We all recognize the regenerative capacities of the countless characters from Marvel and DC, and Wolverine is one the first that comes in my mind when speaking of immortality (or "unkillability"?), after Deadpool and many others.  There is already a , and although it is also included in X-Men universe, the logic clearly is not the same for both characters. Furthermore, after the events in many X-Men movies - particularly in X-Men Origins or , which in my country is titled "Wolverine: The Immortal" -, we also perceive much more the immortality or semi-immortality after him being shoot in the head or losing part of his members.  I'm a layman when speaking of Marvel comics, but I'm always very curious when it comes to be a subject matter of immortality. For example, what would happen if Wolverine was vertically cut in half? Two wolverine are generated? At least two brains, in ideal conservation ambient, could develop by themselves?  In the real world something similar can happen with non-differentiated stem cells, but I don't know what logic is used when concerning adamantium and tissue regeneration in Marvel universe. I wonder if there are any references that put this topic in evidence in the comics.
Clearly, if Wolverine bleeds on the ground, a new Wolverine doesn't reconstitute himself from the drop. Presumably, if somehow his arm could be severed, that wouldn't happen there, either. If his head were cut off...probably not either.  So it seems that there's a minimum amount of contiguous body mass Wolverine has to have from which he can heal.  Maybe limbs are the limit...are there examples of Wolverine regenerating a lost limb? If so, well, this question. If not, is there anything else in the canon to suggest an answer?
How to implement a follow author (or user) function? Is it possible to implement a Follow button for authors of posts in Drupal 7, so that users can:   Follow any one they wish, and Be notified about the new posts of those authors.   This sounds like 'Mashable Follow button' for authors.  Do you have any suggestions about how to implement this? Plot $|z - i| + |z + i| = 16$ on the complex plane Plot $|z - i| + |z + i| = 16$ on the complex plane   Conceptually I can see what is going on. I am going to be drawing the set of points who's combine distance between $i$ and $-i = 16$, which will form an ellipse. I was having trouble getting the equation of the ellipse algebraically.  I get to the point:  $x^2 + (y - 1) ^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} + x^2 + (y+ 1)^2 = 256$  $2x^2 + 2y^2 + 2\sqrt{x^2 + (y - 1)^2}\sqrt{x^2 + (y + 1)^2} = 254$  It seems like I'm and doing something the hard way. Proving that if $A^n = 0$, then $I - A$ is invertible and $(I - A)^{-1} = I + A + \cdots + A^{n-1}$ Delete with Join in MySQL Here is the script to create my tables:   CREATE TABLE clients (    client_i INT(11),    PRIMARY KEY (client_id) ); CREATE TABLE projects (    project_id INT(11) UNSIGNED,    client_id INT(11) UNSIGNED,    PRIMARY KEY (project_id) ); CREATE TABLE posts (    post_id INT(11) UNSIGNED,    project_id INT(11) UNSIGNED,    PRIMARY KEY (post_id) );   In my PHP code, when deleting a client, I want to delete all projects posts:  DELETE  FROM posts INNER JOIN projects ON projects.project_id = posts.project_id WHERE projects.client_id = :client_id;   The posts table does not have a foreign key client_id, only project_id. I want to delete the posts in projects that have the passed client_id.  This is not working right now because no posts are deleted. How to create a .deb file with dependency? I have a game made with the LÖVE framework and I want to package it into a .deb file so it can be easily installed by others and be easily findable through the Unity dash for example.   I also have the LÖVE framework love_0.7.2.deb which I want to be in the deb file also somehow.  Files that I have:  game_icon.png game .desktop the_game.love love_0.7.2.deb   Could someone explain to me how to do this? Can I get SSH to use an askpass program even if it was run from a terminal? says:  SSH_ASKPASS    If  ssh needs a passphrase, it will read the passphrase from the    current terminal if it was run from a terminal.  If ssh does not    have  a  terminal associated with it but DISPLAY and SSH_ASKPASS    are set, it will execute the program  specified  by  SSH_ASKPASS    and open an X11 window to read the passphrase.   I'd like SSH to use an askpass program even if it was run from a terminal.  