, how would this interact with an magic item that makes an extra-dimensional space?  What happens if you give your familiar a Bag of Holding, Portable Hole, or similar item, and then you dismiss your familiar to its pocket dimension? In particular, such items when put in each other tend to burst and cause problems. Does that happen in this case as well? If not, why not?
The Find Familiar spell states, in part:  As an action, you can temporarily dismiss your familiar. It disappears into a pocket dimension where it awaits your summons.  The question arising from this is:  Is this pocket dimension different for every familiar in the world? Thus being a personal pocket dimension? If this is true, then it should be able to store items given by its master for safe keeping away from anyone else, as long as these are items that a familiar could carry with it. Can the familiar bring small items with it into the pocket dimension for safe keeping?
I know that $$\gcd(a^b-1,a^c-1)=a^{\gcd(b,c)}-1$$ which can be seen by expanding both terms into geometric series. Is there any such simplification for $$\gcd(a^b+1,a^c+1)=?$$  EDIT: When working with polynomials instead of whole numbers, it is $$gcd(x^n+1,x^m+1)=\begin{cases}a^{\gcd(n,m)}+1 &amp; \nu_2(n) =\nu_2(m)\\1 &amp;\text{else}\end{cases}$$ But this gives only a lower bound for the whole-number case. What is the difference between <%# and <%= in ASP.NET MVC? Why does the question's closed status disappear when clicking to load a new edit? I've come across a small issue which occurs when a question has been closed but the question itself or one of its answers are being edited.  Steps to reproduce:   Visit a closed question. (even closed years ago)   In a different tab or browser, edit the question or one of its answers. "an edit has been made to this post; click to load" message will appear above the edited post in the first tab/browser, click it. The .question-status div becomes hidden, which includes the closed notice below a question and, for a duplicate-closed question, the list of duplicates above the question body. This gives the appearance that the question is no longer closed (until you refresh).   Some screenshots:                I know it isn't common/important but it did confuse me when looking at a particularly bad question - I thought it had been reopened.   Is this an intentional feature? If so, what was the thinking behind it?    Another example  Before    After When should I add another answer? I have been member of SO for 4 months. I have a question regarding some functionality. Whenever I give an answer to a question, there is functionality given Add Another Answer. I can comment on my answer at any time and also able to edit answer. Then   Why this functionality given on SO to add another answer?  When should I add another answer?  Why there is requirement of multiple answers? What is limit of adding answers to single question? How is the passage of time handled in Pokemon? I stopped following the Pokemon anime a while ago, but I know the series has had Ash go through many regions, including retreading at least one of them.  When Ash was about to get his eight badge, Misty mentioned that they'd been on the road for a year at that point.  I saw some recent episodes, and Ash looks nearly identical to when he first started as a trainer.  Normally I'd attribute that to cartoon sitcom logic, where the characters never age; but in a series with continuity and references to earlier seasons, this makes no sense.  What's the official word on a timeline for the Pokemon anime? Many-worlds: how often is the split how many are the universes? (And how do you model this mathematically.) Suppose $A,B,C$ are sets. Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$. Suppose $A,B,C$ are sets.  Prove that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$.  I proved that for $x\in A\setminus B$, $x\in A\cup C$ and for $x\in B\setminus A$, $x\in B\cup C$. How do I show that $A\cup C=B\cup C$, and that $A\mathbin\triangle B\subseteq C \iff A\cup C=B\cup C$? Combinatorial Proof Question: Vandermonde convolution identity I'm really iffy on combinatorial proofs in general and now that there is a sum, it's just confused me even more. Can someone try and walk me through this proof?  $$ \binom{m + n}{r} = \sum_{k=0}^r \binom{m}{k}\binom{n}{r - k} $$ What equipment is necessary/preferred for real estate photography? What equipment is necessary for real estate photography, and what is preferred?  I'm starting into real estate photography, and I'm trying to figure out what I need and what I will want to plan on adding to my kit as I go.  I'm thinking especially along the lines of lighting equipment and lenses (focal lengths).  For example, is an ultra-wide actually necessary? Or is a general walk around sufficient. Is momentum conserved in this situation? Suppose we have object A , which has momentum mv. Then it elastically collides with a wall, so it's momentum now -mv, So how is momentum conserved here ? Where are Steam games installed? I want to buy a new notebook in the near future to play games on it. I'm looking forward for the new Steam for Linux client. My question is that where does the Steam install the games, on the home partition, or in the usr folder, or somewhere else? I do not use pre-release (i.e. beta) versions of either Steam or Ubuntu. I've been trying to understand $\sigma$-algebras and how it encodes information in context of filtration. While certain parts seem clear and logical, I can't say I get the whole picture.  I'll try to explain the counter-intuition I get with the classical example of : the probability space $\Omega = \{ HH, HT, TH, TT \}$ and a r.v. $X(\omega)$ equal to the number of heads.  