Can somebody tell me the exact process of installing composer correctly?
Could not open input file: composer.phar
Why is angular velocity of any point about any other point of a rigid body always the same? I have the following question regarding an ideal rigid body.  Firstly, is it always true without any exceptions that the angular velocity of any point about any other point on a rigid body is always the same?  If so, what happens in case of precession, in which two different angular velocities are predefined? I've been looking around on the internet and I can't find anything in depth on how the cable/connection functions. I've only been able to find simple pin out descriptions.  If someone can link me or point me in the right direction, I would be very grateful. To make sure that we are talking about the same, I would like to post the relevant definitions I know first.  Definitions:  A pair $(G, +)$ where $G$ is a set and  $+: G \times G \rightarrow G$  is called a commutative group if it has all of the following three characteristics:   (associativity): $\forall x,y,z \in G: (x+y)+z=x+(y+z)$ (identity element): $\exists e \in G \forall x \in G: e + x = x = x +e$ (inverse elements): $\forall x \in G \exists x^{-1} \in G: x^{-1} + x = e = x + x^{-1}$ (commutativity): $\forall x,y \in G: x+y=y+x$.     A triple $(\mathbb{K}, +, \cdot)$, where $\mathbb{K}$ is a set and  $+: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$  $\cdot: \mathbb{K} \times \mathbb{K} \rightarrow \mathbb{K}$  is called a field if it has all of the following three characteristics:   $(\mathbb{K}, +)$ is a commutative group $(\mathbb{K} \setminus \{0\}, \cdot)$ is a commutative group distributive properties: $\forall x,y,z \in \mathbb{K}: x \cdot (y+z) = (x \cdot y) + (x \cdot z)$ and $\forall x,y,z \in \mathbb{K}: (y+z) \cdot x = (y \cdot x) + (z \cdot x)$     $\mathbb{N^+} := 1, 2, 3, 4, 5 ...$ (all positive integers)  $\mathbb{N}_0 := \{0\} \cup \mathbb{N}$ (the natural numbers with zero)  $\mathbb{N}$ can be both: $\mathbb{N^+}$ or $\mathbb{N}_0$.    My question: Does a field exist that has $\mathbb{N}$ as its set?  I know that $(\mathbb{Q}, +, \cdot)$ is a field and I know that $(\mathbb{N}, + )$ is only a commutative semigroup.   But maybe it's possible to define two mappings $\circ, *$ for $\mathbb{N}$ in such a way that $(\mathbb{N}, \circ, *)$ is a field.    Related:     The smallest field containing the integers is the field of rational   numbers.   Source: en-Wikipedia:   I don't know if this is true. If somebody can prove (or at least scribble the proof of) the quote it would be a good answer, I guess. How can I disable Alt-Scroll zoom in XFCE4 (Xubuntu) In XFCE4 desktop environments, alt+scroll wheel zooms my entire desktop. I'm finding this very annoying because sometimes I flick the scroll wheel to go to the top of a document then press the alt key for a some unrelated keyboard shortcut and find myself zoomed in so far I can't see anything at all. Is there any way to disable this feature? Convert XML/HTML Entities into Unicode String in Python A toplogical space $X$ is said to be second countable   if there exists a countable basis for the topology.      $X$ is separable if there exist a countable dense subset.   Show that a second countable space always is separable.      Give an example of a space that is separable but not    second countable   The first part was easy and i managed to solve it on my own  Assume that $X$ is second countable and let $B_1, B_2, B_3...$ be a countable basis. Construct the set $A$ by taking one element $x_n$ out of every $B_n$. Then $A$ becomes dense because every open set $U$ contain a $B_n$ which in turn contain the element $x_n$ in $A$.    Again, like in the question i posted yesterday, i had trouble to come up with an example but our professor gave an example during the last lecture but i dont fully understand it:  "For an example of a separable but not second countable space let $X$ be the real line provided with the cofinite topology, i.e. where a set is open if its either empty or if its complement is finite.   Then every infinite set is dense and hence $X$ is separable.  But for an arbitrary countable family of open sets $U_1, U_2...$ write $U_i = X$\ $A_i$, where $A_i$ is finite, there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,  such that the open set $X$\ $\{x\}$ is then not able to contain any $U_i$. Consequently $X$ is not second countable."  We asked him to explain it in more detail but his english is poor and nobody understood anything...:)       Def: A subset $A$ of a topological space $X$ is dense in $X$ if for any point $x$ in $X$, any neighborhood of $x$ contains at least one point from $A$.   $1.$ "Then every infinite set is dense"  Take $A$ to be one of these infinite sets. If $x \in A \subset X$ then it trivially fulfills the definition. But if we take a $x \not\in A$ why is it not possible to find a neighborhood in these finite sets that does not contain an element in $A$?  edit: its obvious if it is only open neighborhoods, open sets in the topology, but that is not specified in the definition.  $2.$ I am not fully understanding this either  "But for an arbitrary countable family of open sets $U_1, U_2...$ write $U_i = X$\ $A_i$, where $A_i$ is finite, there is an element $x \in X$\ $(\cup_i Ai)$, because $A_i$ is countable,  such that the open set $X$\ $\{x\}$ is then not able to contain any $U_i$. Consequently $X$ is not second countable."  Is it possible to express this in a simpler way? I am currently in the process of trying to repair a power supply that has had one of its DC rails fail. A resistor in the PSU has exploded and I am trying to find what value it is.    Band 1: White Band 2: Brown Band 3: Silver (Possibly Grey?) Band 4: Gold Band 5: Green    The weird thing is that the middle band appears to be silver, which should be invalid as far as I can tell. If you assume the silver is grey (although this is unlikely, as there still needs to be a grey colour as well and the band is metallic), it produces a value of resistor which is incredibly uncommon to the point it does not exist (91.8 Ohms 0.5%).  