I wonder if somebody could help me understand the following solution:  $\textbf{Problem:}$ Let $F=\{ f \ast f: f \in \mathcal{E}'(\mathbb{R}^n) \} \subset \mathcal{E}'(\mathbb{R}^n)$. Show that $F \neq \mathcal{E}'(\mathbb{R}^n)$ ($f\ast f$ is the convolution of $f$ with it self and $\mathcal{E'} $ is the space of distributions with compact support in $\mathbb{R}^n$).  $\textbf{Solution:}$ Assume the contrary. Then $\delta_1 -\delta_0 \in F$ so there is an $f \in \mathcal{E'}$ such that $f \ast f= \delta_1 -\delta_0$ Since $f \in \mathcal{E'}$ such the Fourier transform of $f$, $\hat{f}$, is complex analytic. Then $ (\hat{f}(\xi))^2= e^{-i\xi}-1 \in \mathcal{O}({\xi}) \implies \hat{f}(0)=0$.  I fail to see why this is a contradiction (also I don't get the last implication). Does anybody understand this?

For a compactly supported distribution $f$, I want to show the non-surjectivity of the map $H:f \mapsto f*f \in \mathcal{E}'$.     My attempt: I have chosen to show that this is not the case by showing that $\text{image } (H)$ does not contain $ \delta_0'$. For instance I restrict myself to $d=1$.   Suppose that $H$ is surjective. Applying the Fourier transform on the convolution, I get $(\hat{f}(z)) ^2 = i z$. The theorem of Paley-Wiener-Schwartz guarantees that $(\hat{f}(z)) $ should be an entire function. However, as it is well-known, no such entire function exists:   If this were true, then $e^{i \pi /4}\hat{f}(z)$ would be the square root function, which cannot be well defined on $\mathbb C$, and even less so entire.    Q1: Is there another way of achieving the result (i.e. with a choice significantly different than $\delta_0'$, and with an approach different than the one using the Fourier transform)?  Q2: Can I generalize this without much effort in the case of $d \ge 1$ (probably using the tensor product)?    Edit: The source is .


I have to show that for all $\sigma \in S_n$ following equation holds: $\sum_{i=1}^n|\sigma(i)-i|=2k$ for $k\in \mathbb{Z}$.  I have no idea how to show it and would be very grateful for any hint.

Let $\sigma \in S_n \;$ then $\; \sum_{i=1}^n|\sigma(i)-i|\;$ is an even number. I tried to show the statement for cyclic permutations first. However, I am not getting anywhere. The statement is trivial for transposition $(k,l)$. That´s all what I have found out so far. Can someone give me a hint why the statement is in fact true for any cyclic permutation ?


While i was initializing a software i required to put my system architecture in a command which is mentioned in the picture. Please suggest how to check my systems architecture in a format given in example of initialization section mentioned in the picture.

I downloaded and installed Ubuntu from the official site. However, I don't know if I installed the 32-bit or 64-bit version.  In Windows 7 I could right click My Computer and it listed which version it was.  Is there an easy way to check in Ubuntu?


I want to stick footer at the bottom of the page even when there is no content. When there is enough content, the footer should move to the end of the page.   I'm learning flexbox and tried doing so using flexbox. But it isn't working.     html {   height: 100%; }  body, #root {   display: flex;   flex-direction: column; }  .app {   text-align: center; }  .header, .footer {   background: #282c34;   flex-shrink: 0;   color: white; }  .main {   flex: 1 0 auto; }  a {   text-decoration: none;   color: #42A5F5;   padding: 20px 0; }

       Logo
Application Name
                Link1         Link2         Link3

Small Description

         <input type="text" />
Please Insert Search Query

Some Random Company

Contact

Twitter

     I want to know why it's not working and the mistake that I'm doing here. I know this question has been asked before. I just want to know, what's wrong with my code so that I can write better CSS after understanding it.  Also, here is the updated

I have a simple 2-column layout with a footer that clears both the right and left div in my markup.  My problem is that I can't get the footer to stay at the bottom of the page in all browsers.  It works if the content pushes the footer down, but that's not always the case.


I have created a document in LaTeX and the document is complete, but the printers have asked for a 3mm white board applied to the PDF, so the overall size is 6mm wider and taller than standard A4.  How would I do this?  I found a similar question but this is for A5 paper instead:   \documentclass[a4paper, twoside, 12pt, hidelinks, final]{article}  \usepackage[top=1in, bottom=1in, left=0.75in, right=0.75in, headheight=35.4pt, footskip=35.4pt]{geometry}  \begin{document}  \clearpage\mbox{}\clearpage  \end{document}```

I have seen the options such as a4paper, a5paper, letterpaper, etc... to go in the square brackets after \documentclass, but what I want is 8.25 inches width by 10.75 inches height for one document and 6 inches width by 9 inches height for another one. How is this done?


