Is $x\sin(\frac{1}{x}),\;0<x<\infty$ uniformly continuous?
Uniform continuity of $f(x) = x \sin{\frac{1}{x}}$ for $x \neq 0$ and $f(0) = 0.$
What causes this corkscrew-shaped object to rise up in a stream of water falling down? How can I reuse existing materials, objects, etc from an existing .blend? I'm writing a tool to report information about .NET applications deployed across environments and regions within my client's systems.  I'd like to read the values of assembly attributes in these assemblies.  This can be achieved using Assembly.ReflectionOnlyLoad, however even this approach keeps the assembly loaded.  The issue here is that I cannot load two assemblies that have the same name from different paths, so naturally I can't compare the same application deployed in different systems.  At this point I'm assuming the solution will involve using temporary AppDomains.  Can someone detail how to load an assembly into another AppDomain, read the attributes from it and then unload the AppDomain?  This needs to work for assemblies on the file system as well as those at URL addresses. Virtual box is not working on windows 10 Are there any disadvantages to using your companions for transporting Power Armor Is there any difference in equals and == for String variables? if ('text'.equals(str))   or   if ('text' == str)   Is there any difference? If yes, then what is it?   Added, after answered.  String q = 'teXt'; system.debug('teXt'.equals(q)); // true system.debug('teXt' == q);      // true system.debug('text'.equals(q)); // false system.debug('text' == q);      // true If $G/Z(G)$ is cyclic, then $G$ is abelian Continuing my work through Dummit &amp; Foote's "Abstract Algebra", 3.1.36 asks the following (which is exactly the same as exercise 5 in ):     Prove that if $G/Z(G)$ is cyclic, then $G$ is abelian. [If $G/Z(G)$ is cyclic with generator $xZ(G)$, show that every element of $G$ can be written in the form $x^az$ for some $a \in \mathbb{Z}$ and some element $z \in Z(G)$]      The hint is actually the hardest part for me, as the quotient groups are somewhat abstract. But once I have the hint, I can write:  $g, h \in G$ implies that $g = x^{a_1}z_1$ and $h = x^{a_2}z_2$, so \begin{align*}gh &amp;= (x^{a_1}z_1)(x^{a_2}z_2)\\\  &amp;= x^{a_1}x^{a_2}z_1z_2\\\ &amp; = x^{a_1 + a_2}z_2z_1\\\ &amp;= \ldots = (x^{a_2}z_2)(x^{a_1}z_1) = hg. \end{align*} Therefore, $G$ is abelian.  1) Is this right so far?  2) How can I prove the "hint"? How can I minimize Thunderbird 60.2.1 to tray after "MinimizeToTray revived" add-on stopped working? After update, my Thunderbird stopped getting minimized with the "MinimizeToTray revived" add-on. Is there any way to minimize my Thunderbird? Any other add-on that works with the new version? What techniques have been used to repeat geometry or other detail to the infinite horizon? Automatic coloring of numbers according to size Scientific talks have their own culture, reporting the results, typically for me, in a table of the following form  1.7333    1.6472    1.5418    1.4682    1.5740 1.1525    1.2878    1.2652    1.3511    1.5102 1.2181    1.1500    1.2220    1.2425    1.5699 0.9786    1.1051    1.1472    1.3096    1.6757 0.9569    0.9190    1.0433    1.2924    1.4867   There is clearly a(n almost perfectly) monotonic pattern for rows and columns. Now for the naked eye, this is not the most quick way of capturing informing.  I would like numbers that bigger in absolute value, to appear differently (either green vs red or light vs dark) than smaller numbers, such that the pattern becomes clear instantly.  Is there a way of programming this in LaTeX? Changing precision on a column in the attribute table without losing data? I have data in an Excel file that I want to join with a shapefile in my project. The Excel file contains values like volumes. The Excel file and the shapefile I want to join it with both contain the same property names.  