I've tried to create German quotation marks in LaTeX, but after compiling with \usepackage[ngerman]{babel}, I get this error:   ! Package babel Error: Unknow option `ngerman'. Either you misspelled it (babel) or the language definition file ngerman.ldf was not found.   The solution  "Partei"' doesn't work at all and the alternative \glqq only works with the \usepackage[ngerman]{babel}...   This is my code:   \documentclass[12pt,a4paper,oneside]{scrartcl} \usepackage[latin1]{inputenc} %\usepackage[ngerman]{babel} \usepackage{graphics} \usepackage{multicol} \usepackage{hyperref} \renewcommand*{\UrlFont}{\normalfont} \usepackage{fancyhdr}   can anybody help me please? I have to turn in my paper next week...  Here's my "extended example":   I want to have German quotation marks. Normally, there are two possibilities to accomplish that:    either you do it this way: "`Partei"' and compile it.  Unfortunately, this does not work any more (it has worked for my last paper, but I didn't change anything in my TeXmaker) you can write: \glqq Partei\grqq and then you get German quotation marks in case you have installed the \usepackage[ngerman]{babel} or \usepackage{german}, but with both usepackaged I get the error described above.    Normally, I work with TeXmaker on Linux Ubuntu.   Do you need any more information? I'm not quite an expert on this field, sorry...
I recently updated my operating system from Ubuntu 13.04 to Ubuntu 13.10. Since then, I am facing difficulties compiling my TeX code in which the french babel package is called. Please note that all my sources were compiling very fine before the system upgrade. However, now, given the following header:   \usepackage[francais]{babel}  I receive the following error:  ! Package babel Error: Unknown option `francais'. Either you misspelled it   What should I do? Please note that if I remove this part of code (\usepackage[francais]{babel}), everything is compiling well. I also tried it with french, and I receive the same result. Thank you in advance for your help.  MINIMUM WORKING EXAMPLE : \documentclass[11pt,a4paper]{article} \usepackage[utf8]{inputenc} \usepackage[francais]{babel} \usepackage[T1]{fontenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb}  \begin{document} This is a MINIMUM WORKING EXAMPLE. \end{document}   ! Package babel Error: Unknow option `francais'. Either you misspelled it
Is every complex (smooth) manifold a scheme? The question in the title doesn't quite make sense. I was always wondering if the scheme is the generalization of manifold. The precise statement should be like following:  If $X$ is a complex (smooth)manifold, is there a scheme $Y$ over $\mathbb{C}$ such that the corresponding analytic space $Y^{an}$ isomorphic to $X$?  I guess if $X$ is projective, then this is Chow's theorem. And I doubt the question has an affirmative answer if $X$ is non-compact (I guess that $X: y-e^x=0$ in $\mathbb{C}^2$ might be an example, but I do not know how to show this).  Edit 1:  There are also example (as @hunter pointed out) of a complex torus but not an abelian variety shows that a compact manifold might not be a variety, however, is it possible to be a scheme ?(according to Hartshorne, the category of varieties is a subcategory of quasi-projective, integral, separated scheme. But could there be some non-quasiprojective scheme($Y$ has to be integral, separated in our case according to GAGA, I guess) )?  Edit 2: Another question I think of is: is (Artin) Stack a generalization of manifold? I really want to know if manifold could be think of some subcategory of some category of with "algebraic objects". Function $f(z)$ such that $|f(z)| > |z|$ for every $z\in\mathbb{C}, 1/f(z)$ $f(z)$ is a function such that $|f(z)| &gt; |z|$ for all $z\in\mathbb{C}$  How would you prove the function $\frac{1}{f(z)}$ is bounded. I think you have to use the $|f(z)|\leq M$ but I'm not sure. No matter how I do it, I always end up with $x = 0, 90, 270$ and $360$. All of those except $270$ is right, but I can't quite figure out how to get the $270$ degrees out of the answer. I've tried using trig identities, I've tried squaring both sides, but I always end up with $$2\sin x\cos x$$ which then leads me to $x = 0, 90, 270, 360$. But $$\sin (270) + \cos (270) = -1$$ so I'm doing something wrong. What does this red icon on my panel mean? This red icon.... It showed up after I upgraded to Ubuntu 12.04LTS; I can't figure out how to ditch it. It seems to be a notification thing.  