Schengen business visa extension in Spain I'm travelling to Spain for business purpose and my visa is valid for just 14 days. Now I have another very urgent business to attend to in the Netherlands. Can I extend my stay for another 10 days to complete my business meeting in the Netherlands?  Can I get an extension in Spain, prior to entering the Netherlands? I haven't left for Spain yet and currently staying in my residence country (India).
Is it possible to extend a tourist Schengen visa (12 more days)? My daughter applied at the Italian Consulate in Philadelphia, and they gave her two entries for 20 days (we thought it was for 20 days each entry). She was in Italy 12 days, then went to England for 1 week and flew to Venice to take a 10-day cruise. She couldn't make it because they said she has only 8 remaining days, but she still has a land tour already paid this week.  What can she do? Can she extend the Schengen visa for 12 more days? Right now, she is in Madrid, but next week she has to fly from Barcelona to Paris to start with the tour. What would happen if she decides to take the tour with the expired visa? She is a US permanent resident.
How can one express frequentist p-value in terms of Bayesian concepts? iPhone asks 'Trust This Computer' again (I am on macOS 10.14.6 and iOS 12.4)   I am certain that I have trusted my Mac before. Without the latter trust, I can also access the phone in iTunes.   I have tried reset "Location and Privacy" on my phone. Then   When connecting the iPhone with iTunes running, it asks for the trust and I trust it.  Reboot my Mac and the phone asks for another trust.   Is there some known malware mimics system behavior and steal data from an iPhone?  (I do have installed some pir*ted software recently. That's my fault.) Let $E_n$ be $$E_n:=\{f\in C[0,1]\mid \text{exist } x_f\in[0,1] \text{ such that } |f(x)-f(x_f)|\leq n|x-x_f|,\, \forall x\in[0,1]\}.$$  How show that $E_n$ is nowhere dense, that is, $\mathrm{int}\  \overline{E_n} = \emptyset$?  Can someone help me? Since upgrading to 13.10, I can't log in to unity desktop. Light dm works correctly, but attempting to log in tries to start the session then drops back to light. I've already dropped to terminal (ctrl+alt+f2) and done this:  sudo apt-get update  sudo apt-get install --reinstall ubuntu-desktop  sudo apt-get install unity   Logging in as a guest session also fails. Logging in to other window managers works with varying degrees of success.  Note: I have Cinnamon 2.0 installed from PPA.  I'm using a 2 monitor setup.  Also of note is that the session prior to my upgrade to 13.10 the background of unity failed to display at all, instead showing what was there in the screen buffer from the previous frame. The entire OS worked correctly otherwise though, so I just ignored it for the session. No other upgrades or even updates were done prior to this occurring.  My upgrade path to 13.10 was basically this: Install 13.04 alongside Windows 7, use ubuntu as a glorified web browser for a while, get updates (in preparation for 13.10), install 13.10.  I also used Unity Tweak Tool to change some aspects of unity, particularly auto-hide.  Any help or ideas would be appreciated, as I'm typing this on my phone :( Do I really need to replace my helmet every 2-3 years if it hasn't been in a crash? Ubuntu 20.04 on DualBoot fail 90% of tries - Initramfs unpacking failed: Decoding failed I just installed Ubuntun 20.04 on a Dell Latitude 7400 on DualBoot. I had to remove the RST as mentioned on this post :  It works but now I need to test 5 or more times to boot. After the grub screen, most of the time i have a black screen. Sometimes I have this error : Initramfs unpacking failed: Decoding failed And then it starts.  here is a file called initramfs.conf cat /etc/initramfs-tools/initramfs.conf MODULES=most  BUSYBOX=auto  COMPCACHE_SIZE=&quot;&quot;  COMPRESS=lz4   DEVICE=  NFSROOT=auto RUNSIZE=10% Stability region for two step Nyström method I have the following two step Nyström method: $u_{k+1}=u_{k-1}+2h\cdot f(u_k)$. I want to know the stability region, so I wrote this as $w_{k+1} = A\cdot w_k$ with $A=\left(\begin{matrix} 2h\lambda &amp; 1 \\ 1 &amp; 0\end{matrix}\right)$ and $w_k = (u_{k},u_{k-1})^t$. I use the sample/test equation $u'=\lambda\cdot u = f(u)$ with $\lambda\in\mathbb{C}$. To find the stability region I must set some limitations on the matrix A.  