In answering , I proposed the following command   \newcommand{\testa}{%   \ifmmode     ^{12}\text{CO(1-0)}%   \else     $^{12}$CO(1-0)%   \fi }   But @egreg pointed out that I should be writing either:  \newcommand{\egrega}{%   \relax   \ifmmode     ^{12}\text{CO(1-0)}%   \else     $^{12}$CO(1-0)%   \fi }   or  \DeclareRobustCommand{\egregb}{%   \ifmmode     ^{12}\text{CO(1-0)}%   \else     $^{12}$CO(1-0)%   \fi }   and proposed I test things out in an array environment to see what happens.  \[\begin{array}{c}   \testa \end{array}\]   Sure enough, \testa results in the following warning:  ! Missing $ inserted. &lt;inserted text&gt;                  $ l.42   \testa  ?    whereas the two command \egrega and \egregb work as expected.  I wasn't happy with just this.  My understanding of robustness is quite shacky.  And, while I've read various posts such as , I still felt a bit in the dark.    So, I tried a different approach.  I tried writing  \[\begin{array}{c}   \ifmmode{Math}\else{NOT MATH}\fi \end{array}\]   Really, I don't why I did this.  I expected to the same error as with the faulty version of the command.  But instead I didn't get an error at all.  Instead, I got the following unexpected and erroneous result:    Clearly something is going very wrong here, and not in any way I anticipated.  If instead I write   \[\begin{array}{c}   \relax\ifmmode{Math}\else{NOT MATH}\fi \end{array}\]   or precede the \ifmmode by anything else to be processed, all seems to go well.  What am I failing to understand here?
I’m struggling with getting \ifmmode to work correctly when used inside an array environment which is inside a displaymath environment.  Here is a minimal example:  \documentclass{article}  \newcommand{\signal}[1]{%   \ifmmode     1#1   \else     2$#1$   \fi }  \begin{document} \begin{displaymath}   \begin{array}{cc}     \signal{a} \signal{b} &amp; \signal{c} \signal{d} \\   \end{array} \end{displaymath} \end{document}   This example generates the following output:        Why doesn’t all invokations of \signal detect math-mode?
Update  appear to have been updated recently (as of writing this the last update appears to have taken place on 3rd of June) to include the following statement:  The -d switch is necessary to upgrade from Ubuntu 18.04 LTS as upgrades have not yet been enabled and will only be enabled after the first point release of 20.04 LTS.  Also note that .  TLDR (as of 15 Aug 2020):   is the canonical tracking document for the first Focal Fossa point-release (20.04.1). It’s a live document. The Ubuntu release team will be updating it as we work on releasing 20.04.1. […] Upgrades from 18.04 to 20.04.1 are still disabled as we are working through a few upgrade blockers   Edit The answers (so far) do not answer the question. The question is pretty simple. Ubuntu.com points to a &quot;FinalRelease&quot;, which turns out to be a &quot;Development Release&quot;, see evidence below. I'm not asking how and when to upgrade. I'm pointing out what seems to me to be an apparent inconcistency (in semantics, linguistics, logic) in the way Ubuntu names or labels its releases. From an answer to the question I'd except someone to pick up and acknowledge this inconsistency. An answer could start with a &quot;Yes, ...&quot; or a &quot;No, ...&quot;, for starters. (Moreover, a good answer could distinguish between the release being in dev stage and/or the upgrade process (e.g. from LTS to LTS) being in dev stage, if appropriate.) This question, as I see it, is not a duplicate. Futher, I think it is a question which deserves a non-patronizing answer. Thanks.   indicates that today (23rd April 2020) the new Ubuntu 20.04 LTS is to be released.  