Can someone please help me get the code below to work but by running biber, not bibtex? I have a biblatex style (nejm) that only works with biber.  I'm currently running RStudio 0.98.501 and this is the first line of the log file.   This is XeTeX, Version 3.1415926-2.5-0.9999.3 (TeX Live 2013) (format=xelatex     2013.5.30)  25 MAR 2014 17:55 entering extended mode    \documentclass{article} \usepackage{polyglossia} \usepackage{fontspec} \usepackage{csquotes} \setmainlanguage{english} \begin{filecontents}{\jobname.bib} @misc{test,   author = {Author, A.},   year = {2001},   title = {Alpha}, } \end{filecontents} \usepackage[backend=bibtex, style=numeric]{biblatex} \addbibresource{\jobname.bib} %\addbibresource{/Users/simon/Documents/simon.bib} \begin{document} Test text   \parencite{test}  \printbibliography  \end{document}
I'd like to create a PDF report with knitr with LaTeX-code in RStudio with biblatex for citations. However, the in-text-citations are always undefined and no bibliography is printed. The same bibliography works perfectly with a pure .tex-document stored in the same folder and containing the same citations, same path to the .bib-file etc. However, as for all references (also with hyperref) I have to compile twice. I thought this might be the problem with knitr (that the compilation is run only once), but the hyperref-references to figures work well. Any idea what might go wrong with the bibliography when using knitr and how I can solve the problem?  MWE:  \documentclass{article}  \usepackage[style=authoryear-comp, maxnames=1, bibencoding=utf8, backend=biber]{biblatex} \bibliography{ZoteroOutput.bib}  \begin{document}  According to the literature \parencite{gonzalez2003} we can assume...  \printbibliography  \end{document}   From the log-file:  Package biblatex Warning: Please (re)run Biber on the file: (biblatex)                mwe (biblatex)                and rerun LaTeX afterwards.
Find the limit: $\lim_\limits{x\to 0}{\frac{\left(1+x\right)^{1/x}-e}{x}}$ Machine Learning - Where is the difference between one-class, binary-class and multinominal-class classification? Where is the difference between one-class, binary-class and multinominal-class classification?  If I like to classify text in lets say four classes and also want the system to be able to tell me that none of these classes matches the unknown/untrained test-data.   Couldn't I just use all the methods that I mentioned above to reach my goal? e.g. I could describe C1, C2, C3 and C4 as four different trainings-sets for binary-classification and use the trained models to label an unknow data-set ...  Just by saying, Training-Set for C1 contains class 1 (all good samples for C1) and class 0 (mix of all C2, C3 and C4 as bad samples for C1).  Is unlabeled-data C1 -> 1 or 0  Is unlabeled-data C2 -> 1 or 0 ... and so on ...  For multinominal classification I could just define a training-set containing all good sample data for C1, C2, C3 and C4 in one training-set and then use the one resulting model for classification ...  But where is the difference between this two methods? (except of that I have to use different algorithms)  And how would I define a training-set for the described problem of categorizing data in those four classes using one-class classfication (is that even possible)?  Excuse me if I'm completely wrong in my thinking. Would appreciate an answer that makes the methodology a little bit clearer to me =) How do I install an older version of an app on an iPhone? Am not doing anything fancy, just trying to update a field on Contact in JS of LWC but getting this exception.  Uncaught TypeError: 'set' on proxy: trap returned falsish for property 'Name' throws at mydomain/auraFW/javascript/mhontaYdOya4Y_lBu7v9yg/aura_prod.js:2:27687  HTML Code:   &lt;template&gt;      &lt;template if:true={wiredContact}&gt;          {wiredContact.Name}          &lt;lightning-input value={wiredContact.Name} onchange={updateName}&gt;&lt;/lightning-input&gt;     &lt;/template&gt;  &lt;/template&gt;   JS:   import { LightningElement ,wire,track,api } from 'lwc'; import myContact from "@salesforce/apex/ContactController.fetchContact";  export default class Myrefreshapextest extends LightningElement {    @track wiredContact;   @wire (myContact)         fetchedContact({error, data}){             if(data){                 console.log(JSON.stringify(data));                 this.wiredContact = data;             }else if (error){                 console.log(error);             }     }      updateName (event){         console.log(JSON.stringify(event.detail.value));         console.log(JSON.stringify(this.wiredContact));         this.wiredContact.Name = event.detail.value;     }  }   Apex:  public class ContactController {      @AuraEnabled(cacheable=true)     public static Contact fetchContact(){         return [SELECT Id,Name FROM COntact LIMIT  1];     } }   On top of my head, am not doing anything wrong, anyone has idea what's wrong with my code?  When I print console.log(JSON.stringify(this.wiredContact)); I get old values so am pretty sure it exists.  