Do I want the whole vector $w_{k+1}$ to go to the zero vector as $k$ goes to infinity, or just the first component that contains $u_{k+1}$? I don't think I can let the second component go to zero, since the matrix just has a $1$ in the bottomleft corner. So I think I want $2h\lambda$ to be negative in the reals (like Forward Euler) , or at least less than 1? I'm unfortunately just guessing at this point.  For (Forward) Euler we have $u_{k+1}=u_k + h\cdot f(u_k) = (1 + h\lambda)u_k$ and so we require that $h\lambda$ does not have a positive real part.  So I guess I also want some anologue for matrices being less than 1? I do not know how to deal with the matrices since the left hand vector not only contains $u_{k+1}$ but also the old $u_k$. Proof for maximal ideals in $\mathbb{Z}[x]$ I have been trying to prove the following theorem:      Every maximal ideal in $\mathbb{Z}[x]$ has the form $(p, f(x))$ where $p$ is prime integer and $f$ is primitive integer polynomial that is irreducible modulo $p$.    Idea: I tried to establish a homomorphism $\phi: \mathbb{Z}[x] \rightarrow \mathbb{F}$. Since $\mathbb{F}$ is a field it has characteristic $p$ and so integer prime p are mapped to 0 in $\mathbb{F}$. Hence $p\in \ker \phi$. Next we consider $\phi': \mathbb{Z}[x] \rightarrow \mathbb{Z_p}[x]$ and pick an arbitrary maximal ideal $M\in \mathbb{Z}[x]$. So, $\phi'(M)$ is maximal as long as $p \in M$ by correspondence. But now I am stuck at this stage and do not know how to proceed. I guess we might have to use primitivity given in problem but dont know how. How to use a flyback diode on a +/-12V motor I've got a motor hooked up to +12V on one side and -12V on the other, being driven in both directions by a Motor controller (). We were getting voltage spikes in the common ground after sending big impulses to the motor, so we put a flyback diode from the negative cable to the positive cable. However, this caused the motor to be less responsive in the reverse direction. How do we correctly prevent voltage spikes in the ground from this type of motor? I have the following sentence:     If T had still been alive, there is the great possibility that either T or C ...   My teacher says that the word "then" must appear after the comma, but I think that it's implied and unnecessary. Who is right? Elevated command line prompt can't access shared drives I map a share from another machine using my user account. I launch an elevated command prompt (cmd.exe, right click, Run as administrator). Navigating to shared drive (Z:) results in:   The system cannot find the drive specified  Now if I open a non elevated command prompt, I can navigate to Z: just fine. I am having trouble comprehending how anyone could come up with this formula:  $$F = \frac{GMm}{d^2}.$$   Could someone walk me through this? Recursively rename all the files without changing their extensions? How to recursively rename all the files in several layers of subdirectories without changing their extensions?  Below's a toned down version (to save room) of what I've got.  For argument's sake, I want all the files to have the same title, yet retain their original extension. There's never more than a single file per directory, so there's no chance of doubling up.  For simplicity, we'll just call them all foo, followed by their current extension.    So just to clarify: Asset\ 1.pdf, Asset\ 1.png, Asset\ 1@4x.png, Asset\ 1.svg Will become: foo.pdf, foo.png, foo.png, foo.svg And so on in that fashion.    I would typically use parameter expansion and a for loop, like:    for f in */*; do mv "$f" "${f%/*}/foo.${f##*.}"; done     But it's not recursive. So I would prefer to use something with find..-exec or similar.    ~/Desktop/Project/Graphics/ ├── Huge │   ├── PDF │   │   └── Asset\ 1.pdf │   ├── PNG │   │   ├── 1x │   │   │   └── Asset\ 1.png │   │   └── 4x │   │       └── Asset\ 1@4x.png │   └── SVG │       └── Asset\ 1.svg ├── Large │   ├── PDF │   │   └── ProjectAsset\ 2.pdf │   ├── PNG │   │   ├── 1x │   │   │   └── ProjectAsset\ 2.png │   │   └── 4x │   │       └── ProjectAsset\ 2@4x.png │   └── SVG │       └── ProjectAsset\ 2.svg ├── Medium │   ├── PDF │   │   └── ProjectAsset\ 3.pdf │   ├── PNG │   │   ├── 1x │   │   │   └── ProjectAsset\ 3.png │   │   └── 4x │   │       └── ProjectAsset\ 3@4x.png │   └── SVG │       └── ProjectAsset\ 3.svg ├── Small │   ├── PDF │   │   └── ProjectAsset\ 4.pdf │   ├── PNG │   │   ├── 1x │   │   │   └── ProjectAsset\ 4.png │   │   └── 4x │   │       └── ProjectAsset\ 4@4x.png │   └── SVG │       └── ProjectAsset\ 4.svg └── Tiny     ├── PDF     │   └── Asset\ 5.pdf     ├── PNG     │   ├── 1x     │   │   └── Asset\ 5.png     │   └── 4x     │       └── Asset\ 5@4x.png     └── SVG         └── Asset\ 5.svg  30 directories, 20 files Peer to Peer: Methods of Finding Peers Are there any known methods of finding peers without using a dedicated central server?  