I am at a loss.  The icon is this one:      I clicked it and read this:     An error occurred, please run Package Manager from the right-click menu or apt-get in a terminal to see what is wrong. The error message was:  'Error: BrokenCount&gt;0'.        This usually means that your installed packages have unmet dependencies. XeLaTeX update [20201004] and item ball in Beamer How Can I Expand A Tilde ~ As Part Of A Variable? When I open up a bash prompt and type:  $ set -o xtrace $ x='~/someDirectory' + x='~/someDirectory' $ echo $x + echo '~/someDirectory' ~/someDirectory   I was hoping that the 5th line above would have went + echo /home/myUsername/someDirectory.  Is there a way to do this?  In my original Bash script, the variable x is actually being populated from data from an input file, via a loop like this:  while IFS= read line do     params=($line)     echo ${params[0]} done &lt;"./someInputFile.txt"   Still, I'm getting a similar result, with the echo '~/someDirectory' instead of echo /home/myUsername/someDirectory. I need to get MODIS NDVI (MOD13A3), but output from EarthExplorer or Daac2Disk HDF format is a numerical range is -2000 to 9996.   But I need for futher work basic range -1 to 1.   Is there any conversion method? Is normality testing 'essentially useless'? A former colleague once argued to me as follows:      We usually apply normality tests to the results of processes that,   under the null, generate random variables that are only   asymptotically or nearly normal (with the 'asymptotically' part dependent on some quantity which we cannot make large); In the era of   cheap memory, big data, and fast processors, normality tests should   always reject the null of normal distribution for large (though not insanely large) samples. And so, perversely, normality tests should   only be used for small samples, when they presumably have lower power   and less control over type I rate.   Is this a valid argument? Is this a well-known argument? Are there well known tests for a 'fuzzier' null hypothesis than normality? Move rigidbody using WASD keys but rotate it based on camera/mouse movement I am very new to unity and my scripts are based on some scripts found on the web.   I want to make a first person view so the camera is attached to my rigidbodys head. Then i want to move my rigidbody based on keyboard input (wasd) and rotate it using the mouse movement.  Until now, i used the following code for moving my rigidbody:  void Update() {      /* calculate rigidbody movement */     movement = new Vector3( Input.GetAxisRaw("Horizontal"), 0, Input.GetAxisRaw("Vertical")).normalized;      /* calculate mouse look to rotate the rigidbody*/     mouseDelta = new Vector2(Input.GetAxisRaw("Mouse X"), Input.GetAxisRaw("Mouse Y"));     mouseDelta = Vector2.Scale(mouseDelta, new Vector2(sensitivity * smoothing, sensitivity * smoothing));     smoothV.x = Mathf.Lerp(smoothV.x, mouseDelta.x, 1f / smoothing);     smoothV.y = Mathf.Lerp(smoothV.y, mouseDelta.y, 1f / smoothing);     mouseLook += smoothV;  }  void FixedUpdate() {         /* apply rigidbody movement from Update() */     transform.Translate( movement * moveSpeed * Time.deltaTime );             /* apply rigidbody rotation from Update() */     transform.localRotation = Quaternion.AngleAxis(mouseLook.x, transform.up);             /* apply camera rotation from Update() */     Camera.main.transform.localRotation = Quaternion.AngleAxis(-mouseLook.y, Vector3.right);    }   I learned today that i dont have to move a rigidbody's transform but now i am totally confused and it seems i have to restart my whole (first!) project.  So when i use MovePosition and MoveRotation, how do i rotate it based on the camera perspective/mouse movement?  I think the movement Vector is ok so i can simply rewrite the FixedUpdate to something like  rigidbody.MovePosition(movement * speed * Time.deltaTime)   but i am totally stuck on how to apply the camera rotation to my rigidbody to use it like  void FixedUpdate() {      /* move the rigidbody */                rigidbody.MovePosition(movement * moveSpeed * Time.deltaTime);     /* rotate the camera */     camera.transform.localRotation = Quaternion.AngleAxis(-mouseLook.y, Vector3.right);      /* but how can i rotate the rigidbody based on the cameras rotation??? */     rigidbody.MoveRotation(?????);  }   I hope its clear what i want, thank you! Difference between Ethernet splitter and switch I have one Ethernet port that is wired directly to the router on another level.  