use 'is' to check int identity performs different between IDE and terminal
Is there a difference between "==" and "is"?
How would you explain JavaScript closures to someone with a knowledge of the concepts they consist of (for example functions, variables and the like), but does not understand closures themselves?  I have seen  given on Wikipedia, but unfortunately it did not help. "I wish I didn't do" vs. "I wish I hadn't done" on the premise that it "wasn't getting any attention" and now it has gotten a bunch of helpful answers which all help me in some ways making it impossible for me to decide which one to award the bounty to.  Shouldn't there be an option to award the bounty equally among selected answerers, especially in this case where was getting no answers whatsoever before the bounty, or am I overthinking it and should award it to the first answer? UK visa refusal on V 4.2 a + c (and sometimes 'e') Many of the UK visa refusals we see here share a common pattern and the prevailing reasons refer to V 4.2 (a) and (c).    I do understand that, while the applicants may describe very different individual circumstances, there's a consistent pattern and, broadly, fall into specific categories:    Credibility (lifestyle, lack of ties, visit history, lies &amp; omissions) Funding (insufficient funds, provenance of funds, funds parking) Sponsorship (family, friends, employer)   Question: What most commonly triggers a UK visa refusal where V 4.2 (a) and (c) are given as the grounds?  Secondarily: Given that there is a clear pattern, to what extent are these refusals predictable? Does sponsorship make a difference? Is there a set of personal circumstances, however abstract, where a refusal is all but guaranteed? For example, why do those we see here on TSE who sought entry for the PLAB or British Army appear to be refused with a common theme? Is there a uniform shortcoming or is it just discrimination?   PLAB: What is about applications for a visa, for the purpose of sitting the Professional and Linguistic Assessments Board (PLAB) exam, that seem to invite refusals. After all, the test is given so that international medical graduates can show that they qualify to practise medicine in the UK. The first part, PLAB 1, can be taken in centres outside of the UK. However, PLAB 2 can be taken only in the UK. Why can't you can't get a visa just to sit the exam, promising that you'll leave immediately after?    British Army: Since Commonwealth citizens are eligible to apply online to join, even those who don't reside in the UK, why is it so difficult to get a visa just to attend an interview to see whether you are suitable? Even with an invitation from the Army, such visa applications seem to be unsuccessful. Isn't a career in the British Army a valid reason?     Lastly:  After such a refusal, what approaches would increase the chance of a successful application? Past participle vs being+past participle Bijection between an open and a closed interval Recently, I answered to this problem:     Given $a&lt;b\in \mathbb{R}$, find explicitly a bijection $f(x)$ from   $]a,b[$ to $[a,b]$.   using an "iterative construction" (see below the rule).  My question is: is it possible to solve the problem finding a less exotic function?  I mean: I know such a bijection cannot be monotone, nor globally continuous; but my $f(x)$ has a lot of jumps... Hence, can one do without so many discontinuities?    W.l.o.g. assume $a=-1$ and $b=1$ (the general case can be handled by translation and rescaling). Let:  (1) $X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$, and  (2) $f_0(x):=\begin{cases}     -x-\frac{3}{2} &amp;\text{, if } -1&lt;x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &amp;\text{,     if } \frac{1}{2}\leq     x&lt;1\\ 0 &amp;\text{, otherwise} \end{cases}$,  so that the graph of $f_0(x)$ is made of two segments (parallel to the line $y=x$) and one segment laying on the $x$ axis; then define by induction:  (3) $X_{n+1}:=\frac{1}{2} X_n$, and  (4) $f_{n+1}(x):= \frac{1}{2} f_n(2 x)$  for $n\in \mathbb{N}$ (hence $X_n=\frac{1}{2^n} X_0$ and $f_n=\frac{1}{2^n} f_0(2^n x)$).  Then the function $f:]-1,1[\to \mathbb{R}$:  (5) $f(x):=\sum_{n=0}^{+\infty} f_n(x)$  is a bijection from $]-1,1[$ to $[-1,1]$.  