I'm doing some experiments with \hypertarget and \hyperlink (from hyperref) used inside some TikZ nodes and I noticed an unexpected behaviour.  Please consider the following example code:  \documentclass{article} \usepackage{tikz} \usepackage{graphicx} \usepackage{hyperref}  \begin{document}  \hypertarget{test}{This is the target}  \begin{tikzpicture}   \node[rotate=90]      at (0,0)      {\hyperlink{test}{This is the link?}};   \node      at (4cm,0)      {\rotatebox{90}{\hyperlink{test}{This is the link}}}; \end{tikzpicture}%  \end{document}   The output:    As you can see, when the rotate option is used to rotate a node and its contents, the hyperlink doesn't get affected; the text rotates but the clickable area remains unaffected. If the rotation is performed using a \rotatebox applied to the node contents, the hyperlink gets transformed appropriately.  I know that this behaviour has been noticed before and some alternatives (included the use of \rotatebox, \adjustbox or making the whole tikzpicture the contents of the hyperlink) have been suggested in the answers to , but I am not interested here in known workarounds; I want to know the reason behind the behaviour and whether or not it is possible to to get the hyperlinks to obey the rotate option for nodes.
I have a TikZ picture with a node containing a minipage with an itemize environment; each item is a hyperref to other sections of the document.   When scaling the image (for example with \begin{tikzpicture}[scale=.7, transform shape] I've a strange behavior: links in itemize are not in the right place: they present a somewhat vertical offset, as if they don't scale as the rest of image.   Here the MWE:  \documentclass[a4paper]{article} \usepackage{tikz} \usepackage[pdftex]{hyperref} \begin{document}  Here something interesting\label{text:interesting-one} \newpage  Here something interesting, too\label{text:interesting-two} \newpage  and here there are very very interesting stuff\label{text:interesting-three} \newpage  \newpage Now the problem: links rectangles have some offset, making links unusable in case of long itemize  \begin{figure}[htb] \begin{tikzpicture}[scale=.7, transform shape] \node at (0,0) (mynode) { \begin{minipage}{4cm} \begin{itemize} \item \hyperref[text:interesting-one]{First link} \item \hyperref[text:interesting-two]{Second link} \item \hyperref[text:interesting-three]{Third link} \item \hyperref[text:interesting-one]{First link again} \item \hyperref[text:interesting-two]{Second link again} \item \hyperref[text:interesting-three]{Third link again} \end{itemize} \end{minipage} }; \end{tikzpicture} \end{figure} \end{document}   And here a snapshot of the problem    Note: This is evident using the default settings of hyperref package, which draws the red rectangle around links, but it took me some time to found it when only the text is coloured, i.e. when using \usepackage[linkcolor=blue,colorlinks=true]{hyperref}
How to end paragraph text in an inverted pyramid shape? In many old books, the final paragraphs of chapters end in an inverted pyramid shape, like this:    I think this is beautiful, so I would like to imitate this for my book. I've found the code for a triangle in this post   \documentclass{article} \usepackage{microtype,shapepar}  \def\triangleupshape{% {0}% {0}b{0}\\% {8.66}t{-5}{10}\\% {17.32}t{-10}{20}\\% {17.32}e{0}% }      \begin{document} \shapepar{\triangleupshape}A ipsum dolor sit amet, consectetuer    adipiscing elit, sed diam nonummy nibh euismod tincidunt ut laoreet dolore magna aliquam erat volutpat. Ut wisi enim ad minim veniam, quis nostrud exerci tation ullamcorper suscipit lobortis nisl ut aliquip ex ea commodo consequat. Duis autem vel eum iriure dolor in hendrerit in vulputate velit esse molestie consequat, vel illum dolore eu feugiat nulla facilisis at vero eros et accumsan et iusto odio dignissim qui blandit praesent luptatum zzril delenit augue duis dolore te feugait nulla facilisi. Lorem ipsum dolor sit amet, consectetuer adipiscing elit, sed diam nonummy nibh euismod tincidunt ut laoreet dolore magna aliquam erat volutpat. \par \end{document}   Which results in this:    But I'd like it to be inverted and the first part of the paragraph need to be full line width. Only the last few lines should end in an inverted pyramid like in the example above.  