combine a list with &&
How do I convert an argument list to a sequence of arguments?
Let $G$ be a finite abelian group, then $G$ has a subgroup of order $n$ if and only if $n\mid G$.  Proof: by Lagrange if $H\leq G$ then $|H|$ divides $|G|$ so this proves one of the implications.  We prove the other implication by strong induction on the order of $G$, for primes this is just Cauchy, so we can take these like base cases. Now let $n$ be a proper divisor of $|G|$, take $p$ a prime in the factorization of $|G|$ (instead of $|G|$ is should be $n$, I had made a mistake when typing the first time), then by cauchy there is an element $x$ of $G$ of order $p$, since $G$ is abelian $\langle x\rangle$ is normal in $G$ so we can look at the quotient group $G/\langle x\rangle$, this group has order $\frac{|G|}{p}$, now take a prime $p_1$ in the factorization of $\frac{|G|}{n}$. By induction the quotient group has a subgroup $E$ of order $G/pp_1$ now look at the preimage of $E$ under the natural homomorphism from $G$ to $G/\langle x\rangle$, it is easy to see this is a subgroup of $G$, we call it $I$, consider the restriction of the natural projection to $I$, this is a surjective homomorphism from $I$ to $E$ with kernel $\langle x\rangle$ so by the first isomorphism theorem $I/x\cong E$ so $\frac{|I|}{|\langle  x\rangle|}=|E|\implies  |I|=\frac{|G|}{pp_1}{p}=\frac{|G|}{p_1}$ Notice $\frac{|G|}{p_1}$ is a multiple of $n$ since $p_1$ was in the fatorization of $\frac{|G|}{n}$. So by induction $I$ has a subgroup of order $n$ which is also a subgroup of $G$. Evaluating $\int_{0}^{1}\frac{x-1}{(x+1)\ln x} dx $ Evaluate the following integral $$\displaystyle I=\int_{0}^{1}\frac{x-1}{(x+1)(\ln x)} \mathrm{d}x $$ My work: I tried it by letting $\displaystyle I(a)=\int_{0}^{1}\frac{(x-1)x^a}{(x+1)(\ln x)} \mathrm{d}x$ and then $\displaystyle I'(a)=\int_{0}^{1}\frac{(x-1)x^a}{x+1} \mathrm{d}x$. Now $\displaystyle I'(a)=\int_{0}^{1}x^a \mathrm{d}x-\int_{0}^{1}\frac{2x^a}{x+1} \mathrm{d}x$ Now if $\displaystyle J(a)=\int_{0}^{1}\frac{x^a}{x+1}\mathrm{d}x$ , then by applying integration by parts, we get the reccurence relation $J(a)+J(a-1)=\dfrac{1}{a}$ and we can solve it then, but the thing is, we neet to find $I(0)$ ,so even if we compute $J(a)$, it wouldn't be defined at $0$ and so would $I(a)$, then how do I find $I'(a)$ by other method? I also tried the substitution $x \to \frac{1}{x}$, which yields $\displaystyle I=\int_{1}^{\infty}\frac{(x-1)}{(x+1)(\ln x)} \mathrm{d}x$ and when I saw their graphs, it clearly doesn't seem that the area under the graph of this function from $0$ to $1$ and from $1$ to $\infty$ are equal. I would appreciate if someone could  continue from my method and other solutions are also welcomed... How can I limit the headphone max volume on OS X? So I hurt my ears from excessively loud music last week at work. I'm well aware of the volume down key of course, but I tend to crank it up too much throughout the day without noticing it.    I need a limiter and it needs to be system-wide, since I listen to various other sound sources aside from iTunes. This isn't really an issue on my phone, since I can set a volume limit in the iOS settings.  Is there an equivalent built-in or third-party fix for OS X Lion? Help me to get command or ansibile Regex to fetch below details Uses of the definite article (the) in generic noun phrases Geometry with Trigonometry From the given figure, prove that      $$\cot\ \theta=\cot\ A+\cot\ B+\cot\ C$$     I could not even start to the first point. Please help me to solve this.  Thanks in advance Convert a Bitmap to GrayScale in Android Sorting seach results in Android app The issue is how can user sort questions for a particular search query?   For example, I would want to search questions for  and want to see newest questions first.  Right now If we write any query in Search Questions... input box then users are not able to select sorting options as search questions input box takes whole width of the screen and sorting options box gets hidden.  I have also tried to select sorting options first and then search for  but still its showing questions without considering sorting option I have selected. Deseasonalizing count data I used stl() in R to decompose count data into trend, seasonal &amp; irregular components. The resulting trend values are not integers anymore. I have the following questions:   Is stl() an appropriate way to deseasonalize count data? Since the resulting trend is not interger valued anymore, can I use lm() to model the trend components? how to concatenate string variables into a third? As a new user, I find it impossible to contribute to the community or to even begin contributing. I know why the reputation system is in place and I respect it, but at the same time it is too hard on new users.  It started with me wanting to upvote a question...     "Vote Up requires 15 reputation"   Just to appreciate a good question that helped me I need 15 rep? OK..., so I find out how I can get reputation points. More or less, I need to ask a question for +5 and add an answer for +10. I don't have any questions at the moment and I don't want to spam a random question, but I did happen to have a slightly different answer to the earlier question that I wanted to upvote. So I try to answer...     