I am trying to compile tex document using biblatex to handle references with no success. I am aware of the cycle Latex > Biber > Latex. I always use pdflatex, therefore it should be  pdflatex main.tex biber main[.bcf] pdflatex main.tex   In the first step, I get e.g. Citation 'citekey' on page 53 undefined which is normal as I have not generated the proper reference text. But upon running biber main I get error  Undefined subroutine &amp;Biber::Config::biber_error called at /var/folders/zv/ds3zw_xs2dv4c1jp7m1t44f80000gt/T/par-6973776168797574616d61/cache-ec7adafccc6b3891989ad6eb09778d44c1fecb97/inc/lib/Biber/Config.pm line 640. Any idea how to fix this?
I'm trying to use biblatex and when I compile my document I get an error like the following:  Windows  data source C:\Users\&lt;username&gt;\AppData\Local\Temp\par-5061756c\ cache-890efc00b3ca6b775c7d44a325c1349fb2a3a3bd\inc\lib/Biber/LaTeX/recode_data.xml not found in .   Mac  data source /var/folders/v2/rld0ls7d2935gkvqv5hfr1x00000gn/T/par-616c616e/ cache-cdd483146f82a9655ce063f848d5139480fbf872/inc/lib/Biber/LaTeX/recode_data.xml not found in .   Linux  data source /tmp/par-6963617269756d/cache-f37ab610b7d79b2720a8ee3732849c6821705520 /inc/lib/Biber/LaTeX/recode_data.xml not found in .
Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian. Prove that if $g^2=e$ for all $g$ in $G$ then $G$ is Abelian.   This question is from group theory in Abstract Algebra and no matter how many times my lecturer teaches it for some reason I can't seem to crack it.   (Please note that $e$ in the question is the group's identity.)  Here's my attempt though...  First I understand Abelian means that if $g_1$ and $g_2$ are elements of a group $G$ then they are Abelian if $g_1g_2=g_2g_1$...  So, I begin by trying to play around with the elements of the group based on their definition...  $$(g_2g_1)^r=e$$  $$(g_2g_1g_2g_2^{-1})^r=e$$ $$(g_2g_1g_2g_2^{-1}g_2g_1g_2g_2^{-1}...g_2g_1g_2g_2^{-1})=e$$  I assume that the $g_2^{-1}$'s and the $g_2$'s cancel out so that we end up with something like,  $$g_2(g_1g_2)^rg_2^{-1}=e$$ $$g_2^{-1}g_2(g_1g_2)^r=g_2^{-1}g_2$$  Then ultimately...  $$g_1g_2=e$$  I figure this is the answer. But I'm not totally sure. I always feel like I do too much in the pursuit of an answer when there's a simpler way.  Reference: Fraleigh p. 49 Question 4.38 in A First Course in Abstract Algebra. Why is $L^{\infty}$ not separable? Is there a specific order to complete Saints Row 2? How can a module detect when the "access denied" page is output? How is it possible for a module to detect when Drupal is outputting the "access denied" page? I know how to do it with Drupal 6; I need to know how to do it with Drupal 7. Find the volume of rotation about the y-axis for the region bounded by $y=5x-x^2$, and $x^2-5x+8$  Here is an image:    Normally I can do this question, but this one is tricky because since we are rotating about the y-axis, and we are quadratic, when I solve for $x$ I get two answers, one positive and one negative.   What I mean is that if you look at the graph $5x-x^2$, then there are two halves, and if you look at the green line, that's one half of the curve (positive one), and the red half is the negative one.  Since we can see both the negative and positive sections in this graph, and both are being rotated, I wasn't sure how to create the equation that I can later integrate.  Usually in the problems I do, the negative doesn't exist, and so it's trivial. In this case I am not sure. Where do deleted items go on the hard drive? What happened to the Yin half of Kurama? Did Naruto get both halves of Kurama after fighting Sasuke, or did the other half of Kurama die when Minato was killed? What do you call a person who does/thinks the opposite of the crowd for just that reason? Such a person does not hold in high regard the decisions taken by a group (not a team) of people. She believes that groups tend to be anarchic and don't serve their purpose.  