I am flying to US with a transit in Amsterdam for 15 hours. I have a Turkish passport and f1 visa for US. Do I need a visa for transit in Amsterdam?
I found many related questions on this site but I am still not sure about the rules. How can I decide if I need a visa to transit? Schengen members as of May 2021 are as follows:  Austria Belgium Czech Republic Denmark (excluding Greenland and the Faroe Islands - but an open border with the Schengen Area is maintained) Estonia Finland France (excluding overseas departments and collectivities) Germany Greece Hungary Iceland Italy Latvia Liechtenstein Lithuania Luxembourg Malta Netherlands (excluding Aruba, Curaçao, Sint Maarten and the Caribbean Netherlands) Norway (excluding Svalbard) Poland Portugal Slovakia Slovenia Spain (except Ceuta and Melilla) Sweden Switzerland
tasks package: items are in two line with 4 columns and 4 items I write some choice questions into beamer frame, after some research, here is my code:  \documentclass[aspectratio=169]{beamer} \usetheme{Berkeley} \usepackage{tasks,CJK} \begin{document} \begin{CJK*}{utf8}{bsmi} \begin{frame} \transwipe 1.Which one you want to choose? \begin{tasks}[counter-format=tsk[A].,debug=true](4) \task A \task B \task C \task D \end{tasks} \end{frame} \end{CJK*} \end{document}   output with pdflatex:  But I don't want the D choice goto the second line, I want them stay in one line, what should I do? I want to repeatedly execute a function in Python every 60 seconds forever (just like an  in Objective C). This code will run as a daemon and is effectively like calling the python script every minute using a cron, but without requiring that to be set up by the user.  In , the solution appears to effectively just  for x seconds. I don't need such advanced functionality so perhaps something like this would work  while True:     # Code executed here     time.sleep(60)   Are there any foreseeable problems with this code? Derivation of survival copula $$ \begin{align}F(x,y) &amp;= P(X\leq x, Y\leq y) \\  &amp;= C(F(x), F(y)) \end{align}$$ The copula of a bivariate distribution is equal to its CDF, $C(u,v) = F(x,y)$. The density of the data's CDF therefore can be expressed as $$f(x,y) = c(u,v) \times f_X(x) f_Y(y)$$ where $c(u,v)$ is the copula density and $f_X(x)$ is a marginal. If $\hat{C}(u,v)$ is the survival copula for determining $$ \begin{align}\bar{F}(x,y) &amp;= P(X&gt; x, Y &gt; y) \\  &amp;= \hat{C}(1-F(x), 1-F(y)) \end{align}$$ how is the first equation of the ordinary copula, $C(u,v)$, linked to the survival copula as $$\hat{C}(u,v) = u + v - 1 + C(1-u,1-v)$$ are $u$ and $v$ merely vectors containing the (ranked) observations of the uniform marginals? A derivation in integrals and probabilities (rather than $F$s) would be good. In English, I sometimes hear a phrase like green is the new black.  Literally, this makes absolutely no sense whatsoever.  What does it mean when someone says X is the new Y, when this makes no sense when taken literally?   gives plenty of examples:          I've found definitions online for specific examples, such as , but overall I don't understand what X is the new Y means, and I can't find it explained anywhere.  What does X is the new Y mean?  As far as I'm aware, this phrase exists only in English. Is there a power saving application similar to Jupiter? Does entropy increase or decrease as our Universe is expanding? Scientists say that entropy of our universe is increasing as it is expanding and our universe is cooling down gradually from the time of its birth. If something is getting cooler and cooler, then how can it become more random (entropy increase) with the passage of time? According to laws of thermodynamics, at absolute zero temperature, the entropy is zero. That means as we go down and down to the temperature scale entropy must decrease, but why is it not happening in the case of our universe? A diagram which is not the torus At page 17 of Munkres' Elements of Algebraic topology it says, referring to fig. 3.6, that "the diagram does not determine [the torus]. It does more than paste opposite edges together":   Is it so? I tried to glue them together, but end up with a normal torus. Are configurations written and read directly to/from .yaml files or still to the database? Background  Drupal 7 and previous versions have attracted criticism for Drupal web site admin configuration being stored in the MySQL database, the criticism being that being in a database means that it cannot be directly managed in git together with the PHP code based development for a Drupal site (custom modules, themes etc.).   Secondly with configuration being in the database along with content, transporting configuration between development, test, staging and live servers as part of development activity is complicated, though Features module (and related) can help.  Reading through:   ...I'm getting the impression that, in Drupal 8, the database is still used to store the config, when for example a site builder creates a new content type, rather than this being written to a .yaml file which is used directly.  