I have to admit I didn't really search in-depth for this feature, but is it possible to "Bookmark" or "Star" answers like in Gmail?
Lots of questions are very generic, &quot;share an experience&quot; kinds of questions. In these cases, I am quite often interested in specific answers that may not have gotten that many upvotes. These questions also tend to elicit lots of responses, so going back to find the answers I'm interested in somewhere in the middle of a sea of responses is a pain. Is there a way to favorite or bookmark these answers?
I understand that the letter "J" is relatively new — perhaps 400–500 years old.  But since there has long been important names that begin with J, such as Jesus, Joshua, Justinian, etc., and which predate the introduction of a special letter, does that mean that the "J" sound predated the letter, or were such famous names spelled and pronounced differently? Does "minor revision" mean "accepted with minor revision"? I just received a decision letter from a journal editor (after three rounds of major revision). The letter shows that the reviewers "have asked for some other minor revisions". After getting the minor revision letter, I immediately completed the revision in accordance with the feedback from the reviewers (the suggested changes just involve adding a few sentences) and immediately resubmitted the revision to ScholarOne. A week has passed since I submitted the revision, but the status in ScholarOne remains unchanged. Is this normal? Can I assume that this basically means "acceptance with minor revision"? Thanks for your insight and sharing. $$\lim\limits_{x\to 1}\frac{\sqrt x-x^2 }{1-\sqrt x}$$  I'm not entirely sure how to go about this, since rationalizing anything doesn't quite work for me. Just looking for some help. SQL Join Differences What's difference between inner join and outer join (left join,right join), and which has the best performance of them?  Thanks! How can I modify my hot sauce to make it Shelf Stable? I would like to be able to can and then sell my own hot sauce. The hot sauce I make is a low acid meat (ground beef) and tomato based sauce. I had sent it in to get a shelf life expectancy, and was advised the acidity level was too low and the only way I could make the sauce in a way that it was shelf stable was if I used a retort or sold it as a frozen product. Neither of these are options I can consider at this time. How to find windows 10 activation Key Where can it be find?  At activation panel i can only find option to change the activation key. The current key is not presented on this screen. Can I use Named Credential merge fields as URL parameters on the endpoint? I have a named credential as follows:  Name   Test_Credential URL    https://www.somesite.com Identity Type    Named Principal Authentication Protocol    Password Authentication Username   foobar Password   mySomeSiteDeveloperKey Generate Authorization Header   FALSE Allow Merge Fields in HTTP Header    TRUE Allow Merge Fields in HTTP Body     TRUE   The service that I am trying to authenticate against uses URL parameters to pass in the developer key and username. Is there any way to use  to pass those in to the URL?   The Named Credential attributes seem not to get decoded until the request is actually sent, as debugging the endpoint, or headers if I set them per the linked doc, show the un-merged data. I can get the callout to go to the right place, and return the expected error because the auth information is not provided. Anybody know of a "show me my HTTP request" site that I could point to to get a better idea of how this works?  I've tried both of these:  req.setEndpoint('callout:Test_Credential/api/action?id={!$Credential.Password}&amp;account={!$Credential.UserName}');  req.setEndpoint('callout:Test_Credential/api/action' + '?id=' + '{!$Credential.Password}' + '&amp;account=' + '{!$Credential.UserName}');   (Edited to remove the part about $Credential vs. $Test_Credential as I was able to determine that it is the former, and see the merged creds by calling out to .) Question:  Let ${a_n}$ be a sequence in $\mathbb{R}$ with $\lim_{n\to\infty}a_n=a$  Prove that  $\lim_{n\to\infty}a_n^2=a^2$  Attempt: Without using any properties of limits (this is a question in a section of the book that came before any list of any limit theorems)  By definition:  $|a_n^2-a^2|&lt;\epsilon$   $|a_n^2-a^2|$ $\leq$ $|(a_n+a)(a_n-a)|$ $\leq$ $|(a_n+a)| |(a_n-a)|$ $\leq$   $(|a_n | + |a|)(|a_n -a|)$ $&lt;$ $\epsilon$  And since I know that $\lim_{n\to\infty}a_n=a$, I'll subsitute that in for $|a_n|^2$ as $a^2$   $\therefore$ $a^2-|a|^2 &lt; \epsilon$ which is true since $a^2-|a|^2$ in my case.  I wonder if this is a possible solution? What is the earliest work considered to be Science Fiction? Updating my appointment without cancelling the previous date If a function has a finite limit at infinity, does that imply its derivative goes to zero? I've been thinking about this problem: Let $f: (a, +\infty) \to \mathbb{R}$ be a differentiable function such that $\lim\limits_{x \to +\infty} f(x) = L &lt; \infty$. Then must it be the case that $\lim\limits_{x\to +\infty}f'(x) = 0$?  