Is there a way to prove this without using Bertrand's postulate?
How to prove that there exists a prime number between $n$ and $ n!$, for all $ n&gt; 2$?    ( gives a much better bound, but this question is about obtaining a self-contained proof.)
I have keypair (u,r), and two pieces of data, X and Y.  I reveal X publicly but keep Y secret. Y has high entropy.  I then concatenate X and Y, into Z.  I encrypt Z, into "encrypt(Z,u) = Q". r is needed to decrypt Q back into Z.  Later, (when it doesn't matter), r will be revealed and everyone will decrypt Q into Z. At that point, everyone will know that, all along, Q "contained" X. Can I prove that Q contains X before r is revealed (including interactive zero-knowledge proofs)? In the addon ArchiMesh there is a Japaneses curtain feature, but, it looks straight and does not have the fold that I want in it. Is there an addon that can do this for me? If not, how can I make a curtain (like in the images below)? Why is there a white space on the left side?  \documentclass{standalone} \usepackage{asymptote} \begin{document} \noindent \begin{asy} size(10cm,10cm); draw((0,0)--(100,100)); draw((0,100)--(100,0)); dot((50,50)); \end{asy} \end{document} Google Voice Recognition on iOS What's the best counter strategy against battlecruisers? How should I prepare my army when attacking someone who has 12+ battlecruisers? (Protoss/Zerg/Terran answers are more than appreciated) How bad for TeX is omitting braces {}, even if the result is the same? Sometimes writing braces and omitting them produces the same result:  \tilde{\psi}  \mathrm{e}^{\mathrm{i} \pi}   vs  \tilde \psi   \mathrm e^{\mathrm i \pi}   Or (taken from Werner's comment below)   \frac{1}{2} vs \frac12   If that's the case, how bad for the quality of my code is omitting them? Of course, if one compiles a pair of formulas, there is no difference. But, since I ignore how TeX works, my point is: if one writes long codes with these habits,   doesn't TeX go somehow crazy because I'm not writing as it expects?   For instance, does it have any effect on how fast are long documents compiled? Any reason to keep them? The system of equations Let $a,b,c\in \mathbb{R}$ that $a^{2}+b^{2}+c^{2}=1$. We want to find $x,y,z,w$ in the following equations:  $$\begin{align} x^{2}+y^{2}+z^{2}+w^{2}&amp;=1 \tag{1}\\ x^{2}+y^{2}-z^{2}-w^{2}&amp;=a \tag{2}\\ xw+yz&amp;=2b \tag{3}\\ yw-xz&amp;=2c.\tag{4} \end{align}$$ In general what is the approach to switch the desktop environment or window manager on Ubuntu? The main difference between stock Ubuntu, Kubuntu and Lubuntu is supposedly the use of Unity vs. KDE vs. LXDE.  However, in order to try various desktop environments and window managers I don't want to have to reinstall the respective flavor from scratch just to get a particular desktop environments or window manager. There probably is an easier method than that.  The question: How can I easily test different desktop environments and respective compatible window managers (say KDE, LXDE, Unity, GNOME, XFCE and StumpWM)? Is past simple ever justified with 'by'? Can anyone shed some light on the following aspect?  By 5 PM I gave him the book.  By 5 PM I had given him the book.  In both the sentences the book was given before 5 PM and no later than 5 PM. But I can't understand then what the difference between them is. a Also is it right to use simple past with 'by' at all? If yes, then why and if not, then why is it incorrect to use simple past with 'by'? Maximum likelihood when usual procedure doesn't work I am trying to get the maximum likelihood estimate for the parameter $p$. The distribution is the following:   $$ f(x\mid p) = \begin{cases} \frac{p}{x^2} &amp;\text{for} \ p\leq x &lt; \infty  \ \\ 0 &amp;\text{if not} \end{cases} $$  The sample has size $n$.  