I found a pretty cool box I would like to replicate and use in my report, however, the source code for this particular box is not available, and although I'm getting close to figuring it out, I've already put a ridiculous amount of time into it, so I'm hoping maybe one of y'all have an easy answer.   The box I'm referencing can be seen in the Table of Contents -- Section 6 -- of the . It's the small green "side by side" box.  I basically just want to reproduce the box width/size, and vertically-stacked text style, change the colors, and enter arbitrary text.      Start with the used \documentclass{... and end the code with \end{document} so that it is compilable with pdflatex. Not all of us use LyX ...    Here's what I have so far:  \documentclass{article} \usepackage[most]{tcolorbox}  \begin{document} \tcbsidebyside [ nobeforeafter, sidebyside adapt=both, size=fbox, drop lifted shadow, bicolor, colback=black, colbacklower=white, colframe=black, colupper=white,  collower=black, fontupper=~\bfseries, fontlower=\bfseries, ] {AC} {H} \end{document}   Here's my alternate approach:   \documentclass{article} \usepackage[most]{tcolorbox}  \begin{document}  \begin{tcolorbox} [ nobeforeafter, sidebyside, size=fbox, skin=bicolor, halign=center, halign lower=center, colframe=black, colback=black,  colbacklower=white,  colupper=white,   collower=black, fontupper=~\bfseries,  fontlower=\bfseries, ] \textbf {AC} \tcblower {H} \end {tcolorbox} \end{document}   Thanks!  ijwmcb

I'm working with tcolorbox, I saw the pdf  and I want to reproduce the little green box in this image:     Can anyone help me please?


So this question is a direct follow on from  I am trying to get an expression for $\frac{d\left<p\right>}{dt}$. I have my working out below, which was thanks to the 's help, (not to mention all the patient people in the mathematics chatroom) I have got to this expression: $$ i\hbar \left( \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx + \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx \right). $$ And now I'm stuck. What can I do about $$ \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx $$ In particular how to deal with $$ \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x}? $$ With the $\partial t$ in play I don't know how solve the integral... As before, I really would appreciate hints to get me moving in the right direction more than an outright solution. My working so far We are given $\left<p\right>$ in the text, which is determined by computing $\frac{d\left<x\right>}{dt}$ and multiplying by $m$. Lets just start with what we know: $$ \left<p\right> = \int{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx. $$ So taking the time derivative, $$ \frac{d\left<p\right>}{dt} = \frac{d}{dt}\int{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx. $$ We probably should at this point pause, and listen to Ted, (). So take out the constants before going any further. We get, $$ -i\hbar \frac{d}{dt}\int\limits_{-\infty}^{+\infty}{         \left(             \Psi^*                 \left[                     \frac{\partial}{\partial x}                 \right]             \Psi         \right)         }dx. $$    Bringing the time derivative into the integral, gives us $$ -i\hbar \int\limits_{-\infty}^{+\infty}{\frac{\partial}{\partial t}\Psi^*\left[\frac{\partial}{\partial x}\right]\Psi}dx. $$ Then differentiating with the product rule yields, $$ -i\hbar \int\limits_{-\infty}^{+\infty}{ \frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x}+  \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx. $$ Now we have two terms in our integral, let's split em up. Let's consider the left handside first, we have $$ \int\limits_{-\infty}^{+\infty}{ \frac{\partial \Psi^*}{\partial t} \frac{\partial \Psi}{\partial x} }dx. $$ By Schrödinger this can become, $$ \int\limits_{-\infty}^{+\infty}{ \left(\frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} + \frac{i}{\hbar}V \Psi^* \right) \frac{\partial \Psi}{\partial x} }dx. $$ Expand to get, $$ \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x}+ \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx. $$ So, just to review, our full problem can be written as, $$ i\hbar \left( \int\limits_{-\infty}^{+\infty}{ \frac{-i\hbar}{2m} \dfrac{\partial^2 \Psi^*}{\partial x^2} \frac{\partial \Psi}{\partial x} + \frac{i}{\hbar}V \Psi^* \frac{\partial \Psi}{\partial x} }dx + \int\limits_{-\infty}^{+\infty}{ \Psi^* \frac{\partial}{\partial t} \frac{\partial \Psi}{\partial x} }dx \right). $$

I am currently working through the problems in , with the goal of hopefully understanding how magnets work. At the moment I am a little stuck on one of the early problems in the first chapter. And I would like a Hint about how to proceed. The problem is to solve the time derivative of the expectation value of momentum of a particle. ie: $$ \frac{d\left<p\right>}{dt}. $$ My working so far Just prior to that question the text gives an expression for momentum, $$\int\limits_{-\infty}^{+\infty}{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx, $$ (Where $\Psi$ is a function satisfying Schrödinger's equation, $i$ is the square root of negative 1, and $\hbar$ is Planck's constant divided by $2\pi$.) So my initial approach was to try and differentiate this with respect to $t$. $$ \frac{d}{dt}\int\limits_{-\infty}^{+\infty}{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx. $$ However I didn't know how to do that, so I asked around in the maths chatroom, and got a lot of help. Specifically  from @RobJohn: $$ \begin{align} \int vv'\,\mathrm{d}x &=vv-\int vv'\,\mathrm{d}x\\ &=\frac12vv+C \end{align} $$ So I suppose it would follow that, $$ \begin{align} \int\limits_{-\infty}^{+\infty}{         \left(             \Psi^*                 \left[                     \frac{\partial}{\partial x}                 \right]             \Psi         \right)         }\,\mathrm{d}x &=  \left[\Psi^* \cdot \Psi\right]_{-\infty}^{+\infty} - \int\limits_{-\infty}^{+\infty} {\Psi\frac{\partial \Psi^*}{\partial x} }\,\mathrm{d}x\\ &= \frac{1}{2}|\Psi|^2 \Big|_{-\infty}^{+\infty}, \end{align} $$ And then, $$ \begin{align} &\frac{d}{dt}\int\limits_{-\infty}^{+\infty}{\Psi^*\left[-i\hbar \frac{\partial}{\partial x}\right]\Psi}dx \\&=-i\hbar \frac{d}{dt}\left( \frac{1}{2}|\Psi|^2 \Big|_{-\infty}^{+\infty} \right). \end{align} $$ But this feels wrong. I'm not sure exactly why. (Maybe it's just because it is the limits of my maths)  For reference, regarding what I understand, my maths is pretty low-level, I took a course in undergrad calculus over 10 years ago and only just passed that at the time, and my physics knowledge is not much further than that of an enthusiastic lay person.


