In my iPhone app, when people are playing it, the screen goes black until you touch it again  after they play it for a couple minutes. Is there any way for me to prevent this from happening?
I'm working on an app that requires no user input, but I don't want the iPhone to enter the power saving mode.  Is it possible to disable power saving from an app?
How could the trained, physically stronger person lose this lightsaber fight? How could     Kylo Ren   lose the fight against      Rey?   I find the outcome totally illogical even though the Force may be strong in the woman. Furthermore, I would expect a man to be physically stronger than a woman. It is like having a very intelligent but uneducated person beating a math Phd in a math exam. Proof by induction of summation inequality: $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac1{2^n}\ge 1+\frac{n}2$ How do you deserialize json properties that are reserved words in apex? This is a follow-up to .  Suppose you want to denote vectors by upright bold letters; then \mathbf{u} works fine, except that if it's a Greek letter you have to use \boldsymbol{\omega} instead.  Question: Is it possible to define a generic vector command in LaTeX which automatically chooses between \mathbf and \boldsymbol, so that one can write just \vect{u} and \vect{\omega} without having to treat the Greek symbols specially by hand?  (Just to be clear: \newcommand{\vect}[1]{\boldsymbol{#1}} is no good in this situation, since \boldsymbol{u} gives bold italic instead of roman.) Can I speed up iCloud photo sync between iPhone and Macbook? I have an iPhone 5 running iOS 9.1 and a Macbook Pro running Yosemite.  Both devices are set up to sync photos to and from iCloud.  When I take a photo on the iPhone (for further processing on the Macbook), it takes an inordinate amount of time until the photo shows up in the Macbook's photos library - sometimes up to 10-20 minutes. Sometimes it seems to work faster if I quit the Photos app on both devices, and start it again.  I often end up sending myself photos through E-Mail because it's quicker. That seems ridiculous!  Is there a way to force a sync on the iPhone, or to speed up the process some other way? Windows 7 Aero Snap two windows horizontally As many people already know, there are several ways to trigger Windows 7 Aero Snap feature. A user can either press Win+{Left,Right,Up,Down} to snap the window to the left or right or maximize or restore. One can also drag the title bar to all the corner except bottom one.  However, is there any keyboard or mouse shortcut to place one window on top half the screen, and another window on bottom half the screen? Please do not tell me to use "Show Windows Stacked" because if I have more than two windows, it will place ALL the windows horizontally, which is not what I want. I want to be able to select the window I want and place it automatically either on top or bottom half the screen.  Thanks!  Edit: Thanks for Moab's suggestion, both on recommend WinSplit and updating my question.  I should make it more clear that I'm looking for a software that will do the functions WinSplit provided, but with on screen input, such as touch screen or pen. On the proof that the product of diagonal matrices is a diagonal matrix? Let $A [a_{i,j}]$  and $B =[b_{ij}]$   be two diagonal matrices in   $M_n(F)$. Let $i \ne j \in \left\{1, \ldots, n \right\}$. Using the   product rule, we obtain      $$(AB)_{ij} = \sum_{k=1}^{n} a_{ik}b_{kj}$$      We claim that $a_{ik}b_{kj} =0$ for all $k \in \left\{1, \ldots, n \right\}$ thus $(AB)_{ij}=0$ for all $i \ne j$ and $AB$ is diagonal. To    prove the claim, note that since $i \ne j$ , we have $i \ne k$ or $    k\ne j$. Thus either $a_{ik} = 0$ (since $A$ is diagonal) or $b_{kj} =  0$ (since $B$ is diagonal), thus in all cases $a_{ik}b_{kj}=0$ and the   claim is proved.   Where does the bit "we have $i \ne k$ or $k\ne j$" come from? I understand that since $i \ne j$, and $k$ and $j$ run over the same set of natural numbers, we also have $i \ne k$. But the "or $k \ne j$" baffles me. I understand that $k \ne j$ implies that we can have $i=k$. But that's from the other direction ($\Longleftarrow).$  Also, when it says $a_{ik}=0$ or $b_{kj}=0$ it's talking about when $i \ne k$ or $k \ne j$ right? Many thanks. I have the CKeditor plugin installed by itself (i.e. not using WYSIWYG module.) Then, it uses a CKEditor library located in sites/all/libraries/ckeditor. This works fine. I'd like to add more plugins to it; namely, the footnotes plugin. (For the moment, please refrain from "just download a new library from ckeditor.com and replace it"; there were other issues/considerations involved.)  Now, I see this everywhere when I try to look for a solution  config.extraPlugins = 'footnotes';   I tried adding that to admin/config/content/ckeditor/edit// -> Advanced Options -> Custom JavaScript configuration, but it white screens the editor.  