Let $R[x,\frac{1}{x}]=\{\sum_{k=-m}^m a_kx^k\mid a_k\in R\}$. Let $\varphi:R[u,v]\to R[x,1/x]$ defined by $$\varphi(p(u,v))=p(x,1/x).$$  What is $\ker \varphi$ ? To me, $\varphi(p(u,v))=0\implies p(x,1/x)=0\implies p\in (u-x,v-\frac{1}{x})$ since $u=x$ and $v=\frac{1}{x}$ are solution. but this ideal look strange... (in fact, it's $v-\frac{1}{x}$ that looks strange).
I am having trouble with the following problem.      Let $R$ be an integral domain, and let $a \in R$ be a non-zero element. Let $D = \{1, a, a^2, ...\}$. I need to show that $R_D \cong R[x]/(ax-1)$.    I just want a hint.  Basically, I've been looking for a surjective homomorphism from $R[x]$ to $R_D$, but everything I've tried has failed. I think the fact that $f(a)$ is a unit, where $f$ is our mapping, is relevant, but I'm not sure. Thanks
Ways of getting three of a kind in a 52 card deck This question has probably been asked before, but just to be clear here, I am NOT asking for the answer, I know the answer.  What i want to know is why my solution is not equivalent to the actual solution.  Three of a kind in a deck means you draw 3 cards of the same rank, and 2 other cards each of different ranks.  So. The solution in my book is the following: (13 C 1)(4 C 3) * (12 C 2)(4 C 1)(4 C 1)  The logic is the following:  1) Choose 1 out of the 13 ranks, then for the 4 cards in the chosen rank, pick 3.  2) Choose a combination of 2 ranks from the remaining 12, then for the 4 cards in each rank, choose 1.  Total combinations: 54912  My solution is the following:  ((13 C 1)(4 C 3)) * ((12 C 1)(4 C 1)) * ((11 C 1)(4 C 1))  1) My logic for the 3 of a kind is the same as the solution.   2) Choose 1 of the remaining 12 ranks, and 1 card from the 4 cards in the rank.  3) Choose 1 of the remaining 11 ranks, and 1 card from the 4 cards in the rank.  Total combinations: 109824  My solution says that the combinations are exactly double of the book's solution.  I'm just very confused as to why this is. If anyone could explain this to me I would be very appreciative, thank you. Top Answers in network profile shows incorrect score post of mine has a score of 13, but the  shows that it has 14:  This seems to be a recur of  bug; @Catija's  says it's been fixed.  Can we fix that (again)? I've noticed that man pages and other documents formatted by Unix utilities often use double backticks `` followed by double single quotes '' to wrap quoted phrases instead of the double quote character ". Single quotes are similarly replaced. Why is this?  Here are a couple examples, from the man page for grep:   To find all occurrences of the pattern `.Pp' at the beginning of a line:         $ grep '^\.Pp' myfile   The apostrophes ensure the entire expression is evaluated by grep instead  of by the user's shell.  The caret `^' matches the null string at the  beginning of a line, and the `\' escapes the `.', which would otherwise match  any character.   The grep utility is compliant with the IEEE Std 1003.1-2008 (``POSIX.1'')  specification. For $n \geq 2$, prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$ For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$  We need to use induction.   The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:   $P(n_o)$ $( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$   Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.  Suppose,  $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.  $\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$  The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.  $(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$  $\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$  $\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$  $\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$  $\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$  $\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$  We need to demonstrate that   $(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$  holds true for $n \geq 2$  The question is do I just substitute a number that is greater than two after I did the induction? How to export/backup Notes in macOS in one go I have a lot of Notes on my machine which I don't want to loose. It's not logical to export them one by one. Is there any way to export them in one go, which would make me either use them on my new   I learned that I could backup ~/Library/Group Containers/group.com.apple.notes but I am not really sure if this would be helpful. This is what I get as for this directory.  $ du -sh ./* 4.0K    ./Library 256K    ./NoteStore.sqlite  32K    ./NoteStore.sqlite-shm 3.0M    ./NoteStore.sqlite-wal   I am running macOS Mojave. Reading the book on which I'm learning English, I've come across a weird (for me) omitting the "to" in sentence: "I'd better go meet with him then."  Say, if it were "I'd better go to meet with him then.", would there be any difference between the two sentences? When I should prefer using one instead the other?  P.S. Sorry for my bad English. If I made a mistake somewhere, please correct me. Soft Thresholding CUDA implementation $(z_{2n})_{n\in \mathbb{N}}$ and $(z_{2n+1})_{n\in \mathbb{N}}$ converges to same limit. Show $(z_n)_{n\in \mathbb{N}}$ converges to the same limit. $(z_n)_{n\in \mathbb{N}}$ is a complex valued sequence such that subsequences $(z_{2n})_{n\in \mathbb{N}}$ and $(z_{2n+1})_{n\in \mathbb{N}}$ converges to same limit.  How could I show that the sequence $(z_n)_{n\in \mathbb{N}}$ converges to the same limit using $\epsilon$, $N$ definition?    For every $\epsilon &gt;0$,  $|z_{2n}-l|&lt;\epsilon$ whenever $n&gt;N_1$, and   $|z_{2n+1}-l|&lt;\epsilon$ whenever $n&gt;N_2$.  We want to show that for every $\epsilon &gt;0$, there exists $N$ such that $|z_{n}-l|&lt;\epsilon$ whenever $n&gt;N$. I tried something like  $|z_{n}-l|\le|z_n-z_{2n}|+|z_{2n}-l|$ but wasn't sure what to do next On which Canary island is it possible to sleep on the beach? We are planning a two-night trip to the Canary Islands. The cheapest options include Lanzarote and Las Palmas for flights. Travel dates are last week of November, so I guess weather should be okay. I have never been to these islands. I'd like to sleep on the beach in a tent or mattresses. Is that possible, safety, legality, and weather? It would also be nice to know how populated with tourists the islands are. Suppose your git history looks like this:  1 2 3 4 5  1–5 are separate revisions. You need to remove 3 while still keeping 1, 2, 4 and 5. How can this be done?  Is there an efficient method when there are hundreds of revisions after the one to be deleted? Convert /etc/network/interfaces to netplan for floating IP I am new to the netplan on Ubuntu 18.04, just discovered it exists today. I have an interface that I am trying to add to create a floating IP on google cloud. It is based on this example:   I tested the example, and it works, but now I try to do the same on ubuntu, and I don't know how to convert this:  cat &lt;&lt; EOF &gt;&gt; /etc/network/interfaces auto eth0:0 iface eth0:0 inet static     address 10.190.1.1     netmask 255.255.255.255 EOF   to netplan.  The output for ls /etc/netplan is '50-cloud-init.yaml'. and the output for cat /etc/netplan/*.yaml is:  network:     version: 2     ethernets:         ens4:             dhcp4: true             match:                 macaddress: 42:01:0a:8e:00:3e             set-name: ens4   my ifconfig:  # ifconfig  ens4 ens4: flags=4163&lt;UP,BROADCAST,RUNNING,MULTICAST&gt;  mtu 1460     inet 10.142.0.62  netmask 255.255.255.255  broadcast 0.0.0.0     inet6 fe80::4001:aff:fe8e:3e  prefixlen 64  scopeid 0x20&lt;link&gt;     ether 42:01:0a:8e:00:3e  txqueuelen 1000  (Ethernet)     RX packets 9430  bytes 1635180 (1.6 MB)     RX errors 0  dropped 0  overruns 0  frame 0     TX packets 13383  bytes 1513428 (1.5 MB)     TX errors 0  dropped 0 overruns 0  carrier 0  collisions 0   What is the equivalent of what he said to put in the file, for netplan? Calculating projected area of polygons with field calculator in ArcMap? How do I press Ctrl+Alt+Delete in VirtualBox? I'm running VirtualBox