I am trying to construct a macro to use the editor field of @thesis entries to assign the supervisors of the thesis. I used the following post as a starting point  but this prints out the supervision and joint supervision strings in bold fonts (not sure why) instead that using the same fnt than the rest of the reference.  Here an MWE taken from  that I cannot get to output what the post intended to help doing as is provided. Perhaps the first would be to help me understand why this code does not print the supervisors and then how to give the supervision and joinsupervision strings an italic rather than bold font:  \documentclass[british,12pt,a4paper]{article} \usepackage{filecontents} \begin{filecontents*}{\jobname.bib} @thesis{geer,         author       = {de Geer, Ingrid},         title        = {Earl, Saint, Bishop, Skald~-- and Music},         type         = {phdthesis},         institution  = {Uppsala Universitet},         date         = 1985,         subtitle     = {The Orkney Earldom of the Twelfth Century. A Musicological             Study},         location     = {Uppsala},         supervisor       = {James Oint and Stan Upervisor},     }      @thesis{loh,         author       = {Loh, Nin C.},         title        = {High-Resolution Micromachined Interferometric Accelerometer},         type         = {mathesis},         institution  = {Massachusetts Institute of Technology},         date         = 1992,         location     = {Cambridge, Mass.},         supervisor       = {Stan Upervisor},     } \end{filecontents*} \usepackage[utf8]{inputenc} \usepackage{babel} \usepackage{csquotes} \usepackage{xpatch} \usepackage[style=authoryear,backend=biber]{biblatex} \usepackage{hyperref}   \begin{filecontents*}{english-thesis.lbx}     \ProvidesFile{english-thesis.lbx}[2014/06/14 english for thesis]     \InheritBibliographyExtras{english}     \NewBibliographyString{supervision,jointsupervision}     \DeclareBibliographyStrings{%         inherit           = {english},         supervision       = {{under the supervision of}{under sup\adddotspace of}},         jointsupervision  = {{under the joint supervision of}{under joint sup\adddotspace of}},     } \end{filecontents*}  \DeclareLanguageMapping{english}{english-thesis}  \newbibmacro*{thesissupervisor}{%     \ifnameundef{editor}{}{%         \ifnumgreater{\value{editor}}{1}         {\bibstring{jointsupervision}}         {\bibstring{supervision}}         \printnames{editor}}}  \xpatchbibdriver{thesis} {\printfield{type}} {\printfield{type}     \newunit     \usebibmacro{thesissupervisor}} {\typeout{yep}} {\typeout{no}}  \addbibresource{\jobname.bib} \nocite{*} \begin{document}     \printbibliography \end{document}
In the biblatex doc, it says ( Version 1.7, p.11):     thesis      A thesis written for an educational institution to satisfy the   requirements for a degree. Use the type field to specify the type of   thesis.      Required fields: author, title, type, institution, year/date      Optional fields: subtitle, titleaddon, language, note, location,   month, isbn, chapter, pages, pagetotal, addendum, pubstate, doi,   eprint, eprintclass, eprinttype, url, urldate   I would like to know if there is any "canonical" way to add the name of the supervisor(s) in a field of the .bib file (no .bbl manipulation please, I want my bibliography to be usable "as it" in several documents).  The expect result could be something like  :     Meand Myself. “A brilliant result, explained”. PhD thesis**, under the   joint supervision of Pr. Toto and Pr. Tata**. Université Paris 24,   Nov. 2013.   The "under the (joint) supervision" could be a string that would be declined in four ways:   under the supervision of  under the joint supervision of and the abridged forms
How to show that the commutator subgroup is a normal subgroup It is suggested as an exercise in Serge Lang's book "Algebra" to show that the commutator subgroup $G^c$ of a group $G$ is a normal subgroup.   I'd like to do that but I am afraid I need help,   I think the first thing I need to figure out is how a general element in the commutator subgroup looks like, so that I can check that the defining condition for normality is satisfied.  That is, supposing for a moment that a general element in $G^c$ is denoted by $g$, I need to show that $aga^{-1} \in G^c,$ for all $a \in G$.  But here I get stuck, first because I am unsure how to write a general element in $G^c$ - a simple product in $G^c$ is of the form $xyx^{-1}y^{-1}aba^{-1}b^{-1}$ where $a,b \in G$. I cannot see a way to simplify this - I am sure there is one, but somehow I am blind today.  The second thing then is, even if one tries out the conjugation of a simple element like $xyx^{-1}y^{-1}$ in $G^c$, again not simplification offers itself easily I think .. what am I missing ?  