I tried adding it to the config.js in sites/all/libraries/ckeditor like below, but it had no effect.  CKEDITOR.editorConfig = function( config ) {   config.extraPlugins = 'footnotes'; };   I can't really add it to a ckeditor.config.js file, dump that file in my theme, and choose "Yes" for "Load ckeditor.config.js from the theme path", since (for content creation/editing) it uses the admin theme which is different from the theme we're developing.  So, I'm at a loss. The only other thing I can think of is to use a hook in the module we have for the site, but I'm not sure if that's write/what hook would be called.  Is it just me, or does it seem like clear documentation on how to manage CKEditor in Drupal is severely lacking? Any help would be appreciated.  UPDATE: After some testing, it looks like the "Custom JavaScript configuration" is supposed to work, but it is possibly broken for CKEditor 4.4+ I read something (can't seem to find it now) that suggested the plugin/widget import was broken on 4.4+, so I changed to 4.3 and the extraPlugins worked. Now, I wasn't able to get footnotes working (which is what kicked this whole thing off), but some other plugins that previously failed during the same steps now work on 4.3. So, there's that. I'm wondering now if it's something with footnotes itself. Probability Number A greater Number B Given $a \in \{1,2,...,250\}$ and $b\in \{0,1,...,1000\}$  What is the probability that $a &gt; b$?  I know there already is  question with a very similar question, however it's over an interval of $\mathbb{R}$ so I hope there is another way opposed to the Integral used in the formal solution.  How does one go about calculating said probability? When is the graph of the divisor relation on $[n]$ planar? A recently-deleted question asked to show that if $G(n)$ is the graph on $[n]=\{1,\dots,n\}$ where $x$ and $y$ are adjacent if $x|y$ or $y|x$, then $G(n)$ is planar.  This is false if $n\geq150$, for the subgraph induced by $\{2,3,5,90,120,150\}$ is isomorphic to $K_{3,3}$.  My question is, what is the largest value of $n$ such that $G(n)$ is planar?  Of course, various software packages would make short work of this, but I'd be interested in a verbal argument, if someone can give one. "I think him to be about 50" or "I think he is about 50"? Is it possible to remove "Show Desktop" from the Alt+Tab application switcher? How do I remove a directory and all its contents? In bash all I know is that   rmdir directoryname   will remove the directory but only if it's empty. Is there a way to force remove subdirectories? Detecting absolute motion inside a box This is not a contradiction and I know it is impossible but still consider a thought experiment by me and point out if something is wrong. See the following picture and then the explanation follows.    Rest frames are easy to understand. They are just for clarification. Lets move on to the moving frames. The velocity of the box is $0.1\ c$. Now a photon is emitted (I am not taking a light ray to avoid complication in discussion). After emission the box also moves a certain distance ahead. So the photon takes more than one second to reach the wall. But even the source moves that much distance ahead. When the light is reflected back, the source and the wall again move forward as the box is moving. But after reflection the wall has no role to play. We are concerned about the source. The source moves a certain distance ahead and therefore light takes less than one second to reach back. But the difference of the first case and the second case is not the same. To explain, I will give some equations.  Moving frame 1:  Velocity of the box: $0.1\ c = 30,000km/s$  Time taken by light to reach the wall $=\frac{330,000\ \text{km}}{300,000\ \text{kms}^{-1}}= 1.1s$  Distance moved ahead by the box from original position: $30,000\ \text{km}$  Moving frame 2:  distance moved ahead by the box from original position  $&gt;30,000\ \text{km}$ as the box also moves ahead after reflection.  let $d$ be the distance moved ahead by the box after reflection.  Time taken by light to reach the source: $\frac{270,000\ \text{km}\ -\ d}{300,000\ \text{kms}^{-1}} &lt;0.9s$  Therefore we see that this could possibly determine the absolute motion.  So in total we see that the light will take lesser time to reach the source. Please correct me if I am wrong and give your opinion why is it so.  