A friend got a brand new iPhone 11 just over three weeks ago. It's running iOS 13.0 (17A577), so presumably it hasn't ever had a software update.  When my friend tries to update iOS to the current version (13.2.2), the "Verifying update..." dialog appears for a while, then the   A dialog then appears, saying:     Unable to Install Update      An error occurred installing iOS 13.2.2      Retry      Remind me later   Retrying ends up with the same result.  iOS 13 has various fairly serious bugs, including super-slow tap responsiveness (several seconds passing before taps on the screen register).  Is there any way to get past this error and get my friend's iOS updated?
I have an iPhone 6 128 GB which is not jailbroken. The current version of iOS on the phone is 8.1.3 and yesterday I wanted to update to iOS 8.2.  The download went through without any problem but while installing, it gave the error:     Unable to Install Update   An error occurred installing iOS 8.2   Tried installing after switching Off and ON and also tried Hard Reset (Holding Home button and On/off switch).  Same error displays every time.  Also the Touch ID has stopped working and I find that the Settings - Touch ID &amp; Passcode - iPhone Unlock is OFF.  When I try to switch it ON, the system displays the error     Failed   Unable to complete Touch ID setup. Please go back and try again.   Is anyone facing this problem?  I live in India.
I have the two PGF nodes foo and bar positioned in a row.  \documentclass{article} \usepackage{tikz} \usetikzlibrary{positioning} \begin{document}   \begin{tikzpicture}[node distance=0.2cm,mynode/.style={rectangle,draw}]     \node[mynode] (foo)                {foo};     \node[mynode] (bar) [right=of foo] {bar};   \end{tikzpicture} \end{document}   How can I create a third node below those two nodes with the width of foo, bar and the node distance between them? Is there a way to do this with the fit library or is this the wrong approach? I know that I can make certain blocks only placeable on a specified block, but what is the command to give an item that can only BREAK certain blocks? Android EditText listener for cursor position change How does gravity work on Serenity? When the good ship Serenity is traveling in deep space, the crew experiences what looks like an Earth-normal (or Earth That Was-normal) level of gravity. Is it ever explained how the ship is able to generate artificial gravity? The ship doesn't spin, so  is out.  is a possibility, but I think the gravity in the ship pulls in a direction that is perpendicular to the direction the ship is usually traveling in, not opposite, like this method would require.   So: how does she pull it off?   Extra credit: how does she pull it off in the episode ? Serenity's engine is totally dead, to the extent where not even life support systems are online. Yet Mal still has to valiantly haul himself around on the floor to replace the compression coil, so whatever creates the artificial gravity on Serenity is clearly still going. How? Why do we say that the Earth moves around the Sun? How to check if click event is already bound - JQuery "Newtonian limit" property of special relativity Books say that special relativity is indistinguishable from Newtonian mechanics when the speed of the primed frame ($v$) is small compared to the speed of light ($c$). This is what I mean by the "Newtonian limit" property of special relativity. (I don't know the correct name for it). But the transformation of the time coordinate  $$t' = \gamma \left(t + \frac{vx}{c^2}\right)$$  involves the spatial quantity $x$ which, if large enough, could balance out the smallness of $v/c^2$. So why doesn't this observation mess up the "Newtonian limit" property? How to get a cross-origin resource sharing (CORS) post request working Gauge symmetries and elementary particles The  (disclaimer: I don't know this wikipedia entry) is usually mentioned as the reason why gravitons may not be composite particles. I do understand the proof of the theorem, but not the previous conclusion.   The theorem states that in an interacting and Poincare invariant quantum field theory, there are not massless, spin-2 particles unless there exists a gauge symmetry — which makes the energy-momentum tensor non-covariant (actually covariant up to a gauge transformation) under Lorentz transformations in the Fock space. So the immediate conclusion of the theorem is that the existence of a massless, spin-2 particle (like a graviton) requires linearized diffeomorphisms.  My question is: why do linearized diffeomorphisms imply that gravitons are elementary particles? Or, more in general, why  the particle corresponding to a gauge field must be elementary (I know that a gauge symmetry must be exact, but why this implies that the corresponding particle must be elementary?). Using tikzcd in newcommand (for inline arrows) I’m using tikz-cd for diagrams, and have a medium-sized diagram that I want to re-use several times.  