An example would be - for outings with friends, this person will not like planning it with the group since they will mess it up.  Therefore, this person has convinced herself that when more people are involved with something, it will be worse. She distrusts crowds instinctively basically. Stack overflow links not working Help with completing the proof: Any unbounded sequence is either infinitely large or has a convergent subsequential limit Given $\{x_n\}$ is an unbounded sequence, prove:   $$ \lim_{n\to\infty}x_n = \infty\ \  \text{or}\ \ \lim_{n_k \to \infty} = A $$   In other words any unbounded sequence either infinitely large or has a convergent subsequence.     Everything before udpate section is wrong.  Part 1. Proving $|x_n| &gt; M \implies \lim_{n\to\infty} x_n = \infty$  $\Box$ Start with the definition of an unbounded sequence: $$ \forall M &gt; 0 \in \Bbb R \ \exists N \in \Bbb N: \forall n &gt; N\implies |x_n| &gt; M $$  Suppose that: $$ \lim_{n\to\infty}x_n \ne \infty $$ That means: $$ \forall n\in\Bbb N\ \exists C &gt;0 \in\Bbb R : |x_n| &lt; C $$  Now taking $M^\prime = \max\{M, C\}$ we have that: $$ \begin{cases} \begin{align} \exists N \in \Bbb N : &amp;\forall n&gt;N \implies &amp;|x_n| &gt; M^\prime \\ &amp;\forall n \in \Bbb N: &amp;|x_n| &lt; M^\prime \end{align} \end{cases} $$  So we have arrived to a contradiction which means the assumption is wrong and: $$ \lim_{n\to\infty}x_n = \infty $$ $\Box$     Part 2. It's not clear to me how to prove that if unboundedness does not imply infinity limit then it must imply the existence of a convergent subsequence.   Two questions is my mind:   Is part 1 correct?  How do I proceed with the second part?     I'm using the following definition. $$ \lim_{n\to\infty}x_n = +\infty \stackrel{\text{def}}{\iff} \forall \epsilon &gt; 0 \exists N \in \Bbb N :\forall n &gt; N \implies x_n &gt; \epsilon \\ \lim_{n\to\infty}x_n = -\infty \stackrel{\text{def}}{\iff} \forall \epsilon &gt; 0 \exists N \in \Bbb N :\forall n &gt; N \implies x_n &lt; -\epsilon \\ $$ A sequence is considered infinitely large when: $$ \lim_{n\to\infty}x_n = \infty\ \ \text{when}\ \ \lim_{n\to\infty}x_n = + \infty \ \ \text{or} \ \ \lim_{n\to\infty}x_n = - \infty $$    Update.   Looks like i've messed up a lot of things. I will try once again. As shown in comments and answers the statement holds only in case: $$ \lim_{n\to\infty} |x_n| = \infty \iff \forall \epsilon &gt; 0 \exists N \in \Bbb N: \forall n &gt; N \implies |x_n| &gt; \epsilon $$  Consider the following: $$ \lnot P = \lnot\left(\lim_{n\to\infty}|x_n| = \infty \right) \iff \exists \epsilon &gt; 0\ \forall N_1 \in \Bbb N : \exists n &gt; N_1 \land |x_n| &lt; \epsilon $$  But on the other hand it is given that $x_n$ is unbounded: $$ Q = \forall M &gt; 0\ \exists N_2 \in \Bbb N : |x_{N_2}| &gt; M $$  Construct a negative expression for boundedness: $$ \lnot P = \exists M &gt; 0\ \forall N_2 \in \Bbb N : |x_{N_2}| &lt; M $$  If $S = \lnot P \implies \lnot Q$ then $S = P \lor \lnot Q$  Let $\epsilon = M$, choose $N = \max\{N_1, N_2\}$ then both statements are true and: $$ \exists \epsilon &gt; 0\ \forall N = \max\{N_1, N_2\}: \exists n &gt; N \land |x_n| &lt; \epsilon $$  Now either $P$ is true, which would mean $\lim |x_n| = \infty$, or $\lnot Q$ is true which would mean the sequence is bounded and hence contains a convergent subsequence. Inequality about infinite cardinals The following is a question I'm interested in solving:    The idea is to solve it without using AC. As @Asaf Karagila pointed out in the comment section:  "Even if $\kappa$ is an infinite cardinal, you can't assume that $\kappa \geq \omega$ without using AC".  