The problem is, when I join the two, I get too many decimals in the attribute table. I get six, but I only want it to be one.  I tested to change the format of the cells to one decimal, and then I saved the file as *csv. But, somehow I seem to loose data in the process? It should be 42 rows that contains data in the column i am working with, but I end up with only like 10?  Another thing is that my property names contains the letters Å. Ä and Ö, perhaps this complicates things?  I read the question , but it doesn't mention the problem with the data loss.  My question was tagged with , but I didn't write what version I'm using. I use QGIS 2.2.0. Would it help to upgrade to the latest version? Show that some monomial ideal is primary Show that $I=(X_{k_1}^{a_1},...,X_{k_s}^{a_s})$ is $(X_{k_1},...,X_{k_s})$-primary.   I noticed  that $$\sqrt{({X_{i_1}}^{a_1},...,{X_{i_k}}^{a_k})}=\sqrt{({X_{i_1}}^{a_1})+\cdots+({X_{i_k}}^{a_k})}=\sqrt{\sqrt{(X_{i_1})^{a_1}}+\cdots+\sqrt{(X_{i_k})^{a_k}}}=\sqrt{(X_{i_1})+\cdots+(X_{i_k})}=\sqrt{(X_{i_1},...,X_{i_k})}=(X_{i_1},...,X_{i_k}).$$ I have to show that $I$ is $P$-primary (where $P=(X_{k_1},...,X_{k_s})$), i.e., if $u v\in I$ then $u\in I$ or $v\in P$. I tried to prove that if $u\notin I$ and $v\notin P$ then $uv\notin I$. (There is a similar question, but the answer is too hard for me.) I am stuck in a question where energy conservation is failing and momentum conservation is correct. I think I might be doing something wrong, that's why I'm asking this question. The problem is as follows:     A bullet of mass $25\,g$ is fired horizontally into a ballistic pendulum of mass $5.0\,kg$ and gets embedded in it. If the centre of pendulum rises by a distance of $10\,cm$, find the speed of the bullet.   (H.C. Verma, Centre of Mass, Q47)     Let the mass of the bullet be $m=25\,g$. Let the mass of the pendulum be $M=5\,kg$. Let the peak height be $h=0.1\,m$. Let the initial and final velocities be $u$ and $v$ respectively.  Method 1: (Momentum Conservation)  $$mu=(M+m)v \implies v=\frac{mu}{M+m}\\ \text{Also, }\frac{1}{2}(M+m)v^2=(M+m)gh\\ \implies u^2=2\biggl(\frac{M+m}{m}\biggr)^2gh\\ \implies u=201\sqrt2 \text{ m/s}$$  Method 2: (Energy Conservation)  Since both the masses move together after the collision, and because the velocity at highest point is null, therefore: $$\frac{1}{2}mu^2=\frac{1}{2}(M+m)v^2=(M+m)gh\\ \implies u^2=2gh\biggl(\frac{M+m}{m}\biggr) \implies u=\sqrt{402}$$    To summarize myself, I am curious about the following: Why do the results differ? Shouldn't the energy be conserved as well as the momentum? Since it should be, then what is the flaw in my calculations? Automatically convert quotations in the form of "abc" to become ``abc" When does a Square Matrix have an LU Decomposition? There are tons of questions I could maybe post this as an answer to, but before choosing the best one, I'll just post this as an announcement question that can be linked elsewhere. The short story From now on, intra-word emphasis is possible in our Markdown version. That means you can now write  The &quot;V&quot; is for Model**View**Controller. That's un*fricking*believable.  and it renders like this:  The &quot;V&quot; is for ModelViewController. That's unfrickingbelievable.  This only works for asterisks, not for underscores. It works in posts and comments. The reasons There are two big reasons for this. For one thing, this is more or less the behavior of , which we eventually . &quot;More or less&quot; because CommonMark's emphasis behavior is still in flux, and supports a few things that we don't support yet, but at least with this change we're closer. The other reason is languages that don't even have the concept of &quot;words separated by spaces&quot;. See  and  for two examples. This change allow posts in those languages to emphasize words without having to resort to HTML tags. To use an example from the first linked post,  **北京**和**上海**是直辖市。   