Say we let $f$ be a real-valued function, let $S\subseteq \text{dom}(f)\subseteq R$, let $a\in \bar{S}$, and let $L\in R$.  We know the $\delta -\epsilon$ condition for $\lim_{x\to a} f(x)=L$ is:  $$\forall \epsilon &gt;0: \exists \delta &gt;0: \forall x\in S: |x-a|&lt;\delta \to |f(x)-L|&lt;\epsilon.$$  So given all that, how can we show $\lim_{x\to 0}x \cdot \sin(\frac 1 x )=0$ by proving the condition I wrote above, for $a=0$, $S=(0,\infty)$, $f(x)=x\cdot \sin(\frac 1 x)$ for all $x\in (0,\infty)$, and $L=0$?
How to find limit as $x \to 0$ of $x\sin(1/x)$?    For $x^2\sin(1/x)$, I know it's $0$ since by the Squeeze theorem, $-x^2 \le x^2\sin(1/x) \le x^2$, but for $x\sin(1/x)$, I run into some problems when applying Squeeze theorem.
Subset data frame based on number of rows per group Finding the error in this proof that 1=2 I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof:  If $x = y$, then  $$ \begin{eqnarray} x^2 &amp;=&amp; xy \nonumber \\ x^2 - y^2 &amp;=&amp; xy - y^2 \nonumber \\ (x + y)(x - y) &amp;=&amp; y(x-y) \nonumber \\ x + y &amp;=&amp; y \nonumber \\ 2y &amp;=&amp; y \nonumber \\ 2 &amp;=&amp; 1 \end{eqnarray} $$    My best guess is that the error starts with the line $2y = y$. If we accept that $x + y = y$ is true, then   $$ \begin{eqnarray} x + y &amp;=&amp; y \\ x &amp;=&amp; y - y \\ x &amp;=&amp; y = 0 \end{eqnarray} $$  Did I find the error? If not, am I close? I am looking for instructions on how to install cedarview graphics drivers on Ubuntu 13.04 32bit.  My graphics hardware is a GMA3600 aka Power VR SGX545.  EDIT: It seems that they are not in Raring yet, but only Precise:   EDIT2: they seem to be closed source. I'm wondering if there's any log which stores information about a particular program such as when the program has started and closed manually on Windows such as MS Word, or Chrome. Any ideas? How do I read from /proc/$pid/mem under Linux? The  tells me that /proc/$pid/mem “can be used to access the pages of a process's memory”. But a straightforward attempt to use it only gives me  $ cat /proc/$$/mem /proc/self/mem cat: /proc/3065/mem: No such process cat: /proc/self/mem: Input/output error   Why isn't cat able to print its own memory (/proc/self/mem)? And what is this strange “no such process” error when I try to print the shell's memory (/proc/$$/mem, obviously the process exists)? How can I read from /proc/$pid/mem, then? Can you overload controller methods in ASP.NET MVC? I'm curious to see if you can overload controller methods in ASP.NET MVC.  Whenever I try, I get the error below.  The two methods accept different arguments.  Is this something that cannot be done?     The current request for action 'MyMethod' on controller type 'MyController' is ambiguous between the following action methods: Which set is unwell-orderable? In textbook it says that every well-orderable set is equipotent to an initial ordinal number. However, of course unwell-orderable cannot equipotent to any ordinal number, but is there any set is unwell-orderable? By axiom of regularity we know that every set is well-founded, so can we conclude that every set is well-orderable? If not, which set is unwell-orderable? Convex function as supremum of countably many affine functions Let $f:\mathbb{R}^n\to \mathbb{R}$ be a convex function. It is well-known that (see e.g.~) that there exists a set $A$ such that  $$ f(x)=\sup_{a\in A}A(a)x+b(a), \quad \forall x\in \mathbb{R}^n, $$ where $A(a)\in\mathbb{R}^{n\times n}$ and $b(a)\in \mathbb{R}^n$.  I was wondering, is there any reference saying we can choose $A$ as a countable set? Ball to Ball Collision - Detection and Handling With the help of the Stack Overflow community I've written a pretty basic-but fun physics simulator.    You click and drag the mouse to launch a ball.  It will bounce around and eventually stop on the "floor".  My next big feature I want to add in is ball to ball collision.  The ball's movement is broken up into a x and y speed vector.  I have gravity (small reduction of the y vector each step), I have friction (small reduction of both vectors each collision with a wall). The balls honestly move around in a surprisingly realistic way.  I guess my question has two parts:   What is the best method to detect ball to ball collision? Do I just have an O(n^2) loop that iterates over each ball and checks every other ball to see if it's radius overlaps? What equations do I use to handle the ball to ball collisions? Physics 101 How does it effect the two balls speed x/y vectors?  What is the resulting direction the two balls head off in?  How do I apply this to each ball?     Handling the collision detection of the "walls" and the resulting vector changes were easy but I see more complications with ball-ball collisions.  