This question is protected to prevent "thanks!", "me too!", or spam answers by new users. To answer it, you must have earned at least 10 reputation on this site.   Now... I know this doesn't apply to every question, but at the same time, this means I'm forced to answer yet another question, and I don't want to go around looking for random questions and spam irrelevant answers...  Now I'm stuck. So I try to talk to some other users in the chat rooms about how to go about getting reputation points and how they first started. I entered into a "casual chat room".     You must have 20 reputation on Meta Stack Overflow to talk here.   Yes... I can understand why this is enforced, because we don't want new users to spam chat rooms, but all I wanted to do was ask for help. So I can't even talk to anyone about my newbie problems. So okay, I'll just settle for upvoting comments in the question that helped me. I don't see an upvote hover-over button so I can't even do that.    So now I'm stuck at 1 reputation point, unable to cross that 15 rep barrier, and even after posting this question I'll only have 6 rep points and still unable to do anything, and I'll be forced to look for questions to answer. Yes, I should be contributing to the community, but I shouldn't need to begin answering or making questions to upvote questions and answers. This only spurs more spam/irrelevant answers if anything.   Very, very, very new-user unfriendly. Google Drive files opening in new window vs new tab? This might be a duplicate of  but that question is 5 years old and unanswered.  When I'm in drive.google.com and I open a G Suite file (either double-click or select &amp; press enter) in Firefox it opens in a new window whereas in Chrome it opens in a new tab.   Why do they do that? Is this a browser-specific quirk or a Google setting (that I haven't found)? Is there a way to control the behavior? I'd like them both to open in tabs, but presumably the same answer would be relevant in the opposite direction too. I am looking for a book that goes through the mathematical aspects of neural networks, from simple forward passage of multilayer perceptron in matrix form or differentiation of activation functions, to back propagation in CNN or RNN (to mention some of the topics).  Do you know any book that goes in depth into this theory? I've had a look at a couple (such as Pattern Recognition and Machine Learning by Bishop or Deep Learning by Goodfellow, Bengio and Courville) but still have not found a rigorous one (with exercises would be a plus). Do you have any suggestions? I am trying to install Microsoft Office 2013 on Ubuntu 12.04 LTS (32-bit). I have downloaded an iso file of said product and installed wine1.4-i386 using the apt-get command on the terminal.  The question is I don't know the steps how to install it. I can't get much tutorials on Internet how MSOffice2013 can be installed on Ubuntu.  If any versions of wine isn't an option, what other options can be used to run Microsoft Office 2013? How is the recommended next badge determined exactly? According to , the     "Progress to next badge" bar automatically selects the badge you're closest to earning (or that's most logical to go for next)   While "the badge you're closest to earning" is pretty easy to determine: it's the bar with the most progress, I'm wondering what "most logical to go for next" means. Because right now on Meta Stack Exchange, I can track two badges: Illuminator and Legendary. My progress for Illuminator is 66.40% and for Legendary it's 66.67%, yet Illuminator is shown as the recommended next badge:    On some child metas where I barely have any activity, Student is shown as recommended next badge with a progress of 0/1, but I do have cast some votes so I have a small progress towards e.g. Civic Duty. So obviously there is some other logic at work, but what? Is this correct: "Our listeners are what make X"? Why two inverters cascaded in multiplexer control signal terminals? Why cascaded inverters are used in the multiplexer when it is possible to realize the circuit using a single inverter. The indefinite article applied to a group of things I would like to install Ubuntu  for almost 120 users. But it's very difficult in using USB and drives.  Any suggestions? I have the following diagram:  \documentclass{article} \usepackage{amsmath,amsfonts,graphicx} \usepackage{algpseudocode} \usepackage{tikz, nth} \usetikzlibrary{arrows.meta,decorations,decorations.pathreplacing,calc,bending,positioning, chains}  \begin{document} \begin{tikzpicture}[scale=0.6]         \draw[thick,]  (0,8) -- (0,0) node[left] {0};         \draw[thick,]  (0,0) -- (8,0) node[right] {8};         \draw[thick,]  (8,8) -- (8,0) node[left] {};         \draw[thick,]  (8,8) -- (0,8) node[left] {8};          \draw[thick,]  (4,0) -- (4,8) node[left] {};          \fill (1, 4) circle[radius=2.5pt] node[below]{$A$};         \fill (2, 6) circle[radius=2.5pt] node[below]{$B$};         \fill (3, 2) circle[radius=2.5pt] node[below]{$C$};         \fill (4, 8) circle[radius=2.5pt] node[below]{$D$};         \fill (7, 3) circle[radius=2.5pt] node[below]{$E$};         \fill (6, 1) circle[radius=2.5pt] node[below]{$F$};         \fill (5, 7) circle[radius=2.5pt] node[below]{$G$};         \fill (3, 8) circle[radius=2.5pt] node[below]{$H$}; \end{tikzpicture} \end{document}   How can I position the caption of the node $D$ both below and left, so that the line doesn't obscure the caption?