Does a RAID 1 setup by software (win) break once you move both drives to another Win PC? Is there any way to bind the drives together so that the RAID setup stays, even after changing motherboard/cpu?

I have 2 identical drives in an old PC running as a mirror RAID (RAID1) using the Windows 7 software RAID. These drives are data-only drives and are not being used to boot an operating system.  How do I safely move these drives to a new PC?  The new PC is also running Windows 7.  There is a Windows Computer Management help article titled "Move Disks to Another Computer", which talks about dynamic disks, but I find it a bit confusing. It's not clear if all of the steps given are necessary for dynamic disks (of which mirrored disks are a type), and the step that is clearly for dynamic disks mentions menu items that don't exist for mirrored drives.  Basically, I'd appreciate any knowledge or experience in this area.


I have very easy question that I'm hoping someone can assist me with:  I ran an example logistic regression using this R code:       hours <- c(0.5, 0.75, 1, 1.25, 1.5, 1.75, 1.75, 2, 2.25, 2.5, 2.75, 3, 3.25, 3.5, 4, 4.25, 4.5, 4.75, 5, 5.5)         pass <- c(0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 1, 1, 1, 1)         data <- data.frame(hours, pass)         mylogit <- glm(pass ~ hours, data = data, family = "binomial") #Activates the logistic regression model         summary(mylogit) #Summary of the model      Call:     glm(formula = pass ~ hours, family = "binomial", data = data)      Deviance Residuals:           Min        1Q    Median        3Q       Max       -1.70557  -0.57357  -0.04654   0.45470   1.82008        Coefficients:                 Estimate Std. Error z value Pr(>|z|)       (Intercept)  -4.0777     1.7610  -2.316   0.0206 *     hours         1.5046     0.6287   2.393   0.0167 *     ---     Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1      (Dispersion parameter for binomial family taken to be 1)          Null deviance: 27.726  on 19  degrees of freedom     Residual deviance: 16.060  on 18  degrees of freedom     AIC: 20.06      Number of Fisher Scoring iterations: 5      round(exp(cbind(OR = coef(mylogit), confint(mylogit))),3)                 OR 2.5 % 97.5 %    (Intercept) 0.017 0.000  0.281     hours       4.503 1.698 23.223   I know that by taking the exponent of the log-odds/coefficients for hours the odds of passing increase by a factor of 4.503 for a one-unit change in hours.  However, given that the explanatory variable (hours) is continuous, what is considered a 'one-unit change' i.e. going from 1 to 2 hours as one unit? or from 1.75 to 1.76 hours as one unit?  Also, is this interpretation of one-unit the same for regular OLS regression as well? I'm seeking to better understand the rules R applies to creating its regression coefficients.

I'm somewhat new to using logistic regression, and a bit confused by a discrepancy between  my interpretations of the following values which I thought would be the same:    exponentiated beta values  predicted probability of the outcome using beta values.    Here is a simplified version of the model I am using, where undernutrition and insurance are both binary, and wealth is continuous:  Under.Nutrition ~ insurance + wealth   My (actual) model returns an exponentiated beta value of .8 for insurance, which I would interpret as:     "The probability of being undernourished for an insured individual is .8 times the probability of being undernourished for an uninsured individual."   However, when I calculate the difference in probabilities for individuals by putting in values of 0 and 1 into the insurance variable and the mean value for wealth, the difference in undernutrition is only .04. That is calculated as follows:   Probability Undernourished = exp(β0 + β1*Insurance + β2*Wealth) /                              (1+exp(β0 + β1*Insurance + β2*wealth))   I would really appreciate it if someone could explain why these values are different, and what a better interpretation (particularly for the second value) might be.     Further Clarification Edits As I understand it, the probability of being under-nourished for an uninsured person (where B1 corresponds to insurance) is:  Prob(Unins) = exp(β0 + β1*0 + β2*Wealth) /               (1+exp(β0 + β1*0+ β2*wealth))   While the Probability of being under-nourished for an insured person is:  Prob(Ins)= exp(β0 + β1*1 + β2*Wealth) /            (1+exp(β0 + β1*1+ β2*wealth))   The odds of being undernourished for an uninsured person compared to an insured person is:  exp(B1)   Is there a way to translate between these values (mathematically)? I'm still a bit confused by this equation (where I should probably be a different value on the RHS):  Prob(Ins) - Prob(Unins) != exp(B)   In layman's terms, the question is why doesn't insuring an individual change their probability of being under-nourished as much as the odds ratio indicates it does?  In my data, Prob(Ins) - Prob(Unins) = .04, where the exponentiated beta value is .8 (so why is the difference not .2?)


