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I have a background in QFT, GR and differential geometry at the level of a master student in theoretical physics. I would like to touch the area of CFT. I know the textbook of Philippe Di Francesco. It may be too big for the beginner like me. Are there some good introductory lectures or textbooks adapted to the needs of beginners? How to add a custom field to the media screen (image/gallery)? I’m trying to add a custom field to image and gallery edit screen.  I’m aware there are similar questions here, although answers either don’t work with the current version or they add field only to upload screen while I need a field to be visible in all edit screens (see below).  For this, I believe, you’ll have to mess with backbone.js as attachment_fields_to_edit only adds to upload screen.  Adding field with attachment_fields_to_edit    ↑ This is upload image screeen. Here Style is added with attachment_fields_to_edit filter, and my_field is added with ACF plugin.   But they are missing in edit screen in the post    ↑ Click edit on the actual post    ↑ No style &amp; my_field fields!  The question  How to add fields to still have them on the edit screen? Ideally answer will include adding fields to gallery edit screen if it’s a similar process  This question is really important for me so I’ll be adding a 100 rep bounty when it’s available.   Thanks! Multiple panel figure, with caption in one of the panels? I want to put three figures inside a multi-figure with a single caption. If you imagine a 2x2 table, I want to place the three figures in three of the table cells, and the caption in the remaining cell.   Each subfigure should have its own label, and small captions (like (a), (b), and (c), to number the figures inside the panel). Then inside the caption for the whole figure, I want to reference the subfigures, but without using the figure number (that is, just (a) instead of Fig.3(a)), but outside the panel, I'll reference the subfigures as Fig.3(a), etc.  For example,  \begin{figure} \centering     \begin{subfigure}{.4\textwidth}         \centering         \includegraphics[width=\linewidth]{fig_a.pdf}         \caption{}\label{fig:fig_a}     \end{subfigure} %     \begin{subfigure}{.4\textwidth}         \centering         \includegraphics[width=\linewidth]{fig_b.pdf}         \caption{}\label{fig:fig_b}     \end{subfigure} %     \begin{subfigure}{\textwidth}         \centering         \includegraphics[width=\linewidth]{fig_c.pdf}         \caption{}\label{fig:fig_c}     \end{subfigure} \caption{Some general caption of all the figures. In (\subref{fig:fig_a}) you can see a green square....} \end{figure}   But this puts two figures ((a) and (b)) side by side on the first row, then (c) alone on the second row, and the caption appears below. The second row has much space unused. The following is a picture of what I expect to get:    How can I do this? Here's what is happening; everytime I go to Stack Overflow (stackoverflow.com, NOT meta) with Firefox, it gives me the "Welcome back" tab that pops up from the top. Whenever I click in the "click here to refresh the page" link, all it does is refresh the page without logging me back in.  Even if I try to log in using the "log in" button at the top, all it does is pop up the tab that says "Welcome name here, you're being redirected." The problem is I'm not being logged in, not to mention that it keeps refreshing the page over and over again without giving me a choice to log in with Facebook.  For some reason, I was capable of logging in to Meta Stack Overflow using Firefox. I am still unable to log in to Stack Overflow. Any ideas? Can anyone explain how one would convert one voltage range to another? I'm going to need to convert a range of -10 to +10 volts into a range of 0 to +5 volts. How would this be accomplished? Thanks!   Let me say  something more. Inside of a torpedo there are gyroscopes. This gyroscope give -10 volts when the torpedo have 180 degrees of deviation to the west and 10 volts when the deviation is to the east. This signal is DC, so i need to change this range a range: 0 to 5 v DC . Thanks Suppose that $X_1,X_2,...,X_n$ is a random sample from a normal distribution, $X_i\sim N(\mu,9)$.  Find the UMVUE (uniformly minimum variance unbiased estimator) of $P(X\le c)$ where $c$ is a known constant.  Do this by finding the conditional distribution of $X_1$ given $\bar{X}={\bar{x}}$ and apply the Rao-Blackwell theorem with $T=u(X_1)$, where $u(x_1)=1$ if $x\le c$ and zero otherwise.  