A source is emitting photons towards all directions. An observer sitting at millions of light years away can receive the photon from all the places. There will be infinite directions from where the photons can be observed. Is it possible if the source is emitting photons towards infinite (not huge) directions?
It would seem that far-away stars are at such a distance that I should be able to take a step to the side and not have the star's photons hit my eye. How do stars release so many photons to fill in such great angular distances?
For which $n$ does $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$ imply $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$ I'm having trouble finishing a problem on an old national competition.  As the title states, the question says asks:  Given $a,b,c \neq 0,a+b=c$ such that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}$,  Find all integers $n$ such that $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}=\frac{1}{a^n+b^n+c^n}$.  I know that if it is true for $n$, it is also true for $-n$ since $(a^n+b^n+c^n)(\frac{1}{a^n+b^n+c^n})=1$ and $(a^n+b^n+c^n)(\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n})=1$ and the statment follows for $-n$ given that we can make the substitution for $n$. Thus I found out that $n=-1$ satisfies the equality.  I also know that it fails for all even $n$. If $n$ were even, $(a^n,b^n,c^n)$ would all be positive, and Titu's Lemma asserts that $\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}\geq \frac{9}{a^n+b^n+c^n}$. Thus $n$ must be odd.  I also observed that $(2,-1,1)$ was a solution, and from here noted that this particular solution satisfies all odd $n$.  I have been trying to prove that all other solutions satisfy all odd $n$ with no success.  Any help would be greatly appreciated. Thanks! When new users edit a post, the review is posted to a review queue and other reputable users can review the edit and choose to approve or reject it. However, I feel that a lot of people do it pretty quickly.    to . He was fixing broken links in the answer, and as soon as I saw it I knew it would have been a good edit. However, only one reviewer found so, and three others rejected it. I ended up manually doing the edit, and poor Onato didn't get any credit for it (in fact, he got an undeserved 'rejected' for it).  The point I want to bring is that, first and foremost, the person who wrote the answer probably knows best if an edit is acceptable or not. Could there be a time frame where the author of the answer (or question) can overturn a review decision?    EDIT: one more  was rejected by overeager reviewers as well. I absolutely disagree with the notion that the community knows better. The community repeatedly proves that it only wants Internet points.    EDIT: I told a commenter to fix something himself in one of my posts, but . I went back and applied the changes myself. It's not credited to him and his stats got a rejected edit now (which really should be an approved edit). The reason ("This edit is incorrect or an attempt to reply to or comment on the existing post") is blatantly wrong and shows once more that the community has much less insight into edit quality than the people who wrote the answers in the first place. It's true that the UI didn't make it easy for them, but those guys clearly didn't read the comment thread; I did, because as the author, I was notified.    EDIT: This time around, a user fixed a blatant (though small) mistake in one of my accepted answers. One hour later, another, unrelated 10k user , even though it had been approved by the community. I went back to undo his undo.  This goes on again to show that an active user is the best judge for the quality of an edit, and that since I can have the last word on edits on my own posts, I should be allowed to retroactively overturn a community edit vote and accordingly credit the person who proposed the change.    EDIT I wrote a  years ago. Some guy found it and tried to fix it but the community thought that it wasn't worth it. I think that the changes are great and once more I'll integrate them by myself into my post. Unfortunately for this user, he won't be credited for it. I still don't understand why I can't have a binding vote on edits on my posts. "One of the children who was" vs. "one of the children who were" In the construction "one of the [plural noun] who ...", should the verb agree with "one" or "[plural noun]"?   