I just installed MacTeX on a new machine. I'm running into a difficulty I hadn't had before.   I have a .cbx file inside ~/Library/texmf/tex/latex/biblatex/cbx/ that has the same name as another .cbx file inside /usr/local/texlive/2017/texmf-dist/tex/xelatex.   If I run LaTeX on a file calling the corresponding citation style (\usepackage[citestyle=name-of-cbx-file]{biblatex}) the .cbx file inside the texmf tree inside ~/Library is given priority. If I run xelatex on the same file, however, the one that is inside texmf-dist is given priority.   This is the output of kpsewhich -expand-var='$TEXINPUTS':  .:{{}/Users/apc/Library/texlive/2017/texmf-config,/Users/apc/Library/texlive/2017/texmf-var,/Users/apc/Library/texmf,!!/usr/local/texlive/texmf-local,!!/usr/local/texlive/2017/texmf-config,!!/usr/local/texlive/2017/texmf-var,!!/usr/local/texlive/2017/texmf-dist}/tex/{kpsewhich,generic,}//   Shouldn't xelatex and latex be searching for biblatex style files in the same way? If so, why is xelatex not finding the files?
I've just made a minor alteration to a file from a TeX package.  I copied it into my local texmf tree and made the change there.  When I tried kpsewhich &lt;filename&gt; to test that it was finding my local version, it returned the original one and not the new one.  In actual fact, the original file was from my local texmf tree already so I renamed the old one something simple.  Now kpsewhich &lt;filename&gt; returned the system version of the file.  To check that the file was there, I tried kpsewhich -all &lt;filename&gt; and got the correct two versions of the file: my modified one and the system one (in that order).  If I try this with another file that I know exists in both places, then kpsewhich &lt;filename&gt; returns my local file.  Here's the various invocations and their results:  ~% kpsewhich pgfsys-pdftex.def /usr/local/texlive/2012/texmf-dist/tex/generic/pgf/systemlayer/pgfsys-pdftex.def ~% kpsewhich -all pgfsys-pdftex.def  /Users/stacey/Library/texmf/tex/latex/bzr/pgfsys-pdftex.def /usr/local/texlive/2012/texmf-dist/tex/generic/pgf/systemlayer/pgfsys-pdftex.def ~% kpsewhich tqft.sty               /Users/stacey/Library/texmf/tex/latex/bzr/tqft.sty ~% kpsewhich -all tqft.sty /Users/stacey/Library/texmf/tex/latex/bzr/tqft.sty /usr/local/texlive/2012/texmf-dist/tex/latex/tqft/tqft.sty   Is it something special about the .def extension?  Is there some command that I forgot to run?  Are there some other options to kpsewhich that will help me debug this?
Word that means "outdated name" Is there a word that means "outdated name"? For example:   Record, although very little music is on vinyl Film, although most movies are digital Horsepower, although no one uses horses as a metric anymore Phone, although it's mostly used to access the web   and so on. How to set JAVA_HOME to point to a JDK installation? I installed sun-java6-jdk.  I need to find its home directory. JAVA_HOME must point to a JDK installation, not JRE. $a+1,a-1$ invertible for nilpotent element Given a ring $R$ (with unity), prove that if $a \in R$ is nilpotent, then $a+1,a-1$ are both invertible.   Suppose $a^n=0$. Then $1=1-a^n=(1-a)(1+a+\ldots+a^{n-1})$, so $1-a$ is invertible.  If $n$ is odd we can use $1+a^n=(1+a)(1-a+\ldots+a^{n-1})$.  But what if $n$ is even? The following time-frequency correspondence illustration is in .    Also a gif animation version (if the animation doesn't show, please open the following image in a new window):    This picture has so much explanatory power, and I would like to replicate it in TikZ for future use.  Here is what I came up with:  \documentclass{minimal} \usepackage{tikz}  \begin{document} \begin{tikzpicture} [x={(1cm,0.5cm)},z={(0cm,0.5cm)},y={(1cm,-0.2cm)}] \draw[-&gt;,thick,blue!90] (0,6.5,0) -- (6.2,6.5,0) node[right] {Frequency}; \draw[-&gt;,thick,red!90] (0,0,0) -- (0,6.5,0) node[below] {Time}; \draw[-&gt;,thick] (0,0,0) -- (0,0,2) node[above] {Magnitude}; \foreach \x in {0.5,1.5,2.5,3.5,4.5,5.5}{   \draw[blue!50] (\x,0,0)   \foreach \y in {0,0.02,...,6.28}{     -- ({\x},{\y},{sin(\x*\y*(157))/sqrt(2*\x)})   };   \draw[blue!90, thick] (\x,6.5,0) -- (\x,6.5,1/\x); }  \end{tikzpicture} \end{document}   This is the output so far.    