I am getting warning "Empty Bibliography" using bibtex at backend. It's been 2 days I am unable to resolve this problem I tried to compile latex 2 time after bibtex compilation. My references are in bib file and I am using jabref.  \documentclass[conference]{IEEEtran} \IEEEoverridecommandlockouts \usepackage{amsmath,amssymb,amsfonts} \usepackage{algorithmic} \usepackage{graphicx} \usepackage{textcomp} \usepackage[backend = bibtex, sorting=none]{biblatex} \usepackage{xcolor} \def\BibTeX{{\rm B\kern-.05em{\sc i\kern-.025em b}\kern-.08em     T\kern-.1667em\lower.7ex\hbox{E}\kern-.125emX}}   \addbibresource{facial expression recognition.bib} \begin{document}  \title{}  \cite{Ekman1971}   \maketitle  %\begin{abstract} %This document is a model and instructions for \LaTeX. %This and the IEEEtran.cls file define the components of your paper [title, text, heads, etc.]. *CRITICAL: Do Not Use Symbols, Special Characters, Footnotes,  %or Math in Paper Title or Abstract. %\end{abstract}  %\begin{IEEEkeywords} %component, formatting, style, styling, insert %\end{IEEEkeywords}  \section{Introduction}    \printbibliography \end{document}
Referencing bib files with spaces in the filename I am using a Mac with TeX Live and Mendeley for automatically generating .bib files, some of which have spaces in the file names. I would like to know how to reference such files from LaTeX as using "" or \ doesn't seem to work.
Can commuting matrices $X,Y$ always be written as polynomials of some matrix $A$? Consider square matrices over a field $K$. I don't think additional assumptions about $K$ like algebraically closed or characteristic $0$ are pertinent, but feel free to make them for comfort. For any such matrix $A$, the set $K[A]$ of polynomials in $A$ is a commutative subalgebra of $M_n(K)$; the question is whether for any pair of commuting matrices $X,Y$ at least one such commutative subalgebra can be found that contains both $X$ and $Y$.  I was asking myself this in connection with frequently recurring requests to completely characterise commuting pairs of matrices, like . While providing a useful characterisation seems impossible, a positive anwer to the current question would at least provide some answer.  Note that in many rather likely situations one can in fact take $A$ to be one of the matrices $X,Y$, for instance when one of the matrices , or more generally if its  (so coincides with the characteristic polynomial). However this is not always possible, as can be easily seen for instance for diagonal matrices $X=\operatorname{diag}(0,0,1)$ and $Y=\operatorname{diag}(0,1,1)$. However in that case both will be polynomials in $A=\operatorname{diag}(x,y,z)$ for any distinct values $x,y,z$ (then $K[A]$ consists of all diagonal matrices); although in the example in  the matrices are not both diagonalisable, an appropriate $A$ can be found there as well.   I thought for some time that any maximal commutative subalgebra of $M_n(K)$ was of the form $K[A]$ (which would imply a positive answer) for some $A$ with minimal polynomial of degree$~n$, and that a positive answer to my question was in fact instrumental in proving this. However I was wrong on both counts: there exist (for $n\geq 4$) commutative subalgebras of dimension${}&gt;n$ (whereas $\dim_KK[A]\leq n$ for all $A\in M_n(K)$) as shown in , and I was forced to correct  in the light of this; however it seems (at least in the cases I looked at) that many (all?) pairs of matrices $X,Y$ in such a subalgebra still admit a matrix $A$ (which in general is not in the subalgebra) such that $X,Y\in K[A]$. This indicates that a positive answer to my question would not contradict the existence of such large commutative subalgebras: it would just mean that to obtain a maximal dimensional subalgebra containing $X,Y$ one should in general avoid throwing in an $A$ with $X,Y\in K[A]$. I do think these large subalgebras easily show that my question but for three commuting matrices has a negative answer.  Finally I note that  to the cited MO question mentions a result by Gerstenhaber that the dimension of the subalgebra generated two commuting matrices in $M_n(K)$ cannot exceed$~n$. This unfortunately does not settle my question (if $X,Y$ would generate a subalgebra of dimension${}&gt;n$, it would have proved a negative answer); it just might be that the mentioned result is true because of the existence of $A$ (I don't have access to a proof right now, but given the formulation it seems unlikely that it was done this way).  OK, I've tried to build up the suspense. Honesty demands that I say that I do know the answer to my question, since a colleague of mine provided a convincing one. I will however not give this answer right away, but post it once there has been some time for gathering answers here; who knows somebody will prove a different answer than the one I have (heaven forbid), or at least give the same answer with a different justification. I have several big files with some measurements.   