So actually it seems that the role of the yaml file(s) is a much more purer (than Features) vehicle for the configuration. Whereas with Features it was PHP code which was harder to read what the configuration was, including if doing diffs between commits where config had been updated resulting in a new version of the Features-based config export. By that I mean not that PHP itself is hard to read, as I'm familiar enough with it, having written custom modules, hooks, themes. What I mean is that to express a configuration in PHP is perhaps not the most legible way to do it. So I can see the rationale of using yaml. The difference is that yaml will be read by a parser to apply a config. The complexity of applying the config is in the parser, which most folks need not concern themselves with most of the time, so the yaml can more cleanly describe the config. whereas in Features, there is no parser / yaml (or other script) separation. The descripton of the config and the means to apply it are necessarily in the same place as PHP code which is executed directly, so it's bound to be a bit harder to read the config, particularly if doing git commit diffs.  So there's the rationale for yaml, a Declarative Programming approach to storing configuration management in Git, contrasted with Features which is an Imperative Programming approach. And next what I need to know is this the case that Drupal 8 does still store config in the database does still and yaml files are the means for transporting config (replacing what Features did in D7 and before)?  (By the way I am a big fan of Drupal and spend most of my time at work, working with it for the sites that my organisation produces and maintained and am also involved in the community. So this question is not a stealth campaign against Drupal in favour of other platforms. It is a genuine question to try to dispel the apparent mystique around Drupal 8 and the new config management system)  Update @Berdir's answer (accepted) is a great definitive summary as to the state of Drupal 8 regarding configuration management.  I recently discovered that using yaml for config in Drupal 8 is a concept that has been developed for Drupal 7 with the  module, also read a  on it which includes comparing features.   Update   Is a definitive answer about making config active from DB storage, which corroborates with @Berdir's answer. proof that p-value is uniformly distributed when the null hypothesis is true this is a follow up to the following response:   the response makes sense if $P=F(T)$ but what if the hypothesis you're testing requires $P$ to be calculated differently for example $P=1-F(T)$ ? How do I transfer programs from my old computer to my new Windows 7 computer? I know about the Easy Transfer tool included in Windows 7 that will transfer all my files but I want to transfer my programs. Also, I don't want to buy any gimmicky program if I don't have to. What are the most commonly used and basic Apache htaccess redirects? If $\lim_{x\to\infty}(f(x)+f'(x))=L$ show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$ Let $f:(0,\infty) \to R$ be differentiable. Suppose that $\lim_{x\to\infty}(f(x)+f'(x))=L$. Show that $\lim_{x\to\infty} f(x) = L$ and $\lim_{x\to\infty} f'(x) = 0$. (Hint: Write $f(x) = e^xf(x)/e^x$ and use l’Hopital’s Rule.)    My working for $\lim_{x\to\infty}f'(x)=0$:  For $\lim_{x\to\infty}f'(x) = 0$, I let $f(x) = e^xf(x)/e^x$ and applied quotient rule which then cancels off $e^{2x}$ and I'm left with $\lim_{x\to\infty}(f(x)+f'(x)-f(x))$. Can I then equate this with $\lim_{x\to\infty} \left(f(x)+f'(x)\right) - \lim_{x\to\infty}f(x)$ which then gives $L-L=0$? Is this step correct? I have a model similar to below with a continuous outcome, continuous predictor and a categorical variable. the goal of this analysis is to make inferences on the group variable. The issue is that y is bounded by x in the form of y >= -x. When the model makes predictions you can see in the plot below that the predicted value is outside the possible range. See the last plot.   How can this be remedied? A custom loss function to penalize going beyond the bound and use gradient descent regression? Or is there some other way? Thank you.   x =   rep(seq(-10,10,1),each=5)  y =  rep(0,length(x) ) weights = sample( seq(1,1000,1) ,length(x)) groups = rep( letters[1:5], times =length(x)/5  )   ## the outcome y has the relationship y &gt;= -x for(i in 1:length(x)){   y[i] = sample(  seq(-x[i],-x[i]+ ifelse(x[i]&lt;0 ,                                            sample(seq(1,10,.1),1),                                           sample(seq(5,40,.1),1) ),1)                        ,1) }   dat =data.frame(x = x, y = y, weights = weights, group = groups) model = lm(dat = dat, y ~ x + group + group * x, weights = weights) dat$FIT= predict(model, dat, allow.new.levels = TRUE)  library(ggplot2) ggplot(data = dat, aes(x = x, y = y, color = group))+geom_point( aes(size = weights))+ geom_line(data = dat, aes(x = x, y = FIT, color = group))    #data for new predictions newdata = data.frame( x = seq(-100,100,1) ,                        group = rep("c", length( seq(-100,100,1) )),                       bound = -seq(-100,100,1)                       )  ## when pridicting new data the outcome y has the relationship y &gt;= -x ## but the predicted vales are below the bound. See how red line is below blue to the left of zero new_preds  = predict(model, newdata, allow.new.levels = TRUE) ggplot(data = newdata, aes(x = x, y = new_preds, color = group))+geom_line()  +   geom_line(data = newdata, aes(x = x, y = bound, color = "green")) I've changed the meta description tag in many of my site pages. Should I select Fetch as Google->Fetch->Add to Index->URL and all linked pages in GWT?  