Why can't I focus a textarea using buttons loaded in with AJAX?
Event binding on dynamically created elements?
$L^p$ spaces and counting measure currently I am working on the following two exercises as a revision for my exam.  Let $\mu$ be the counting measure on $\mathbb N$.   Show that if $1 \le p &lt; s &lt; \infty$ then $f \in L^p$ implies $f \in L^s$ and $\|f\|_s \le \|f\|_p$. Show that if $f \in L^p$ for some $p \ge 1$ then $\lim_{s \to \infty} \|f\|_s = \|f\|_\infty$.   My attempts so far:   It is $f \in L^s$ iff $\sum_{k = 1}^\infty |f(k)|^s$ converges. Since $f \in L^p$ we know that $\sum_{k = 1}^\infty |f(k)|^p$ converges. Thus for any $\varepsilon &gt; 0$ there is $N \in \mathbb N$ s.t. \begin{align*} k \ge N \Longrightarrow |f(k)|^p &lt; \varepsilon \Longrightarrow |f(k)| &lt; \varepsilon^{1/p} \end{align*} We have $\forall k \ge N$ that $|f(k)|^s &lt; \varepsilon^{s/p}$. Originally I wanted to use the comparison test now, showing that \begin{align*} \sum_{k = N}^\infty \varepsilon^{s/p} &lt; \infty, \end{align*} but that does not work since \begin{align*} \sum_{k = N}^\infty \varepsilon^{s/p} = \varepsilon^{s/p} \underbrace{\sum_{k \ge N} 1}_{= \infty},  \end{align*} so I don't know how to go on here. Let $K := \|f\|_\infty$. We need to show that $\forall \varepsilon &gt; 0$ $\exists M \in \mathbb N$ s.t. \begin{align*} s \ge M \Longrightarrow \left|\left(\sum_{k = 1}^\infty |f(k)|^s\right)^{1/s} - K\right| &lt; \varepsilon. \end{align*} Choose $M$ s.t. $M &gt; p$. Then $s &gt; p$ and from part (i) we know that \begin{align*} \left(\sum_{k = 1}^\infty |f(k)|^s\right)^{1/s} \le \left(\sum_{k = 1}^\infty |f(k)|^p\right)^{1/p}. \end{align*} But how can I use this? In preparation for the eventual deployment of full HTTPS support on Stack Exchange, and also to make the currently available "alpha" HTTPS support more usable, all links from Stack Exchange sites to other Stack Exchange sites should be .  Currently, at least the following places still feature links with explicit http: protocol prefixes:   the page footer, the "Hot Network Questions" and "Community Bulletin" modules in the sidebar,  (the site switcher menu is OK, though),   and, of course,   lots and lots of user-supplied links in post and comment content.   I've posted this feature request with the intent to provide a single central place for tracking these issues, rather than having them scattered across multiple meta threads.  If you find more places on SE where protocol-relative links should be used, please report them below.  I fully expect this request, in its full sense, to be treated as  for the time being; however, these issues do need to be fixed at some point before our HTTPS support is ready for "prime time".  Ps. Yes, the next release of  (v1.10) is going to include a general client-side fix for most of these issues.  In the mean time, the  plugin effectively provides the same functionality, and much more besides. The space $c_0$ of sequences that converge to 0 is not reflexive When Frodo put on the One Ring, both the Nazgul and Sauron himself found Frodo. However when Bilbo had the ring and possibly used it many times, Sauron sought Gollum to get a hint on where to look for the ring and send the Nazgul after it.   Sauron sought Gollum before Bilbo wore the ring at his birthday party, slightly after that time, he was very effective at locating Frodo when Frodo put on the ring. So I guess it's not a matter of Sauron gaining any extra powers with time.  Also Bilbo seemed to use the ring without it bothering him at any point, unlike Frodo who seemed to be very troubled when using the ring. How is that possible? Do dual American and Mexican citizens need passports to travel between the US and Mexico? I have 2 children that were born in Mexico. They have a birth certificate abroad. So this is dual citizenship. Do they need passports?  My kids were born in Durango, Mexico. I am American. Their father is Mexican. They were both given birth certificates abroad. They are now 16 and 14. They want to go visit grandparents and aunts and uncles in Mexico.  Will they need to get a passport to enter Mexico?  Will they need it to fly? English equivalent for the Persian proverb "The mountain just gave birth to a mouse" I'm looking for an idiom or expression to describe a well-known person/ organization/ politician/ government  whose achievements in a given situation are smaller than what they had claimed or promised to be.  