It looks this way:  N 12344;PE 9.9999999;... #S 0 0 31 44 75 130 165 196... #S_+ "2 5 2 3 3 1 1 2 3 1 2 2...  N 12345;PE 9.9999999;... #S 0 0 34 57 84 133 152... #S_+ "1 0 1 1 2 3 0 0 0...  N 12346;PE 9.9999999;... #S 0 0 31 44 73 140 169... #S_+ "3 3 4 0 0 2 1 2 4...  N 25104;PE 9.9999999;... #S 0 0 36 52 102 108 145... #S_+ "1 1 0 1 0 0 3 0 1...  N 25105;PE 9.9999999;... #S 0 0 32 58 88 130 143...   Sample is here:   The file is much bigger but I replaced part of the data with "..." to make it shorter.  I need to somehow replace the line ends before "#S" - in fact simply merge the "N" line with the following two ones into one line (or with the following three ones so I can get rid of the blank lines). Expect output like this:  N 12344;PE 9.9999999; #S 0 0 31 44 75 130 165 196 #S_+ "2 5 2 3 3 1 1 2 3 1 2 2... N 12345;PE 9.9999999; #S 0 0 34 57 84 133 152 #S_+ "1 0 1 1 2 3 0 0 0... N 12346;PE 9.9999999; #S 0 0 31 44 73 140 169 #S_+ "3 3 4 0 0 2 1 2 4... N 25104;PE 9.9999999; #S 0 0 36 52 102 108 145 #S_+ "1 1 0 1 0 0 3 0 1... N 25105;PE 9.9999999; #S 0 0 32 58 88 130 143...   Is this possible to achieve using some command-line utility in linux?   My knowledge is quite limited in this area so I would appreciate any help.  thanks It is asked to solve the PDE  $$u_x + u_y + u = e^{x+2y}$$  My attempt: We have that $$\frac{dx}{1}=\frac{dy}{1}=\frac{du}{e^{x+2y}-u} $$  $\Rightarrow \frac{dy}{dx}=1 \Rightarrow y=x+C_1 \iff C_1=y-x$  $\Rightarrow \frac{du}{dx}= e^{x+2y}-u \iff \dot{u}+u=e^{x+2y} \iff \\ \frac{d}{dx}(e^x u) = e^{2x+2y} \iff u = e^{-x} ( \frac{1}{2}e^{2x+2y} + C_2)$  As $u$ is constant along its characteristics:  $$u(x,y)=\frac{e^{-x}}{2}(e^{2x+2y} + 2C_2(y-x))$$  where $C_2$ is any function of one variable. But wolfram gave the following answer  $$u(x,y)=\frac{e^{-x}}{4}(e^{2x+2y} + 4C_2(y-x))$$  as you can check   What am I doing wrong? Integrable - martingale Suppose $X_{n}^{(1)}$ and $X_{n}^{(2)}$ are martingales w.r.t. a filtration $F_{n}$. I have do determine whether $Z_{n}=e^{X_{n}^{1}+X_{n}^{2}}$ is a martingale, submartingale or supermartingale. I thought that because $X_{n}^{1}+X_{n}^{2}$ is a martingale I can use Jensen to say that $E[Z_{n}|F_{n-1}]\geq e^{E(X_{n}^{(1)}+X_{n}^{(2)}|F_{n-1})}=Z_{n-1}$. But I have to check whether $Z_{n}$ is integrable and I don't know how. Example of a function such that $\varphi\left(\frac{x+y}{2}\right)\leq \frac{\varphi(x)+\varphi(y)}{2}$ but $\varphi$ is not convex What is this style of text-as-image called? I'm seeing a trend (especially on pinterest) of these posters where the image and subject is the text. THe text itself isn't really that important either, it's mainly the typefaces used and composition.  Does this style have a name to it? Is it just called text-as-image? Could you tell me how to prove that if two series $ \sum_{n=0} ^{\infty}x_n, \sum_{n=0} ^{\infty} y_n$ are convergent, then $\sum_{n=0} ^{\infty}(\alpha \cdot x_n + \beta \cdot y_n)$ is also convergent and its sum equals $\alpha \cdot \sum_{n=0} ^{\infty} x_n + \beta \cdot \sum_{n=0} ^{\infty} y_n$? How do you prove that these four definitions of the operator norm are equivalent? $$\begin{align*} \lVert A\rVert_{\mathrm{op}} &amp;= \inf\{ c\;\colon\; \lVert Av\rVert\leq c\lVert v\rVert \text{ for all }v\in V\}\\ &amp;=\sup\{ \lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert\leq 1\}\\ &amp;=\sup\{\lVert Av\rVert\;\colon\; v\in V\text{ with }\lVert v\rVert = 1 \}\\ &amp;=\sup\left\{ \frac{\lVert Av\rVert}{\lVert v\rVert}\;\colon\; v\in V\text{ with }v\neq 0\right\}. \end{align*}$$ How to copy files via terminal? Photo Map tip in QGIS 3 on windows How do I boot from ISO file stored on hard disk? How to perfectly align UV coordinates automatically? I have this setup:   The mesh is only the visible top side, with a mirror modifier for the bottom part. Now I want to unwrap it like in the picture. I want the inner circle (red) to go to the top of the UV map...  The problem is that no matter what automatic unwrapping I use, it never works as intended.  The closest I've gotten was by using "follow quads", but that skewed the UV map a lot:   I used the result from the second image and manually adjusted the vertices with a lot of scaling and "align auto", but its very far from perfect as you can see in the first image.  Here's what I want exactly:   The uv vertices should be distributed evenly (horizontally) The middle line (yellow) should be on a "natural" y height (so there's no stretching)   I hope it's clear what I want to achieve. I'm sure there must be some automatic way to perfectly align and distribute the UVs in this way.  