I have created a path that is a line segment. It only has two endpoint nodes. How do I create a new node at the exact midpoint of the segment? Thanks.
In Inkscape when I apply a dash stroke, then set a mid marker shape, it does not show anything.
Possible Duplicate:          I know that   $$\lim_{n\rightarrow\infty}\sqrt[n]n=1$$  and I can imagine that $n$ grows linearly while $n$th root compresses it exponentially and therefore the result is $1$, but how do I calculate it? Glassware with scratched/etched surfaces generate more bubbles (, ). Why these rough surfaces allow easier heterogeneous nucleation? Let $\{X_n\}_{n=1}^\infty$ be a sequence of random variables converging to a random variable $X$. Does this fact implies that $\lim\limits_{n\to \infty} EX_n = EX$?  If not, is there any other sufficient condition? Let $f: \mathbb R^2 \to \mathbb R$ defined as   $f(x,y)=(x^2+y^2)\sin(\dfrac{1}{\sqrt{x^2+y^2}})$ if $(x,y) \neq (0,0)$, $f(x,y)=(0,0)$ if $(x,y)=(0,0)$  Prove that $f$ is differentiable at $(0,0)$ but that the partial derivatives of $f$ are not continuous at the origin.  The attempt at a solution:  I had some problems at the first part of the exercise:  I've calculated $\dfrac{\partial f}{\partial x}(0,0)$ and $\dfrac{\partial f}{\partial y}(0,0)$, both exist and are equal to $1$.  So, to prove $f$ is differentiable at the origin, I have to show that   $\lim_{(x,y) \to (0,0)} \dfrac{|f(x,y)-(f(0,0)+&lt;\nabla f(0,0),(x,y)&gt;)|}{\|(x,y)\|}=0$ But $\lim_{(x,y) \to (0,0)} \dfrac{|f(x,y)-(f(0,0)+&lt;\nabla f(0,0),(x,y)&gt;)|}{\|(x,y)\|}$ equals  $\lim_{(x,y) \to (0,0)} \dfrac{|(x^2+y^2)\sin(\dfrac{1}{\sqrt{x^2+y^2}})-(x+y)|}{\|(x,y)\|}$. I don't know how to show this limit equals $0$.  For the second part (I've only analyzed $\dfrac{\partial f}{\partial x}$ since both cases are the same), what I did was try to show that $\lim_{(x,y) \to (0,0)} \dfrac{\partial f}{\partial x}(x,y) \neq \dfrac{\partial f}{\partial x}(0,0)$   I thought of considering the curve $\phi(t)=(t,0)$, so   $\lim_{t \to 0} (f\circ \phi)(t)=\lim_{t \to 0} 2t\sin(|t|^{-1})-\dfrac{t\cos(|t|^{-1})}{|t|^{-1}}$. But $\lim_{t \to 0^{+}} \dfrac{t\cos(|t|^{-1})}{|t|^{-1}}=\lim_{t \to 0^{+}} \cos(\dfrac{1}{t})$ and this limit doesn't exist, so the original limit doesn't exist. This shows the partial derivative with respect to $x$ at the point $(0,0)$ is not continuous.   I would appreciate if someone could help me to finish the first part and tell me if what I did in the second part of the problem is correct. First time using bevel tool, doesn't bevel evenly. What am i doing wrong? How exactly is Arda supposed to be an ancient Earth? As most Tolkien fans know, Arda (Tolkien's world) is supposed to be Earth in its past, and all its stories are supposed to have happened long ago on this very same planet. Of course, this is fiction, so this is not supposed to be actually true.  But, if one wanted to roll with it... how could it possibly make sense from a geographical standpoint? All maps of Middle Earth (before or after the destruction of Beleriand and Numenor) just don't match anything on Earth, even when considering continental drift.  Of course, one could simply say "the Valar reshaped it (again)", but not only this is a bit unsatisfactory, it also doesn't make much sense in context, as Valar are well known for not wanting to meddle with Arda unless absolutely needed, and even then, as little as possible.    Update:  Hey, please calm down, guys. Mine was not a rant at all. And I know perfectly well this is fiction (and I've stated it plainly). The sense of my question was: "If one wanted to follow this line of thought and assume Arda should be related to Earth as Tolkien said, then how could this relationship be explained?". After the top bar changes in the , when a non logged-in user click the Sign Up button, instead of navigating to the corresponding page it displays the openid related code in the screen.  Screenshots for reference: I didn't like the skill I chosen for the first follower I found. Is it possible to respec his skill later on? The union of a strictly increasing sequence of $\sigma$-algebras is not a $\sigma$-algebra How does vendor restocking work in Fallout 4? I've been looking for 2mm EC ammo and according to the  it is supposed to restock every 48 hours at Arturo. However he has not gotten any new 2mm EC ammo for at least 10 restocks.   Do vendors restock on a random selection of items with some being more likely than others? Is there a max limit to how many different items a vendor will sell so that no new items will be added to stock after this limit is reached? Is there a level limit on what items can be restocked? I bought 2mm EC ammunition once from Arturo so I assume I'm high enough level if that's the case? How to do logistic regression subset selection? I am fitting a binomial family glm in R, and I have a whole troupe of explanatory variables, and I need to find the best (R-squared as a measure is fine). Short of writing a script to loop through random different combinations of the explanatory variables and then recording which performs the best, I really don't know what to do. And the leaps function from package  does not seem to do logistic regression.  