If the function $f(x)=\frac{ln(x)}{x}$ where $x&gt;0$, has a maximum at $(e,\frac{1}{e})$ compare the numbers $e^{\pi}$ and ${\pi}^e$.  ***$ln(x)$ is the natural logarithmic function i.e the logarithmic function with base $e$.
How can I  compare (without calculator or similar device) the values of $\pi^e$ and $e^\pi$ ?
Lie Algebra of U(N) and SO(N) I need to install more applications - can I use my external HD (samsung T5) and add an additional applications folder? I have loads of space on it and it the speeds are great. What is the difference between "till" and "until"? What is the difference between till and until?    When to use till or until? Please explain with examples. According to :     The electron has no known substructure. Hence, it is defined or assumed to be a point particle with a point charge and no spatial extent.   Does point particle mean the particle should not have a shape, surface area or volume?  But when I searched Google for the "electron shape" I got many results (like  and ) that says electrons are round in shape. The use of nominative "whom" (as in “persons whom it is foreseeable are likely to...”) From  page 48 of Law: A Very Short Introduction, by Raymond Wacks:     In other words, you owe a duty to persons whom it is foreseeable are likely to be harmed by your conduct.   To try to parse this, temporarily overlook 'it is foreseeable'. Then persons is the subject of the verb phrase are likely to be harmed. So , should the relative pronoun for persons be who instead? Is my parse wrong?   Are the (deleted) comments right: that this style predominates in law texts? If this is correct grammar, what is the general rule? Is it commonly used in other contexts? Do any clones refuse Order 66? When the clones are issued      telling them to turn on the Jedi,    do any of them refuse to carry out the order? remove oldest files I`m trying to delete old files from directory and leave only 3 newest files.   cd /home/user1/test  while [ `ls -lAR | grep ^- | wc -l` &lt; 3 ] ; do      rm `ls -t1 /home/user/test | tail -1`     echo " - - - "  done   something is wrong with conditional statement. Jamaica Inn episode 5 from BBC learning English drama section:  The dialogue says:     Mary: My aunt is in bed.   She doesn't say -      My aunt is in the bed.   That section is for teaching English.  Are they using incorrect English? In Avatar: The Last Airbender, Aang's eyes glow when he goes into the Avatar State. This seems to be the case with all Avatars, but Aang and the other airbending Avatars tattoos glow too.      Is there any reason for why airbending Avatars tattoos glow as well as their eyes? I generated a key and copied it to my target:  ssh-keygen ssh-copy-id username@hostname   I left the the location of the key as default and did not set any passphrase.  For a small while after that, I was able to ssh username@hostname without being prompted for a password, and then it stopped working for some reason. When training a neural network using the back-propagation algorithm, the gradient descent method is used to determine the weight updates. My question is: Rather than using gradient descent method to slowly locate the minimum point with respect to a certain weight, why don't we just set the derivative $\frac{d(\text{Error})}{dw}=0$, and find the value of weight $w$ which minimizes the error? Also, why are we sure that the error function in back-propagation will be a minimum? Can't it turn out the error function is a maximum instead? Is there a specific property of the squashing functions that guarantees that a network with any number of hidden nodes with arbitrary weights and input vectors will always give an error function that has some minima? Note: The legacy Skype for Linux application (v4.*) is now discontinued and was removed from Canonical’s partner repositories, so this question and its answers are obsolete. Please refer to  of our .  