Prove this inequality for all reals $x$ and $y$:  $$|x-y|≥|x|-|y|$$
I've seen the full proof of the Triangle Inequality  \begin{equation*} |x+y|\le|x|+|y|.  \end{equation*} However, I haven't seen the proof of the reverse triangle inequality:  \begin{equation*} ||x|-|y||\le|x-y|. \end{equation*} Would you please prove this using only the Triangle Inequality above?  Thank you very much.
If I write the following class:  public class Example {        int j;       int k;        public Example(int j, int k) {            j = j;            k = k;       }        public static void main(String[] args) {            Example exm = new Example(1,2);            System.out.println(exm.j);            System.out.println(exm.k);       }  }   The program compiles, but when I run the program, the main method will print out two 0s. I know that in order to say that I want to initialize the instance variables in the constructor I have to write:  this.j = j; this.k = k;   But if I don't write it, then which variable is evaluated (or considered) in the constructor (on the left and on the write hand side of the expressions)? Is is the argument or the instance variable? Does it make a difference?  Are there other cases where the use of this is obligatory? For $u:\mathbb{R}\times [0,1]$ with boundary conditions $u(0,x)=\cos (2\pi x)$ and $u(t,0)=u(t,1)$.   Solve $u_t=u_{xx}+u_x$.  I had this on an exam and tried to write $u$ as a product of two single variate functions and convert to an ode using the usual methods but things got messy and I couldn't finish. I also tried to write $u$ as a Fourier series in $x$ with coefficients $a_k(t)$ depending on $t$, but this also didn't seem to yield anything.  Is there a more clever way to approach this, possibly with Fourier series? I heard from a friend there may be a Fourier transform method since the equation was "homogeneous in momentum space."   Thanks! How to safely upgrade from an older Ubuntu version to a newer one? How can I safely upgrade an Ubuntu installation from one release to another? What should I do before upgrading to ensure it goes smoothly, and is there any clean-up I need to do afterwards?  In particular, do I need to do anything special if I am using packages from PPAs, if I have a graphics card that needs drivers, or if I'm running a custom Linux kernel?  This is a general question , but as example we take Ubuntu 12.04 and Ubuntu 12.10 I have a small question regarding using sudo with output redirect &gt;. To enable IP forwrding, someone can use the command:  echo 1 &gt; /proc/sys/net/ipv4/ip_forward   Executing this command will give permission denied as it requires root privileges. However, executing the same command with sudo gives also permission denied error! It seems that output redirect &gt; does not inherit the permissions of the preceding command echo. Is this right?  As a workaround I do:  echo 1 | sudo tee /proc/sys/net/ipv4/ip_forward   Is this the best way to do it? Am I missing something?  Please, note that this is an example and it applies to all commands that use output redirect. This is problem 6.2 from the 3rd edition of Principles of Mathematical Analysis.      Problem 6.2: Suppose $f\geq 0$, f is continuous on $[a, b]$, and $\int_a^b f(x) \, dx = 0$. Prove that $f(x)=0$ for all $x \in [a, b]$.   I'm looking for a critique of my proof. It's a pretty easy problem, but I am always wary of making too bold of assumptions, especially on these low level/fundamental proofs. I'll be using Rudin's notation and refer to theorems from the text (If I should include the text of each theorem, feel free to leave a comment... I'm lazy but could probably use the TeX practice :p)  Proof: Assume, for contradiction, that $f&gt;0$. Then, for any partition $P$ we have the Lower Riemann Sum: $L(P, f)=\sum_{i=1}^n m_i \, \Delta x_i$. At least one $\Delta x_i$ must be positive, since $a &lt; b$, and each $m_i$ must be positive since we have $f&gt;0$ by assumption, so certainly $\sup f &gt; 0$. That means $L(P, f)&gt;0$. Thus, we have: $$0 = 0(b-a) &lt; L(P,f)\leq \sup L(P, f) = \inf U(P,f) =L$$ where the last string of equalities holds because our function is continuous on a compact interval, so is integrable by theorems 6.8 and 6.6. So our integral has value $L&gt;0$. This is in contradiction to our given assumption that $\int_a^b f(x) \, dx = 0$, so we must have that $f=0$ on $[a, b]$. $\Box$  So, I am wondering if my proof is correct (and is presented well). Also if someone could enlighten me as to what Rudin means when he say "Compare this with exercise 1," I'd be appreciative. Is it a hint or is there something else he expects you to notice? there are a lot of things I could compare :)...  