Travelling with Fiktionsbescheinigung
Can I travel to France with Fiktionsbescheinigung (Fiction Certificate) of Germany
How to get vertex normals from a shape-key in Python? I'm trying to create an exporter for a format, which includes shape key support. This format requires both vertex coordinates and normals to be specified. This should be logical, because the vertex selected on the image below won't have the same normals:     Comparing it to this one:    But I actually haven't found any way to export shape key normals. Both Bmesh Vertex and ShapeKey Vertex only store the coordinates.  Is there any way to get them? Including package to create MATLAB labels using LaTeX Allow the author to delete their own question after closure, if other rules permit it I've read through a few "I want to delete my own question" threads on  already, and haven't found any that particularly cover this case.  I asked a question recently on :    This was voted closed as a duplicate of this question:    Since the latter question was obviously an appropriate duplicate, and my question had no votes or answers whatsoever, I decided I would just delete it.  But then, I was greeted by this error:     To allow for possible reopening, you may delete in 2 days     This all happened within hardly 20 minutes of me posting the question in the first place.  I fail to understand the point of this.  Perhaps if there were some up-votes or answers to the question, it might make sense.  However, as it stands, we now have a question with:   No votes whatsoever (up or down). No answers. Only a single (and largely irrelevant) comment not made by the OP. Five 3k+ rep users agreeing it is duplicate and therefore superfluous. The OP expressly desiring the question to be deleted, implying:   The OP has probably reviewed the duplicate question, and found their answer there. The question holds no further value to the OP. The OP agrees (with the close-voters) that the question provides no added value to the community.    At this point, what good reason could there be for disallowing the OP to delete their own question?  Could the system be modified to permit this? How to make DNS server propagate faster if you are a site owner? It's often the case that I want to change name servers because my site is messed up.  Well, it takes a very long time for the name server to propagate.  Is there something in DNS, or registrar or anything I can do to speed that up.  Flushing my dns server using  and it rarely works Why are Perl 5's function prototypes bad? What can I do if I forgot my Windows password? I got a brand new Windows 7 machine, installed the operating system, created one account and forgot its password. What can I do?  There is no external CD, the operating system is loaded from somewhere inside the machine.  I already tried to remember passwords and tried all candidates with all possible combinations of caps lock, num lock etc. How can I eliminate condensation on my toilet tank? My toilet is getting a great deal of condensation on the tank and dripping onto the floor. What can be done to address condensation on the exterior of a toilet tank? Let $G$ be a group. Show that a nonempty subset $H$ is a subgroup of $G$ if any only if $ab^{-1} \in H $ for any $a,b \in H$.  The forward direction is quite easy. Suppose $H$ is a subgroup. Then by closure, $ab \in H$ for any $a,b \in H$. Every element has an inverse. Hence, if $b \in H $, then $b^{-1} \in H$. Hence, by closure again, $ab^{-1} \in H$.   Backward direction, suppose $ab^{-1} \in H $ for any $a,b \in H$. Let $a=b$. Then we have $bb^{-1}=1_H\in H$. Let $a=1_H$ , we have $1_Hb^{-1}=b^{-1} \in H$. I don't know how to show closure.  Reference: Fraleigh p. 58  Question 5.45 in A First Course in Abstract Algebra PHP regex delimiter, what's the point? How do I get the path of the directory in which a  script is located, inside that script? I want to use a Bash script as a launcher for another application. I want to change the working directory to the one where the Bash script is located, so I can operate on the files in that directory, like so: $ ./application Can a lingering injury that requires "magical healing of 6th level or higher" be healed by an upcast lower-level spell? The  table in the DMG (p. 272) contains some wounds that can only be fixed by powerful healing magic, for instance:     Horrible Scar. You are disfigured to the extent that the wound cannot be easily concealed. You have disadvantage on Charisma (Persuasion) checks and advantage on Charisma (Intimidation) checks. Magical healing of 6th level or higher, such as heal and regenerate, removes the injury.   