We Iranians have a proverb that literally means      The (roaring) mountain just gave birth to a mouse!    Its  etymology is like this:  Once a volcanic mountain started roaring, hissing, and bursting bubbles. The people were waiting to see what would happen after all those huge roarings and loud booms. Finally, the volcano activity stopped and the people saw just a tiny mouse came out of there! (rather than a monster that was expected to come). So they said mockingly:     "Heh! After all those huge roarings, the (big) mountain just gave birth to a (tiny) mouse!"    Notes:   The (roaring) mountain -symbol for an apparently big(=important) and strong person-here is figured like a woman who is screaming and shouting while is delivering her baby. We use this proverb for mocking an important person/organization/ government, etc ironically when they have used broad advertisements or have spent much money or resources over doing something noticeable, but the result/ achievement is less noticeable than it was claimed or promised to be.   Is there any equivalent for this Persian proverb?  I just found ; would it convey the same connotation? How to parse data in JSON format? Is Adzerk safe / legitimate? Since 17.10 I got familiar with Wayland and I really need the per-monitor-scaling!  How can I start a wayland session after dist-upgrade to Bionic? There is no option in GDM available out-of-the-box.  I do not see what is described as     The Wayland session will still be available, pre-installed Let $A \subset B$ be a finite ring extension. Fix a ring homomorphism $\nu : A \to \Omega$, where $\Omega$ is an algebraically closed field.   I want to show there exists a non-zero homomorphism $v : B \to \Omega$ of $A$-modules.  I want show that the space $V$ of all the homomorphisms $v$ in (i) is a non-zero finite dimensional vector space over $\Omega$. Show that the algebra $B$ acts on this space. Show that eigenvectors of this action correspond to homomorphisms of algebras $\lambda : B \to \Omega$ that extend the homomorphism $\nu$. Deduce that the set of such homomorphisms $\lambda : B \to \Omega$ is finite and non-empty.    We can assume that $B = Ab_1 + ... + Ab_n$ where $b_1,...,b_n \in B$. The chose $v(b_i)=\omega_i \in \Omega-\{0\}$ and then define $$ v( a_1 b_1 + ... + a_n b_n ) = \nu(a_1) v(b_1) + ... + \nu(a_n) v_n = \nu(a_1) \omega_1 + ... + \nu(a_n) \omega_n \ .$$ This is an $A$-module homomorphism. Am I correct? For the second part, that $V$ is not-zero follows from the first part. That it is finite dimensional follows from the fact that $v_i(b_j)=\delta_{i,j}$ (Kronecker's delta) with $i=1,...,n$ spans $V$ and hence $\dim V \le n &lt; \infty$. I define the action of $B$ over $V$ to be $$b \bullet v = \left( x \mapsto v(b)v(x) \right) \mbox{  for all } b,x \in B \ .$$ This seems to be the natural action to define. Am I correct? However, I haven't managed to solved the rest (the eigenvector stuff and the relation to $\nu$). prove that : $\cos x \cdot \cos(x-60^{\circ}) \cdot \cos(x+60^{\circ})= \frac14 \cos3x$ I should prove this trigonometric identity. I think I should get to this point : $\cos(3x) = 4\cos^3 x - 3\cos x $ But I don't have any idea how to do it (I tried solving $\cos(x+60^{\circ})\cos(x-60^{\circ})$ but I got nothing) According to the special theory of relativity, distant simultaneity depends on the observer's reference frame.  And, according to the quantum theory, in the case of two entangled particles, a measure on one of the particles simultaneously affects the second one. Under which reference frame is this simultaneous? Filling HTML  dropdown list in JSP with values fetched from database in Servlet Proof for $A \cup B = B$ if and only if $A \subset B$ I am trying to define a proof for  $A \cup B = B$ if and only if $A \subset B$  I started with:  $$A \cup B$$ $$x \in A \lor x \in B$$   but I am unsure on how to continue. Alternative ways to say "if and only if"? There are some scenarios about which I would like to get some confirmation:   when defining a concept A,      We call A, if ... [definition of   concept A]   Does "if" here mean equivalence instead of just sufficiency? Is it incorrect to replace "if" with "if and only if"?    For precisely what condition is    B satisfied?   Does "precisely" mean asking for necessary and sufficient condition for B? Some other ways to say "if and only if" / "necessary and sufficient"?   