I want to make a tikz-picture of a Hasse diagram and by problem is that the lines between nodes stop a bit short of the border of the node and I can't figure out how to fix this.  Here is my code:  \begin{tikzpicture}[scale=\tikzscale, source/.style={draw,thick,rounded corners,inner sep=2pt}] \node[source] (min) at (0,0) {$\emptyset$}; \node[source] (A1) at (-2,1) {$a$}; \node[source] (A2) at (-1,1) {$b$}; \node[source] (A3) at (0,1) {$h$}; \node[source] (A4) at (1,1) {$c$}; \node[source] (A5) at (2,1) {$d$}; \node[source] (B1) at (-3.5,3) {$ab$}; \node[source] (B2) at (-2.5,3) {$ah$}; \node[source] (B3) at (-1.5,3) {$bh$}; \node[source] (B4) at (-0.5,3) {$ad$}; \node[source] (B5) at (0.5,3) {$bc$}; \node[source] (B6) at (1.5,3) {$ch$}; \node[source] (B7) at (2.5,3) {$dh$}; \node[source] (B8) at (3.5,3) {$cd$}; \node[source] (C1) at (-3,5) {$abh$}; \node[source] (C2) at (-1.5,5) {$adh$}; \node[source] (C3) at (0,5) {$abcd$}; \node[source] (C4) at (1.5,5) {$bch$}; \node[source] (C5) at (3,5) {$cdh$}; \node[source] (max) at (0,6.5) {$abcdh$}; \draw[thick] (min) -- (A1); \draw[thick] (min) -- (A2); \draw[thick] (min) -- (A3); \draw[thick] (min) -- (A4); \draw[thick] (min) -- (A5); \draw[thick] (A1) -- (B1); \draw[thick] (A1) -- (B2); \draw[thick] (A1) -- (B4); \draw[thick] (A2) -- (B1); \draw[thick] (A2) -- (B3); \draw[thick] (A2) -- (B4); \draw[thick] (A3) -- (B2); \draw[thick] (A3) -- (B3); \draw[thick] (A3) -- (B6); \draw[thick] (A3) -- (B7); \draw[thick] (A4) -- (B5); \draw[thick] (A4) -- (B6); \draw[thick] (A4) -- (B8); \draw[thick] (A5) -- (B4); \draw[thick] (A5) -- (B7); \draw[thick] (A5) -- (B8); \draw[thick] (B1) -- (C1); \draw[thick] (B1) -- (C3); \draw[thick] (B2) -- (C1); \draw[thick] (B2) -- (C2); \draw[thick] (B3) -- (C1); \draw[thick] (B3) -- (C4); \draw[thick] (B4) -- (C2); \draw[thick] (B4) -- (C3); \draw[thick] (B5) -- (C3); \draw[thick] (B5) -- (C4); \draw[thick] (B6) -- (C4); \draw[thick] (B6) -- (C5); \draw[thick] (B7) -- (C2); \draw[thick] (B7) -- (C5); \draw[thick] (B8) -- (C3); \draw[thick] (B8) -- (C5); \draw[thick] (C1) -- (max); \draw[thick] (C2) -- (max); \draw[thick] (C3) -- (max); \draw[thick] (C4) -- (max); \draw[thick] (C5) -- (max); \end{tikzpicture}     Can anybody help me? Thanks!