For example, which of the two following sentences is grammatically correct? Or are both acceptable?        She was one of the several children who was sold at the auction    She was one of the several children who were sold at the auction. I'd like to update the label of an operator when it's called. But this :  def execute(self,context):     self.bl_label = "good night"   ... apparently doesn't work if the bl_label property has already been defined. Am I wrong ? Let $U\subset\mathbb R^3$ be an open set, and $f:U\to \mathbb R$ be a smooth function. Suppose that the level set $S=f^{-1}(\{0\})$ is non-empty, and that at each $p\in S,$ the gradient $\overrightarrow \nabla f(p)$ is not the zero vector. Then $S$ is a smooth two-dimensional surface in $U$, and $p\mapsto \overrightarrow \eta(p)=\frac{1}{||\overrightarrow \nabla f(p)||}\overrightarrow \nabla f(p)$ defines a smooth unit-length normal vector field along $S$. At each $x\in U,$ write $H(f)_{(x)}$ for the $3\times 3$ Hessian matrix specified by $$(H(f)_{(x)})_{ij}=\frac{\partial^2f}{\partial x_i\partial x_j}(x).$$   Show that, at each $p\in S$, the second fundamental form $II_p: T_p(s)\times T_p(s)\to \mathbb R$ is the symmetric bilinear map    $$II_p(\overrightarrow v,\overrightarrow w)=\frac{-1}{||\overrightarrow \nabla f(p)||}\overrightarrow v\cdot H(f)_{(p)}\overrightarrow w,$$for all $\overrightarrow v ,\overrightarrow w \in T_p(s)$.       (Here, we view the tagent space $T_p(S)$ as the two-dimensional subspace $(span\{ {\overrightarrow \eta(p)}\})^{\bot}$ of $\mathbb R^3$.      Edit: Actually my question is why the second fundamental form under the usual definition can be written in this way.      Definition: The quadratic form $II_p$, defined in $T_p(S)$ by $II_p(v)=-&lt;d  N_p(v),v&gt;$ is called the second fundamental form of $S$ at $p$, where $dN_p:T_p(S)\to T_p(S)$ is the differential of the Gauss map.   Hopefully, I express this problem explicitly. I was just wondering how to prove this statement.   I took a diffrential geometry class last semester, and when I organized my notes this morning, I found this statement, but there was no proof...  Looking forward to an understandable explaination. Thanks in advance.     Edit 2:Furthermore, show that, at each point $p\in S$, the expression   $$\phi_p(z)=det\pmatrix{-H(f)_{(p)}-zI_{3\times 3} &amp; \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)&amp; 0}$$   (the underlying matrix here is $4\times 4$) defines a second-degree polynomial whose roots $\lambda_1$ and $\lambda_2$ are $||\overrightarrow \nabla f(p)||k_1$ and $||\overrightarrow \nabla f(p)||k_2$, where $k_1$ and $k_2$ are the principal curvatures of $S$ at $p$.      Also, if a non-zero vector $\pmatrix {\overrightarrow v \\c}$ lies in the kernel of the $4\times 4$ matrix $$\pmatrix{-H(f)_{(p)}-\lambda_jI_{3\times 3} &amp; \overrightarrow \nabla f(p)\\\ \pm \overrightarrow \nabla f(p)&amp; 0},$$   then $\vec v$ is a non-zero element of $T_p(S)$ and lies in the "principal direction" corresponding to $K_j$. If I have the following enum:  public enum ReturnValue{     Success = 0,     FailReason1 = 1,     FailReason2 = 2     //Etc... }   Can I avoid casting when I return, like this:  public static int main(string[] args){     return (int)ReturnValue.Success; }   If not, why isn't an enum value treated as an int by default? Who did Khan get his Starfleet insignia from? Find all integer solutions to Diophantine equation $x^3+y^3+z^3=w^3$ Compute all integer solutions to the equation  $$x^3+y^3+z^3=w^3$$ What is the best way to implement constants in Java? I've seen examples like this:   public class MaxSeconds {    public static final int MAX_SECONDS = 25; }   and supposed that I could have a Constants class to wrap constants in, declaring them static final. I know practically no Java at all and am wondering if this is the best way to create constants. How does the garage system work in GTA V? Find a polynomial f(x) of degree 5 such that 2 properties hold. I have been trying to find a polynomial $f(x)$ such that these $2$ properties hold:   $f(x)-1$ is divisible by $(x-1)^3$ $f(x)$ is divisible by $x^3$   To start, I set $f(x) =ax^5 + bx^4 + cx^3 + dx^2 + ex + f$. This is divisible by $x^3$, so $d, e, f $  must be $0$. So polynomial $f(x) = ax^5 + bx^4 + cx^3$  This minus $1$ is divisible by $(x-1)^3$. So I used synthetic division and got that the remainder of $ax^5 + bx^4 + cx^3$ divided by $(x-1)$ is $a+b+c-1$. But it should divide out evenly, so $a+b+c-1 = 0$.  