So a few days I ago I thought I would make my display name slightly more unique by changing the last three characters to Latin Letter Small Capitals:  C - U+0043: Latin Capital Letter C  ᴏ - U+1D0F: Latin Letter Small Capital O  ʀ - U+0280: Latin Letter Small Capital R  ʏ - U+028F: Latin Letter Small Capital Y  Surely the change is completely pointless.  That said, I seem to have confused the Android application for Stack Exchange. Here's a screenshot from my LG G3:    The darn O is missing!  I would hope that I can keep my cosmetic name change and not have to be known as  "Cry" :). Is this is a bug, or have I picked a non-standard Latin Letter Small Capital O?
App version: 1.0.35 Device: Nexus 7 Problem: Users with weird/unusual name won't display on the Android app, there will just be a blank space where the username should be. For example,  user is showing as:   On a question:   On their profile:
In linguistics, is there a term describing this phenomenon, i.e., when the syllables of two words are slurred together in the spoken language?  They are not contractions. While contractions are acceptable in any register, this combination of words is very informal and hardly ever found in formal writing.  kinda (kind of)   sorta (sort of)   coulda (could have)   shoulda (should have)   lotta (lot of)   oughta (ought to)   betcha (bet you)   lemme (let me)   tseasy (it's easy)   willya (will you)  Inasmuch as English Language speakers (just like the speakers of any natural language) have a tendency to join word sounds in speech, examples abound and a complete list would be hard to produce. I’m looking for a word or phrase describing this linguistic phenomenon as it occurs in speech. How to prove the Fibonacci sum $\sum \limits_{n=0}^{\infty}\frac{F_n}{p^n} = \frac{p}{p^2-p-1}$ Is it possible to save `bc` command line history? bash has a handy file .bash_history in which it saves the history of commands and on the next execution of bash the history is populated with saved commands. Is it possible to save bc command history to a file in the same way and then load it on startup so that bc history is preserved? I tried reading  and it mentions readline and libedit. From ldd /usr/bin/bc I see mine uses readline and readline has  and read_history functions. Is this functionality implemented in bc or to do it I'll need to patch bc? UK visa: unsigned application form Is the change in the Lagrangian always a total derivative for symmetry transformations of the action? Let $\Omega\subset\mathbb R^n$ and consider an arbitrary functional $$ S\colon C^k(\Omega)\to\mathbb R $$ that is local, $$ \phi\overset S\mapsto \int_\Omega L(\phi,D\phi,\dots,D^k\phi) $$ for $L$ some $n$-form taking values in the $k$-jet space $J^k\phi$. We may call $S$ the action functional, and $L$ its associated Lagrangian (although we do not assume that $\delta S[\phi]=0$, i.e., we work off-shell).  There are two notions of a symmetry that are typically discussed:   Lagrangian symmetries, that is, transformations of the form $\phi\to\phi'$ that satisfy $$ L-L'=\mathrm d\Theta $$ for some $(n-1)$-form $\Theta$, and where $L'(J^k\phi)=L(J^k\phi')$. Action symmetries, that is, transformations of the form $\phi\to\phi'$ that satisfy $$ S-S'=0 $$ where $S'[\phi]=S[\phi']$.   If we assume that the boundary conditions are such that $\Theta|_{\partial\Omega}=0$, then it is clear that Lagrangian symmetries are also action symmetries.  If we assume that $\Omega$ is "nice-enough" (e.g. star-shaped) then I believe that any action symmetry is also a Lagrangian symmetry, although I would like to have a more precise statement about this fact. In other words, my question is     Are these two notions of symmetry equivalent? Under what conditions?   Moreover, how much can we weaken the assumptions? For example, can we take $\Omega$ to be simply-connected instead of star-shaped? Can we take $\phi$ to only have weak derivatives instead of strong derivatives? Adding Facebook Online Account - "application you are trying to use doesn't exist or has been disabled" How to preserve bookmarks when rearranging pages of a PDF file with tools like pdftk? I am using pdftk to rearrange pages of a pdf file with bookmarks/outlines, but after that, the output file lost the bookmarks of the original.   My command is pdftk in.pdf cat 1-22 43 23-42 44-end output out.pdf.  