When I make an edit and then get the text "This edit will be visible only to you until it is peer reviewed" am I just supposed to guess the URL for the peer review page? Can't you just link me there direct from the question? Am I missing something?
When a user without editing privileges suggests an edit to the post, a banner gets displayed until it is reviewed. Can the banner also provide a link to the Suggested edit? For example, I made an edit to a post on Unix&amp;Linux.SE. Can the banner also link to my ?     I know I can access the link through my Profile → Activity → Suggestions. But linking it directly on the face of the post would have immense benefits:  I have been a prolific editor on Ask Ubuntu (my first encounter with Stack Exchange Network) with a lot of Rep derived through Suggestions. I didn't get to know about these suggested edit links until very late (I guess I already had got about 400 edit suggestions approved by the time this came to my knowledge).  This means I didn't know about  when it mattered the most.  I never got to know why my suggestions got rejected because of my ignorance.  This would help reduce the barrier between Reviewers and Editors.  This would also help in case of Anonymous edits. Right now, there is no way an anonymous editor can access it unless the edit gets approved and he comes back to go through the Revision page of that particular post in order to access the Suggested Edit Review link. If the edit gets rejected - it's the end of the road for that anonymous editor. I just made an anonymous edit on one of the SE sites where I don't have an account and already feel like it's the end of the road for me since I just can't access the review link unless it gets approved.    Possibly also expand the banner to mention that they can also access the link to the suggestion via Profile → Activity → Suggestions.
Why is $(2, 1+\sqrt{-5})$ not principal? Why is $(2, 1+\sqrt{-5})$ not principal in $\mathbb{Z}[\sqrt{-5}]$?   Say $(2,1+\sqrt{-5})=(\alpha)$, then since $2\in(2,1+\sqrt{-5})$ we have $2\in (\alpha)$, so $\alpha\mid2$ in $\mathbb Z[\sqrt{-5}]$. Writing $2=\alpha\beta$ in $\mathbb Z[\sqrt{-5}]$ and taking norms, $4=N(\alpha)N(\beta)$ in $\mathbb Z$, so $N(\alpha)\mid4$ in $\mathbb Z$. Similarly, since $\sqrt{-5}\in(\alpha)$ we get $N(\alpha)\mid5$, thus $N(\alpha)$ is a common divisor of $4$ and $5$, therefore $N(\alpha)=(1)$, so $\alpha$ is a unit. But that means $1\in (2,1+\sqrt{-5})=(\alpha)$, so it must be the whole ring, but it cannot to reach a contradiction, so how can I find an element, which is not in $(2, 1+\sqrt{-5})$.  Or is there a simpler method ? (Maybe not UFD would imply not PID) Suppose I have a program named any_program.exe and my operating system drive is C:. The location of the program is D:\Any_Folder\any_program.exe  How do I start/execute that program via command prompt in Windows 8?  I have tried the command line START any_program.exe, but it shows me an error that      Windows cannot find 'any_program.exe'. Make sure you typed the name correctly, and then try again.   By the way, it worked perfectly in Windows 7. And, if I type START notepad.exe or START firefox.exe (Firefox is not installed in C: drive), it works in Windows 8. Leaving the States on Visa Waiver and coming back I'm travelling on the ESTA visa waiver, I'm now in the States but would like to go to Costa Rica for a bit (say three weeks) then come back to the States. All will be within my 90 day limit.  Is that allowed?  (I have my onward ticket home.) "Misaligned partition" - Should I do repartition (how?) how I can continue this limit resolution?  The limit is:   $$ \lim_{n \to \infty}{\sqrt[n]{\frac{n!}{n^n}}} $$  This is that I have done:  I apply this test: $ \lim_{n \to \infty}{\sqrt[n]{a_n}} = \frac{a_{n+1}}{a_n} $  Operating and simplifying I arrive to this point: $$ \lim_{n \to \infty}{\frac{n^n}{(n+1)^n}} $$  I've done something wrong? Thanks! ${!FOO} performs a double substitution in bash, meaning it takes the (string) value of FOO and uses it as a variable name. zsh doesn’t support this feature.  Is there a way to make this work the same in bash and zsh?  Background:  I’ve got a list of environment variables, like  PATH MAIL EDITOR   and want to first print the variable names and afterwards their values.  This works in bash but not zsh:    for VAR in LIST do         echo $VAR         echo ${!VAR} done   It should be somehow possible “the old way” with eval, but I can’t get it to work:  for VAR in LIST do         echo $VAR         echo `eval \$$VAR` done   I’m never going to understand why I can’t simply do arbitrary deep substitutions like ${${VAR}} or even ${${${VAR}}} if need be, so an explanation for that would be nice, too. A question about causality and Quantum Field Theory from improper Lorentz transformation Related post   In Peskin and Schroeder's QFT p28, the authors tried to show causality is preserved in scalar field theory.  Consider commutator $$ [ \phi(x), \phi(y) ] = D(x-y) - D(y-x) \tag{2.53} $$ where $D(x-y)$ is the two-point correlation function,  $$D(x-y):= \langle 0 | \phi(x) \phi(y) | 0 \rangle  = \int \frac{d^3 p}{ (2\pi)^3} \frac{1}{ 2E_{\mathbf{p}}} e^{-ip(x-y)}\tag{2.50}$$  P&amp;S argued that each term in the right-hand-side of (2.53) is Lorentz invariant, since $$\int \frac{d^3p }{ (2\pi)^3} \frac{1}{2E_{\mathbf{p}}} = \int \frac{ d^4 p }{ (2\pi)^4} (2\pi) \delta(p^2-m^2)|_{p^0&gt;0} \tag{2.40}$$ is Lorentz invariant.  