I have two questions:   How to produce that red superposed sine wave of all the blue sine waves? I don't know if there is a sum function or I have to use loop yet again? How to make the camera projection in TikZ more similar to the perspective in that wikipedia illustration?   Any suggestion and tweaking of the parameters I used in the sample drawing are welcome as well.  Thanks in advance!    Update 1:  Here is a new version using tikz, more readable than the first one. Yet the superposition of the sine waves are done manually...I still don't know how to use foreach to produce a sum.  \begin{tikzpicture}[x={(1cm,0.5cm)},z={(0cm,0.5cm)},y={(1cm,-0.2cm)}] \draw[-&gt;,thick,black!70] (0,6.5,0) -- (6.2,6.5,0) node[right] {Frequency}; \draw[-&gt;,thick,black!70] (0,0,0) -- (0,6.5,0) node[below right] {Time}; \draw[-&gt;,thick] (0,0,0) -- (0,0,2) node[above] {Magnitude}; \foreach \y in {0.5,1.5,...,5.5}{ \draw [cyan!50, domain=0:2*pi,samples=200,smooth]   plot (\y,\x, {sin(4*\y*\x r)/\y }); \draw[blue, ultra thick] (\y,6.5,0) -- (\y,6.5,1/\y); } \draw [red, thick, domain=0:2*pi,samples=200,smooth]  plot (0,\x, {sin(4*0.5*\x r)/0.5 + sin(4*1.5*\x r)/1.5 + sin(4*2.5*\x r)/2.5 + sin(4*3.5*\x r)/3.5 + sin(4*4.5*\x r)/4.5 + sin(4*5.5*\x r)/5.5} ); \end{tikzpicture}   The result is as follows: Probability of an even number of sixes We throw a fair die $n$ times, show that the probability that there are an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$. For the purpose of this question, 0 is even.   I tried doing this problem with induction, but I have problem with induction so I was wondering if my solution was correct:   The base case: For $n=0$, our formula gives us $\frac{1}{2}[1+(\frac{2}{3})^0] =1$. This is true, because if we throw the die zero times, we always get zero sixes.  Suppose it's true for $n=k$. Then the odds of an even number of sixes is $\frac{1}{2}[1+(\frac{2}{3})^n]$, and thus the odds of an odd number of sixes is $1 - \frac{1}{2}[1+(\frac{2}{3})^n]$.    For $n=k+1$, there are two ways the number of sixes are even:  a. The number of sixes for $n=k$ was even, and we do not throw a six for $n=k+1$: $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n]$  b. The number of sixes for $n=k$ was odd, and we throw a six for $n=k+1$: $\frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$  So the probability $p$ for an even number of sixes at $n=k+1$ is $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$  I have two questions   How do I get from $ \frac{5}{6} \cdot  \frac{1}{2}[1+(\frac{2}{3})^n] + \frac{1}{6}(1 - \frac{1}{2}[1+(\frac{2}{3})^n])$ to $\frac{1}{2}[1+(\frac{2}{3})^n]$? I seem to have done something wrong, I can't get the algebra correct, I get $p = \frac{1}{3}[1+(\frac{2}{3})^n] + \dfrac{1}{6}$ Other than that, is my use of induction correct? Is it rigorous enough to prove the formula? Where to ask Question that can be flagged too broad or opinion based I am looking for information on Heinrich Märte (my grandfather) and his family. He was born on July 1, 1912 in Sipplingen and died On April 19, 1944 when the military plane crashed in Geierswalde because its engine failed.  I am not sure if his family knows that he fathered my dad. I am hoping someone knows of him or his family. I would love to know how he met my grandmother or any further information on him.   We have always had two photo's of him, but never knew his name. We have only just found out about him and it has opened so many more questions.  How can I determine Heinrich's parents and siblings, and then locate any living family members? What is a current replacement of gnome-open command (generic open of files based on the type)?  Before:   gnome-open mydoc.pdf  # opened PDF in default application   Now:  gnome-open The program 'gnome-open' is currently not installed.  You can install it by typing: sudo apt-get install libgnome2-0 If a linear operator has an adjoint operator, it is bounded Cannot find Background Images feature in Blender 2.8 Cannot find Background Images feature to add in a photo in Blender 2.8  Where is it? Reading through my textbook I came across the following problem and I am looking for some help solving it.  