Duplicates in Oracle Full Join
What is the difference between "INNER JOIN" and "OUTER JOIN"?
Ubuntu stuck/freeze on install I've been trying to replace my current Windows 10 installation with Ubuntu 16.04. My computer is a Lenovo Thinkpad T470. I've formatted my 15GB USB-dongle. Downloaded the software Rufus, and burned the ISO onto the USB. I've tried the settings "MBR Partition for BIOS or UEFI", "MBR Partition for UEFI" and "GPT partition for UEFI". I've tried using memtest86+ to test if I have any memory errors, I have not. I've tried disabling Secure boot, and Fast startup. I've tried using other software than Rufus, and I've tried entering the "nomodeset" in the boot options, as well "acpi=off", nothing works.   When I boot from my USB. I choose "Install Ubuntu" from the grub menu. And then select my language in the installer. It is from this point on, it gets problematic. Sometimes when I try to connect to my WIFI, it connects successfully, but when I press the "continue" button, the installer freezes and nothing happens. Sometimes I get past this point, but it then freezes later on in the installation progress - typically right when I press the "Install now" button.   It should also be mentioned that, if I press the "Quit" button in the install. I get to a black screen with a lot of text saying something like squashfs error.   Can someone please help me? When I tried booting memtest86+ it only worked when I changed my boot mode from UEFI to Legacy. Could this be it? Are there any downsides to installing an operating system in legacy?   If anyone has any other suggestions, it would be greatly appreciated!  Thank you :-) Find all solutions to the equation $(z+1)^6=z^6$. Express the solutions in simplified rectangular form. Hint: Clearly $z=0$ is not a solution. $$(z+1)^6=z^6 $$ $$\frac{(z+1)^6}{z^6 }=1$$ $$\left(\frac{(z+1)}{z }\right)^6=1$$ $$\left(\frac{(z+1)}{z }\right)^6=\cos(2*\pi*k)+i\sin(2*\pi*k)$$ $$\frac{(z+1)}{z }=\left(\cos(2*\pi*k)+i\sin(2*\pi*k)\right)^{1\over6}$$ $$1+\frac{1}{z }=\cos({2*\pi*k\over6})+i\sin({2*\pi*k\over6})$$  $$\frac{1}{z }=(\cos({\pi*k\over3})-1)+i\sin({\pi*k\over3})$$ $$\frac{1}{(\cos({\pi*k\over3})-1)+i\sin({\pi*k\over3}) }=z$$ I dont know how else to solve this. A lot of help is appreciated. Also i know that k should go from 1-5. Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice? Is there any example of a non-measurable set whose proof of existence doesn't appeal to the Axiom of choice?  What would it imply if there was such an example?  EDIT: For instance, maybe this will help understand the kind of example I had in mind, it is known that an important feature to determine that the AC is needed in the Banach-Tarski case is the non-transitivity of rotations on Euclidean space, one might find an example of a transitive group for some given space keeping the rest equal and make the AC unnecessary?  I guess then it might be said that this transitivity will be in this particular case equivalent to the AC or some amount of it, but I guess that it would be important to show this if it hadn't been realized before. \pause in tikzpicture breaks footline I've discovered that you can \pause within a tikzpicture, which works well until I switched the theme to one with a footline.  I've created a "test case" that illustrates the problem.  \documentclass{beamer} \providecommand\thispdfpagelabel[1]{} % Not sure what this does but our installation requires it. \usepackage{tikz} \usetheme{Madrid} % Has a footline.  \begin{document}   \begin{frame}     \frametitle{test}     \begin{tikzpicture}       \node at (0, 1) {Hello};       \pause       \node at (0, 0) {World};     \end{tikzpicture}   \end{frame} \end{document}   This creates two slides, but the footline is only drawn on the second (the headline/title appears on both). Is there anything I can do about this? Sum of infinite series $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}$ Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc.. the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and   that is equal to $n*n!