Proof: i. First of all, note that $\{ X_n\}_{n\in \mathbb{N}}$ is a pairwise disjoint covering of $]-1,1[\setminus \{ 0\}$. Moreover the range of each $f_n(x)$ is $f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$.  ii. Let $x\in ]-1,1[$. If $x=0$, then $f(x)=0$ by (5). If $x\neq 0$, then there exists only one $\nu\in \mathbb{N}$ s.t. $x\in X_\nu$, hence $f(x)=f_\nu (x)$. Therefore $f(x)$ is well defined.  iii. By i and ii, $f(x)\lesseqgtr 0$ for $x\lesseqgtr 0$ and the range of $f(x)$ is:  $f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$,  therefore $f(x)$ is surjective.  iv. On the other hand, if $x\neq y \in ]-1,1[$, then: if there exists $\nu \in \mathbb{N}$ s.t. $x,y\in X_\nu$, then $f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$ (for $f_\nu (x)$ restrited to $X_\nu$ is injective); if $x\in X_\nu$ and $y\in X_\mu$, then $f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$ (for the restriction of $f_\nu (x)$ to $X_\nu$ and of $f_\mu(x)$ to $X_\mu$ have disjoint ranges); finally if $x=0\neq y$, then $f(x)=0\neq f(y)$ (because of ii). Therefore $f(x)$ is injective, hence a bijection between $]-1,1[$ and $[-1,1]$. $\square$ How to send AT commands to modem once the connection is established? I have a zte 3g modem. I use carrier provided dialer for connection establishment. Once the ppp connection is active, i would like to send some AT commands(for ex. Query signal strength, AT+CSQ). But the dialer i use locks the /dev/ttyUSB0 port, which is the command port to send AT commands for my modem. So is there any way, to send the commands, once the connection is active?    Edit: I also tried the additional port /dev/ttyUSB1. But the port is flowing with random data from the modem. A sample is given below.  T^PREFMODE??                                                                     ^PREFMODE:8                                                                       OK                                                                               TC                                                                               ^DSDORMANT:1                                                                      +CSQ:19, 99                                                                       OK                                                                               T^SYSINFO                                                                        ^SYSINFO:2,3,0,4,255                                                              OK                                                                               TT^SYSINFO?                                                                      ^SYSINFO:2,3,0,4,255     I tried adding my commands, i even got output. But the response is very poor. Most of the times, my AT commands went unnoticed. Do Stack Exchange sites allow editing of comments and posts from other users? Possible Duplicate:         Is this normal behavior in Stack Exchange sites that after just signing up one can edit not only his/her own comments / posts, but also others' comments / posts?  If so, then the Q&amp;A does not look safe to me. Imagine someone maliciously modifying a post and bringing bad reputation on a Q&amp;A member.  Can you check this and perhaps fix it? I'm trying to convert some LaTeX into HTML but my code listings don't preserve their code coloring. Why is that and how do I fix it?  This is what it looks like:    Side note: I'm also wondering why the caption on the table was removed.  Edit  Here is an example listing with colors:  \documentclass{article} \usepackage{color} \usepackage{listings} \lstset{     basicstyle=\footnotesize\ttfamily,     language=[Sharp]C,      keywordstyle=\color[rgb]{0,0,1.0}\bfseries,       commentstyle=\color[rgb]{0.133,0.545,0.133}\bfseries, }  \begin{document} \begin{lstlisting} int x = 10;  // This is a comment string str = "hello"; \end{lstlisting} \end{document}   This produces the following HTML when doing: htlatex myfile.tex:  &lt;!DOCTYPE html PUBLIC "-//W3C//DTD HTML 4.01 Transitional//EN"     "http://www.w3.org/TR/html4/loose.dtd"&gt;   &lt;html &gt;  &lt;head&gt;&lt;title&gt;&lt;/title&gt;  &lt;meta http-equiv="Content-Type" content="text/html; charset=iso-8859-1"&gt;  &lt;meta name="generator" content="TeX4ht (http://www.cse.ohio-state.edu/~gurari/TeX4ht/)"&gt;  &lt;meta name="originator" content="TeX4ht (http://www.cse.ohio-state.edu/~gurari/TeX4ht/)"&gt;  &lt;!