I am using bootstrap for my simulation.  The number of the population is flexible for each case, and the sample size is decided by a certain percentage. For example, I have a 10,000 population, and I decide to use 10% for each iteration of bootstrap, so the sample size is 1,000.  In practice, I found it is hard to decide how many times to run the bootstrap is enough. With less simulation, the results appear insufficiant, while with a large number of simulation they are purely redundant.  May I know if there is a method that can help me to decide the number of iterations to run?
I wonder if someone knows any general rules of thumb regarding the number of bootstrap samples one should use, based on characteristics of the data (number of observations, etc.) and/or the variables included?
I want to show without using Sylow theorem that     Any group of order $2p$ has a subgroup of order $p~$($p$ being a prime)   My attempt: Since $|G|=2p,$ even $\exists~a\neq e\in G$ such that $a^{-1}=a.$ Then $H=\{e,a\}\leq G.$ Let $S$ be the set of all left cosets of $H$ in $G.$ Now $S$ forms a group w.r.t. the composition $g_1Hg_2H=(g_1g_2)H.$ The mapping $\sigma:S\to G:gH\mapsto g$ is an $1-1$ homomorphism. Then $S\simeq\sigma(S)\leq G.$ Thus $G$ has a subgroup of order $p.$  Please tell me whether I'm right!     I'm skecptical as I didn't use that $p$ is prime. Show vote counts on activity>comments page It would be helpful to show vote counts on the activity > comments page, and also allow sorting by votes.  This would make it easy to see what comments other people thought were valuable or at least amusing. Is it possible to add images to comments? I was wondering if it is possible to post a comment with an image.  Recently, during a question about data-visualization in Cross Validated, there was a controversy about using one type of chart for another.   I had posted a question about pie-charts, after reading some references I decided to build an example and clarify the point of using waffle-chart vs pie-chart. In a way, my example was not an answer it was more properly a comment on waffle-charts. But it seems to be impossible to post a comment with an image (Maybe, there is a way but I can't figure out how).  I don't know if it is relevant to post images in comments in other SO sites. So, if there is no way to post images in comments... why not making it possible? And if there is a way to post an image with a comment, how I can do it? More than one 'Open xxxxx here as Administrator' I want the option to open a Command window or Powershell window as Administrator from the drive and directory context menus in Windows.   I can have either by creating and populating the special 'runas' registry keys shown below, but is there a way I can have both listed on the context menu and have both run as Administrator?  [HKEY_CLASSES_ROOT\Drive\shell\runas] ... [HKEY_CLASSES_ROOT\Drive\shell\runas\command] ... [HKEY_CLASSES_ROOT\Directory\shell\runas] ... [HKEY_CLASSES_ROOT\Directory\shell\runas\command] ... [HKEY_CLASSES_ROOT\Directory\Background\shell\runas] ... [HKEY_CLASSES_ROOT\Directory\Background\shell\runas\command] ... 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: Labelling figures in the tikz package Consider the following piece of code, which uses the tikz-feynman package:  \usepackage{tikz-feynman, contour} \tikzfeynmanset{compat=1.1.0}  \begin{document} \begin{center}     \begin{tikzpicture}[baseline={(current bounding box.center)}]     \begin{feynman}         \diagram [vertical'=a to b] {             {   [edges={draw=none}]                 a -- b [label=0:\(\mu_{3}\)]                 -- c [label=-180:\(\mu_{4}\)]                 -- d [label=-180:\(\mu_{1}\)]                 -- a [label=0:\(\mu_{2}\)],             },             {   [edges={photon}]                 a -- ap [particle=\(p_{2}\)],                 b -- bp [particle=\(p_{3}\)],                 d -- dp [particle=\(p_{1}\)],                 c -- cp [particle=\(p_{4}\)],             },         };         \diagram* {             (c) -- [fermion] (b)                 -- [fermion] (d)                 -- [fermion] (a)                 -- (c),         };     \end{feynman}     \end{tikzpicture}     %     \feynmandiagram[baseline={(current bounding box.center)}, vertical'=a to b,] {         {   [edges={fermion}]             a -- b [label=0:\(\mu_{3}\)]             -- c [label=-180:\(\mu_{4}\)]             -- d [label=-180:\(\mu_{1}\)]             -- a [label=0:\(\mu_{2}\)],         },         {   [edges={photon}]             a -- ap [particle=\(p_{2}\)],             b -- bp [particle=\(p_{3}\)],             d -- dp [particle=\(p_{1}\)],             c -- cp [particle=\(p_{4}\)],         },     };     %     \begin{tikzpicture}[baseline={(current bounding box.center)}]     \begin{feynman}         \diagram [vertical'=a to b] {             {   [edges={draw=none}]                 a -- b [label=0:\(\mu_{3}\)]                 -- c [label=-180:\(\mu_{4}\)]                 -- d [label=-180:\(\mu_{1}\)]                 -- a [label=0:\(\mu_{2}\)],             },             {   [edges={photon}]                 a -- ap [particle=\(p_{2}\)],                 b -- bp [particle=\(p_{3}\)],                 d -- dp [particle=\(p_{1}\)],                 c -- cp [particle=\(p_{4}\)],             },         };         \diagram* {             (a) -- [anti fermion] (b)                 -- [anti fermion] (d)                 -- [anti fermion] (c)                 -- (a),         };     \end{feynman}     \end{tikzpicture}     \end{center} \end{document}   This piece of code draws the following figure:    I would like to label each figure. I do not want a caption, but just a label beneath each diagram. The label should say Diagram I, Diagram II, Diagram III.  How do I implement this into my code? I read an essay about prime numbers. In it the author suggests that a natural first question to ask is,      What is the probability of a random natural number being prime?   but proceeds to dismiss it as "not making sense". I wonder what's wrong with the question. Is its meaninglessness related to the fact that there is an infinite number of naturals? How to promote algebraic expressions to operators in quantum mechanics? Okay, I know that in quantum mechanics the quantum observable is obtained from the classical observable by the prescription   $$ X \rightarrow x,\quad P \rightarrow -i\hbar\frac{\partial}{\partial x} $$   in the position basis. Now my questions are:    What if $x$ or $p$ appears in the denominator in a classical expression?  How to promote this to a quantum expression? What would be the meaning of division by an operator?   My expression likely contains a mixture of $x$ and $p$. For e.g., it could contain terms like $$\frac{p}{x^2}$$ or $$\frac{xp}{(x^2 + a^2)^{3/2}}.$$    How to resolve products of non-commuting operators like $x$, $p$ in a satisfactory way? Diagonalisation and characteristic polynomial Let $A\in M_3(\Bbb R)$ which is not a diagonal matrix. Let $P$ be a polynomial (in one variable), with real coefficients and of degree 3 such that $P(A) = 0$. Pick out the true statements:  a. $P = cP_A$ where $c \in\Bbb R$ and $P_A$ is the characteristic polynomial of $A$;  b. if $P$ has a complex root (i.e. a root with non-zero imaginary part), then $P = cP_A$, with $c$ and $P_A$ as above;  c. if $P$ has a complex root, then $A$ is diagonalizable over $\Bbb C$.  I guess b and c are true although I am unable to disprove a. Because A could satisfy $(X-A)$ but this is not a polynomial. Now if $P$ has a complex root, then its conjugate too is its root as the polynomial has  real coefficients. As we all know An odd degree polynomial has always a real root, what we have is three distinct roots. Since $A$ satisfies the polynomial the only difference between the characteristic polynomial and $P$ has to be a constant. Why are swap partitions discouraged on SSD drives, are they harmful? I often read that one should not place swap partitions on a SSD drive, as this may harm the device. Is this true? Can you please explain the reason to me?  Because I otherwise would have thought that placing swap on an SSD is the best choice, as it's much faster than HDDs and therefore swapping RAM contents to the SSD is not as slow as it would be with the HDD... Duplicate flag lists "you have already raised this flag" after flagging as "Unclear what you're asking" For $D$ a GCD domain , let $a,b,x \in D \setminus \{0\}$ , then is it true that $\gcd (ax,bx)=x \cdot \gcd (a,b)$? Why is a home fiber connection asymmetric? I recently switched from an ADSL to a fiber connection and was surprised to see that the bandwidth is asymmetric (1 Gbps down / 250 Mbps up).  While this is understandable in ADSL, I do not understand why fiber at home is asymmetric as well? It is not a limitation of the technology (I have symmetric fiber at the office, and always had) So assume $Y$ and $X$ are exponentially distributed with parameters $y_1$, and $x_1$ respecitively.  assume c is a constant.  I am having huge trouble to understand the integration of the following expression.  $P(Y&lt;c/u(X))$  $=\int_{t}^{\infty}f_X(x)\int_{0}^{c/u(x)}f_Y(y)dydx +\int_{0}^{t}f_X(x)\int_{c/u(x)}^{\infty}f_Y(y)dydx$  where t is the cross-point that $u(x)$ change sign    here $c/u(x)$ is given by the plot below, t is the point crossing the zero:   Confusion:   I don't understand the second integration of the second term "$\int_{c/u(x)}^{\infty}f_Y(y)dydx$", this isn't right because Y an X only defined for y>0, and x>0. So it's the first quadrant in this plot. I've been looking around on the internet and I can't find anything in depth on how the cable/connection functions. I've only been able to find simple pin out descriptions.  