Let's say I have an infinitely long rod which I use to pass information to my friend on the other end. If the rod moves towards my friend, it means I sent a 1 and if the rod moves away from my friend, it means I sent a 0. Using this rod to transfer binary information to my friend, can I send information that travels faster than the speed of light? Or in other words, can I talk with my friend instantly irrespective of the distance between me and my friend?  Is this theoretically possible? My assumption is that push and pull of the rod is instant. Is it really instant or the push and pull also travels at the speed of light limiting the speed of my communication?
Is it possible for information (like 1 and 0s) to be transmitted faster than light?   For instance, take a rigid pole of several AU in length. Now say you have a person on each end, and one of them starts pulling and pushing on his/her end.   The person on the opposite end should receive the pushes and pulls instantaneously as no particle is making the full journey.  Would this actually work?
Matrix exponential for Jordan canonical form Let $X$ be a real $n \times n$ matrix, then there is a Jordan decomposition such that $X = D+N$ where $D$ is diagonalisable and $N$ is nilpotent.  Then, I was wondering whether the following is correct.  $e^{tX}(x) = \sum_{k=0}^{m} \frac{t^k N^k}{k!} \left(e^{t \lambda_1} \alpha_1 v_1+..+e^{t \lambda_n} \alpha_n v_n \right).$  Here $x = \sum_{i=1}^{n } \alpha_i v_i$ and $v_i$ are the eigenvectors of the diagonalisable matrix, $\lambda_i$ are the eigenvalues of $D$ and $m$ is the degree up to which $N^k$ is still non-zero.  Is this correct or am I doing something wrong? Cause I could not find a general equation for this matrix exponential, so I tried my best. (Thus, I am only asking for a verification or correction of this answer.)  If anything is unclear, please let me know. Cardinality of the set of all real functions of real variable How does one compute the cardinality of the set of functions $f:\mathbb{R} \to \mathbb{R}$ (not necessarily continuous)? We know that $221 = 17*13$. So we can check if the system has roots to both of those equations separately, which it does:  $x^{5} \equiv 2$ mod $13$ has the solution $6 + 13n$ and $x^{5} \equiv 2$ mod $17$ has the solution $15 + 17n$.   I got these numbers from wolfram, I have no idea how to solve this problem WITHOUT a calculator. And even after finding these numbers. How would one obtain a solution modulo $221$? I was thinking the Chinese Remainder Theorem but I am under the assumption that CRT only applies to problems with powers of $x$ which are $1$.  Thanks. I have a bathroom door that is quite loud when operated. And when someone goes to the bathroom at night, it innevatibly wakes someone up (and baby room is just across from the bathroom!) there are 3 components of noise, and I was wondering if there is an alternative to fixing them:   door bangs against the jamb (this is fixed with simple foam bumpers) latch mechanism makes a loud CLUNK when door is pushed closed (I was able to dampen it just a bit by putting foam inside strike plate, but it is still quite loud) lock button also makes a CLUNK when the knob is turned and it is released (this can be battled by replacing with a mechanism that needs to be manually opened [twist instead of push] but it is less convenient, and I would prefer something that automatically unlocks)   Any suggestions as to a knob that has a dampening of the latch and lock button? I have only found one youtube video of my ideal handle, but there is no information on how it works, where to buy it, or anything else: Record Animation In BGE I am having trouble getting the Record Animation setting to work in the BGE. Is it possible to record animated characters that are controlled using BGE's logic bricks with some script? In my script I change the positions of the characters bones. What is the reasoning behind restricting the number of migration paths? On the religion and culture stackexchange sites, many, many users will initially ask a question in a place that's kind of off-topic there, but are fully on-board elsewhere.    Christianity.SE, for example, gets a lot of general philopsophical questions that would be better on Philosophy, historical questions that might be better on History.SE, and hermenutical questions that are better on BH.SE.  It is, however, unreasonable in my mind for a user with a question to be required to have domain expertise on which sub-site each has. Indeed, even amongst users with a high amount of knowledge, there can be disagreement.  Likewise, , but as just about every one of those is in beta, the user would have to be very saavy (or would have to find that post) in order to nail it every time.  That said, recommending that a question be moved is much harder than it needs to be.  I have seen business rule logic that says, for example,    migration paths only exist to graduated sites (ergo politics, money, history, and quantitative finance wouldn't qualify) migration paths seem to be limited to 5.   Anything not on a migration path requires a moderator (and not just a high rep user) to get involved.  So, what is the reasoning behind this?  Note, I'm not asking for a solution (although the obvious solution would be to allow migration paths to any established site, beta or not, i.e. a VTC as off topic b/c it belongs on another site should allow you to pick any other site) - I just want to understand the business purpose in restricting migration paths to a small, predefined number for everyone other than a mod. Does new Star Wars continuity only affect post-Jedi EU? I have read  a couple of times.     In order to give maximum creative freedom to the filmmakers and also preserve an element of surprise and discovery for the audience, Star Wars Episodes VII-IX will not tell the same story told in the post-Return of the Jedi Expanded Universe.   