Power Supply Adapter compatability
Choosing power supply, how to get the voltage and current ratings?
Solve $$u_{xx}-3u_{xt}-4u_{tt}=0$$ where $u(x,0)=x^2$ and $u_t(x,0)=e^x$.  My workings so far: I have factored the differential equation in the following way: $$(\delta_x-4\delta_t)(\delta_x+\delta_t)=0$$ where $\delta_x=\frac{\delta}{\delta x}$ etc. Now if we let $v$ be the solution to $(\delta_x+\delta_t)u$ then we have the following two equations: \begin{eqnarray*} (\delta_x+\delta_t)u=u_x+u_t=v\\ (\delta_x-4\delta_t)v=v_x-4v_t=0 \end{eqnarray*} Now for $v$ we simply find $$v=h(t+4x)$$ where $h$ is an arbitrary function of one variable. Now what remains is find $u$ such that $$u_x+u_t=h(t+4x)$$ I am stuck here, I thought about making a change of variables $\zeta=x+t$ and $\eta=x-t$ and thus using the product rule to show that $u_x=u_{\zeta}+u_{\eta}$ and $u_t=u_{\zeta}-u_{\eta}$ and thus $u_x+u_t=2u_{\zeta}$ and we need to solve $$u_{\zeta}=h(t+4x)$$ (I left out the factor 2 because $h$ is an arbitrary function). Do I simply integrate now and conclude $$u=f(\zeta)h(t+4x)+g(\eta)=f(x+t)h(t+4x)+g(x-t)$$ This seems wrong to me...  Some help would be greatly appreciated! Do we have any other evidence other than red shift? Is anyone trying to disapprove inflation? Also, can red shift occur because of large distance, not only because two objects are going farther from each other? Is there a quasistatic process that is not reversible? I have seen several questions and good answers on the link between reversible and quasistatic processes, such as  or . However, these questions only adress one side of the problem : a reversible process is necessarily quasistatic.  I am interested in the other side of the equivalence : is there a process that is quasistatic, yet not reversible ? It looks to me that an irreversible process cannot be made perfectly quasistatic.  The  article about quasistatic processes takes as an example the very slow compression of a gas with friction. As the compression occurs very slowly, the transformation is quasistatic, and the friction makes it irreversible. I am not convinced by this example : if you press on the piston with a vanishingly small force you will have to reach the threshold of the Coulomb law for solid friction before moving the piston anyway. It makes the process non-quasi-static, however small the Coulomb threshold might be.  Another example I've heard of is the reaction between a strong acid and a strong base. It is always an irreversible process, and you could make it quasistatic by adding very small drops of base into acid at a time. But by trying to do that, you would inevitably reach a limit to the size of the drop imposed by surface tension.  Even if "reversible" and "quasistatic" mean very different things, is it true to consider that in practice, a reversible process and a quasistatic process is essentially the same thing ? Working with large numbers in PHP How can I make an overset symbol at the same size and height as \bar except with a different symbol? Apologies for the cryptic title, but I was not sure how best to phrase the question in one sentence. I am essentially trying the create a vector symbol with \rightharpoonup as the top instead of the arrow.  The output to the following code will feature an overset harpoon over x, next to a simple \bar{x}. Observe that on the left, the top symbol is significantly higher and bigger than on the right one.  \documentclass{article}  \usepackage{amsmath,amsfonts,amsthm,amssymb}  \newcommand{\vect}[1]{\overset{\rightharpoonup}{#1}}  \begin{document} $$ \vect{x} \quad \bar{x} $$ \end{document}   I would like to create something that looks in every way like the right hand symbol, except using something that looks like \rightharpoonup at the top.  If this command sequence already exists in LaTeX, please use that as an answer, because I have searched for and failed to find it so far. Is Site Traffic Data Gone for Good After Fixing Previous Bug? There was a previous bug affecting site traffic statistics showing -zero- for months (). This has been fixed in that it is now graphing the data.  It appears that previous data may be lost? The graphs below are from . I don't have access to any other sites' graphs to see if they are showing similar results. Sometimes writing braces and omitting them produces the same result:  \tilde{\psi}  \mathrm{e}^{\mathrm{i} \pi}   vs  \tilde \psi   \mathrm e^{\mathrm i \pi}   Or (taken from Werner's comment below)   \frac{1}{2} vs \frac12   If that's the case, how bad for the quality of my code is omitting them? Of course, if one compiles a pair of formulas, there is no difference. But, since I ignore how TeX works, my point is: if one writes long codes with these habits,   doesn't TeX go somehow crazy because I'm not writing as it expects?   For instance, does it have any effect on how fast are long documents compiled? Any reason to keep them? Black holes are thought to originate when a star burns out and collapses to a smaller physical size. This compresses it's matter, but would that increase it's gravity? And would it be enough to capture light? If that star's light could have been observed prior to it's collapse, how could it's burning out and collapse enable it to become a black hole and capture light? Case-insensitive attribute-value selector with Jquery Why was Sirius ever convicted? Sirius himself says at some point in Goblet of Fire that Crouch pushed through legislation to have him convicted and thrown in Azkaban without a trial.  