There are a lot of people telling how to make a bootable disc to other people who must understand a lot more than I do.  I try to read those answers and I say "OK, now tell me how to make bootable usb for dummies.  It would make Ubuntu much more popular if you did not need to be a computer hacker to figure out how to install it.  Why do you think Windows is popular?  You put in the disk and click install and 1 hour later you are surfing the web and emailing.  I need Ubuntu to be much easier to install than it is now!  Why not make a file to download that you copy to your USB then just reboot?  I would like to try an alternative to windows if someone could do a much simpler explanation to make a bootable USB, or even a bootable dvd.  Please help.  Thanks for your time!
How do I make an Bootable Ubuntu USB containing the the latest version of Ubuntu?  For completeness, how would I be able to do this on Ubuntu, Windows, OS X or another Linux distro?
Seeking methods to solve: $\int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt$ Seeking Methods to solve the following two definite integrals:  \begin{equation} I_(n) = \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt \qquad J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt \end{equation}  For $n \in \mathbb{R},\:n \gt 1$  The method I took was to take the following integral:  \begin{equation}  \int_0^\infty \frac{t^k}{\left(t^n + 1\right)^m}\:dt = \frac{1}{n}B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \end{equation}  Where: $0 \leq k \lt n$. Here let $m = 1$.   with respect to $k$ and taking the limit as $k \rightarrow 0^+$ (via the ), i.e.   \begin{align} \lim_{k \rightarrow 0+} \frac{d}{dk} \left[ \int_0^\infty \frac{t^k}{t^n + 1}\:dt \right] &amp;= \lim_{k \rightarrow 0+} \frac{d}{dk} \left[\frac{1}{n}B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n} \right) \right] \\ \lim_{k \rightarrow 0+}  \int_0^\infty \frac{t^k \ln(t)}{t^n + 1}\:dt  &amp;= \lim_{k \rightarrow 0+}  \left[\frac{1}{n^2} B\left(1 - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\left[\psi^{(0)}\left(\frac{k + 1}{n} \right) - \psi^{(0)}\left(1 - \frac{k + 1}{n} \right)\right] \right] \\ \int_0^\infty \frac{\ln(t)}{t^n + 1}\:dt&amp;= \frac{1}{n^2} B\left(1 - \frac{1}{n}, \frac{1}{n}\right)\left[\psi^{(0)}\left(\frac{1}{n} \right) - \psi^{(0)}\left(1 - \frac{1}{n} \right)\right] \\ &amp;=- \frac{\pi^2}{n^2}\operatorname{cosec}\left(\frac{\pi}{n}\right)\cot\left(\frac{\pi}{n} \right) \end{align}  Which is our expression for $I_n$. Taking the same approach but differentiating twice with respect to $k$ we arrive at our expression for $J_n$:  \begin{equation}  J(n) = \int_0^\infty \frac{\ln^2(t)}{t^n + 1}\:dt = \frac{\pi^3}{n^3}\operatorname{cosec}\left(\frac{\pi}{n} \right)\left[\operatorname{cosec}^2\left(\frac{\pi}{n}\right) + \cot^2\left(\frac{\pi}{n}\right) \right] \end{equation}  And in fact we may generalise:  \begin{equation}     \int_0^\infty \frac{\ln^p(t)}{\left(t^n + 1\right)^m}\:dt = \lim_{k\rightarrow 0}\frac{d^p}{dk^p}\left[\frac{1}{n} B\left(m - \frac{k + 1}{n}, \frac{k + 1}{n}\right)\right] \end{equation}  Where $p \in \mathbb{N}$  This method however was just an extension of another integral. I'm curious, if I had just started with $I_n, J_n$ what alternative methods could be used? How to select distinct field values using Solr? How delete an application that is bundled with macOS? Backstory:   I am running macOS Sierra and use TextEdit a lot. The problem is that I hate the Versions save schema that causes the universal Save As shortcut of CmdShiftS to instead create a copy of the document. I also hate how Auto-Save makes permanent changes to documents I don't want to modify when I'm tentatively modifying a work-in-progress. Since the introduction of these nuisance features in Lion I have always backed up and then replaced TextEdit.app with the legacy 1.6 version from Snow Leopard.  