now renders like this:  北京和上海是直辖市。  Possible issues In the 99% cases, this should only make things better; I've only found very few posts where this change causes unintended bolding/italicizing. And even in those cases, note that the rendered versions of questions and answers are not re-rendered until a post is edited, so this will not affect anything in posts retroactively. Unlike posts, comments are rendered , so this immediately applies to all comments. In most cases this doesn't change anything or actually fixes things. Should you discover a rare case where this breaks an important comment badly, we can always edit it to fix. There's one side effect that I'd like to mention: If you find a great quicksort implementation on GitHub that has a bug that causes a crash when sorting Windows versions, and the author doesn't want to fix it, you may be inclined to say  f**k this s**t   (i.e., &quot;fork this sort&quot;). With intraword emphasis, this would however look like this:  fk this st  So you would have to escape the asterisks (one of them would actually be enough). I'm not worried about this case too much, since if you must put censor-worthy words into a post, you might as well have to jump through a little hoop. Some details This being Meta, there are probably lots of people interested in a more detailed description of the new behavior (don't feel bad if you're not one of them; just skip this section). Here are the major rules regarding intraword-emphasis:  As mentioned, it only applies to asterisks, not underscores. Underscores  this restriction was in place at all; asterisks are much less of a problem.  If a single or double asterisk is supposed to be interpreted as intra-word emphasis, it cannot have punctuation on one side; it actually must be inside a word. An example of an edge case this solves is  It's a web*-based solution. *see Wikipedia for a definition of &quot;web&quot;  where the asterisks after &quot;web&quot; could cause unintended italicizing if there was a second asterisk in the same paragraph somewhere.  For italics, the &quot;inside a word&quot; part of rule 2 is even a bit stricter. For bolding, any word or digit character is allowed; for italics, digits don't work either. That's because the asterisk is often used as a multiplication symbol, so  The game worked in 320200 and 640480 resolution.  wouldn't look right without this restriction.  A minor addition that I made that's only tangentially related to intra-word emphasis is that a sequence of more than three underscores or more than three asterisks will no longer take part in any emphasis. This shouldn't really cause any problems, but fixes some (e.g. the underline in the second blockquote in , while not really visible, is a weird combination of bold and italic underscore characters; after this change, it would just be (more) underscore characters). This behavior also matches CommonMark. Run a Windows Service as a console app How do I prove/disprove that $F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$, where $F_n$ is the $n$th Fibonacci number? I was looking at the Fibonacci sequence:  $$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,\ldots, F_n : F_n + F_{n - 1} = F_{n + 1} \land F_0 = 0$$  Here I noticed something; $\qquad F_n^2 + F_{n\pm1}^2 = F_{2n\pm1} : n\in \mathbb{N}$  For example; $$\ \ F_3^2 + F_{3 + 1}^2 = F_3^2 + F_4^2 = 2^2 + 3^2 = F_{2\times3 + 1} = F_{6 + 1} = F_7 = 13$$ $$F_3^2 + F_{3 - 1} = F_3^2 + F_2^2 = 2^2 + 1^2 = F_{2\times3 - 1} = F_{6 - 1} = F_5 = 5$$ However, no matter how true this can seem, I cannot seem to prove/disprove it.    