It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy:  $$ \begin{align} \lim_{x \to +\infty} f'(x) &amp;= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\ &amp;= \lim_{h \to 0} \frac1{h}(L-L) \\ &amp;= \lim_{h \to 0} \frac{0}{h} \\ &amp;= 0 \end{align} $$  In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is  already, but I think this is different in two aspects. First, that question assumes that $\lim\limits_{x \to +\infty}f'(x)$ exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case. What I have now is:  /testfor @a {SelectedItemSlot:0,Inventory:[{Slot:0b,id:"diamond_pickaxe"}]}   How can I use the /testfor command to test a specific slot for a specific item that is enchanted? What word (or how do you phrase things) do you use when the ostensible word for the class is the same as the word for a subset of the class?  For example, in the United States, there are many brands of sugared, carbonated soft drink that one can buy: Coca-Cola (or Coke), Pepsi, RC Cola, Sprite, 7-Up, A&amp;W or Mug Root Beer, etc). The first two are kinds of 'coke' or 'cola' (but no one says 'cola').  , one asks for any type of such drink as 'coke'. How does one then ask for the particular kind of 'coke' that is 'Coke' (pronounced the same)? Does one reduplicate and say a 'Coke coke', does one repeat with emphasis (or without), does one use a different word like 'coca-cola' (co-cola), or what?  The (perceived) difficulty (surely those in the South are able to get the drink they want somehow) isn't limited to there. Where the drink is called 'soda', if you ask for a 'coke', you'll still sometimes have the choice among Coca-Cola, Pepsi, etc. "Can I have a Coke?" the response might be "Is Pepsi OK?". "Sure, Pepsi, Coke, whatever." You don't use 'cola', you just ask for the particular kind and hope the listener will help differentiate.   Given these multiple strategies for disambiguating, what do you do say to get a 'Coke' in the southern US? are there any other examples of a hypernym being the same as the hyponym, and so creating a disambiguation problem? How can I move all the window controls to the right (or left)? Downloaded Ubuntu tweak and set window controls to the right side. Only some programs follow the rule. Pidgin as an example has window controls on the right. But Chrome, Thunderbird, Geany, still have controls on the left.  How can I move ALL window controls(close/min/max) to the right (or left)? Does GCM (or GHASH) only provide 64-bit security against forgeries? In a  a doubt was voiced about my answer, which claims GCM to requires $2^{128}$ for a successful forgery. The doubt was that the square root needs to be taken meaning the security would be $2^{64}$.  So of course I immediately checked the relevant security theorem ():  $$\mathbf{Adv}^{\text{auth}}_{\operatorname{GCM}[\operatorname{Perm}(n),\tau]}(\mathcal A)\leq \frac{0.5(\sigma+q+q'+1)^2}{2^n}+\frac{q'(\ell_A+1)}{2^\tau}$$  with $\sigma$ being the total plaintext size in blocks, $q$ being the total number of encryption queries, $q'$ being the total number of decryption queries, $n$ being the size in bits of the underlying permutation, $\ell_A$ being maximum authenticated input length in blocks and $\tau$ being the tag size in bits.  Now clearly from this we can see that performing $2^{n/2}=2^{64}$ queries yields a strong enough advantage for a break.  Now my question is: When talking about "n-bits of security" or "needs $2^n$ operations to break", do we usually talk about "online" security, ie queries to an oracle can be done and cost $1$ or "offline" security where no oracles are available or do we need to make this decision depending on the situation?  TL;DR: Does GCM offer 64 or 128 bit security? I have seen this a lot but i ran into this question today:    And the answer (while good) has 83 upvotes which seemed a little over the top.  I see dozens of answers of this caliber every week that only seem to get a few upvotes.    Was there some sort of SO policy that encouraged people to upvote more during this period?  Am I just imagining that answers from this period received more votes than comparable ones now? AES Mix Column Transformation Section 3.4.3 "The MixColumns Step"(page 39) of , it states     The columns of the state are considered as polynomials over   $\operatorname{GF}(2^8)$ and multiplied modulo $x^4 + 1$ with a fixed polynomial $c(x)$.      The polynomial $c(x)$ is given by   $c(x) = \mathtt{03} \cdot x^3 + \mathtt{01} \cdot x^2 + \mathtt{01} \cdot x + \mathtt{02}$      Let $b(x) \equiv c(x) \cdot a(x) \pmod{x^4 + 1}$. Then         How this polynomial multiplication modulo $x^4+1$ is converted to Matrix Multiplication? If $f^2$ is an analytic function then so is $f$ I want to show the following:  If $f(z) $ is a continuous  function on a connected open subset of the complex plane and $f(z)^2$ is an analytic function, then $f(z)$ is analytic.  Clearly if $f(z) \neq 0$ then $$\frac{f(z+h)-f(z)}{h}=\frac{f(z+h)^2-f(z)^2}{h}.\frac{1}{f(z+h)+f(z)}$$  which shows $f^\prime(z) $ exists if $f(z)\neq0$.  What do I do when $f(z) =0$? Sorry, I'm still a newbie to blender, but why are some edges visible only in edit mode, but not in wireframe view? is it a bug or am I missing something here?  edit mode: Edges visible. Object mode, wireframe view: Pooff, they're gone. what can i do to fix this? / thanks Coconut Water from White vs Brown Coconuts