The problem is, when I try to estimate it by the procedure I know, I would have to estimate the likelihood function and obtain the derivative of the log-likelihood. We'd have:   $$ L(p; x) = \frac{p^n}{\prod_{i=1}^{n} x^2_i}$$  $$ \ln L(p;x) = n \ln(p) - \sum_{i=1}^{n} \ln(x_i^2) $$   For the derivative:  $$ l'(p;x) = \frac{n}{p} = 0$$   And I am stuck because it has no solution for $p$. How do I evaluate this?  Thanks!   EDIT:  So in this case I can use the indicator variable to write:   $$ L(p;x) = \frac{p^n}{\prod_{i=1}^{n}x_i^2} I_{(x_i \geq p)}$$   for $i = 1,2, \ldots n$ in the indicator variable. So the "closest" non-null value of $x \in X$ to $p$ is $min(X_1, \ldots X_n)$. Is that the point? Why is my query suddenly slower than it was yesterday? [Salutations]  (check one)  [ ] Well trained professional, [ ] Casual reader, [ ] Hapless wanderer,   I have a (check all that apply)  [ ] query [ ] stored procedure [ ] database thing maybe     that was running fine (if applicable)   [ ] yesterday [ ] in recent memory [ ] at some point    but is suddenly slower now.  I've already checked to make sure it's not being blocked, and that it's not the victim of some long running maintenance task, report, or other out of band process.  What is the problem, what should I do, and what information can I provide to get some help?  [*Insert appropriate closing remarks*] Is there a term for a word that serves as its own antonym? For example:   &quot;transparent&quot; can mean both &quot;obvious&quot; and &quot;hidden&quot;  &quot;aught&quot; can mean both &quot;all&quot; and &quot;nothing&quot;    Is there a term for a word which has multiple definitions that are antonyms? Can a spellcaster use spell slots above their normal limit if their ability scores allow bonus spells of that level? This question is inspired by discussion , about how Heighten Spell works with Reserve Feats.  The question that I am asking here deals with whether or not players can cast spells using spell slots for spell levels that they don't have access to yet.    I feel like an example would explain this concept best.  Take Xylitol, the first-level wizard with an intelligence of 18.  He has 5 0th-level spells per day and 3 1st level spells per day from his class.  He also has 1 bonus 1st level spell per day from his high intelligence.  The  table says that he also has 1 bonus spell of each 2nd, 3rd, and 4th levels.  The question is this:  Can Xylitol cast 1st level spells using the bonus slots for his 2nd, 3rd, and 4th level spells?  He clearly can't cast spells that are inherently higher level than 1st, since he can't learn them, but can he cast 1st level spells using bonus 2nd, 3rd, and 4th level slots?  This has come up on these boards before, .  What I'm looking for here is a clear rules citation that clarifies this one way or the other.  The wording in the section on ability scores makes it quite clear that characters cannot cast spells that are actually higher level than their class allows them to cast, but it is not clear that these higher level slots are barred until the appropriate spell level is unlocked.  If a character has access to a spell level that they can't use because their ability score isn't high enough, they can freely use those slots for lower level spells, so why can't they use their bonus spell slots for lower level spells?  As KRyan's answer says, the Basics section covers this.  Based on questions and comments both here and elsewhere, that section seems to be worded ambiguously enough that it's caused confusion.  I'd like to find another, more clear wording to this rule if at all possible.  Answers from either D&amp;D 3.5 or Pathfinder are acceptable, but please note which one your answer applies to.  N.B. I'd like to note that I personally don't feel that players should have access to higher level slots like this.  