I am searching for terms to replace “a grieving mother,” or “a childless mother.”  Is one also considered a grieving mother if she has no other children, in other words, her only child died?  This is from a story in which a mother, who lost a child saves a stranded girl, which reminds her of her daughter whose death the mother caused. However, the mother is unaware the girl is on the run from her father whose space fleet is on its way toward our solar system.
Is there a word that describes a parent whose child has died?   Along the lines of "orphan", "widow", and "widower", is there a single word for a parent who has lost a child (of any age)?
is a typical example for , though it is said that long time no see has been accepted by the mainstream English speakers.  Recently, there is a  movement in the , and a few new Chinglish words have been invented to demonstrate the characteristics of the Mainland.  The most outstanding examples (in my opinion) are:   shitizen - a citizen without citizen rights freedamn - the freedom for shitizens (no freedom) democrazy - the democracy for shitizens (no democracy) smilence - the speeches under the freedamn of speech () z-turn - to make effort in vain ( (Pinyin))   Here are my questions. Has there ever been any neologism movement in the history of English? What are the criteria to adopt new words into English? What is the best way to implement constants in Java? I've seen examples like this:   public class MaxSeconds {    public static final int MAX_SECONDS = 25; }   and supposed that I could have a Constants class to wrap constants in, declaring them static final. I know practically no Java at all and am wondering if this is the best way to create constants. Shape of the universe? Why is there no 32 bit ISO for the new Ubuntu 18.04? Why is there no 32 bit ISO for the new Ubuntu 18.04? I played 1st and 2nd edition AD&amp;D a lot back in the day. But when 3rd edition came out, I dropped out of the hobby. Partly I just couldn't be bothered to re-learn the rules. The conversion systems, previews and material for 3rd edition I was seeing in Dragon and Dungeon magazines made it look like a very major change.  I recently picked up the new 5th edition starter set on a whim. Obviously the starter set rules are very stripped down in comparison to the full rules, but nevertheless, I was amazed by how comfortable and familiar the system felt. The only thing that felt like a truly radical change was the way things like thief skills and the like were now handled with a d20 instead of a percentage system.  There's plenty of information online (and here) comparing editions of the game in an incremental manner. But I'd be interested in a breakdown of how 2nd and 5th edition compare directly, specifically looking at whether there are broad areas of similarity. in -> inner  out -> outer / (outter?)  What is the history or set of rules behind why 'inner' doubles the 'n' but 'outer' doesn't double the 't'? Looking for title & author - short story - possibly called "Arms Race" - about rediscovery of manual arithmetic short story - possibly called "Arms Race" - about rediscovery of manual arithmetic.   The hero rediscovers manual calculation - e.g., how to do long division with pencil and paper - in a world where everyone has ben using calculators for aeons. Brought before the president to demonstrate this, the great line in the story is the president, after using his calculator to check the hero's answer, says, "My God -- he's right." A great national project to exploit this advance over the enemy is obviously necessary... I installed Ubuntu 12.04 a month ago and am using it till now. I failed to notice that all this time there was no sound at all while running Ubuntu, even while playing a game in Wine. The weird thing is that only the startup sound comes when I log in (Indian/African drum tone), then comes the utter silence.  I tested both Digital Output (S/PDIF) and the speakers in the sound settings but can hear nothing.  Any help? I'm trying to get through Spivak's Calculus on my own and even though I kinda understand induction I'm not so sure that's the case when it comes to complete induction. So I tried to do a starred problem which involves using it. But I'm not sure that my proof is valid, can someone check this for me?  So the problem is that I have to prove that     For every natural $n&gt;1$, if $n$ is not prime we can write it as a product of primes.   I'm starting with a base case. For $n=2$, $n$ is prime, so the assumption holds.  