Acronyms with ampersand symbol? is it possible to use the ampersand symbol &amp; inside acronyms? \begin{acronym} \acro{ACC S&amp;G}{Adaptive Cruise Control Stop and Go} \end{acronym}  And inside my text i used: \acs{ACC S&amp;G}  The error that occurs is missing \endcsname inserted. It doesn't work. Also I tried it with \&amp; without any success.
How to escape ampersand in \acro command? I am trying to use the acronym package to create an acronym list. The problem is that one of the acronyms I need to use contains an ampersand ('&amp;').  But when I try to compile the .tex file I get this error:  ! Missing \endcsname inserted. &lt;to be read again&gt;                     \&amp; l.2 ...&amp;M}[\AC@hyperlink{M\&amp;M}{M\&amp;M}]{I like them}   Here is an example:  \usepackage{amsmath} \usepackage{acronym}  \begin{document}  \begin{acronym} \acro{M\&amp;M}{I like them} \end{acronym}  \end{document}   So how do I use an ampersand in an acronym using the acronym package?  More generally, how is it possible to apply special formatting to a single acronym? For example  \acro{\textit{IAF}}{Italic acronym frenzy}
What Lagrangian counterterms might be needed in a $\phi^ 6$ theory in 3D? Moving a doorbell transformer Can a doorbell transformer that is on the outside of a wall be place inside the wall behind the drywall. I’m concerned about it getting too hot. Mine is in the kitchen by the ceiling lights Detecting constexpr with SFINAE Should I use the first or last bits from a SHA-256 hash? Suppose $ \alpha, \beta>0 $. Compute: $ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $ Suppose $ \alpha, \beta&gt;0 $. Compute:   $$ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $$   Here is what I do: $$\begin{align} \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &amp;= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}\sin (yx)dy\\ &amp;=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}\sin(yx)dx\\ &amp; \\ &amp;   \qquad\text{let $ yx=u $}\\ &amp; \\ &amp;=\int_{\alpha}^{\beta}\frac{1}{y}dy\int_0^{\infty}\sin u du\\ &amp;=\int_{\alpha}^{\beta}\frac{1}{y}dy\left( -\cos u|_{\infty}+\cos u|_0 \right)\\ &amp;=\log\frac{\beta}{\alpha}(-\cos(\infty)+1)\\ &amp;=\log\frac{\beta}{\alpha}-\cos(\infty)\log\frac{\beta}{\alpha} \end{align}$$  But $ \cos(\infty) $ does not exist right? Does it mean the integral actually diverse?  Edit: The question comes from   Who can point out my mistake in the above deduction? Extra blue line appearing in the top bar (Firefox only) Whenever we click on the Inbox or the Achievements in the top bar, an extra blue line appears near by them. This is reproducible on Stack Overflow and Anime &amp; Manga(as I use only those) but not on Meta Stack Overflow. Here is a snapshot of that.    I'm using Firefox 25.0.1 on Win7 Pro.  Edit: Just to be double sure, I restarted the FF and still I'm able to see it. Here is another snapshot, with the 2 distinct sites. Is there any advantage in marketing my product for £99.99 instead of £100? Why is my membership length only "Six months"? I was viewing the  and I noticed that I've apparently only had my account for six months, which is news to me... I opened my first site account here in February of 2014.  My account is much older than this, what's causing this bug?  To add some detail... On the per-site list, it does line up the &quot;member for X&quot; text with my actual membership time on that site. For example, when I look at the Meta Stack Exchange leaderboard, it shows me as a member for 2 years, 10 months. This is expected.  Update. With some work, today, I've managed to exceed my Interpersonal Skills hat count of with my Meta Stack Exchange hat count. This has changed my account age to match my MSE account age...  Still not reflective of my actual age... but better. The Time That 2 Masses Will Collide Due To Newtonian Gravity My friend and I have been wracking our heads with this one for the past 3 hours... We have 2 point masses, $m$ and $M$ in a perfect world separated by radius r.  Starting from rest, they both begin to accelerate towards each other. So we have the gravitational force between them as:  $$F_g ~=~ G\frac{Mm}{r^2}$$  How do we find out at what time they will collide?  What we're having trouble with is this function being a function of $r$, but I have suspected it as actually a function of $t$ due to the units of $G$ being $N(m/kg)^2$.  I've tried taking a number of integrals, which haven't really yielded anything useful.  Any guidance?  