Since there exists a continuous Lorentz transformation in the spacelike interval $(x-y)^2&lt;0 $ such that $(x-y) \rightarrow - (x-y) $ and $D(y-x)=D(x-y)$, (2.53) equals zero in the spacelike interval. In timelike interval, since such continuous Lorentz transformation does not exist, (2.53) is non-zero in general.  My question is, consider a non-continuous Lorentz transmation in the timelike interval, $PT$, namely time reversal times parity transformation. I can also let  $(x-y) \rightarrow - (x-y) $. Why (2.53) in the timelike interval is non-zero?  I guess $PT$ will let (2.40) go to $p^0&lt;0$ branch. But I am not sure if it breaks the Lorentz invariant of (2.40) and (2.50). Motivation behind the definition of ideal class group Using awk to sum the values of a column, based on the values of another column I am trying to sum certain numbers in a column using awk. I would like to sum just column 3 of the "smiths" to get a total of 212. I can sum the whole column using awk but not just the "smiths". I have:  awk 'BEGIN {FS = "|"} ; {sum+=$3} END {print sum}' filename.txt   Also I am using putty. Thank you for any help.  smiths|Login|2 olivert|Login|10 denniss|Payroll|100 smiths|Time|200 smiths|Logout|10 Upgrading straight handlebars to drop handlebars What are the rules for animal mating? Cyclic Group Generators of Order $n$ How many generators does a cyclic group of order $n$ have? I know that a cyclic group can be generated by just one element while using the operation of the group.  I am having trouble coming up with the generators of a group of order $n$.  Any help would be great!  Thanks! HTML 5 Video stretch How to run a cron job using the sudo command Is it possible to run a  job which needs the  command?  Like:   sudo rm somefile Network reputation graph becomes nervous when clicking quickly, backs away slowly I'm using the built in Screen Sharing tool in El Capitan to control an iMac from a Macbook Pro in full screen mode. Everything works beautifully except that many special keys and gestures are intercepted by the local Mac and not sent to the shared Mac. Specifically:   Ctrl-Arrow (Desktops and Mission Control) F3 (Mission Control) F4 (LaunchPad) Three finger swipe (Switch between Desktops, Mission Control, Exposé) Two finger swipe from left side (Notification Center) Pinch with four fingers (Launchpad)   I can't find any way to switch between desktops on the remote machine other than manually opening the Mission Control app from the Finder or Dock. Disabling the gestures on the local machine doesn't help; they still don't get sent to the remote machine.  Is there any way to get special keys and gestures passed to the shared machine? Could anybody clear it to me whether these two sentence are grammatically equal? If not, what is the difference?     1) They must use some methods...    2) They are to use some methods... Dismiss flags on the same posts with different validities Sometimes a post has both valid flags and invalid flags. As far as I can tell, all the flags on the same post have to be dismissed with the same validity.  The scenario I encountered (on ):   A low-quality question is closed. The asker improves the question and flags for reopening. A moderator reopens the question, but doesn't dismiss the flag yet. Someone else flags the question as “off topic”.   Since I agree with the reopening, and I don't consider the question off-topic, I would like to dismiss the first flag as valid and the second flag as invalid. But there's only one set of buttons for both flags.  And just now, two flags on a closed question: one that says “delete it”, the other one that says “reopen it”. The correct course of action is neither (the ideal outcome would be for the question to be edited then reopened). I should provide different feedback to the two flaggers, but I can't.  Unless I've missed something, please allow multiple flags on the same post to be dismissed separately. problem with using loop inside the tabular environment Why does Ridders' method work as well as it does? I've just read section 9.2.1 in Numerical Recipes Ed. 3 (Press et al. 2007), which describes Ridders' method of root finding. I understand that allowing for some curvature of the function by linearizing three points with an exponential will probably get you a better approximation and more rapid convergence, but I have no idea how they get that it converges quadratically, and the motivation behind choosing an exponential (rather than e.g. a polynomial) is not clear to me.  So, specifically, I'm wondering (1) why linearize with an exponential, and (2) why the method converges quadratically.  I tried looking for Ridders' original paper "A new algorithm for computing a single root of a real continuous function", but the only copy I could find was in a journal that my school's library does not give me free access to.  UPDATE:  I've realized that the fact that an exponential is strictly positive means that weighting the three points with an exponential won't change their signs. Hence, the root of the linearization will have to land inside the original bracket.  Polynomials, rationals, sinusoids, and logarithms all have positive and negative bits, so using them probably runs the risk of getting bad interpolated roots. One of these squared might be usable, but maybe non-monoticity gives some sort of undesirable effect. Since this is a complete list of all simple functions that I am aware of, exponentials seem to be the only easily available weighting function with this property.  A constant function is also technically single-signed and monotone, but it can't be used to linearize stuff.