Any help or suggestions would be greatly appreciated. Question about Angle-Preserving Operators This an exercise out of Spivak's "Calculus on Manifolds".  Edit:  There was a typo in the exercise as is noted below in the answers.  The statement has been edited to reflect this.  Given $x,y\in\mathbb{R}^{n}$, the angle between $x$ and $y$ is defined by  $$\angle(x,y) = \arccos\left(\frac{\langle x,y \rangle}{|x|\cdot |y|}\right),$$ where $\langle x,y \rangle$ denotes the standard Euclidean inner product.  A linear operator $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ is said to be angle-preserving if $\angle(T(x),T(y)) = \angle(x,y)$ for every $x,y\in\mathbb{R}^{n}$.  The exercise as stated:  Let $\{x_{1},\dots, x_{n}\}$ be a basis for $\mathbb{R}^{n}$.  Then suppose that $\lambda_{1}, \dots, \lambda_{n}\in \mathbb{R}$ are such that $Tx_{j} = \lambda_{j}x_{j}$ for each $j = 1,\dots, n$.  Then $T$ is angle-preserving only if (not if and only if!)$|\lambda_{i}| = |\lambda_{j}|$ for every $1\leq i\leq j\leq n$.  I'm having problems with the $(\Rightarrow)$ direction.  My best attempt (which seems to lead nowhere) is to suppose that $|\lambda_{j}|\neq |\lambda_{k}|$.  Then by assumption, \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; = \arccos\left(\frac{\langle Tx_{j},Tx_{k} \rangle}{|Tx_{j}|\cdot |Tx_{k}|}\right)\\ &amp; = \arccos\left(\frac{\langle \lambda_{j}{x_{j}},\lambda_{k}{x_{k}} \rangle}{|\lambda_{j}{x_{j}}|\cdot |\lambda_{k}{x_{k}}|}\right)\\ &amp; = \arccos\left(\frac{\lambda_{j}\lambda_{k}\langle {x_{j}},{x_{k}} \rangle}{|\lambda_{j}|\cdot|\lambda_{k}|\cdot|{x_{j}}|\cdot |{x_{k}}|}\right)\\ &amp; = \arccos\left(\text{sign}(\lambda_{j})\text{sign}(\lambda_{k})\frac{\langle {x_{j}},{x_{k}} \rangle }{|{x_{j}}|\cdot |{x_{k}}|}\right)\\ \end{align*} may also be calculated as \begin{align*} \angle(Tx_{j},Tx_{k}) &amp; =  \angle(x_{j},x_{k})\\ &amp; = \arccos\left(\frac{\langle x_{j},x_{k} \rangle}{|x_{j}|\cdot |x_{k}|}\right). \end{align*}  Then since $\arccos$ is injective, I believe I can make the jump that $\text{sign}(\lambda_{j})\text{sign}(\lambda_{k}) = 1$, which does not resemble the conclusion that I should arrive at.  Note:  I wasn't sure what tag to put this under, so anyone who knows better please feel free to adjust.  Thanks for any help you can give. Problem 20-9 in Lee's Introduction to Smooth Manifolds (2nd Ed.) asks to prove that for a Lie group $G$, and for $X, Y\in \text{Lie}(G)=\mathfrak g$, there is $\epsilon&gt;0$ and a smooth map $Z:(-\epsilon, \epsilon)\to \mathfrak g$ such that   $$(\exp tX)(\exp tY) = \exp\left(t(X+Y) + \frac{1}{2}t^2[X, Y] + t^3 Z(t)\right)$$   What we want to show is that the first and second term in the Taylor expansion of $\exp^{-1}\left((\exp tX)(\exp tY)\right)$ are $t(X+Y)$ and $\frac{1}{2}t^2[X, Y]$ respectively.  Calculating the first term is easy since we just need to differentiate $\exp^{-1}((\exp tX)(\exp tY))$ with respect to $t$ at $t=0$ once, and this gives the desired result.  But for the second term we need to take the second derivative at which I am stuck. jQuery child of clicked element I've got a list of links which have a click event attached to them, I need to get the ID from the child A link. So in the example below if I clicked the first list element I'd need google retuned.   I've tried '$this a' but can't quite work out the syntax.  $("ul li").click(function(event){   $("input").val($(this).html());              });     &lt;ul&gt;     &lt;li&gt;&lt;a href="http://www.google.com" id="google"&gt;Google&lt;/a&gt; &lt;/ul&gt; Under TeX, when I write a line too long, it prints a fat black box at the end of the line.  That is probably a feature, but I dislike it.  How to turn that OFF? Is there a command-line option or something else?   I usually use LaTeX, but I have a complicated file which happens to be plain TeX. I am a new user, who has just been told to run / execute a command. The answer has some text with a grey background like this text here or  sometimes this text here.   However, I don't know what "run" or "command" means in relation to Ubuntu.  How do I do this?    This is intended for use as a Q&amp;A you can link to when you tell someone to run a command. How to add "Refresh" option to right click menu in Lubuntu? I have many users at the place I work, expressing the notion of switching to Ubuntu.  Most of them, for whatever weird reason wanting to know if its possible to have refresh added when you right click on the desktop.   Is that possible? Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$ Allowing line break at ',' in inline math mode? In the inline math mode ($...$), if the formula is too long, LaTeX will try to break it on operators, e.g.  very long text followed by a very long equation like $a+b+c+d+e+f+g+h+i+j+k+l$ etc   may be rendered as  very long text followed by a very long equation like a+b+c+d+e+f+g+h+i+ j+k+l etc   However, the break won't happen if they are separated by commas, e.g.  very long text followed by a very long equation like $a,b,c,d,e,f,g,h,i,j,k,l$ etc   will overflow the page like  very long text followed by a very long equation like a,b,c,d,e,f,g,h,i,j,k,l etc   How to make LaTeX able to insert line breaks after a comma too? Why isn't it possible to combine vendor-specific pseudo-elements/classes into one rule set?