I'm thinking of running a Pathfinder campaign for Skype (text only) and I don't have a group of players. How can I go about recruiting people for play-by-chat games instead of in-person? Are there any online sites in which you can look for people that are looking to play in an online play-by-chat RPG Campaign?
I'd really like to learn how to play pen-and-paper RPGs.  I know that it takes a few players to actually sit down and play, but I'm not at all sure where to find more gamers.  How do I find existing groups to join? Or, are there any resources online that may help me find existing RPG groups or players?
What makes a bread either close or open crumbed/textured? What causes the texture of bread to be open, ie with lots of large holes, or close, ie a regular, uniform tender crumb with no large holes? How can I achieve either? I am flying from Los Angeles to Bangalore. I have a flight option from LAX to Minneapolis and from there to Paris. I have only 45 minutes at Minneapolis before my next flight. Is this too risky a schedule? I am not an American, so I guess I would need to cross emigration too? I have brand new machine.  I want to install Ubuntu and one more Linux distributions on same machine. When the machine is up I want to have the option to choose which Linux to load.   Is it possible?  How to do it? I can't play Pokemon GO! What's wrong? Pokemon GO isn't working at all for me right now. I'm in one of the following circumstances and can't do anything:   My game is displaying a "Try Again" message I'm stuck on the Loading Screen When I sign in I get a message that the servers are experiencing problems. I see the map but nothing is loaded, and my team color is wrong The "loading" Pokeball just ends up spinning and I can't catch or interact with things When I try to catch a Pokemon, the Pokeball just sits on my screen doing nothing Gym opponents will drop to 1 HP and won't take any more damage or attack When I open a pokestop, the pokestop image and the name of the pokestop does not load.   What's causing this? Is there a way to fix it? How can I add a program to the context menu of all files? Haar Measure on Semidirect Product of Unimodular Groups Let $M, N$ be closed subgroups of a given locally compact topological group such that the product $P = MN$ is semidirect, with $M$ normalizing $N$.  Assume second countable, sigma compact, whatever so that Radon products coincide with product measures.  To make life easy, assume $M, N$ are unimodular.  How do we construct a Haar measure on $P$? Can I use old Minolta AF lenses on my new Canon? I just recently bought my first DSLR, a Canon Rebel t2i/550d. I use it mostly for video and it works wonderfully. I bought the kit lens with it, but I am looking to expand my lens collection. I found a set of old Minolta lenses from my dad, and I was wondering if I could use them on my new camera. The lenses are: a Sigma UC zoom 28-105 f4-5.6, a Minolta Maxxum zoom 70-210mm, a Minolta Maxxum zoom 28-58mm and a Kalimar 500mm lens. I know about the crop factor and that the auto focus won't work, but that does not bother me. Is there an adapter for Minolta AF/Sony to Canon EF mount? Any help is appreciated. Thanks! I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!  Enough backstory. Time for math. The problem is as follows:  Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.  I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.  Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions! How to build a custom kernel? What is dependency injection? There have been several questions already posted with specific questions about , such as when to use it and what frameworks are there for it. However,  What is dependency injection and when/why should or shouldn't it be used? How does Spider-Man climb walls? Sum of $k$-th powers Another point of view that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is cyclic. I was thinking that the product of groups $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/2\mathbb{Z}$ is not cyclic, but $\mathbb{Z}/2\mathbb{Z} \times\mathbb{Z}/p\mathbb{Z}$ is cyclic if p is an odd prime. Then I found logic that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is certainly cyclic if $m$ and $n$ are coprimes.  I searched on google a proof of that, to compare with mine and the answer is that $\mathbb{Z}/m\mathbb{Z} \times\mathbb{Z}/n\mathbb{Z}$ is isomorphic to $\mathbb{Z}/mn\mathbb{Z}$ (which is clearly cyclic) $\iff$ $m$ and $n$ are coprimes.  But I got another proof and I want to know what you think about it :     I claim that $([1]_m,[1]_n)$ generates $\mathbb{Z}/m\mathbb{Z}$   $\times\mathbb{Z}/n\mathbb{Z}$, where $[a]_d$ is the canonical   projection of $a\in \mathbb{Z}$ into $\mathbb{Z}/d\mathbb{Z}$.       Let's take multiples of $([1]_m,[1]_n)$ until we get $m \cdot$   $([1]_m,[1]_n) = ([0]_m,[m]_n)$ (where I suppose $m=\min(m,n)$...   otherwise, consider $\mathbb{Z}/n\mathbb{Z}$   $\times\mathbb{Z}/m\mathbb{Z}$ and $\min(m,n)=n$).       