Calculate huge number mod $p$
Mod of numbers with large exponents [modular order reduction]
Not every open set in $\mathbb{R}^n$, for $n\geq2$, is a disjoint union of countable open balls. I know that every open set in $\mathbb{R}$ can be decomposed as the union of countably many disjoint open intervals and I'm trying to show that this property no longer holds in higher dimensions. For this I'm trying to show that there is no countable, pairwise disjoint open balls whose union is the open square $(0, 1)^2 ⊂ \mathbb{R}^2$. I know that it is because of the corners, but I don't know how to write this in mathematical terms. Thanks for any help. In my cover letter I want to write that I am a graduate student in statistics. But I am not sure about the "in". So what is correct?     I am a graduate student in statistics.      I am a graduate student of statistics. Can you contract the main verb '[I] have' in a sentence? Why does everything spin? The origin of spin is some what a puzzle to me, everything spin from galaxies to planets to weather to electrons.  Where has all the angular momentum come from? Why is it so natural?  I was also thinking do photons spin? we always think of the wave as a standard 2d sin wave but could this rotate in 3d? What implications would this have?   And what about spacetime how does all the spinning effect?  This has always been avoided in all lectures and classes I ever went to. I've seen sign in methods using  as an authentication provider, for example : (AWS Cloud9)        Is it possible to authenticate into Stack Exchange using this method? At the moment there's other methods such as OpenID, Google, Yahoo and Facebook, but it would be interesting to also have the GitHub authentication. Help needed with definite integral for $\lim_{n\rightarrow \infty}\sum_{k=1}^{n} \frac{1}{n+k}$ Centre of Lie Algebra $sl_2(\mathbb{F})$ For $L=sl_2(\mathbb{F})$ i.e. matrices with trace zero,  what is the centre i.e. $Z(L)$= {$x\in L : [x,y]=[y,x] \ \forall\ y \in L$}.  I will have to find matrices $A \in L$ such that $AB=BA$ for all $B \in L$. But how to appoach this considering trace is zero here. On the paragraph explaining arithmetic expansion,  uncovers 2 different ways of evaluating an expression, the first one uses $(( EXPRESSION )) and the second one uses $[ EXPRESSION ]. The two ways seem pretty similar as the only difference I have found is:     $[ EXPRESSION ] will only calculate the result of EXPRESSION, and do no tests:   Yet, I am intrigued because the same document recommends using $[ EXPRESSION ] rather than $(( EXPRESSION )).     Wherever possible, Bash users should try to use the syntax with square brackets:   Why would you want that if less tests are being done? Time Dilation inside a hollow shell Assuming I have a hollow shell with total mass $M$ and radius $r$. On the surface, the gravitational time dilation would be  $$\tau=t \cdot \sqrt{1-\frac{v_{esc}^2}{c^2}}$$  where  $$v_{esc} = \sqrt{\frac{2 \cdot G \cdot M}{r}}$$  but inside the shell there would be no gravitational field (Newton's shell theorem and Birkhoff's theorem).  But still, the escape velocity required to escape to infinity would be the same as on the surface, since inside the shell you could move without any accelerating or decelerating forces acting on you until you reach the surface, from where you would get pulled backwards.  So is the time dilation inside the hollow shell relative to a field free observer at infinity    zero (I assume it's not) or  the same as on the surface (my best guess), or something completely different?   I found some not really duplicate but related threads on the interior of black holes which did not really focus on the math, but I am more behind the calculations in terms of $M$ and $r$. Proving $f' = - \frac{\partial{F}/\partial{x}}{\partial{F}/\partial{y}}$ I am trying to prove that, if $f:\mathbb{R}\rightarrow \mathbb{R}$ and $F:\mathbb{R}^2\rightarrow\mathbb{R}$ are differentiable, $F$ satisfying $F(x,f(x))=0$ and $\partial{F}/\partial{y}\ne 0$ then $$f'(x)=-\frac{\partial{F}/\partial{x}}{\partial{F}/\partial{y}}$$  This appears in a section on the chain rule for many-variables functions, but without knowing the details of $F$ I find it very hard to prove much about it.  To the best of my ability to reason about it, I suppose I could say  $$ \frac{\partial{F}}{\partial{x}}=\frac{\partial{F}}{\partial{x}}\frac{\partial{x}}{\partial{x}} + \frac{\partial{F}}{\partial{y}}\frac{\partial{y}}{\partial{x}}$$  but for more than one reason I doubt that what I have written is correct.  Yet I'm not exactly sure what's wrong about it, nor can I guess what else I could do.  I tried finding the partial with respect to $y$ but got nothing more useful looking than this.  I also tried writing out limit definitions and saw no useful path forward through that. Has Java 3D changed much since 2005? I'm thinking of picking up the book called "Killer Game Programming" by Andrew Davidson, but its pretty old, now. I know Java, and I've been messing with jMonkeyEngine for a few days, but I'm curious to see if I would benefit from the 3D chapters of this book.   