Prove that,  $$\binom{3n}{r}=\binom{3n-1}{r} + \binom{3n-1}{r-1}$$  In a previous problem I solved the following but I wasn't sure how to use it to prove the question above. Prove,  $$\binom{n}{r}=\binom{n}{n-r}$$  Solution:  left Side,  $$\binom{n}{r}=\frac{n!}{(n-r)!r!}$$  right side,  $$\binom{n}{n-r}=\frac{n!}{(n-(n-r))!(n-r)!} = \frac{n!}{(n-r)!r!}$$  Therefore completing the proof,  but how do I solve my first question? thanks! The complex general linear group is a subgroup of the group of real matrices of twice the dimension and with positive determinant.  Let us decompose complex matrices $M$ as $M=A+iB$, where $A,B$ are real matrices. Now consider the correspondence $$f(A+iB)=\begin{pmatrix} A &amp; -B \\ B &amp; A\end{pmatrix}.$$  If $\det f(M)=|\det M|^2$ for square matrices, then we would have $GL(n,\mathbb C)\subseteq GL_+(2n,\mathbb R)$ with the identification $M\to f(M)$, which is an injective homomorphism. In other words, the complex general linear group would be a subgroup of the group of real matrices of twice the dimension and with positive determinant.  How is $\det f(M)=|\det M|^2$? LM317 getting hot Should an LM317 (specifically a ) rated for 500ma output current really be getting hot to touch when only pulling 170ma at 5V?  Package is SOT-223, input is 9V.  Is this normal? Terminal not displaying colors for directory listings Remove votes to close question X as a duplicate of Y if question Y is closed as a duplicate of X I just happened across , which has one close vote suggesting that it be closed as an exact duplicate of .  This suggestion is perfectly correct: these are exactly identical questions asked by the same user.  However, there's not much point in closing question A as a duplicate of B, since question B has already been closed as a duplicate of A.  Yet the close vote remains, doing nothing but wasting reviewers' time.  This is not the first time I've seen this happen, so I'd like to suggest that the Stack Exchange software should do something about it.  In particular, I'd like to suggest the following feature:     When a question Y is closed as an exact duplicate of another question X, any votes or flags to close X as a duplicate of Y should be automatically discarded.   Actually, I'm kind of surprised that the SE software doesn't do that already.  If anyone else thinks that this constitutes an actual bug, feel free to retag it. Show that the countable product of metric spaces is metrizable I have a div 200 x 200 px. I want to place a 50 x 50 px image right in the middle of the div.   How can it be done?  I am able to get it centered horizontally by using text-align: center for the div. But vertical alignment is the issue.. Why doesn't \makeatletter work inside \newcommand? After hours of banging my brain against my skull, I narrowed down an issue I'm having to the fact that \makeatletter doesn't work inside \newcommand. Commands that I call after \makeatletter are registered as undefined, as if \makeatletter didn't do anything.  Why is this the case?  In the example below, I define two commands. Calling one of them from the document body causes the undefined error, while the other doesn't.  \documentclass[11pt]{book} \usepackage{kvsetkeys}  \newcommand{\commandThatFails}{     \makeatletter \comma@parse{}{} \makeatother }  \makeatletter     \newcommand{\commandThatWorks}{         \comma@parse{}{}     } \makeatother  \begin{document}  Hi there.  \commandThatFails \commandThatWorks  \end{document} Should I use 'was' or 'were' in this example?     I was always delighted when my brother or one of my sisters was/were asked to do them. In school, we just started learning about trigonometry, and I was wondering: is there a way to find the sine, cosine, tangent, cosecant, secant, and cotangent of a single angle without using a calculator?  Sometimes I don't feel right when I can't do things out myself and let a machine do it when I can't.  Or, if you could redirect me to a place that explains how to do it, please do so.  My dad said there isn't, but I just had to make sure.  Thanks.