Im quite confused over the use of the radical. I can understand its definition in $\mathbb{R}$, with $\sqrt[n]{x}$ being defined for $x&gt;0$ if $n$ an even integer greater than one and $x$ any real number if $n$ an odd integer greater than one. I believe its referred to as the principal $n$th root for even $n$ as it outputs the positive root rather than the negative root. In $\mathbb{R}$ there are no square roots of negative numbers, so on this domain it's undefined. I also know it defines the same function of $x$ as $x^{1/n}$ for $n$ integer greater than $1$.  In $\mathbb{C}$, a principal $n$th root function can be defined using the same radical notation which for me is confusing. The fact there are so many ways to define the principal $n$th root function in $\mathbb{C}$ makes it worse, and is $x^{1/2}= \sqrt{x}$ in $\mathbb{C}$ or do you have to define it that way after you have defined exactly what you mean by $\sqrt{x}$. Should this principal $n$th root function in $\mathbb{C}$ be viewed as a different function altogether from the one in $\mathbb{R}$, or just an extension of the domain?  Thanks
This is really two questions.   Is there a definition of the principal $n$th root of a complex number? I can't find it anywhere. Presumably, the usual definition is $[r\exp(i\theta)]^{1/n} = r^{1/n}\exp(i\theta/n)$ for $\theta \in [0,2\pi)$, but I have yet to see this anywhere. Is this because it has bad properties? For instance, according to this definition is it true that for all complex $z$ it holds that $(z^{1/a})^b = (z^b)^{1/a}$?
How do I access the terminal without logging in to the desktop? Editing polygons simultaneously in ArcMap? I have two polygons:    How can I reshape both Polygon 1 and Polygon 2 at the same time?  Will the area of each polygon change automatically in the table after reshaping?  I'm using ArcMap 10.1. Right at the beginning of the game, Meowth comes into your room and attempts to wake you up. At this point, I assume I have to press some button to get out of bed, but the D-Pad or A button doesn't do anything, nor any of the other buttons. How do I get out of bed to start the game? Shell programming temp=$1 vs temp=1 What's the difference between temp=$1 and temp=1? Why do I need the dollar sign?  For example:   #!/bin/bash  temp=$1 cell=$((($temp-32)*5/9))  echo $cell Flex: Figure out when the running swf was compiled? Why does Li-Po battery voltage drop under load? I have very low background in electronics.  I have a Li-Po battery connected to an electric motor. The question is:   Why does battery voltage drop under load? Why is each cell of a Li-Po battery limited to 3.0 V (discharging)? Why is each cell of a Li-Po battery limited to 4.2 V (charging)? Unable to stylize raster data as pseudocolor render type on MAC OSX In a raster layer's Style Properties, I can choose pseudocolor render type, and choose a color ramp and classify the raster.      However, when I click the 'Apply' button, or click 'OK', the visualization of the raster never changes.  I was able to output a 'colorized' version of the same raster using GRASS plugin.  I also reinstalled QGIS 2.8, and was able to adjust and see a raster stylings using the steps I outlined above.  So, can anyone confirm this as an issue, or confirm that there's a mistake somewhere in my process?  I should note that I'm on a Mac, 10.10.4, and I'm using the QGIS provided by , 2.10.1-1 .  Update 1: Here's a screenshot of the rest of my settings.      I have Grayscale set to off in the Style setting.  Also, I don't believe grayscale/color-blind Preview is available on the Mac version of QGIS.  They're does't appear to be the option where it , in the Menu Bar under View / Preview / Photocopy (grayscale) etc.  Either way-- I added a shapefile to show color.  Update 2: I've experimented with setting color interpolation to discrete, linear, and exact, but still am not able stylize using Singleband pseudocolor.  I've also confirmed that the 'Render' box on the the main map view is checked.  Also, I am able to adjust Color Rendering settings, like Hue, Blending Mode, etc., and see the results when clicking 'Apply' or 'OK'.  Update 3: As other Mac users are confirming this, I think it may be a bug. I've created a bug report at the official . The new 17.10 only has 64 bit ISOs. However 32 bit is still supported, there just are no live ISOs for it.  I have a 32 bit machine with WIFI, and I need to install from a USB disk. I tried the "mini.iso" but it doesn't have WIFI support.  What options do I have?  (PS I don't want another distro, and I can't use ethernet for installation, must be WIFI). Has anyone had any success using twitters typeahead.js within visualforce pages? I am trying to get  to work within salesforce visualforce pages.  I have a basic vf page from  example, but it does not respopnd, I suspect there may be some "unknown" workaround needed? Does anyone know?  &lt;apex:page standardController="Contact" showHeader="false" sidebar="false" docType="html-5.0"  standardStylesheets="false"&gt;  &lt;script src="https://ajax.googleapis.com/ajax/libs/jquery/1.10.2/jquery.min.js"&gt;&lt;/script&gt; &lt;script src="http://twitter.github.io/typeahead.js/releases/latest/typeahead.bundle.js"&gt;&lt;/script&gt;      &lt;div id="the-basics"&gt;   &lt;input class="typeahead" type="text" placeholder="States of USA"/&gt; &lt;/div&gt;  &lt;script&gt;     var substringMatcher = function(strs) {       return function findMatches(q, cb) {         var matches, substrRegex;          // an array that will be populated with substring matches         matches = [];          // regex used to determine if a string contains the substring `q`         substrRegex = new RegExp(q, 'i');          // iterate through the pool of strings and for any string that         // contains the substring `q`, add it to the `matches` array         $.each(strs, function(i, str) {           if (substrRegex.test(str)) {             // the typeahead jQuery plugin expects suggestions to a             // JavaScript object, refer to typeahead docs for more info             matches.push({ value: str });           }         });          cb(matches);       };     };      var states = ['Alabama', 'Alaska', 'Arizona', 'Arkansas', 'California',       'Colorado', 'Connecticut', 'Delaware', 'Florida', 'Georgia', 'Hawaii',       'Idaho', 'Illinois', 'Indiana', 'Iowa', 'Kansas', 'Kentucky', 'Louisiana',       'Maine', 'Maryland', 'Massachusetts', 'Michigan', 'Minnesota',       'Mississippi', 'Missouri', 'Montana', 'Nebraska', 'Nevada', 'New Hampshire',       'New Jersey', 'New Mexico', 'New York', 'North Carolina', 'North Dakota',       'Ohio', 'Oklahoma', 'Oregon', 'Pennsylvania', 'Rhode Island',       'South Carolina', 'South Dakota', 'Tennessee', 'Texas', 'Utah', 'Vermont',       'Virginia', 'Washington', 'West Virginia', 'Wisconsin', 'Wyoming'     ];      $('#the-basics .typeahead').typeahead({       hint: true,       highlight: true,       minLength: 1     },     {       name: 'states',       displayKey: 'value',       source: substringMatcher(states)     }); &lt;/script&gt; Let $G$ be a group, $H\subseteq G$ a subgroup and $a\in G$ an element of the group. Is it possible that $aHa^{-1} \subset H$, but $aHa^{-1} \neq H$?   If $H$ has finite index or finite order, this is not possible. Which mathematical operation does the right hand rule for current come from? I am currently wondering about the famous .  Where does it come from mathematically that when you point with your thumb in the direction of the current, your curved fingers will point in the direction of the $B$-field? In other words, which mathematical operation does it come from? Probably it has to do with some vector product or Stokes theorem application, but I am not quite sure about it. How do I add a DOM element with jQuery? Graphics issues after/while installing Ubuntu 16.04/16.10 with NVIDIA graphics Do indexes work with "IN" clause "Newtonian limit" property of special relativity Books say that special relativity is indistinguishable from Newtonian mechanics when the speed of the primed frame ($v$) is small compared to the speed of light ($c$). This is what I mean by the "Newtonian limit" property of special relativity. (I don't know the correct name for it). But the transformation of the time coordinate  $$t' = \gamma \left(t + \frac{vx}{c^2}\right)$$  involves the spatial quantity $x$ which, if large enough, could balance out the smallness of $v/c^2$. So why doesn't this observation mess up the "Newtonian limit" property? What would be the the mathematical equivalent of this excel formula? =PMT() I need to create a JavaScript form that does the same calculation as this =PMT() function.   mortgage = 220000 rate= 4.75% term = 30   The example formula I have is =PMT(4.75%/12,30*12,220000*-1)  What would be the equation for this function? I tried to look up this function but it doesn't explain it very well. I spotted a  (10k-only by now) and when I went to flag it, the flag dialog told me that I'd already flagged it as VLQ:    The thing is, I'd never seen this post before. And a look at the flag history in my user profile confirms that I haven't flagged a post since yesterday:    Certainly, well before the 2015-02-04 05:36:21Z time of Mr. Miracle's post.  I was able to reproduce this on SO from the VLQ review queue: when you're presented with an answer, click on link at the right side of the page, then click flag. You get the "you have already raised this type of flag" message on either the "it is not an answer" or "it is very low quality" options, I gather depending on the original flag that pushed the post into the VLQ queue. I was readng Wikipedia article (English one) about radiation pressure because there is something I still cannot figure out. As I understand it, radiation pressure emerges from conservation of momentum. Photons or electromagnetic waves possess momentum and when they are absorbed, reflected or even emitted, the aborber/reflector/emitter experiences a pressure that is proportional to the irradiance (in watts per square meter). OK. In the case of a perfect reflector that does nothing but reflecting incoming radiation, it would seem the reflector is pushed and the radiation is reflected in opposite directions. But the radiation has not lost anything, it has only changed direction. And I do not understand how something can be set in motion (the reflector), which amounts to do work, yet the source of this work does not lose energy. I mean, if the radiation, after a U turn, meets a second reflector, it would have pushed two reflectors a way, yet it would continue in its original direction as if nothing happened...? What am I missing here? Should not photons lose something? The indefinite article applied to a group of things Realtek RTL8723BE Wi-Fi incredibly weak