I am a Russian citizen. I have Schengen visa type C valid till 01/03/2020 and I am going to change my name and I will keep my surname as it is.   Can I travel with my new passport with new name and old passport which has a valid Schengen visa in the new year to Europe?
I'm an Indian National. I have a Type C Schengen Business Visa which is valid from 26-11-2015 to 26-11-2020. I received it when I intended to travel to Netherlands, although I have not used it.   When I applied for a spouse endorsement, the passport on which the visa is stamped was cancelled and a new passport was issued.  Now, I plan to travel to Italy on business from July 8th to 16th.    Can travel with the same visa in old passport, or should I cancel the existing visa and apply for a new one?  I contacted the Netherlands Embassy call center (not VFS Netherlands) which responded that I can use it without any issues. However, they have not confirmed it by email.   I asked about any immigration issues in Italy and was told me that, if I have a valid itinerary and invitation letter from the office, there should be no problem. I was also told that Italian immigration authorities are able to contact the Dutch, if necessary.   I emailed the Italian Embassy in Mumbai (visa.mumbai@esteri.it) and await its response.
How does extra damage work for critical hits? What does this notation mean? $f(x) \equiv A$? If $\lim_{x\rightarrow\infty}f(x) = A$, prove that $f(x)\equiv A$.  Any help would be appreciated! I just don't even understand the question. We let a>0 and b>0 and  we have $\sum_{n=0}^{\infty}b^n-a^n$. We have to find all values a,b>0 where $\sum_{n=0}^{\infty}b^n-a^n$ is convergent. I think I can use that $$\sum_{n=0}^{\infty} b_n &lt; \infty , \sum_{n=0}^{\infty} a_n &lt; \infty \implies \sum_{n=0}^{\infty} (b_n-a_n) &lt; \infty. $$. And I think $\sum_{n=0}^{\infty}b^n$ and $\sum_{n=0}^{\infty}a^n$ is convergent if a&lt;1 and b&lt;1? It will make sense. But how can I show it? Prove Uncountable set minus a countable set is uncountable Let B be a countable set, and let A be an uncountable set. Prove that A-B is uncountable jsp:useBean scope Once in a while you wish to address a comment to someone in particular, so you type @ and then you pause ... because the username is  or .  Short of using copy-paste to get their username right (which is even more cumbersome on a mobile device), is there another way around this? In a few cases I've noticed that the system will attempt to "promote" a comment to a reply, but I'm not sure if that's very reliable.  Should user names at least start with a normal letter?  Update  The auto-complete feature seems to normalize Zalgo'ish usernames, so typing the letter L will trigger the suggestions as expected; this is of course knowing that the name actually spells "lawrence", which may not always be obvious.  Knowing the way of the symbol ninja would surely help to type Ω using only the elbows as well, but not everyone is blessed with the same keyboard layout.  Suggestion  It would be nice to have a reply feature like this:        When ↳ reply is clicked, it will automatically populate the comment box with @username pre-filled in.   Additional notes  To clarify, I'm not suggesting to put the "reply" on all comments, as proposed , because in most cases the owner of a post is already notified of a new comment; this would strictly be to effect cross-commenter communication. How do I find out which program and process ID accesses a given IP address in Windows? Is it possible to catch, which program is accessing a specific IP address?   I found my computer is flooding and wish to check addresses one by one.  May be it is possible to set up some audit for this? Is there a way to find out if I need a transit visa for a layover in the UK? How to make partitions mount at startup? I've been going through  and have a question on the inductive proof for the . Recurrent relation:  $J(1) = 1$ $J(2n) = 2J(n) - 1$ $J(2n+1) = 2J(n) + 1$  Closed form hypothesis:  $J(2^m + l) = 2l + 1$, for $m \ge 0$ and $0\le l&lt;2^m$  Inductive Proof (for the even case):  $J(1)=1$ $J(2^m+l)= 2J(2^{m-1}+l/2)-1=2(2l/2+1)-1=2l+1$  How do the authors get from $2J(2^{m-1}+l/2)-1$ to $2(2l/2+1)-1$? Is there any advantage to using pedal straps over clipless pedals? I'm trying to identify fantasy book series I encountered some time ago. Here are things I remember:   it was a fantasy setting but with some s-f elements; with many worlds and possibility to travel between them; resembling  setting a little, the two main races were non-human and probably hostile to each other,  a labyrinth (?), which somehow played an important role in the plot, a protagonist was some kind of emissary and was sent on a mission to another worlds  time period: 80s - 90s    Does anyone know what it is the name of the series? How to obtain optimal hyperparameters after nested cross validation? I'm not too sure how to describe the effect that I'm going for or what it is even called.  