Capitalization and hyphenation of proper noun declensions I'm transcribing some speech and I came across     One of the accusations that certain non-Orthodox Christians level against the Orthodox is that we worship idols.   However, I am not certain on how to style the word "non-Orthodox" properly.  Is there a reference I can consult here? Is there one right way, or are multiple forms acceptable?  For example:   "Nonorthodox" makes sense because 'nonorthodox' is a normal word, but should be capitalized as part of a proper noun. "non-Orthodox" makes sense because "Orthodox Christian" is the base term, but Nonorthodoxy is not a specific class of Christianity.
Hyphens in words with prefixes like neo-Nazism I have a question regarding the use of hyphens in words with certain prefixes. For example,     Neo-Nazism   Neo-Confucianism   Neo-liberal     are words with a hyphen, but     Neoplatonism   Neocolonialism   are words without hyphens. I'm especially confused with neo-Confucianism and neoplatonism: one contains  a hyphen and a capital letter and the other doesn't.  Is there a specific rule for hyphens in words with prefixes or is it purely arbitrary?
Smooth transition between two materials Do I need a Japanese transit visa when arriving at Narita and departing from Haneda? How do you square an ideal? For some of you, this question is going to seem extremely basic.  I think I understand what an ideal is. Such as $\langle 2, 1 + \sqrt{-5} \rangle$, it consists of all numbers in this ring of the form $2a + (1 + \sqrt{-5})b$.  But then what is $\langle 2, 1 + \sqrt{-5} \rangle^2$? My first thought was $\langle 4, -4 + 2 \sqrt{-5} \rangle$, but that seems wrong somehow. I also had something in the back of my mind saying $\langle 2 \rangle$, but I'm not sure about that one either.  Then I thought about trying to figure out $\langle 2, 1 + \sqrt{-5} \rangle \langle 2, 1 + \sqrt{-5} \rangle$ when I realized I don't actually understand how to multiply ideals to begin with.  I'm only using $\langle 2, 1 + \sqrt{-5} \rangle$ as an example (though that does draw in one question identified as similar that looks much more relevant than all the questions identified as "Questions that may already have your answer"). In a principal ideal domain, would it be correct to think that $\langle a \rangle \langle b \rangle = \langle ab \rangle$?  Any help would be much appreciated. I am trying to create a vase that looks similar to this:    If there is another post out there I will happily remove this one or redirect it there.  Specifically, I am trying to create the creased effect in the geometry that looks like it is twisting around the vase.  I have tried using Checker Deselect and extrude->scaling the verts but that really screws up my mesh.  Link to my model if it helps: I can seem to run scripts (.sh) with and without them being set as executable. So where exactly this matters? Hovering over featured moves the questions down a bit When I hover over featured tab, both on the front page and in questions, the questions move down by one pixel. Here's a small screen capture demonstrating the issue:        I see it in Chrome 34 and Opera 20, but I don't see it on IE11 or Firefox 28. I'm getting frequent inaccurate location as the same location, can I make this stop? I need to be able to merge two (very simple) JavaScript objects at runtime.  For example I'd like to:  var obj1 = { food: 'pizza', car: 'ford' } var obj2 = { animal: 'dog' }  obj1.merge(obj2);  //obj1 now has three properties: food, car, and animal   Does anyone have a script for this or know of a built in way to do this?  I do not need recursion, and I do not need to merge functions, just methods on flat objects. This should be a fairly trivial problem. I'm trying to open an ofstream using a std::string (or std::wstring) and having problems getting this to work without a messy conversion.  std::string path = ".../file.txt";  ofstream output;  output.open(path);   Ideally I don't want to have to convert this by hand or involve c-style char pointers if there's a nicer way of doing this? How to use SSH to run a local shell script on a remote machine? Questions with many upvotes should not be closable I see a lot of good, well-writen and constructive questions (that are probably not only good for me) closed, by various reasons which I fail to digest, so I propose the idea that questions with n upvotes should not be closable.  