Some references that summarize some standard terminologies in Mathematics  such as this one?  Thanks and regards! Residue theorem with $\frac{1}{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) d\theta$. The following integral should be doable using the residue theorum:     $$\frac1{2\pi}\int_{0}^{2\pi}e^{\cos\theta}\cos(n\theta) \,d\theta$$ How do you determine the winner of a hand? In Texas Hold'em poker, there are some cases where a particular card, often called kicker, acts as a tiebreaker between players to determine who wins the pot, or if the pot has to be shared.  For some combinations such as three or four of a kind, there is no doubt: the kicker is the 5th card of the chosen combination, and determines the winner.  Example     Alice has 7 and 10   Bob has 7 and K   On the board are 7, 7, 2, 3, 4   In that case, both have a three of a kind. But because Bob has a king and Alice only a 10, he wins.  Most of poker websites explaining the rules clearly mention the role of the kicker applying to three of a kind and double pairs. But much fewer say something about colors, straights and full houses, and I found many contradicting answers. I also asked the question to some used online poker players and they also gave me contradicting answers.  I have tried to search for official tournament rules, but most of them only explain what happens with bad behaviors, bad deals, incorrect or confusing betting, showing cards when you shouldn't, acting when it's not your turn, etc. without mentioning subtleties about combinations at all.  To simplify my question, I will take three examples; I think it's better to start with examples before getting to the general answer if one exists. So, what's happening in the following 3 examples? Could you give a more generalized answer?  Example 1 - Flushes     Alice has 2&clubs; and 3&clubs;   Bob has 4&clubs; and 5&clubs;   On the board are 6&clubs;, 8&clubs;, 10&clubs;, 2&hearts;, 5&diams;   Contradicting answer 1: it's a tie, because the highest card included in the flush is the 10, which everybody chooses to include in their 5 showdown cards. Contradicting answer 2: Bob wins, because he has the greatest private card that is part of the flush  Example 2 - Straight     Alice has 6 and K   Bob has 6 and J   On the board are 4, 5, 7, 8, 10   Contradicting answer 1: it's a tie, because the greatest card in the straight is the 8 for both Alice and Bob Contradicting answer 2: Alice wins, because she owns an extra king, compared to the jack of Bob  Example 3 - Full house     Alice has 3 and 7   Bob has 3 and 6   On the board are 3, 3, 2, 2, 5   Contradicting answer 1: it's a tie, because one is supposed to choose only five cards to make a combination, and a full house is already five cards. There couldn't be any kicker, and thus their showdowns are strictly equals. Contradicting answer 2: Alice wins because of her extra 7, compared to the 6 of Bob  Thank you very much for your answers.  Note: I'm unable to post next to you, so I edit my own post; strange not be able to answer to an answer.  Ok, So if I summarize what you are saying :   In the flush case, Bob wins because at some point, their hand differs. Technically, I can continue comparing the highest private card to decide who wins. I had it correct. In the straight case, if the highest card of the straight is public, then it's technically always a tie, no matter what the players had as second private card (asuming that only one of the two cards was part of the straight). In a full house if both players have the same triplet and the same pair, it's always a tie, no matter what the players had as second private card (assuming again that only one of the two was part of the full house).   Please confirm me that what I summarized is correct.  Thank you for your answer. Is there a closed form expression for $\text{n}^{\text{th}}$ coefficient of Maclaurin series of  $$\ln(1-x+x^2)$$ What are nodes, and how do I use them? I'm a bit new to Blender, and in trying to learn more, I've stumbled across nodes.  What are nodes, and how do I use them? How do I get the size of a directory on the command line? I tried to obtain the size of a directory (containing directories and sub directories) by using the ls command with option l. It seems to work for files (ls -l file name), but if I try to get the size of a directory (for instance, ls -l /home), I get only 4096 bytes, although altogether it is much bigger.