When I use TikZ tree with nodes with rounded corners, the connecting lines do not touch the nodes (near corners) but end where would rectangle end. Is there an easy way around it?  \documentclass{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} [sibling distance=100]    \node[rectangle, draw, rounded corners = 3]{ }   child { node { } }   child { node { } }   ; \end{tikzpicture} \end{document}
Deciding whether two metrics are topologically equivalent in the space $C^1([0,1])$ Consider the space $C^1([0,1])$ and the function $d:C([0,1])\times C([0,1]) \to \mathbb R$ defined as $d(f,g)=|f(0)-g(0)|+sup_{x \in [0,1]}|f'(x)-g'(x)|$. Decide whether the metrics $d$ and $d_{\infty}$ are topologically equivalent in $C^1([0,1])$ (where $d_{\infty}=sup_{x \in [0,1]}|f(x)-g(x)|)$  My attempt at a solution:  If two metrics are topologically equivalent, then they have the same convergent sequences. Honestly, I couldn't do anything. I am trying to define a sequence of functions $\{f_n\}_{n \in \mathbb N}$ such that $f_n \to f$ in, for instance, $(C^1([0,1]),d)$ but $f_n \not \to f$ in $(C^1([0,1]),d_{\infty})$. Could it be this two metrics are topologically equivalent? If this is the case, how could I prove it? If not, I would appreciate any hint to find an adequate sequence of functions that works for what I am trying to prove. This is a special case of , which, I think, might have drawn little attention because it was too general. In this question, I would like to consider a specific case.  Take a classical system given by the action $$ S[x] = \int dt \left[ \frac{\alpha}{2} \ddot{x}^2 + \frac{\beta}{2} \dot{x}^2 + \frac{\gamma}{2} x^2 \right]. $$  This action is Lagrangian, but it is not what we are usually dealing with in physics, because the Lagrangian contains second time derivatives.  The equations of motion are:  $$ \alpha \ddddot{x} - \beta \ddot{x} + \gamma x = 0. $$  This can be solved as usual by employing Fourier transform:  $$ x(t) = a_1 e^{i \omega_1 t} + a_1^{*} e^{-i \omega_1 t} + a_2 e^{i \omega_2 t} + a_2^{*} e^{-i \omega_2 t}, $$  where $\omega_1^2$ and $\omega_2^2$ are the two roots of $\alpha \omega^4 + \beta \omega^2 + \gamma$, given by  $$ \omega_{1,2}^2 = \frac{- \beta \pm \sqrt{\beta^2 - 4 \alpha \gamma}}{2 \alpha} .$$  Now by definition the phase space is the space of solutions of equations of motion. In this case it is parametrized by two complex co-ordinates $a_{1,2}$, meaning that it is 4-dimensional.  I would like to know if there's a natural way to associate the symplectic structure (Poisson bracket) on this phase space. I.e.,  $$ \{a_1, a_1^{*}\} = ? $$ $$ \{a_1, a_2^{*}\} = ? $$ $$ \{a_2, a_1^{*}\} = ? $$ $$ \{a_2, a_2^{*}\} = ? $$ $$ \{a_1, a_2\} = ? $$ $$ \{a_1^{*}, a_2^{*}\} = ? $$ HTML 5 Video stretch I'm searching for a really simple and beautiful proof that the sequence $(u_n)_{n \in \mathbb{N}} = \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ converges. At first I want to know if my answer is OK.  My try: $\lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} - \log (n) = \lim\limits_{n\to\infty} \sum\limits_{k=1}^n \frac{1}{k} + \sum\limits_{k=1}^{n-1} \log(k)-\log(k+1) = \lim\limits_{n\to\infty} \frac{1}{n} + \sum\limits_{k=1}^{n-1} \log(\frac{k}{k+1})+\frac{1}{k}$     $ = \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ Now we prove that the last sum converges by the comparison test: $\frac{1}{k}-\log(\frac{k+1}{k}) &lt; \frac{1}{k^2} \Leftrightarrow k&lt;k^2\log(\frac{k+1}{k})+1$ which surely holds for $k\geqslant 1$      As $ \sum\limits_{k=1}^{\infty} \frac{1}{k^2}$ converges $ \Rightarrow \sum\limits_{k=1}^{\infty} \frac{1}{k}-\log(\frac{k+1}{k})$ converges and we name this limit $\gamma$ q.e.d PrintWriter and PrintStream never throw IOExceptions How do you join on the same table, twice, in mysql? What does 'Ign' , `Get` or 'Hit' mean when running an apt-get update? Apache doesn't log remoteIP when RemoteIPHeader X-Forwarded-For is present Are there any free/open source alternatives to Adobe After Effects? I've been a illustrator and graphic designer for a number of years and now I'm looking to go more into motion graphics and visual effects. Is there anything that is open source/free (not a demo, education version or trial) that is as complex and feature-rich as Adobe After Effects but for Linux? I've tried Cinlerra, kdenlive, openshot and Blender and whilst they are good at video editing (and 3d modelling in Blender's case) they fall short in effects and compositing features.  I've also tried , but the project recently closed and the last update was quite buggy.  