$ Finally I want to prove the general equality: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$ I'm trying to search and replace a string in all files matched by grep: grep -n 'foo' * will give me output in the form: [filename]:[line number]:[text]  For each file returned by grep, I'd like to modify the file by replacing foo with bar. Does the concept of infinity have any practical applications? I know what you're thinking: "of course it has, for example, it can be used to tell you how many times you can go around a circle". But that isn't really true, now is it? You'd be dead or the world would go under long before an infinite amount of loops had been reached.  Are there any practical applications for the concept of infinity? Is it a useful concept in maths at all?  I know that Donald E. Knuth has argued that for all practical purposes, a very, very large number has the same effect as infinity, in his book "Things a Computer Scientist Rarely Talks About" (can't remember the exact quote, nor find it online, unfortunately).  Examples are appreciated. Why is RSA encryption significantly faster than decryption? I am designing an implementation of RSA . I recorded computation times in Java by using System.currentTimeMillis(). It returned an encryption time of 0.05 ms and decryption time of 0.55 ms, or a ratio of 1:11.   This seems like a large difference. Is it supposed to be this way?  //here my key has 256 bits  for (;;) {             long begin = System.currentTimeMillis();              for (int i = 0; i &lt; num; i++) {              decrypt();             }              long end = System.currentTimeMillis();              long time = end - begin;              if (time &gt;= 10000) {                 System.out.printf("Average Encryption takes: %.2f ms\n",                         (double) time / num);                 break;             }              num *= 2;         }  p = BigInteger.probablePrime(128, random); q = BigInteger.probablePrime(128, random); N=(p.subtract(one)).multiply(q.subtract(one)); e=BigInteger.probablePrime(32, random); d=e.modInverse(N); private void encrypt() {  C= M.modPow(e,N);  }  private void decrypt() {     RM = C.modPow(d, N);     } Why is Python 2.7 still the default Python version in Ubuntu? According to official documentation of Python, Python2.7 support will end near in future.     DEPRECATION: Python 2.7 will reach the end of its life on January 1st, 2020. Please upgrade your Python as Python 2.7 won't be maintained after that date. A future version of pip will drop support for Python 2.7. More details about Python 2 support in pip, can be found at    Why is Ubuntu not using Python3 as default in their upcoming version?  $ cat /etc/os-release  NAME="Ubuntu" VERSION="19.04 (Disco Dingo)" ID=ubuntu ID_LIKE=debian PRETTY_NAME="Ubuntu 19.04" VERSION_ID="19.04" HOME_URL="https://www.ubuntu.com/" SUPPORT_URL="https://help.ubuntu.com/" BUG_REPORT_URL="https://bugs.launchpad.net/ubuntu/" PRIVACY_POLICY_URL="https://www.ubuntu.com/legal/terms-and-policies/privacy-policy" VERSION_CODENAME=disco UBUNTU_CODENAME=disco     $ python Python 2.7.16 (default, Apr  6 2019, 01:42:57)  [GCC 8.3.0] on linux2 Type "help", "copyright", "credits" or "license" for more information. How to prove $\bar{m}$ is a zero divisor in $\mathbb{Z}_n$ if and only if $m,n$ are not coprime Let us consider the ring $\mathbb{Z}_n$ where $\bar{m}\in\mathbb{Z}_n$  Could anyone help me prove that $\bar{m}$ is a zero divisor in $\mathbb{Z}_n$ if and only if $m,n$ are not coprime  So far I have:  Assume $\exists \bar{a}: \bar{m} \bar{n}=n \mathbb{Z} \Rightarrow$ for some $b\in\mathbb{Z}:am=bn$  I then assumed $n,m$ were coprime and attempted to use $\exists a&#39;,b&#39;\in \mathbb{Z}:a&#39;m+b&#39;n=1$ to come to a contradiction, however haven't found one, the most promising thing I have found so far is that  $a=n(a&#39;b+ab&#39;)$ and $b=m(a&#39;b+ab&#39;)$ $\Rightarrow n|a$ and $m|b$ I've recently been learning about differential equations, and my teacher has been giving some particularly difficult examples out for those of us who finish early. He gave us the following differential equation: $$x=y\frac{dy}{dx}-\left(\frac{dy}{dx}\right)^{2}$$ With the hint "Differentiate with respect to $y$, then let $\frac{dy}{dx}=p$". I've solved it after the lesson and would like someone to check whether or not I've done so correctly.    