I was wondering how I could keep the bookmarks while rearranging pages? Cardinality of cartesian product of an infinite set with N Here the problem 1.4.26 from Sohrab, Basic Real Analysis  Duplicate  This are not duplicate of this question:    the answer gives no proof  despite the title is not really connected  only finite set   Problem  Show that, if $A$ is an infinite set, then $|A\times\mathbb{N}|=|A|$. Hint: Let $\mathcal{F}$ denote the set of all bijective maps $f:S\times\mathbb{N}\to S$, where $S\subset A$. Since $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$, we have $\mathcal{F}\neq \varnothing$ (Why?) Show that Zorn's Lemma can be applied in $\mathcal{F}$ to produce a maximal bijection $h:B\times\mathbb{N}\to B$, with $B\subset A$, and that we must have $B=A$, by examining the cases where $S\setminus B$ is finite or infinite.  Proof  Let $\mathcal{F}$ denote the set of all bijective maps $f:S\times\mathbb{N}\to S$, where $S\subset A$. Since $|\mathbb{N}\times\mathbb{N}|=|\mathbb{N}|$ (see exercise 1.4.10(i)), we have $\mathcal{F}\neq \varnothing$ (From Exercise 1.4.22(d) we know that for a infinite set $A$, we have $|A|\geq\aleph_0$, and $|\mathbb{N}|=\aleph_0$, thus there is an injection from $g:\mathbb{N}\to A$, and, by defining $S:=g(\mathbb{N})\subset A$, we get a bijective map (composition of two bijective maps) $S\times\mathbb{N}\to\mathbb{N}\times\mathbb{N}\to\mathbb{N}\in\mathcal{F}$).  If we consider the inclusion as partial order, then every chain $S_i\subset A$, $i\in I$, has $A$ as upper bound and applying Zorn's Lemma, we get the existence of a maximal $\bar{S}$ and hence a bijection $h:\bar{S}\times\mathbb{N}\to\bar{S}$, with $\bar{S}\subset A$.  Now we prove that $\bar{S}=A$. $R:=A\setminus \bar{S}$. $\bar{S}$ is infinite, since the example of element of $\mathcal{F}$, that we gave above is bijective to $\mathbb{N}$. We have to do this. The second is trying to follow the hint and is incomplete:   Consider that $A\neq\bar{S}$, then there is $r\in R:=A\setminus\bar{S}$. We chose an element $s\in\bar{S}$ and define the bijection $h':\bar{S}\cup\{r\}\times\mathbb{N}\to \bar{S}\cup\{r\}$ as follow  $h'(a,n)=\begin{cases}h(s,2n-1)&amp;\text{for }a=s\\ r&amp;\text{for }a=r,n=1\\ h(s,2n-2)&amp;\text{for }a=r,n\neq 1\\ h(a,n)&amp;\text{otherwise}\end{cases} $  Since $\bar{S}\subsetneq \bar{S}\cup\{r\}$ we get a contradiction to the maximality of $\bar{S}$. Now either $R$ is finite or infinite. 1) $R$ finite, let say $R=\{r_1,\cdots,r_n\}$. Then chose $n$ elements $s_1,\cdots,s_n\in\bar{S}$ and we define the bijection $h':A\times\mathbb{N}\to A$ as follow $h'(a,n)=\begin{cases}h(a,2n-1)&amp;\text{for }a\in\{s_1,\cdots,s_n\}\\ a&amp;\text{for }n=1,a\in\{r_1,\cdots,r_n\}\\ h(s_i,2n-2)&amp;\text{for }a=r_i\\ h(a,n)&amp;\text{otherwise}\end{cases}$  In case $n&gt;0$ we get a contradiction with $\bar{S}$ being the maximal element.   Thus we have only to exclude the case 2) $R$ infinite. HOW TO DO THAT?   Question  Above I tried two proofs. The first part is the same. The two second parts are given by the two bullets.  The second proof tries to follow the hint.  My questions are:   - Is my proof correct?   - In which direction should go the proof from the hint? Alert on edit after a downvote I wonder if it would be possible to extend the current notification system.   Namely, if I downvote a question and that thing had an edit, I'd be happy to see that in my inbox. As possibly the cause for downvote was cleaned up and it should be undone. I want to convert .mov to .mp4! I saved the video from iMovie but it is in .mov format! How do I convert it to .mp4?  I'm using a MacBook and I want to share it to my phone, but my phone can't open .mov format video. What do you call somebody who asks a question and somebody who answers a question? Maven artifact and groupId naming Use mathematical induction to prove that any integer $n\ge2$ is either a prime or a product of primes. Use  strong mathematical induction to prove that any integer $n\ge2$ is either a prime or a product of primes.    I know the steps of weak mathematical induction... Basis step: $p(n)$ for $n=1$ or any arbitrary $n_0$ ... show that it is true Inductive hypothesis: $p(n)$ for $n=k$ ... Assume that it is true for $n=k$ Inductive step: $P(n)$ for $n=k+1$ ... Show that this is true for $n=k+1$ $A$ and $B$ are $n\times n$ matrices and $A+B=AB$. I have an interesting proof that this implies $A$ commutes with $B$, but the proof only works when $||B|| \lt 1$. I'm looking for a way to salvage the proof so that it works for all $B$.  $$A=AB-B$$ $$A=(A-I)B$$ by repeatedly substituting $A$ with $(A-I)B$ on the RHS, $$A= (A-I)B^N - \sum_{i=1} ^{N-1} B^i$$  Since the space of $n\times n$ matrices is Banach (hence complete) under the standard operator norm, the limit $\sum_{i=1}^{\infty}B^i$ is well defined given $||B|| \lt 1$. Hence, taking the limit as $N \rightarrow \infty$, $$A =  -\sum_{i=1}^{\infty}B^i$$  Now the LHS commutes with B, hence so does the RHS.    