${}^{16}\mathrm{O}$ which is also a double magic isotope, as 'anna v' pointed out. For it, the Gamow factor is smaller by $e^{-4}\simeq 0.02$, and emission should still be viable. I read this book once when I was in middle school. A girl solves a murder or a mystery on a spaceship. It was a thin paperback books.   All I can remember is that there was an alien, and there was a movie star or actress or famous makeup artist on the ship traveling with another woman. By the end of the book the murderer was the woman the actress was traveling with who forced her to use her makeup skills to disguise her, and she was really some kind of alien. Get index of each capture in a JavaScript regex I've just seen that the reason of closing is missing in the reviewing page of closing votes.    Is it a temporary error or it's a bug?  EDIT: Seems that only when there are audits the reason is missing.  If I cannot upload image I will make an ASCII art:  Should this question be closed as: ? (less)  ┌────────────┐ ┌─────┐ ┌────┐ ┌────┐                                              │ Leave Open │ │Close│ │Edit│ │Skip│                                              └────────────┘ └─────┘ └────┘ └────┘ Leave Open if this question should not be closed  Close if this question should be closed for any reason; consider leaving an  explanatory comment before voting  Edit if this question could be improved so that it does not need to be closed  Skip if you are not sure and want to go to the next question Solution to German Tank Problem Is there a formal mathematical proof that the solution to the  is a function of only the parameters k (number of observed samples) and m (maximum value among observed samples)? In other words, can one prove that the solution is independent of the other sample values besides the maximum value? Suppose a gambler starts with one dollar and plays a game in which he or she wins one dollar with probability $p$ and loses one dollar with probability $1 - p$. Let $f_{n}$ be the probability that he or she first becomes broke at time $n$ for $n = 0, 1, 2\ldots$. Find a generating function for these probabilities.  Is it binomial distribution $?$ and then use the $\operatorname{mx}\left(s\right) = rx\left(\mathrm{e}^{s}\right)$ of the binomial distribution $?$. Help$\ldots!$. I am correcting my quiz. According with Mumford the answer is yes, but there are some obscure points in the proof.  We know that there exists a very ample line bundle on $A$ since every abelian variety is projective. Hence, via some corollaries of the Theorem of the Cube, we get an ample line bundle whose restriction to the kernel of $n_A$ is both the trivial bundle and very ample. This should imply that the dimension of $\ker n_A$ is $0$.  Furthermore, Mumford says that the fact that the dimension of the kernel is $0$ implies the surjectivity of $n_A$.  Is anybody able to clarify these two points to me?  Thanks in advance!