Some strange problem occurred to me, when writing a report. I found some similar questions but none of the solutions work for me. Until today, my bibliography worked fine, today everything crashed. I was working with biber but I can not get it working anymore, so I had to switch to bibtex now. Here is my MWE: main.tex: \documentclass[a4paper,11pt]{article}  \usepackage[     backend=biber, % or bibtex     style=phys,     sorting=none, ]{biblatex} \addbibresource{references.bib}  \begin{document}  Hi there, this is me.~\cite{Nobody06}  \printbibliography \end{document}  and the references.bib: @misc{ Nobody06, author = &quot;Nobody Jr&quot;, title = &quot;My Article&quot;, year = &quot;2006&quot; }   I am using VScode and its recipes (pdflatex-bibtex-pdflatex x2 or latexmk). When using biber as a backend I get undefined references error, while when using bibtex I just get Using fall-back BibTeX(8) backend: (biblatex) functionality may be reduced/unavailable. warning but it compiles successfully. The strangest thing is that both options works fine on overleaf. When I try to compile it by hand in the command line pdflatex main and then biblatex main I still get undefined references. When doing biber main I get Cannot find 'main.bcf', so I am wondering which is the correct combination? Relevant versions: pdfTeX 3.14159265-2.6-1.40.21 (TeX Live 2020) BibTeX 0.99d (TeX Live 2020) kpathsea version 6.3.2 biber version: 2.15
I am on a Mac with MacTex 2020 and ran into a strange problem with Biber. Working on a document all was fine, I did not change the bibliography (says my git) and only fixing typos (not even adding cites) and suddenly latexmk stopped working after a Biber run, while that looked like  INFO - This is Biber 2.14 INFO - Logfile is 'doc.blg' INFO - Reading 'doc.bcf' INFO - Found 31 citekeys in bib section 0 INFO - Processing section 0 INFO - Globbing data source 'literature.bib' INFO - Globbed data source 'literature.bib' to literature.bib INFO - Looking for bibtex format file 'literature.bib' for section 0 INFO - LaTeX decoding ... INFO - Found BibTeX data source 'literature.bib'   so harmless in my view. Then with --debug I additionally get in the end.  INFO - LaTeX decoding ... warning: overriding existing definition of macro "jun" warning: overriding existing definition of macro "jul" warning: overriding existing definition of macro "sep" warning: overriding existing definition of macro "jan" [...] warning: overriding existing definition of macro "nov" INFO - Found BibTeX data source 'literature.bib' Unicode::UCD: failed to find unicore/version in /var/folders/hm/gt63cn0j2r7bwbzw7c1hww1r0000gn/T/par-626572676d616e6e/cache-2f8a664c3c933a44d096bcc8d018a5e92bced897/inc/lib /var/folders/hm/gt63cn0j2r7bwbzw7c1hww1r0000gn/T/par-626572676d616e6e/cache-2f8a664c3c933a44d096bcc8d018a5e92bced897/inc CODE(0x7fd026aeb5a8) CODE(0x7fd026aebb90) at /var/folders/hm/gt63cn0j2r7bwbzw7c1hww1r0000gn/T/par-626572676d616e6e/cache-2f8a664c3c933a44d096bcc8d018a5e92bced897/inc/lib/Biber/Input/file/bibtex.pm line 1011.   It seems unrelated to the library itself since deleting a few entries (going through deleting one half, then the other), the problem persists. It does not happen for an empty literature, but as soon as a first entry is added (just author/journal/title/year/volume/pages - no utf8-only chars) this error message appears.  what does this last error message mean?  How can I solve this?
Proving $\lim_{n\to\infty} a^{\frac{1}{n}}=1$ by definition of limit When is 6 a square in Zp To find when $6$ is a square I divided $6$ in to $2$ and $3$ by The Legendre symbol.  $$\frac{6}{p} = \frac{2}{p} \times \frac{3}{p}$$  I know that we can multiply the cases of when $2$ and $$ are squares because then it is (1)(1) by Legendre.   $2$ is a square when $p \equiv 1 \pmod{8}$   $3$ is a square when $p \equiv 1 \pmod{12}$  so $6$ is a square when $p \equiv 1 \pmod {24}$ ($96$, but the $lcm(8,12)= 24$)   but how do I find the cases when $2$ and $3$ are both NOT squares? because then $6$ would be a square then as well, $(-1)(-1) = 1$ by Legendre. Let $A$ be a $m \times n$ matrix with entries from some field $F$. Define the determinant rank of $A$ to be the largest possible size of a nonzero minor, i.e. the size of the largest invertible square submatrix of $A$. It is true that the determinant rank is equal to the rank of a matrix, which we define to be the dimension of the row/column space.  It's not difficult to see that $\text{rank} \geq \text{determinant rank}$. If some submatrix of $A$ is invertible, then its columns/rows are linearly indepedent, which implies that the corresponding rows/columns of $A$ are also linearly indepedent.  Is there a nice proof for the converse? proving of inequality $(n!)^2 \leq n^n\cdot n!\leq (2n!)$, where $n\in \mathbb{N}$ How can we prove the inequality $(n!)^2 \leq n^n\cdot n!\leq (2n!)$, where $n\in \mathbb{N}$  $\bf{My \; Try}$::  $1\leq n$  $2\leq n$  $3\leq n$  ....  ....  $n\leq n$  So $1\cdot 2 \cdot 3 ..n \leq n^n$  So $n!\leq n^n\Rightarrow (n!)^2 \leq n^n\cdot n!$  But  did not understand how can we prove $n^n\cdot n! \leq (2n!)$  Help me  Thanks Convert NTFS to FAT32 without losing data I understand that you can  without formatting or backing up the data and later restoring - is the same thing possible in the other direction?  I want to convert my NTFS portable hard drive to FAT32 so that my PS3 will recognise it.  If it's not possible from within Windows, are there any free programs which will allow me to do it? What does template  mean? Orthogonal matrix and orthonormal columns If GCD $(a_1,\ldots, a_n)=1$ then there's a matrix in $SL_n(\mathbb{Z})$ with first row $(a_1,\ldots, a_n)$ Let $n \geq 2$. Let $a_1, a_2, \ldots, a_n$ be $n$ integers such that $\gcd\left(a_1, a_2, \ldots, a_n\right) = 1$. Prove that there exists a matrix in $\operatorname{SL}_n\left(\mathbb{Z}\right)$ whose first row is $\left(a_1, a_2, \ldots, a_n\right)$.   