Apologies for the noobie question, I'm just stumped after hours of trying to understand why this is and coming up short :). Calculate the limit including binomial coefficients $\lim\limits_{n\to\infty}{\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{1}{5})^{n}(\frac{4}{5})^{k-n}}$  It's clear that we can simplify the limit a little bit, after which we get:  $\lim\limits_{n\to\infty}{(\frac{1}{4})^{n}\sum\limits_{k=n}^{5n}{k-1 \choose n-1}(\frac{4}{5})^{k}}$  I could further simplify the expression, but I feel like there's a more elegant solution.   Give me a hint, please Let $X$ be any distribution with defined mean, $\mu$, and standard deviation, $\sigma$.  The central limit theorem says that  $$ \sqrt{n}\frac{\bar{X} - \mu}{\sigma}  $$ converges in distribution to a standard normal distribution.  If we replace $\sigma$ by the sample standard deviation $S$, is there a theorem stating that  $$ \sqrt{n}\frac{\bar{X} - \mu}{S}  $$ converges in distribution to a t-distribution?  Since for large $n$ a t-distribution approaches a normal, the theorem, if it exists, may state that the limit is a standard normal distribution.  Hence, it would seem to me that t-distributions are not very useful - that they are useful only when $X$ is roughly a normal.  Is this the case?    If it is possible, would you indicate references that containing a proof of this CLT when $\sigma$ is replace by $S$?  Such a reference could preferably use measure theory concepts.  But anything would be great to me at this point. No association bonus for Area51? My Area51 account crossed 200 rep today, but still other accounts don't reflect the change in reputation. When will the changes be applied? or is there no association bonus for Area51 account? How to burn or mount an ISO file? I have downloaded an ISO file recently.  How do I burn it to a CD or DVD or mount it? Please give me feedback on my answer to this question. Question: For all $ n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} $   is equal to $ \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} $. Answer: False, since;  Let $a=1,  b=-1$  and $n=1$  ,$k=0$   , using Binomial expression.  $$(a+b)^n=  \sum   \binom{n}{k}  a^{k}b^{n-k}$$  Then $(1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}$  Thus :For all $n\geq1:  \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n}   \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}$  . Examples:        Why, I'd love to.   Why, of course!      I get the concept of starting a sentence with a word not completely related to the overall response, but this one seems to be a particularly odd choice&nbsp;— is there any semantic meaning I'm missing? Electorate badge rules? I just got the  badge on StackOverflow, which is nice, but I noticed a number of top users aren't yet listed (inluding Jon Skeet and Marc Gravell, who would certainly qualify).  This made me wonder how the badges are retrospectively applied. Is it just that the display is heavily cached and the badge is awarded when the user logs in again or something? Analogy: "as if" vs. "as though" This is something that confuses me from time to time. When making an analogy in literature, is it better to use the phrase "as if" or "as though", or is it completely a style thing?  E.g.      She looked frazzled, as if she had just gotten off of a roller coaster.   or     She looked frazzled, as though she had just gotten off of a roller coaster. Did Harry Potter lose Voldemort's powers when he lost his connection to Voldemort? Harry noted to Dumbledore that he had a number of things in common with Voldemort. And it was later revealed that some of Voldemort's power was transferred to Harry when Voldemort tried to kill him. So, what powers did Harry get from Voldemort (i.e. Parseltongue and others) and did they all leave him when he stopped being a horcrux?    Update In the Chamber of Secrets Dumbledore states that Harry has many qualities that Salazar Slytherin values in his hand picked students. Parseltongue - resourcefulness - determination - a certain disregard for rules.    So, does anyone know what specific powers Harry got from Voldemort and is there anyway to tell if he lost anything else besides Parseltongue? "Can/may/will you help me with this?" Which word to use when we ask for help?  Some conditions:     We know that the person asked is able to do it. We don't know if the person asked is able to do it. I define a shell keybinding in my ~/.bashrc like so:  bind -x '"\e\C-w":"/usr/bin/reset"'   This does make Ctrl-Alt-w launch the reset executable. However, it leaves the terminal in a state that resembles the one you get into by pressing Ctrl-v. So, for example, if immediately after resetting in this way I press Ctrl-p, instead of invoking the action of that shortcut (repeat previous command) I get ^P printed.  Any idea what's causing this problem and how to fix it?  EDIT: I am attaching a Python program that reproduces the messed-up terminal state I am trying to fix with the reset. I encounter this condition very often while debugging multi-threaded applications:  import pdb, thread, time  def interrupt():     time.sleep(2)     pdb.set_trace()  thread.start_new_thread(interrupt, ()) raw_input('? ')   To reproduce the messed-up state, just run the code above, and press Enter when you see the debugger invoked (it will print a bunch of text, ending with pdb.set_trace(), about two seconds after you run the program and the initial prompt appears). How to automatically install entire Google Web Fonts collection? Why do different humans look different? Although farmers appear to be able to tell their cows apart, cows look very much alike to me. And this similarity in appearance seems to be a general trait across the animal kingdom: one individual of one animal species looks much like another.  Why do human facial and body shapes vary so much?  I understand that human beings live in many different environments (e.g. hot and cold) and have adapted to these (e.g. through lighter or darker skin color), but there seems to be much variation that has no apparent evolutionary purpose. So does the difference have a purpose in itself? That is, did we evlove to vary in appearance, and what is the purpose of this variation?