with Windows Vista as host OS and Windows Server 2008 as guest OS. To login to Windows Server I have to press Ctrl+Alt+Delete, but my host OS reacts on that so I can not send it to VirtualBox. How can I do it? Are there instances where citing Wikipedia is allowed? For example, suppose you're writing an article about triangles, and you want to include a short proof of Pythagoras' theorem, but you can't quite remember how to derive it, so you pop over to:  and check the proof there. It works mathematically, makes sense and is as far as you can tell, correct, so you include it. Now the proof on Wikipedia doesn't have a citation because probably someone has been taught it and copied it over to Wikipedia, but it's still correct. The only thing you could cite for the proof would be Wikipedia, would a citation be ok or should it just be copied down without a citation?  Edit: Ok, there have been many responses. Most with a firm "no", but some giving the reason: "Just get the proof from a book" since it's so common, but suppose then it wasn't a common proof of something. Maybe it's not very well known or something but suppose there is a small Wikipedia page with the correct mathematical proof. How much effort are people going to go to to find a proof in a book when there's a (correct (which we know, becuase it's maths and we can check it)) proof on Wikipedia? Why is it so bad to say: "I found this proof on Wikipedia, and the maths checks out so it doesn't matter who wrote it, it is correct, but that's where I found it." Or if you do find it somewhere else why not say: "Proof taken from Wikipedia and verified by the proof shown in "Triangles and their properties, Nature, 2014, p113 etc..." An inequality about Stirling numbers For all positive integers $n,r,k$ that $n=rk$, how to prove the inequality $$(r-1)!^k\le \frac{c(n,k)}{S(n,k)}\le (n-k)!$$ $c(n,k)$ is unsigned Stirling numbers of the first kind and $S(n,k)$ is Stirling numbers of the second kind.  Any help would be appreciated! I've been playing the heck out of this game for the past few days, and I noticed some players had some nice shiny golden guns.  What conditions do I need to fulfill in order to unlock these weapons? (Are the they tied to the story mode, Amiibo, etc.)? How to calculate the rotation matrix between 2 3D triangles? I need to calculate the rotation matrix and the translation vector between 2 given triangles in Euclidean space.  This is really just a rotation and a translation, the lengths of sides of the triangle are equal.  Coordinates of the points are real. $P_1, P_2, P_3$ are the 3 points of the 1st triangle, $Q_1, Q_2, Q_3$ are the 3 points of the 2nd.  My idea was: I translate the center of the triangles to the point of origin with $t_1 := \frac{P_1+P_2+P_3}{3}$, $P'_i = P_i-t_1$ and  $t_2 := \frac{Q_1+Q_2+Q_3}{3}$, $Q'_i = Q_i-t_1$ and then I only need to solve the set of 9 linear equations with 9 unknown (the 9 entries of the rotation matrix $R$):  $$ Q'_i = R.P'_i $$ Afterwards I could calculate the translation with  $$ t = t_2 - R.t_1 $$ The problem is: after I translate the triangles to the point of origin the points are linearly dependent ($P_3 = -P_1 - P_2$)  So in fact I only have 6 equation with 9 unknown, which allows multiple solutions for $R$. But I need "one" precise solution, because the triangles are just parts of two identical models, which I need the rotation for.   I think the approach will work, if I use 4 points instead of 3.  But can I somehow calculate the rotation matrix with "only" 3 points? Do "owed" and "due" mean the opposite? How to show $n\sum_{i=1}^n {x_i^2} \ge (\sum_{i=1}^n{x_i})^2$ This is something that always annoys me when putting an A4 letter in a oblong envelope: one has to estimate where to put the creases when folding the letter. I normally start from the bottom and on eye estimate where to fold. Then I turn the letter over and fold bottom to top. Most of the time ending up with three different areas. There must be a way to do this exactly, without using any tools (ruler, etc.).