So clearly the parameter that we're trying to estimate given the problem description can be written as $\Phi\left( \frac{c-\mu}{3} \right)$ where $\Phi$ represents the CDF of the standard normal distribution.  However, I get stuck when I need to try and find the conditional distribution.  Without knowing the joint distribution or the other conditional distribution, how can I possibly find the conditional distribution of $X_1$ given $\bar{X}={\bar{x}}$? How can I wrap text around a bottom-right div? Every time I try to do something seemingly-simple in CSS, it doesn't work.  I have a content div that contains a 460x160 image. All I want to do is position the image at the bottom-right corner and wrap my text around it.  &lt;div id="contents"&gt;     &lt;img src="..." /&gt;     text text text text text text ... &lt;/div&gt;   So I want it to look like:  ------------------ | text text text | | text text text | | text text -----| | text text |    | ------------------   Doing it with a top-left or top-right image is cake:  #contents img { float:right; }  ------------------ | text text |    | | text text -----| | text text text | | text text text | ------------------   Now how do I push that to the bottom? The best I've come up with so far are:  #contents img { float:right; margin-top: 90%} // really needs to be 100%-160px  ------------------ | text text      | | text text      | | text text -----| | text text |    | ------------------   In this case the text does not print in the margin, so there is white space above the image.  #contents { position:relative; } #contents img { position:absolute; right:0; bottom:0; }  -or-  // move the img tag under the text in the html and: #contents img { float:right; margin-top:-160; }  ------------------ | text text text | | text text text | | text text -----| | text text | te | ------------------   In this case the text prints over or under the image.  So... how can I accomplish this? Is there a way I can see all the versions that are in the archives that I have configured in sources.list. I can see the last version of each archive withapt-get policy, but how can I see them all?  Is there any way that this can also include PPA and anything in sources.list.d? Multivariable limit proof: $\lim\limits_{(x,y)\rightarrow (0,0)}\frac{\left|x\right|^a\left|y\right|^b}{\left|x\right|^c + \left|y\right|^d} = 0$ Duplicate user in reputation league So I understand that this has happened in the past: , but all of the &quot;duplicate&quot; posts are marked as  so here is a new bug report: On the , I am listed twice:  While that's fun and all, it isn't exactly right. How do I fix this "E: Type '*' is not known on line * in source list ..." update error? No sound and flash videos play super fast after daily update I just updated my 11.10 Ubuntu from the regular auto updater. After I restarted my laptop I have no sound and the videos in YouTube (and other websites) play super fast.  Not sure if it has something to do with the case, but since last night I get hd0 out of disk error before the grub screen and on the login screen I have some delay before I can type my password. In the console I see errors like these: status: {DRDY err}.  I could say the errors have something to do with the sound problem and the flash player, but if this happened when I got the hd0 error, not now after the regular daily update.  Here is the update history log from /var/log/apt/history.log   Start-Date: 2012-04-03  10:04:39  Commandline: aptdaemon role='role-commit-packages' sender=':1.71' Upgrade: python-aptdaemon.gtkwidgets:amd64 (0.43+bzr697-0ubuntu1, 0.43+bzr697-0ubuntu1.2),  libvlc5:amd64 (2.0.1+git20120329+r134-0~r35~oneiric1, 2.0.1+git20120402+r146-0~r36~oneiric1),  python-aptdaemon:amd64 (0.43+bzr697-0ubuntu1, 0.43+bzr697-0ubuntu1.2),  vlc-nox:amd64 (2.0.1+git20120329+r134-0~r35~oneiric1, 2.0.1+git20120402+r146-0~r36~oneiric1),  vlc-plugin-notify:amd64 (2.0.1+git20120329+r134-0~r35~oneiric1, 2.0.1+git20120402+r146-0~r36~oneiric1),  aptdaemon:amd64 (0.43+bzr697-0ubuntu1, 0.43+bzr697-0ubuntu1.2),  python-aptdaemon-gtk:amd64 (0.43+bzr697-0ubuntu1, 0.43+bzr697-0ubuntu1.2),  vlc:amd64 (2.0.1+git20120329+r134-0~r35~oneiric1, 2.0.1+git20120402+r146-0~r36~oneiric1),  aptdaemon-data:amd64 (0.43+bzr697-0ubuntu1, 0.43+bzr697-0ubuntu1.2),  vlc-data:amd64 (2.0.1+git20120329+r134-0~r35~oneiric1, 2.0.1+git20120402+r146-0~r36~oneiric1),   python-aptdaemon.gtk3widgets:amd64, 0.43+bzr697-0ubuntu1,0.43+bzr697-0ubuntu1.2),  libvlccore5:amd64 (2.0.1+git20120329+r134-0~r35~oneiric1, 2.0.1+git20120402+r146-0~r36~oneiric1),  vlc-plugin-pulse:amd64 (2.0.1+git20120329+r134-0~r35~oneiric1, 2.0.1+git20120402+r146-0~r36~oneiric1) End-Date: 2012-04-03  10:05:15 Exploiting JavaScript's eval() method ImageView without xml definition Difference between wouldn't and didn't What is the difference between would not and did not?  