-- html --&gt;  &lt;meta name="src" content="LaTeX4.tex"&gt;  &lt;meta name="date" content="2012-07-20 12:30:00"&gt;  &lt;link rel="stylesheet" type="text/css" href="LaTeX4.css"&gt;  &lt;/head&gt;&lt;body  &gt;    &lt;!--l. 12--&gt;    &lt;div class="lstlisting" id="listing-1"&gt;&lt;span class="label"&gt;&lt;a   id="x1-2r1"&gt;&lt;/a&gt;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;int&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;x&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;=&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;10;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;br /&gt;&lt;span class="label"&gt;&lt;a   id="x1-3r2"&gt;&lt;/a&gt;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;br /&gt;&lt;span class="label"&gt;&lt;a   id="x1-4r3"&gt;&lt;/a&gt;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;//&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;This&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;is&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;a&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;comment&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;br /&gt;&lt;span class="label"&gt;&lt;a   id="x1-5r4"&gt;&lt;/a&gt;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;string&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;str&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;=&lt;/span&gt;&lt;span  class="cmtt-8"&gt;&amp;#x00A0;&lt;/span&gt;&lt;span  class="cmtt-8"&gt;"&lt;/span&gt;&lt;span  class="cmtt-8"&gt;hello&lt;/span&gt;&lt;span  class="cmtt-8"&gt;"&lt;/span&gt;&lt;span  class="cmtt-8"&gt;;&lt;/span&gt;     &lt;/div&gt;  &lt;/body&gt;&lt;/html&gt;   Which looks like this:    I would want the HTML result to look like this: What will at actually execute? Code Formating in Comments Closest points on two line segments I am looking for a general formulation to find the closest points on two line segments. What I was thinking about is to define our lines as:  $$ P1 + s  (P2-P1)$$  $$ Q1 + t  (Q2-Q1)$$  Where $P1 , P2, Q1$ and $Q2$ are the beginning and the end points on each segment.  Now we should go through an optimization problem as: $\min f(s,t)$ such that $0&lt;s&lt;1$ and $0&lt;t&lt;1$.  Where $f(s,t)$ is the point-to-point distance function.  Is there any straight forward solution? In Dummit and Foote, Exercise 15.18 (p. 727 3rd edition) one is asked to prove that for a commutative ring with unity $R$, and $f$ a non-nilpotent element of $R$, that the localization $R_f$ is isomorphic to $R[x]/(xf - 1)$.  I have done this, using the strategy outlined in the problem, using the universal property of $R_f$ proved in Theorem 36. The problem I have is this:  If $R_f \cong R[x]/(xf - 1)$ and $\phi:R[x] \to R_f$ is a surjective homomorphism, it stands to reason that $\text{ker}(\phi)$ should be $(xf - 1)$. I cannot show this directly. The surjective homomorphism I have in mind is, for:  $p(x) = a_0 + a_1x + \cdots + a_mx^m$  $\phi(p) = \dfrac{a_0}{1} + \dfrac{a_1}{f} + \cdots + \dfrac{a_m}{f^m}$  It is clear to me that $\phi$ annihilates $(xf - 1)$, so that ideal is contained in the kernel. What I cannot show is the other containment, or equivalently:  If $\phi(g) = 0_{R_f}$ then $g(x) \in (xf - 1)$. The trouble I am running into is that $R$ might be a "bad" ring, containing a lot of zero-divisors, $f$ might even be a zero-divisor (being non-nilpotent is a considerably weaker property), so the usual polynomial tricks of the division algorithm or using the irreducibility of $xf - 1$ aren't much use, here.  I feel I am missing something obvious....what is it? How to keep processes running after ending ssh session? Let's say I launch a bunch of processes from a ssh session. Is it possible to terminate the ssh session while keeping those processes running on the remote machine? $(\mathbb R, \oplus)$ is a group. Define a multiplication with which we get a field. Where $a \oplus b = a + b +1$ I have this problem from the book Real Analysis of Miklos Laczkovich ():     Let $\oplus : \mathbb R \times \mathbb R\to \mathbb R $, where $(a,b) \mapsto a + b +1$. Then,  $(\mathbb R, \oplus)$ is a group. Define a multiplication with which we get a field.   