\begin{eqnarray*}   % \nonumber % Remove numbering (before each equation)     (a, b) &amp;=&amp; \{ x \in X | a &lt; x &lt; b \} \\     (a, b] &amp;=&amp; \{ x \in X | a &lt; x \leq b \} \\     [a, b) &amp;=&amp; \{ x \in X | a \leq x &lt; b\} \\     [a, b] &amp;=&amp; \{ x \in X | a \leq x \leq b \}.   \end{eqnarray*}   What's error?   Missing number, treated as zero. &lt;to be read again&gt;                     a l.281     [a, b]                  &amp;=&amp; \{ x \in X \mid a \leq x \leq b \}.
I tried to find a topic here on this (possible problem) but I didn't.   I reported it at  but Tim Hoffmann pointed me that it is not a problem of editor. After tested with nano and pdflatex I'm posting here.  The minimal code produces errors. OK, we can solve it using \\{}.   But is this a problem with TeXlive 2014 or there are some good reasons to have that behavior.  \documentclass{report} \begin{document} test\\ [foo]  test\linebreak [foo]  test\newline [foo] \end{document}      ! Missing number, treated as zero.                        f l.3 test\ [foo]
Biblatex citation order I want to write batch or something, that will write output of top into a file on login.  I did top &gt;&gt; output-file, but it contains some strange character!  Can somebody give a simple tutorial about how to write batch file in Linux? Determine the number of functions $f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}$satisfying the condition that $f(1)+f(2)+...f(1999)$ is odd. Problems:Determine the number of functions $$f: \{1,2,3....,1999\}\to \{2000,2001,2002,2003\}$$ satisfying the condition that $f(1)+f(2)+...f(1999)$ is odd.   My Attempt: Each integer in domain has $4$ choices and therefore the total number of functions is $f$ is $4^{1999}.$ Since there are an equal number of functions that yield odd or even result, we can directly write that the number of functions satisfying the above condition to be $2*4^{1998}.$ I am unsure about this last claim and would like to prove it (if it is true.)Also is the answer to this problem correct?  Edit $1$: I tried to proceed further in the following manner:   For a given mapping $f$ let the number of integers assigned to $2000,2001,2002$ amd $2003$ be $p,q,r$ and $s$ respectively, where $p,q,r,s\in \mathbb{Z}^+$. Therefore the sum in question can be written as $f(1)+f(2)+...+f(1999)=2000p+2001q+2002r+2003s.$ Clearly the sum will be odd iff $q+s$  is odd. Also note that $p+q+r+s=1999\Rightarrow p+r$ is even. On the other hand if we let $q+s$ as even then $p+r$ must be even. Now the number of values of $p,q,r$ and $s$ for which the above two lemmas hold is the same and therefore if we let $f$ to be the function which maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $q+s$ is even and $g$ to be the function  hich maps elements form $\{1,2,3,...,1999\}$ to $\{2000,2001,2002,2003\}$ such that $p+q$ is even we will get a bijection.(Maybe!) Regression: Transforming Variables What is the format for e-mail headers that display a name rather than the e-mail? I think I remember reading somewhere that the Baire Category Theorem is supposedly quite powerful. Whether that is true or not, it's my favourite theorem (so far) and I'd love to see some applications that confirm its neatness and/or power.  Here's the theorem (with proof) and two applications:    (Baire) A non-empty complete metric space $X$ is not a countable union of nowhere dense sets.  Proof: Let $X = \bigcup U_i$ where $\mathring{\overline{U_i}} = \varnothing$. We construct a Cauchy sequence as follows: Let $x_1$ be any point in $(\overline{U_1})^c$. We can find such a point because $(\overline{U_1})^c \subset X$ and $X$ contains at least one non-empty open set (if nothing else, itself) but $\mathring{\overline{U_1}} = \varnothing$ which is the same as saying that $\overline{U_1}$ does not contain any open sets hence the open set contained in $X$ is contained in $\overline{U_1}^c$. Hence we can pick $x_1$ and $\varepsilon_1 &gt; 0$ such that $B(x_1, \varepsilon_1) \subset (\overline{U_1})^c \subset U_1^c$.   Next we make a similar observation about $U_2$ so that we can find $x_2$ and $\varepsilon_2 &gt; 0$ such that $B(x_2, \varepsilon_2) \subset \overline{U_2}^c \cap B(x_1, \frac{\varepsilon_1}{2})$. We repeat this process to get a sequence of balls such that $B_{k+1} \subset B_k$ and a sequence $(x_k)$ that is Cauchy. By completeness of $X$, $\lim x_k =: x$ is in $X$. But $x$ is in $B_k$ for every $k$ hence not in any of the $U_i$ and hence not in $\bigcup U_i = X$. Contradiction. $\Box$    Here is one application (taken from ):  Claim: $[0,1]$ contains uncountably many elements.  Proof: Assume that it contains countably many. Then $[0,1] = \bigcup_{x \in (0,1)} \{x\}$ and since $\{x\}$ are nowhere dense sets, $X$ is a countable union of nowhere dense sets. But $[0,1]$ is complete, so we have a contradiction. Hence $X$ has to be uncountable.     And here is another one (taken from ):  Claim: The linear space of all polynomials in one variable is not a Banach space in any norm.  Proof: "The subspace of polynomials of degree $\leq n$ is closed in any norm because it is finite-dimensional. Hence the space of all polynomials can be written as countable union of closed nowhere dense sets. If there were a complete norm this would contradict the Baire Category Theorem." food safety regarding re freezing twice cooked chicken Regression with only categorical variables Is it possible to conduct a regression if all dependent and independent variables are categorical variables? According to the documentation  in order to retrieve a list of picklist values for an object we need to pass in a Record Type Id which can be retrieved using the getObjectInfo method.   The issue is that these are both wire functions and I cannot figure out how to chain these methods.  