Booleans are extremely slow
why Booleans slow my computer more than expected
MCMC in a frequentist setting I have been trying to get a sense of the different problems in frequentist settings where MCMC is used. I am familiar that MCMC (or Monte Carlo) is used in fitting GLMMs and in maybe Monte Carlo EM algorithms. Are there more frequentist problems where MCMC is used? In the season 8 episode finale of Game of Thrones the highest important lords met with Grey Worm to free      Tyrion and Jon. While it was pretty clear why they wanted to free Jon it was unclear to me why they also asked for Tyrion.   To make things even more strange, they asked about his opinion of a potential solution which becomes the actual final one in the end      (choose king and nominate someone).   I thought they (besides Sansa and Bran) don't like him and wanted to judge him? Or am I missing something? Hypernym for "audiobook", "podcast", "speech", etc.? I am looking for a short and concise term (preferably one word) that can be used as a hypernym for essentially everything that contains (audio) speech (e.g. audiobooks, podcasts, recorded news bulletins, recorded performances of comedians,..).  Does such a term exist? I've thought about 'speech', but that seems to be more adequate for e.g. political speeches, and not so much for audio books and the like. But maybe I'm wrong. We want to encrypt / encode URL parameters on our pages. I tried to encrypt a parameter using  with AES algorithm as explained in the   %%[SET @encData=EncryptSymmetric("test", "AES", @null, "1111", @null, "0000000000000000", @null, "00000000000000000000000000000000") ]%% AES: %%=v(@encData)=%%   The result of this function is      AT1r0irnzM+Ax1j8zN+zAw==   Then I tried to decrypt the value using an , but I always received an error message.  I suppose there is a difference between AES algorithm in Marketing Cloud and AES used in the online tool. Indeed if I encrypt the same word with an online tool (128-bit), I received this result      l4Yo3sFofFD1EMmtoh397w==   Do you have any deep information regarding this function? What could be the difference? If I recall it correctly, this book has two main characters, both male, one of them big and well muscled. During the novel, they found evidence that the current human race descends from ancient Martian astronauts who traveled to Earth in prehistory, causing the extinction of prehistoric Earth humans, (which explains the Missing Link in human evolution).  Later in the novel the big guy travels back in time (no idea of how or why) to Prehistory.  And near the end of the book, the other main character, during an archaeological field research, finds remains of a human being who lived thousand of years ago, a human with large bones, and who holds an item that could not be made in that era (a gold plaque? A metal rod?)  That's all I can remember. I know it is not from the ABC, Asimov, Bradbury or Clarke.  Thanks in advance. We have a directory and want to protect that from remove and rename, but we need to be able to rename, remove and create contents. What we can do? SSH connection problem with "Host key verification failed..." error I can connect to another Ubuntu machine in my LAN via SSH. On both of then PC's I installed   but from another Ubuntu computer I can not connect to my PC via SSH and I got this error:     Host key verification failed... Why is the edit button disabled? $P$ is a prime ideal of $R$ iff $R/P$ is an integral domain. : $P≠R$ Here is the theorem and proof in my textbook.     Thm.  Let $P$ be an ideal of a commutative ring $R$ with identity $1$.  Then $P$ is a prime ideal of $R$ if and only if $R/P$ is an integral domain.   Proof.   Suppose that $P$ is a prime ideal of a commutative ring $R$ with $1$.  Then $P$≠$R$ implies $1+P≠0+P$. Hence $R/P$ is a commutative ring $R$ with identity.  Assume that $(a+P)(b+P)=0+P$.  Then $ab+P=0+P$ and $ab∈P$.  By the definition of a prime ideal P we get $a∈P$ or $b∈P$.  That is, $a+P=0+P$ or $b+P=0+P$.  Thus $R/P$ is an integral domain.   Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$ and $R/P$ is a commutative ring $R$ which has no zero divisors.  Hence $P≠R$.  Assume $ab∈P$.  Then $ab+P=0+P$ and $(a+P)(b+P)=0+P$.  Since $R/P$ is an integral domain, we get $a+P=0+P$ or $b+P=0+P$.  So $a∈P$ or $b∈P$.  Thus $P$ is a prime ideal.    I am having problems understanding proof with "Conversely, if $R/P$ is an integral domain, then $1+P≠0+P$.".  I know that $R/P$ has an identity. (cause it is ID)  But, why doesn't "$P$" have an identity?  Isn't there a case $P=R$?  Please help me understanding. I think $P≠R$ should be with condition of Thm..  Thank you in advance. I found this odd relationship, $x^2 = \sum_\limits{k = 0}^{x-1} (2k + 1)$. I saw a commercial on televison for preparing spaghetti without boiling the noodles, just place them in water for about on hour.  Can this be done? Should questions having lots of downvotes be downvoted further? Is down-voting an already heavily down-voted/closed question kind of like kicking a dead horse? Is this bad practice? Where does gravity get its energy from? I would like to know where gravity gets its energy to attract physical bodies?  I know that the law of conservation states that total energy of an isolated system cannot change. So gravity has to be getting its energy from somewhere, or else things like hydropower plants wouldn't be able to turn the power of the falling water into a spinning rotor.  Just to be clear, Lets create an example:  Lets say we have two objects with equal mass close to each other. So gravity does its job and it pulls each other closer, this gets turned into kinetic energy. This is where I'm lost. According to the law of conservation energy can't be created or destroyed and the kinetic energy comes from the gravitational pull so where does the gravitational pull gets its energy.  