The problem:  Today when I tried to delete TextEdit.app from my Applications folder using this Terminal command  sudo rm -rf /Applications/TextEdit.app   I got this error  rm: TextEdit.app: Operation not permitted   I was surprised to find that I can't kill it, even with sudo.  How can I actually get rid of an application that is bundled with macOS?  What I've tried so far:  I followed the instructions in , but it didn't work. I also read , which almost accomplishes the same solution by means of disabling Versions and Auto-Save, but then I don't get the benefit of legacy TextEdit's temporary backup saving feature. See the part in the answer stating "This effectively disables the old-style, pre-Lion autosaving completely".   So I think deleting the newer TextEdit v1.12 and replacing it with the legacy v1.6 is still the best option, however I can't get my system to delete it. Any help appreciated. queries with db_select don't respect integers How is every subset of the set of reals with the finite complement topology compact? While we know that every finite set is compact, how to prove that every infinite subset of $\mathbf{R}$, the set of all real numbers, is also compact in the finite-complement topology (also called the co-finite topology)?   What is the situation in the case of an arbitrary infinite set $X$ with the finite-complement topology? Fish vs Fishes for plural use Calculation of pressure from flow rate of water Anybody kindly help me to find how to calculate pressure in bar from flow rate. I have a pipe and from that I am transferring water at a constant flow rate of 5ml/min. At this flow rate, with a 0.5 cm diameter pipe, what will be the pressure generated by water? Should Dalvik Cache be wiped in Android OS that supports ART? When switching custom ROMs (in ART), is it necessary to wipe the Dalvik Cache?  Wiping the Cache is necessary, but should wiping Dalvik (which doesn't exist in Android Lollipop and afterward) be necessary?  What impact will it cause if we wipe Dalvik (something not present)?  In ART, what does the Dalvik cache contains? (is it an empty folder?) Lightning component : Get the id of a record In a lightning component I display a list of articles (Article__c custom object) so I ca see, for example, the name of the article, the price, and a button that I'd like redirect to another component which display the details of the article selected.   For now I'm just trying to get the id of the article, in a basic alert in javascript, but I doesn't work : it says the id is null.   Here is a part of the component :  &lt;aura:handler name="init" value="{!this}" action="{!c.doInit}" /&gt; &lt;aura:attribute name="wrappers" type="ListeArticlesWrapper" /&gt; &lt;aura:attribute name="articles" type="Article__c" /&gt;  &lt;aura:attribute name="Detail" type="String"/&gt; &lt;aura:registerEvent name="navigate" type="c:NavigateToDetail"/&gt;  &lt;div class="page"&gt;     &lt;h1&gt;Liste des articles&lt;/h1&gt;     &lt;div class="slds"&gt;         &lt;table class="slds-table slds-table--bordered slds-table--cell-buffer"&gt;             &lt;thead&gt;                 &lt;tr class="slds-text-heading--label"&gt;                     &lt;th scope="col" class="head"&gt;&lt;/th&gt;                     &lt;th scope="col" class="head"&gt;Nom&lt;/th&gt;                     &lt;th scope="col" class="head"&gt;Type&lt;/th&gt;                     &lt;th scope="col" class="head"&gt;Prix&lt;/th&gt;                     &lt;th scope="col" class="head"&gt;&lt;/th&gt;                 &lt;/tr&gt;             &lt;/thead&gt;             &lt;tbody&gt;                 &lt;aura:iteration items="{!v.wrappers}" var="wrap"&gt;                     &lt;tr&gt;                         &lt;td class="cell"&gt;                             &lt;ui:inputCheckbox value="{!wrap.selected}" /&gt;                         &lt;/td&gt;                         &lt;td class="cell"&gt;                            &lt;ui:outputText value="{!wrap.art.Name}" /&gt;                         &lt;/td&gt;                         &lt;td class="cell"&gt;                             &lt;ui:outputText value="{!wrap.art.Type__c}" /&gt;                         &lt;/td&gt;                         &lt;td class="cell"&gt;                             &lt;ui:outputText value="{!wrap.art.Prix__c}" /&gt;                         &lt;/td&gt;                         &lt;td class="cell"&gt;                             &lt;button class="slds-button slds-button--neutral slds-not-selected"                                      onclick="{!c.getDetail}" data="{!wrap.art.Id}"&gt;Details&lt;/button&gt;                         &lt;/td&gt;                     &lt;/tr&gt;                 &lt;/aura:iteration&gt;             &lt;/tbody&gt;         &lt;/table&gt;   The associated controller js (with the function getDetail)  ({ doInit : function(component, event, helper) {     helper.init(component, event); },  getArticles : function(component, event, helper) {     helper.getArticles(component, event); },  getDetail : function(component, event, helper){     var id = event.target.getAttribute("data-data") || event.target.parentNode.getAttribute("data-data")     alert('detail article : id = ' + id); //it gives me null } })   Thanks !  EDIT  And the wrapper class in case you wonder what is the "art" :  public with sharing class ListeArticlesWrapper { //wrapper pour récupérer les articles selectionnés @AuraEnabled public Article__c art {get; set;}  @AuraEnabled public Boolean selected {get; set;} } Material setings in properties window is missing I have added a sphere and I was able to add material and give it a color. Now when I add a second sphere (separate object), the material panel is just blank. For some reasons, homomorphism is a very hard area for me to make improvements. I've been hitting the brick wall for almost 2 hours.   Prove that no homomorphism exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$.  Assume a homomorphism $\phi$ exists from $Z_{16}\oplus Z_2$ onto $Z_4\oplus Z_4$. Then $\phi$ is an isomorphism.  Note that the $\left | \ker \phi \right |=2$ The $\ker \phi$ is also a normal subgroup for $Z_{16}\oplus Z_{2}$. We want possible normal subgroup of order 2. I.e., 2 elements in each normal subgroups.  By Lagrange's theorem, the order of each element in a group divides the order of a group so the possible order of the elements are 1 or 2. If the elements has order 1, then the Ker \phi cannot have 2 elements.  Hence, $\ker\phi$ = $\left \{ (0,0),(8,0) \right \}$ ,$\left \{ (0,0),(0,8) \right \}$,$\left \{ (0,0),(8,1) \right \}$, possibly.  By the First isomorphism theorem:  $\phi: Z_{16}\oplus Z_2 \rightarrow Z_2\oplus Z_2$  $\Psi: G/\ker \phi \rightarrow \phi\left ( Z_16\oplus Z_2 \right )=Z_2\oplus Z_2$  $g\ker \phi  \mapsto \phi(g)=\Psi(g\ker \phi)$  Any help is appreciated.  Edit: I know that $(2,0)$ has order $8$ in $Z_{16}\oplus Z_{2}$ but any elements in $Z_{4}\oplus Z_{4}$ does not have order $8$ which would have solved the question at the outset. But I would like a different route. What is the difference between == and ===?  How exactly does the loosely == comparison work? How exactly does the strict === comparison work?  What would be some useful examples? Python List vs. Array - when to use? $ p(x)\in\mathbb{R[X]} $ is a polynomial of degree $n$ with no real   roots. Show that: $$\int\limits_{-\infty}^{+\infty}\dfrac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n^{3/2}\pi.$$   It's easy to see that the degree of $ p$ has to be even. For $n=2$ this integral is at most $2\pi$. For $n&gt;2$ the maximum value of this integral is obtained when all the imaginary parts of the roots of $p(x)$ tend to $0$, but I couldn't go further.    Any help would be appreciated, thanks.    Edits by David Speyer: It seems very likely now that the optimum bound is $n \pi$, not $n^{3/2} \pi$. As pointed out in the comments below, and further in , this is the value we get in the limit where $p$ has double roots on the real axis.   