See &quot;condensed screenshot&quot; here:  Indeed, it seems that that's what happened, see  and :  However, when following the instructions to upgrade (from 18.04 LTS to 20.04 LTS) found  and elsewhere (edit: actually , too!), even prior to the July point release, I'm a bit confused to see that the instruction is to use update-manager -d which also points to the &quot;Development Release&quot; (see screenshot below) rather than the &quot;FinalRelease&quot; (see screenshot and link above). Is the &quot;Final Release&quot; (which I'm after) the/a &quot;Development Release&quot;? Why isn't it called something like &quot;Production Release&quot; or &quot;Final Release&quot;? When will it no longer be referred to as &quot;Development Release&quot;? In July? I'm confused :( Every finitely generated algebra over a field is a Jacobson ring Knowing that the polynomial ring in $n$ variables over a field $k$ is a Jacobson ring, how can we prove from it that every finitely generated $k$-algebra is a Jacobson ring?   EDIT: We define a ring to be Jacobson if every prime ideal is an intersection of maximal ideals. Padding is missing under the &quot;Communities&quot; border line, and the list is also showing &quot;Communities (0)&quot; (maybe the reason is due to caching as it shows 0 Communities):  Generally, padding is present: I am fairly sure that I understand the how Monte Carlo integration works but I am not understanding the formulation of how it is used to estimate Pi.  I am going by the procedure outlined in the 5th slide of this presentation   I understand the preliminary steps.  Pi is equal to 4 times the area of a quarter of the unit circle.  And the area of the top-right quarter of the unit circle centered at (0,0) is equivalent to the integral of the curve that is the top-right quarter of the unit circle in $0&lt;x&lt;1$ and $0&lt;y&lt;1$.    What I don't understand is how this integral is  $\iint I((x^2+y^2)&lt;1)P(x,y)dxdy$  where $P(x,y)$ is uniformly distributed in the unit square around the quarter circle (i.e. it is always equal to 1 if $0&lt;x&lt;1$ and $0&lt;y&lt;1$ and 0 otherwise).  So this would mean that  $I((x^2+y^2)&lt;1)P(x,y)$ is the function that is the top-right quadrant of the unit circle at $0&lt;x&lt;1$ and $0&lt;y&lt;1$ but I do not understand how this is true since the indicator function can only be 1 or 0.  I understand that it is probably written in this way to make Monte Carlo sampling easy (i.e. it is an expectation so just sample from $P(x,y)$ and get the average of the samples applied to $I((x^2+y^2)&lt;1)$) but it just does not make intuitive sense to me why that integral represents the area under that curve.  Could someone provide an intuitive explanation of this.  Maybe show how that integral was derived in a step-by-step way?  EDIT:  I was able to gain a better understanding by relating the expectation to an area.  I will explain it here in case it helps anyone.  First start with relating Pi to the area of the top-right quadrant of the unit circle  $\pi=4\times A_{tr}$  Then we place the top-right quadrant into the unit square.  And under a uniform distribution over the unit square, the area of the circle quadrant is proportional to the probability of obtaining a sample from it.  It follows that the following equality holds  $P(x^2+y^2&lt;1)=\frac{A_{tr}}{A_{square}}$  and $A_{square}=1$ so  $P(x^2+y^2&lt;1)=A_{tr}$  And substituting into the original equation  $\pi=4\times P(x^2+y^2&lt;1)$  and it is also true that $P(x^2+y^2&lt;1)=E[I(x^2+y^2&lt;1)]$ which is equal to the original double integral.  So I understood it by relating the area to a probability then relating that probability to an expectation that is equivalent to the integral.  Let me know if I have made any mistakes. Bijection between an infinite set and its union of a countably infinite set I have $A$ as an infinite set and $S$ as a countably infinite set, (so that means there exists a one-to-one correspondence between $S$ and $\mathbb{N}$).  How do I show that there always exists a bijection between $A$ and $A\cup S$? Can it be done by showing that there is a bijection from $A$ to $S$ or from $A$ to itself? I am lost on this one.   Oh, and can it be possible that there is no bijection if it is between $S$ and $A\cup S$? What about a map that maps $\mathbb{N}\to \mathbb{N}\cup\mathbb{R}$? I've been battling the following two questions for more than a day today.  Write a regular expression (comprised of {a, b}) that contain at least two b and do not contain abb. Write a regular expression (comprised of {a, b}) that contain abba and do not contain baa.  