An alternative would be to find a homomorphism of $G$ whose kernel is precisely $G^c$ - here I tried to think of this as a map $G \times G \to G$ but whatever I cook up is not a homomorphism.  Thanks for your hints !! Why is it a bad practice to lock the object we are going to change? Can you concentrate on a special ability while also concentrating on a spell? There exist several abilities that have the text similar to, this effect lasts while you maintain concentration (as if you were concentrating on a spell). One such example is the Bard College of Glamour: Mantle of Majesty ability.     At 6th level, you gain the ability to cloak yourself in a fey magic   that makes others want to serve you. As a bonus action, you cast   command, without expending a spell slot, and you take on an appearance   of unearthly beauty for 1 minute or until your concentration ends (as   if you were concentrating on a spell). During this time, you can cast   command as a bonus action on each of your turns, without expending a   spell slot.   Another example is the Cleric Trickery Domain: Channel Divinity: Invoke Duplicity ability.     Starting at 2nd level, you can use your Channel Divinity to create an   illusory duplicate of yourself.      As an action, you create a perfect illusion of yourself that lasts for   1 minute, or until you lose your concentration (as if you were   concentrating on a spell).   While you are concentrating on maintaining such an ability would you also be able to maintain concentration on a spell that requires concentration?  There does not seem to be any RAW that limits this however the limitation around concentrating on a second spell exists which may suggest that the intent is that a character cannot concentrate on 2 effects at the same time unless explicitly stated. I have made a script that will sort files according to their extension and place them in the proper folder. For example, place abc.jpg in the directory jpg.  #!/bin/bash #this script sorts files according to their extensions oldIFS=$IFS IFS=$'\n' (find . -type f) &gt; /tmp/temp for var in `cat /tmp/temp` do name=`basename "$var"` ext=`echo $name | cut -d'.' -f2- | cut -d'.' -f2- | cut -d'.' -f2- | cut -d'.' -f2- | cut -d'.' -f2- | cut -d'.' -f2- | cut -d'.' -f2-` mkdir -p $ext mv "$var" $ext/ 2&gt; /dev/null done IFS=$oldIFS   problem with this script:    it involves use of IFS, it is said to avoid use of IFS, as much as possible it does not sorts file without file extensions  it will sort files like abc.tar.bz in folder named bz, but however such a file should go in tar.bz folder see line 9 of my script; if any file contain more no. of dots(in its name) than no. of cut -d'.' -f2- in the script than if will result in file name taken in extension part. for example, a file named i.am.live.in.india.and.i.study.computer.science.txt will be placed in folder named study.computer.science.txt   you may also suggest any tweaks to make this script more smaller and neat. Conformally Map Region between tangent circles to Disk SQL Server Query Plan Hash Collision I'm wondering if anyone can shed light on a worry of mine that a query_plan_hash collision could cause a query to be executed as an entirely different query.  The hash is a 16 digit Hex that sp_help sys.dm_exec_query_stats says is a binary. Therefore it's only a 64 bit hash, and a collision seems very probable (considering SHA1 [160 bit] was just verified to have had collisions).  Would the plan_hash and the query_plan_hash both have to collide for this (query to be executed as an entirely different query) to happen?  I'm also curious if there is a setting somewhere in SQL Server that will allow us to change this hash to SHA2-512 (to reduce the possibility of a collision). Our data is very important.  I searched high and low via Google and Stack Exchange forums. What is the most efficient way to convert a MySQL query to CSV in PHP please?  It would be best to avoid temp files as this reduces portability (dir paths and setting file-system permissions required).  The CSV should also include one top line of field names. Can Raise Dead revive a character with zero maximum Hit Points? So, a character died from Draining Kiss of a succubus and his Max HP is 0. Can he be restored by Raise Dead, or does he need a higher level spell? Or a Greater Restoration cast on his lifeless body before Raise dead? I'm probably interested in RAW interpretation, since the character is in Adventurerer's league. Conjugate subgroup strictly contained in the initial subgroup? Let $G$ be a group, $H\subseteq G$ a subgroup and $a\in G$ an element of the group. Is it possible that $aHa^{-1} \subset H$, but $aHa^{-1} \neq H$?   