Is there a way to do this? Thanks in advance for your help. How do I set a user environment variable? (permanently, not session) What is the difference between == and equals() in Java? Ubuntu 12.04 boots to TTY1 Alrighty, I have perused the threads already in existence and am finding none of them seem to have the exact combination of variables I've got going.  I'm working on a computer using Ubuntu 12.04. Currently, it boots straight into tty1  Ubuntu 12.04.4 LTS [computer name] tty1  [computer name] login: Password:   Login leads to:  Last login: Wed Mar 19 13:47:11MDT 2014 on tty1  Welcome to Ubuntu 12.04.4 LTS (GNU/Linux 3.2.0-32generic-pae i686   The owner of the computer stated that this issue happened after a recent update, if that gives anyone a hint as to the specific location of the graphics failure.   I'm capable of booting into the GUI utilizing startx, but am frankly at a loss of where to go from here. Years of Windows has molded me into a GUI heavy kinda person, so, if possible, work off the assumption that I've no idea what you're talking about and break it down to the simplest of steps.  Thanks! In the Avengers movie, why is Hulk's behaviour inconsistent? In the Avengers movie, when      Bruce Banner is on the helicarrier, he becomes angry and transforms into the Hulk. Black Widow happens to be next to him and even though both are on same team, he attacks her and then Thor.  He becomes hostile to everyone on the ship.  But later on, in the fight against Chitauri troops, Hulk is very comfortably self-controlled.    Why is that?    The only thing I could think of is that it has been many days (a year?) since the Hulk had gotten angry, so he couldn't control himself.    But that's lame to be so puny when Bruce Banner was compared to Stephen Hawking in intelligence. Operator Exponential $e^A e^B = e^{A+B}$ This question comes from an exam in my functional analysis class.  Suppose $X$ is a Banach space, and $T \in B(X,X)$ is a bounded linear operator on $X$. For any non-negative integer $n$, let $$S_n=\sum_{k=0}^n \frac{1}{k!} T^k$$ where $T^k$ is the composition of $T$ with itself $k$ times and $T^0=I$.  We can show that for any integer $k&gt;0$, $\Vert T^k \Vert \le \Vert T \Vert ^k$. Then we can show that $S_n \in B(X,X)$ and there is some $S \in B(X,X)$ such that $S_n \to S$. We write $S = e^T$ for this operator.  Finally, We were asked to show that $e^T$ has an inverse, and that it is $e^{-T}$. My thought is to prove the following claim first: if $A,B \in B(X,X)$ and $AB = BA$, then $e^A  e^B = e^{A+B}$. If the claim is true, it follows that $e^T e^{-T}=I$.  The claim can be proven provided that the product series can be computed with the Cauchy rule.  $$e^Ae^B=\sum_{i=0}^{\infty}\frac{A^i}{i!}\sum_{j=0}^{\infty}\frac{B^j}{j!}=\sum_{k=0}^{\infty}\sum_{l=0}^{k}\frac{A^lB^{k-l}}{l!(k-l)!}$$ $$=\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{l=0}^{k}\frac{k!}{l!(k-l)!}A^lB^{k-l}= \sum_{k=0}^{\infty}\frac{1}{k!