Please assume that we are now within a directory containing 13 files that were named as follows:  p1 p2 p3 ... p11 p12 p13   Through bash, when we execute the following command:  for i in p*; do echo $i; done   The star * wildcard seems to be expanded without an ascending order:  p1 p10 p11 ... p7 p8 p9   Yet for example, the following piece of commands expand the numerical values in an ascending order:  for i in p{1..13}; do echo $i; done   Is there any bash settings to alter this default star * wildcard behaviour?  If not, could you please tell us what is the motivation behind this preference for such an expansion?  PS: For a similar question, please refer to: . Yet herein I wonder if we can change the default behaviour of bash, and if so, how?
I have a directory filled with files with names like logXX where XX is a two-character, zero-padded, uppercase hex number such as:  log00 log01 log02 ... log0A log0B log0C ... log4E log4F log50 ...   Generally there will be fewer than say 20 or 30 files total.  The date and time on my particular system is not something that can be relied up on (an embedded system with no reliable NTP or GPS time sources).  However the filenames will reliably increment as shown above.  I wish to grep through all the files for the single most recent log entry of a certain type, I was hoping to cat the files together such as...  cat /tmp/logs/log* | grep 'WARNING 07 -' | tail -n1   However it occurred to me that different versions of bash or sh or zsh etc. might have different ideas about how the * is expanded.  The man bash page doesn't say whether or not the expansion of * would be a definitely ascending alphabetical list of matching filenames.  It does seem to be ascending every time I've tried it on all the systems I have available to me -- but is it DEFINED behaviour or just implementation specific?  In other words can I absolutely rely on cat /tmp/logs/log* to concatenate all my log files together in alphabetical order?
What is the antiderivative of $x^{x^{x^{x...}}}$? What is $\int{x^{x^{x^{x^{\dots }}}}dx}$? I managed to get it down to a complicated series after using $u$-substitution, but I want to see other methods of approach, or if my answer was right.   First I let $$u=x^{x^{x^{x^{\dots }}}}$$, and $du$ would be, in terms of $u$, $$du=  \frac{u}{\frac{1-ln(u)}{u}e^{ln(u)/u}}$$ This makes the integral of $$\int{x^{x^{x^{x^{\dots }}}}dx}=\int \frac{1 - ln(u)}{u}{e^{ln(u)/u} }du$$ Now making $w = ln(u), dw$ becomes $1/u$ and the integral becomes $$\int e^{w/e^{w}}(1-w)dw$$ From there, I distributed and made $2$ integrals of series, using the Taylor polynomial for $e^x$. From there, I integrated and back substituted. I was only trying to find the indefinite integral.   I'm still a junior in high school, so let me know if the answer is beyond my scope of understanding. Uncaught SyntaxError: Unexpected token with JSON.parse I have many tags marked as "interesting" and others as "ignored". (I use Java at work, I'm a Cocoa guy by night, any anything dealing with Windows/.NET/C# simply doesn't apply to me.) When tags are added to the list in the right navbar, they appear in the order in which they were added. There are several problems with this:   It can be difficult to visually locate a tag quickly, unless you have memorized its location. The tags appear as the same gray boxes like everywhere on SO, but with a small "x" afterward, and they wrap based on size. This tends to leave a ragged right margin and wasted space. Ordering based on when the tag was added is not always the most intuitive default. Various users might prefer to view them alphabetically, by popularity, recent activity, etc.   It would be nice to introduce a tabbed interface to switch between various orderings, and remember the value in the cookie, profile, etc. Coupled with this, perhaps users could specify a custom ordering that works best for them? (Currently the workaround is to copy all the tag names, remove them all, and re-add them in the desired order, but this is obviously annoying.)  Thoughts? fsck error on boot: /dev/sda6: UNEXPECTED INCONSISTENCY; RUN fsck MANUALLY I dual boot Ubuntu and Windows 7. I was browsing on FireFox in Ubuntu and all of a sudden I get some error message, so I rebooted.  Now when I try to boot into Ubuntu I get this message:  fsck from util-linux 2.26.2 /dev/sda6 contains a file system with errors, check forced. /dev/sda6: Inodes that were part of a corrupted orphan linked list found.  /dev/sda6: UNEXPECTED INCONSISTENCY; RUN fsck MANUALLY.         (i.e., without -a or -p options) fsck exited with status code 4 The root filesystem on /dev/sda6 requires a manual fsck  Busybox v1.22.1 (Ubuntu 1:1.22.0-15ubuntu1) built in shell (ash) Enter 'help' for a list of built-in commands.  (initramfs) _   How can I fix this? Motivated by :   Which finite groups are the group of units of some ring? Which finite groups are the group of units of some finite ring? Which finite abelian groups are the group of units of some commutative ring? Which finite abelian groups are the group of units of some finite commutative ring?   It  that even for fields this is not simple to answer.  I expect the general question (Which groups are the group of units of some ring?, with no finiteness hypotheses) is even harder. The energy of a moving object is $E = mv^2\;.$ That is it increases with velocity squared.  I walk at say 3 miles per hour, or lets round that down to 1 meter per second for a slow walk. I weigh less than $100~\mathrm{kg}\;,$ but lets just round that up to $100~\mathrm{ kg}$ for convenience (it is just after Christmas).  So when I walk along the pavement, I have $100~\mathrm{kg\; m^2 s^{-2}}$, 100 joules of kinetic energy.  Now I get on a passenger jet, which is cruising at around 500 knots, call that 250 meters per second.  In my seat I have $100\times 250^2 = 6250000$ joules of kinetic energy. But when I walk down the aisle I have $100\times 251^2 = 6300100$ joules of kinetic energy. The difference between these is: 50100 joules.  It feels the same to me, walking down the pavement as walking down the aisle of the plane. I didn't have to make some huge effort to get up to speed on the plane, yet I needed 500 times the energy to do it.  How is this possible, and where did the energy come from? Uk tourist visa - duration of stay My mother is given a UK tourist visa for 2 years with a duration of stay as 180 days. It is a c type multiple entry visa. Is this 180 days is the total stay she can take or is it duration of stay per visit.  Thanks in advance for your reply. What is the fastest way to parse & split XML content with huge file size (800MB UP) into several xml files in Java How did the English language come to this?       The play was awful.     Is the complete opposite of     The play was awesome.   But if you break it down to awe followed by ful or some, it doesn't make sense at all. Can someone shed a light on this?  Is there a rule to this, or is it one of those things you need to just memorize? How do I figure out the right size font / text for a given read distance? To avoid printing single letters at varying sizes and measuring how far away I can see it from a distance is there a formula for determining what size font for a particular read distance? Is it better to take the midpoint between the ascender line and x-height or to go purely off of the x-height (since going purely off the ascender line could leave lowercase letters difficult to read)? How does vertical height play into this? Would I need to then figure the hypotenuse distance instead of a straight distance?  For example if I'm shooting for a 20' read distance but the sign is to be placed about 30' up should I use a distance of  35' instead of 20'?  I'm shooting for a 20' read distance for the headline and then possibly a 10' read distance for additional content depending on how it all fits on the active area. After swapping the positions of the hour and the minute hand, when will a clock still give a valid time? At 12 o'clock, the hour hand and minute hand of the clock can be swapped, and the clock still gives the same time, but at 6 o'clock, it can not be swapped. So in what cases when we swap the hour and the minute hand position does a clock still give a valid time? Prove or the disprove the existence of a limit of integrals Let $A \subset \mathbb{R}^2$ be the annulus defined, in polar coordinates, as  $$ A = \{ (r, \theta) \in [0, \infty) \times [0, 2\pi) : 2 \leq r \leq 3 \} \, .$$ Let $F: A \to A$, $F:(r, \theta) \mapsto (r, r \theta \text{ (mod } 2\pi))$. Let $g : A \to \mathbb{R}$ be a continuous function. Discuss the existence of the limit $$ \lim_{n \to \infty} \int_A g \circ F^{(n)}(x) \, \text{d}x \, ,$$ where $F^{(n)} = F \circ F \circ \dots \circ F$ denotes the composition of $F$ with itself $n$ times.  