However, when I try using a definition like:  \newcommand{\myarrow}{$\begin{tikzcd} A \ar{r} &amp; B \end{tikzcd}$   I get an error:  ERROR: Package pgfbasematrix Error: Single ampersand used with wrong catcode.   Is it possible to put a tikz-cd diagram into a defined command somehow? Notation of the second derivative - Where does the d go? In school I was taught that we use $\frac{du}{dx}$ as a notation for the first derivative of a function $u(x)$. I was also told that we could use the $d$ just like any variable. After some time we were given the notation for the second derivative and it was explained as follows: $$ \frac{d(\frac{du}{dx})}{dx} = \frac{d^2 u}{dx^2} $$ What I do not get here is, if we can use the $d$ as any variable, I would get the following result: $$ \frac{d(\frac{du}{dx})}{dx} =\frac{ddu}{dx\,dx} = \frac{d^2 u}{d^2 x^2} $$ Apparently it is not the same as the notation we were given. A $d$ is missing. I have done some research on this and found some vague comments about &quot;There are reasons for that, but you do not need to know...&quot; or &quot;That is mainly a  notation issue, but you do not need to know further.&quot; So what I am asking for is: Is this really just a notation thing? If so, does this mean we can actually NOT use d like a variable? If not, where does the $d$ go? I found  related question, but it does not really answer my specific question. So I would not see it as a duplicate, but correct me if my search has not been sufficient and there indeed is a similar question out there already. I am having a problem running the CDC change capture job against some of my test databases.  The error message I'm getting is the following:     Executed as user: NT AUTHORITY\NETWORK SERVICE. Cannot execute as the database principal because the principal "dbo" does not exist, this type of principal cannot be impersonated, or you do not have permission. [SQLSTATE 42000] (Error 15517).   How can I fix this? Do users get notifications when @ was edited in Almost every time I turn my XBox on and start up Fallout 4, I am unable to use the "Continue" option because the most recent save is corrupt.  I instead use the "Load" option. Here, I can see that the last save I manually performed is perfectly fine, and load it with no problem. However, there is an additional, corrupt save with the same time stamp immediately after it.  This doesn't happen quite every time, but often enough that I'm surprised when there's no corrupt save waiting for me to delete it.  I've tried a few things to solve this, including exiting to the game's main menu before turning the XBox off, going out to the guide before turning the XBox off, turning it off immediately after saving, and waiting a while after saving.  Why do I get all these corrupt saves? Do others have this problem? How do I fix it? Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.   Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.   Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2&lt;n^7-n^3=\text{r.h.s}.$ This means that either $p&gt;n^2$ or $p&lt;n^2$ must hold.  Assuming $p&gt;n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p&gt;n^2)$.  Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.  Finally, we get $(5,2)$ as the solution to the given equation.      But how do I conclude this is the only solution possible? Also, why does'nt $p&lt;n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem. SFDX Pull fails because of AppMenu/AppSwitcher Does $f(x) = f(2x)$ for all real $x$, imply that $f(x)$ is a constant function? If a Continuous function $f(x)$ satisfies $f(x) = f(2x)$, for all real $x$, then does $f(x)$ necessarily have to be constant function? If so, how do you prove it? If not any counter examples? In propositional logic, the statement "If pigs can fly, then elephants can lay eggs." validates to true because the antecedent validates to false.  In other words, given $a \rightarrow b$, if a is false, the entire statement is true. Why?  Just because the antecendent is false doesn't mean that another fact depends on it, right? How to undelete a directory that was just deleted with "rm -R" command (in cloud drive)? I deleted an important directory with rm -R command. This is an very populer host data. It was located /var/zpanel/hostdata/ayrintilih After remove ayrintilih directory has gone everything.  I tried to undelete with ext4magic program. But it need to another hdd volume for recovery.  Error message:   ERROR: can not use "/root/recoveried" for recover directory. It's the same filesystem : "/dev/dm-0"   My ISP just replied my request:  Unfortunately there is no way to add another Volume to a virtual Instance.  If it would be a dedicated Server, we could do it, but not in a Cloudserver.  You could try to minimize the partition and create another one, that would  count as another Volume. We also recommend to have allways a  Backup for your Important files. Is there a correct resistance value for I2C pull-up resistors?