Following the hint, $$\frac{dx}{dy}=\frac{1}{p}=p+y\frac{dp}{dy}-2p\frac{dp}{dy}$$ Which rearranges to $$p-\frac{1}{p}+y\frac{dp}{dy}=2p\frac{dp}{dy}$$ Multiplying through by $\frac{dy}{dp}$, $$\left(p-\frac{1}{p}\right)\frac{dy}{dp}+y=2p$$ Rearranging into the form $\frac{dy}{dp}+f(p)y=g(p)$, and writing $p-\frac{1}{p}=\frac{p^{2}-1}{p}$ $$\frac{dy}{dp}+\frac{p}{p^{2}-1}y=\frac{2p^{2}}{p^{2}-1}$$ Our integrating factor is $$e^{\int\frac{p}{p^{2}-1}dp}=e^{\frac{1}{2}\ln(p^{2}-1)}=\sqrt{p^{2}-1}$$ So we have $$\sqrt{p^{2}-1}\frac{dy}{dp}+\frac{p}{\sqrt{p^{2}-1}}y=\frac{2p^{2}}{\sqrt{p^{2}-1}}$$ The left hand side is, by design, $\frac{d}{dp}(y\sqrt{p^{2}-1})$, and the right hand side can be integrated as follows: $$\int \frac{2p^{2}}{\sqrt{p^{2}-1}}dp=\int\frac{2(p^{2}-1)+2}{\sqrt{p^{2}-1}}dp=\int 2\sqrt{p^{2}-1}dp+\int \frac{2}{\sqrt{p^{2}-1}} dp$$ These can both be solved using the substitution $p=\cosh(u)$, and gives $$y\sqrt{p^{2}-1}=p\sqrt{p^{2}+1}-\cosh^{-1}(p)+\alpha$$ Where $\alpha$ is an arbitrary constant. Dividing, we finally have $$y=p+\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}$$  Here I was stuck for a long time, until I realised that we also have the equation we began with: $x=py-p^{2}$, which we can solve as a quadratic in $p$ to also get $$p=\frac{y\pm \sqrt{y^{2}-4x}}{2} \implies p^{2}=\frac{y^{2}+y^{2}-4x \pm 2y\sqrt{y^{2}-4x}}{4}$$  Thus, $$y=\frac{y\pm \sqrt{y^{2}-4x}}{2}+\frac{\cosh^{-1}\left(\frac{y\pm \sqrt{y^{2}-4x}}{2}\right)+\alpha}{\sqrt{\frac{y^{2}-2x \pm y\sqrt{y^{2}-4x}}{2}-1}}$$  This is as good as I can do - we have a relation between $y$ and $x$ - am I done? Is this the correct answer? Is there an easier way?     Thank you for your time.    Following the suggestion by @Valentin $$\frac{dy}{dp}=1+\frac{\frac{1}{\sqrt{p^{2}-1}}\sqrt{p^{2}-1}-p\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}=1+p\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}$$ Hence, $$dx=\frac{dy}{p}=dp\left(\frac{1}{p}+\frac{1-\frac{\cosh^{-1}(p)+\alpha}{\sqrt{p^{2}-1}}}{p^{2}-1}\right)$$ Add an unnmbered section to the table of Contents in Lyx Opening a non-utf8 encoded text file I sometimes need to open text files that are encoded in EUC-KR.    man gedit gives:      --encoding           Set the character encoding to be  used  for  opening  the  files           listed on the command line.    This can be used to open specific files with the desired encoding:  gedit --encoding euc-kr readme.txt   However, each file to be opened requires a similar command to be typed.  I can run iconv on the file(s), but then I have to deal with different file versions.    Do I have better options for dealing with non-utf8 text files? When I slam my hand onto my keyboard, only a maximum of 6 keys will be registered, I've tested this for the past few minutes and was unable to get more than 6. Is this the same on all keyboards? If not, what would be the distinguishing factor, If yes: why 6 specifically?  For clarification: I do not have a usecase for this, I am simply curious Initializing Manipulator value within DialogInput (version 10.1 only) Motley fool says you can make $15,978 more per year with Social Security. Is this for real? The Motley fool has an ad campaign that says you can make $15,978 more per year on Social Security by doing 17 minutes of work using a legal loophole.  I don't want to link to their ad (although, you can find it by searching ).  What are they talking about here?  Is there any truth to this?  If someone could explain it to me without me having to watch the entire ad video or subscribe to their newsletter, I'd appreciate it. C++: delete vs. free and performance I'm looking for some sample code that will sort the list items in an HTML list by alphabetical order.  Can anyone help?  Here is a sample list for people to work with:  &lt;ul class="alphaList"&gt;     &lt;li&gt;apples&lt;/li&gt;     &lt;li&gt;cats&lt;/li&gt;     &lt;li&gt;bears&lt;/li&gt; &lt;/ul&gt; "A/An" preceding a parenthetical statement When a/an precedes a parenthetical aside (sometimes seen in informal/conversational writing), should the vowel rule depend on the first word in parentheses, or the next word in the "regular" flow of the sentence?     I need a (memorable) idiom (preceding an m word; use a)   or     I need an (memorable) idiom (preceding an i word; use an) Decode Base64 data in Java I have an image that is Base64 encoded. What is the best way to decode that in Java? Hopefully using only the libraries included with Sun Java 6.