By the way: If you look closely you can see that I already placed a uv seam (aligned with the green y axis in the screenshot) but that didn't really help. When is an elected moderator shown as "departed" on the election page? On Philosophy, one of the elected moderators has since deleted their account, and they're shown as "Departed" on :        However, , one of the elected moderators still has their "previous" flair shown in place, despite the fact that they've already resigned and left the Network completely in the wake of the recent events.        , the same happens (old flair remains), but without the clickable &lt;a&gt; around the image, and with a noticeable :        This looks inconsistent, so when exactly will a placeholder image showing "Departed" replace the flair for a departed election winner? Stance on repost question instead of editing to invalidate answers? Regarding , made by me, and general questions in this matter. I did receive decent answers on the question that was asked, but it wasn't asked clear enough by me I now see. I do not want to invalidate the answers by completely restructuring the question. Should one post a new question ignoring the other? or Should one refer to their old post and writing a short &quot;because x I reposted this...&quot;? I've heard Google's algorithm is based on PageRank, and that it has something to do with the links to your site. Can someone explain it, and more importantly, how I can increase my site's PageRank? Android and getting a view with id cast as a string How did Ron speak Parseltongue and re-enter the Chamber of Secrets? Although 'The Deathly Hallows' was overall an excellent conclusion to Harry Potter's saga, there were moments within the book that felt infuriatingly lazy, like Rowling was whipping the plot along regardless of the holes she tore in it. For me, the lowest point was when Ron gargles at the entrance of the Chamber of Secrets- and succeeds!- so he can get some Basilisk fangs to destroy a Horcrux (which are conveniently vulnerable to this venom).  This is, by itself, a shockingly slapdash deus ex machina, but even worse is how it is completely inconsistent with the guidelines Rowling herself set. First and foremost, Parseltongue is not just a language, but a magical ability that can only be learned through hereditary or magical means (Potter being the sole example of the latter). This is firmly established by the fact that even Dumbledore, the greatest wizard of all time and a polyglot, is completely unable to speak it despite understanding it perfectly.  Without linguistic knowledge nor inherent ability, the chances of making any kind of coherent sentence (let alone the right one!) would be incredibly slim even in the case of mundane languages; it would be like gibbering at a Filipino in the hope that what you said was fluent Tagalog and didn't insult their mother.  Other baffling issues somewhat related to my question are thus; even if Ron had somehow unlocked the gate, how did he make his way down? In the second novel, Lockheart's foolery demolished the main tunnel to the Chamber, making it impossible to access or leave; the fact that the teeth are still there also implies that the Chamber wasn't excavated and the rubble cleared in the interval. Granted, he could have tried Apparating, but then he wouldn't have needed to fake Parseltongue to begin with. Zombies - Why do they need to feed on living animals/humans? Does %* in batch file mean all command line arguments? Pagination go to first page if i'm on last post I have a custom post type and used get_next_post() and get_previous_post() to create a simple pagination. However i need to create a pagination loop (if that is the correct term). If i'm on the first post i need to browse back to the last post and if i'm on the last post i need to browse to the first post. Is there a simple way to do this that i haven't seen but Google, this site and the codex has not been very helpfull :/  The current code i have is. I've tried to use wp_list_pages and other functions but with little luck.  $next_post = get_next_post(); $prev_post = get_previous_post(); $nav = array(     "next_post" =&gt; array(         "url"       =&gt; get_permalink($next_post-&gt;ID),         "id"        =&gt; $next_post-&gt;ID,         "titill"    =&gt; get_field("vinstri_titill", $next_post-&gt;ID)." - ".get_field("vinstri_undirtitill", $next_post-&gt;ID)     ),     "prev_post" =&gt; array(         "url"       =&gt; get_permalink($prev_post-&gt;ID),         "id"        =&gt; $prev_post-&gt;ID,         "titill"    =&gt; get_field("vinstri_titill", $prev_post-&gt;ID)." - ".get_field("vinstri_undirtitill", $prev_post-&gt;ID)     ), );