The next step is $([1]_m,[m+1]_n)$. This pair is not equal to   $([1]_m,[1]_n)$, since $m+1 \not\equiv1$ mod $n$ ($m$ and $n$ are   supposed to be coprimes). We can now continue adding $([1]_m,[1]_n)$   to the results.      The last step is precisely reached when you got   $([mn]_m,[mn]_n)=([0]_m,[0]_n)$. At this point, the fact that $m$ and $n$ are coprimes is absolutely necessary because if they are not coprimes, $lcm(m,n)\ne m\cdot n$ and   $([lcm(m,n)]_m,[lcm(m,n)]_n)=([0]_m,[0]_n)$ but all the $m \cdot n$ possible pairs are not reached, only the $lcm(m,n)$ are ! How to get a vintage/dreamy look like this photo?   It has some yellow color cast and seems like extra blur added in post process? Let $A \subset B$ be a finite ring extension. Fix a ring homomorphism $\nu : A \to \Omega$, where $\Omega$ is an algebraically closed field.   I want to show there exists a non-zero homomorphism $v : B \to \Omega$ of $A$-modules.  I want show that the space $V$ of all the homomorphisms $v$ in (i) is a non-zero finite dimensional vector space over $\Omega$. Show that the algebra $B$ acts on this space. Show that eigenvectors of this action correspond to homomorphisms of algebras $\lambda : B \to \Omega$ that extend the homomorphism $\nu$. Deduce that the set of such homomorphisms $\lambda : B \to \Omega$ is finite and non-empty.    We can assume that $B = Ab_1 + ... + Ab_n$ where $b_1,...,b_n \in B$. The chose $v(b_i)=\omega_i \in \Omega-\{0\}$ and then define $$ v( a_1 b_1 + ... + a_n b_n ) = \nu(a_1) v(b_1) + ... + \nu(a_n) v_n = \nu(a_1) \omega_1 + ... + \nu(a_n) \omega_n \ .$$ This is an $A$-module homomorphism. Am I correct? For the second part, that $V$ is not-zero follows from the first part. That it is finite dimensional follows from the fact that $v_i(b_j)=\delta_{i,j}$ (Kronecker's delta) with $i=1,...,n$ spans $V$ and hence $\dim V \le n &lt; \infty$. I define the action of $B$ over $V$ to be $$b \bullet v = \left( x \mapsto v(b)v(x) \right) \mbox{  for all } b,x \in B \ .$$ This seems to be the natural action to define. Am I correct? However, I haven't managed to solved the rest (the eigenvector stuff and the relation to $\nu$). How to find a starting point for root finding in complex plane? The function $f(x) = x^2 +1$ has zeros in the complex plane at $x = +i$ and $x = -i$. Is there a real starting point for complex Newton's method such that the iterates converge to either of these zeros? Is there a complex starting point too?  I know that Newton's method for finding roots is $x_{n+1} = x_n - \frac {f(x)}{f'(x)}$. But I don't know how to find the proper starting point. Do I lose anything converting to DNG? I'd like to convert all my ARW (Sony) files to DNG for several reasons, but the only thing keeping me back from doing it is the fear of losing useful metadata during the conversion. I know that the image it's self is completely safe during the conversion but what about proprietary maker noteS? Will those stay too? And even if they do, are they actually useful and can be actually used by something other then Sony's own RAW converter?  P.S. I'm using Ubuntu so I'd prefer to use KipiPlugins' DNG converter because from what I understand it does everything the Adobe converter does, but if I'm wrong I can use Adobe's instead.  EDIT:  It looks like the LensID does carry over! With the ARW and the DNG files converted by Kipiplugins' built in converter, info is stored as "Exif.Sony.0x___". With Adobe's official converter, it moves the info to more logically named fields in the XMP, such as "LensID" and "Lens".  Now that I've figured it out, I think I might use Adobe's Official converter, just because I prefer how it reorganizes the Exif Data.  I'm not fully convinced though and I'd like to hear others take on all this. Why does the Moon not revolve around the Sun directly? The sun pulls on the moon with a force that is more than twice the magnitude of the force with which the earth attracts the moon. Why, then, doesn’t the sun take the moon away from the earth? Prove that there exist two integers such that i - j is divisible by n. Here's the full question:  Prove that, for any $n + 1$ integers, $\{x_0, x_1, x_2, . . . , x_n\}$, there exist two integers $x_i$ and $x_j$ with $i \neq j$ such that $x_i − x_j$ is divisible by $n$.  Now, the integers aren't necessarily consecutive, positive, or without repeats. I've tried breaking this into cases, such as "$x_i = x_j, \dfrac{0}{n} = 0$" etc. But I don't think it's possible to exhaust the cases.  I feel like there should be an easy way to contradict this one, by saying that $x_i - x_j$ isn't divisible by $n$. But I just wasn't getting anywhere with that one.  I thought about using induction, but I would have just as hard a time proving this for $n$ integers before $n+1$ integers, so I don't think it would be possible for me to do it that way.  If anyone could help out, that would be wonderful. I'm pretty stuck here. Getting No module name exifread error from QGIS Photo2Shape plugin?