But in the exercise there are no hints. Could anyone help me? Let $a,b,c\in\mathbb{R}$ such that $\sqrt{a^2+b^2}=1&lt;c$.  Find $I:=\int_{0}^{2\pi} \frac{1}{a\cos(t)+b\sin(t)+c}dt$.  So, I set $z=e^{it}$ and then $dz=ie^{it}dt$ and so $\frac{dz}{iz}=dt$. Therefore,  $\begin{align*} I &amp; =\frac{1}{i}\int_{|z|=1} \frac{dz/z}{a\cdot\frac{z+z^{-1}}{2}+b\cdot\frac{z-z^{-1}}{2i}+c} \newline &amp;=\frac{2}{i}\int_{|z|=1} \frac{dz}{a\cdot(z^2+1)-ib\cdot(z^2-1)+2cz} \newline &amp;=\frac{2}{i}\int_{|z|=1} \frac{dz}{(a-ib)z^2+2cz+a+ib}\newline &amp;= \frac{2}{i}\int_{|z|=1} \frac{dz}{(z-z_+)(z-z_-)}\end{align*}$  Where $z_{\pm}=\frac{-2c\ \pm\ \sqrt{4c^2-4(a+ib)(a-ib)}}{2(a-ib)}=\frac{-c\ \pm\ \sqrt{c^2-(a^2+b^2)}}{(a-ib)}=\frac{-c\ \pm\ \sqrt{c^2-1}}{(a-ib)}$.  Therefore, if we define $f(z)=\frac{1}{z-z_-}$, we get that $f$ is analytic in $\{z:|z|\leq 1\}$ since  $$|z_-|=\Bigg|\frac{-c-\sqrt{c^2-1}}{a-ib}\Bigg|=\bigg|c+\sqrt{c^2-1}\bigg|&gt;|c|&gt;1$$  Finally from Cauchy's Integral Formula:  $$I=\frac{2}{i}\cdot 2\pi i\cdot f(z_+)=4\pi \frac{1}{z_+-z_-}=\frac{2\pi(a-ib)}{\sqrt{c^2-1}}$$ Which is not always real... So my question is what is the problem.  Moreover, sometimes it is correct, e.g. $a=1,b=0,c=2$ gives the correct answer $\frac{2\pi}{\sqrt{3}}$. Invoking JavaScript code in an iframe from the parent page Basically, I have an iframe embedded in a page and the iframe has some  routines I need to invoke from the parent page.  Now the opposite is quite simple as you only need to call parent.functionName(), but unfortunately, I need exactly the opposite of that.  Please note that my problem is not changing the source  of the iframe, but invoking a function defined in the iframe. Execute sudo without Password? network manager says "device not managed" I used to be able to share my network connection by editing the Auto Eth0 to share with other computers in the Ipv4 section and now I upgraded my laptop to 11.10 .  Now are there no wired connections in the network manager. When I click my connections applet at the top of the tray it also says wired device not managed!  when I type ifconfig -a  daniella@daniella-1001HA:~$ ifconfig -a eth0      Link encap:Ethernet  HWaddr 90:e6:ba:83:d6:fc             UP BROADCAST MULTICAST  MTU:1500  Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:0 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000            RX bytes:0 (0.0 B)  TX bytes:0 (0.0 B)           Interrupt:45   eth0:avahi Link encap:Ethernet  HWaddr 90:e6:ba:83:d6:fc             inet addr:169.254.13.40  Bcast:169.254.255.255  Mask:255.255.0.0           UP BROADCAST MULTICAST  MTU:1500  Metric:1           Interrupt:45   lo        Link encap:Local Loopback             inet addr:127.0.0.1  Mask:255.0.0.0           inet6 addr: ::1/128 Scope:Host           UP LOOPBACK RUNNING  MTU:16436  Metric:1           RX packets:0 errors:0 dropped:0 overruns:0 frame:0           TX packets:0 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:0            RX bytes:0 (0.0 B)  TX bytes:0 (0.0 B)  wlan0     Link encap:Ethernet  HWaddr 00:25:d3:90:43:41             inet addr:192.168.0.6  Bcast:192.168.0.255  Mask:255.255.255.0           inet6 addr: fe80::225:d3ff:fe90:4341/64 Scope:Link           UP BROADCAST RUNNING MULTICAST  MTU:1500  Metric:1           RX packets:12881 errors:0 dropped:0 overruns:0 frame:0           TX packets:12867 errors:0 dropped:0 overruns:0 carrier:0           collisions:0 txqueuelen:1000            RX bytes:8851081 (8.8 MB)  TX bytes:2343199 (2.3 MB) Show that if $\sup\big\{\sum\lvert\, f(a)\rvert\big\}  0\}$ is countable. Let $f:A \to \mathbb R$ and suppose that $$ \sup\Big\{\sum_{a\in F}\lvert\, f(a)\rvert :  F\text{ is finite subset of }A\Big\} &lt; \infty $$ then the set $\{ a \in A : f(a) &gt; 0\}$ is countable.  My try: Define these element of ${ a ∈ A : f(a) &gt; 0}$, then $∑_{n=1}^\infty\lvert\, f(a_n)\rvert$ is determined, regardless of the method of arrangement.