If $H$ has finite index or finite order, this is not possible. Does a capacitor store charge? There're some easy concepts which I don't get exactly in my mind. I'm afraid I've been studying these things for two years of my engineering but they still bother me. Capacitor is one of them. Can someone explain?    What does a capacitor do? Does it store charges? If so, then how does it do?   I have searched it on Google and Yahoo but didn't find any helpful thing there (for me). So I will be glad if I got my problem solved here.  P.S. I hope that the question would not again be an off-topic, as it always does and moreover people don't suggest then where to go. It's a real sad thing. Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:  $$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$  I really have no idea why this statement is true. Can someone please explain why this is true and if possible show how to arrive at one given the other? Possible to do a MySQL foreign key to one of two possible tables? If $f^2$ is an analytic function then so is $f$ I want to show the following:  If $f(z) $ is a continuous  function on a connected open subset of the complex plane and $f(z)^2$ is an analytic function, then $f(z)$ is analytic.  Clearly if $f(z) \neq 0$ then $$\frac{f(z+h)-f(z)}{h}=\frac{f(z+h)^2-f(z)^2}{h}.\frac{1}{f(z+h)+f(z)}$$  which shows $f^\prime(z) $ exists if $f(z)\neq0$.  What do I do when $f(z) =0$? How do I migrate an SVN repository with history to a new Git repository? I need to add a coloured line before my caption; this is what I used:  \usepackage[labelfont=bf]{caption}  \DeclareCaptionFormat{myformat}{\hrulefill #1  #2 #3} \captionsetup[figure]{format=myformat}   but the figure turns out to be line this:    and not a horinzontal line which what I want  2) I want the line to be coloured RGB [152 204 204] with thickness about the same as \top rule. Can anyone show me how to do this? Does the Symmetric difference operator define a group on the powerset of a set? $G$ is the set of all subsets of a set $A$, under the operation of $\triangle\;$: Symmetric Difference of sets.  $A$ has at least two different elements.  I need to check if this is a group, and if it does to show if the group is abelian and/or finite.  Associativity - easy from Symmetric Difference. Identity element - empty group. Inverse element - each element is inverse to itself. Am I right?  abelian? finite? Show that a connected metric space is $\epsilon$-chainable for $\epsilon>0$ Show a connected metric space (X,d) is $\epsilon$-chainable for $\epsilon &gt;0$       $\epsilon$-chainable Definition      (X,d) is $\epsilon$-chainable if given any two points $a,b \in X$, $\exists$ a finite set $\{ x_0,x_1,...,x_n\}$ of points in X with $x_0 = a$, $x_n=b$ and $d(x_{n-1},x_n)&lt;\epsilon$, $\forall$ n      Clopen Definition      A set is clopen $iff$ its boundary is empty      Connected Metric Definition      Let $(X,d)$ be a metric space. It is connected $iff$ it has no proper subsets which are clopen      The following are equivalent:         $X$ is connected   The only subsets of $X$ that are clopen are $\emptyset$ and $X$   $\nexists$ non-empty, disjoint, open sets $A$ and $B$ such that $X=A\bigcup B$   $\nexists$ non-empty, disjoint, closed sets $A$ and $B$ such that $X=A\bigcup B$         Theorem      Let $(X,d)$ be a metric space, and let $S$ be any subset of $X$. Then $S$ is closed $iff$ $S^c$ is open.     So how would someone show this exactly?  Let the set $A$ be the set of all elements in $(X,d)$ that can be reached from $a\in X$ in a finite number of steps with length less than $\epsilon$.  Does this mean that there should be no isolated points? (an isolated point would be at distance infinity from any other point, or distance zero from itself, since $\epsilon &gt;0$ we don't want this?)  Knowing that $X$ is connected $\implies$the only subsets of X that are clopen are $\emptyset$ and $X$  Also, by construction of A;   A is the set of all $\epsilon$-chains between any element and $a$  And, so $A^c = \emptyset$ if $A$ was $\epsilon$-chainable (because A has to have a path from all elements to a, for this to be true, the complement wouldn't have anything in it)  Leaving only $A$ as the remaining clopen set  Does this make any sense? Or should an alternative be used, and how to formalize it.  Thanks. Evaluation of $\sum_{n=0}^{\infty}\frac{1}{(n^4+n^2+1)n!}$ A possible incorrect application of Law of Large numbers Word for a person being used I'm looking for a word to describe someone who is being used. I want a noun to describe this person, not a verb or or adjective.  Maybe like a pushover.