There seems to be a general consensus, specifically voiced by Fudge but also implicit in Crouch pushing through this legislation, that Sirius was considered one of Voldemort’s innermost, most dangerous, and most trusted right hands. Easily as inner-circle as the Lestranges, for example.  It seems quite unlikely to me that someone so close to Voldemort should not be a Death Eater—apart from Fenrir Greyback (who I would think is much less inner-circle), I do not recall hearing of any of Voldemort’s important servants not being Death Eaters.  As far as canon describes, all Death Eaters have the Dark Mark magic-tattooed on their arm, to be used as a way for Voldemort to summon them to him. Considering that Karkaroff, who had definitely burnt all his Death Eater bridges, still had the Dark Mark, I am assuming that getting rid of the Mark is not possible: once it’s on, it stays on.  Yet, as Hermione points out in Order of the Phoenix (ch. 16, “In the Hog’s Head”), Sirius does not have the Dark Mark.  So why was he so swiftly convicted and locked up? I mean, I know they were eager to be seen to be doing something (as Fudge keeps pointing out is his excuse for most of the miscarriages of justice he commits)… but seriously, did no one think to check if Sirius actually had a Dark Mark on his arm or not? Did the Ministry of Magic not even know at the time that all Death Eaters had the Mark? Or were they just happy to accept—without the scrutiny of a full trial—that a non-Death Eater would commit mass murder for a Voldemort who was presumed defeated? Possessive 's for things How do I tell if ACR will work with my camera and how do I get the correct version for Photoshop? I am a beginner photographer with a Canon 7d and MacBook Pro. I'm not sure if Adobe camera Raw is compatible with either and really need advice on how to go about downloading the correct version. I did a fresh install of Ubuntu 16.04. /dev/md1 contains an ext4 filesystem that is mounted at /:  /dev/md1 on / type ext4 (rw,relatime,data=ordered)   Now I want to set the automatic filesystem check interval to one month, but I'm getting a strange error that I find no useful information about:  &gt; tune2fs -i 1m /dev/md1 tune2fs 1.42.13 (17-May-2015) tune2fs: Filesystem has unsupported read-only feature(s) while trying to open /dev/md1 Couldn't find valid filesystem superblock.   Rebooting didn't help. On another machine running Ubuntu 12.04, the same command works like a charm.  Does anyone have an idea what's wrong?  The command tune2fs -l /dev/md1 works fine, by the way. Here is its output:  tune2fs 1.42.13 (17-May-2015) Filesystem volume name:   &lt;none&gt; Last mounted on:          / Filesystem UUID:          38366a40-bbe1-476b-acfe-d271be6335eb Filesystem magic number:  0xEF53 Filesystem revision #:    1 (dynamic) Filesystem features:      has_journal ext_attr resize_inode dir_index filetype needs_recovery extent 64bit flex_bg sparse_super large_file huge_file dir_nlink extra_isize metadata_csum Filesystem flags:         signed_directory_hash Default mount options:    user_xattr acl Filesystem state:         clean Errors behavior:          Continue Filesystem OS type:       Linux Inode count:              8380416 Block count:              33521664 Reserved block count:     1676083 Free blocks:              32419430 Free inodes:              8312762 First block:              0 Block size:               4096 Fragment size:            4096 Group descriptor size:    64 Reserved GDT blocks:      1024 Blocks per group:         32768 Fragments per group:      32768 Inodes per group:         8192 Inode blocks per group:   512 Flex block group size:    16 Filesystem created:       Thu May 31 10:01:27 2018 Last mount time:          Thu May 31 10:49:41 2018 Last write time:          Thu May 31 10:15:14 2018 Mount count:              7 Maximum mount count:      -1 Last checked:             Thu May 31 10:01:27 2018 Check interval:           0 (&lt;none&gt;) Lifetime writes:          5782 MB Reserved blocks uid:      0 (user root) Reserved blocks gid:      0 (group root) First inode:              11 Inode size:               256 Required extra isize:     32 Desired extra isize:      32 Journal inode:            8 First orphan inode:       4325378 Default directory hash:   half_md4 Directory Hash Seed:      9d55d989-2c2b-4ad8-a29e-53c8798ec68e Journal backup:           inode blocks Checksum:                 0x3d9f8d23 Math people:  The title is the question. I am convinced that uncountable sets exist, thanks to Cantor's diagonal proof.  It is not intuitively clear to me that uncountable ordinals or cardinals should exist.  What axioms of set theory are needed to accept their existence?  I think I have read here that the Axiom of Choice is not necessary.  I am reluctant to accept as "true" any theorems that rely on the Axiom of Choice.  EDIT: This question may be a duplicate: I just found  .  