I am still trying to think this through: My reasoning being, that as $\kappa &gt; n \forall n \in \omega$, then $\kappa \geq \sup\{n:n\in \omega\}=\omega$ as $\omega$ is a limit ordinal.  Where does my reasoning fail? And also, a detailed solution would be very appreciated. Resetting crontab file to default Word for killing off or attempting to kill off an entire bloodline? Why are the solutions of polynomial equations so unconstrained over the quaternions? Does a controlled mount share its rider's turn? From "Controlling a Mount" on p.198 of the Player's Handbook:     The initiative of a controlled mount changes to match yours when you mount it. It moves as you direct it, and it has only three action options; Dash, Disengage, and Dodge. A controlled mount can move and act even on the turn that you mount it.   Does this mean that the rider and mount share a single turn? Program counter goes to the end of code and 8051 Restarts I have an 8051 MCU which I use keil to program it with C. My problem is that the micro-controller restarts at the end of some functions while they should return.  Using a debugger I tried to check it line by line. And I figured out the the MCU is not restated right after the end of the function. The PC(progam counter) goes to somewhere else (sometime somewhere that there is not any code to execute) and then it increases one by one until the end of code. It is very strange for me because I write my code in C and never seen the type of error previously.    I think this problem might be related to the compiler settings and thus I post the settings here:   Assembler settings: XR GEN DB EP NOMOD51  Compiler settings:  DB OE BR LARGE OPTIMIZE(11, SIZE) DEFINE(SILABS_PLATFORM_WMB912) DEFINE(SILABS_WMCU_SI106x)  Linker settings: CODE REMOVEUNUSED CLASSES(BIT(0x20.0-0x2F.7), DATA(D:0-D:0x7F), IDATA(I:0-I:0xFF), XDATA(X:0-X:0xFFF), CODE(C:0-C:0xFBFF), CONST(C:0-C:0xFBFF))  I tried both large and small settings but the problem remains. Kindly help me if you know the reason. Clicking on tags in global search results is broken If you search for something on "all sites" (using the searchbox at the top of , you'll reach the search results, e.g.  .  If you click directly on a tag on that page, e.g.        You get 404 Page Not Found error, e.g.     Steps to reproduce   Go to  and enter e.g. "keyforge" into the "Search all sites" field. Arrive at  Click on the "keyforge" tag below a search entry. Reach  (404 page not found)   Expected behaviour  Reach   System Info  Happens in both Firefox and Vivaldi browsers on Windows 10. Say I had;   \begin{equation}   \begin{cases}      x = 0 \\      y = 0 \\      z = 0     \end{cases}     \qquad \text{and} \qquad     \begin{cases}      x = a \\      y = a \\      z = b     \end{cases}   \end{equation}   How could I insert a brace ( } ) in the second 'cases' to say x and y are equal to a. (i.e. } = a ).   I hope this is clear, if not I can sketch it out - just let me know!  Thanks in advance! Expected value as a function of quantiles? I was wondering where there is a general formula to relate the expected value of a continuous random variable as a function of the quantiles of the same r.v. The expected value of r.v. $X$ is defined as: $E(X) = \int x dF_X(x) $ and quantiles are defined as :  $Q^p_X = \{x : F_X(x) = p \} =F_X^{-1}(p) $ for $p\in(0,1)$.  Is there for instance a function function $G$ such that:  $E(X) = \int_{p\in(0,1)} G(Q^p_X) dp $ Negation in English