Let's assume that for some $n &gt; 1$ and also all numbers $p &lt;= n$ the assumption holds.  Then $n+1 = cd$. If $n+1$ is not prime, then $c &lt; n+1$ and $d &lt; n+1$, but as they are natural numbers, we can also write that $c &lt;= n$ and $d &lt;= n$. But we assumed that if $p &lt;= n$, and $p$ is not prime, we can write it as a product of primes. So if $c$ or $d$ are not primes, we can write them as a product of primes. That means that $n+1$ can be written as a product of primes if it's not prime, which completes the proof. Bound on the difference of two determinants Let $A$ and $B$ be two real, $n\times n$ matrices. Using Hadamard's inequality, it is not hard to show that $$ \left|\det A - \det B \right| \leq \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}. $$ Where $\|A\|_2=\sqrt{\sum_{i,j}a_{ij}^2}$.  From this, I can derive a sup bound, for example  $$ \left|\det A - \det B \right| \leq n^{n+1} \|A-B\|_{\infty} \max (\|A\|_{\infty}^{n-1},\|B\|_{\infty}^{n-1}). $$ Where $\|A\|_\infty=\sup_{i,j}|a_{ij}|$.   The constant $n^{n+1}$ is not the best bound possible : any reference (or proof) for a better (or the best) one? I show below that one can obtain $n^2(n-1)^{n-1}$, but that isn't much better. I just tried  $10^5$ random matrices on Maple and obtained a maximal constant (much) smaller  than one : this is not a proof, but it looks like there is room for improvement nevertheless.    Just for completeness (and in case someone sees a factor I missed), to get the first bound, writing $A=[A_1,\ldots,A_n]$ in terms of its column vectors, an expansion shows \begin{eqnarray*} \det A &amp;=&amp; \det (A_1 -B_1,A_2,\ldots,A_n) + \det (B_1,A_2,\ldots,A_n) \\   &amp;=&amp; \sum_{j=1}^n \det (B_1,\ldots, B_{j-1}, A_j -B_j,A_{j+1},\ldots,A_n) \\ &amp;&amp; + \det B,  \end{eqnarray*} Thus by Hadamard's inequality, \begin{eqnarray*} \det A -\det B &amp;\leq&amp;  \sum_{j=1}^n \prod_{i=1}^{j-1} \|B_i\|_{2}\prod_{i=j+1}^{n} \|A_i\|_{2} \|A_j-B_j\|_{2} \\ &amp;\leq&amp;  \|A-B\|_{2} \sum_{j=1}^n \|B\|^{j-1}_{2}\|A\|^{n-j}_{2} \\  &amp;=&amp; \|A-B\|_{2} \frac{\|A\|_{2}^n -\|B\|_{2}^n}{\|A\|_2 -\|B\|_2}. \end{eqnarray*}  The second bound is just that $x^n -y^n\leq n \max(|x|^{n-1},|y|^{n-1}) |x-y|$ and $\|A\|_2 \leq n\|A\|_\infty$.    Another approach is calculus, namely, to write that $\det B - \det A = f(1)-f(0)$, with $f(t)=\det(A + t(B-A))$.   By the mean value theorem $f(1)-f(0)\leq \max |f^\prime (t)|$.   We can compute that $$f^\prime(t) =  {\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)$$  (if I did not mess up, using the formula for the differential of a determinant, where Cofm means the matrix of Cofactors).   Then, it should deliver something better, if there is a nice way to bound it. The simplest thing is to use Cauchy-Schwarz, namely $$ \left|{\rm trace}\left({\rm Cofm}(A +t(B-A))(B-A)\right)\right|\leq \|B-A\|_{2} \|{\rm Cofm}(A +t(B-A))\|_{2} $$ and then, by lack of a better idea,  $$ \|{\rm Cofm}(A +t(B-A))\|_{2}\leq n \max_{ij} |{\rm Cof}_{i,j}(A +t(B-A))|, $$ and brutally, $|{\rm Cof}_{i,j}(A +t(B-A))|\leq ((n-1) \max(\|A\|_\infty,\|B\|_\infty))^{n-1}$ gives a slightly better constant, namely $$ n^2(n-1)^{n-1} &lt;n^{n+1} $$ but that still seems a very rough way to bound a determinant, as it is never sharp, since to attain this bound all coefficients should be equal, and therefore the cofactor would be zero. They are of the same order in $n$, and I suspect this order is wrong. How can I view the Android internal partition table? I've rooted my device, put it in recovery mode, and have linked it to PC by .  How can I view the list of storage devices and partitions?  Note: fdisk doesn't work. Why does the sum of the reciprocals of factorials converge to $e$? I've been asked by some schoolmates why we have $$ \sum_{n=0}^\infty \frac{1}{n!}=e.$$ I couldn't say much besides that the $\Gamma$ function, analytic continuation of the factorial, is defined with an integral involving $e$. Then I also know that actually $$ \sum_{n=0}^\infty \frac{x^n}{n!}=e^x.$$ Is there a reason for these facts?  P.S. I added the tag "intuition", please remove it if you think it is not pertinent. This is what my  looks like when it first loads:    But if we unselect math.SE, we get a y-axis rescaling mess:    I'm using Firefox. I'm in China, so many things involving Google and Imgur are blocked, and I use  to stop the browser attempting to load unloadable Google objects.  