Exercise 6.1: Suppose $\alpha$ increases on $[a, b]$, $a \leq x_0 \leq b$, $\alpha$ is continuous at $x_0$, $f(x_0)=1$, and $f(x)=0$ if $x\neq x_0$. Prove that $f$ is Riemann-Stieltjes Integrable and that $\int f \, d\alpha = 0$ What is Renly's basis for removing Joffrey? Related question:   I get that he considers himself a better king than Stannis, but why isn't he ?  Did Renly know about Joffrey being illegitimate? I always thought he didn't and was distrustful that Cersei would accept Ned as Regent.  I think in both the books and the series, Renly finds out Joffrey is illegitimate from Stannis' letters, but rewatching that clips makes me think Renly was with Ned, Varys and LF in the plan to arrest Joffrey and Cersei.  [looks up something]  Oh also . So Renly did know? How can I check if a single character appears in a string? Let A,B be in $M_{mxn}(\mathbb{R})$  Prove that if rank(A + B) = rank(A) + rank(B), then col(A) ∩ col(B) = {0}  I started with a proof by contradiction, since we know that rank(A + B) ≤ rank(A) + rank(B). I assume that if rank(A + B) &lt; rank(A) + rank(B), then col(A) ∩ col(B) = {0}. I realize that we don't have to do the > case. However I am a little confused on how to show col from rank. I have not done much proofs before this and need some guidance. I know that for a simple graph such as this :  node - node - node -node   Removing the first and last vertex will not disconnect the graph. So far, I can think of two cases: Where there are leaves, or nodes with degree of 1. Where there are no leaves, which means all nodes have degrees 2 or more.   In the first case, removing the leaf will not disconnect the graph. In the second case, removing any should not disconnect the graph. Problem is, I don't know how to formulate this into a good proof. If $H \leq G$ and $[G:H]! \leq |G|$ then $G$ is not simple $P_n(x):=1+ \sum_{m=1}^n\dfrac{x^m}{m!}$ has no real root for even $n$ and exactly one real root for odd $n$ This is a problem from Introduction to Representation Theory by Etingof et al. I have no idea how to approach it. My friend wrote a program that uses Gaussian elimination to find a basis, which starts with x, y, [x, y], [y, [x, y]], ... but freezes once it gets to elements with 9 or more occurrences of x, y.  I expect that there is some way to characterize the nonzero elements of this Lie algebra (i.e., elements that are "immune" to being shown to be zero through repeated application of the Jacobi identity) which could be turned into a way to enumerate them. I don't see what it is, though.  It's also totally unclear to me how to show that some set we know spans the Lie algebra (proven using some criterion as described in the previous paragraph) is actually linearly independent, unless we find a faithful representation... but I have no idea how to find a faithful representation for an infinite-dimensional Lie algebra.  Hints would be greatly appreciated.  Edit: After looking at the partial output, we can form the "straightforward" conjecture that:   If $n \geq 4$ is even, then there is one basis element of grade $n$, namely $\mathrm{ad}([x, y])^{(n-4)/2}[y, [y, [y, x]]]$. If $n \geq 5$ is odd, then there are two basis elements of grade $n$, namely $[x, z]$ and $[y, z]$ where $z$ is the single basis element of grade $n-1$.   I also have a different conjecture, which is mutually exclusive with the above: that if we flatten out all nonzero elements into the form ${\rm ad}(e_1) {\rm ad}(e_2) \ldots {\rm ad}(e_{n-1}) e_n$ using the Jacobi identity, then in the sequences $e_1, \ldots, e_{n-1}, e_n$ and $e_1, \ldots, e_n, e_{n-1}$, any span containing two x's must contain at least two y's and any span containing five y's must contain at least two x's, otherwise we can use the Jacobi identity to rearrange the element somehow to show that it vanishes. If this is true, then the long-range behaviour must be:   For $n \geq 8$ there is exactly one basis element of grade $n$, namely $[... [y, [x, [y, [y, [x, y]]]]] ...]$ (the pattern $y x y$ repeats).   I still don't see an approach to proving that none of the elements in this sequence actually vanishes, though.  Edit 2: Here is the hint given by Etingof in an email:     No, there are no published solutions, since these problems are used as   a homework for classes. This is indeed a nontrivial problem.      