Cite footnote not showing in figure environment For citing I'm using biblatex with the chicago style, and I want to use \autocite within the caption for a figure. However, even though the footnote number is shown within the caption, the corresponding footnote text doesn't appear on the same page nor on any page in the whole document.  I'm aware that you have to use \protect for this kind of matter, but it still doesn't seem to affect the result.  Here the simplified MWE of my LaTeX code segment:  \documentclass{report} \usepackage{geometry} \usepackage[T1]{fontenc} \usepackage[utf8]{inputenc} \usepackage[english,ngerman]{babel} \usepackage{graphicx} \usepackage{varioref} \usepackage[pdfborder={0 0 0}]{hyperref} \usepackage{cleveref} \usepackage{etoolbox} \usepackage[style=authoryear-icomp,autocite=footnote]{biblatex} \usepackage{csquotes} \usepackage{lipsum}  \addbibresource{example.bib}  \begin{document}  \section{some section} \lipsum[1]  \subsection{some subsection} \begin{figure}[h!] \centering %\includegraphics[width=.5\textwidth]{...} \caption[here some text]{here some more text.\protect\autocite[Vgl.][11]{dirac1981principles}} \label{fig:figureone} \end{figure}  \end{document}   And here the according bibliography file:  % Encoding: UTF-8  @Book{dirac1981principles,   title     = {The Principles of Quantum Mechanics},   publisher = {Clarendon Press},   year      = {1981},   author    = {Paul Adrien Maurice Dirac},   series    = {International series of monographs on physics},   isbn      = {9780198520115},   keywords  = {physics}, }  @Comment{jabref-meta: databaseType:bibtex;}     I already tried the solution mentioned in the thread , and this is what I came up with:  \begin{figure}[h!]     \centering     %\includegraphics[width=.5\textwidth]{...}     \caption[here some text]{here some more text.\protect\footnotemark}     \label{fig:figureone} \end{figure} \footnotetext{\cite[Vgl.][11]{dirac1981principles}}   But this is not what I first intended, because I want to change the cite style (by the use of biblatex arguments) dynamically, without any manual interference for each occurrence of cites within figure environments.
Using \footnote in a figure's \caption Maybe this is an easy one, but I struggled with this now too long :)  I want to have a footnote in a caption of a figure, see the example.   \begin{figure}[!ht]   \caption{a figure caption\footnote{where i got it from}}   \label{somelabel}   \begin{center}        \pgfuseimage{...}   \end{center} \end{figure}   The compilation error reads as follows  ! Argument of \@caption has an extra }. &lt;inserted text&gt;              \par  l.192 ...i got it from}}   The actual tex code for my figure with the answer of Leo  \pgfdeclareimage[width=6cm]{aba.medcenter}{aba.medcenter}  \begin{figure}[!ht]   \begin{minipage}{\textwidth}     \caption[Medcenter Monthly Medication System]{Medcenter\textsuperscript\textregistered Monthly Medication System\footnote{Quelle Bild: http://www.amazon.com/dp/B000RZPL0M}}     \label{aba.medcenter}     \begin{center}       \pgfuseimage{aba.medcenter}     \end{center}   \end{minipage} \end{figure}   leads to the error  ! LaTeX Error: Command \itshape invalid in math mode.   in the same line. If I comment out the foot note, everything compiles fine. \textsuperscript\texttrademark isn't the problem, too.
If $f^3=\rm id$ then it is identity function Let $f:\mathbb{R}\to \mathbb{R}$ be a continuous function such that $f^3(x)=f\circ f\circ f(x)=x$ for all $x$.  How can I prove $f$ is the identity function? I've noticed this a few times on the Feed activity on the Android app, but never thought anything of it.  Why does this question (How To auto enable checkbox?) appear as "Unanswered" when there is an icon which denotes there are 5 answers, one of which has been marked as the answer?  Dragging down to refresh doesn't seem to clear the unanswered flag either.  Device: Nexus 5 Check all dependencies of a Record Type in the Organisation before deactivating it How can I ensure that old documents will compile in the future? Let $f$ be a measurable function on a measure space $X$ and suppose that $fg \in L^1$ for all $g\in L^q$. Must $f$ be in $L^p$, for $p$ the conjugate of $q$? If we assume that $\|fg\|_1 \leq C\|g\|_q$ for some constant $C$, this follows from the Riesz Representation theorem. But what if we aren't given that such a $C$ exists? Why teach linear algebra before abstract algebra? Is there a reason why most undergraduate curriculums put linear algebra before abstract algebra?  I'm asking this because personally it seems to be much easier to understand the architecture behind linear algebra as supposed to simply solving problems after a course in abstract algebra. From what I know, it is unknown whether $x_n=x_{n-1}^2 + 1$ has a closed form. Is there a recurrence relation which is known to have no closed form with a proof of inexistence?  Assuming a closed form is a non recursive description using the elementary operations of addition multiplication and power, or, assuming any other good definition of "closed form".  Edit  There are similar questions out there, but the answers are a little bit going around the question.   So is there one with proof or is it unknown? And if it depends on the definition of a closed form, then what are the (or some) options? How strong can a closed form definition be to still have a recursive relation that can be proved to not being able to have its form?  