Does there exist $A$ of infinite order in $\{ A \in GL_2(\mathbb{R}) : A^T = A^{-1} \}$?
Matrix Group induction proof and order of elements question
Solve $f(x)f(2x^2) = f(2x^3+x)$ The full question is to solve $f(x)f(2x^2) = f(2x^3+x)$, prove there are at most one solution per degree of $f(x)$ and then find all the solutions. (Here $f(x)$ is assumed to be polynomial.)  So far I have proven that $(x^2+1)^n$ works for even degree polynomials, and I'm pretty near certain there are no odd solutions.  Is the following a valid argument for there only being at most 1 solution per degree? Let $K$ be the degree of $f(x)$ then $f(x)f(2x^2) = f(2x^3+x)$ produces $3k+1$ linearly independent equations with only $k+1$ independent variables. Therefore for any degree $K$ polynomial there can be at most 1 solution.  Finally, is there any elegant way to prove that no odd solutions exist? I have some basic data on emission reductions and cost per car:  q24 &lt;- read.table(text = "reductions  cost.per.car     50  45     55  55     60  62     65  70     70  80     75  90     80  100     85  200     90  375     95  600     ",header = TRUE, sep = "")   I know that this is an exponential function, so I expect to be able to find a model that fits with:      model &lt;- nls(cost.per.car ~ a * exp(b * reductions) + c,           data = q24,           start = list(a=1, b=1, c=0))   but I'm getting an error:   Error in nlsModel(formula, mf, start, wts) :    singular gradient matrix at initial parameter estimates   I've read through  on the error I'm seeing and I'm gathering that the problem is probably that I need better/different start values (the initial parameter estimates makes a little more sense) but I'm not sure, given the data that I have, how I'd go about estimating better parameters. Why would I use manual camera controls instead of the automatic modes? I'm very new to digital photography, so I still haven't gotten the hang of all this exposure and ISO stuff, etc. I have a Canon EOS T3.   My question is how big of a difference does it make using manual settings rather than auto settings for everything? I've taken some pretty nice photos with all auto settings, and I'm not sure if it could have turned out any better using manual settings. I know that there are some people who have been in the game for years upon years, and to them they can notice the difference between photos taken with manual vs auto settings, but for my situation, do you think it really makes a difference? Is my camera smart enough to determine the best settings for me?  Also, lets say manual settings are better (which I'm sure they are, since they give you greater control). How often are you supposed to adjust your settings? Between every shot, considering they are unique shots? Or does it not matter about the object, but rather the lighting environment? So if I'm in a forest at 2:00 PM on a sunny bright day and I take photos for one hour, should I be constantly adjusting my settings between every unique shot to get the best out of it? Or do I usually just find one setting that works for the environment I'm in? I appreciate the mechanisms which help bring attention to new answers to old questions - front page sorting, and the moderation tool explicitly designed for it. But I wonder if we can do more. I'm particularly interested in addressing answers whose validity has changed. In the worst case, they were correct when they were posted, but no longer are, so they mislead all the current users who find their way to them looking for an answer. More benignly, perhaps the best answer was good, but slightly flawed.  At this point, a helpful user may post a new answer, which is much better than existing upvoted and accepted answers. It can be difficult to overcome that momentum. What if the default sorting method took this into account? A naive suggestion: if a new answer has a large positive vote velocity, and an old answer has a negative vote velocity, perhaps the default sorting should place the new answer above the old answer. This could help make new answers visible to users who need a correct answer, and also help garner more votes to increase the confidence of the sorting decision.  I know this would be a fairly low-impact change, but nothing is permanent, and I'm afraid of answers with 100 upvotes being pinned at the top for years after they've become obsolete.  Edit: I would welcome other proposals that accomplish the same goal - my naive suggestion wasn't intended to be a complete solution, just one thing that a better sorting method would do. Does the regularization of a divergent infinite sum yield a unique value?  I.e. do different regularization schemes acting on the same infinite sum produce the same exact value independent of the regulator?  What, exactly, do these values mean?  Or what are they?  My understanding is that they are not "convergent" values.  (Sorry in advance for being a physicist. View approximate number of users online Is there any way to view the number of users online at any given time (or even an approximation)?  This could be really useful to provide an indication of whether your question is likely to get answered any time soon.  It could also provide a bit of an indication for why your question is getting very low views - if there are only a few users online, then that could be the reason for little interest in a question, whereas if there are lots of users online and the question still doesn't garner many views, then maybe it needs to be rephrased or something.  I know that this can't be used definitively, but it could be used as a rough guide, and could also just be an item of interest.  Perhaps it could show the number of users that have been seen in the past 5 / 10 minutes? (Shouldn't be too hard, since each user page already shows the last time that they were seen). Cantrips seem to be the only type of spell that cares only about your class level. However, in the multiclassing section they are not mentioned, does this mean that cantrips use your character level, instead of your class level?  Does this mean a Warlock 2 / Fighter 15 can cast an Eldritch Blast with 4 beams just like a level 17 Warlock? Bounding lines around tax nodes I am trying to draw lines around a number of nodes. I very much like the definition of \convexpath definition given , but it is not working well on my setup (MacTex 2014 with all available updates).   