I cannot find a way to attack this problem. Here's what I tried: if $g \circ F^{(n)}$ converged, I think we could take the limit under the integral sign. So I tried to show the convergence of $F^{(n)}$. If it converged, then the limit function must send every point to a fixed point of $F$. So I determined the fixed points of $F$. But I don't think $F^{(n)}$ converges. Is the Moonbeam spell amazing, or are we doing it wrong? The description of the moonbeam spell (PHB pg. 261) says:  When a creature enters the spell's area for the first time on a turn or starts its turn there, it is engulfed in ghostly flames that cause searing pain... On each of your turns after you cast this spell, you can use an action to move the beam 60 feet in any direction.  We read this to mean a creature takes damage from moonbeam when any of the following conditions are met:  It begins its turn inside the moonbeam It enters the moonbeam on its turn The creature finds itself inside the beam on another character's turn, no matter how it got there  The creature will not take damage in the following instances:  A creature moves from one part of the beam to another (obviously it had to take damage, but it won't take more damage) A creature is in the beam after having been inside the beam already on that turn.  Following these assumptions, our druid cast a moonbeam and did damage to two enemies, Bob and Frank. On their turn, they took damage again, killing Bob. Frank left the beam. On the druid's next turn, she moved the beam in a circular motion, not exceeding 60 total feet of movement, and in the process the beam came in contact with a dozen enemies and ended on Frank. All dozen enemies received damage, as they were inside the beam at some point during the turn. On Frank's next turn, he again took damage for the fourth time and died. On each of the druid's turns, she moved the moonbeam around like this until the fight ended (it did not exceed the time limit and she made all concentration checks). Is this a legitimate use of moonbeam? It seems awfully powerful for a level 2 spell. What is the sum of the 'second half' of the harmonic series?  $$\lim_{k\to\infty}\sum\limits_{n=k+1}^{2k}{\frac{1}{n}} =~ ?$$  More precisely, what is the limit of the above sequence of partial sums? Checking if Ubuntu package contains desired update I have the following shortcut which opens an SSH instance. How can I modify it so that when SSH closes, the terminal stays open?  [Desktop Entry] Terminal=true Type=Application Name[en_US]=ssh Icon=/logo_sprite.png Exec=gnome-terminal --geometry=... --window-with-profile=... --title=... -e 'bash -c "ssh -t -i ~/dsa.key cle@12.210.9.17 sudo -s"' Why is the space between words in my source code listing so big? I used the following code to list my java code, but the it looks like there's too much space between words. What can I do?  % source code listing \usepackage{listings} \usepackage{color} \usepackage{xcolor}  \usepackage{caption}  % citation style \usepackage{apacite} \DeclareCaptionFont{white}{\color{white}} \DeclareCaptionFormat{listing}{\colorbox{gray}{\parbox{\textwidth}{#1#2#3}}}  \begin{document} \begin{lstlisting}[label=some-code,caption=bind to service call] bindService(new Intent(this, Recognizer.class), recConnection, Context.BIND_AUTO_CREATE); \end{lstlisting} \end{document}   The snapshot of what I got. The code not completely shown, and space too wide, in function bindService. Who is the eponymous Jedi in Return of the Jedi? Prove that if $a$ and $b$ are relatively prime, then $\gcd(a+b, a-b) = 1$ or $2$ Prove that if $a$ and $b$ are relatively prime, then $\gcd(a+b, a-b) = 1$ or $2$.  I started off by putting $\gcd(a+b, a-b) = d$. This implies that there are two relatively prime integers $x_1, x_2$, such that   $dx_1 = a+b$   $dx_2 = a -b$  Adding the first equation to the second gives us: $d(x_1 + x_2) = 2a$, and subtracting the second from the first gives us $d(x_1 - x_2) = 2b$. This implies that $d\mid2a, d\mid2b \implies \gcd(2a, 2b) = d \implies d = 2\cdot \gcd(a,b) \implies d = 2$.   This is based on the fact that $\gcd(x_1 + x_2, x_1 - x_2) = 1$, which is in fact the statement of the problem. How do I prove this? Can I somehow use the Euclidean Algorithm for this. bash: /usr/bin/locale-check: No such file or directory error after i downgraded from ubuntu 18.10 to 18.04