However, I want to run a desktop and a server both off this one Ethernet port.  Can anyone tell me the difference between an Ethernet splitter and a switch?    Also, will either the splitter or switch slow down the connection as opposed to just one connection? If $z$ lies on the circle $|z-1|=1,$ then the value of $\frac{z-2}{z}$ is If $z$ lies on the circle $|z-1|=1, $ then the what is the value of $\frac{z-2}{z}$ ($z\ne 0 \ \text{and} \ 2$ ) is?  What I am aware of : the equation $|z-1|=1$ represent a circle with  centre $(1,0)$ and radius $1$. Also $$|z-1|=1 \implies (z-1)^2=1 \\ \implies  z^2-2z+1=1 \implies z(z-2) =0$$ $\implies z = 0; z =2 $ I think this is wrong.. please suggest thanks.. Rosewill RNX-N600UBE Connectivity issue on Ubuntu 17.04 I assume the difference in the number of Stack Exchange sites/communities is a mistake... correct? On my Toshiba Encore 8 Tablet I noticed that the CPU usage is very high. In Taskmgr / ProcessExplorer, I can see that OneDriveSetup.exe is hogging 1 CPU cores heavily for hours.    How an I solve this hang and finish the OneDrive Update? Speed of light in a gravitational field? Find all continuous functions $f$ satisfying   $$\int_0^xf=(f(x))^2+C$$ for some constant $C \neq 0$, assuming that $f$ has at most one $0$.  I have a question about the solution to this problem. It says that   Clearly $f^2$ is differentiable everywhere, its derivative at $x$ is $f(x)$ (I'm assuming that this is because the left hand side is differentiable, the right hand side must be differentiable as well). So $f$ is differentiable at $x$ whenever $f(x) \neq 0$, and  $$f(x)=2f(x)f'(x),$$ so $f'(x) = \frac{1}{2}$ at such points. Thus, $f= \frac{1}{2}x+b$ for some $b$ on any interval where it is non-zero; if $f$ has a zero, this gives two possible solutions, with two possible values of $b$, but since $f$ is assumed continuous, they must be the same. So we need $$\int_0^x(\frac{1}{2}t+b)dt=(\frac{1}{2}x+b)^2+C$$  so we must have $b=\sqrt{-C}$ or $-\sqrt{-C}$, leading to two solutions for $C \lt 0$.  I don't understand the bolded statements. First, how does the differentiability of $f^2$ guarantee that $f$ is differentiable, and why only at $x$ where $f(x) \neq 0$?  Next, why are there two possible solutions if $f$ has a zero, but since $f$ is continuous they must be the same?  Finally, does the last statement mean that if $C \gt 0$, then there is no solution, since we're restricted to real numbers here?  I would appreciate it if anyone answers my questions. Let $(a_n)_{n}\in\ell^2:=\ell^2(\mathbb{R})$ be a fixed sequence. Consider the subspace $$C=\{(x_n)_{n}\in\ell^2 : |x_n|\le a_n\text{ for all }n\in\mathbb{N}\}.$$  According to the book [Dunford and Schwartz, Linear operators part I, page 453] $C$ is compact in the $\ell^2$-norm, but there is no proof. How can I show that $C$ is indeed compact in $\ell^2$ ? Why doesn't an Environment texture appear when the world has a Volume Scatter shader? I'm working on lighting a scene in which I need to have both an Environment texture for the background and Volume Scatter for light cones, but when I check the volume section in World setting and choose the Volume Scatter, the environment becomes black background (but the lighting works as I want).  I'm using Blender 2.72  How can this be fixed? Create model from XYZ data points I'm trying to create a 3D heat map model out of data that I have from a csv file. I have a few thousand data points with XYZ coordinates.  I'm able to import the data into blender as a point cloud with , however I would like to create a model with solid faces. Is there a way when importing the data to somehow join the points to create faces instead of generating thousands of separate points?  Currently, this what I have in terms of point cloud  The end product would look something like this (minus the texture) and obviously the model wouldn't be as regular as this one  Any help is greatly appreciated! Just posted a bug report, and found another one :p    (lame hand-drawn cursor coz windows doesn't capture the real cursor)  Oh yeah, I'm on Firefox.