On occasion, I have to connect to servers, where there's some delay in showing a password prompt (maybe due to network issues, maybe due to attempted reverse DNS lookups, …). I get annoyed and switch to something else, and forget about the attempted connection. (Insert joke about attention span of a goldfish.) When I finally get back to it, the prompt's timed out and even a correct password would just result in a closed connection.  Keys would be one solution, but not every system I use has my usual SSH keys. However, I usually use Ubuntu systems, and Ubuntu has an SSH askpass program installed by default.  If an askpass window popped up, however, I'd be immediately aware of it. That's a good enough compromise for me, if only I can get it to work. I'm having trouble with this functional equation, to find all functions $f:\mathbb R \to \mathbb R$ for which the following is valid for all $x,y\in \mathbb R$: $f(x) + f(x+f(y)) = y + f(f(x) + f(f(y)))$ I haven't been able to pinpoint any solution, nor to prove surjectivity or injectivity. So 0 progress. Tried with $x=y=0, x=0, y=0, x=-f(y), y=f(x)$, and even with some assumptions on bijectivity, but nothing helped. I don't know the source but seems like a pretty difficult problem, maybe even impossible to solve. I'm struggling to understand the concept of quantum entanglement. I've distilled my understanding into an analogy, and I need your help to validate it. Here it is:  Let's say I receive two envelopes. Both envelopes have this written on them:     Hello! Open this letter...   If the paper inside is red, then other envelope has blue paper.   If the paper inside is blue, then other envelope has red paper.   So, if I open one letter, and the paper inside is red, then I know the other letter contains the blue paper.  Is this equivalent to quantum entanglement? The fact that I will know the color of the other paper when I open (i.e. observe) one envelope?  In theory, if one of the letters "appeared" in a different galaxy, I would still know the color of the paper inside that letter, just by opening my letter, instantly, correct?  Before I open the letter, is the paper inside each envelope in superposition (??), i.e. 50/50, red or blue?  In this analogy, I am assuming that what's written on the letters is true. Is this a correct assumption in quantum physics? I think so yes?  In real quantum physics, would I be able to change the color of one paper, so that I change the color of the other one? No right? This is a  about Hairpin NAT (Loopback NAT).   The generic form of this question is:  We have a network with clients, a server, and a NAT Router. There is port forwarding on the router to the server so some of it's services are available externally. We have DNS pointing to the external IP. Local network clients fail to connect, but external work.   Why does this fail? How can I create a unified naming scheme (DNS names which work both locally and externally)?   This question has answeres merged from multiple other questions. They originally referenced FreeBSD, D-Link, Microtik, and other equipment. They're all trying to solve the same problem however. On some icons I am seeing a two blue arrows at the top the right pointing towards each other. I first noticed it on the icons on folders which I archived which happened randomly. I archived the folder and the text went blue as expected, then when I went to the folder again the text was black and the folder icon has these arrows. When I just recently installed Office 2007, I noticed the same arrows on the icons for the programs. OS: Windows 10 Pro 64-bit (the latest update: 1115) Application: Microsoft Office 2007 suite and file explorer How to prove that any (integer)$^{1/n}$ that isn't an integer, is irrational? Is my proof beneath perfect and complete?  I wanted to prove that for any nth root of an integer, if it's not an integer, than it's irrational: $$\begin{cases} m,n\in \mathbb{N}\\\sqrt[n]{m}\notin \mathbb{N} \end{cases}\implies \sqrt[n]{m}\notin \mathbb{Q}.$$  I start by assuming that $m^{\frac 1n}$ is rational and non-integer. So there exist  co-prime integers $a,b$ so that $$\sqrt[n]{m}=\frac{a}{b}$$ $$\implies  m=\frac{a^n}{b^n}\in\mathbb{N}.$$ But since $a$ and $b$ have no common factor, $a^n$ and $b^n$ also have no common factor. So: $$\frac{a^n}{b^n}\notin\mathbb{N},$$ a contradiction. Is 'efficate' a word in English? I routinely hear the word "efficate" being used.  For example, "The most powerful way to efficate a change in the system is to participate."  