For the first part I wrote a basic one to start off with that doesn't handle the must two-b: b* OR a* OR (a+ AND b)* OR a*  I ran a few &quot;truth tables&quot; inputs over it and it seems ok. But the moment I add the requirement, things go haywire. Mostly because when pushing in those &quot;b&quot;s, you can suddenly use the OR to break things up. Same thing on the second one.... Any help would be greatly appreciated! Norah What do the nomodeset, quiet and splash kernel parameters mean? What do Kernel command line parameters (also also known as boot options) mean?  When you go to Ubuntu (or better, GRUB) options, you have things like nomodeset, quiet and splash. What do these options mean?  I am also interested in other parameters and would be interested to see descriptions of any kernel parameter (like acpi), but these are not essential. In Stein's Fourier Analysis i'm having trouble Justifying the following calculations for the Fourier Coefficients in $(1.)$  $(1.)$  Let $f(\theta) = \theta$ for $ -\pi \leq \theta \leq \pi$ in the case when $n \neq 0$:  $$f(n)=\frac{1}{2 \pi}\int_{-\pi}^{\pi}\theta e^{-in \theta} d \theta = \frac{1}{2 \pi}[-\frac{\theta}{in}e^{-in \theta}]_{-\pi}^{\pi} + \frac{1}{2 \pi in} \int_{-\pi}^{\pi}e^{-in\theta}d \theta = \frac{(-1)^{n+1}}{in} $$  Then the Fourier Series of $f$ is given by:  $$ f(n) = \sum_{n \neq 0}  \frac{(-1)^{n+1}}{in}$$ I just learned that using CBC encryption with an IV which is predictable is not secure.  From what I understand, using certain plain texts, and then guessing the IV that it uses, the attacker can verify if the IV he guessed was right. How does this qualify as an attack? How can I generate and print a 32 bit sequence of 0s and 1s in Qiskit or IBM Q experience quantum lab using quantum random number generator implementation? Do you say IN one of the computers or ON one of the computers? Which is right? Polygamma reflection formula Why am I not burned by a strong wind? In convex optimization, the feasible region is convex if equality constraints $h(x)$ are linear or affine, and inequality constraints $g(x) \leq 0$ are convex. Does this mean that if $h(x)$ is nonlinear, then the optimization problem is non-convex? Is there any special case where $h(x)$ is nonlinear and the feasible region is still convex? Why should the first person pronoun 'I' always be capitalized? What is the equation for the tension on the ends of a cable suspended at different heights? I'm looking for an equation to find the tension on the ends of a cable suspended between two poles (one higher than the other) with no load but the cable itself.    I determined that the tension would be different on each end, that the shape of the suspended cable would be a  truncated at one end, and that the following would be the variables necessary in the equation:   cable weight cable length vertical distance between ends horizontal distance between ends   Some links that could be helpful: Unity Performance: How many Materials per Object? Is the closure of path-connected set path connected? Why does $(\frac{p-1}{2}!)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$? Assume $p$ is prime and $p\ge 3$.  Through experimentation, I can see that it's probably true. Using Wilson's theorem and Fermat's little theorem, it's equivalent to saying $2^2 4^2 6^2 \cdots (p-1)^2 = (-1)^{\frac{p+1}{2}}$ mod $p$, but I can't figure out any more than that. I'm trying to split text in a JTextArea using a regex to split the String by \n However, this does not work and I also tried by \r\n|\r|n and many other combination of regexes. Code:  public void insertUpdate(DocumentEvent e) {     String split[], docStr = null;     Document textAreaDoc = (Document)e.getDocument();      try {         docStr = textAreaDoc.getText(textAreaDoc.getStartPosition().getOffset(), textAreaDoc.getEndPosition().getOffset());     } catch (BadLocationException e1) {         // TODO Auto-generated catch block         e1.printStackTrace();     }      split = docStr.split("\\n"); }