When I try to update my notification email, it doesn't seem to work on any SE site. I fill in my new address at "Send emails to," then click on "Save email settings." After that, the "Save email settings" button is gray and nothing seems to happen. If I leave the preferences page and come back, the old email address is still there. Any idea what this problem might be?  This seems to be similar to, but not the same as, this problem:  By clicking on "edit," I can change my email address, but then when I go to preferences, my notification email is still the same. Are these supposed to be two independently controllable email addresses?

The email for getting notifications can be changed by clicking the "preferences" tab in your profile. After entering a new email address, the "Save email settings" button becomes clickable:    Clicking the button turns it grey:    So all looks well. But if you don't confirm the email by clicking on the link sent to your email inbox, the next time you go back to the preferences, you will see your old email address with no indication that it changed.  You can subscribe to the weekly newsletter from the "edit" tab of your profile. If you click the option and save, the next time you visit the edit page before confirming your email you get this message:    Once you confirm the email, the resend link disappears so you know the subscription is set up:    Ideally, the two systems would be unified so that all email subscriptions happen the same way. But short of that, it would be very helpful to indicate that an email address is unconfirmed for notification emails just like newsletters.


Often questions from math.stackexchange.com get into MultiCollider's "hot questions".  Formuli is often main things in such questions, but they doesn't get rendered in the dropdown, instead I see something like this:    It's tricky to judge whether the question is interesting without viewing the formula.  Can formulas be displayed in MultiCollider?  P.S. I'm surprised that there seems to be no such question already asked on the Meta nor on meta.math.SE.

I guess not, but just in case it's easily fixed: could the hot questions and SE profile support MathJax to render question titles and summaries? On most SE sites, the SE MultiCollider does not support MathJax, and  the global SE profiles:   Of course, the specific site itself supports  just fine:   I know support for this on the regular sites , "as it is an extremely heavy dependency". But maybe it's still doable just for rendering the titles and summaries? Or maybe only on the global SE profile? (Careers might like this too? Never used that.)


All question I have looked to doesn't help me to solve my problem. Here's my code :  \documentclass[a4paper,10pt]{article} \usepackage[utf8]{inputenc} \usepackage[frenchb]{babel} \usepackage{color} \usepackage{colortbl} \usepackage{longtable} \usepackage{lmodern} \usepackage[table]{xcolor} \usepackage{multirow} \usepackage{pdflscape}  \usepackage{hyperref} \newcommand*{\thead}[1]{\multicolumn{1}{|c|}{\bfseries #1}}  \begin{document}   \newcommand{\mc}[3]{\multicolumn{#1}{#2}{#3}} \begin{landscape} \begin{longtable}{|>{\raggedright\arraybackslash}p{3cm}|p{7cm}|c|p{7cm}|} \hline \cellcolor{gray!30} \centering \bfseries Flux de\par substance étudié &  \cellcolor{gray!30} \centering  \bfseries Mécanisme \par environnemental &  \cellcolor{gray!30} \bfseries Unité &  \multicolumn{1}{|p{7cm}|}{\cellcolor{gray!30} \centering \bfseries Remarques\par(RECORD)} \\ \cline{1-4} \cellcolor{green!50}  & % \cellcolor{green!50} \'Emission  de CO2 liée à l'épandage de chaux & \cellcolor{green!50}  & \\ \cline{2-2} \cellcolor{green!50} \multirow{-2}{*}{CO2} &  \cellcolor{green!50}![enter image description here][1] \'Emission  de CO2 liée à l'épandage d'urée & \cellcolor{green!50} \multirow{-2}{*}{$kg/ha$} & \\ \hline \end{longtable} \end{landscape} \end{document}   Here's what I get :   Table cell color overlaps cell border but I want the borders to be always visible. The answers given to questions I read doesn't helped as the people who give the solution doesn't explain well the code they give.  How can I solve my problem?

When I change the background color of a cell in a table, the color covers the borders, however I want the horizontal lines to remain.  I have a simple command \cellcolor[gray]{0.9} in my table. Can anybody help me with this problem?


I have attempted to reinstall previous purchases via the software center. I am aware that some of the trashier devs (super meat boy) have decided to drop support for Linux, so the software center doesn't allow me. It tells me that it doesn't have a package for my version of Ubuntu. Fine.  However, all of my other games (and there are quite a lot of them) harbor no complaints with the Software Center. If I try to install any of these, it correctly adds the private ppa to my sources.list. However, upon updating apt, I always get a 401, regardless of the package. This is a fresh install.  Why am I unable to use any of my bought games on 13.04?

After a reinstall of Ubuntu 12.04 ,it's impossible to install commercial games, e.g Oil Rush. It gives this error:  Failed to fetch https://private-ppa.launchpad.net/commercial-ppa-uploaders/oilrush/ubuntu/dists/oneiric/main/binary-amd64/Packages  The requested URL returned error: 401 Failed to fetch https://private-ppa.launchpad.net/commercial-ppa-uploaders/oilrush/ubuntu/dists/oneiric/main/binary-i386/Packages  The requested URL returned error: 401 Some index files failed to download. They have been ignored, or old ones used instead.   I bought these games in 11.10, but I installed these games in 12.04 before... How can I solve this problem?