ie: If I have peers which are disconnecting and reconnecting to the internet but getting a new IP address each time, and I want to connect to them without setting up a dedicated server to register with.  I was thinking about using peers email address to send a manifest of connected peers periodically, with some sort of timecode, negating the need for a dedicated server.  This would be a fallback if none of the peers could be connected to after trying all the previously known peer addresses.  But existing models of finding peers would be preferable. Agreement in "[Singular Noun] Is/Are [Plural Noun]"? My fish's native habitat is rice fields. My fish's native habitat are rice fields.   Which one is correct? I'm pretty sure it's the first, since 'is' modifies 'habitat,' but it still sounds weird... Questions about proving $\lnot \lnot a = a$ This is all in the context of a Hilbert system with modus ponens $A, A\to B \vdash B$ and axioms:  Axiom $1$: $A \to (B \to A)$   Axiom $2$: $(A \to (B \to C)) \to ((A \to B) \to (A \to C))$   Axiom $3$: $(\lnot B \to \lnot A) \to (A \to B)$  So I tried to prove double-negation elimination by doing the following:  \begin{array}{ccc} 1 &amp; \{ \lnot \lnot a \} \vdash \lnot \lnot a &amp; \text{Premise}\\ 2 &amp; \{ \lnot \lnot a \} \vdash \lnot \lnot a \to (\lnot \lnot \lnot \lnot a \to \lnot \lnot a)  &amp; \text{ Axiom } 1\\ 3 &amp; \{ \lnot \lnot a \} \vdash \lnot \lnot \lnot \lnot a \to \lnot \lnot a  &amp; \text{ Modus Ponens } 1, 2\\ 4 &amp; \{ \lnot \lnot a \} \vdash (\lnot \lnot \lnot \lnot a \to \lnot \lnot a) \to (\lnot a \to \lnot \lnot \lnot a)  &amp; \text{ Axiom } 3\\ 5 &amp; \{ \lnot \lnot a \} \vdash \lnot a \to \lnot \lnot \lnot a  &amp; \text{ Modus Ponens } 3, 4\\ 6 &amp; \{ \lnot \lnot a \} \vdash (\lnot a \to \lnot \lnot \lnot a) \to (\lnot \lnot a \to  a)  &amp; \text{ Axiom } 3\\ 7 &amp; \{ \lnot \lnot a \} \vdash \lnot \lnot a \to  a  &amp; \text{ Modus Ponens } 5, 6\\ 8 &amp; \{ \lnot \lnot a \} \vdash a  &amp; \text{ Modus Ponens } 1, 7\\ 9 &amp; \vdash \lnot \lnot a \to a  &amp; \text{ Deduction Theorem } \\ \end{array}  And then I continued on, to prove double negation introduction:  \begin{array}{ccc} 10 &amp; \vdash \lnot \lnot \lnot a \to \lnot a &amp; \text{Double Negation Elimination}\\ 11 &amp; \vdash (\lnot\lnot \lnot a \to \lnot a) \to (a \to \lnot \lnot a) &amp; \text{ Axiom } 3\\ 12 &amp; \vdash a \to \lnot \lnot a &amp; \text{ Modus Ponens } 10, 11\\ \end{array}  My questions:   Am I allowed to invoke the deduction theorem in the middle of a proof like this (step $9$)? Is this considered a context change? Is context-changing allowed? Are the proofs even right/correct? Assuming it's all correct: Are we able to combine them to say that $a \to \lnot \lnot a$ and $\lnot \lnot a \to a$, therefore $a \iff \lnot \lnot a$ or is this symbol not allowed? What ultimately let us say $\lnot \lnot a = a$, the use of the equal sign here? Reason for the discreteness arising in quantum mechanics? What is the most essential reason that actually leads to the quantization. I am reading the book on quantum mechanics by Griffiths. The quanta in the infinite potential well for e.g. arise due to the boundary conditions, and the quanta in harmonic oscillator arise due to the commutation relations of the ladder operators, which give energy eigenvalues differing by a multiple of $\hbar$. But what actually is the reason for the discreteness in quantum theory? Which postulate is responsible for that. I tried going backwards, but for me it somehow seems to come magically out of the mathematics. What does \x mean in C/C++? A graph $G$ is bipartite if and only if $G$ can be coloured with 2 colours How to show $(a^b)^c=a^{bc}$ for arbitrary cardinal numbers? One of the basic (and frequently used) properties of cardinal exponentiation is that $(a^b)^c=a^{bc}$.  What is the proof of this fact?  As Arturo pointed out in his comment, in computer science this is called .