My Attempt: $$\because (a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + b^2 + 2ab : (a,b)\in \mathbb{Z}$$ $$\implies F_n^2 + F_{n\pm 1}^2 = (F_n + F_{n\pm 1})^2 - 2F_nF_{n\pm 1}$$ $$\implies F_n^2 + F_{n - 1}^2 = F_{n + 1}^2 - 2F_nF_{n - 1} : F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_n^2 + F_{n + 1}^2 = F_{n + 2}^2 - 2F_nF_{n + 1} : F_n + F_{n + 1} = F_{n + 2} \iff F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_{2n\pm 1} + 2F_nF_{n\pm 1} = F_{n + 2\lor1}^2 : F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1} \tag{+2 if +, +1 if -}$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1} - F_{n\pm 1}^2}\sqrt{F_{2n\pm 1} - F_n^2}= F_{n + 2\lor1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{(F_{2n\pm 1} - F_{n\pm 1}^2)(F_{2n\pm 1} - F_n^2)} = F_{n + 2\lor1}^2$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1}(F_{2n\pm 1} - F_n^2 - F_{n\pm 1}^2) - (F_nF_{n\pm 1})^2} = F_{n + 2\lor1}^2$$.  $$\because F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{-(F_nF_{n\pm 1})^2} = F_{n + 2\lor1} \neq F_{2n\pm 1} + 2\sqrt{\big(-(F_nF_{n\pm 1})\big)^2}$$ $$\therefore F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} : i = \sqrt{-1}$$  I have looked over every step and cannot see where I am wrong. The only part that must be wrong is that:  $$F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} \iff F_n^2 + F_{n\pm1}^2 = F_{2n\pm 1}$$  I can seperate $F_{2n\pm 1}$ as $F_n^2 + F_{n\pm 1}^2$ (if true) so I turn the equation into:  $$F_n^2 + F_{n\pm 1}^2 + 2F_nF_{n\pm 1}i = F_{n + 2\lor1}$$ $$\therefore (F_n + F_{n\pm 1})\bigg(\frac{F_n}{F_{n\pm 1}} + \frac{F_{n\pm 1}}{F_n} + 2i\bigg) = F_{n + 2\lor1}$$ Now I know that the greater the value of $n$, one of the fractions in the above equation (depending on which sign is used in $F_{n\pm 1}$) will get closer to the golden ratio $φ = \frac{1 + \sqrt{5}}{2}$ but $i = \sqrt{-1}$ is not a "real" number, although $F_n$ is.  I then searched for other formulae for finding $F_n$ and came across the one below (being Binet's Formula):  $$F_n = \bigg\{\frac{φ^n - ψ^n}{φ - ψ} : ψ = φ - \frac{\sqrt{20}}{2} = \frac{1 - \sqrt{5}}{2} \ \land \ φ - ψ = \sqrt{5}\bigg\}$$  But I do not know what how to turn that into $F_{n\pm 1}$. Could you please show me where I went wrong (because I must have gone wrong somewhere), and also if you wish, could you please show me how to turn the above fraction into $F_{n\pm 1}$ (if possible).  I only used the $(a + b)^2$ expansion rule because my skill level in mathematics is not particularly high compared to most of the users on this site (especially when logging into the MSE and finding fancy symbols like $\nabla$ and $\partial$ and $\sum$ which all look Greek to me, however the only "fancy" thing I know is that $i = i^5 = \cdots = \sqrt{-1}$) so if you used any other techniques in trying to prove this formula, could you please show me step by step?  Much appreciated. Thank you in advance.    Edit: I believe this question is a duplicate of this other question $\longrightarrow$   and I can go here to serve as a little hint $\longrightarrow$ View only the new entries in a growing log file Injectivity and surjectivity of a recursive function Let $f:\mathbb{N} \to \mathbb{Q}^+$ defined as follows : $$\begin{cases} f(0) = 0 \\ f(2n) = \dfrac{1}{f(n)+1} \\ f(2n+1) = f(n)+1 \end{cases}\,.$$ It is asked to prove the injectivity then the surjectivity of $f$.  After examining first values taken by $f$, I proved by induction that $f(2^k)=\dfrac{F_k}{F_{k+1}}$ where $(F_n)$ is the fibonacci sequence. Also one notes that if $m$ and $n$ have different parity then we can't have $f(m) = f(n)$ since $f(2\mathbb{N}) \subset (0,1)$ and $f(2\mathbb{N}+1) \subset [1,+\infty)$. Any ideas on how to finish the proof of injectivity and any thoughts about surjectivity are welcome. Thanks