Skype 4.3 is just released with considerable improvements. As a regular Skype user I want to use the latest Skype on my Ubuntu 14.04, but there is no updates from Update Manager. Even in Skype website only Skype 4.3 is available only for Ubuntu 12.04. Is there anyway to use this latest Skype on Ubuntu 14.04 or how much we have to wait for the update? Get all non-unique values (i.e.: duplicate/more than one occurrence) in an array Now, that Adobe has stopped issuing new releases for Linux desktops, the only option for users is to use Google Chrome, for those who want to use the latest Flash updates. I use Chromium. Is there any way to get Chromium use the Flash from Google Chrome? The reason I as this question is that I have noticed both these browsers run in the background, so technically, the browsers' plugins are available to use. Intersect (analysis) of polygons generating empty output I am reading the Dummit and Foote's proof of Cauchy's theorem for abelian groups. On the highlighted line, how do you conclude from $y\notin N$ and $y^p\in N$ that $\langle y^p \rangle \neq \langle y \rangle$ and that $|y^p|&lt;|y|$?     Proposition 21.   If $G$ is a finite abelian group and $p$ is a prime dividing $|G|$, then $G$ contains an element of order $p$.      Proof:   The proof proceeds by induction on $|G|$, namely, we assume the result is valid for every group whose order is strictly smaller than the order of $G$ and then prove the result valid for $G$ (this is sometimes referred to as complete induction).   Since $|G| &gt; 1$, there is an element $x \in G$ with $x \neq 1$.   If $|G| = p$ then $x$ has order $p$ by Lagrange’s Theorem and we are done.   We may therefore assume $|G| &gt; p$.      Suppose $p$ divides $|x|$ and write $|x| = pn$.   By Proposition 2.5(3), $|x^n| = p$, and again we have an element of order $p$.   We may therefore assume $p$ does not divides $|x|$.      Let $N = \langle x \rangle$.   Since $G$ is abelian, $N \trianglelefteq G$.   By Lagrange’s Theorem, $|G/N| = |G|/|N|$ and since $N \neq 1$, $|G/N| &lt; |G|$.   Since $p$ does not divide $|N|$, we must have $p \mid |G/N|$.   We can now apply the induction assumption to the smaller group $G/N$ to conclude it contains an element $\bar{y} = yN$, of order $p$.   Since $y \notin N$ ($\bar{y} \neq \bar{1}$) but $y^p \in N$ ($\bar{y}^p = \bar{1}$) we must have $\langle y^p \rangle \neq \langle y \rangle$, that is, $|y^p| &lt; |y|$.   Proposition 2.5(2) implies $p \mid |y|$.   We are now in the situation described in the preceding paragraph, so that argument again produces an element of order $p$.   The induction is complete.      (Original image .) Enumeration within self defined environment My self defined environment having a symbol at the beginning and at the end works fine for text without enumeration. When including enumerate the symbols are not there where desired. Is it possible to have the first symbol directly in front of &quot;1.&quot; i. e. in the same line and the ending symbol in the same line as the last line of the enumeration? With \ignorespacesafterend and \unskip I did not succeed. \documentclass{article} \usepackage{amssymb} \newenvironment{myenv}{$\blacktriangleright$}{\hfill$\blacktriangleleft$}  \begin{document} There is a text.  \begin{myenv} And an example \end{myenv}  Here is another text.  \begin{myenv} \begin{enumerate} \item A \item B \end{enumerate} \end{myenv}  \end{document} How can I make Devanagari fonts render properly in Ubuntu 16.04? Devanagari fonts appear broken like this:     It should appear this way    I have the same problem in browsers and office applications.  Is there any way to fix it? I wonder whether there is any automatic way of determining (at least roughly) the Big-O time complexity of a given function?  If I graphed an O(n) function vs. an O(n lg n) function I think I would be able to visually ascertain which is which; I'm thinking there must be some heuristic solution which enables this to be done automatically.  Any ideas?  Edit: I am happy to find a semi-automated solution, just wondering whether there is some way of avoiding doing a fully manual analysis. Show that if $a^{n-1} \equiv 1 \pmod{n}$ for all $a$ such that $\gcd(n,a) = 1$, then $a^{n} \equiv a \pmod{n}$ for all $a$. In Burton's Elementary Number Theory, when introducing absolute pseudoprimes (also called Carmichael numbers), the author states that they are composite numbers $n$ such that $a^{n-1} \equiv 1 \pmod n$ for all integers $a$ with $\gcd(a,n) = 1$.   When showing that 561 is an absolute pseudoprime, he then concludes that $a^{561} \equiv a \pmod{561}$ for all $a$.   I searched a bit on the internet, and it seems that these two are actually equivalent definitions of a Carmichael number. However I have not been able to find a proof; the fact that the implication in the title of the post is stated so plainly in my textbook makes me think it is quite simple to see, but I can't seem to understand the argument.