Also,  doesn't work and hasn't been updated in many many years, so don't mention it ;-) Textbook on the Geometry of Special Relativity I am looking for a textbook that treats the subject of Special Relativity from a geometric point of view, i.e. a textbook that introduces the theory right from the start in terms of 4-vectors and Minkowski tensors, instead of the more traditional "beginners" approach. Would anyone have a recommendation for such a textbook ?  I already have decent knowledge of the physics and maths of both SR and GR ( including vector and tensor calculus ), but would like to take a step back and expand and broaden my intuition of the geometry underlying SR, as described by 4-vectors and tensors. What I do not need is another "and here is the formula for time dilation..." type of text, of which there are thousands out there, but something much more geometric and in-depth.  Thanks in advance. The problem:  Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $|f(t) - f(x)| \leq |t - x|^2$ for all $t, x$. Prove $f$ is constant.  I believe I have some intuition about why this is the case; i.e. if $t$ and $x$ are very close, ($|t - x| = \epsilon$), then $\sqrt{\epsilon}$ will converge to 0, so $|f(t) - f(x)|$ will also converge to zero if you keep taking $t$ and $x$ closer and closer to each other. However, how do I formalize this argument? Thanks. Screen (not Lock Screen) Timeout Stuck at 1 Minute I recently got a new laptop (Lenovo T440p with Windows 8.1 Pro) and was having the following issue: the screen would lock (screen would go black and require a password to get back in) after 1 minute. I would prefer something more like 10 or 15 minutes. So, I tried the obvious changes (check the screensaver settings, power management settings for sleep, turning off the screen, etc.), with no success. I did, however, notice a rather odd behavior: if I changed the screensaver timeout from 1 minute to anything else (say, 15 mins), clicked OK, and re-opened the screensaver settings window, it would be back to 1 minute. Odd.  At that point, I turned to Google. I found posts ( in particular was helpful) suggesting all sorts of causes and resolutions:   If you have Cisco AnyConnect VPN client (which I did), try uninstalling it. I uninstalled, rebooted, no change. Open the User Accounts control panel and, on the left side of the screen, click Reset Security Policies. My User Accounts control panel, however, has no such link. Reboot into a repair console and run "sfc /scannow". It did indeed find and repair some issues, but didn't fix the screen timeout problem. Edit the Group Policy to force a particular timeout time; the setting showed up correctly (15 mins) in screensaver settings, power management settings, etc., but the screen would still go black and lock after 1 minute. In the registry, manually change the screensaver timeout and the "turn off the screen" timeout. No luck. What do we mean when we say "for all n where n is an element of..." When we write "for all n where n is an element of A, P(n) is true." do we mean for all n, if n is in A then Pn or for all n, n is in A and Pn? What is the difference between "I have a doubt with ...." & "I suspect there is a problem"? How to burn or mount an ISO file? I have downloaded an ISO file recently.  How do I burn it to a CD or DVD or mount it? My Android device screen got broken and I can't see what I'm doing. That's the only wrong thing with my device, the rest is apparently working OK. Is there any way to send my device screen to my computer? I don't want to work blindly anymore! How can you draw more detailed/smoother images in pygame? I often see questions on Stack Exchange sites which I presume are written by non-native English speakers who use the word "doubt" in place of the word "question". Is this a case of misunderstanding the correct meaning or are people being taught that this is correct usage? Is there a closed form expression for $1 + x + x^{4} + x^{9}+x^{16}+x^{25} +..+x^{k^2}$? We all know what the sum of a geometric series is  $1 + x + x^2 + x^3 + ... + x^k = \frac{x^{k+1} - 1}{x - 1}$ .  I was wondering if similar formulas exist in case the exponents  form some other sequence, for example are squares or cubes of positive integers? What is a pull up and pull down? Can some one explain this terminology please I think I understand it but not completely sure.   I think pull down is where you place a resistor between +V and the other component and pull up is where you place the resistor between 0v and the component.    If I am completely wrong then let me know!