At times $0$, $1$ and $2$ the available information is represented using $\sigma$-algebras $\mathcal{F}_0=\{\emptyset,\Omega\}$, $\mathcal{F}_1=\{\emptyset, \Omega, \{HH,HT\},\{TH,TT\}\}$ and $\mathcal{F}_2=\{\emptyset, \Omega,\{HH,HT\},\{TH,TT\},\{HH\},\{HT\},\{TH\},\{TT\}\}$.  One can notice that $X(\omega)$ is not measurable with respect to $\mathcal{F}_0$ and $\mathcal{F}_1$, because $X^{-1}((\frac{3}{2}; +\infty))=\{HH\}$. To me it is quite surprising: intuitively $X$ makes perfect sense at all times. In particular it has an expected value at time $0$, which I interpret as that the probability and value of all outcomes $\{\omega\}$ can be computed. How do you think of a non-measurable function?  Here's another way of expressing the same confusion. The most natural choice of $\sigma$-algebra in a finite discrete case is $\mathcal{F}=2^\Omega$, and it is implicitly used in all elementary probability problems. However, this choice of $\mathcal{F}$ does not reflect the fact that some information is known or unknown, conditional probability does. Does it mean that the statement "$\sigma$-algebra is known information" make sense only in conditioning? Why is it convenient then? Mirror modifier. Mirrored half is not aligned, but rotated I was working on my character's hand when I realized half of the body was suddenly deleted. I am trying to recreate the other half, but it just won't mirror properly in the center. I tried moving the origin to the 3D cursor while it was completely centered, but the mirror modifier is still applied at an angle. What does "if not" mean in the given sentence A Polygon is inscribed in a circle $\Gamma$ A regular polygon P is inscribed in a circle $\Gamma$. Let A, B, and C, be three consecutive vertices on the polygon P, and let M be a point on the arc AC of $\Gamma$ that does not contain B. Prove that  $MA\cdot MC=MB^2-AB^2$  I inscribed the polygon P in the unit circle and let B=1. Past this, I'm not really sure how to proceed. It might be helpful to have M at 1 as well but given that the polygon would have the n-th roots of unity in it B seemed like a good choice. What would be a good way to proceed from here? I am new to Ubuntu and want to install VMWare Player. What is the step by step procedure to install that? This is a  about DoS and DDoS mitigation.   I found a massive traffic spike on a website that I host today; I am getting thousands of connections a second and I see I'm using all 100Mbps of my available bandwidth.  Nobody can access my site because all the requests time out, and I can't even log into the server because SSH times out too!  This has happened a couple times before, and each time it's lasted a couple hours and gone away on its own.  Occasionally, my website has another distinct but related problem: my server's load average (which is usually around .25) rockets up to 20 or more and nobody can access my site just the same as the other case.  It also goes away after a few hours.  Restarting my server doesn't help; what can I do to make my site accessible again, and what is happening?  Relatedly, I found once that for a day or two, every time I started my service, it got a connection from a particular IP address and then crashed.  As soon as I started it up again, this happened again and it crashed again.  How is that similar, and what can I do about it? How come if $\ i\ $ not of the following form, then $12i + 5$ must be prime? I know if $\ i\ $  of the following form $\ 3x^2 + (6y-3)x - y\ $ or $\ 3x^2 + (6y-3)x + y - 1, \ \ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$,  then $\ 12i + 5\ $ must be composite number, because:  $12(3x^2 + (6y-3)x + y - 1) + 5 = 36x^2 + (72y - 36)x + (12y - 7) = (6x + 12y - 7)(6x  + 1)$  How come if $\ i\ $ not of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $,  then $12i + 5$ must be prime? $\ x,y \in \mathbb{Z}^{+},i \in \mathbb Z_{\ge 0}$.  For example: $\ 5 = \ 3*1^2 + (6*1-3)*1 - 1\ $ ,when $\ x = y = 1\ $, as proved before,$\ 5*12+5\ $ must be composite number;  $\ 0, \ 1. \ 2.\ 3.\ 4\ $ can't of the following form $\ 3x^2 + (6y-3)x - y\ $ and $\ 3x^2 + (6y-3)x + y - 1\ $,  $\ 12*0 + 5 = 5\ $, $\ 5 $ is prime,$\ 12*1 + 5 = 17\ $, $\ 17 $ is prime, and so on, Simplest or nicest proof that $1+x \le e^x$ The elementary but very useful inequality that $1+x \le e^x$ for all real $x$ has a number of different proofs, some of which can be found online. But is there a particularly slick, intuitive or canonical proof? I would ideally like a proof which fits into a few lines, is accessible to students with limited calculus experience, and does not involve too much analysis of different cases. Installing Guest Additions causing problems I am trying to install Guest Additions in VirtualBox with Windows XP host and Ubuntu 12.04 guest. I followed the above steps:  Devices-&gt; Insert Guest Additions CD Image. After that I locate the folder /media/VBoxAdditions4.3.4_91027/ and run sudo ./VBoxLinuxAdditions.run   I want to be able to create shared folders and adjust the resolution of the screen. I achieved to do the first issue, but I couldn't do the second one. I didn't get any message, however it is not possible to change the resolution of the screen and I cannot switch to fullscreen. What am I doing wrong?  I found in the instal log the following message: unknown version of the X Window System. I've tried the solution of the proposed link but I got the following message:  Building dependency tree        Reading state information... Done Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies: virtualbox-guest-x11 : Depends: xorg-video-abi-11                     Depends: xserver-xorg-core (&gt;= 2:1.10.99.901) E: Unable to correct problems, you have held broken packages.   I tried also to install with the command sudo apt-get install virtualbox-guest-dkms but nothing is changing. During the installation of dkms I got the message: Error! Bad return status for module build on kernel 3.13.0-32-generic(i686).