This is the positive part of the affine Kac-Moody algebra of type   A_2^2, which is the twisted loop algebra for sl(2). So what you need   to do is to relate this Lie algebra to loops into sl(2). I see from   your post that you are on the right track. If $I_n=\int _0^{\pi }\sin^{2n}\theta \:d\theta $, show that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, and hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $  Hence calculate $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$  I knew how to prove that $I_n=\frac{\left(2n-1\right)}{2n}I_{n-1}$, but I am not very good at English, what does it mean Hence $I_n=\frac{\left(2n\right)!}{\left(2^nn!\right)2}\pi $ do we need to prove this part as well or is it just a hint to use? and for the other calculation to find $\int _0^{\pi }\:\sin^4t\cos^6t\:dt$ is there something in the first part that I can do to help me solve this question because otherwise it becomes very long and it's part of the question. Fred and George Weasley are known for their trouble-making and general &quot;solemnly swearing they are up to no good&quot;. Their antics/adventures give something to smile about and show us not all wizards are as serious as Professor McGonagall and Hermione. This gives them a reputation of being less intelligent and indeed they only achieved three OWLs. However, we also see the twins have quite a brilliant side! Being able to open and successfully run a shop in Diagon Alley without having completed their Hogwarts education is no easy take (IMO, anyway). Their creations helped students get out of class, our heroes out of tight spot, and even stop our enemies execute an evil plot. Ginny may have said it best:  &quot;We’ve been hearing explosions out of their room for ages, but we never thought they were actually making things,&quot; said Ginny. &quot;We thought they just liked the noise.&quot; Harry Potter and the Goblet of Fire  Do we know how many things Fred and George invented? How do I prove if $\sup\{ (n!)^{1/n} | n \in\Bbb N\} = +\infty$?  I don't know how to properly explain it Compatible version of "Adobe Shockwave Player?" Is there a Ubuntu compatible version of the "Adobe Shockwave Player?" I'm running Firefox and when I try to bring up one of my School's websites it says missing plugin. When I click on "Install missing plug-in" it says none found, so when I click on manually install plug-in it takes me to Adobe Shockwave Player installation, but there is no ubuntu version to install??? On the Firefox plug-in finder service it says "Unknown Plugin (application/x-director) How do you return 1 value per row of the max of several columns:  TableName  [Number, Date1, Date2, Date3, Cost]   I need to return something like this:  [Number, Most_Recent_Date, Cost]   Query? How to define a Bijection between $(A^B)^C$ and $A^{B \times C}$ What does a linear transformation do to a vector in $\mathbb{R}^2$ $\newcommand{\Reals}{\mathbb{R}}$Doing some linear algebra. This question has me intrigued. I imagine it causes the vector to rotates the $x$ axis. But I mean, is there more to it?  $$ T_{A(\theta)}: \Reals^2 \to \Reals^2,\qquad A(\theta) =  \left[\begin{array} {rr}\cos \theta &amp; -\sin \theta \\   \sin \theta &amp; \cos \theta \\ \end{array}\right]. $$ What, geometrically, does the linear transformation $T_{A(\theta)}$ do to a vector in $\Reals^2$? Coordinating conjunction immediately followed by parenthetical — Is a comma needed or not? states:     If a dependent clause, or an introductory phrase requiring to be set off by a comma, precedes the second independent clause, no comma is needed after the conjunction.   Strunk's example illustrates the point:     The situation is perilous, but if we are prepared to act promptly, there is still one chance of escape.   However, I would be tempted to put a comma after the "but," like this:     The situation is perilous, but, if we are prepared to act promptly, there is still one chance of escape.   So, I know you're thinking, "Well, Strunk's Rules of Usage says that you're wrong. No comma."   But is Rules of Usage really the best source? I'd like to see this rule in a more contemporary and authoritative source, like the Chicago Manual of Style.  There are many instances —  — where the comma is included before a parenthetical comment that comes after a coordinating conjunction.  Any authoritative source on the issue?