}(A+B)^k= e^{A+B}$$  But why can the product series be summed in the Cauchy way? I know for real-number series, by Cauchy's theorem, if $\sum_{n=1}^{\infty} a_n$ and $\sum_{n=1}^{\infty} b_n$ are absolutely convergent to $A$ and $B$, respectively, then we can add $a_i b_j$ in any way, and the resulting series will converge to $AB$. Does this proposition still hold for commutable operators? (It would be greatly appreciated if ideas of proof or reference is suggested.) What is the Hausdorff dimension of the set of irrationals on $\mathbb R^1$ I am sorry if this is a dump question but I can't find any reference for this. What is the Hausdorff dimension of the set of irrationals on $\mathbb R^1$? I speculate this is $1$ but don't know how to prove it rigorously. Difference between $\sum_{i=1}^{k}{\frac{s_i}{kn_i}}$ and $\frac{\sum_{i=1}^{k}{s_i}}{\sum_{i=1}^{k}{n_i}}$ I am getting confused at how to calculate the average probability.  Suppose we repeat a kind of binary survey $k$ times each of which was done on a completely separate sample group.  For each $i^{th}$ group, $i=1,2,3,...,k$, Let $n_i$ be the number of samples and $s_i$ be the number of positive results. With this we know that the probability $p_i$ of the positive result for the $i^{th}$ group is $s_i/n_i$  With my ignorance, I happened to use two ways to calculate the average probability in my work wrongly assuming that they are the same:  $$\frac{p_1+p_2+p_3+...+p_k}{k}$$ and $$\frac{s_1+s_2+s_3+...+s_k}{n_1+n_2+n_3+...+n_k}$$  I don't know which is the correct way to calculate the average probability. So could you please explain the difference of these two and when to use which? I am trying to solve the following exercise:     Prove that complex multiplication does not extend to a multiplication   on $\mathbb R^3$ so as to make $\mathbb R^3$ into a real division   algebra.   I am aware of Frobenius' theorem that there are only three finite dimensional real associative division algebras. But the exercise would be pointless if the theorem was assumed.   Let $x,y \in \mathbb R^3$. Since $x y$ has to extend the complex multiplication:  $$xy = (x_1y_1-x_2y_2, x_1y_2 + x_2 y_1, ?)$$  I have no idea how I can proceed from here. Could someone tell me how to do this? It should be easy as the other exercises I did so far were also easy. Cruise ship in the Schengen area: Do the days spent at sea count toward the limit for short visits? I am planning on spending 90 days in Schengen zone countries.  I will be taking a cruise between Lisbon and Barcelona.  Do the days at sea count against the 90 days allowed for staying in the Schengen zone? All my college years I have been using public, and would like to know the difference between public, private, and protected?  Also what does static do as opposed to having nothing? In XFCE4 desktop environments, alt+scroll wheel zooms my entire desktop. I'm finding this very annoying because sometimes I flick the scroll wheel to go to the top of a document then press the alt key for a some unrelated keyboard shortcut and find myself zoomed in so far I can't see anything at all. Is there any way to disable this feature? how to reset a lost admin password? Read Committed Snapshot Isolation vs Read Committed - pros and cons? Given the functional dependencies of a relation is there more then one way to normalize it? How to make `cd dir/filename` take me to dir/? I would find it very convenient to be able to use cd with a file argument.  cd myDirectory/anyname.anyExtension would be equivalent to cd myDirectory/  What would be the best alias or function to achieve this behavior ?  EDIT: Sorry I didn't mention it in the 1st place: I use zsh How can I disable the "Open File Security Warning" when I want to run downloaded executable files? Deriving the joint probability density function from a given marginal density function and conditional density function We have a multivariate normal vector $\mathbf{X} \sim \mathcal{N}(\boldsymbol{\mu}, \boldsymbol{\Sigma})$, where $\mathbf{X} =  \left[ \begin{matrix} X_1 \\ X_2 \\ \dot\\\\ \dot\\\\ X_n \end{matrix} \right]$,    $\boldsymbol{\mu} =  \left[ \begin{matrix} \mu_1 \\ \mu_2 \\ \dot\\\\ \dot\\\\ \mu_n \end{matrix} \right]$, and  $\boldsymbol{\Sigma} =  \begin{bmatrix} \sigma_1^2 &amp; \sigma_1\sigma_2 &amp; \dots &amp; \sigma_1\sigma_n   \\ \sigma_2\sigma_1 &amp; \sigma_2^2 &amp; \dots &amp; \sigma_2\sigma_n  \\ \dot\\\\ \dot\\\\ \sigma_n\sigma_1 &amp; \sigma_n\sigma_2  &amp; \dots &amp; \sigma_n^2 \end{bmatrix} $ is the covariance matrix.  