The book was published was in 2005. Has Java changed to the point where this book would now be obsolete, assuming it was up to date at time of publication? I am having some issues trying to show multiple fields for a single label in QGIS. In QGIS2.6, attempting to display labels for a feature consisting of multiple fields results in an empty label if one of the fields is null. So for example, in the label expression dialog:  concat("Temp Site",'\n',"Notes",'\n',"Function")   will function normally unless one of the fields is null, then it results in no label being shown. Using the '||' operator instead in the expression dialog also results in the same behaviour:  "Temp Site" || '\n'  ||  "Notes" || '\n'  ||  "Function"   Is there a way to display the labels even if one of the fields is null? How to use GeoPandas buffer function to get buffer zones in kilometers? How to solve this integral? $I=\int{e^x \Big(2\log x+\frac{1}{x}+\frac{1}{x^2}\Big)dx}$ $$I=\int{e^x \Big(2\log x+\frac{1}{x}+\frac{1}{x^2}\Big)dx}$$ I have tried to use many methods of substitution  But I am aware of a formula  $$I=\int{e^x[f(x)+f'(x)]dx}=e^xf(x)+C$$  but can this formula be applied to the above equation OED says:     doobie: a marijuana cigarette    Origin unknown. A relationship with dobby has been suggested.    dobby/dobbie: A silly old man, a dotard, a booby. Dialectal.      First citations:    1967   J. B. Williams Narcotics &amp; Hallucinogenics. Dubbe, Negro slang for a marijuana roach.    1982   A. Maupin Further Tales of City 95,  I smoke a doobie at lunch.   But I just chanced upon  in Yahoo Answers (don't laugh! :) where someone posted:     "It's obvious that Shaggy's a , I'm sure he's been feeding Scoob a bit of his hash and whatnot."    Scoob = , cartoon "pet" dog of constantly-hungry hippie/village idiot Shaggy (from 1969)   I'm not entirely convinced that 1967 dubbe was necessarily a significant factor in the appearance of  since around 1970 (the hippie era, some might say).  So I'd like to know whether it's credible to claim the usage arose (primarily?) through association with the TV show (specifically, the two characters therein closely associated with marijuana). How is Cycles different from Blender Internal? How does Cycles work and how is it different from Blender Internal?  Which is better? POSTFIX main.cf permission denied Drupal 7. Custom module. I have a content type 'RFI' with a handful of fields I created with Drupal UI. I want a user to: fill out a form &amp; click submit while drupal: saves fields to database, creates a new node of 'RFI' and populates the fields of the node with the content provided by the form's corresponding fields.  From what I understand:  Build a form with fields (taking from the 'DBTNG examples' module just for clarity):  function rfi_build_form($form, &amp;$form_state) { $form = array();  $form['add'] = array( '#type'  =&gt; 'fieldset', '#title' =&gt; t('Add a person entry'), ); $form['add']['name'] = array( '#type'  =&gt; 'textfield', '#title' =&gt; t('Name'), '#size'  =&gt; 15, ); $form['add']['surname'] = array( '#type'  =&gt; 'textfield', '#title' =&gt; t('Surname'), '#size'  =&gt; 15, ); $form['add']['age'] = array( '#type'  =&gt; 'textfield', '#title' =&gt; t('Age'), '#size'  =&gt; 5, '#description' =&gt; t("Values greater than 127 will cause an exception. Try it - it's a    great example why exception handling is needed with DTBNG."), ); $form['add']['submit'] = array( '#type'  =&gt; 'submit', '#value' =&gt; t('Add'), ); return $form; }   Here the form fields are saved to database with database API.  function rfi_build_submit($form, &amp;$form_state) { global $user;  // Save the submitted entry. $entry = array( 'name'    =&gt; $form_state['values']['name'], 'surname' =&gt; $form_state['values']['surname'], 'age'     =&gt; $form_state['values']['age'], 'uid'     =&gt; $user-&gt;uid, ); $return = dbtng_example_entry_insert($entry); if ($return) { drupal_set_message(t("Created entry @entry", array('@entry' =&gt; print_r($entry, TRUE)))); } }   Now here is how I know how to create a node of RFI type. How do I take the values of the fields above [name], [surname], [age] and have them related to corresponding fields in my content type?  function rfi_build_type($form, &amp;$form_state){ $node = new stdClass(); $node-&gt;type = 'rfi'; $node-&gt;name = ''; $node-&gt;surname = ''; $node-&gt;age = ''; $node-&gt;language = 'en'; $node-&gt;uid = 1; node_save($node); }   I was thinking that since the fields have been saved already, that I could use a database query to get these values, but intuition tells me I'm overlooking something.  Thank you! What is the definition of "creature" and is it used consistently? What happens, if a rocket is filled with a vacuum instead of high pressured air? Suppose you put wheels under a compressed air tank    so that it can move horizontally to the right and to the left. Suppose there is a nozzle on the right hand side of the tank (in the picture on the backside, if you like).  If the tank is filled with high pressured air and you open the nozzle, the tank will move to the left. This is how rockets work and one can easily imagine that.   What happens when the tank is filled with a vacuum and you open the nozzle?  My first impression was, that the tank is moving to the right but someone told me, that it would be stand still. Is this true? What is an intuitive explanation for that?