In Landau-Lifschitz, following expansion is given,  We have, $$L(v'^2)~=~L(v^2+2\textbf{v}\cdot\epsilon+\epsilon ^2)$$ expanding this in powers of $\epsilon$ and neglecting powers of higher order, $$L(v'^2)~\approx~L(v^2)+\frac{\partial L}{\partial (v^2)}2\textbf{v}\cdot\epsilon$$  $L$ is Lagrangian. and $\textbf{v}'=\textbf{v}+\epsilon$. I am unable to follow what expansion this exactly is. It does not look like Taylor expansion.
I was reading a bit of Landau and Lifshitz's Mechanics the other day and ran into the following part, where the authors are about to derive the kinetic energy of a free particle. They use the fact that the Lagrangian of this particle must be the same (or at most, differ by the total time derivative of a function of co-ordinates and time) in different inertial frames.     We have $L'=L(v'^2)=L(v^2+2\mathbf{v}\cdot\boldsymbol{\epsilon}+\boldsymbol{\epsilon}^2)$. Expanding this expression in powers of $\boldsymbol{\epsilon}$ and neglecting terms above the first order, we obtain    $$L(v'^2)=L(v^2)+\frac{\partial L(v^2)}{\partial (v^2)}2\mathbf{v}\cdot\boldsymbol{\epsilon}.$$   I think I'm ok with all the physics in this section. What I don't get is just the part I quoted above (so maybe this post is better suited for the math site, but since this is book is so physics-y I thought I'd post it here). My math is pretty rusty, so I'm not really sure- how do the authors expand the function to arrive at the above expression? It reminds me a bit of a Taylor expansion, but not very much. What's the process used to arrive at it?
Description of varieties in $\mathbb{P}^2\times \mathbb{P}^1$ If $[x:y]$ are coordinates of $\mathbb{P}^1$ and $[X:Y:Z]$ are coordinates of $\mathbb{P}^2$, what do the following varieties look like?   $V(x^2X+y^2Y+xyZ)\subset \mathbb{P}^2\times \mathbb{P}^1$ $V(x^2X^2+y^2Y^2+xyZ^2)\subset \mathbb{P}^2\times \mathbb{P}^1$   As of now, I don't quite know how to visualize these varieties. A detailed geometric picture would be especially welcome. How can I check for an empty/undefined/null string in JavaScript? I saw , but I didn't see a JavaScript specific example. Is there a simple string.Empty available in JavaScript, or is it just a case of checking for ""? How to change the shadow color in Blender Cycles? Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.   While trying to do this exercise I noticed two facts.  First, $g$ and $f(g)$ have the same order because $o(f(g))|o(g)$ and, applying $f$ again and using $f^2=\text{id}_G$, $o(f(f(g)))=o(g)|o(f(g))$ and, once the order of any element is $\geq 1$, it follows that $o(g)=o(f(g))$. Also, it's easy to see that $g$ and $f(g)$ commute.  Second, there cannot exist such an automorphism if the order of $G$ is even, because $f(e)=e$ and we can form pairs like $\{g,f(g)\}$ with $f(g)\neq g$ that are invariant under $f$, i.e., $f(\{g,f(g)\})=\{g,f(g)\}$. But once the order of $G$ is even, proceeding with the construction of the pairs, we'll end up with just one element $\neq e$, so we must have some $\gamma\in G\setminus{e}$ with $f(\gamma)=\gamma$, contradicting the hypothesis.  Let $n$ be the order of $G$ and let's fix an enumeration of the elements of $G$, say $G=\{g_1,\ldots,g_n\}$. My approach was the following. For each $\sigma\in S_n$, let $x_{\sigma}=\prod_{i=1}^ng_{\sigma(i)}$. Then we have that $f(x_{\sigma})=x_{\sigma^{\prime}}$. If it's shown that $f(x_{\sigma})=e\;\forall\;\sigma\in S_n$, then we'd have $x_{\sigma}=e\;\forall\;\sigma\in S_n$, which implies that $G$ is abelian. The problem is that $S_n$ has $n!$ elements while $G$ has $n$ elements, so there repetitions among the $x_{\sigma}$, and we cannot apply directly the reasoning of the last paragraph.  Is this the right way, or is there an easier manner to solve this? Selecting an ID with a colon in it with jQuery The following problem was on a math competition that I participated in at my school about a month ago:      Prove that the equation $\cos(\sin x)=\sin(\cos x)$ has no real solutions.   I will outline my proof below.  I think it has some holes.  My approach to the problem was to say that the following equations must have real solution(s) if the above equation has solution(s):  $$ \cos^2(\sin x)=\sin^2(\cos x)\\ 1-\cos^2(\sin x)=1-\sin^2(\cos x)\\ \sin^2(\sin x)=\cos^2(\cos x)\\ \sin(\sin x)=\pm\cos(\cos x)\\ $$  I then proceeded to split into cases and use the identity $\cos t = \sin(\frac{\pi}{2} \pm t\pm y2\pi)$ to get   $$ \sin x=\frac{\pi}{2} \pm \cos x\pm y2\pi\\ $$  and the identity $-\cos t = \sin(-\frac{\pi}{2}\pm t\pm y2\pi) $ to get   $$ \sin x=- \frac{\pi}{2}\pm \cos x \pm y2\pi.\\ $$  where $y$ is any integer.  I argued that $y=0$ was the only value of $y$ that made any sense (since the values of sine and cosine remain between $-1$ and $1$).  