No, this is not an actual homework problem, we're just 2 math/physics/computer people who are very bored at work :) I am upgrading my PC, but I'm sticking to my old HDD and gfx card. I would like to avoid reinstalling Win 8. Is this possible? Or, at least, is it possible to use Windows 8's refresh pc option?  UPDATE  It worked like a charm. Windows acted as if it was the first boot and displayed the "Getting devices ready %  " progress bar. I am not sure if this proves you can do it in every scenario, but my new gear is relatively modern (some gigabyte board with two PCIe 3.0 slots and USB 3.0 support + i5), so I guess we can agree the reinstall is unnecessary in most cases. Q: Will there be, or have there been, any “late blooming” students in the school who come into their magic potential as adults, rather than as children? By the way, I loved meeting you, and hearing you speak, when you came to Anderson's in Naperville. I can hardly wait until you tour again. JKR: Ahhh! I loved the event at Anderson’s. It was one of my favorites. That is completely true. No, is the answer. In my books, magic almost always shows itself in a person before age 11; however, there is a character who does manage in desperate circumstances to do magic quite late in life, but that is very rare in the world I am writing about. (src: )  Who was JKR referring to in “there is a character”? Clearly, it not a student; so the most obvious guess of &quot;Neville Longbottom&quot; is wrong – both due to “no” answer regarding the question itself, and the fact that Neville bloomed in his 6th/7th year, well before you can legitimately term “quite late in life”. If $u$ is harmonic and bounded in $0 < |z| < \rho$, show that the origin is a removable singularity This is a reworking of a previous question  which was marked as a duplicate. Some nice folks have referred me to solutions to similar problems. I still have a couple of questions, since one of the solutions used the Poisson kernel, which is not the desired solution in the text, because the Poisson kernel has not been introduced yet. The other solutions were for problems that were similar, but whose solutions were only partially applicable, as I understand them.  I'm working on a problem from p. 166 of Lars Ahlfors' Complex Analysis:     If $u$ is harmonic and bounded in $0 &lt; |z| &lt; \rho$, show that the origin is a removable singularity in the sense that $u$ becomes harmonic in $|z|&lt;\rho$ when $u(0)$ is properly defined.   My Solution So Far:   Here's what I've got for a non-Poisson kernel solution: We want to find a harmonic conjugate for $u$ in the punctured disk, and then we can apply big Picard to show that $0$ is a removable singularity. The problem is that the punctured disc is not simply connected.   If we take $\sigma = (u \circ \exp)(z): L \to \mathbb{C}$, where $L$ is the left half-plane, now we have $\sigma$ harmonic, and defined in a simply connected set $L$, so it has a harmonic conjugate $\tau$. Now we wish to take $v = (\tau \circ \log):\{0 &lt; |z| &lt; \rho\}\to \mathbb{C}$, but in order to do so we need $\tau$ to be $2\pi i$-periodic. This is where my proof breaks down.  One of the similar problems which readers referred me to used the following strategy: note that since $\sigma$ is $2\pi i$-periodic, and $\sigma, \tau$ satisfy the Cauchy-Riemann equations, we have that $\nabla\tau(z) = \nabla\tau(z + 2 \pi i)$ for all $z\in L$, or in other words, $\tau(z) - \tau(z+2\pi i)$ is a constant function, say, equal to $\alpha \in \mathbb{R}$. Then, the solution suggests, we can set $$g(z) = u + iv - \frac{c}{2\pi}z,$$  in order to make $g$ $2\pi i$-periodic, and then we can use $(g\circ \log)$ on $\{0 &lt; |z| &lt; \rho\}$. But the reason why this won't work for my purposes is: now we have changed $u$! On the other hand, we can't simply subtract $\frac{c}{2\pi}\Im{(z)}$, and try to leave $u$ untouched, because now $g$ is not analytic. How can we get around this?  Another strategy entirely might be to try to use the mean-value property. Ahlfors discusses that a lot in the preceding section, and his main example is the punctured disc. Since $u$ is harmonic, we know that the average over circles centered at zero is a constant... How to calculate the following sum: How may one calculate  $$\lim_{n\to\infty} \ \left(\left(\sum_{k=1}^{n} \frac{1}{3k-1}\right) - \frac{\ln n}{3}\right) \ ?$$ Java Server Socket not connecting Absolutely continuity and Lipschitz-like continuity I'm trying to figure out if the following is true or false:  Let $\mu$ and $\nu$ be finite measures on Borel subsets of $[0,1]$,and $\nu \ll \mu$ ($\nu$ is absolutely cts with respect to $\mu$). Does there exist always a $C \in \mathbb{R}$ such that $\forall A \subseteq [0,1]$ that is Borel, we have $\nu(A)$ $\lt$ $C \mu (A)$.  