I am currently working on the sketch below using pfgplots.   \begin{tikzpicture}   \begin{axis}[     hide axis,     scale=2,     view={120}{40},     xmin=-4,xmax=4,     ymin=-4,ymax=4,     zmin=-2,zmax=10,     trig format plots=rad,   ]   \addplot3 [ surf, colormap/bone, domain=-4:4, domain y=-4:4,               samples=30, samples y=30,               variable=\u, variable y=\v,               point meta=u*v]             ( {u}, {v}, {cos(u) + cos(v)} );    \addplot3[black, dashed, thick, variable=\t, domain=-1:1] ( {1}, {t}, {cos(1) + cos(t) - 2});   \addplot3[black, dashed, thick, variable=\t, domain=-1:1] ( {-1}, {t}, {cos(1) + cos(t) - 2});   \addplot3[black, dashed, thick, variable=\t, domain=-1:1] ( {t}, {1}, {cos(1) + cos(t) - 2});   \addplot3[black, dashed, thick, variable=\t, domain=-1:1] ( {t}, {-1}, {cos(1) + cos(t) - 2});    \addplot3 [ surf, domain=-1:1, domain y=-1:1,               samples=30, samples y=30,               variable=\u, variable y=\v,               point meta=u*v]             ( {u}, {v}, {cos(u) + cos(v)} );    \addplot3[black, thick, variable=\t, domain=-1:1] ( 1, t, {cos(1) + cos(t) + 2});   \addplot3[black, thick, variable=\t, domain=-1:1] ( -1, {t}, {cos(1) + cos(t) + 2});   \addplot3[black, thick, variable=\t, domain=-1:1] ( {t}, 1, {cos(1) + cos(t) + 2});   \addplot3[black, thick, variable=\t, domain=-1:1] ( {t}, -1, {cos(1) + cos(t) + 2});    \addplot3[black, thick, variable=\t, domain=-1:1] ( 1, {t}, {cos(1) + cos(t)});   \addplot3[black, thick, variable=\t, domain=-1:1] ( -1, {t}, {cos(1) + cos(t)});   \addplot3[black, thick, variable=\t, domain=-1:1] ( {t}, 1, {cos(1) + cos(t)});   \addplot3[black, thick, variable=\t, domain=-1:1] ( {t}, -1, {cos(1) + cos(t)});    \addplot3[black, thick, variable=\t, domain=0:2] ( 1, 1, {2*cos(1) + t});                \addplot3[black, thick, dashed, variable=\t, domain=0:2] ( 1, 1, {2*cos(1) - t});    \addplot3[black, thick, variable=\t, domain=0:2] ( 1, -1, {2*cos(1) + t});                \addplot3[black, thick, variable=\t, domain=0:2] ( 1, -1, {2*cos(1) - t});    \addplot3[black, thick, variable=\t, domain=0:2] ( -1, 1, {2*cos(1) + t});                \addplot3[black, thick, variable=\t, domain=0:2] ( -1, 1, {2*cos(1) - t});    \addplot3[black, thick, variable=\t, domain=0:2] ( -1, -1, {2*cos(1) + t});                 \end{axis} \end{tikzpicture}   The black lines should represent a "box" that goes across the surface. The problem is that when I draw these black curves with \addplot3 I would like to get the "arch" only, but it draws a line between the endpoints as well so I end up with all these straight lines in my plot. In addition, it does not draw dashed curves. As to draw the blue surface I took an example from another post, I tried playing with the options of the axis to see which of them were necessary and if they fixed my problem, but I didn't succeed.  Thank you very much in advance!
I'm working on creating an illustration that I will need in many, different guises in differential geometry/manifold theory etc. Following on from a previous (answered) question of mine (, I want to show how two coordinate lines in the plane originate from two curves on a surface via the chosen map:  \begin{tikzpicture}     \begin{axis}[             name=mfd,             declare function={                     f(\x,\y)=10-(\x^2+\y^2);             },                   declare function={                     c_x(\t)=(cos(\t)+(sin(5*\t)/10))/3+1;             },                   declare function={                     c_y(\t)=(sin(\t))/2-1;             },                   declare function={                     c_z(\t)=f(c_x(\t),c_y(\t));             },           ]                    \addplot3[surf,domain=-2:2,domain y=-2:2,]{f(x,y)};             \addplot3[black,opacity=1.0,variable=t,domain=0:360,dashed,thin] ({c_x(t)},{c_y(t)},{c_z(t)});             \addplot3[black,opacity=1.0,variable=t,domain=-2:2,thin] (t,-1.2,{f(t,-1.2)});             \addplot3[black,opacity=1.0,variable=t,domain=-2:2,thin] (0.8,t,{f(t,-1.2)});             \addplot3[black,opacity=1.0,only marks,mark=text,text mark=$\cdot$] (1,-1,{f(1,-1)}) coordinate (a);     \end{axis}     \begin{axis}[             at={($(mfd.north east)+(1cm,-2cm)$)},             anchor=north west,             declare function={                     c_x(\t)=(cos(\t)+(sin(5*\t)/10))/3+1;             },                   declare function={                     c_y(\t)=(sin(\t))/2-1;             }            ]                    \addplot[variable=t,domain=0:360]({c_x(t)},{c_y(t)});             \addplot[black,opacity=1.0,only marks,mark=text,text mark=$\cdot$] (1,-1) coordinate (b);             \addplot[variable=t,domain=0.6:1.4](t,-1.2);             \addplot[variable=t,domain=-1.5:-0.5](0.8,t);     \end{axis}     \draw [-stealth, shorten &lt;= 3pt, shorten &gt;= 3pt] (a) to[bend left] (b); \end{tikzpicture}   However, I have a couple of problems:   The curves representing the coordinate axes pulled back from the plane via the coordinate map insist on being closed off with a straight line - is there a way to avoid that? I want to draw all the black curves with a thin, dashed line, but I probably don't understand how to set those options. What am I doing wrong?   I apologise if these questions are rather elementary, but I find the relevant manuals difficult to search through; they seem to be meant for reading cover to cover rather than reference.
Can I get cheaper plane tickets by using a VPN? In a Youtube video by ADVChina, the speaker is recommending using a VPN in China for various reasons, and to sign up using his referral code.  He says there are various practical uses including "You could also go and get cheaper plane tickets by hopping around different countries." and shows a screenshot of a flight website called KAYAK, with the VPN software changing his country to Italy.  Is it possible to get cheaper plane tickets by using a VPN in this way? I've never heard of KAYAK, but if I use something like Skyscanner, is it common for tickets to be priced differently depending on which country the user's IP is from?  Is it because of different taxes in different countries? So could it be illegal to get discounts like this? Or is there some pricing strategy from airlines to charge different amounts depending on the country of purchase? How to install Eclipse? After deleting all nodes, how do I reset the node ID to start from 1 again? I generated tons of dummy content for testing, and then deleted them all.  As nid is auto-increment, newly created node will have a nid of last created nid +1.  How can I make nid start from 1 again? Take for example  both at SO and SU. Now that there's , (new) users asking a LaTeX question should at least be informed that there is a new Stack Exchange site dedicated to TeX and thus increases their chance of an answer. This should happen before the question is actually posted and ideally offer the option to post the question directly there instead (without the need to copy-paste).  LaTeX is just an example, there's also , ,  etc., and it is obvious that this can only be decided on a case-by-case basis which is why I certainly do not suggest migrating questions automatically when these tags are used. But informing users of that possibility would probably help the SE network in general. I'm trying to migrate a MySQL-based app over to Microsoft SQL Server 2005 (not by choice, but that's life).  In the original app, we used almost entirely ANSI-SQL compliant statements, with one significant exception -- we used MySQL's group_concat function fairly frequently.  group_concat, by the way, does this: given a table of, say, employee names and projects...  SELECT empName, projID FROM project_members;   returns:  ANDY   |  A100 ANDY   |  B391 ANDY   |  X010 TOM    |  A100 TOM    |  A510   ... and here's what you get with group_concat:  SELECT      empName, group_concat(projID SEPARATOR ' / ')  FROM      project_members  GROUP BY      empName;   returns:  ANDY   |  A100 / B391 / X010 TOM    |  A100 / A510   So what I'd like to know is: Is it possible to write, say, a user-defined function in SQL Server which emulates the functionality of group_concat?   I have almost no experience using UDFs, stored procedures, or anything like that, just straight-up SQL, so please err on the side of too much explanation :) This is an idea I have seen around, but I have not seen a feature request for it, so here it is.  Downvotes that have no comments are frustrating and difficult to manage.  This is a way to help improve that:   Have a downvote initially cost more to cast for the downvoter (5 points for example) When a comment (from the downvoter) on the downvoted question/answer gets an upvote, you get your points back.   All points (ie all 5) could be refunded or just up till the traditional -1 that is in place now (ie 4 points refunded).    This would encourage good comments and downvotes for real reasons.  (Because comments for questions/answers that are bias related will not get the upvotes as readily.)  This could/should be limited to non-subjective tags.  (ie tags where a downvote just because you "don't like it" is not appropriate.) Cyberpunk/Post-Apocalyptic Sci-Fi series from the 90s or 00s based in Manhattan A generic list of anonymous class In C# 3.0 you can create anonymous class with the following syntax  var o = new { Id = 1, Name = "Foo" };   Is there a way to add these anonymous class to a generic list?  Example:  var o = new { Id = 1, Name = "Foo" }; var o1 = new { Id = 2, Name = "Bar" };  List&lt;var&gt; list = new List&lt;var&gt;(); list.Add(o); list.Add(o1);   Another Example:  List&lt;var&gt; list = new List&lt;var&gt;();  while (....) {     ....     list.Add(new {Id = x, Name = y});     .... } What are logic gates? Problems with booting the system I have yesterday updated packages and today the system cannot boot properly. After seeing the Kubuntu logo there is a black screen which shows repeatedly.     Stopped User Manager for UID 121.      Removed slice User Slice of gdm.      Created slice User Slice of gdm.       Starting User Manager for UID 121...      Started Session c3 of user gdm.       Started User Manager for UID 121.   Please see the screenshot below How do I get the current location of an iframe? Restricted logins with automatic full permission in databases My scenario: I have some groups with full - and inappropriate - access to all Sql Server instances. I want to change that.  My Plan: Give full and automatic access to all databases (it is customer databases and this groups must be allowed to change any property if necessary).  Restrict permissions on Instance Level, because this groups should not be allowed to do any unexpected change on instance properties.  I don't want to create a user in databases, because our external customer could drop the user ( they have db_owner permission )  Someone have any idea? Tks I'm flying to Tokyo. The flight was sold through Finnair. The first leg is operated by American Airlines. The second and final legs are operated by Finnair.  When I checked in online, it said that some airports on my journey (didn't say which one) don't allow mobile boarding passes?? Meaning I will need a physical copy.  Unfortunately I don't have easy access to a printer. Will I still be able to print all my boarding passes at the airport even though I already checked in online? proof that translation of a function converges to function in $L^1$ Let $f \in L^1(\mathbb{R})$, for $a\in \mathbb{R}$ let $f_a(x)=f(x-a)$, prove that: $$\lim_{a\rightarrow 0}\|f_a -f \|_1=0$$  I know that there exists $g\in C(\mathbb{R})$ s.t $\|f-g\|_1 \leq \epsilon$, this is also true for $f_a$ and $g_a$.  Now I have the next estimation:  \begin{align*} \|f_a-f\|_1 &amp;= \int_{\mathbb{R}} |f(x)-f(x-a)| dx \\ &amp;= \int |f(x)-g(x)+g(x)-g(x-a)+g(x-a)-f(x-a)| \\ &amp;\leq \|f-g\|_1 + \|f_a-g_a\|_1 + \int |g(x)-g(x-a)| \end{align*} my question is: I can argue that $\lim_{a\to0} \int |g(x)-g(x-a)| = \lim_{a\to0} \lim_{T\to\infty} |g(x_0)-g(x_0-a)| 2T$, now I think that I can change the order of the limits here, but I am not sure why?  P.s $x_0$ is some point in $[-T,T]$, and the above is valid from the intermediate integral theorem for continuous functions, right?  Thanks. Number of triangles formed by $m$ lines Problem:    How many triangles do $m$ lines form if   a) Every two lines intersect and no three lines intersect at one point.  b) There are $n$ lines among $m$ lines that are parallel to each other. No other line is parallel to these $n$ lines, and no other two lines are parallel to each other. Again no three lines intersect at one point.  Thank you. This is a  about  Related:      I have a question regarding capacity planning. Can the Server Fault community please help with the following:   What kind of server do I need to handle some number of users? How many users can a server with some specifications handle? Will some server configuration be fast enough for my use case? I'm building a social networking site: what kind of hardware do I need? How much bandwidth do I need for some project? How much bandwidth will some number of users use in some application? Permission Denied To Delete File First of all, I'm not looking for an answer on how to remove the files. I am looking for a reason why this doesn't work - and a suggestion on how to remove the files without sudo.  I have a directory in my Desktop that has a few inner directories (don't judge the naming - I just extract what I need from files I'm given) and after getting the files I try and remove the files with rm -rf Bengal\ Tiger\ Creative/, I am surprised to see the following warning for each image.     