Is there some trick to extend the method to work with any $B$ where $||B|| \geq 1$? I know that the result is true due to another simple algebraic proof, but I'd like to see if this method can be salvaged. Under what conditions is $AA^T$ invertible? Given a matrix $A$ with dimensions $m \times n$, is $B=AA^T$ invertible if and only if the rows of $A$ are linearly independent?  So far, I've tried writing A as row vectors,   $$A = \begin{bmatrix}v_1\\ v_2\\ \vdots \\ v_m\end{bmatrix}$$  where $B_{i,j} = (v_i v_j) $  The Wikipedia page on the  suggests that the condition I stated above is necessary and sufficient, but does not provide a proof. I am looking to implement a biplot for principal component analysis (PCA) in JavaScript. My question is, how do I determine the coordinates of the arrows from the $U,V,D$ output of the singular vector decomposition (SVD) of the data matrix?  Here is an example biplot produced by R:  biplot(prcomp(iris[,1:4]))     I tried looking it up in the  but it's not very useful. Or correct. Not sure which. Quotient ring of Gaussian integers $\mathbb{Z}[i]/(a+bi)$ when $a$ and $b$ are NOT coprime The isomorphism $\mathbb{Z}[i]/(a+bi) \cong \Bbb Z/(a^2+b^2)\Bbb Z$ is , when the integers $a$ and $b$ are coprime. But what happens when they are not coprime, say $(a,b)=d&gt;1$?      — For instance if $p$ is prime (which is not coprime with $0$) then $$\mathbb{Z}[i]/(p) \cong \mathbb{F}_p[X]/(X^2+1) \cong \begin{cases} \mathbb{F}_{p^2} &amp;\text{if } p \equiv 3 \pmod 4\\ \mathbb{F}_{p} \times \mathbb{F}_{p} &amp;\text{if } p \equiv 1 \pmod 4 \end{cases}$$ (because $-1$ is a square mod $p$ iff $(-1)^{(p-1)/2}=1$).  — More generally, if $n=p_1^{r_1} \cdots p_m^{r_m} \in \Bbb N$, then  each pair of integers $p_j^{r_j}$ are coprime, so that by CRT we get $$\mathbb{Z}[i]/(n) \cong \mathbb{Z}[i]/(p_1^{r_1}) \times \cdots \times \mathbb{Z}[i]/(p_m^{r_m})$$  I was not sure how to find the structure of $\mathbb{Z}[i]/(p^{r}) \cong (\Bbb Z/p^r \Bbb Z)[X] \,/\, (X^2+1)$ when $p$ is prime and $r&gt;1$.  — Even more generally, in order to determine the structure of $\mathbb{Z}[i]/(a+bi)$ with $a+bi=d(x+iy)$ and $(x,y)=1$, we could try to use the CRT, provided that $d$ is coprime with $x+iy$ in $\Bbb Z[i]$. But this is not always true: for $d=13$ and $x+iy=2+3i$, we can't find Gauss integers $u$ and $v$ such that $du + (x+iy)v=1$, because this would mean that $(2+3i)[(2-3i)u+v]=1$, i.e. $2+3i$ is a unit in $\Bbb Z[i]$ which is not because its norm is $13 \neq ±1$.  — I was not able to go further. I recall that my general question is to known what $\mathbb{Z}[i]/(a+bi)$ is isomorphic to, when $a$ and $b$ are integers which are not coprime (for instance $a=p^r,b=0$ or $d=(a,b) = a^2+b^2&gt;1$).  Thank you for your help! This is a question thats been bothering me a while. I don't even know if it makes sense or not (like if it is a physics question or becoming a philosophical one). But here it goes. The crux of my question basically is that we all know that we can't see light (like in its photon or electromagnetic wave form) directly when it is traveling past us. However, we also know that the way we see objects is by light reflecting off them. This then means that we are "seeing" the light reflecting from the object which then sends the signal to our brain saying that we are seeing a particular object. We know that both light traveling past us and light reflected from objects are made of photons (so they are the same kind)? So then my question is that what is happening to the photon of a light after it is reflected from the objects, that causes us to see it or the object, but on the other hand we can't see light as it is directly traveling past us. Can we switch terminals at Heathrow (LHR) for shopping only? We will be flying into and out of Terminal 5 at Heathrow. Now, my wife has taken fancy to a certain duty-free store located at Terminal 3. Assuming we factor in enough time before boarding our outbound flight from Terminal 5, are we allowed to enter Heathrow from Terminal 3, shop and then shuttle over to Terminal 5? If $\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$ then $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$ This is a question from the book Methods of Real Analysis by R. R. Goldberg.     If $(s_n)$ is a sequence of real numbers and if $$\sigma_n=\frac{s_1+s_2+\cdots+s_n}{n}$$ then prove that: $\operatorname{{lim sup}}\sigma_n \leq \operatorname{lim sup} s_n$.   I don't have any idea how to start working on this problem. Please help. Thanks.