With walls I simply had to take the negative of the appropriate x or y vector and off it would go in the correct direction.  With balls I don't think it is that way.  Some quick clarifications: for simplicity I'm ok with a perfectly elastic collision for now, also all my balls have the same mass right now, but I might change that in the future.    Edit: Resources I have found useful  2d Ball physics with vectors:  2d Ball collision detection example:     Success!  I have the ball collision detection and response working great!  Relevant code:  Collision Detection:  for (int i = 0; i &lt; ballCount; i++)   {       for (int j = i + 1; j &lt; ballCount; j++)       {           if (balls[i].colliding(balls[j]))           {             balls[i].resolveCollision(balls[j]);         }     } }   This will check for collisions between every ball but skip redundant checks (if you have to check if ball 1 collides with ball 2 then you don't need to check if ball 2 collides with ball 1.  Also, it skips checking for collisions with itself).  Then, in my ball class I have my colliding() and resolveCollision() methods:  public boolean colliding(Ball ball) {     float xd = position.getX() - ball.position.getX();     float yd = position.getY() - ball.position.getY();      float sumRadius = getRadius() + ball.getRadius();     float sqrRadius = sumRadius * sumRadius;      float distSqr = (xd * xd) + (yd * yd);      if (distSqr &lt;= sqrRadius)     {         return true;     }      return false; }  public void resolveCollision(Ball ball) {     // get the mtd     Vector2d delta = (position.subtract(ball.position));     float d = delta.getLength();     // minimum translation distance to push balls apart after intersecting     Vector2d mtd = delta.multiply(((getRadius() + ball.getRadius())-d)/d);        // resolve intersection --     // inverse mass quantities     float im1 = 1 / getMass();      float im2 = 1 / ball.getMass();      // push-pull them apart based off their mass     position = position.add(mtd.multiply(im1 / (im1 + im2)));     ball.position = ball.position.subtract(mtd.multiply(im2 / (im1 + im2)));      // impact speed     Vector2d v = (this.velocity.subtract(ball.velocity));     float vn = v.dot(mtd.normalize());      // sphere intersecting but moving away from each other already     if (vn &gt; 0.0f) return;      // collision impulse     float i = (-(1.0f + Constants.restitution) * vn) / (im1 + im2);     Vector2d impulse = mtd.normalize().multiply(i);      // change in momentum     this.velocity = this.velocity.add(impulse.multiply(im1));     ball.velocity = ball.velocity.subtract(impulse.multiply(im2));  }   Source Code:     If anyone has some suggestions for how to improve this basic physics simulator let me know!  One thing I have yet to add is angular momentum so the balls will roll more realistically.  Any other suggestions?  Leave a comment! Spatial query to return multiple values I have a woodland scheme advice layer that has a polygon for each area of advice (brown layer in the picture). I also have a polygon layer of UK local authorities (red boundary layer in the picture). I would like to perform a spatial query which returns the local authority name for each woodland advice scheme. This would work fine if each woodland advice scheme was found in just one local authority boundary. However many are found across a boundary of two different authorities. For instance, one scheme sits across the boundaries of the authorities of Fife and Angus. The scheme is the yellow selected polygon in the picture and the red line is the authority boundary. How do I get the spatial join to return a value of Fife; Angus (or something similar to identify two values) instead of just Fife or Angus. PhD in Europe without scholarship from non-European homeland Can you terminate more than one common on the furnace C terminal? Covariance of Poisson and Conditional Binomial RV's Problem Statement  Let $X$ and $Y$ be random variables such that $X \sim \text{Poisson}(\lambda)$ and $Y|X \sim \text{Binomial}(x+1,p)$. Find $\text{Cov(X,Y)}$.  Attempt at a Solution  I would like to be able to write $Y$ as $X + Z$ where $Z$ is Poisson and independent of $X$, since then by  the variance is easy to compute. To this end, write $Y|X$ as a sum of independent binomials: $ Y|X = Z_1|X + Z_2|X, $ where  $Z_1|X \sim \text{Binomial}(1,p)=\text{Bernoulli}(p)$ and $Z_2 | X \sim \text{Binomial}(x,p).$ Then $Z_2 \sim \text{Poisson}(\lambda p)$, which we can obtain by directly computing the pdf: \begin{aligned} P(Z_2 = z) &amp;= \sum_x P(Z_2=z, X=x)\\ &amp;= \sum_x P(Z_2=z\mid X=x)P(X=x) \\ &amp;= \sum_{x=y}^\infty \frac{e^{-\lambda} \lambda^x}{x!}\frac{x!}{z!(x-z)!}p^z(1-p)^{x-z} \\ &amp;= \frac{(\lambda p)^z}{z!}e^{-\lambda p} \sum_{x=z}^\infty \frac{[\lambda(1-p)]^{x-z} e^{-\lambda(1-p)}}{(x-z)!