Can the necessary magical healing described here be a lower-level spell (such as cure wounds) that is upcast to 6th level? Or must it be a healing spell that is already 6th level or above before upcasting? I have an iPhone 6 128 GB which is not jailbroken. The current version of iOS on the phone is 8.1.3 and yesterday I wanted to update to iOS 8.2.  The download went through without any problem but while installing, it gave the error:     Unable to Install Update   An error occurred installing iOS 8.2   Tried installing after switching Off and ON and also tried Hard Reset (Holding Home button and On/off switch).  Same error displays every time.  Also the Touch ID has stopped working and I find that the Settings - Touch ID &amp; Passcode - iPhone Unlock is OFF.  When I try to switch it ON, the system displays the error     Failed   Unable to complete Touch ID setup. Please go back and try again.   Is anyone facing this problem?  I live in India. Question involving discriminant of quadratic congruence in elementary number theory. Consider a complete metric space $(X,d)$ and $T\colon X\to X$. Suppose there exists $n\in\mathbb{N}$ such that the n-th power of $T$ is $q$-contractive. Show that then $T$ has exactly one fixed point $\overline{x}\in X$.      The n-th power is defined inductively: $$ T^{n+1}(x):=T(T^{n}(x)), n\in\mathbb{N}. $$ And for a function $T\colon X\to X$, $q$-contractive means  $$ \exists 0\leq q&lt;1~\forall~x,y\in X: d(T(x),T(y))\leq q d(x,y). $$    Now to the proof.  I think I have to show, that $T$ is q-contractive, because then it follows with Banach, that $T$ does have exactly one fixed point. So I have to show, that there exists a $0\leq q&lt;1$ so that for all $x,y\in X$ it is $$ d(T(x),T(y))\leq q\cdot d(x,y). $$ And most likely I have to use the q-contractivity of $T^n$, i.e. that there exists a $0\leq q &lt;1$ so that for all $x,y\in X$ it is $$ d(T^n(x),T^n(y))\leq q\cdot d(x,y). $$  Can you help me? Customize equation numbering for Equation environment? I want to manually label equation number ? For example I want to label this equation 3.1 instead of (1) which automatically pops up. Is it also possible in split environment as well or not ?      \begin{equation}     \log \mu_{ijk}= \lambda + \lambda_i ^I + \lambda_j ^J+\lambda_k ^K+\lambda_{ij} ^{IJ}+\lambda_{ik} ^{IK}+\lambda_{jk} ^{JK}+\lambda_{ijk} ^{IJK} \label{eq:3.1}     \end{equation} How can I export a list containing e-mail addresses for everyone that has sent me an e-mail?  I have all of the messages archived. Jquery, changing background image with timer Looping through array and removing items, without breaking for loop How to find the sum $$\sum^n_{k=1} (k^2+k+1)k!$$  What I tried as follows :   $$\sum^n_{k=1} (k^2+k+1)k!$$  =$$\sum^n_{k=1} (k^2)k!+ \sum^n_{k=1} (k)k! + \sum^n_{k=1} (1)k!$$  Now we can write $\sum^n_{k=1} (k^2) $ as $$\frac{n(n+1)(2n+1)}{6}$$ but what to do with $k!$ please guide thanks....... MSO Review Queue Privileges are Misleading When I visit Stack Overflow, I see the review option in the bar at the top of the screen.    However when on MSO, I don't see it. I assumed this was because I didn't have enough reputation to access the queue. I was curious what the requirement was on meta, so I moseyed on over to the privileges page. To my surprise, I saw that the privilege is listed as only requiring 500 rep and informs me I already have it:    Confused, I manually typed in meta.stackoverflow.com/review to see what would happen. The page did in fact load, but as you can see meta has no "First Posts" or "Late Answers" queue.    Seeing on that page that a minimum of 2k reputation is required to access any queues on MSO, I report the following bug:  1) On the privileges list on MSO, the 500 rep privilege of "Access first posts and late answers review queues" should not appear.  I also have the following feature request:  2) On the privileges list on MSO, it would be useful if instead there was a 2k privilege that informed you that you have access to the "Suggested Edits" and "Low Quality Posts" queues.