In Egypt, I applied for the General Medical Council identity check and the visa was refused two times.  The first time, I was going as a visitor, but they said that I didn't show that I have sufficient funds to cover the cost of my travel. I did include a letter from my employer showing my and bank statement.  The second time, I asked my father to add some funds to my bank account and I included another bank statement with funds in US dollars where my salary is transferred.   I stated that my father will help me to cover the costs of my visit, noted that he had transferred money to me, and included his bank statement.    I was refused, saying that I must explain my own circumstances and my purpose of travel, and that my finances would not be sufficient. It was also noted that there was no evidence that my father was giving me the money.   My concern is that my GMC date expires in 45 days. Is there anything I can do?
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?
I have a .txt file with numbers ordered like this (on the same row):  106849_01373    106849_01967    106850_00082    23025.7_01059      I would like to convert them like that:  106849_01373 106849_01967 106850_00082 23025.7_01059   I have no idea which command to use. Can somebody help me with this? Before asking my question, I'll use an example to clarify the idea.  Say that a company used to sell horses and then they turn into a sheeps business. While some of the old pages can be mapped to the new pages using a 301 redirect, because they're really related in the content and the company don't want to lose users and link juice, there are a lot of pages strictly related to horses that they want to shutdown.  What's the best practice? Is there a way to say to users and Search Engines the content you were looking for won't be available any more?  Basically a 301 redirect of the closed pages to new ones seems not so good, because you'll show a content totally unrelated to the searched one.  P.S: I'm aware that probably the best fallback is to simply map those pages to the homepage but I hope in a smarter solution. Wires do not connect in Circuitikz I've been trying to use relative coordinates to make my drawings more precise but it seems I don't know how the command \node works since I've been having this problem while connecting a circuit element to another:    I'm trying to eliminate the blank between the two wires.  This is the minimal code:  \documentclass{book} \usepackage[siunitx,RPvoltages]{circuitikz} %Circuit Schematics/Diagram \usetikzlibrary{arrows,shapes,calc,positioning} \begin{document}     \begin{circuitikz}             \path (0,0) coordinate (C);             % Comparatore             \path (C) + (0,-1) + (1,0) coordinate (com) node[above]{};             \path (com) + (1.5,-1) node[op amp] (comp) {};             % Collegamenti Comparatore             \node (G1) at ([shift=({-1.3cm,0cm})]comp.-) {};             \draw (G1) -- (comp.-);             \node (G2) at ([shift=({-1.3cm,0cm})]comp.+) {};             \draw (G2) -- (comp.+);             \draw (C) -- (G1); % Collegamento Rettificatore - Comparatore         \end{circuitikz} \end{document}   Thanks for any help! Prove that a power series $\sum_{n=0}^{\infty}{{a_n}{z^n}}$ which converges for any $z \in \mathbb{N}$, converges for any $z \in \mathbb{C}$. PROBLEM  Prove that a power series $\sum_{n=0}^{\infty}{{a_n}{z^n}}$ which converges for any $z \in \mathbb{N}$, converges for any $z \in \mathbb{C}$.  MY ATTEMPT  Since the power series $$\sum_{n=0}^{\infty}{{a_n}{z^n}}$$ converges for any $z \in \mathbb{N}$, then $$\sum_{n=0}^{\infty}{{a_n}{z^n}}$$ can be considered (?) to be a polynomial in $\mathbb{C}$.  Hence, $$\sum_{n=0}^{\infty}{{a_n}{z^n}}$$ is entire, and therefore continuous in $\mathbb{C}$ (since differentiability implies continuity).  It follows that $$\sum_{n=0}^{\infty}{{a_n}{z^n}}$$ converges for any $z \in \mathbb{C}$.  QUESTION     I am aware that, while I may have (some) ideas for a (general) proof, admittedly there are several grey areas (i.e., gaps) that need to be filled.  Would you be kind enough as to fill in these gaps, by providing useful hints? I've just got my first follower, and I'm wondering if I can get a second person to follow me. Is the game limited to one or two follower(s) or does it allow for a large number of followers? It really seems handy to me to have some followers protect me when I draw attention from a larger group, or when I try to steal something where I really shouldn't... $n$ choose $k$ where $n$ is less than $k$ I am working on parameter estimation and one of the estimators involves a summation of $_nC_k$ ($n$ choose $k$) expressions. For some iterations, I need to compute expressions like $_0C_1$, $_0C_2$, etc.  In general how do we compute $_nC_k$ when $n$ is less than $k$? Do we still use the formula $\frac{n!}{(n-k)!k!}$ and use the gamma function to compute the negative factorial?  Thanks! For a presentation with Beamer, I would like to have a slide where a long list of formulas (maybe mixed with text) scrolls automatically within a certain part of the slide. This should look like e.g. closing credits in movies.   I have tried the idea from  based on animateinline from the , but whenever I use environments like \begin{equation} ... \end{equation} or commands like \textcolor{red}{...} I get an error:  TeX capacity exceeded, sorry [main memory size=3000000]. \end{frame}   Ideally, the formulas/the text should be given within the document itself, but if a solution with an external PDF works fine, I'll take it!  