I'm hoping other users can check whether or not this issue is localized to my system, and if it's not just me, here's the bug report. Making the background of an image transparent in Gimp I have some jpg images and I need for each of them to have a transparent background.  I have GIMP as a graphic tool.   I need to put them in a web page and I need to see just the main element and have a transparent background. Custom text color in C# console application? Consider the following code:  foo () {     echo $* }  bar () {     echo $@ }  foo 1 2 3 4 bar 1 2 3 4   It outputs:     1 2 3 4       1 2 3 4   I am using Ksh88, but I am interested in other common shells as well. If you happen to know any particularity for specific shells, please do mention them.  I found the follwing in the Ksh man page on Solaris:     The meaning of $* and  $@  is   identical  when  not  quoted  or  when  used  as a parameter   assignment value or as a file name. However, when used as  a   command argument, $* is equivalent to ``$1d$2d...'', where d   is the first character of the IFS variable,  whereas  $@  is   equivalent  to  $1 $2 ....   I tried modifying the IFS variable, but it doesn't modify the output. Maybe I'm doing something wrong? Why shouldn't root be allowed to login via ssh? why should root access be disabled for ssh? I always hear that's for security reasons. But I don't get it. What's different to login as non-root and then sudo su -? What's the preferred way to handle tasks that need privileged permissions?  Thanks  Thomas So I decided to install my copy of Ableton (v9.1) on my computer now that I've upgraded to a machine that isn't ancient. HOWEVER Despite WineHQ reporting that Ableton is platinum-certified, I can't get the dern thing to open.  It crashes out just as the splash screen pops up.  Sometimes I see it, sometimes not.  It tells me that some catastrophic error has occurred, and it needs to close.  I've tried this on two different systems running two different versions of wine  (One runs the up-to-date public distro 1.6.2, while the machine I actually want to install it on is running the 1.7.19)  I don't know what to do.  I'd really just like to get this working, as my friends really like the look and feel of Ableton, and I really don't like LMMS (Linux-y free Audio Software).  Any idea what's going on here? I've found a couple of other people who seem to be having the same problem as me, but no dice on any solutions.  None provided, really.  Specs: Ubuntu 14.04 LTS  8 Gig ram  i7-2670QM CPU @ 2.20GHz × 8  Lots of free space... yadda yadda yadda   Error message: The following is the first part of Proposition 2.9 in "Introduction to Commutative Algebra" by Atiyah &amp; Macdonald.     Let $A$ be a commutative ring with $1$. Let $$M' \overset{u}{\longrightarrow}M\overset{v}{\longrightarrow}M''\longrightarrow  0\tag{1} $$ be sequence of $A$-modules and homomorphisms. Then the sequence (1) is exact if and only if for all $A$-modules $N$, the   sequence  $$0\longrightarrow \operatorname{Hom}(M'', N)  \overset{\overline{v}}{\longrightarrow}\operatorname{Hom}(M,  N)\overset{\overline{u}}{\longrightarrow}\operatorname{Hom}(M', N)  \tag{2} $$ is exact.   Here, $\overline{v}$ is the map defined by $\overline{v}(f)=f\circ v$ for every $f\in\operatorname{Hom}(M'', N)$ and $\overline{u}$ is defined likewise.   The proof one of direction, namely $(2)\Rightarrow (1)$ is given in the book, which I am having some trouble understanding. So assuming (2) is exact sequence, the authors remark that "since $\overline{v}$ is injective for all $N$, it follows that $v$ is surjective". Could someone explain why this follows?  Given that $\overline{v}$ is injective, we know that whenever $f(v(x))=g(v(x))$ for all $x\in M$, we have $f=g$. I am not sure how we conclude from this surjectivity of $v$.  Thanks! I am trying to implement the following K-map with a table    AB C     00 01 11 10       ______________     0| 1  0  1  0     1| 0  1  1  0   I implemented it like this:   \begin{table}[htdp] \begin{center} \begin{tabular}{c|c|c|c|c|}  &amp;00 &amp; 01&amp; 11&amp; 10 \\ \hline 0&amp;1&amp;1&amp;1&amp;1\\ 1&amp;0&amp;0&amp;0&amp;0\\ \end{tabular} \end{center} \label{default} \end{table}   However on the top row I have vertical lines that I'd like to not see. I went with this approach because if i was to use a one row with the information so that I wouldn't have to worry about aligning the headings. So is there a way I can get the vertical lines not to appear in the heading row?