I am DMing for my family (a total of 7 other people) and have never played D&amp;D or any other Tabletop RPG before. We will be playing D&amp;D 5e. I know a lot of the basics from listening to The Adventure Zone podcast and reading various posts here. I also only have the player manual. I do not yet have the Dungeon Master's Manual or the Monster Manual. Are there any tips or ideas anyone can give me for our first campaign on Saturday?
I'm a 16-year-old male, and I have been interested in tabletop RPGs for a long time, especially D&amp;D. I sorta dove into D&amp;D really fast and hard by buying the Red box, Dungeon Master Kit, Monsters Vault, Players Handbook, Rules book, and even the Shadowfell Campaign not quite knowing what I'd get myself into. I've done tooonnsss of research and even watched live games on Youtube.   For the most part, I'm basically on my own, My friends are somewhat interested in it and want to play, but expect me to Teach/Dungeon Master the game for them, I'd really wish to do this, but need more insight and expertise. I've played the solo Campaign in the Red box and even thought about joining a local group from a popular gaming store in Phoenix, AZ called Imperial Outpost. But I'm not sure if they would take a newbie like myself. Any advice?
Derive Fourier transform of sinc function We know that the Fourier transform of the sinc function is the rectangular function (or top hat). However, I'm at a loss as to how to prove it. Most textbooks and online sources start with the rectangular function, show that   $$\int_{-\infty}^\infty \text{rect}(x)e^{i\omega x}dx=\int_{-1/2}^{1/2}e^{i\omega x}dx=\left.\frac{e^{i\omega x}}{i\omega}\right\vert_{-1/2}^{1/2}=\text{sinc}(\omega/2)$$  and then just invoke duality and claim that the Fourier transform of the sinc function is the rectangular function. Is there any way of deriving this directly? i.e., starting with the sinc function?   I've tried, but I'm not sure as to how to proceed. I know that the sinc is not Lebesgue integrable and only improper Riemann integrable. Some vague recollection of these being important in the Fourier transform hinders my thought process. Can someone clear things up for me? Find all natural numbers $n &gt; 1$ and $m &gt; 1$ such that $1!3!5!\cdots(2n - 1)! = m!$   I have been thinking about coming up with some inequalities which would narrow the possible range of pairs $(n, m)$, however the best I have been able to find so far are $m \ge 2n - 1$ and $m \lt n^2$, which is clearly not enough.  Update. The question has tag combinatorics as it is from a book about combinatorics, so there must be at least partly combinatorial solution. How can I obtain the limit of the series     $$\sum_{n=1}^{\infty} \frac{2^{[\sqrt{n}]}+2^{-[\sqrt{n}]}}{2^n}$$   Where $[\ \ ]$ is . What is the difference between the memory usage report in System Monitor and the one by free? Can Mjolnir "stick" to anything? Let $S$  be a square of unit area. Consider any quadrilateral whose $4$ vertices lie on each side square $S$. Let the length of the sides of this quadrilateral be $a,b,c,d$. Then prove that $$2 \leq a^2+b^2+c^2+d^2 \leq 4$$     This problem appeared in IIT JEE $1997$ (re-exam).I really do not know know how to approach this problem properly.I have managed to prove $a^2 + b^2 + c^2 +d^2  \leq 4$ but not the initial inequality. A few days ago I was recalling some facts about the $p$-adic numbers, for example the fact that the $p$-adic metric is an ultrametric implies very strongly that there is no order on $\mathbb{Q}_p$, as any number in the interior of an open ball is in fact its center.  I know that if you take the completion of the algebraic closure of the $p$-adic completion you get something which is isomorphic to $\mathbb{C}$ (this result was very surprising until I studied model theory, then it became obvious).  Furthermore, if the algebraic closure is of an extension of dimension $2$ then the field is orderable, or even real closed. Either way, it implies that the $p$-adic numbers don't have this property.  So I was thinking, is there a $p$-adic number whose square equals $2$? $3$? $2011$? For which prime numbers $p$? How far down the rabbit hole of algebraic numbers can you go inside the $p$-adic numbers? Are there general results connecting the choice (or rather properties) of $p$ to the "amount" of algebraic closure it gives? SELECT permission on view querying table from another database Resetting crontab file to default Is there a constructive way to exhibit a basis for $\mathbb{R}^\mathbb{N}$? As  says:     [...] the kinetic energy of a non-rotating object of mass $m$ traveling at a speed $v$ is $\frac{1}{2}mv^2$.   Why does this not increase linearly with speed? Why does it take so much more energy to go from $1\ \mathrm{m/s}$ to $2\ \mathrm{m/s}$ than it does to go from $0\ \mathrm{m/s}$ to $1\ \mathrm{m/s}$?  My intuition is wrong here, please help it out! How to draw circle in html page? I am doing an animation. The images I have attached show 2 different render layer view node results.    Image one shows a cube on layer 1, and 2 icosphere on layer 2. I have layer 2 set to fast blur. I have to use transparent film and therefore I use the Alpha over nodes in composition.  