I tried to compile , but what I get is the following, which is clearly not what I should be getting. Any ideas would be much appreciated. So, I've finally obtained the Werewolf form. After returning out of the form, fully naked...  I seem to be stuck in third person view! How can I fix this? Scrolling or pressing F does not work.  I want to get back in first person view...    Mouse wheel allows to zoom in/out but not get me into the first person view but rather behind and slightly to the right of the character. The command animcam locks me at a certain distance, doing it again brings me back into the mode that I'm stuck in third person view. Hitting F after this doesn't work. tfc doesn't help me either, although that command is nice to take screenshots... Align cout format as table's columns I'm pretty sure this is a simple question in regards to formatting but here's what I want to accomplish:  I want to output data onto the screen using cout.  I want to output this in the form of a table format.  What I mean by this is the columns and rows should be properly aligned.  Example:  Test                 1 Test2                2 Iamlongverylongblah  2 Etc                  1   I am only concerned with the individual line so my line to output now (not working) is  cout &lt;&lt; var1 &lt;&lt; "\t\t" &lt;&lt; var2 &lt;&lt; endl;  Which gives me something like:  Test                 1 Test2                  2 Iamlongverylongblah         2 Etc                  1 Difference between confidence intervals and prediction intervals For a prediction interval in linear regression you still use $\hat{E}[Y|x] = \hat{\beta_0}+\hat{\beta}_{1}x$ to generate the interval. You also use this to generate a confidence interval of $E[Y|x_0]$. What's the difference between the two? ...where each object also has references to other objects within the same array?  When I first came up with this problem I just though of something like   var clonedNodesArray = nodesArray.clone()   would exist and searched for info on how to clone objects in javascript. I did find a  on StackOverflow (answered by the very same @JohnResig) and he pointed out that with jQuery you could do  var clonedNodesArray = jQuery.extend({}, nodesArray);   to clone an object. I tried this though, this only copies the references of the objects in the array. So if I   nodesArray[0].value = "red" clonedNodesArray[0].value = "green"   the value of both nodesArray[0] and clonedNodesArray[0] will turn out to be "green". Then I tried  var clonedNodesArray = jQuery.extend(true, {}, nodesArray);   which deep copies an Object, but I got "too much recursion" and "control stack overflow" messages from both Firebug and Opera Dragonfly respectively.  How would you do it? Is this something that shouldn't even be done? Is there a reusable way of doing this in Javascript? I need a graduate level, mathematically rigorous textbook on multivariate analysis and inference to enhance myself. I've been reading the  and doing the problems within, but I need a book with other focuses. Topics like famous distributions (Wishart, Wilks lambda, etc.), hypothesis testing, some theory on estimation (point, interval), and other more modern materials are welcome. I've checked , but the OP there was looking for something useful for his psychological analysis, which is not exactly what I want here. Currently I have the , so I would also like to hear the comments about this book and its exercises. Thank you. Why is $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz$ an integer? Let $f$ be holomorphic in an open $\Omega \subset \mathbb{C}$ and $\gamma$ a closed curve in $\text{int}(\Omega)$, along which $f$ is never zero.   Are these hypotheses enough to claim $\frac{1}{2\pi i}\int_{\gamma}\frac{f'(z)}{f(z)}dz$ is an integer? If not, what are the necessary and sufficient conditions for that?  I can prove that's an integer for some particular cases, for example when $\Omega$ is convex and  $\gamma$ is a circle. But I've seen people claming this in many other contexts, like when $\Omega$ is an annulus around $0$ and $\gamma $ is an arbitrary curve inside it. This isn't obvious to me at all.  Thanks! I'm having difficulty with exercise 1.43 of Lang's Algebra. The question states     Let $H$ be a subgroup of a finite abelian group $G$. Show that $G$ has a subgroup that is isomorphic to $G/H$.   Thinking about this for a bit, the only reasonable approach I could think of was to construct some surjective homomorphism $\phi\colon G\to K$ for $K\leq G$, and $\ker\phi=H$, and then just use the isomorphism theorems to get the result.  After a while of trying, I've failed to come up with a good map, since $H$ seems so arbitrary. I'm curious, how can one construct the desired homomorphism? This is just the approach I thought of, if there's a better one, I wouldn't mind seeing that either/instead. Thank you. Convert timestamps with offset to datetime obj using strptime We've drastically simplified the user deletion process! Alignment of the titles in a ToC and \Roman{section} There is a problem with the alignment of the numbering of the ToC: When we use   \renewcommand{\thesection}{\Roman{section}}   the numbers encroach on the titles. When the command is commented out, the numbers are all aligned [from the first number on the left] but the titles do not begin at the same position.  What I want is to use \Roman{section} but with the numbers being displaced after V (or 9 in arabic) so that the first letters of the  titles are all aligned !  An illustration:  %\RequirePackage{fix-cm} \documentclass[a4paper, 11pt, twoside]{article}  \renewcommand{\thesection}{\Roman{section}}  \begin{document}   \tableofcontents \normalfont %\cleardoublepage  \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY} \section{QWERTY}  \end{document} Why does default auto.arima stop at (5,2,5)? The function auto.arima in the forecast package of R is a powerful tool to identify the best ARIMA(p,d,q) model for a given data series. In the   of the function, they report the following:  auto.arima(y, d=NA, max.p=5, max.q=5, max.order=5, max.d=2, start.p=2, start.q=2, ...)   According to the above function it seems that, by default, auto.arima tries to find the best model among all the combination having:   p equal or lower than 5 q equal or lower than 5 d equal or lower than 2 order equal or lower to 5, i.e. p + q + d  ≤ 5.   However, why does the default code stops at these conditions? Is there any particular reason why p, q and d should be equal or lower than 5, 5 and 2 and their sum lower than 5? Proof of $\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$