I'm creating a HashMap as follows:  private HashMap <String, JSONArray> HiveMap; HiveMap = new HashMap <String, JSONArray>();   and also a private class variable to store some calculated data which will be put into the HashMap after the calculations are done:  private JSONArray hiveList;   After hiveList is calculated, I put the hiveList into the created HashMap (HiveMap):  HiveMap.put(hiveNode,hiveList);   When I print out HiveMap now, the output I get is as expected, with both the fields hiveNode and hiveList existing in the Hashmap. <"hiveNode":"hiveList">  After this is done, I assume that the data I put into the HashMap from hiveList is going to persist, so I clear the hiveList array using hiveList.clear();  But when I print the Hashmap after clearing hiveList, I see that the data from hiveList is gone from the HashMap as well.  Printing the HashMap after clearing hiveList results in: <"hiveNode": >  I don't understand this behavior, I would appreciate it if someone could shed some light on this.

I always thought Java uses pass-by-reference. However, I've seen a couple of blog posts (for example, ) that claim that it isn't (the blog post says that Java uses pass-by-value). I don't think I understand the distinction they're making. What is the explanation?


I am running Debian Buster. On my desktop settings/ Power, suspend and hibernate are deactivated. As expected the computer never sleeps/ hibernates. The problem: When working remotely, and the computer is idle then it enters into hibernation and it is unresponsive. The only way to get it back is to restart the computer locally. It looks 'sleep' and 'hibernate' are configured differently when the machine is used locally or remotely. Any idea what is going on? and how to solve it? thanks

I have a machine that I access only via SSH that I just updated to Debian 10 a couple of days ago. Since the update it seems to be going to sleep when it is inactive. This has never happened with previous updates like from 7 to 8 or from 8 to 9. It seems like maybe the sleep settings have reverted to a default state. How can I view and edit the power and sleep settings in the command line? Any guidance much appreciated. Thanks!


This is more of a conceptual doubt. I am learning to use vectors in C++.  While iterating through a vector, I could do it in two ways:  vector<int> temp; for (int j = 0; j < 10; j++){          temp.push_back(j); }  int sum1 = 0; int sum2 = 0;  //Method 1: almost treating it like an array for (int i = 0; i < temp.size(); i++){     sum1 = sum1 + temp[i]; }  //Method 2: using an iterator vector<int>::iterator it; for(it = temp.begin(); it < temp.end(); it++) {          sum2 = sum2 + *it; }   Both methods worked fine and yielded expected results. However, I have noticed that most of the suggested codes (on stackexchange, etc) use  iterators. Is there any specific reason for that or is it just out of convenience?

Take the following two lines of code:  for (int i = 0; i < some_vector.size(); i++) {     //do stuff }   And this:  for (some_iterator = some_vector.begin(); some_iterator != some_vector.end();     some_iterator++) {     //do stuff }   I'm told that the second way is preferred. Why exactly is this?


I'm creating a 4 x 6 meter (~13 x 20 feet) outdoor billboard advertisement that includes an image and vector logo (rasterized) in Photoshop. I was working using a smaller size for design purposes, but now the client has approved a final design and wants to go to print. I need to know what DPI/PPI I should export it as.   Additionally, I was given advice by a friend to make it at 30 DPI and work in centimeters (so 400 x 600 cm) then let the print house blow it up to 4 x 6 meters, is this true?

(This question is meant to be a resource to direct others to given that this is a common question.)  Given the task of creating an image that will be printed at a certain size, what pixel dimensions and resolution should I be setting my file at?


I am using bind for internal networking it is working good,  now I am a another requirement.  bind9 resolving the domain myweb.example.com from ip 192.168.1.1 port 80  now I want to resolve domain myweb.example.com:82 from 192.168.1.2:82.  is that possible?

Greetings all,  I have a server with two tomcat services (one being run on port 80, the other 8080).   What I'm wanting to achieve is to have two different DNS records for these two ports, so they'll be accessible via:  http://example1/ http://example2/   Instead of  http://example1/ http://example1:8080/   I know this isn't possible using DNS A records, but does anyone know of another solution I could look into?  Thanks,  Andrew


Can anyone explain me how to stop recursive triggers with an example?

I have two objects A(parent) and B(child) related by a lookup relationship. I  need to update records in object A using data loader. There is an after update trigger on object A which updates some fields on related records of object B. On object B as well there is an after update trigger which updates a field on parent object A. How can i handle recursion.


EDITED  John is playing a game on $n$ days, each day being independent.  On each day $i$, his probability of success is $p_i$.   We have $\frac{1}{n} \sum_{i=1}^n p_i = p$, and typically, the standard deviation $\sigma$ of these $p_i$ is small. $\sigma$ is known.  So I have a succession of $Bernoulli(p_i)$. I want to model the probability of having k success after n trials, using $p$ and $\sigma$ for instance.  I can approximate this with a Binomial distribution, but even though it's close it still leads to biases. Any thoughts on how to improve this?

If 20 independent Bernoulli trials are carried out each with a different probability of success and therefore failure. What is the probability that exactly n of the 20 trials was successful?  Is there a better way of calculating these probabilities rather than simply summing together the combinations of success and failure probabilities?


This is the question i've been given along with the solution i have written, however, could someone explain why showing $z \notin V$ shows any metric space is a Hausdorff space

I wish to show that any metrizable space $(X,\mathcal{T})$ is Hausdorff  Proof attempt:  Let $d$ be the metric that generates the topology on $X$. Pick two points $x,y \in X$, we wish to produce two disjoint open sets that separates $x,y$.  Let $B_\epsilon(x)$ and $B_\delta(y)$ be two metric balls containing $x,y$ respectively. Suppose that  $B_\epsilon(x) \cap B_\delta(y) \neq \varnothing$       Stuck Here: Hmm...How should I adjust $\epsilon, \delta$ so that these   balls are separated?   Idea: Reduce $\epsilon$ by half. If they are still intersecting...reduce $\delta$ by half. Continue ad infinitum         Is there a more satisfying solution i.e. closed form expression for   reduced $\epsilon, \delta$ so they are no longer intersecting. Thanks!


