How to transform symmetric matrix to diagonal?
$\mathbf L\mathbf D\mathbf L^\top$ Cholesky decomposition
Change password on root user and user account If I change the password for the root user is that going to automatically change the password for the user account?   I found this link for root:    How do I change the user account password? Using Windows 10 in a virtual machine. Can it be done? I am currently running 8.1. I'd like to try out Windows 10 and if I like it upgrade. If not, I'll keep running as I am.   I've never used a virtual machine before: can Window 10 be used from a virtual machine as a "sandbox", and if so how? TikZ: turning edges into arrows using "child" notation I'm using TikZ for the first time to draw a simple Kripke model, and I'm using the child notation to produce new nodes. How can I turn my edges into arrows?  \begin{tikzpicture}   \node [circle,draw] (q0) {}   child {node [circle,draw] (q1) {$p$}     child {node [circle,draw] (q4) {$p$}       child {node [circle,draw] (q7) {$p\land q$}}       child {node [circle,draw] (q8) {$p\land q$}}       child {node (q9) {} edge from parent[-&gt;, dashed]}}     child {node [circle,draw] (q5) {$q$}}     child {node (q6) {} edge from parent[dashed]}}   child {node [circle,draw] (q2) {$p$}}   child {node (q3) {} edge from parent[dashed]}; \end{tikzpicture}   I know how to change the color or thickness of all edges by passing an option at the beginning of \begin{tikzpicture}, but I can't turn them into arrows.   Also, while we're at it, is there a way to label these edges? Or am I left with simply using node notation for all my nodes?  Thanks for all your help, Piotr Why can't I mine diamond, gold ore, or redstone ore? Normal blocks I can mine fine, but when it comes to diamond and gold ores, the block just gets "destroyed" (as in you can't pick it up). I'm playing on multiplayer with no spawning monsters. Parse JSON using Python? The title says it all. I want to resize windows with an right click while holding the alt key, instead of the new default alt+middle click.  I have found no way to enable this behaviour in 13.04. The gconf-Key suggested in this , doesn't seem to exist any longer. How do i test capacitors? I've just gotten through replacing capacitors on a trio of dead LCD screens (nothing's blown up yet, so far) - they either had one or two capacitors on their inverter circuit SLIGHTLY bloated, and not quite leaky. I ended up replacing all capacitors of the same brand/'colour', even the ones that looked fine, in case.  Now, checking a bad resistor is simple - i can use a standard multimeter to test it, and i tend to check my solders with the continuity testing option of the multimeter.  How would i test a capacitor ? Is there some standard, common way to test one? Tips for getting really good answers on SO The way the askers write questions is a factor in determining how good the answers we get are. Obviously, if you ask a clear, direct question it is easier for a someone to answer you.  But, sometimes I've done that and my questions have still been overlooked.  I've noticed some question-posting methods that aren't exactly obvious, but do help me get better answers.    Some tricks I've learned...   Even if my question is very complicated, I try to write a short question.  People get turned off by longer questions, probably because they think its not worth their time (relative to other questions with equal reputation potential).  After I get some traffic and votes on my question, and maybe some comments like "Can you post the code?" or "Clarify this..." I might post more, but by then I am already getting attention to my question.  So, in summary, even if I know the answer requires a long question, I don't make it long until I've already gotten people to look at it. I use more generic tags when possible.  People are more likely to have c# and linq as a favorite tag, than MethodCallExpression, and I tend to get more views if I use at least one or two general purpose tags along with my more specific tags. I've noticed that if I give my question the performance tag, then SO heavy-weights like Marc Gravell, Jon Skeet and Eric Lippert are more likely to answer my question (especially if I use it in conjunction with a C# tag).  Since having this realization, if I can think of a way to rephrase my question to sound like it is related to C# performance, I will word it as such and tag it accordingly.   "Be polite. SO Users HATE when people be impatient, rude or impolite." - Benny   So,  Does anyone else have any tips for getting better answers and question-recognition on SO?  I think it would be really helpful if everyone contributed their own observations and techniques here.  P.S. This question is probably too long. Here's a module with castellated pads attached to an eval board:  (source: ) Looking closely at the wires interfacing with the castellated pads, they're clearly soldered to the eval board. However, they don't look like they're soldered to the module, but rather they're being pressed against the pads with the tension in the wire. It also looks like the free end of the loops are protruding down into a channel cut in the PCB. Do these kind of connectors/metal loops have a name? If $\int_{0}^{x}f(t)dt\rightarrow \infty$ as $|x|\rightarrow \infty\;,$ Then every line $y=mx$ Intersect $ y^2+\int_{0}^{x}f(t)dt=2$ If $f$ is a continuous function with $\displaystyle \int_{0}^{x}f(t)dt\rightarrow \infty$ as $|x|\rightarrow \infty\;,$ Then show that every line $y=mx$ Intersect the curve $\displaystyle y^2+\int_{0}^{x}f(t)dt=2$  $\bf{My\; Try::}$ Put $y=mx$ in $$\displaystyle y^2+\int_{0}^{x}f(t)dt=2\;,$$ We get $$\displaystyle m^2x^2+y^2+\int_{0}^{x}f(t)dt=2.