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zero matrix to the power of 0 Why $0^0=I$?   I'd tried prove that considering $N^0$ where N is a Nilpotent matrix and then using the Cayley -Hamilton theorem  Thanks in advance. What is projection of WGS84 in GIS software? WGS84 is a reference ellipsoid that determines coordinates of every point on Earth using latitude, longitude and height about the its surface.  This ellipsoid is often used as a base to make maps. Though, to make a map on a flat paper or a flat digital screen you need a projection too. An example of a popular projection would be UTM.   Mainstream GIS software give plethora of projection alternatives to display your data layers. Among them, they also let you display your data using WGS84 only.  I think they are using a certain projection but what projection is that? For US federal tax, whether a taxpayer is subject to AMT is determined by a complicated way in Instructions for Form 1040, Line 45, which involves an incomprehensible "worksheet to see if you should fill in form 6251".  By simplification and approximation, is there a rule of thumb for an estimate of a salary value of a taxpayer, so that a salary above the value might cause the taxpayer to pay AMT with a big possibility, and a salary below the value might not with a big possibility?  Why do the following two sources seem to differ greatly in estimates of the boundary salary value:    says that  the possible boundary salary value for a single taxpayer  to pay AMT is $53,900.  says that      Taxpayers with small incomes fall within the AMT exemption limits.   That leaves the upper middle class to face the tax, especially those   with incomes between $200,000 and $500,000    Thanks. Question statement is as in the title:     $\varphi:G\to H$ is a Lie Group Homomorphism, then $\ker \varphi$ has Lie algebra $\ker D_e \varphi$   Since $\varphi$ is a Lie group homomorphism, in particular it is continuous. As $H$ must be Hausdorff, every point set is closed, and hence $\{e\}$ is closed in $H$, which implies that $\ker\varphi = \varphi^{-1}(e)$ is closed as well.  So $N=\ker\varphi$ is an embedded Lie subgroup of $G$. Therefore, the inclusion $j:N\to G$ is a Lie Group homomorphism, and induces a Lie algebra homomorphism $D_e j :T_e N \to T_e G$. So I only need to show that $D_e j$ is in fact an isomorphism between $T_e N$ and $\ker D_e \varphi$.  But I'm stuck at this point. I know that for $X\in T_e N$, we have, for $f\in C^{\infty}(H)$,  $$ D_e\varphi (D_e j(X))(f)=D_e j (X)(f\circ \varphi)=X(f\circ \varphi\circ j)=0$$  So $D_ej(T_e N) \subset \ker D_e \varphi$.  At this point, I can either show inverse inclusion: that is, for every $X\in \ker D_e \varphi$, we can find $Y\in T_e N$ such that $D_e j (Y)=X$, or for all $f\in C^{\infty}(G)$, $Y(f\circ j)=X(f)$  Or I can somehow show that dimensions of the two spaces are the same.   But I can't proceed from here.  It'd be great if someone could help me Login required moving to Super User on iPad When moving around the sites on Stack Exchange using Safari on my iPad (not the Stack Exchange App), I am fine when moving to any site other than SuperUser.  When logged in, I can switch sites and stay logged in, but when I switch to Super User I need to log in again. If I choose not to log in and switch to another site, I am logged in there. Is there a problem with the login handling for Super User? If it helps, I log in with a Stack Exchange account, not through Facebook or Google. Update After getting fed up with the problem, I took the advice given in the answer below which was to use Chrome, and the problem still exists. The same goes for Firefox. Have I missed something?  As you can see from the screenshots below, there are no cookie settings to be found. Chrome for iOS settings page  Privacy settings page  Content settings page  Google app settings page I’m studying a PhD in Theoretical Physics (quantum field theory, to be specific), but I would like to learn more about black holes (and related areas. Does anyone have any suggestions for textbooks/available notes on these subjects for people without much background in them? Obviously I’m very comfortable with mathematics and physics in general so I’m not looking for “pop science” material. I'd like to emulate HTML's infinite page height using LaTeX/XeTeX, i.e. have the PDF pages grow higher until a manual \newpage is issued.  This could be emulated by setting a very large page height and using pdfcrop on the output (or just setting the page height manually), but I'd like a pure TeX variant and if possible support for footnotes etc at the bottom of the cropped page. New Google SEO benefit with SSL - with only a few sections in SSL, or always on? Ubuntu open to black screen with blinking underscore character after release upgrade I upgraded my Ubuntu OS from 12.10 to 13.04 and after the process of upgrading it via Software Updater, it requested me to restart the computer. I rebooted and on boot I proceeded to a purple screen with Ubuntu text and 5 dots below it and continued to a blank black screen with a blinking underscore character (_).  I have all my files in there and I can't access them. I tried the grub boot menu edit that other questions pointed out, but it didn't seem to do anything. Describing integral closure of quadratic number fields I'm facing the following problem.      Let $p$ be a prime and $ K=\mathbb{Q}(\sqrt{p}) $. I'm trying to find the integral closure of $ \mathbb{Z} $ in $ K $.   I don't really know where to start. I've managed to prove that $ \mathbb{Z}[\sqrt{p}] $ is in the integral closure and I'm almost sure $ \mathbb{Q} $ is not.   