I tried with @track and @api, but same response. Can anyone shed some light? What factors determine enemy levels or: How do enemy levels scale to the player character's? When I finished Normal Mode at around Level 30 and went back to the  (still at Normal mode), I found that all enemies there are now very easy to kill, meaning they didn't level up (or hit a level cap).  When I went to  as a Level 36 character in True Vault Hunter Mode, I noticed that the enemies are the same level as me.  What factors determine enemy levels?   Specifically, I would like to know:       Are enemies affected by what level a character accepted a mission?      Or is it what level a character is when they first entered an area?      How does Normal Mode, True Vault Hunter mode, and the Ultimate True Vault Hunter Mode affect enemy levels?     Are there enemies that will scale to the player level, as the player levels up? Any enemy leveling differences specific to the DLC areas?         EDIT:   There is now  that explains how this works generally: "In playthrough 1, outside of the DLC, there's no such thing as scaling. Enemy levels are determined solely by the area they're in and your mission." Can anyone expound on that, and comment or provide an answer on what are the specific enemy levels on specific areas and/or missions?    states that "enemies do not actually "scale", or at least from what we know of..." and lists specifics for enemy levels on certain areas in the game (sans DLC areas). Can anyone expound on that and/or add specifics for enemy levels in the DLC missions or areas in TVHM (PT 2.0) and/or UTVHM? The  update has just been released, can anyone provide info about enemy levels on that mode, including the DLC areas or missions? I have a Sony VAIO S series with Windows 8 preinstalled, and I installed Ubuntu 12.10 on a new partition. When the PC boots Grub is display, and I can choose between Windows 8 (didn't worked at the beginning, had to change Grub settings) and Ubuntu 12.10. When I select Ubuntu 12.10 everything is OK, but when I use Windows 8, and I restart Grub isn't displayed anymore (Windows 8 boots immediately).  I booted a Ubuntu 12.10 LiveUSB, used efibootmgr, and found out that Windows 8 (each time it boots) changes EFI default bootloader from Grub to Windows Boot Manager.  How do I change this in Windows 8? I read something about bcdedit, but it does not work or I don't know how to use it... I understand that in a lot of big-lab fields it is common for the principal investigator to append their name to a paper even if they did not write the paper, design the experiment, or collect data since they spend energy securing funding, and managing the whole lab. What about for small labs?  What are the requirements for a supervisor to be included as an author on a paper, as opposed to just appearing in the acknowledgements? If you are working on your own projects independently of your supervisor, but using funding provided by your supervisor (how does this change when the funding provides resources versus just your salary), are you suppose to add them as authors or just acknowledge the source of funding? My laptop (running Win10) repeatedly "steals" the bluetooth connection for my headphones while they're connected to my phone. The only way I've been able to prevent the headphones' connection from switching to the laptop is to play some media or make a phone call immediately after reconnecting to my phone, or to unpair the headphones from the laptop.  The headphones are Plantronics Backbeat Go 2, and the specifications from the Plantronics product page give the following information for the Bluetooth entry:      Bluetooth v2.1:   A2DP 1.2, AVRCP, HFP v1.5, HSP v1.1   Thanks in advance! How to open pdf file in Windows Phone 8? counting number of specific string in each row I have a SNP table looks like this  A N N N N N N N N N N N N C N N N C N N N N N N N N N N N N N N N N N N N C N N N C N N N N N N N N N N N N N N N N N N C C N N N N N N N N N N C C N N N C C N N N N N N N N N N N N N N N N N   I want to count total number of N in each row? any idea how to do that Let A and B be sets. Show that A is a subset of B if and only if for any set C, one has A union C is a subset of B union C. Can you verify my proof if it is right?  Let $A$ and $B$ be sets.  (a) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ union $C$ is a subset of $B$ union $C$. (b) Show that $A$ is a subset of $B$ if and only if for any set $C$, one has $A$ intersect $C$ is a subset of $B$ intersect $C$.  Edited Solution :  (a)  (=>) Given A is a subset of B, this implies A U B = B. Given a set C, A U B U C = B U C. Since A is a subset of B, this implies A U C is a subset of B U C.      Solution 1 :(&lt;=) Given A U C is a subset of B U C , For A U C , x is a member of A U C . For B U C, x is a member of B U C. Suppose x is a member of A but not a member of B, meaning A is not a subset of B. Therefore , A U C is not a subset of B U C. Hence, A must be a subset of B.      Solution 2 :(&lt;=) Given A U C is a subset of B U C, For A U C, x is a member of A U C and for B U C, x is a member of B U C. Suppose C is a disjoint set. Since A U C is a subset of B U C, it follows that A is a subset of B.  (b) (=>) Since A is a subset of B , it follows that A U B = B. [A intersect C] is a subset of A and  [B intersect C ] is a subset of B by definition of intersection. Let x be member of A and member of C. Since [ A intersect C ] is a subset of A, x must be in A. Let x be a member of B and member of C. Since [ B intersect C ] is a subset of B, x must be in B. Therefore, A intersect C is a subset of B intersect C.  (&lt;=) Since A intersect C is a subset of B intersect C, suppose we let Set C be a subset of B and a subset of A as well. C is a subset of A implies A intersect C is equals to C. C is a subset of B implies B intersect C is equals to C as well. Since A intersect C is equals to B intersect C, A is a subset of B. Equicontinuity on a compact metric space turns pointwise to uniform convergence I know that   If $\{f_n\}$ is an equicontinuous sequence,  defined on a compact metric space $K$, and for all $x$, $f_n(x)\rightarrow f(x)$, then $f_n\rightarrow f$ uniformly.  I'm having trouble proving this.  I see the same problem  but am having trouble following the proof, particularly with part (3).  Can someone guide me through a proof of this result? In Delphi, is TDataSet thread safe? I'd like to be able to open a TDataSet asynchronously in its own thread so that the main VCL thread can continue until that's done, and then have the main VCL thread read from that TDataSet afterwards. I've done some experimenting and have gotten into some very weird situations, so I'm wondering if anyone has done this before.  I've seen some sample apps where a TDataSet is created in a separate thread, it's opened and then data is read from it, but that's all done in the separate thread. I'm wondering if it's safe to read from the TDataSet from the main VCL thread after the other thread opens the data source.  I'm doing Win32 programming in Delphi 7, using TmySQLQuery from  as my TDataSet descendant. Where to ask question about hardware? My question is similar to  except it's about VMware Workstation Player instead of VirtualBox.  How do I disable mouse integration? I want the VM to "capture" my mouse. One option is to pass my mouse directly to the VM, but that's more difficult if I need to exit the VM (I want to just use Ctrl+Alt).  I'm using "VMware Workstation 15 Player (Non-commercial use only)" On Mac, are there keyboard shortcuts to create a new Space? Or at least any way to automate this process with keyboard shortcuts? I want to not have to use the mouse to create Spaces. I want to use the keyboard for that. As a Graduate Mathematics student, my interest lies in Number theory. I am curious to know if Number theory has any connections or applications to physics. I have never even heard of any applications of Number theory to physics. I have heard Applications of linear algebra and analysis to many branches of physics, but not number theory.  Waiting forward in receiving interesting answers! Adding a Method to an Existing Object Instance I've read that it is possible to add a method to an existing object (i.e., not in the class definition) in Python.   I understand that it's not always good to do so. But how might one do this? Theorem. If the identity is written as the product of $r$ transpositions, $id=τ_1τ_2\dots τ_r$, then $r$ is an even number. Proof. We will employ induction on $r$. A transposition cannot be the identity; hence, $r &gt; 1$. If $r = 2$, then we are done. Suppose that $r &gt; 2$. In this case the product of the last two transpositions, $τ_{r−1}τ_r$, must be one of the following cases: $(ab)(ab) = id\\ (bc)(ab) = (ac)(bc)\\ (cd)(ab) = (ab)(cd)\\ (ac)(ab) = (ab)(bc),$ where $a$, $b$, $c$, and $d$ are distinct. The first equation simply says that a transposition is its own inverse. If this case occurs, delete $τ_{r−1} τ_r$ from the product to obtain $id=τ_1τ_2···τ_{r−3}τ_{r−2}$. By induction $r − 2$ is even; hence, $r$ must be even. In each of the other three cases, we can replace $τ_{r−1} τ_r$ with the right-hand side of the corresponding equation to obtain a new product of $r$ transpositions for the identity. In this new product the last occurrence of $a$ will be in the next-to-the-last transposition. We can continue this process with $τ_{r−2} τ_{r−1}$ to obtain either a product of $r − 2$ transpositions or a new product of $r$ transpositions where the last occurrence of $a$ is in $τ_{r−2}$. If the identity is the product of $r − 2$ transpositions, then again we are done, by our induction hypothesis; otherwise, we will repeat the procedure with $τ_{r−3}τ_{r−2}$. At some point either we will have two adjacent, identical transpositions canceling each other out or $a$ will be shuffled so that it will appear only in the first transposition. However, the latter case cannot occur, because the identity would not fix $a$ in this instance. Therefore, the identity permutation must be the product of $r − 2$ transpositions and, again by our induction hypothesis, we are done. Question: That is the proof written in the book, but unfortunately after more than $10$ times reading I don't understand the last two paragraph (i.e., from &quot;In each of the other three cases&quot;). Is there another simpler proof or could someone please clarify what is going on? Thank you. Is the gravitational constant $G$ a fundamental universal constant?