If that energy isn't being recycled from some where else then that means you have just created energy, therefore breaking the law of conservation. I would like to see a full how-to guide on how to install Ubuntu. I have a very strange behavior on an Android device (Nexus 7) when trying to access local network applications. Instead of getting the actual IP of the machine on LAN, the Android device gets the public IP, which means that Chrome, Firefox or any other browser simply shows the web page of the router.  I have an internal DNS server which handles the local networks. Doing ping from a PC works correctly:  $ ping s.pelicandd.com  PING pelicandd.com (192.168.1.15) 56(84) bytes of data. 64 bytes from front.pelicandd.com (192.168.1.15): icmp_seq=1 ttl=64 time=0.524 ms 64 bytes from front.pelicandd.com (192.168.1.15): icmp_seq=2 ttl=64 time=0.578 ms ^C   On an HTC One device (accessed with adb shell), it works well too:  shell@m7:/ $ ping s.pelicandd.com  PING pelicandd.com (192.168.1.15) 56(84) bytes of data. 64 bytes from front.pelicandd.com (192.168.1.15): icmp_seq=1 ttl=64 time=2.56 ms 64 bytes from front.pelicandd.com (192.168.1.15): icmp_seq=2 ttl=64 time=27.8 ms ^C   However, this is what I get when doing the same ping from Nexus 7:  shell@flo:/ $ ping s.pelicandd.com  PING pelicandd.com (90.78.26.42) 56(84) bytes of data. 64 bytes from LFbn-1-2441-42.w90-78.abo.wanadoo.fr (90.78.26.42): icmp_seq=1 ttl=64 time=2.56 ms 64 bytes from LFbn-1-2441-42.w90-78.abo.wanadoo.fr (90.78.26.42): icmp_seq=2 ttl=64 time=8.63 ms ^C   Instead of resolving to the internal IP, it resolves to the public one.  The network configuration is—except the IP address of the device—exactly the same on both devices: IP settings are set to Static, and DNS servers are the internal ones. Both Android devices are connected through Wi-Fi (unlike the PC). A major difference, however, is that Nexus 7 is using Android 6.0.1, while HTC One uses Android 5.0.2.  There is no nm-tool, dig or nslookup on Nexus 7. The device is not rooted.  The problem existed since I bought the device a few weeks ago, so it could hardly be the issue with DNS cache.  What can I do to further inspect this issue? $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$  According to my textbook the limit equals $2$.  What I have tried:  Using the squeeze theorem:  $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 2} $$ $$ 0 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le 0 $$  I have also tried to use the squeeze theorem with two other equations and obtained different values:  $$ \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 - 1}{\sqrt{x^2 +y^2 + 1}} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2 + 1}{\sqrt{x^2 +y^2 + 1} - 2} $$ $$ -1 \le \lim\limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} \le -1 $$ Minecraft crashes on launch with EXCEPTION_ACCESS_VIOLATION, Problematic frame: ig4dev32.dll or ig4dev64.dll or ig4icd32.dll or ig4icd64.dll Can someone explain a stock's "bid" vs. "ask" price relative to "current" price? In my online brokerage account, I want to buy a particular stock and I see the following:  Bid:  13.20 x200           Ask:  13.27 x1,000   the current stock price is 13.22.   Can someone explain what the bid and ask prices mean relative to the current price?  If I buy 1000 shares, why would I pay more? $_SERVER["REMOTE_ADDR"] gives server IP rather than visitor IP Can the real and imaginary parts of $\dfrac{\sin z}z$ be simplified? I have calculated the real and imaginary parts of $\dfrac{\sin z}z.$ I've obtained  $$\begin{eqnarray} \frac{\sin z}z&amp;=&amp;\frac{\sin(x+iy)}{(x+iy)}\\ &amp;=&amp; \frac{\sin(x)\cosh(y)+i\cos(x)\sinh(y)}{x+iy}\\ &amp;=&amp; \frac{x\sin(x)\cosh(y)+y\cos(x)\sinh(y)}{x^2+y^2}+i\frac{x\cos(x)\sinh(y)-y\sin(x)\cosh(y)}{x^2+y^2}. \end{eqnarray}$$  Before calculating it, I hoped I could use this in a solution to a certain problem, but I would need to differentiate the real and imaginary parts of $\dfrac{\sin z}z$ with regard to $x$ and $y$. But what I'm getting is very complicated. I understand that it could be that that's just the way it is. But maybe there are some simplifications to be done either in the functions or in their derivatives? I've never really used hyperbolic functions before so I'm not sure I'm not missing something.  Added. The problem I'm working on is the following.     Prove that the solutions of the equation $$\tan z=z$$ are all real.   I will sketch the solution I know here.  It can be calculated that  $$\tan(x+iy)=\frac{\sin(2x)}{\cos(2x)+\cosh(2y)}+i\frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}.$$  Plugging this to the equation, we get  $$\begin{eqnarray} x&amp;=&amp;\frac{\sin(2x)}{\cos(2x)+\cosh(2y)}\\ y&amp;=&amp;\frac{\sinh(2y)}{\cos(2x)+\cosh(2y)}. \end{eqnarray}$$  From this we can get   $$\sin(2x)\cdot2y=\sinh(2y)\cdot2x.$$  By plotting the functions $\sin,$ $\mathrm{id}$ and $\sinh$ (and proving the right inequalities we can see from the plots), we can see that this is only possible when $xy=0$. But when we plug $x=0$ to the second equation, we can show that the only solution is $y=0$, which ends the proof.  What strikes me in this solution is the fact that $\dfrac{\sin z}z$ is used. (Not explicitly in my statement.) This function has another connection to the problem:     The non-zero solutions of the equation $\tan z=z$ are exactly the zeros of the derivative of $\dfrac{\sin z}z.$   This is easy to prove. I wondered if there shouldn't be a solution to the problem which uses this fact, given that $\dfrac{\sin z}z$ appears in the solution I know.