Update  Because it is a big question, and I don't want to open a new small one, I think a good representative small question about this subject is:  Is there a recurrence relation with a domain of the natural numbers, described by elementary functions, which has no closed form defined as:  A closed form of a recurrence relation with a domain of natural numbers is a function $f$ that can be described as a composition of a constant number of elementary functions without reccursion. i.e, $f(n) = f_1(f_2( \dots (f_k(n))) \dots) $ where $f_i$ is elementary for all $1 \leq i \leq k$   I think this definition is the most intuitive and basic.  For example, the relation $x_n = x_{n-1}+1$ has a closed form in this sense: $$f(n) = f_1(f_2(f_3(f_4(n))))$$  Where $f_1$ is division, $f_2(n)=(n,2)$, $f_3$ is multiplication and $f_4(n) = (n,n+1)$   Is there a recurrence relation which has no closed form in this sense? What are the most important things I need to do to encourage Google Sitelinks? My website used to have sitelinks and now it doesn't. It's very possible that it's due to changing the website to a sidebar design instead of having an "interstitial" type landing page which limited the number of choices, but I'm not sure.  Here is how sitelinks might look for a site:    What are some things that I can do to improve my chances of getting sitelinks? Mysterious cord running from wall. Safe? My wife and I are preparing to sell our home and are discovering some weird things. Our home was built in 1958, so it is older. We found this in a corner that had some furniture in front of it.   We are wondering    What kind of wiring is this?  Is it safe to handle or remove? HTML encoding strikes again (but with a twist) - Chat messages inbox notifications aren&#39;t looking good I'm struggling to understand the concept of quantum entanglement. I've distilled my understanding into an analogy, and I need your help to validate it. Here it is:  Let's say I receive two envelopes. Both envelopes have this written on them:     Hello! Open this letter...   If the paper inside is red, then other envelope has blue paper.   If the paper inside is blue, then other envelope has red paper.   So, if I open one letter, and the paper inside is red, then I know the other letter contains the blue paper.  Is this equivalent to quantum entanglement? The fact that I will know the color of the other paper when I open (i.e. observe) one envelope?  In theory, if one of the letters "appeared" in a different galaxy, I would still know the color of the paper inside that letter, just by opening my letter, instantly, correct?  Before I open the letter, is the paper inside each envelope in superposition (??), i.e. 50/50, red or blue?  In this analogy, I am assuming that what's written on the letters is true. Is this a correct assumption in quantum physics? I think so yes?  In real quantum physics, would I be able to change the color of one paper, so that I change the color of the other one? No right? Boot drops to a (initramfs) prompts/busybox Should I use a canonical link which refers to the current URL with pagination? How can I assign different random values to each row in a SELECT statement? Please look at this code:  create table #t1(   id int identity (1,1),   val varchar(10) );   insert into #t1 values ('a'); insert into #t1 values ('b'); insert into #t1 values ('c'); insert into #t1 values ('d');   Now, whenever you execute this  select *,      ( select top 1 val from #t1 order by NEWID()) rnd  from #t1 order by 1;   you will get a result with where all rows have the same random value. e.g.  id          val        rnd ----------- ---------- ---------- 1           a          b 2           b          b 3           c          b 4           d          b   I know a way using a cursor to loop throw the rows and get different random values, but that is not performant.   A clever solution to this is   select t1.id, t1.val, t2.val from #t1 t1     join (select *, ROW_NUMBER() over( order by NEWID()) lfd from #t1) as t2 on  t1.id = t2.lfd    But I simplified the query. The real query looks more like  select *,      ( select top 1 val from t2 where t2.x &lt;&gt; t1.y order by NEWID()) rnd  from t1 order by 1;   and the simple solution doesn't fit. I'm looking for a way to force repeated evaluation of  ( select top 1 val from #t1 order by NEWID()) rnd    without the use of cursors.  Edit: Wanted output:  perhaps 1 call  id          val        rnd ----------- ---------- ---------- 1           a          c 2           b          c 3           c          b 4           d          a   and a second call  id          val        rnd ----------- ---------- ---------- 1           a          a 2           b          d 3           c          d 4           d          b   The value for each row just should be a random value independent from the other rows  Here is the cursor version of the code:  CREATE TABLE #res ( id INT, val VARCHAR(10), rnd VARCHAR(10));  DECLARE @id INT DECLARE @val VARCHAR(10) DECLARE c CURSOR FOR SELECT id, val FROM #t1 OPEN c FETCH NEXT FROM c INTO @id, @val WHILE @@FETCH_STATUS = 0 BEGIN     INSERT INTO #res     SELECT @id, @val, ( SELECT TOP 1 val FROM #t1 ORDER BY NEWID()) rnd      FETCH NEXT FROM c INTO @id, @val END CLOSE c DEALLOCATE c  SELECT * FROM #res I'm rotating an object on two axes, so why does it keep twisting around the third axis? I see questions come up quite often that have this underlying issue, but they're all caught up in the particulars of a given feature or tool. Here's an attempt to create a canonical answer we can refer users to when this comes up - with lots of animated examples! :)    Let's say we're making a first-person camera. The basic idea is it should yaw to look left &amp; right, and pitch to look up &amp; down. So we write a bit of code like this (using Unity as an example):  void Update() {     float speed = lookSpeed * Time.deltaTime;      // Yaw around the y axis using the player's horizontal input.             