I want to know the basic about  with basic examples.
Since the equation of mechanics are of second order in time, we know that for $N$ degrees of freedom we have to specify $2N$ initial conditions. One of them is the initial time $t_0$ and the rest of them, $2N-1$ are initial positions and velocity. Any function of these initial condition is a constant of motion, by definition. Also, there should be exactly $2N-1$ algebraically independent constants of motion.  On the other hand, Noether's procedure gives us integrals of motion as a result of variational symmetries of the action. These integrals of motion are also conserved but they are not always $2N-1$ in number. In consequence, we classify the system by their integrability.  So, what is the difference between the  and ? Why do non- have less integrals of motion when they should always have $2N-1$ constants of motion?
Constructing a continuous function whose graph seems 'special' I've been reading through Zorich's "Analysis I" book recently, and I came across this nice little exercise.  Let $f: [0,1]\to \mathbb R$ be a continuous function such that $f(0)=f(1)$. Show that    for any $n\in \mathbb N$ there exists a horizontal closed interval of length $\frac 1n$ with endpoints on the graph of this function; if the number $\ell$ is not of the form $\frac 1n$ there exists a function of this form on whose graph one cannot inscribe a horizontal chord of length $\ell$.   The first part can be proven like this: Consider $g: [0,(n-1)/n] \to \mathbb R$ given by $g(x) = f(x) - f(x+1/n)$. Then   $$\sum_{k=0}^{n-1} g(k/n) = 0$$  and therefore either all of these points are zero or there exists both a point where $g$ is positive and a point where $g$ is negative. By continuity, there must then also be a point where $g = 0$. So we are done.  Now, the second statement seemed rather counterintuitive, and I have given it some time now, but don't see a counterexample for $\ell &lt; 1/2$.  (For $\ell &gt; 1/2$ the function $f(x) = \sin(2\pi x)$ will do.)  Can anyone help me out?  Cheers, While editing some ad copy, and I was given the sentence, "Defending a lawsuit can be a big cost for your business." (My italics.)  I keep thinking the proper word to use is "expense" rather than "cost" as in: "Defending a lawsuit can be a big expense for your business." But I'm just not sure I'm correct.   Can anyone help with more definitive guidance? Can "cost" and "expense" be used interchangeably in a sentence like this?   Thank you! In a non-inertial reference frame, the centripetal force is balanced by a centrifugal force. Therefore, shouldn't the net acceleration be zero? Yet, you still feel an acceleration outwards? Why is that? Passing a variable to sed Recursive grep vs find / -type f -exec grep {} \; Which is more efficient/faster? I'm actually at the cemetery at White Orchard. There is a white question mark and some little red arrows.  What are this little red arrows?  Do they point to something dangerous? Does the shebang determine the shell which runs the script? I am trying to prove that any conjugate of $a$ has the same order as $a$.  Let $G$ be a group and let $a \in G$.  An element $b \in G$ is called a conjugate of $a$ if $b=xax^{-1}$.  My professor gave us a hint.  Prove $a^k=e$ iff $b^k=e$ for all $k \in \mathbb{Z}$ implies $o(a) = o(b)$.  My first move is to $b^k = (xax^{-1})^k = (x^{-1})^ka^kx^k = (x^{-1})^kex^k = (x^{-1})^kx^k=x^{-k+k}=x^0$  I am not sure that I am doing this correctly?  I am trying to get $b^k=e$.  However when I raise $(xax^{-1})^k$, I am not sure if I what I have done is correct?  I am kind of confused here. TL;DR: how does an Ubuntu flavor iso installation ( for example Lubuntu ) differ from installing a desktop package via apt-get ? (for instance sudo apt-get install lubuntu-desktop)  Context for the question:   Some time ago I answered a question on AskUbuntu , where I tested in virtual machine removing unity-desktop and installing ubuntu-gnome-desktop. That makes the boot logo ( plymouth ) change, desktop changes, different packages get installed (gnome-network-manager instead of network-manager , if I am not mistaken). It appeared that in effect I've switched to different Ubuntu flavor, even though I started out with the standard Ubuntu system. Hence the question: what differentiates the pre-made Ubuntu flavor iso from installing a package ? Is it all the same base underneath the hood or does installation from iso have specific configurations ? If it's possible to install a flavor via package manager, why then separate flavor iso's exist ? (obvious answer is user convenience, but I think there has to exist other reasons. Canonical is a business , and it makes not a lot of sense for a business to provide additional flavor iso , if a sysadmin or end-user can get the same thing via package manager by themselves). I own a Panasonic DMC-G81 (same as G80/G85) with in-body Image Stabilization.  