Will Google recrawl the whole site faster than it'd normally?  This option can be used maximum of 10 times in a month. Can there be any side affects when doing this way in case only a few of site pages have changed? I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is: $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$ How do I use large bitmaps in .NET? I'm trying to write a light-weight image viewing application. However, there are system memory limitations with .NET.  When trying to load large bitmaps (9000 x 9000 px or larger, 24-bit), I get a System.OutOfMemoryException. This is on a Windows 2000 PC with 2GB of RAM (of which 1.3GB is used up). It also takes a lot of time to attempt loading the files.  The following code generates this error:  Image image = new Bitmap(filename); using (Graphics gfx = this.CreateGraphics()) {     gfx.DrawImage(image, new Point(0, 0)); }   As does this code:  Stream stream = (Stream)File.OpenRead(filename); Image image = Image.FromStream(stream, false, false); using (Graphics gfx = this.CreateGraphics()) {     gfx.DrawImage(image, new Rectangle(0, 0, 100, 100), 4000, 4000, 100, 100, GraphicsUnit.Pixel); }   Also, it is enough to do just this:  Bitmap bitmap = new Bitmap(filename); IntPtr handle = bitmap.GetHbitmap();   The latter code was intended for use with GDI. While researching this, I found out that this is in fact a memory issue where .NET tries to allocate twice as much as is needed in a single contigous block of memory.    I know from other applications (Internet Explorer, MS Paint etc.) that it IS possible to open large images, and rather quickly. My question is, how do I use large bitmaps with .NET?  Is there anyway to stream them, or non-memory load them? Having recently introduced an overload of a method the application started to fail. Finally tracking it down, the new method is being called where I did not expect it to be.  We had  setValue( const std::wstring&amp; name, const std::wstring&amp; value );  std::wstring avalue( func() ); setValue( L"string", avalue ); std::wstring bvalue( func2() ? L"true", L"false" ); setValue( L"bool", bvalue ); setValue( L"empty", L"" );   It was changed so that when a bool value is stored we use the same strings (internal data storage of strings)  setValue( const std::wstring&amp; name, const std::wstring&amp; value ); setValue( const std::wstring&amp; name, const bool&amp; value );  std::wstring avalue( func() ); setValue( L"string", avalue ); setValue( L"bool", func2() ); setValue( L"empty", L"" ); &lt;&lt; --- this FAILS!?!   The problem with L"" is that it is implicitly casting and previously it was happy to be a std::wstring, but not it prefers to be a bool. The MSVC compiler does not complain, or raise warning, so I'm worried that even if I "fix" the setValue( L"empty", L"" ); to be  setValue( L"empty", std::wstring() );   somebody else may come later and simply use setValue( L"empty", L"" ); and have to track down this issue again.  We thought to use explicit on the method but it is not a valid keyword for this usage. Is there some way to get the compiler to complain about this, or otherwise prevent the issue?  Otherwise I'm thinking to change the name of the method which takes a bool to ensure it can't make an incorrect guess. Show that $\lim _{r \to 0} \|T_rf−f\|_{L_p} =0.$ I am having a hard time with the following real analysis qual problem.  Any help would be awesome.  Suppose that $f \in L^p(\mathbb{R})$, where $1\leq p&lt; + \infty$. Let $T_r(f)(t)=f(t−r)$. Show that $\lim_{r \to 0} \|T_rf−f\|_{L_p} =0$, that is $$ \lim_{r\to 0} \left( \int_{\mathbb R} |f(t-r) - f(t)|^p \mathrm d t\right)^{1/p} =0.$$ Why doesn't Cantor's diagonal argument also apply to natural numbers? In my understanding of Cantor's diagonal argument, we start by representing each of a set of real numbers as an infinite bit string.  My question is: why can't we begin by representing each natural number as an infinite bit string? So that 0 = 00000000000..., 9 = 1001000000..., 255 = 111111110000000...., and so on.  If we could, then the diagonal argument would imply that there is a natural number not in the natural numbers, which is a contradiction. I know Americans commonly use &quot;damn&quot; or &quot;damned&quot; to describe things. Sometimes, more appropriately, it's even &quot;darn&quot; or &quot;darned&quot;. For example,  This damn/damned computer is too slow.  However, I can't wrap my head around when to use &quot;damn&quot; and when to use &quot;damned&quot;. Or is only one of them correct?