} \\ &amp;= \frac{(\lambda p)^z}{z!}e^{-\lambda p}. \end{aligned} However, I cannot get the same approach to simplify for $Z_1|X$, which instead involves the CDF of $X$. I also tried this approach on $Y|X$ directly, hoping to obtain a sum in the form of the expectation of $X$, but I can't seem to reduce it. Even if I could, I'd still need to compute $E[XY]$ for the covariance. I suspect there may be a better approach than grinding through several more sums. How do I convert materials from Blender Internal to Cycles? I know this may be a tough or confusing question, but I have numerous existing materials I created and reuse for a project, which were all prepared for Blender Internal.  Now, I want to start the transition to Cycles, but I've hit a roadblock. How can I transition my materials without completely starting over? Or should I just forget it and start with fresh materials? My friend recently received a new digital drawing tablet (Wacom Intuos Pro). Among other features, there is a hole that resembles a charm hole found on mobile phones:    Other parts of the package (e.g. pen) do not have such holes, so it isn't possible to attach them using a band. What is the purpose of this hole? Is it just for attaching various charms and such? If $\mathrm{Tr}(A)=0$ then $T=R^{-1}AR$ has all entries on its main diagonal equal to $0$ Prove that if $A$ is a square matrix and $\mathrm{Tr}(A)=0$, then there exists an invertible matrix $R$ such that the matrix $T=R^{-1}AR$ has all entries on its main diagonal equal to $0$.   It seems like the formula $A=S\Lambda S^{-1}$, but maybe it does not help.  Thanks so much. A function takes every function value twice - proof it is not continuous I want to prove the following nice statement I've found:  A function $f: [0,1] \rightarrow \mathbb{R}$ takes every function value twice - proof it is not continuous  I've already found an answer to my question but one thing is still not clear to my mind. The answer I found was :   Whole post:  "Assume that you have a function $f$ that does take every real value exactly twice. Let $f(a) = f(b) = 0$ with $a &lt; b$. Let $c \in (a,b)$ be given, and assume without loss of generality $f(c) &gt; 0$. Then for every $0 &lt; \epsilon &lt; f(c)$ there exist $d \in (a,c)$ and $e \in (c,b)$ such that $f(d) = f(e) = \epsilon$. So on the intervals $[0,a)$ and $(b,1]$ our function cannot go any higher than zero. That means that $f$ should take every positive real value on the interval $[a,b]$. However the restriction of $f$ to $[a,b]$ is continuous from a closed interval to $\mathbb{R}$ and thus bounded. Contradiction."  Now everything is clear here except for:  "However the restriction of $f$ to $[a,b]$ is continuous from a closed interval to $\mathbb{R}$ and thus bounded. Contradiction."  Well I think know what Arthur is telling there: Because of the boundary there is always at least one function value that can't be taken twice by the function, e.g. just like the peak of $-x^2$.  But that is exactly where I was stuck before I started researching that issue. Is it really possible to just say it the way Arthur did? I thought that I'd have to further somehow prove the statement that boundary leads to that contradiction.    I hope that it is clear what I was trying to express here...  Anyways as always - thank you for your help!  FunkyPeanut VPN server on Google Compute Engine with OpenVPN App Store is the wrong language I'm very new to classical propositional logic, and my lecturer is using a system with the following axioms:  A1. X→(Y→X)  A2. (X→(Y→Z))→((X→Y)→(X→Z))  A3. (¬Y→¬X)→(X→Y)  Use uniform substitution and Modes Ponens (MP) as rule of inference. Deduction Theorem also allowed.  I'd like to deduce the following:  (a) X, Y ⊢ ¬(X→(¬Y)) (conjunction, but expressed in terms of NOT and IMPLIES)  (b) X→((X→Y)→Y)  (c) (X→Y)→(((¬Y)→X)→Y)  (d) (((¬Y)→X)→Y)→(X→Y)  Remark: The last two are to show that the axiom (A3) can be replaced with a different axiom:  B3. (¬Y→¬X)→(((¬Y)→X)→Y).  Solution: Bram28 proves (b) and (c) below. I demonstrate (a) and (d).  (d) is alright; first, I claim that X, ((¬Y→X)→Y) ⊢ Y.  Axiom (A1) gives ⊢X→(¬Y→X) so that ((¬Y→X)→Y)⊢X→Y by Hypothetical Syllogism (HS), which Bram28 shows to be valid.  Now that we have ((¬Y→X)→Y)⊢X→Y just apply the deduction theorem to get what we want: ⊢((¬Y→X)→Y)→(X→Y).  (a) is slightly trickier. Axiom (A1) gives ⊢ Y→(X→Y) so that X, Y ⊢ (X→Y).  Next, by part (b), we have as a theorem that ⊢X→((X→¬Y)→¬Y).  Therefore X, Y ⊢(X→¬Y)→¬Y.  But ⊢(¬¬(X→¬Y))→(X→¬Y) by DN Elim, which Bram28 shows to be valid.  By HS, X, Y ⊢(¬¬(X→¬Y))→¬Y.  By Axiom (A3) and MP, X, Y ⊢(Y→¬(X→¬Y)).  Also, rather trivially X, Y ⊢Y.  Finally by MP we conclude X, Y ⊢¬(X→¬Y) as required.