Y is a random scalar of the form $Y = \beta_0 + \boldsymbol{\beta^TX} + \epsilon$, where $\epsilon \sim \mathcal{N}(0, \sigma_\epsilon^2)$  How can we go about deriving the joint probability density function(pdf) of the random vector $(Y, \boldsymbol{X^T})$?  The marginal multivariate pdf of $\boldsymbol{X}$ is given by  $ pdf(\boldsymbol{x}) = \frac{1}{\sqrt{(2\pi})^n. \begin{vmatrix} \boldsymbol{\Sigma} \end{vmatrix}}.exp\left[ \frac{-1}{2}\boldsymbol{(x-\mu)}^T \boldsymbol{\Sigma^{-1}}\boldsymbol{(x-\mu)} \right]$  The conditional probability density function of $Y \,\mid \, \boldsymbol{X=x}$ is given by  $pdf(y\mid \boldsymbol{X=x}) = \frac{1}{2\pi\sigma_\epsilon}.exp\left[  \frac{-1}{2\sigma_\epsilon^2}(y - \beta_0 - \boldsymbol{\beta^Tx})^2  \right]   $  To compute the joint probability density function, we can use the fact that   $pdf(y, \boldsymbol{x})=  pdf(y\mid \boldsymbol{X=x}).pdf(\boldsymbol{x})$  I am unable to do the witty manipulation required in the algebra to arrive at the result the joint distribution will have a probability density function of a normal distribution. All questions previously asked seem to address the problem of deriving the conditional density function given the joint probability density function, for example,   Edit:  Multiplying the pdfs, we get $pdf(y, \boldsymbol{x}) = \frac{1}{\sqrt{(2\pi})^n. \begin{vmatrix} \boldsymbol{\Sigma} \end{vmatrix}}. \frac{1}{2\pi\sigma_\epsilon}.exp\left[ \frac{-1}{2}\boldsymbol{(x-\mu)}^T \boldsymbol{\Sigma^{-1}}\boldsymbol{(x-\mu)} + \frac{-1}{2\sigma_\epsilon^2}(y - \beta_0 - \boldsymbol{\beta^Tx})^2 \right]$   Like the comment below mentions, most manipulations dealing with normal pdfs involve completing squares; but since this specific algebraic expression contains $\boldsymbol{x}$ in both terms within the exp term, I'm unsure on how to go about completing the squares. Validity of conditional statement when the premise is false. I have been trying to use Sidecar in the public betas of macOS Catalina and iPadOS 13.  I have a 2019 Macbook Pro, and a 3rd generation iPad Pro, so I know I have supported devices.  Both devices are on the same Wifi network, and both have Bluetooth enabled.  Both devices are logged into iCloud with the same AppleID.  Both devices have two-factor authentication enabled.  I have tried this both with and without a USB-C cable connection between the two devices.  The Sidecar preference pane in System Preferences does appear, and when I go into it I can see in the "Select Device" dropdown that my Mac can "see" my iPad Pro.  However, when I try to connect to it, I get the following after a pause:    I keep  for troubleshooting tips, but none have been useful for me thus far.  FWIW, I have been a "good citizen" and reported my problems to Apple via the Feedback Assistant apps on both devices, so in case this is a temporary bug, I've reported it.  I'm just asking here in case someone else has encountered this issue.  Edit: Eventually, Sidecar started working for me (sometimes) after some of the later public betas came out.  I'm not sure exactly which version(s) of Catalina and/or iPadOS made the difference for me.  So far it has worked without any timeouts in the final versions. I didn't need to downgrade and then re-upgrade as the submitted answer suggests, although I'm glad that has worked for some.