It seems likely that  moving the roots of $p$ away from the real axis can only make the integral less. Write the roots of $p$ as $a_j \pm i b_j$, so $p(x) = \prod ((x-a_j)^2 + b_j^2)$ and  $$\frac{p'(x)}{p(x)} = \sum \frac{2 (x-a_j)}{(x-a_j)^2 + b_j^2}.\quad (\ast)$$  So making the $b_j$ larger tends to make $p'(x)/p(x)$ smaller, which makes the integral smaller. But this argument is not rigorous, because the terms of $(\ast)$ can have both positive and negative sign, so it could be that making the individual terms closer to $0$ makes the absolute value of $(\ast)$ larger.  I don't see how to beat this issue easily.  Thus, I'm putting up a bounty for proving or disproving   $$ \int_{-\infty}^{+\infty}\dfrac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n\pi.$$ The laptop is dual booting with windows 7 and ubuntu 11.10.  For 11.10 to boot, I had to add 'nolapic' in GRUB boot options.  I downloaded 12.04LTS LiveCD yesterday and made a CD and a bootable USB stick.  With both USB and LiveCD, occasionally I can boot into 12.04.  But many times, the boot process stuck at some point.  For USB boot, sometime it works.  But other times, it seems to stuck right after seeing the message:  NMI watchdog enabled. takes one hw-pmu counted [0.408590] #2   The number in [] changes.  Some times it stuck at #2, and sometimes at #3.  For LiveCD boot, sometime it works.  But when it stuck, it stuck at different places.  At times, the boot process go directly to a blank screen with a blink '_' at up left corner.  Sometime it went a bit further, after showing options to try/install ubuntu then stuck with a blank screen with blink '_'.  Anything I can do to get around of this? Agent Smith, in the first Matrix movie (The Matrix) talks about how the machines originally created a paradise setting and humans didn't accept it, so it went back and created a more realistic one that humans would accept.  The human bodies, as we've seen, are kept in tanks and the body is kept asleep.  But why did the Matrix need to provide any kind of simulation for the brain to experience?  Why couldn't it have just kept all humans in a medically induced coma?  Or performed something equivalent to a lobotomy on every human?  Why did the Matrix need to spend the resources to keep the humans in good mental shape? If $n&gt;1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.  If you know , then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.  Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong. Make gnome-terminal show the command running as title I want the name of command running , e.g unzip to be visible through the title bar of gnome-terminal , but it seems to be impossible , if the application running doesn't set a title explicitly , even though I choose 'Replace initial title' option in the profile dialog. I am trying to determine if the sum $\sum_{n=1}^\infty \frac{n+4^n}{n+6^n}$ is convergent or divergent.   In attempting to solve this problem I began by seeing if I could do the divergence test - in this case it is not helpful. The terms $4^n,6^n$ look like geometric series type terms but the $n$ in the numerator and denominator are troublesome. I first tried factoring $n$ out of the expression:  $$\sum_{n=1}^\infty \frac{n+4^n}{n+6^n} = \sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$$  Next I tried comparing to the series $$\sum_{n=1}^\infty \frac{\frac{4^n}{n}}{\frac{6^n}{n}} = \sum_{n=1}^\infty \frac{{4^n}}{{6^n}} = \sum_{n=1}^\infty{\left(\frac{2}{3}\right)}^n$$  This is a convergent geometric series as $r = \frac{2}{3}\lt1$  now I reason that the series $\sum_{n=1}^\infty \frac{1+\frac{4^n}{n}}{1+\frac{6^n}{n}}$ is neither smaller not greater than the series $\sum_{n=1}^\infty{(\frac{2}{3})}^n$ Am I able to conclude anything in this case by a direct comparison test or can I only conclude that the first series is convergent if the second is greater than the first?  Thanks How does passive perception work Total noob question, but the passive perception is confusing. I don't understand how to calculate it or use it, and I need help.