Why isn't there a proper solution on internet on this sum:  $$\sum_{i=1}^{n} \frac{1}{i}$$  Can you help me with your solution?

How to find the sum of the following series? $$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$ This is a harmonic progression. So, is the following formula correct? $\frac{(number ~of ~terms)^2}{sum~ of~ all ~the~ denominators}$ $\Rightarrow $ if $\frac{1}{A} + \frac{1}{B} +\frac{1}{C}$ are in H.P. Therefore the sum of the series can be written as : $\Rightarrow \frac{(3)^3}{(A+B+C)}$ Is this correct? Please suggest.


Ok so Blender was working perfectly fine for months now until today I tried to render a weapon with a chrome material, only to find out that it rendered without the material! I already tried resetting Blender to all default settings and I've quadruple checked to be sure non of the objects are hidden. Yes, the lighting is very bright too, but that doesn't show either. See pictures below.  (Edit): Guys, I'm not stupid. I already tried the background idea. I looked around for an hour for solutions before posting this question and found a few people having the same problem, (including the thread that this is a "duplicate" of) but none of the suggestions people gave them fixed my problem. Even if there is no background, the object should at least be interacting with the light somewhat. Example:     That's a render I made the week before I started having issues, I'm not very proud of it but it proves my point. So what's below doesn't make any sense.    Even with a different color.    But when I use a dryer, non glossy material like leather or just the default diffuse mat everything shows up fine.    Plz help ;__;

I'm new to blender (and 3D modelling) so please use dummy terms if you answer!  I have applied 3 materials one each to 3 different objects and only 1 renders the same as the material preview. I've attached a screenshot to show you what I mean. The silver material should be on the top ball and the brass material should be on the cylinder holding the ball up. I've no idea why it's not working when the copper material on the spiral wire worked. I'm using cycles render if that helps?


I'm reading Quantum field theory in a Nutshell and I find a very interesting calculation that leads to the gravitational interaction between 2 light beam.   Is this kind of interaction permitted in general relativity? (not necessarily a quantum gravity theory or the like) Is there a energy tensor $T_{\mu\nu}$ asscociated with a photon so that it creates $R_{\mu\nu}\ne 0$? It's awesome if someone can give a simple derivation in the framework of general relativity.

In general relativity, light is subject to gravitational pull.  Does light generate gravitational pull, and do two beams of light attract each other?


I am new to C but I'm currently working on a project and I am hitting a strange problem.   I have the following piece of code:  int insertID = 0; asprintf(&inboundSql, "INSERT INTO DataTable VALUES (%i, %i, '%s', '%s', %i),"         "(%i, %i, '%s', '%s', %i), (%i, %i, '%s', '%s', %i), (%i, %i, '%s', '%s', %i),"         "(%i, %i, '%s', '%s', %i), (%i, %i, '%s', '%s', %i)",         dataRow, D_DATE, callLogSearchData[dataRow].date, epochBuffer, insertID++,         dataRow, D_TIME, callLogSearchData[dataRow].time, epochBuffer, insertID++,         dataRow, D_APARTY, callLogSearchData[dataRow].aParty, epochBuffer, insertID++,         dataRow, D_BPARTY, callLogSearchData[dataRow].bParty, epochBuffer, insertID++,         dataRow, D_DURATION, durationBuffer, epochBuffer, insertID++,         dataRow, D_RESULT, callLogSearchData[dataRow].cleardownCause, epochBuffer, insertID++);   When I compile the code, I get the following:  warning: operation on insertID may be undefined  Even though I get the above warning, my code is working as expected so I don't understand what the problem is. I guess it thinks there's something wrong with doing insertID++ but I can't see why that should be a problem.   Thanks for your help.

Can it be assumed a evaluation order of the function parameters when calling it in C ? According to the following program, it seems that there is not a particular order when I executed it.   #include <stdio.h>  int main() {    int a[] = {1, 2, 3};    int * pa;      pa = &a[0];    printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa), *(pa++),*(++pa));    /* Result: a[0] = 3  a[1] = 2    a[2] = 2 */     pa = &a[0];    printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa++),*(pa),*(++pa));    /* Result: a[0] = 2  a[1] = 2     a[2] = 2 */     pa = &a[0];    printf("a[0] = %d\ta[1] = %d\ta[2] = %d\n",*(pa++),*(++pa), *(pa));    /* a[0] = 2  a[1] = 2 a[2] = 1 */  }


Discs are naturally occurring shapes in the world, because any soft object experiencing rotation on a single axis will stretch itself into a disc shape such as pizzas and galaxies due to centrifugal forces caused by rotation. Galaxies and pizzas are "soft" in the sense that the tensile forces holding the structure together are not strong enough to resists being stretched by rotation. Rocks on the other hand, do not stretch when rotating because the average tensile forces between the atoms of a rock are strong enough to resist centrifugal forces, given that rotation is not too fast. If Earth is mostly fluid, why does it not stretch as it rotates?