For example:     I called my girlfriend but she wouldn't stop.   I called my girlfriend but she didn't stop. I'm confused about the use of symbols $\vdash$ and  $ \models$. Reading the answers to  and  I see that:  the turnstile symbol $ \vdash $ denotes syntactic implication. Then $ S \vdash \psi$ means that $\psi$ can be derived from the formulae in $S$  the double turnstile $ \models $, denotes semantic implication. Then the notation $A \models B$ means simply that B is true in every model in which A is true.  So I understand that, e.g., $(ZFC+ A) \vdash B$ means that $A \Rightarrow B$ is a theorem in ZFC and, in this sense $\vdash$ is a mathematical symbol for a theorem. While $(ZFC+ A) \models B$ means that $ B$ is true in every model of ZFC in which $A$ is true, and, in this sense $\models$ is not a mathematical but a metamathematical symbol.  I understand correctly? And those are the only uses of the two symbols? Text highlighting in revisions is wrong For example, when watching revision 2 in  in the side-by-side mode, word "answer" marked as deleted on the left hand side and as added again on the right hand side:        If we switch to the side-by-side markdown mode no "answer" word is highlighted at all (correctly): Is infinity a real number?  If not, why not?  I want some very good arguments.   Thanks.  $$\rightarrow\leftarrow\Huge\Huge\Huge\boldsymbol\infty$$ Partition on sublists beginning with a certain marker elements I'm reading the proof of the multiplicative property of $$s_k(n)=\sum_{d|(n,k)}f(d)g\bigg( \frac kd\bigg)$$ The book wrote that in order to understand the proof, we need to know if $a,b,k,m$ are integers such that $(a,k)=(b,m)=1$, then $$(ab,km)=(a,m)(b,k).$$  I have written proof, but I don't know is it consistent, and this seems used more tools than it necessarily has to.   My proof: Let $$S_1=\{d: d|(a,m), d&gt;0\}\\S_2=\{d: d|(b,k), d&gt;0\}\\S=\{d:d|(ab,km), d&gt;0\}.$$ I will prove the statement by showing that the function $\phi:S_1\times S_2\to S$ defined as $$\phi(d_1,d_2)=d_1d_2$$ is bijective, and complete the proof by connecting the maximal elements from these partial ordering sets.  Obviously, for $d_1\in S_1, d_2\in S_2$, this tells us $d_1 $ is both divisor of $a$ and $m$, and $d_2$ is both divisor of $b$ and $k$, hence $d_1d_2$ is both divisor of $ab$ and $km$, so we can tell $\phi(S_1\times S_2)\subseteq S$.  To show one-to-one, if we have $c_1,d_1\in S_1$ and $c_2,d_2\in S_2$ such that $$\phi(c_1,c_2)=\phi(d_1,d_2)$$ then $$c_1c_2=d_1d_2$$  In particular, $c_1|d_1d_2$, we need to show $c_1|d_1$. Indeed, if $(c_1,d_2)&gt;1$, this implies $a,k$ have common factor larger than $1$, contradicts the assumption. Therefore $c_1|d_1$. A similar argument implies $d_1|c_1$, therefore $c_1=d_1$. Again a similar argument implies $c_2=d_2$, which shows one-to-one.   To show onto, let $d\in S$, since $d|ab$, we can split $d$ into $d_1d_2$ such that $d_1|a$ and $d_2|b$. Since $(a,k)=1$, we see $(d_1,k)=1$, by property of $S$ we see also $d_1|m$, therefore $d_1\in S_1$, a similar argument implies $d_2\in S_2$. So every number in $S$ has a preimage in $S_1\times S_2$  This completes the proof. Is there any mistake in the proof?     Edit: A proof about the statement: If $d|ab$ then there are $d_1,d_2$ such that $d=d_1d_2$ and $d_1|a, d_2|b$.  Let $ab=p_1^{e_1}\cdots p_r^{e_r}$ for distinct prime factors $p_1,\cdots,p_r$, then we can write $$a=p_1^{x_1}\cdots p_r^{x_r}\\b=p_1^{y_1}\cdots p_r^{y_r}$$ where some of $x_i, y_i$ can be zero, and $x_i+y_i=e_i$ for all $i=1,\cdots,r$.  Since $d|ab$, we can write $$d=p_1^{f_1}\cdots p_r^{f_r}$$ with $f_i\leq e_i$ for all $i=1,\cdots,r$. Then since $$f_i\leq x_i+y_i$$ Split into two cases:   Case 1: $f_i&gt; x_i$, then we can choose $g_i=x_i$ and $h_i=f_i-x_i$.  Case 2: $f_i\leq x_i$, then we can choose $g_i=f_i$ and $h_i=0$.  Then the choice $$d_1=\prod_{i=1}^r p_i^{g_i}, \quad d_2=\prod_{i=1}^r p_i^{h_i}$$ will prove the statement.