So this is pretty minor in the grand scheme of things. But, well it's still a bug.    The "add comment" link is correctly greyed out, but it is accompanied by a "requires 50 reputation" link that &mdash; at 83,921 rep &mdash; I've never seen before.  Presumably this shows up pre-login? If so, it's now started showing up for everybody on deleted questions.
So I recently saw add comment (requires 50 reputation) on a deleted answer. After looking around a bit, it seems to be on all deleted posts now (only deleted posts, normal posts still say just add comment).  I'm sure it wasn't there before (I think it just said add comment (unclickable) or possible nothing).  It is also on , presumably the root cause is the same.  And yes, I do have more than 50 reputation (in case anyone was wondering).  Just to clarify, requires 50 reputation obviously shouldn't be there for users with 50 or more reputation. Choices:   Display nothing Unclickable add comment - probably not ideal Unclickable add comment (disabled for deleted posts) or something along those lines   FYI - You may be somewhat confused right about now if you can't  yet.
${\rm Aut}(G)$ is cyclic $\implies G$ is abelian I would appreciate if you could please express your opinion about my proof. I'm not yet very good with automorphisms, so I'm trying to make sure my proofs are OK. Proof: Since ${\rm Aut}(G)$ is cyclic, ${\rm Aut}(G)$ is abelian. Thus for any elements $\phi, \psi \in{\rm Aut}(G)$ and some elements $g_i \in G$, $\phi\psi(g_1g_2)=\phi\psi(g_1)\phi\psi(g_2)=g_3g_4=\phi\psi(g_2)\phi\psi(g_1)=\phi\psi(g_2g_1)=g_4g_3$. Hence, $g_3g_4=g_4g_3\implies G$ is abelian. The proof seems to be quite straightforward, but I'd rather ask for advice. RegEx: Grabbing values between quotation marks in my earlier question you can find at:   i describe my error:     I'm using an Asus Zenbook UX562F with an integrated GTX 1050 Max-Q.   After the installation, I installed Nvidia-390 and updated the kernel   with ukuu to 4.20.17 to fix no sound issue.      After the reboot, everything went well and I did:      apt-get update apt-get upgrade + J      All was done well, so I restarted and on the next boot after selecting   Ubuntu in GRUB it gave me a long black screen (violet screen) where   nothing changes after so much time.      I tried to add nomodeset and noveau.modeset=0 single and together but   didn't help so I don't know what to do.   After investigating deeper by installing packages and try to boot i found out that the package which is causing the error is:  intel-microcode   when i purge and block this package with:  sudo apt-get purge intel-microcode sudo apt-mark hold intel-microcode   before upgrading wuth  sudo apt-get upgrade   my system is able to boot normal.  The problem i see is that as i understand it the package protects my intel cpu  Intel® Core™ i7 8565U 1.8 GHz   is vulnearable to meltdown attacks which get fixed by the package.  What can i do to fix the vulnearbility and also get my system to a bootable state?  Thanks for your help Vimu  Edit: I used the manual update technique mentioned at (in german):    and i used the files from the intel github repo:    with the following code mentioned in the tutorial to update the firmware microcode files  sudo cp intel-ucode/06-8e-09 /lib/firmware/intel-ucode/ echo 1 | sudo tee /sys/devices/system/cpu/microcode/reload   i overwrote the old microcode firmware with the newest available on the intel repo  it leada to a bootable system  will this be the best possible fix without upgrading the intel-microcode package provided by apt? Setting up LVM Snapshot as a backup/restore point in ubuntu. Why? Well, as I see it, with all the great usefulness of Linux, its biggest drawback is with its mixed up way of installing apps. All the dlls go to one directory. All the configuration files to another. And the binaries are also put in a bunch. Yes, Windows works the same way more or less, but with the nature of system/server software installed on Linux, being that most of the software installed is only about 95% mature at any given time, messing up your system twice a month is very probable.. Of course, we can backup the system before and after each installation, but that takes a million years. System backups (as oppose to just data/files backup) requires you to turn off you computer and use some form of bootable live CD and a backup media. And, if the drive is more than 15% full, it might take a long time to do the backup, even with modern hardware. So, how can this be done? How do I change the location of macOS screen captures? As a developer, I take a lot of screen captures. I often take screen shots to share with Stack Exchange members in chat.  The problem is my Desktop is so cluttered with screen captures that they're now overlapping! I'd like to save all screenshots in a dedicated folder or directory.  