When turned on everything is just fine, but when turned off, the sensor is wobbling around in the body. When no lens attached, it's clearly visible moving about 1/2 cm in the body. I didn't find anything about it in the internet. Shouldn't the sensor be in a kind of parking position, should I worry about the sensor ?  I'm not sure if that is the usual behavior, maybe someone with the same or similar body can have a look. Counter example for a result of intersection of subspaces I am struggling with this question from Halmos's text, please ignore the imperative language.  "Suppose that $L, M$  and $N$ are subspaces of a vector space. Show that the equation    $$L \cap (M + N) = (L \cap M) + (L \cap N)$$   is not necessarily true.  Since each of these subspaces has origin in them, clearly there intersections could not be empty. I wasn't able to formulate an example where this result did n't hold. Any help would be highly appreciated. Idiom for a speech with redundant information Let's assume there's this random company (XYZ)--that few are aware of--is selling stocks and they're doing really well. If their stock price is at $50 and no one is buying, will it still go up?  Do stock prices only go up when people are buying? I've been reading a lot about closures and I think I understand them, but without clouding the picture for myself and others, I am hoping someone can explain closures as succinctly and clearly as possible.  I'm looking for a simple explanation that might help me understand where and why I would want to use them. Let $G$ be a finite semi-group with identity such that it has only one idempotent.Is $G$ a group?  It only remains to show that for any $a\in G$ $\exists b\in G$ such that $ab=ba=e$ where $e$ is the identity of $G$  Also $e$ is the only idempotent of $G$ .How to proceed next? Mount cifs Network Drive: write permissions and chown I have access to a cifs network drive. When I mount it under my OSX machine, I can read and write from and to it.   When I mount the drive in ubuntu, using:  sudo mount -t cifs -o username=${USER},password=${PASSWORD} //server-address/folder /mount/path/on/ubuntu   I am not able to write to the network drive, but I can read from it. I have checked the permissions and owner of the mount folder, they look like:  4.0K drwxr-xr-x  4 root root    0 Nov 12  2010 Mounted_folder   I cannot change the owner, because I get the error:  chown: changing ownership of `/Volumes/Mounted_folder': Not a directory   When I descend deeper into the network drive, and change the ownership there, I get the error that I have no permission to change the folder´s owner.  What should I do to activate my write permission? How do I create an Excel (.XLS and .XLSX) file in C# without installing Microsoft Office? What are the proposed theoretical explanations for the EmDrive? Did you know that :  Map&lt;Object,Object&gt; m1 = new HashMap&lt;Object, Object&gt;(); Map&lt;Object,Object&gt; m2 = new HashMap&lt;Object, Object&gt;(); System.out.println("m1.equals(m2) = "+m1.equals(m2)); System.out.println("m1.keySet().equals(m2.keySet()) = "             +m1.keySet().equals(m2.keySet())); System.out.println("m1.entrySet().equals(m2.entrySet()) = "             +m1.entrySet().equals(m2.entrySet())); System.out.println("m1.values().equals(m2.values()) = "             +m1.values().equals(m2.values()));   would output :  m1.equals(m2) = true m1.keySet().equals(m2.keySet()) = true m1.entrySet().equals(m2.entrySet()) = true m1.values().equals(m2.values()) = false   This is caused by the fact that AbstractCollection (which HashMap$Values inherits from) does not overrides #equals().  Do you have an idea why this is so ? Is the EmDrive, or "Relativity Drive" possible?