Thus, I'm having some trouble finding any help through Google.    I want to accomplish something like this "SLOPED" text below, but have the bottoms of the letters follow the slope as well instead of being horizontal.    I have tried using the free transform tool, skewing, and using perspective, but I don't like the effect they are resulting in.  All of the tools that I just mentioned result in the appearance of the word being angled away from the viewer (perspective).  I want more of this style effect, whereas the letters are not skewed horizontally, but scaled evenly along a path that I lay out.  What is the name of this text effect?  Can anyone point me in the right direction to a tutorial or know how to accomplish this themselves?  I know that I could go through and do this letter by letter, but I was really hoping to find an easier method.  EDIT I have removed many elements of my design and quickly threw some sample text in it, but this should give you an idea of what I am trying to accomplish.  I want a mixture of the above and below effects. OpenLayers 3: Cross-Origin Request Blocked: The Same Origin Policy disallows Can the Help action be used to give advantage to a specific ally's attack (rather than just the next ally who attacks the target)? Latest NVIDIA driver on Ubuntu 16.04 Will factory resetting my device remove the Android upgrade? I have upgraded my phone to Android 2.1. I need to reset my phone but I don't want to lose the upgrade. Is it possible to do this? How? Why is the empty set considered an interval? What is the definition of an interval and why is the empty set an interval by that definition? The words in the node of tree I want to plot this graph, but I don't know how to code the words in the nodes.     Here is my code:  \documentclass{article} \usepackage{amsmath} \usepackage{titletoc} \usepackage{tikz} \usetikzlibrary{trees,positioning,fit}  \begin{document}{\allowdisplaybreaks[4]}  \tikzstyle{bag} = [text width=2em, text centered] \tikzstyle{end} = []  \begin{center} \begin{tikzpicture}[sloped] \node (a) at ( 0,0) [bag] {$\ln r_0$}; \node (b1) at ( 3,1.5) [bag] {$\ln r_{10} + \sigma_1\sqrt{\Delta t_1}$}; \node (b2) at ( 3,-1.5) [bag] {$\ln r_{10}$}; \node (c1) at ( 6,3) [bag] {$\ln r_{20} + 2\sigma_2\sqrt{\Delta t_2}$}; \node (c2) at ( 6,1.5) [bag] {$\textrm{Mean} = \ln r_{20} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2}$\\$\textrm{Mean Change} = \mu_{11}\Delta t_2 = \ln\dfrac{r_{20}}{r_{10}} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2} - \sigma_1\sqrt{\Delta t_1}$}; \node (c3) at ( 6,0) [bag] {$\ln r_{20} + \sigma_2\sqrt{\Delta t_2}$}; \node (c4) at ( 6,-1.5) [bag] {$\textrm{Mean} = \ln r_{20} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2}$\\$\textrm{Mean Change} = \mu_{11}\Delta t_2 = \ln\dfrac{r_{20}}{r_{10}} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2} - \sigma_1\sqrt{\Delta t_1}$}; \node (c5) at ( 6,-3) [bag] {$\ln r_{20}$}; \draw [-&gt;] (a) to node [below] {} (b1); \draw [-&gt;] (a) to node [above] {} (b2); \draw [-&gt;] (b1) to node [below] {} (c1); \draw [-&gt;] (b1) to node [below] {} (c2); \draw [-&gt;] (b1) to node [below] {} (c3); \draw [-&gt;] (b2) to node [below] {} (c3); \draw [-&gt;] (b2) to node [below] {} (c4); \draw [-&gt;] (b2) to node [below] {} (c5); \end{tikzpicture} \end{center}  \end{document}       Update code use  \draw [-&gt;] (a) to node [below] {} (b1.south east); \draw [-&gt;] (a) to node [above] {} (b2.north east); \draw [-&gt;] (b1.south east) to node [below] {} (c1.west); \draw [-&gt;] (b1.south east) to node [below] {} (c2.west); \draw [-&gt;] (b1.south east) to node [below] {} (c3.west); \draw [-&gt;] (b2.north east) to node [below] {} (c3.west); \draw [-&gt;] (b2.north east) to node [below] {} (c4.west); \draw [-&gt;] (b2.north east) to node [below] {} (c5.west);