The Radon-Nykodym theorem seems to almost suggest that is true, if I knew it was bounded, then I could say $\nu (A) \le \int_{A} g \ d\mu \le \int_{A} C \ d\mu \le C\mu (A)$, where $C$ is the bound. However, I don't know if it is bounded and in fact, I could take an unbounded function that is still integrable on $[0,1]$, such as $\frac{1}{\sqrt x}$.  I can't even be sure if it is true or false.   Thank you! At work we were discussing this as my boss has never heard of normalization. In Linear Algebra, Normalization seems to refer to the dividing of a vector by its length. And in statistics, Standardization seems to refer to the subtraction of a mean then dividing by its SD. But they seem interchangeable with other possibilities as well.   When creating some kind of universal score, that makes up $2$ different metrics, which have different means and different SD's, would you Normalize, Standardize, or something else? One person told me it's just a matter of taking each metric and dividing them by their SD, individually. Then summing the two. And that will result in a universal score that can be used to judge both metrics.  For instance, say you had the number of people who take the subway to work (in NYC) and the number of people who drove to work (in NYC).   $$\text{Train} \longrightarrow x$$ $$\text{Car} \longrightarrow y$$  If you wanted to create a universal score to quickly report traffic fluctuations, you can't just add $\text{mean}(x)$ and $\text{mean}(y)$ because there will be a LOT more people who ride the train. There's 8 million people living in NYC, plus tourists. That's millions of people taking the train everyday verse hundreds of thousands of people in cars. So they need to be transformed to a similar scale in order to be compared.  If $\text{mean}(x) = 8,000,000$  and $\text{mean}(y) = 800,000$  Would you normalize $x$ &amp; $y$ then sum? Would you standardize $x$ &amp; $y$ then sum? Or would you divide each by their respective SD then sum?  In order to get to a number that when fluctuates, represents total traffic fluctuations.  Any article or chapters of books for reference would be much appreciated. THANKS!  Also here's another example of what I'm trying to do.   Imagine you're a college dean, and you're discussing admission requirements. You may want students with at least a certain GPA and a certain test score. It'd be nice if they were both on the same scale because then you could just add the two together and say, "anyone with at least a 7.0 can get admitted." That way, if a prospective student has a 4.0 GPA, they could get as low as a 3.0 test score and still get admitted.  Inversely, if someone had a 3.0 GPA, they could still get admitted with a 4.0 test score.  But it's not like that. The ACT is on a 36 point scale and most GPA's are on 4.0 (some are 4.3, yes annoying). Since I can't just add an ACT and GPA to get some kind of universal score, how can I transform them so they can be added, thus creating a universal admission score.  And then as a Dean, I could just automatically accept anyone with a score above a certain threshold.  Or even automatically accept everyone whose score is within the top 95%.... those sorts of things.  Would that be normalization? standardization? or just dividing each by their SD then summing? What will reduce my rendertime? What would reduce the render time? Should I get a better GPU or more RAM? How to Deserialize XML document How to make a cyclorama I can create a collection of pictures giving a 360-degree view from a certain spot, just taking a bunch of photographs from the same spot but in a different direction.  However, what software could I use (on Windows) to stitch these images together to get one big image?  I prefer something that would stitch them together without showing any seams. I've made some high resolution pixel art in PS by drawing it down at the pixel level and then increasing the image size with the nearest neighbor setting to keep it crisp. In the web browser, however, I need to display these images as icons with a dynamic size, so I can't just size the images correctly to begin with, and the broswer seems to blur the images when rendering them smaller, while vector images seem to stay more crisp as they shrink.   So I'm wondering if there's a way to convert the pixel art neatly into vector. I'm aware of the trace function in illustrator, but it seems to do a rough job iirc.