$$ We can write it as $$\displaystyle 2\int_{0}^{x}2m^2tdt+\int_{0}^{x}f(t)dt=2$$ So $$\displaystyle \int_{0}^{x}\left[2m^2t+f(t)\right]dt-2 =0$$ Now Let $$g(x) = \displaystyle \int_{0}^{x}\left[2m^2t+f(t)\right]dt-2.\;,$$ When $x\rightarrow \infty\;,$ Then $g(x)\rightarrow \infty.$ (Where $g(x)$ is an Continuous function.) Now I did not Understand How can i solve that, Help me Thanks I'm looking for an elegant way to show that, among non-negative numbers, $$ \max \{a_1 + b_1, \dots, a_n + b_n\} \leq \max \{a_1, \dots, a_n\} + \max \{b_1, \dots, b_n\} $$  I can show that $\max \{a+b, c+d\} \leq \max \{a,c\} + \max \{b,d\}$ by exhaustively checking all possibilities of orderings among $a,c$ and $b,d$.  But, I feel like there should be a more intuitive/efficient way to show this property for arbitrary sums like the one above. By considering:  $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^1}{n^{2}} = \frac 1 2$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^2}{n^{3}} = \frac 1 3$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^3}{n^{4}} = \frac 1 4$$  Determine if this is true: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}} = \frac 1 {{m+1}}$$  If it is, prove it.  If it is not, evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{{m+1}}$. How do I change the computer name? Is there a fairly simple way to divide a circle into equal parts using Adobe Illustrator CC? Holomorphic function has at most countably zeros What ensures big holes in my bread? It was suggested to me recently that Annihilator sacrifices (such as 's) are only chosen during resolution, rather than while they're going on the stack.  Do I choose those sacrifices during resolution? What part of the comprehensive rules backs that up?  I can't find any clarity on the internet, in the rules, on the MTG Salvation wiki for Annihilator, or in the ability's rulings. It seems reasonable, but I'd like to know what's behind it. Plus, if I can choose and then later  some stuff, all the better. How to find a whole word in a String in java A problem with my reasoning in a problem about combinations I was given the following problem to solve:     A committee of five students is to be chosen from six boys and five   girls. Find the number of ways in which the committee can be chosen,   if it includes at least one boy.   My method was $\binom{6}{1}\binom{10}{4}= 1260$, using the logic of choosing $1$ boy, then choosing the rest. This was wrong, as the answer was $\binom{11}{5}-\binom{5}{5}= 461$.  The correct answer's logic was committee with no restrictions – committee with no girls.  Why was my method wrong? Please help... I want to plot this graph, but I don't know how to code the words in the nodes.     Here is my code:  \documentclass{article} \usepackage{amsmath} \usepackage{titletoc} \usepackage{tikz} \usetikzlibrary{trees,positioning,fit}  \begin{document}{\allowdisplaybreaks[4]}  \tikzstyle{bag} = [text width=2em, text centered] \tikzstyle{end} = []  \begin{center} \begin{tikzpicture}[sloped] \node (a) at ( 0,0) [bag] {$\ln r_0$}; \node (b1) at ( 3,1.5) [bag] {$\ln r_{10} + \sigma_1\sqrt{\Delta t_1}$}; \node (b2) at ( 3,-1.5) [bag] {$\ln r_{10}$}; \node (c1) at ( 6,3) [bag] {$\ln r_{20} + 2\sigma_2\sqrt{\Delta t_2}$}; \node (c2) at ( 6,1.5) [bag] {$\textrm{Mean} = \ln r_{20} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2}$\\$\textrm{Mean Change} = \mu_{11}\Delta t_2 = \ln\dfrac{r_{20}}{r_{10}} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2} - \sigma_1\sqrt{\Delta t_1}$}; \node (c3) at ( 6,0) [bag] {$\ln r_{20} + \sigma_2\sqrt{\Delta t_2}$}; \node (c4) at ( 6,-1.5) [bag] {$\textrm{Mean} = \ln r_{20} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2}$\\$\textrm{Mean Change} = \mu_{11}\Delta t_2 = \ln\dfrac{r_{20}}{r_{10}} + \dfrac{3}{2}\sigma_2\sqrt{\Delta t_2} - \sigma_1\sqrt{\Delta t_1}$}; \node (c5) at ( 6,-3) [bag] {$\ln r_{20}$}; \draw [-&gt;] (a) to node [below] {} (b1); \draw [-&gt;] (a) to node [above] {} (b2); \draw [-&gt;] (b1) to node [below] {} (c1); \draw [-&gt;] (b1) to node [below] {} (c2); \draw [-&gt;] (b1) to node [below] {} (c3); \draw [-&gt;] (b2) to node [below] {} (c3); \draw [-&gt;] (b2) to node [below] {} (c4); \draw [-&gt;] (b2) to node [below] {} (c5); \end{tikzpicture} \end{center}  \end{document}       Update code use  \draw [-&gt;] (a) to node [below] {} (b1.south east); \draw [-&gt;] (a) to node [above] {} (b2.north east); \draw [-&gt;] (b1.south east) to node [below] {} (c1.west); \draw [-&gt;] (b1.south east) to node [below] {} (c2.west); \draw [-&gt;] (b1.south east) to node [below] {} (c3.west); \draw [-&gt;] (b2.north east) to node [below] {} (c3.west); \draw [-&gt;] (b2.north east) to node [below] {} (c4.west); \draw [-&gt;] (b2.north east) to node [below] {} (c5.west);