rm: cannot remove ‘Bengal Tiger Creative/Bengal Tiger Creative/images/customer-services-landscape.jpg’: Permission denied   Below is some code outputs.  tree  nathan@nathan-pc:~/Desktop$ tree . ├── Bengal Tiger Creative │   └── Bengal Tiger Creative │       └── images │           ├── awards.jpg │           ├── blank.gif │           ├── bullet1-light-green.gif │           ├── bullet2-light-green.gif │           ├── bullet3-light-green.gif │           ├── bullet4-light-green.gif │           ├── customer-services-landscape.jpg │           ├── desktop.ini │           ├── header-img1.jpg │           ├── header-img2.jpg │           ├── header-img4.jpg │           ├── header-img5.jpg │           ├── softhr.gif │           ├── Thumbs.db │           ├── tiger_main-img1.jpg │           ├── tiger_main-img2.jpg │           ├── tiger_main-img3.jpg │           ├── tiger_main-img4.jpg │           ├── Vanquis-benefits-icons_iphone-green.gif │           └── white-arrow-green.gif   ls -laR  nathan@nathan-pc:~/Desktop$ ls -laR Bengal* Bengal Tiger Creative: total 12 drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:30 . drwxr-xr-x 6 nathan nathan 4096 Aug 10 11:34 .. drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:34 Bengal Tiger Creative  Bengal Tiger Creative/Bengal Tiger Creative: total 12 drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:34 . drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:30 .. dr-xr-xr-x 2 nathan nathan 4096 Aug 10 11:36 images  Bengal Tiger Creative/Bengal Tiger Creative/images: total 356 dr-xr-xr-x 2 nathan nathan   4096 Aug 10 11:36 . drwxrwxr-x 3 nathan nathan   4096 Aug 10 11:34 .. -rwxrwxrwx 1 nathan nathan  14910 Mar 18 12:29 awards.jpg -rw-rw-r-- 1 nathan nathan   1093 Mar 21  2014 blank.gif -rw-rw-r-- 1 nathan nathan   1978 May  8  2014 bullet1-light-green.gif -rw-rw-r-- 1 nathan nathan   1969 May  8  2014 bullet2-light-green.gif -rw-rw-r-- 1 nathan nathan   2416 May  8  2014 bullet3-light-green.gif -rw-rw-r-- 1 nathan nathan   2449 May  8  2014 bullet4-light-green.gif -rw-rw-r-- 1 nathan nathan 113138 Apr 14  2014 customer-services-landscape.jpg -r--r--r-- 1 nathan nathan    142 Mar 25 11:58 desktop.ini -rw-rw-r-- 1 nathan nathan  14312 Mar 23 11:39 header-img1.jpg -rw-rw-r-- 1 nathan nathan   1480 Mar 23 11:41 header-img2.jpg -rw-rw-r-- 1 nathan nathan   1278 Mar 23 11:41 header-img4.jpg -rw-rw-r-- 1 nathan nathan   1168 Mar 23 11:41 header-img5.jpg -rw-rw-r-- 1 nathan nathan   1703 Mar 21  2014 softhr.gif -rw-rw-r-- 1 nathan nathan  88064 Aug  4 10:39 Thumbs.db -rw-rw-r-- 1 nathan nathan  23607 Mar 23 11:56 tiger_main-img1.jpg -rw-rw-r-- 1 nathan nathan  23400 Mar 23 11:56 tiger_main-img2.jpg -rw-rw-r-- 1 nathan nathan   6970 Mar 23 11:56 tiger_main-img3.jpg -rw-rw-r-- 1 nathan nathan   8996 Mar 23 11:57 tiger_main-img4.jpg -rw-rw-r-- 1 nathan nathan    275 Jul 31  2014 Vanquis-benefits-icons_iphone-green.gif -rw-rw-r-- 1 nathan nathan   1821 Jul  9  2014 white-arrow-green.gif   Doing chmod 777 and then delete does not work.  Doing rm -rf * inside the directory does not work.  Doing sudo rm file does work.   I can easily remove these files using sudo, however, I cannot see why they cannot be removed without sudo. As a relatively new Ubuntu user, I would like to learn why.  Thanks in advance, Nathan Why does the set of all singleton sets not exist? Checking if file is checked out through CSOM Is there a way to check if a file is checked out by another user? I am going trhough a list file by file and would like to avoid the "file is opened by another user" error. Something like this would be nice:  Guid id = new Guid(slideLibrary); // ID of Slide-Library. List list = ctx.Web.Lists.GetById(id); ListItemCollection items = list.GetItems(CamlQuery.CreateAllItemsQuery()); ctx.Load(items); // Loading all the fields System.Security.SecureString pwd = new System.Security.SecureString(); foreach (char c in "pwd".ToCharArray()) pwd.AppendChar(c);  ctx.Credentials = new SharePointOnlineCredentials(login, pwd); ctx.ExecuteQuery();  // Iterating through all the entries. foreach (var item in items) {     String currentID = item["ID"].ToString(); // Getting the ID of the current item.      if (!item.IsCheckedOut()) // This is the part I am looking for.     {        // Do stuff     }     else     {        Console.WriteLine("Item is blocked, skipping it.")     } }   I have yet not been able to find something fitting. Can you trip as part of a Tome of Battle Maneuver?