In macOS, how do I change the default location where screen captures are saved?  These are the screen shots taken with CommandShift3 or CommandShift4 keyboard shortcuts.  I would prefer a solution which doesn't require installing any software. Microsoft Edge keeps taking over PDF file open/association Does the integral $\int_{1}^{\infty} \sin(x\log x) \,\mathrm{d}x$ converge? I am working on vector auto-regression (VARs) and impulse response function (IRFs) estimation based on panel data with 33 individuals over 77 quarters.  How should this type of situation be analyzed?  What algorithm's exist for this purpose?  I would prefer to conduct these analyses in R, so if anyone is familiar with R code or a package designed for this purpose that they could suggest, that would be especially helpful. Flames with no gravity? I was watching  today, and noticed this: in some moment the space station changed orbit and the people inside experienced zero-gravity. At that moment, a candlestick passed floating in the air, with the candles burning ().   But in our environment the flames go upward because the hot air is less dense, and that effect should disappear when no gravity is present, am I right? How would a candle flame look then? C Scanf suddenly stopped reading in values A geometric reason why the square of the focal length of a hyperbola is equal to the sum of the squares of the axes? When I teach conics, I give a simple geometric argument for the Pythagorean relation for ellipses. (By "Pythagorean relation" I mean the fact that the square of the focal length $c$ is the difference of the squares of the semimajor and semiminor axes, i.e. $c^2=|a^2-b^2|$ for the ellipse defined by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.) The picture argument goes as follows: just draw the obvious isosceles triangle from the two foci to one of the two points at the end of the minor axis. We see two right triangles; in each, one leg is the focal length $c$ and the other leg is the semiminor axis, while the hypotenuse is the semimajor axis. The result follows from the Pythagorean theorem applied to this right triangle.  But when it comes to the Pythagorean relation for hyperbolas, I can't think of an equally convincing picture to draw. Instead I have to give a rather inelegant algebraic argument for the relation $c^2=a^2+b^2$ (for the hyperbola defined by $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, with focal length $c$), directly from the definition of the hyperbola. The messy details are below.*  Now, obviously there is an asymmetry between the argument for the ellipse and the argument for the hyperbola: the ellipse argument is almost trivial and doesn't involve working straight from the locus definition, but the hyperbola argument is tedious and does require working straight from the definition. So, my question:     Is there an equally persuasive picture I could draw to convince high   school students of the Pythagorean relation for hyperbolas -- without having to get technical with points at infinity, say?   The natural triangle to draw is the right triangle with legs from the center to a vertex and from that vertex to an asymptote, which has legs of length $a$ and $b$; but then the problem becomes why the length along the asymptote should be equal to $c$. I doubt there is a purely (Euclidean) geometrical reason because this triangle lies on the asymptote, not on the hyperbola itself; I suspect some sort of limiting procedure will be required.    [*] The tedious algebraic details:     The foci lie at $(\pm c,0)$.   Say $a&gt;b$. Then hyperbola is the locus of points $(x,y)$ such that       $$\left|\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}\right|=2a$$      Assume for simplicity the expression between the absolute value bars   is positive. (The algebra is even more tedious if we don't make such a   simplifying assumption.) Then moving the right root to the other side   and squaring gives      $$(x+c)^2+y^2=4a^2+(x-c)^2+y^2+4a\sqrt{(x-c)^2+y^2}$$      Simplifying gives      $$xc-a^2=a\sqrt{(x-c)^2+y^2}$$      and upon squaring again we have      $$x^2c^2-2xca^2+a^4=a^2(x^2-2xc+c^2+y^2)=a^2x^2-2xca^2+c^2a^2+a^2y^2$$      which reduces to      $$x^2(c^2-a^2)-y^2a^2=a^2(c^2-a^2)$$      Defining $b^2:=c^2-a^2$ brings the equation into the form   $x^2/a^2+y^2/b^2=1$, and the Pythagorean relation follows from the   definition of $b$. Without using any non-standard (Windows included) utilities, is it possible to download using the Windows command line?  The preferred version is Windows XP, but it's also interesting to know for newer versions.  To further clarify my question:   It has to be using HTTP The file needs to be saved Standard clean Windows install, no extra tools   So basically, since everybody is screaming , I want simple Wget functionality, without using Wget. Amy: What day is it today?    