I apologize if this is so. With biblatex, is it possible to replace one author of an author list by a bibnamedash or idem string? With biblatex, the author can be replaced by a dash (or “idem”) in the bibliography. Is it possible to have this replacement also if there are multiple authors, and the first of these multiple authors is the same as the author of the book listed before? (It is somewhat difficult to describe, see the attached example instead, where “Buchautor, Hans-Wilhelm” should be replaced in book3 as well.)   I assume that it should be possible by customising \DeclareNameFormat, but I don’t know how.  Minimal example:  \documentclass{scrartcl} \usepackage[T1]{fontenc} \usepackage[latin1]{inputenc} \usepackage[english]{babel} \usepackage{csquotes} \usepackage{filecontents} \begin{filecontents}{\jobname.bib} @BOOK{book1,   author = {Buchautor, Hans-Wilhelm},   title = {Irgendein Buch},   location = {Buch am Wald},   date = {2000} } @BOOK{book2,   author = {Buchautor, Hans-Wilhelm},   title = {Ein anderes Buch},   location = {Berlin},   date = {2001} } @BOOK{book3,   author = {Buchautor, Hans-Wilhelm and Zweitautor, Zoran},   title = {Noch ein Buch},   location = {London and New York},   date = {2010} } \end{filecontents}  \usepackage[ style=authortitle, backend=bibtex8 ]{biblatex} \bibliography{\jobname}  \renewcommand*{\bibnamedash}{%   \bibstring[\mkbibbold]{idem\thefield{gender}}}  \begin{document} \nocite{*}  \printbibliography \end{document}   Edit 15/02/2011 17:14   I should have added that there was a feature request for biber which pointed back to biblatex. See .  Edit 21/06/2011 9:26  Is this somehow possible now with biblatex 1.4/1.5? I played a bit with uniquename/uniquelist, but I could not get it to work. System freezes completely with Intel Bay Trail My Android device won't connect. I've never got MTP to work, I've only ever got it to connect by USB mass storage, which was how it was working most recently. But now, since 13.04, it seems that option has been taken away entirely.  I need to be able to connect by USB to transfer files.  When I plug in the device I get an error that says Unable to open MTP device '[usb:002,003]':    After some web searching, I found that I might need a program called go-mtpfs, so I tried that, and it failed:  $ go-mtpfs /media/MyAndroid 2013/07/16 16:41:43 compiled against libmtp 1.1.5 Device 0 (VID=04e8 and PID=685e) is a Samsung Galaxy Y. 2013/07/16 16:41:43 found device Samsung: Galaxy Y (04e8:685e) @ bus 2, dev 3 :  LIBMTP PANIC: Unable to find interface &amp; endpoints of device 2013/07/16 16:41:43 rdev.open failed: open: open returned nil   How do I get my USB device to just connect, automatically, like it used to? Skipping acquire of configured file 'main/binary-i386/Packages' as repository 'xxx' doesn't support architecture 'i386' Getting below error message while issuing :  sudo apt-get update  Get:1 http://us.archive.ubuntu.com/ubuntu xenial InRelease [95.8 kB] Ign:2 http://repo.mongodb.org/apt/debian wheezy/mongodb-org/3.2 InRelease                                                                                                             Ign:3 http://dl.google.com/linux/chrome/deb stable InRelease                                                                                                                          Hit:4 http://ppa.launchpad.net/canonical-x/vulkan/ubuntu xenial InRelease                                                                                                 Hit:5 http://repo.mongodb.org/apt/debian wheezy/mongodb-org/3.2 Release                                                                             Hit:6 http://us.archive.ubuntu.com/ubuntu xenial-security InRelease                                                   Ign:7 http://dl.google.com/linux/talkplugin/deb stable InRelease                                                      Hit:8 http://us.archive.ubuntu.com/ubuntu xenial-updates InRelease                                                    Ign:9 http://linux.dropbox.com/ubuntu wily InRelease                                            Hit:10 http://ppa.launchpad.net/numix/ppa/ubuntu xenial InRelease                               Get:12 http://dl.google.com/linux/chrome/deb stable Release [782 B]       Hit:13 http://dl.google.com/linux/talkplugin/deb stable Release                   Ign:14 https://apt.dockerproject.org/repo ubuntu-wily InRelease           Hit:15 https://apt.dockerproject.org/repo ubuntu-wily Release             Get:16 http://dl.google.com/linux/chrome/deb stable Release.gpg [181 B]   Hit:17 http://linux.dropbox.com/ubuntu wily Release                                   Get:20 http://dl.google.com/linux/chrome/deb stable/main amd64 Packages [1,191 B] Fetched 98.0 kB in 0s (118 kB/s)                                  Reading package lists... Done N: Skipping acquire of configured file 'main/binary-i386/Packages' as repository 'http://dl.google.com/linux/chrome/deb stable InRelease' doesn't support architecture 'i386'   This cuts across various releases of ubuntu Prove inequality using Schwarz's Lemma Typeset an = with an ! above I've seen an equals sign with a small exclamation mark on top being used to say "equals because of data that was given in the problem", for example for a boundary condition or initial value in a differential equation.  A rough picture of what I mean:       ! f(x) = 4   How do I typeset this in LaTeX? I've tried $=^!$ (which doesn't produce the desired result) and looked through the "relations" section of the  without success.