Does this describe material set after Jedi, or published after?  Is it also saying that only things published as "Legends" are now non-continuity?  I have read that it is only films + Clone Wars + Rebels that are now in the main continuity but can't see this sourced by Lucas Film Story Group. Does anyone have a reference for this? I have a Java bean like this:  class Person {   int age;   String name; }   I'd like to iterate over a collection of these beans in a JSP, showing each person in a HTML table row, and in the last row of the table I'd like to show the total of all the ages.  The code to generate the table rows will look something like this:  &lt;c:forEach var="person" items="${personList}"&gt;   &lt;tr&gt;&lt;td&gt;${person.name}&lt;td&gt;&lt;td&gt;${person.age}&lt;/td&gt;&lt;/tr&gt; &lt;/c:forEach&gt;   However, I'm struggling to find a way to calculate the age total that will be shown in the final row without resorting to scriptlet code, any suggestions? Why was my "shopping list" question closed? My question was closed for being a "shopping list question", even though it was very much related to the topic of the site. Why? points out that certain key and block lengths were a requirement for the AES submissions:     The candidate algorithm shall be capable of supporting key-block combinations with sizes of 128-128, 192-128, and 256-128 bits.   My question is this; how can having a key length that is greater than the block size enhance security?  If we look at ECB mode for simplicity, when the key length is equal to the block length we have a permutation relationship between the set of all plaintexts and the set of all ciphertexts.  Once you increase the key length beyond that, you start to introduce collisions. In fact, you only introduce collisions from then on don't you? Collisions in the sense that a key will create the same ciphertext-plantext relationship as another key.  This would lead me to believe that only part of the key-space is unique, and only that part of key-space needs to be brute-forced?  I know that I am clearly incorrect, I just don't understand why. How does an internet server respond to a request from a private IP? Show that $u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$ is in $W^{1,n}(U)$, where $U=B(0,1)\subset\mathbb{R}^n$. The entire problem statement is:  Let $n&gt;1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by  $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$  My attempt at the proof is as follows:  To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$  Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$  Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n&lt;\infty.$$  Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$  And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.  Thank you for any help or feedback! Windows 8 disk usage is 100% Why aren't hash values for "nothing" (empty) defined as all zeros? Any decent hash algorithm will produce basically "random like" values for any given input including the empty input.  For example MD5 produces d41d8cd98f00b204e9800998ecf8427e on an empty input.  I wonder, however, why the MD5 algorithm wasn't defined to have a final step where the output is XORed with d41d8cd98f00b204e9800998ecf8427e. This should not affect the cryptographic strength of the hash algorithm, and it would have the convenient property that a hash of all zeros represented a zero input.  Would it simply just because adding that step would slow the hash down a bit? event.returnValue is deprecated. Please use the standard event.preventDefault() instead "How to [...]?" and "Where to [...]?" Questions that are not questions. Is this defensible? Adding a question mark to the end of a "How to" or "Where to" sentence appears to be quite common. Here are two examples from this very site:  How to punctuate a list of questions? ()  Where to put the periods when using a parenthetical sentence? ()  In my mind, the question mark doesn't belong at all, and it bothers me.  Am I being unreasonable? Is this practice defensible? When in Demon Form, are demons vulnerable to mortal frailties? When a Demon is in Demonic Form, does he gain any intrinsic protection against mundane concerns like breathing or falling unconscious? In particular, when in Demonic Form do/are demons:  need to breath? immune to poisons? fall unconscious? (ie do they stay conscious on full lethal?) suffer from wound penalties for having damage in last 3 boxes? need to consume food/water? need to sleep?  The last two are only really a concern to demons who have burnt their last cover. However this can happen in play, and is pretty significant for any burnt NPCs. Is seems odd that a swarm of glass beetles clinging together into the shape of a man might have any issues with being poisoned. But I can't find it clearly stated, whereas it is clearly stated in Werewolf (the Forsaken). What's a (big) word or phrase to describe a book that is so good that I cannot put down when I start to read it? What's a word for someone who is constantly asking for same thing? Proving $\frac{x}{y} +\frac{y}{z} + \frac{z}{x} \ge 3$ for positive $x,y,z$ Suppose $x,y,z$ are real positive numbers, prove that: $$\dfrac{x}{y} +\dfrac{y}{z} + \dfrac{z}{x} \ge 3$$ with equality when $x=y=z$.  Can someone help me find an easier solution? I started with assume only two are equal, without loss of generality, $x=y$. Then we get $$1 + \dfrac{x}{z} + \dfrac{z}{x}$$ $$1 + \dfrac{x^2 + z^2}{xz} = 1+ \dfrac{(x-z)^2 + 2xz}{xz} = 1 + 2 + \dfrac{(x-z)^2}{xz} \ge 3$$ To prove the general result I considered using a derivative with respect to $y$ to show that, starting with $x=y$ and increasing $y$ by a tiny amount, will show the equation's derivative as positive. I haven't gotten this to work yet. Any ideas?