This would be in my opinion very much in the spirit of "", and unless I'm wrong, it would be beneficial for the users to choose by themselves whether the site should have or not x question. The ones which are low quality never get many upvotes anyway.  The only cons I can think of are... DB space? And at a minimum there's the pro of more popular pages with more ads (for the admins), and questions people like (for the users).  I'm also counting for , in the lines of Google's "", but questions  are more like what I'm aiming at (probably that question has not many upvotes for this though). If $\psi h$ is in $L^2$ for all $h\in L^2$, must $\psi$ be essentially bounded? If $\psi:[a,b]\to\mathbb C$ is a (measurable) function such that $\psi h$ is in $L^2[a,b]$ for all $h\in L^2[a,b]$, then must $\psi$ be essentially bounded?  The converse direction is clear: If $\psi$ is essentially bounded, then $\|\psi h\|_2\leq \|\psi\|_\infty\|h\|_2$ for all $h\in L^2$, and thus the multiplication operator $M_\psi:h\mapsto \psi h$ is defined and bounded.  The question here is whether, without prior assumptions on $\psi$, simply having $M_\psi(L^2)\subseteq L^2$ implies that $\psi$ is bounded. What is the radius of convergence of $\sum z^{n!}$? How to find the radius of convergence of $\sum z^{n!}$?   I'm used to applying the ratio test to power series of the form $\sum a_{n}z^{n}$, but for a different power of $z$, I am a bit stumped. What about $\sum z^{2n+a}$ for another example? Where $a\in \mathbb{R}$. Dimensionality of null space when Trace is Zero This is the fourth part of a four-part problem in Charles W. Curtis's book entitled Linear Algebra, An Introductory Approach (p. 216).  I've succeeded in proving the first three parts, but the most interesting part of the problem eludes me.  Part (a) requires the reader to prove that $\operatorname{Tr}{(AB)} = \operatorname{Tr}{(BA)}$, which I was able to show by writing out each side of the equation using sigma notation.  Part (b) asks the reader to use part (a) to show that similar matrices have the same trace.  If $A$ and $B$ are similar, then  $\operatorname{Tr}{(A)} = \operatorname{Tr}{(S^{-1}BS)}$  $= \operatorname{Tr}(BSS^{-1})$  $= \operatorname{Tr}(B)$,  which completes part (b).  Part (c) asks the reader to show that the vector subspace of matrices with trace equal to zero have dimension $n^2 - 1$.  Curtis provides the hint that the map from $M_n(F)$ to $F$ is a linear transformation.  From this, I used the theorem that $\dim T(V) + \dim n(T) = \dim V$ to obtain the dimension of the null space.  Part (d), however, I'm stuck on.  It asks the reader to show that subspace described in part (c) is generated by matrices of the form $AB - BA$, where $A$ and $B$ are arbitrary $n \times n$ matrices.  I tried to form a basis for the subspace, but wasn't really sure what it would look like since an $n \times n$ matrix has $n^2$ entries in it, but the basis would need $n^2 - 1$ matrixes.  I also tried to think of a linear transformation whose image would have the form of $AB - BA$, but this also didn't help me.  I'm kind of stuck...  Many thanks in advance! Prove by induction that $n! > 2^n$ Suppose that when $n=k$  $(k\geq4)$, we have that $k!&gt;2^k$.  Now, we have to prove that $(k+1)!\geq2^{k+1}$ when $n=(k+1)$ $(k\geq4)$.  $$(k+1)! = (k+1)k! &gt; (k+1)2^k  \text{ (since }k!&gt;2^k)$$  That implies $(k+1)!&gt;2^k2$, since $(k+1)&gt;2$ because of $k$ is greater than or equal to 4.  Therefore $(k+1)!&gt;2^{k+1}$  Finally  we may conclude that $n!&gt;2^n$ for all integers $n\geq4$    Solved, thank you! Positive operator is bounded For a real Banach space $X$ let $A:X\rightarrow X^*$ be a positive operator in the sense that $(Ax)(x)\geq 0$ for all $x\in X$. Show that $A$ is bounded.   I don't know how to do that, maybe it's an application of the closed graph theorem? Add Word Limit to Posts Keep values selected after form submission Unable to SSH to my virtual machine via Windows Power Shell ISE Algorithm to return all combinations of k elements from n I want to write a function that takes an array of letters as an argument and a number of those letters to select.   Say you provide an array of 8 letters and want to select 3 letters from that. Then you should get:  8! / ((8 - 3)! * 3!) = 56   Arrays (or words) in return consisting of 3 letters each.