Therefore, the above equations become   $$ \sin x=\frac{\pi}{2} \pm \cos x\implies \sin x \pm \cos x=\frac{\pi}{2}\\ $$  and   $$ \sin x=- \frac{\pi}{2}\pm \cos x\implies \cos x\pm\sin x= \frac{\pi}{2}.\\ $$  Then, by a short optimization argument, I showed that these last two equations have no real solutions.  First, does this proof make sense?  Second, if my proof makes sense, then I feel that it was not very elegant nor simple.  Is my approach the best, or is there a better (i.e., more elegant, shorter, simpler) proof? How to configure Ubuntu as an LDAP client? How to configure Ubuntu 12.04 as LDAP client with PAM modules? Coupon Collector Problem with Batched Selections Is it ever acceptable to have a  in your C or C++ application?  What if you allocate some memory and use it until the very last line of code in your application (for example, a global object's destructor)? As long as the memory consumption doesn't grow over time, is it OK to trust the OS to free your memory for you when your application terminates (on Windows, Mac, and Linux)? Would you even consider this a real memory leak if the memory was being used continuously until it was freed by the OS.  What if a third party library forced this situation on you? Would refuse to use that third party library no matter how great it otherwise might be?  I only see one practical disadvantage, and that is that these benign leaks will show up with memory leak detection tools as false positives. First, … Secondly, … &c Take it for granted that enumerations have secondly, …; thirdly, …; finally, … Is there any reason, except tradition, to prefer the traditional first to firstly? Mouse cursor doesn't change to pointer when hovering "full site" on mobile When (accidentally) clicking "mobile" in the footer, one gets the mobile site in a regular browser too. To switch back, one can click "full site" in that same footer:    However: when hovering it, the mouse is changed as to select text, rather than to indicate one can click (even though it's clickable).  All other links do change the mouse into a pointer.  (Not a big deal, except when somehow  getting into the mobile mode, I guess.) Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G has subgroup of order $\operatorname{lcm}(n,m)$. Limit Ordinal Question how to prove this result.     $\alpha$ is limit ordinal if and only if $\beta&lt;\alpha$ implies $\beta+1&lt;\alpha$ for any $\beta$.   I am getting confused because of the definition of limit ordinal. it says     $\alpha=\sup\{\beta: \beta&lt;\alpha\}=\bigcup\alpha$    for sup I am ok but for $\alpha=\bigcup\alpha$ how ?? "Actually" vs "really" Restricing domains for universal and existential quantifiers I am currently using Rosen's "Discrete Mathematics and Its Applications" (7th ed.) for my discrete mathematics course. We recently talked about quantifiers, more specifically the universal and existential quantifier. What confuses me is when we started talking about restricting the domain (i.e. instead of all real numbers, only all real positive numbers), where does the conditional statement come from when restricting the domain of a universal quantifier and where does the conjunction come from when restricting the domain of a existential quantifier?  This link did help:   However, the part that confuses here is how are the domains restricted in the answer given in the link above? Include reviews made outside of /review in progress towards the badge The recent changes to /review has gotten everyone interested in cleaning up the muck. At present, only posts "reviewed" (by clicking on the review link) and actions taken (up/down/close/edit/flag) from within /review count towards the badge.   However, it is entirely possible that one catches a poor question outside of /review (e.g., on the front page/tag page, etc.) and "reviews" it and takes some action on it. However, these actions do not count for the badge, at present. In all likelihood, this question would've also been in the /review queue and if one were fortuitous enough to have reviewed it there, these actions would've counted towards the badge.   Would it be possible to include reviews made outside of /review to count if the question was also in the review queue? Note that this shouldn't apply if the question dropped out of the queue (either by inaction or after it was edited and improved by previous reviews). How do I correctly tag my questions? Here are some tips for tagging questions. These guidelines will help you more accurately tag your questions, which in turn will help them get more attention and lead to better answers faster.     For more information, see "" in the . How to get glare on "Quick Smoke" (Fire & Smoke explosion) How to prove $|a^k|=n/\gcd(n,k)$ whenever $|a|=n$? How can I create my own Action using Photoshop? Can anyone guide me with a simple example?