Since the gcd of the integers $a_1,\ldots, a_n$ is $1$, there exists weights $x_i \in \mathbb{Z}$ such that $a_1x_1+\cdots+ a_nx_n=1$. My two ideas are (a) to brute force construct an $n\times n$ matrix with first row $a_1,\ldots ,a_n$ and to construct the remaining rows such that the determinant is $\sum a_ix_i=1$ or (b) to use induction.  (a) (Constructive) This is tedious since once I find a way to construct the remaining $n-1$ rows to ensure that $a_1x_1$ appears in the determinant, I am not sure how to modify these $n-1$ rows to ensure  that only the terms $a_ix_i$ appear in the cofactor expansion. If such a matrix exists, I'd like to see it.  (ii) (Non-constructive) If I proceed by induction then the base case $n=2$ is settled since I can choose the 2nd row to be $-x_2, x_1$ so that the determinant is $a_1x_1-a_2(-x_2)=1$. However, I'm not sure how to use the inductive hypothesis to show that if I can construct such an $n\times n$-matrix then I can construct an $\left(n+1\right) \times \left(n+1\right)$-matrix with the desired property. In particular, if the gcd $(a_1,\ldots ,a_{n+1})$ is $1$, it is not necesarry that the gcd of any $n$ of these terms is $1$, so induction may not even apply here.  How can I construct such a matrix or prove that one exists (without necessarily constructing it)? I have an older iPod touch which which supports up to iOS 6.1.6 and nothing newer.  This device is equipped with a Linea Pro barcode scanner and previously had an application called "" installed.  Recently I was attempting to do some work on a web application I wrote and realized that I needed to update to a newer version of Web Barcode - but unfortunately the Apple ID saved on my device was deactivated a few years ago.  So I went to settings and updated it with my current Apple ID.  Then again tried to update the App from the Apple store.  Still no dice - it was asking me for the password on the old account.  So then I figured I'd just remove it and reinstall from scratch.  Deleted the app and went to the store but now I am met with a message that the new version requires at least iOS 8.0.  I feel I followed a very logical series of steps to update this application and while I can understand that I won't be able to install the newest version, how can I get the old version back?  The device is completely useless without the barcode scanning application and the Apple store is now refusing to install any version at all on my device.  What can I do to get the original version of this application reinstalled? Change audio output depending on which one is on I have a PC that is hooked up to my HDTV (via a long hdmi cable) and to a monitor, which is in another room. I have speakers directly plugged to the PC audio out. The PC is next to the monitor and speakers.   I am not sure but I think Windows can play sound on only 1 audio device at a time. And I can only set 1 device as the default output. I can get sound on the TV or speaker depending on which device I set to default. But I would have to do this every time I switch between using my TV or my monitor!   Is there some way to configure such that sound plays through the TV if the TV is on, else it plays in the speakers? If this is not possible, then the next best alternative would be to get the sound to play on both the devices at the same time. Rules when using a Joker I had on my rack among other tiles, a black 9, 1 x blue 10, and a black 12.  On the table was a set of 3 x 10s including a Joker(no blue 10), and other sets of tiles including an available black 11.  Can I replace the joker with my blue 10 and then set out :  my black 9, the joker(representing a black 10), a black 11 from the table, and my black 12? I have a training set with about 3000 positive instances and 3000 negative instances. But my test data set is pretty much un-balanced. The positive set only has 50 instances and negative has 1500 instances.This causes the precision very low. Are there any approaches to solve this problem? I use SVM to build classifier. I am having trouble getting onto a server in Minecraft PE on iOS. When I tap on the person's server it just says invalid name:    I have tried turning off and restarting the iPad, going to home screen and logging back into the server but I have had no luck and no change. Is there a reason why this is so? I often hear people saying that "I am" is the shortest sentence in the English language.  I know that there are also discussions about sentences using the imperative mood such as "Go." that would be shorter, but my question is this:  Why would we (the people saying "I am" is a full sentence) not accept "I'm" as a complete sentence?  Is there some unwritten rule about contractions that says "I'm" wouldn't be correct? Illustrator Vector Graphics Appearing Pixellated Expected value of x in a normal distribution, GIVEN that it is below a certain value Using .NET, how can you find the mime type of a file based on the file signature not the extension Which is more correct:     Apropos of your earlier comment, I decided to....   or      Apropos your earlier comment, I decided to...   Actually, apropos is so fancy a word, even I, a word maven if I do say so myself, am not entirely sure when to use it.  If I may add to my preposition question, are there any constructions where it would be appropriate to use something like with regard to and not apropos? How can we evaluate this series  $$\sum_{n=1}^\infty \frac{n^a}{b^n}?$$  Here $a$ and $b$ are positive integers. If $b=1$ then series will be diverging, in other cases, it will be converging, but how to find this sum?  Isn't there a simple solution which doesn't involve Stirling numbers? Where can I write a temp file from ASP.NET?