The question I am working on is, "Consider the following. (a) Find the angular speed of Earth's rotation about its axis. rad/s (b) How does this rotation affect the shape of Earth?"  I am fully capable of solving part (a); however, I am not sure how to describe the effect earth's rotation on its shape. I tried to search my textbook for the answer, but could not find anything. Is there an actual effect on the shape?


Is it really impossible to not have Steam track your time playing multi-player games?  This seems like an obvious privacy issue to me, no? Is it at least possible to not share them? Also, is there a way to not have them displayed as you log in and look at your own profile?  Is there no way to let people use the other Steam friend features, but not have your playing hours be shown to them?

Is it possible to avoid recording the time you play CS:GO on Steam? My friend and I had a deal that we would spend the same amount of time playing it so it would be fair. Now I have like 30 hours more on record.   I want to play more and get better but I don't want everyone to see I spend almost all day playing it.  Can I play without getting my time recorded on Steam?


So, I didnt know doing this sudo chmod -R 777 / would completely destroy everything i cant access to ssh neither sftp i dont have any backup there is work of months... btw im using a dedicated of OVH, os ubuntu 20.04

I accidently ran  sudo chmod 755 -R /   instead of  sudo chmod 755 -R ./   I stopped it after few seconds, but now there is some problems such as  sudo: must be setuid root   How can I revert permissions back?


I am using below code to list all the custom object  Integer count = 0;  for ( Schema.SObjectType o : Schema.getGlobalDescribe().values() ) {     Schema.DescribeSObjectResult objResult = o.getDescribe();      if(objResult.isCustom())        {          count++;         //system.debug(objResult.getLabel());     } } system.debug('Total custom object :'+ count);     // USER_DEBUG [11]|DEBUG|Total custom object :576   But when i see the system overview is different ? How to get this count match with system overview count. ?  SetUp -> System overview     Custom object - 515  Custom setting -113  Finally got to know apex code gives all installed object which is not counted in UI data.

I queried the below in the service console to get the custom object count. But I got one object difference. i.e "PersonAccount" .  When i check the ID its not displaying in the UI page.     select id, DeveloperName, Description, ExternalName, ManageableState,   NamespacePrefix, SharingModel from CustomObject where ManageableState   != 'installed'   // Output //Make sure you checked the tooling API checkbox when u run the query in console  //Count 514      Setup-> System overview -> custom object count (513)    1 difference in object count.  Object name is  "PersonAccount"   Why is it showed here?


Every system sound mutes spotify. On Ubuntu 16.04 it was working fine. Video:

Since I updated Ubuntu from 16.x to 17.10 and then to 18.04 I don't have any sound. There is also a problem with videos in web browser (they look like a slideshow, still no sound) and Spotify, where it looks like music is loading but nothing happens.


Is there any way to drop 230V AC  to 200 V AC by using resistor? I want to drop voltage before applying to a 12 0 12V 500mA transformer. If yes what will be its rating?

I have been looking around for an easy way to convert 12V to 5V. I have seen some people saying that a simple resistor is all that is needed.  $$ Volts = Ohms \cdot Amps $$ $$ Amps  = \frac{Volts}{Ohms} $$ $$Ohms  = \frac{Volts}{Amps}$$  So applying a resistor will diminish the voltage of the circuit. That should mean that an appropriately sized resistor could simply be placed in the path of a 12V circuit, converting it to 5v.   If this is the case how would one reduce amps? Would series vs parallel make a difference in this area?   I have seen designs that include a regulator IC and some capacitors, but if a simple resistor/fuse/diode setup will do the trick, I would really prefer that.


I have a folder structure like this   main_folder | |--done |  | |  |--test1 |  |--__init__.py | |---check.py   __init__.py:  class Tries(object):     def __init__(self):         print "Test"   check.py:  from done.test1 import Tries Tries()   Error:  --------------------------------------------------------------------------- ImportError                               Traceback (most recent call last) <ipython-input-8-10953298e1df> in <module>() ----> 1 from done.test1 import Tries  ImportError: No module named done.test1    I am not able to import modules from the nested folder. Is there any way to do this.  Edit:  After Salva's answer I changed my structure like this   . ├── check.py |--__init__.py(no content) └── done     ├── __init__.py(no content)     └── test1         └── __init__.py <-- this files contains your Tries class   Same error is thrown now also.

How do I import a Python module given its relative path?  For example, if dirFoo contains Foo.py and dirBar, and dirBar contains Bar.py, how do I import Bar.py into Foo.py?  Here's a visual representation:  dirFoo\     Foo.py     dirBar\         Bar.py   Foo wishes to include Bar, but restructuring the folder hierarchy is not an option.


So I'm trying to create n Tasks and run them, but I'm not sure what I'm doing wrong  class Program {     static void Main(string[] args)     {         int n = 5;         Task[] taskList = new Task[n];         for (int i = 0; i < n; i++)         {              taskList[i] = Task.Factory.StartNew(() => doStuff("Task" + i));         }         Task.WaitAll(taskList);          Console.WriteLine("All threads complete");         Console.ReadLine();       }     static void doStuff(string strName)     {         for (int i = 1; i <= 3; i++)         {             Console.WriteLine(strName + " " + i.ToString());             Thread.Sleep(1000);         }     } }   When running this I'm getting next(Also I don't get why it's task 5, i max number is 4):  But if I'm creating Tasks manually like this          Task task1 = Task.Factory.StartNew(() => doStuff("Task1"));         Task task2 = Task.Factory.StartNew(() => doStuff("Task2"));         Task task3 = Task.Factory.StartNew(() => doStuff("Task3"));         Task.WaitAll(task1, task2, task3);   Everything is running as intended  What am I doing wrong?