transform.Rotate(0f, Input.GetAxis("Horizontal") * speed, 0f);      // Pitch around the x axis using the player's vertical input.     transform.Rotate(-Input.GetAxis("Vertical") * speed,  0f, 0f); }   or maybe  // Construct a quaternion or a matrix representing incremental camera rotation. Quaternion rotation = Quaternion.Euler(                         -Input.GetAxis("Vertical") * speed,                          Input.GetAxis("Horizontal") * speed,                          0);  // Fold this change into the camera's current rotation. transform.rotation *= rotation;   And it mostly works, but over time the view starts to get crooked. The camera seems to be turning on its roll axis (z) even though we only told it to rotate on the x and y!    This can also happen if we're trying to manipulate an object in front of the camera - say it's a globe we want to turn to look around:    The same problem - after a while the North pole starts to wander away to the left or right. We're giving input on two axes but we're getting this confusing rotation on a third. And it happens whether we apply all our rotations around the object's local axes or the world's global axes.  In many engines you'll also see this in the inspector - rotate the object in the world, and suddenly numbers change on an axis we didn't even touch!    So, is this an engine bug? How do we tell the program we don't want it adding extra rotation?  Does it have something to do with Euler angles? Should I use Quaternions or Rotation Matrices or Basis Vectors instead? How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$? How to calculate $\lim_{x\to 1^+} \log (x)^{\log(x)}$ ? i know that its "1", but why? How can i calculate this?  Thank you very very much =) How can a "proper" function have a vertical slope? Plotting the function $f(x)=x^{1/3}$ defined for any real number $x$ gives us:   Since $f$ is a function, for any given $x$ value it maps to a single y value (and not more than one $y$ value, because that would mean it's not a function as it fails the vertical line test). This function also has a vertical tangent at $x=0$.   My question is: how can we have a function that also has a vertical tangent? To get a vertical tangent we need 2 vertical points, which means that we are not working with a "proper" function as it has multiple y values mapping to a single $x$. How is it possible for a "proper" function to have a vertical tangent?  As I understand, in the graph I pasted we cannot take the derivative of x=0 because the slope is vertical, hence we cannot see the instantaneous rate of change of x to y as the y value is not a value (or many values, which ever way you want to look at it). How is it possible to have a perfectly vertical slope on a function? In this case I can imagine a very steep curve at 0.... but vertical?!? I can't wrap my mind around it. How can we get a vertical slope on a non vertical function? Possible Duplicate:           What book would you recommend for an undergraduate wanting to learn the basics of topology? I've come across several books that claim to give an introduction to the subject, but I wasn't sure which one(s) to pick.  Here are some of the books I came across:   Introduction to Topology - Bert Mendelson Basic Topology - M.A. Armstrong First Concepts of Topology - William G. Chinn Topology Without Tears - Sidney A. Morris Topology - James Munkres   They don't go over the exact same topics, some are more in-depth than others, some seem to target slightly different audiences etc. What would be right for me (not necessarily from that list)? To give some more background, I want to learn topology because I feel it will be somewhat helpful in the field of Computer Vision. How do you unit test private methods? Prove that the set of palindromes are not regular languages using the pumping lemma. Firstly I pick a language $xyz$ where $x = \epsilon$, $y = (abb)^{k}$, $z = (bba)^{k}$ where $|y| \ge$ the number of states in the automaton representing my language. Then $xyz = (abb)^k(bba)^k$ is a palindrome.  Now I split $y$ into $uvw$, where $v \ne \epsilon$ contains a state visited more  than once. If the language is regular then $xuv^iwz$ is within the language $\forall i \ge 0$.  I choose $i = 2$, then $uv^2w = (abb)^n$ where $n &gt; k$.  Then $xuv^2wz = (abb)^n(bba)^k$ where $n &gt; k$, which is not a palindrome.   Firstly, is my reasoning for this particular proof correct?  I'm unsure about how to properly apply the pumping lemma to this problem. And secondly the question asks me to state a proof for palindromes in general, whereas I only attempted to prove it for a single case. Is there an example I can choose which proves the conjecture for all palindromes? I assume not, seeing as I can't think of how that could be represented using regular expressions.