Any ideas how that might work? I'm looking for the title of a short story by Ursula Le Guin about a scientist (?) who is given a drug that slows the timescale of his perception enormously, so that he can perceive that an alien (?) forest is actually behaving as a sentient being, reacting to current events, but at an enormously slow (from a normal human point of view) timescale. It might be in , but I don't remember the title (and it doesn't jump out at me from the list of titles) and don't have access to that collection right now. It is not &quot;The Word for World is Forest&quot;, nor &quot;Direction of the Road&quot; (from ), nor &quot;Vaster than Empires and More Slow&quot;, the latter two of which look at similar plant perception/time-scale issues. Update: it turns out that the description was right, but I had most of the details wrong: not Le Guin, not alien, not a scientist ... Are there any examples of subspaces  of $\ell^{2}$ and $\ell^{\infty}$ which are not closed? Prove using natural deduction $((P \land Q) \rightarrow R) \vdash (P \rightarrow R) \lor (Q\rightarrow R)$ How do I prove the following using Natural Deduction?  $$((P \land Q) \rightarrow R) \vdash (P \rightarrow R) \lor (Q\rightarrow R)$$   My current approach:  So instead of proving $(P \rightarrow R) \lor (Q\rightarrow R)$, I figure I just need to prove $(P \rightarrow R)$ and use disjunction introduction.  The problem is my hypothesis requires $P$ and $Q$. How do I "create" this Q? I tried using the Law of the Excluded middle, but it does not seem to work.  Any help or insights is deeply appreciated. What can a wizard take instead of a familiar at 1st level? The familiar is brilliant feature for a wizard, but not for a prestige multiclassing one. Such a wizard stops upgrading his familiar once getting off the main road. So if he still wants to utilize his familiar, he should take the obtain familiar feat, but this would made the basic familiar feature redundant.  So what options can a wizard take instead of gaining a familiar at 1st level? Saleforce Too many future calls: 11 There are two syntaxes for command substitution: with dollar-parentheses and with backticks. Running top -p $(pidof init) and top -p `pidof init` gives the same output. Are these two ways of doing the same thing, or are there differences? Can you connect to a LAN server without being on the same internet connection Segmentation fault on large array sizes Find   $$\lim _{x\to \infty }x\Big(\ln\left(x+1\right)-\ln x\Big)$$  Here's how I do it:  $$x\Big(\ln(x+1)-\ln x\Big) = x\Bigg(\ln(x(1+\frac{1}{x}) - \ln x\Bigg)$$ $$x\Big(\ln x + \ln(1+\frac{1}{x}\Big)-\ln x =x\ln\Big(1+\frac{1}{x}\Big)$$ $$\ln\Bigg( \Big(1+\frac{1}{x}\Big)^x \Bigg) \rightarrow \ln 1^\infty = 0$$  What am I doing wrong? The answer is supposed to be $1$, but I get $0$. What is the origin of "a cut above/below the rest"? Where did the phrases "a cut above the rest" and "a cut below the rest" come from? I'm supposing they came from the same source. I feel it something very obvious, but I don't know what. Why is the accountsd process eating so much CPU? The following is a rigorous proof of De Moivre's theorem by means of mathematical induction. The theorem put simply is that:  Any complex number, $z = a+bi$, on a cartesian plane can be expressed in polar form, where $a=r\cos\theta$ and $b=r\sin\theta$ and $r$ is the absolute distance from the origin to the point $z$. In light of this, the expression for the complex number $z$ in polar form is $z=r(\cos\theta+i\sin\theta)$ In order to find the $n^{th}$ power of $z$, the following rule applies: $$\displaystyle z^n=(r(\cos\theta+i\sin\theta))^n=r^n(\cos n\theta+i\sin n\theta)$$ where $n\in\mathbb N$ (I have confined $n$ to the natural numbers given that I intend on using mathematical induction to prove the theorem.  Base step: We must prove that what is stated is true for $n=1$, hence $$z^1=r^1(\cos 1\cdot\theta+i\sin1\cdot \theta)$$ $$=r(\cos\theta+i\sin\theta)$$ which is true. Next we assume it is true for $n=k$, and therefore proceed to the inductive step. Inductive step: When calculating $z^{n+1}$, it's the same as calculating $z^nz$, hence $$z^{n+1}=(r^n(\cos n\theta + i\sin n\theta))(r(\cos \theta+i\sin \theta))$$ $$=r^nr(\cos n\theta+i\sin n\theta)(\cos\theta +i\sin \theta)$$ $$=r^{n+1}(\cos n\theta\cdot\cos\theta+\cos n\theta \cdot i\sin\theta+i\sin n\theta\cdot\cos\theta+i^2\sin n\theta\cdot\sin\theta)$$ from here, $i^2=-1$, and therefore $+i^2\sin n\theta\cdot\sin\theta$ becomes $-\sin n\theta\cdot\sin\theta$. In addition to this, the following trigonometric identities will be used: $$\sin(n\theta + \theta) = \sin n\theta\cdot\cos\theta+\cos n\theta\cdot\sin\theta$$ $$or$$ $$k\sin(n \theta+\theta)=k\sin n \theta\cdot\cos \theta + \cos n\theta\cdot k\sin\theta$$ $$and$$ $$\cos(n \theta + \theta)=\cos n\theta\cdot\cos\theta-\sin n\theta\cdot\sin\theta$$ by using these, the original equation has now become $$=r^{n+1}(\cos(n \theta + \theta) + i\sin(n \theta + \theta))$$ then, after factoring out $\theta$, the theorem becomes $$=r^{n+1}(\cos((n+1)\theta)+i\sin((n+1)\theta))$$ which completes the proof. Question: is this proof correct all the way through? Or have I missed any conditions? If all is well in the case of this proof, could somebody aid me in proving it for $\{n|n\in\mathbb R,n\geq1\}$ How do you convert a matrix to a cell array of equivalent size?