There are a large number of locations where it's heavily implied that this is not the case, but I feel that finding a clear rules citation here would be helpful.  I can't find one that is clear for either interpretation here. I wonder whether it is possible to draw the three following quadric surfaces (cone, hyperboloids)  x^2+y^2-z^2=0  x^2+y^2-z^2=1  x^2+y^2-z^2=-1  (and the axis)   using pgfplots 3d or TikZ as it is done in the following image Suppose $f$ and $g$ are entire functions, and $|f(z)| \leq |g(z)|$ for all $z \in \mathbb{C}$, Prove that $f(z)=cg(z)$.   My try :  I consider $h(z)=\frac{f(z)}{g(z)}$. If I prove that $h(z)$ is entire, using the fact that $|h(z)|\leq 1$, the result follows immediately from Liouville's theorem.  To prove that $h(z)$ is entire, I have to prove that $h(z)$ has removable singularities at possible zeros of $g(z)$.  Suppose $g(z_0)=0 \Rightarrow |g(z_0)|=0 \Rightarrow |f(z_0)|=0$, BUT what if $z_0$ be a root of of order $k$ of $g(z)$ (i.e. $f^{(n)}(z_0) = 0 $ for all $ 0 \leq n \leq k-1$) and be a zero of order $m$ of $f(z)$ and $k&gt;m$ ?  And if that's not the case, how to rigorously show that $z_0$ is a removable singularity ?  Thank you all in advance. Denote by $\mathbb{N}$ the set of natural numbers and by $2^{\mathbb N}$ the set of all subsets of $\mathbb N$. Let $E$ be some subset of $2^{\mathbb N}$ such that for  every pair of elements in $E$, one is a subset of the other. Clearly, $E$ could be countable by considering $E=\big\{\{1,\cdots, n\}, n\in\mathbb N\big\}$. My question is the following: could we find an example that $E$ is uncountable? Many thanks for the answer! Chrome and Firefox cannot add specific URL to home screen Using same modulus for RSA I know that there exist some attack when using same modulus.      But with a little modification, $m$ is the plain-text $N$ is the RSA modulus $r_1, r_2$ is the random padding $e, s$ is the public exponent $C_e, C_s$ is the cipher-text  encrypt as follow $$С_e = (m + r_1)^e   \bmod N$$ $$С_s = (m + r_2)^s   \bmod N$$  If attacker only knows $C_e, C_s, r_1, r_2, e, s, N$. Is it possible to know $m$?  It seems the  does not work? Why $\sum_{k=0}^m x^k $ is equal to $\dfrac{x^{m+1}-1}{x-1} $? Who can help me proving this (but you can no longer use $\sum$ in the proof):  $$\sum_{k=0}^m x^k=\dfrac{x^{m+1}-1}{x-1}$$ I keep getting this warning whenever I try to run sudo apt-get update.  W: Duplicate sources.list entry http://archive.ubuntu.com/ubuntu/ precise-updates/main i386 Packages (/var/lib/apt/lists/archive.ubuntu.com_ubuntu_dists_precise-updates_main_binary-i386_Packages) W: You may want to run apt-get update to correct these problems   Below is the output from /etc/apt/sources.list file:  deb http://archive.ubuntu.com/ubuntu precise main restricted deb-src http://archive.ubuntu.com/ubuntu precise main restricted  deb http://archive.ubuntu.com/ubuntu precise-updates main restricted deb-src http://archive.ubuntu.com/ubuntu precise-updates main restricted  deb http://archive.ubuntu.com/ubuntu precise universe deb-src http://archive.ubuntu.com/ubuntu precise universe deb http://archive.ubuntu.com/ubuntu precise-updates universe deb-src http://archive.ubuntu.com/ubuntu precise-updates universe  deb http://archive.ubuntu.com/ubuntu precise multiverse deb-src http://archive.ubuntu.com/ubuntu precise multiverse deb http://archive.ubuntu.com/ubuntu precise-updates multiverse deb-src http://archive.ubuntu.com/ubuntu precise-updates multiverse  deb http://archive.ubuntu.com/ubuntu precise-security main restricted deb-src http://archive.ubuntu.com/ubuntu precise-security main restricted deb http://archive.ubuntu.com/ubuntu precise-security universe deb-src http://archive.ubuntu.com/ubuntu precise-security universe deb http://archive.ubuntu.com/ubuntu precise-security multiverse deb-src http://archive.ubuntu.com/ubuntu precise-security multiverse   How do I fix it?