From here, I can't find any other equations involving the three variables. Could someone have any suggestions for how to continue? Thanks in advance. :) I thought the way to undo a \def was to define it to \relax, but that does not appear to work.  Also, am not sure why this produces no output if I try to use \def instead of \let in my attempt to do an undef.  \documentclass{article}  \begin{document}  \def\CaseA{}%  \ifdefined\CaseA%     \def\ProcessCaseA{}%     % Do stuff that uses \ProcessCaseA     %\def\ProcessCaseA\relax% no output if use this, and comment next line     \let\ProcessCaseA\relax% no output if comment this, and use above line \fi  % Now do stuff that requires \ProcessCaseA NOT be defined \ifdefined\ProcessCaseA     ERROR: \textbackslash ProcessCaseA is still defined \else     SUCCESS: \textbackslash ProcessCaseA is not defined \fi  \end{document} What do you wish someone had told you before your first commute? How to add features to a PostGIS layer using Quantum GIS? I am using Quantum GIS and the PostGIS dbms. I am able to call the PostGIS layer into QGIS but i am unable to add feature to it. I am using the method of adding features to a Shapefile but still it is not getting reflected in the layer, and I don't know how to save it to the PostGIS database if I am able to add the feature to the PostGIS layer.  I may not have used some technical term for this problem because i don't know what to call it, I am expressing the problem in my own terms. Change footer in beginning of chapters I want the footer of all pages to be "Page N°X" on the right. I used this:  \usepackage{fancyhdr} \pagestyle{fancy} \fancyhf{} % clear all header and footer fields \fancyhead[R]{\leftmark} \fancyfoot[R]{\footnotesize Página N$^\circ$ \thepage\ } \renewcommand{\headrulewidth}{0pt} \renewcommand{\footrulewidth}{0pt}   It works on every page but the chapter beginning ones. Find the limit of $\lim\limits_{(x,y)\to (0,0)} \frac{x^2y^2}{x^2y^2+(x-y)^2}$ Run TTY script every time I open a GUI terminal Where can I find my .emacs file for Emacs running on Windows? I tried looking for the .emacs file for my Windows installation for Emacs, but I could not find it. Does it have the same filename under Windows as in Unix?  Do I have to create it myself? If so, under what specific directory does it go? Is it true every ideal of $\mathbb{Z}[x]$ is finitely generated? Let $I$ be any ideal of $\mathbb {Z}[x]$, and $I=\left&lt;{f_1,f_2 , ...}\right&gt;$.  Without loss of generality, $f_1$ has minimal (absolute value) coefficient of constant term (if there exist).  Then, let $g$ be any element of $I$ .  We have for some integer $m$ that $g - mf_1$ is element of $I$ (minimal coefficient and division algorithm).  I guess this process terminate finitely, but I don't know how can I do next step.  ( ex. $g-mf_1, g-mf_1-nf_2, ...$)  Please help me and thanks for in advance. Prove existence of disjoint open sets containing disjoint closed sets in a topology induced by a metric. Question: Let $(X, d)$ be a metric space. Let $A$ and $B$ be disjoint subsets of $X$ that are closed in the topology induced by $d$. Prove that there exist disjoint open sets $U$ and $V$ such that $A\subset U$ and $B\subset V$.   Is this proof correct? : By assumption, $A\cap B=\emptyset$, so there is no pair like $(x_A,x_B)$ (i.e., $x_A\in A$ and $x_B\in B$), such that $x_A=x_B$. Because every metric space is Hausdorff, so for every pair $(x_A,x_B)$ there are two disjoint open sets $U_A\ni x_A$ and $V_B\ni x_B$; and thus union of all $U_A$s (which is an open set containing A) are disjoint from union of all $V_B$s (which is an open set containing A).   Thanks a lot for any help.    PS In special case of standard metric topology on $\mathbb{R}$, if $[a,b]$ and $[c,d]$ are disjoint (say $c&gt;b$), intuitively it is apparent that there is a 'space' between $c$ and $b$ so we can have two disjoint open sets, each of them containing $[a,b]$ or $[c,d]$.