If I try code like the following:  getObjectInfo({ objectApiName: ACCOUNT_OBJECT })             .then(result =&gt; {                 return getPicklistValues({recordTypeId: result.defaultRecordTypeId, fieldApiName: TYPE_FIELD});             })             .then(result =&gt; {                 this.picklistValues = result.data             })             .catch(error =&gt; {                 this.error = error;             });   I get an error      render threw an error in 'c:getpicklistexample' [Imperative use is not   supported. Use @wire(getObjectInfo).]   If I try something like   @track rtId;      @wire(getObjectInfo, { objectApiName: ACCOUNT_OBJECT })     handleResult({error, data}) {         if(data) {             this.rtId = data.defaultRecordTypeId;         }     }      @wire(getPicklistValues, {recordTypeId: this.rtId,fieldApiName: TYPE_FIELD})     picklistValues;   Then it throws an internal server error. Anybody have an idea as to how I could retrieve the record type Id and then call to get the pick list values like the documentation suggests?  FYI my imports are:  import { LightningElement, wire, track } from 'lwc'; import { getPicklistValues, getObjectInfo } from 'lightning/uiObjectInfoApi'; import TYPE_FIELD from '@salesforce/schema/Account.Type'; import ACCOUNT_OBJECT from '@salesforce/schema/Account'; Extending a Homotopy from $X \times [0,1]$ to $X \times \mathbb{R}$ What's the most efficient way to select the last n rows in a table without changing the table's structure? What's the most efficient way to select the last n number of rows in a table using mySQL? The table contains millions of rows, and at any given time I don't know how large the table is (it is constantly growing). The table does have a column that is automatically incremented and used as a unique identifier for each row. More than one optional argument for newcommand The \newcommand macro allows to use an optional argument for the first parameter #1 with:  \newcommand{\mycommand}[3][defaultfor1]{blah blah blah}   Is it possible to have more than one option with \newcommand? Can we denied permission for synchronized data extension in Salesforce Marketing Cloud, My requirement is, I don't want to allow users to access synchronized data extension either in contact builder or in Subscribers. Multiple returns from a function I have a GTX 660M and Ubuntu 18.04 64-bit.  According to , the latest graphics driver for this card is version 410.66.  The  says  Current long-lived branch release: `nvidia-410` (410.66)   I have nouveau drivers installed at the moment according to lsmod and, apt-get purge 'nvidia*' does nothing.  I have done add-apt-repository ppa:graphics-drivers and apt-get update and a reboot.  Running ubuntu-drivers devices gives this output  == /sys/devices/pci0000:00/0000:00:01.0/0000:01:00.0 == modalias : pci:v000010DEd00000FD4sv00001028sd00000551bc03sc00i00 vendor   : NVIDIA Corporation model    : GK107M [GeForce GTX 660M] driver   : nvidia-304 - third-party free driver   : nvidia-driver-390 - third-party free driver   : nvidia-340 - distro non-free driver   : nvidia-driver-396 - third-party free driver   : nvidia-driver-410 - third-party free recommended driver   : xserver-xorg-video-nouveau - distro free builtin   Note this line:  driver   : nvidia-driver-410 - third-party free recommended   It looks like everything is ready to go to install driver version 410, but autocompleting apt-get install nvidia- indicates the highest available version is nvidia-390, and trying apt-get install nvidia-410 directly fails.  Why does apt not list the version 410 that the PPA and ubuntu-drivers both say is there?  And if I can't apt-get install nvidia-410, how can I install this version? How are these AES MixColumn multiplication tables calculated? I'm using the mul2, mul3, mul9, mul11, mul13 and mul14 tables for the MixColumn and InvMixColumn steps in AES-128. However, I got these off some Github repository, and now I'm looking for an actual source explaining how these were calculated and so on.  I cannot find anything proper on Google, can anyone help me out? How to pick a specific symbol from a specific font? When should I create a new tag? How do I request a new tag if I don't have enough rep? Can someone please create the tag  on our site? We need to have it because        Foo is a recent/upcoming release / edition / model / episode of a popular, established OS / board game / bike brand / TV show   Foo is a hot new language / ingredient / game   we're currently using some other tag for questions about Foo, but it's causing ambiguity with a separate thing that's also named Foo      This is not a joke post. We get a lot of feature requests that ask for this or that tag to be created, and every time, a discussion ensues wherein people explain that there is no "tag creation department" and that community members create tags just by using them. I'm posting this "question" so that future requests of that type can be closed as a duplicate of this, saving time and eliminating redundant conversations for everyone involved.     See also: How do I create a shortcut for a script, and run it as a .desktop file? I have installed Teamspeak onto my Ubuntu 15.10 system. Teamspeak requires a script to be ran in-order to start. How can I create a symbolic link that is a .desktop so I can open the .desktop file, and have it run the script? I have looked at a few methods, but they all seem like hotfixes. Is there a proper way to do this? Force update from unsigned repository I'm using an unsigned repo in Ubuntu 16.04 from Debian multimedia:  deb http://www.deb-multimedia.org jessie main   To install deb-multimedia-keyring, I'm running:  apt-get update &amp;&amp; apt-get install deb-multimedia-keyring -y   This gives an error:  W: GPG error: http://www.deb-multimedia.org jessie InRelease: The following signatures couldn't be verified because the public key is not available: NO_PUBKEY 5C808C2B65558117 E: The repository 'http://www.deb-multimedia.org jessie InRelease' is not signed. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details.