I can't think of a systematic way to approach this. I would appreciate any help. Which command controls the open file limits? How to prove the convergence of $\sum_{n\geq 0}\frac{n!^d}{(dn)!}$ for $d\geq 2$? How can we find the general formula about the convergence of the serie $\sum_{n=0}^\infty \frac{(n!)^d}{(dn)!}$ for all $d\geq2$ ?  I tried using the d'Alembert Criteria but it doesn't simplify itself enough to prove the convergence. In naive set theory, I believe ∅ = {∅} = {{∅}} is correct, but just wanted to make sure that I understood this correctly.   ∅ is an empty set, so having an empty set as an element of a set that contains nothing else is pretty much the same thing as e.g. marking the number 2 as 2.000000, right? Doesn't the symbol ∅ intrinsically imply that this is an empty set which contains it self? Are comments discouraged in a BibTeX file? biber complains about "junk" in what should be seen rather as commented-out lines in a BibTeX file.  I was wondering if comments are discouraged in BibTeX at all.  \documentclass {article}  \usepackage {filecontents}  \begin {filecontents} {ref.bib}  @book {book1,  author={author 1},  title={title 1}}   book {book2,  author={author 2},  title={title 2}}   @book {book3,  author={author 3},  title={title 3}} \end{filecontents}  \usepackage [backend=biber] {biblatex} \addbibresource {ref.bib}  \begin {document}   \nocite{*}  \printbibliography \end {document}   Here is biber's output:  INFO - This is Biber 2.1 INFO - Logfile is 'mwe.blg' INFO - Reading 'mwe.bcf' INFO - Using all citekeys in bib section 0 INFO - Processing section 0 INFO - Looking for bibtex format file 'ref.bib' for section 0 INFO - Decoding LaTeX character macros into UTF-8 INFO - Found BibTeX data source 'ref.bib' WARN - BibTeX subsystem: /tmp/No7MbpTruX/ref.bib_31209.utf8, line 13, warning: 43 characters of junk seen at toplevel INFO - Overriding locale 'en-US' defaults 'variable = shifted' with 'variable = non-ignorable' INFO - Overriding locale 'en-US' defaults 'normalization = NFD' with 'normalization = prenormalized' INFO - Sorting list 'nty' of type 'entry' with scheme 'nty' and locale 'en-US' INFO - No sort tailoring available for locale 'en-US' INFO - Writing 'mwe.bbl' with encoding 'ascii' INFO - Output to mwe.bbl INFO - WARNINGS: 1 How can I debug my JavaScript code? CPU usage is too high with Yosemite and an external monitor I use Yosemite 10.10.4 on my MacBook Pro (2012).  I used to work with an external monitor: Apple Thunderbolt 27" monitor.  The issue is that, according to Activity Monitor, the kernel_task process consumes more than 600% of CPU even when no tasks are running! The computer becomes totally unusable.   The workaround is to disconnect the external monitor, but I really need this secondary screen to work.  Why didn't Apple fix this issue in the latest Yosemite update?  Is there a fix for this? Let $a_n=\cos(a_{n-1}), L=[a_1,a_2,...,a_n,...].$ Is there an $a_0$ such that $L$ is dense in$[-1,1]?$ Where should I post my sum-up of the answers, as the OP? Sometimes as the OP (question asker), after reading through all the answers, I have my own sum-up about them. I want to tell people and all further readers about my feeling and comprehension about my desired answer and how all the answers helped towards solving my problem.  Where's the correct place to put that information? Things I've considered:   Edit into my question: No because that's not part of the question A comment under the question: I dislike that due to limited formatting in comments and its subject to removal (though the latter won't be the case) A comment under the highest-scored or the accepted answer: Still the same reason above. This is, for the time being, the place I'm posting that. A separate answer: Those thoughts probably don't qualify as an answer Proving that $A+A \not\subset 2A $ using a particular definition for sum of sets Let $X$ de a vector space and $A\subset X$. If we define $A+B :=\{a+b|a\in A, b\in B\}$ I am trying to show that    $$ \begin{equation} 2A \subsetneq A+A, \end{equation} $$  i.e., it might happen that $2A \neq A+A$. I was able to show that $2A \subset A+A$, but I am not able to come up with a counterexample to show that the converse is not necessarily true. If I assume $A+A\subset 2A$ and let $x\in(A+A)$, then by its definition we have $x=a_1 + a_2$ for some $a_1,a_2\in A$ I guess I could naively say that $a_1+a_2 = 2(a'_1+a'_2)$ where $a_i=a_i/2$ for $i=1,2$ and I guess $(a'_1+a'_2)$ might not be an element of $A$ in some concrete case but I am not able to come up with anything and I've spent more time than I would like to admit. Any suggestions?  Thank you. The number $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$. Prove that for every $n\in \mathbb N$, $2^{3^n}+1$ is divisible by $3^{n+1}$ and not divisible by $3^{n+2}$.   I was able to prove that $2^{3^n}+1$ is divisible by $3^{n+1}$ using induction. First, I check whether the statement is true for $n=1$. The statement holds since: $$2^{3^1}+1=9$$ $$3^{1+1}=9$$ Second, I assume that the statement is true for every $n\in \mathbb N$. Finally, I take that $n=n+1$. From that, the original expression becomes $(2^{3^n})^3+1$, which is a sum of cubes, and the same as: $$(2^{3^n}+1)((2^{3^n})^2-2^{3^n}+1)$$ As I assumed that $2^{3^n}+1$ is divisible by $3^{n+1}$, and one of the factors is $2^{3^n}+1$, the entire expression must be divisible by $3^{n+1}$, and so the statement is true for every $n\in \mathbb N$.  Now, I don't know how to prove that $2^{3^n}+1$ isn't divisible by $3^{n+2}$. Please explain how I can do so step-by-step, without using complicated terms.