EDIT 2 for clarification  The box is the compartment in space in which you are moving and I am calculating the time the light takes to reach back the source. I am not even adding velocities. Please correct me and tell me where did I add velocities. For clarification just let me give a simple example. Suppose person A is standing still and B is running towards a light source. Obviously he will see the light before person A even though the speed would be the same. So, what I am doing over here is not increasing the speed of light, but instead decreasing the distance to be covered by it. Disable HTTPS for a domain that shares IP with another domain that is under HTTPS I am running into a weird problem trying to disable HTTPS for one domain, but the other domain has HTTPS. They both are different websites running on a dedicated server, sharing the same IP address for the domains.  However, I have a domain that does not have a SSL certificate, but the other does. I want to disable HTTPS for the non-certified domain while leaving the other domain with HTTPS on.  I have two different separate vhosts in /etc/nginx/sites-enabled/,   domain1.org (no HTTPS)  server {      listen                      80;     server_name                 www.domain1.org domain1.org;     root                        /var/www/domain1.org;     index                       index.html index.php index.htm;      error_log                   /var/www/logs/domain1.errors.log;      [ ... ]   domain2.com (with HTTPS)  server {      listen                      443 ssl;     server_name                 domain2.com www.domain2.com;      # INCLUDE SSL CERTIFICATE     include                     sites-enabled/domain2_ssl_include;       root                        /home/domain2/www/www;     index                       index.html index.php index.htm;      [ ... ]   I've included a listener to the port 443 for domain1.org, but once I do that, I get an aborted error from Firefox saying (when accessing ):     This Connection is Untrusted      You have asked Firefox to connect securely to domain1.org, but we can't confirm that your connection is secure.   I opened the technical data, it appears that the domain1.org is using the domain2.com SSL certificate:     domain1.org uses an invalid security certificate.      The certificate is only valid for the following names:    *.domain2.com , domain2.com        (Error code: ssl_error_bad_cert_domain)   So I added this to domain1.org vhost configuration:  server {      server_name                 domain1.org www.domain1.org;      listen                      443 ssl;      rewrite ^ http://domain1.org permanent; }   Which still gets me prompted with the Firefox error.  Now... when I access to , I get an 400 error from nginx saying this:     400 Bad Request      The plain HTTP request was sent to HTTPS port   I am not entirely sure what to do, disabling the https with ssl off; has no effect. How can I actually disable HTTPS for domain1.org but not for domain2.com?  I am using Ubuntu 12.04 running nginx 1.2.6. In my passport, my name is Firstname middlename1 middlename2 lastname, and in the ticket is LASTNAME/FIRSTNAME MR.  Will be a problem? I'll travel from Brazil to Bali, Indonesia and according to the agency I bought the ticket from, there's no problem and it's common.  The airline is Qatar Airlines and I have looked , which says that the passenger name can't be changed after the booking has finished.  But I don't know if it is a problem. After reviewing ...  I'd like to make an argument for answers that have attained more votes than an accepted answer always be ranked above when answers are being sorted by votes (the default)...     Someone visiting this (older) question expecting a relevant answer would be assuming the big green check mark means that it is the correct answer today when in fact it's the correct answer when the question was originally asked.  Allowing votes to override the asker's time bound response, it allows the best answers to remain at the top and evolve with our ever changing field without requiring the asker to maintain their questions.  This would be a step in the direction of the site's ethos in my opinion, trusting the community to determine the best answer to place at the top for prospective visitors. How to derive the ridge regression solution? I am having some issues with the derivation of the solution for ridge regression.  I know the regression solution without the regularization term:  $$\beta = (X^TX)^{-1}X^Ty.$$  But after adding the L2 term $\lambda\|\beta\|_2^2$ to the cost function, how come the solution becomes   $$\beta = (X^TX + \lambda I)^{-1}X^Ty.$$ How can I read a file even when getting an "in use by another process" exception? Prove that , any primitive root $r$ of $p^n$ is also a primitive root of $p$ For an odd prime $p$, prove that any primitive root $r$ of $p^n$ is also a primitive root of  $p$   So I have assumed $r$ have order $k$ modulo $p$ , So $k|p-1$.Then if I am able to show that $p-1|k$ then I am done .But I haven't been able to show that.Can anybody help me this method?Any other type of prove is also welcomed.