If someone can link me or point me in the right direction, I would be very grateful. the teacher gave me this problem to solve. So i hope you can help me with it. Here is the problem: John poured 150ml ob boiling water in 2 cups made out of polystyrene. Cups are open on the top so heat goes out there. Temperature of the outside air is 20°C. In 1 cup he add 5 mm thick layer of hot oil. He measured starting temperatures and then he measured temperatures again after 10 min. He measured the mass of water and water+oil before and after 10 min.  Here are the resaults:   -cup 1 (only water)  starting temp: 96°C  finnish temp: 64°C  starting mass: 149.4g  finnish mass: 142.42g  -cup 2(with oil on top)  starting temp.: 96°C  finnish temp.:91°C  starting mass: 164.72g  finnish mass: 164.70g  The surface of the both cups is 0.003 m2  Questions: 1.Why did the water covered with oil cooled down slower that the one with no oil? Show with calculations. 2.Why did John use oil with temperature around 100°C and not oil with room temperature? How would the resaults change if he used room temeprature oil? 3. Which other physical values, which are relevant, can you calculate from this informations? Basiccaly i need the physics behing why does this happen. When drawing a line between two nodes in TikZ, you can specify the anchors on the beginning and end nodes in order to control how the line is drawn.  I would like to specify the anchor on the beginning node, and the angle at which the line is drawn, and let TikZ figure out where it will intercept the end node.  The usual use case is when I'm connecting to a "big node" and I would like the lines to be vertical.  I can cheat by placing a coordinate behind the big node, drawing a line to the coordinate instead, and then drawing the big node so that it covers things up.  However this won't work if I need to put an arrow on the end.  I could try to calculate the intercept coordinate on the edge of the big node manually, but that seems against the TikZ spirit.  |- doesn't work in this case.  \documentclass[10pt]{minimal} \usepackage[active,tightpage,pdftex]{preview} \usepackage{tikz}  \begin{document} \begin{preview}  \tikzstyle{RectObject}=[rectangle,fill=white,draw,line width=0.5mm] \tikzstyle{line}=[draw] \tikzstyle{arrow}=[draw, -latex]   \begin{tikzpicture}   %%%%%%%%%%%%%%%%%%%%%%%%   %%  LEFT SIDE   %%%%%%%%%%%%%%%%%%%%%%%%   \draw (1,0) node[RectObject] (Small1A) {A};   \draw (3,0) node[RectObject] (Small1B) {B};   \coordinate (Cheat1A) at (1,2);   \coordinate (Cheat1B) at (3,2);   \path [line] (Small1A.north) -- (Cheat1A);   \path [line] (Small1B.north) -- (Cheat1B);    % need to draw Big Node last to cover up our cheating   \draw (2,2) node[RectObject, inner xsep=1cm] (Big1) {Big Node};    \draw (2,4) node[text width=4cm] (text1) {L: Cheating to control line orientation. Arrows would be hard.};    %%%%%%%%%%%%%%%%%%%%%%%%   %%  RIGHT SIDE   %%%%%%%%%%%%%%%%%%%%%%%%   \draw (7,2) node[RectObject, inner xsep=1cm] (Big2) {Big Node};   \draw (6,0) node[RectObject] (Small2A) {A};   \draw (8,0) node[RectObject] (Small2B) {B};   \path [arrow] (Small2A) -- (Big2);   \path [arrow] (Small2B) -- (Big2);    \draw (7,4) node[text width=4cm] (text2) {R: No cheating. Arrows easy. Can't control line orientation.}; \end{tikzpicture} \end{preview} \end{document}   Here's a side-by-side example of "cheating", which shows what I want to be able to do, and "not cheating", which does not do what I want.    The left side of this example shows what I want to be able to do, except I want to do it without cheating, and I'd like to put arrows on the end of the lines. Why does the new review page sometimes shows questions to vote to close without anybody having voted to close it? What is the purpose of -e in sed command? I can't find any documentation about the sed -e switch, for simple replace, do I need it?  e.g.  sed 's/foo/bar/'   VS  sed -e 's/foo/bar/' Is it logically correct to assert that the expectation of the momentum $$\langle \hat p \rangle=0$$ for any bound state because it is bound to some finite region? What is the physical interpretation of the fact that $$\langle \hat p \rangle=0$$ in an energy eigenstate $\psi_n(x,t)$  but $$\langle \hat p \rangle\neq0$$ in some superposition state $$\psi(x,t)=c_m\psi_m(x,t)+c_n\psi_n(x,t)~?$$ Here $\psi_n(x,t)$ the eigenstates of the Hamiltonian, for example, in the problem of particle in a box (say).