Tom: It's Monday. Hmm... I like Monday.   Amy: What do you have today?   Tom: We have math, science and art.What about you?   Amy: We have English and P.E. What do you have tomorrow?    Tom: We have art and P.E.tomorrow.    Amy: Oh, great.      I can understand that "What about you" means "What do you have today?"  But I wonder whether "What about you" is an elliptical sentence. If it is, what is the full sentence? If it isn't, how to parse this structure? Why is there a discrepancy in disk usage reported by df and du? I have a Linux CentOS server, the OS+packages used around 5GB. Then, I transferred 97GB data from a Windows server to two folders on this Linux server, after calculated the disk usage, I see the total size of the two folders is larger than the disk used size.  Run du -sh on each folder, one use 50GB, the other one use 47GB  But run df -h, the used space is 96GB. (50GB + 47GB + 5GB) > 96GB  Is there any problem? Those two folders contain lots of files (1 million+). Thanks. Is $\sqrt{64}$ considered $8$? or is it $8,-8$? Last year in Pre-Algebra we learned about square roots. I was taught then that $\sqrt{64}=8$ and $\sqrt{100}=10$, which I understood and accepted. I was also taught that $\pm\sqrt{64} = 8,-8$ because both of those numbers squared is 64, which I also get. But this year, with a new school and teacher in a different state, our teacher is telling us that: $\sqrt{64}=8,-8$ and $\pm\sqrt{64}$ also is $8,-8$. The way to get the positive root of something is: $+\sqrt{64}=8$ And these seem to contradict each other. I was always taught that a regular square root returned a positive number and only a positive number, but now my teacher is saying a regular square root gives two numbers, and considering the square root of a number $n$ is defined as $y^2=n$ I see where he is coming from. Upon researching this Wikipedia says:  For example, $4$ and $−4$ are square roots of $16$ because $4^{2} = (−4)^{2} = 16$  And Wolfram MathWorld says:  Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of $9$ are $-3$ and $+3$  But on the other side, Wolfram Alpha, when given &quot;The square root of 9&quot; gives only 3.  So, which is right? Is $\sqrt{64}$ considered $8$? or is it $8,-8$? Adding columns to production tables What's the best way to add columns to large production tables on SQL Server 2008 R2? According to Microsoft's books online:     The changes specified in ALTER TABLE are implemented immediately. If the changes require modifications of the rows in the table, ALTER TABLE updates the rows. ALTER TABLE acquires a schema modify lock on the table to make sure that no other connections reference even the metadata for the table during the change, except online index operations that require a very short SCH-M lock at the end.   (http://msdn.microsoft.com/en-us/library/ms190273.aspx)  On a large table with millions of rows, this can take a while. Is taking an outage the only option? What's the best way to handle this kind of situation? Proof relation between Levi-Civita symbol and Kronecker deltas in Group Theory In order to prove the following identity: $$\sum_{k}\epsilon_{ijk}\epsilon_{lmk}=\delta_{il}\delta_{jm}-\delta_{im}\delta_{jl}$$ Instead of checking this by brute force, Landau writes thr product of Levi-Civita symbols as: $$\epsilon_{ijk}\epsilon_{lmn}=\det\left|    \begin{array}{cccc}       \delta_{il}   &amp; \delta_{im}  &amp; \delta_{in}   \\       \delta_{jl}   &amp; \delta_{jm}  &amp; \delta_{jn}   \\       \delta_{kl}   &amp; \delta_{km}  &amp; \delta_{kn}       \end{array} \right| $$  The proof that the equalty holds is quite straightforward if you consider what values the indices can take. But I've been told that there's a much more profound and elegant demonstration based on the representation of the symmetric group.  Does anybody know this approach based on group theory? How to find range $a_{75}$ of the term of the series $a_n=a_{n-1}+ {1 \over {a_{n-1}}} $ If $a_1=1$ and for n&gt;1$$a_n=a_{n-1}+ {1 \over {a_{n-1}}} $$ $a_{75}$ lies between (a) (12,15) (b) (11,12) (c) (15,18)  Now , in this question, I rewrote, $a_n-a_{n-1} = {1 \over {a_{n-1}}}$, to make a telescopic type and summed the terms to get $$a_n-a_{1}= {1 \over {a_{n-1}}}+ {1 \over {a_{n-2}}}...{1 \over {a_{1}}}$$ And as ${1 \over {a_{r}}} &lt;1$ for all r , I got $a_{75} &lt;76$ , but it's of no help as , the upper bound is much higher. I got no further ideas , on how to solve it, please help. I'm getting the following error when i try and install MSSQL 2008 on Windows Server Standard 2008 R2: If a function $f$ defined on $[a,b]$ and differentiable everywhere, does it mean that its derivative $f'(x)$ is continuous everywhere on $[a,b]$?  My understanding is 'yes'. Because if there were 'a gap' in $f'(x)$, we could integrate back to $f(x)$ and show that where the gap is, there would be a sharp "angle" where $f(x)$ is not differentiable.