Image two shows a cube and 1 icosphere. Image 1 should not be showing 2 icospheres, as the left most one is behind the cube.    Image 2 is from the same render result, and is correcting showing only 1 icosphere. However, that icosphere should be blurred as it is on layer 2. It seems for whatever reason I can not have both the correct composition effect and also have every object's visibility render correctly. I can only either have everything work fine, with no composition, or composition with every emission object showing through every other object in the scene.   I am aware that masks would solve this problem, but because of the complicated setup of object materials and particles, masking produces these very strange and suddenly appearing artifacts. Material Indexing also produces odd artifacts and does not consistently effect the appropriate materials.   This is also a new problem, I did not have this issue in the last blender version. So, material indexing is out, masking is out, object indexing is out.       I want to use the compositor to blur certain layers, but every emission object in the scene shows through. It shows through as though it was on x-ray, regardless if it is behind another or inside another object. The first image is an example of this, the left icosphere is behind the cube, and should not be visible. That is the problem. Why is it visible when it shouldn't be? In Ubuntu 13.04 I used to use the ppa:upubuntu-com/tor64  PPA to install the 64-bit Tor Browser Bundle. However, in Ubuntu 13.10 it doesn't work anymore. How do I install the Tor Browser Bundle in Ubuntu? The Monty Hall problem What is the correct name for a "product function" on a monoid? I created this style that shifts the labels and put them on the bottom of the ybars.  However, I have a problem with the bars that have too short length (for example, see the 8.7 and 5.1 in the example below). I want to check if the values are less than some threshold, or the length of the bar is shorter than the label (number) then change its color.  How can I overlay another label on top of the one in white when the bar is shorter? or maybe change the color of the label to a gradient color? or just overlay the part that is outside of the bar?    \documentclass{standalone}  \usepackage{tikz} \usepackage{pgfplots} \usepackage{pgfplotstable} \usetikzlibrary{calc}  \begin{document} \makeatletter \begin{tikzpicture} \begin{axis}[ ybar, every node near coord/.style={   check for zero/.code={ % If meta=0, make the node a coordinate (which doesn't have text)     \pgfmathfloatifflags{\pgfplotspointmeta}{0}{         \pgfkeys{/tikz/coordinate}     }{}   },   check for zero, color=white, text opacity=1, font=\footnotesize, inner ysep=0.5pt, },% % Bottom nodes  calculate full offset/.code={     \pgfkeys{/pgf/fpu=true,/pgf/fpu/output format=fixed}     \pgfmathsetmacro\testmacro{((\pgfplotspointmeta-\pgfkeysvalueof{/pgfplots/ymin})*10^\pgfplots@data@scale@trafo@EXPONENT@y)*\pgfplots@y@veclength}     \pgfkeys{/pgf/fpu=false} },% nodes near coords bottom/.style={     every node near coord/.append style={         /pgfplots/calculate full offset,         yshift=-\testmacro,         rotate=90, anchor=west,%I need the rotate text here and not in the          %general style, as the ybar redefines the style     } },% ymin=0, nodes near coords={\pgfmathprintnumber[fixed zerofill,precision=1]{\pgfplotspointmeta}}, nodes near coords bottom, ] \pgfplotstableread{ % set real imag mag 1 8.7 2 5.1 3 91.8367  }\mydata \addplot table[x index=0,y index=1,header=false] {\mydata}; \end{axis} \end{tikzpicture} \makeatother \end{document}   * EDIT *  I need to create some special gradient on each label that is below the bar height. For example, I need to adapt something like  to each label that is not showing. However, I cannot find a way to put the macro in the style section of the codes. Or is there any equivalent solution? Which method performs better: .Any() vs .Count() > 0? in the System.Linq namespace, we can now extend our  to have the  and  extension methods.  I was told recently that if i want to check that a collection contains 1 or more items inside it, I should use the .Any() extension method instead of the .Count() &gt; 0 extension method because the .Count() extension method has to iterate through all the items.  Secondly, some collections have a property (not an extension method) that is Count or Length. Would it be better to use those, instead of .Any() or .Count() ?  yea / nae ? How do I set a monitor resolution that is not available in the Display Settings dialog? Is it possible to dual boot two OS's at the same time? Is it possible to dual boot two OS's at the same time?  For example: I'm currently dual booting Ubuntu 11.04 and Windows 7. Ubuntu is running on a single 500gb HDD, Windows 7 is running on 1TB RAID 1 mirror. To switch between the two OS's, a reboot is required. Is it possible to switch between the two OS's without a reboot?  NOTE: I'm fully aware of the capabilities of VirtualBox and VMWare Player - I use both. Please don't answer "Use Virtualization!"  EDIT: If this is not possible can you share why it's not possible? To me, it doesn't seem like it would be that difficult. Keep in mind that my vision is unencumbered by knowledge!