I do not find entries for this word in common English dictionaries, but I do not have an unabridged dictionary.  I have checked the OED (I'm not sure if it is considered unabridged), and it has no entry for "efficate".  It does have an entry for "efficiate", which is used in the same way.  Wordnik has an  for "efficate" with over 1800 hits, thus providing some evidence for the frequency of use.  I personally like the word and find the meaning very clear and obvious when others use it.  If it's not currently an "officially documented" word, perhaps its continued use will result in it being better documented. This is the terminal message I keep getting when trying to update via: sudo apt-get update  $ sudo apt-get update [sudo] password for petros: Reading package lists... Done E: Could not get lock /var/lib/apt/lists/lock - open (11: Resource temporarily unavailable) E: Unable to lock directory /var/lib/apt/lists/   How can I solve it? I would like to count subgroups of each order (2, 3, 4, 6, 8, 12) of $S_4$, and, hopefully, convince others that I counted them correctly.  In order to do this by hand in the term exam, I need a clever way to do this because there can be as many subgroups of a group of order 24 as $2^{23}$.  Do you know how to do this?   (I would be most grateful if you could tell me what part of the answer to the old question answers my question before voting to close.) I am trying to run Notepad as admin so I can edit my hosts file from the command line.  I have tried runas /user:(myusername)\administrator "notepad c:\windows\system32\drivers\etc\hosts" I then input my password and I get     RUNAS ERROR: Unable to run - notepad   c:\windows\system32\drivers\etc\hosts 1327: Account restrictions are   preventing this user from signing in. For example: blank pa sswords   aren't allowed, sign-in times are limited, or a policy restriction has   been enforc ed.   PS: I know if I give permission to my user account I can edit it without running as admin. But I'd like to know how to do this without having to change permissions on the hosts file. How to mount an image file without root permission? Incredible frequency of careless mistakes Toggle wireframe mode with hotkey How do I toggle in and out of wireframe mode with my keyboard? Is there a hotkey for this?  I'm a beginner (obviously). I thought I could find this in the "view" menu, but it either doesn't exist or I have horrendous eyesight. Thanks! Why do objects move when you apply a force that's equal in magnitude to the friction? If the friction on an object is 200 N and you apply 200 N, it seems, the forces should immediately cancel each other out since the friction and your applied force go in opposite directions, leaving you with an object that is motionless. Yet, from what I've read, that's not the case. I feel like I'm misunderstanding something. Could it be I'm wrong about needing to apply 200 N initially? Might the amount of force I actually need be greater, but instantaneously decrease to 200 N? Let $G$ be a finite abelian group, then $G$ has a subgroup of order $n$ if and only if $n\mid G$.  Proof: by Lagrange if $H\leq G$ then $|H|$ divides $|G|$ so this proves one of the implications.  We prove the other implication by strong induction on the order of $G$, for primes this is just Cauchy, so we can take these like base cases. Now let $n$ be a proper divisor of $|G|$, take $p$ a prime in the factorization of $|G|$ (instead of $|G|$ is should be $n$, I had made a mistake when typing the first time), then by cauchy there is an element $x$ of $G$ of order $p$, since $G$ is abelian $\langle x\rangle$ is normal in $G$ so we can look at the quotient group $G/\langle x\rangle$, this group has order $\frac{|G|}{p}$, now take a prime $p_1$ in the factorization of $\frac{|G|}{n}$. By induction the quotient group has a subgroup $E$ of order $G/pp_1$ now look at the preimage of $E$ under the natural homomorphism from $G$ to $G/\langle x\rangle$, it is easy to see this is a subgroup of $G$, we call it $I$, consider the restriction of the natural projection to $I$, this is a surjective homomorphism from $I$ to $E$ with kernel $\langle x\rangle$ so by the first isomorphism theorem $I/x\cong E$ so $\frac{|I|}{|\langle  x\rangle|}=|E|\implies  |I|=\frac{|G|}{pp_1}{p}=\frac{|G|}{p_1}$ Notice $\frac{|G|}{p_1}$ is a multiple of $n$ since $p_1$ was in the fatorization of $\frac{|G|}{n}$. So by induction $I$ has a subgroup of order $n$ which is also a subgroup of $G$.