I met an interesting issue about C#. I have code like below.  List<Func<int>> actions = new List<Func<int>>();  int variable = 0; while (variable < 5) {     actions.Add(() => variable * 2);     ++ variable; }  foreach (var act in actions) {     Console.WriteLine(act.Invoke()); }   I expect it to output 0, 2, 4, 6, 8. However, it actually outputs five 10s.  It seems that it is due to all actions referring to one captured variable. As a result, when they get invoked, they all have same output.  Is there a way to work round this limit to have each action instance have its own captured variable?


Good night everyone,  I just reinstalled ubuntu 19.04 from scratch because I was having problems with it. In the Ubuntu software center, there are two options to install Krita, one from the snap store, and other from the ubuntu-disco-universe. Before I reinstalled ubuntu, I had Krita installed from the disco-universe source, assuming that this 2nd source hasn't changed in reinstallation. Now, after reinstallation, this second source has Krita 4.1 and not 4.2, which is the latest version, but before reinstallation, this second source had the latest Krita. Is there a way to update this ubuntu-disco-universe repository, so it has the newest version?  Thanks for your help.

Why are packages in the official Ubuntu repositories older than the latest (upstream) versions from Debian Sid, PPAs, the authors, etc.?


I keep getting an error while trying to parse a parametrized query. Error: Token in error = @tableName  Seems that everything is ok with the variables, tried passing strings still the same.  using (SqlCeCommand command = new SqlCeCommand("SELECT * FROM @tableName WHERE @columnName = @id", connection))             {                 command.Parameters.Add(new SqlCeParameter("tableName", tableName));                 command.Parameters.Add(new SqlCeParameter("columnName", column));                 command.Parameters.Add(new SqlCeParameter("id", id));   EDIT1:  Yes it might be wrong way of querying the database. For example if you need to access different tables at a different time of runtime whats the most suitable way of doing it. I mean the architecture of the class which accesses the database.  I am trying to use this to access two tables at a different runtime. I just pass these three variables and return the output.  P.S apologise if im breaking the rules.  EDIT2:  Sorry if I didnt make it clear but theres my question.   What architecture(if I can call that) is better for accessing different database tables at a different time while running the application? I mean what would be your solution to that? Is it good or bad and so on.

As the topic suggests I wish to be able to pass table names as parameters using .NET (doesn't matter which language really) and SQL Server.  I know how to do this for values, e.g. command.Parameters.AddWithValue("whatever", whatever) using @whatever in the query to denote the parameter. The thing is I am in a situation where I wish to be able to do this with other parts of the query such as column and table names.  This is not an ideal situation but it's one I have to use, it's not really prone to SQL injection as only someone using the code can set these table names and not the end-user. It is messy however.  So, is what I am asking possible?   EDIT: To make the point about SQL injection clear, the table names are only passed in by source code, depending on the situation. It is the developer who specifies this. The developer will have access to the database layer anyway, so the reason I am asking is not so much for security but just to make the code cleaner.


How to construct a bijection between $[1,6)$ and $(-3,3)$?  I've tried to come up with a proper bijection, but haven't had any luck. Any ideas how to solve it?

How to define a bijection between $(0,1)$ and $(0,1]$?   Or any other open and closed intervals?   If the intervals are both open like $(-1,2)\text{ and }(-5,4)$ I do a cheap trick (don't know if that's how you're supposed to do it): I make a function $f : (-1, 2)\rightarrow (-5, 4)$ of the form $f(x)=mx+b$ by \begin{align*} -5 = f(-1) &= m(-1)+b \\ 4 = f(2) &= m(2) + b \end{align*} Solving for $m$ and $b$ I find $m=3\text{ and }b=-2$ so then $f(x)=3x-2.$  Then I show that $f$ is a bijection by showing that it is injective and surjective.


Today in class my professor mentioned that there are different types of infinity. This confused me at first because I always thought infinity is just infinity. What are the different types of infinity?

Can someone explain to me how there can be different kinds of infinities?  I was reading "" by  and came across the concept of countable and uncountable infinities, but they're only words to me.  Any help would be appreciated.


(I'll try to keep the question simple for now, since I don't really know how to go about this.)  I have to change unordered list elements inside an html file by  1.) Create a DOM fragment ( for each of the new <li> ) in javascript.  2.) Remove the original batch of <li> in the ul.  3.) Append each new DOM fragment into the ul  4.) Make each of the new <li> elements clickable again  Clicking on a <li> element removes the contents of the ul and replaces it with a new batch (the batch varies from 0 <li> to 6 <li> elements (depending on which element you clicked))  The first problem I'm having is that there is no way to create multiple elements clickable in one go (for loops doesnt work) so at the moment im just repeating the functions I have.      $( "#choices" ).children().eq(0).click(function() {         selectChoice(allChoices, $( "#choices" ).children().eq(0).text());     });      $( "#choices" ).children().eq(1).click(function() {         selectChoice(allChoices, $( "#choices" ).children().eq(1).text());     });     // more li elements to make clickable   Second, when you remove a li element and put back a new element, you have to make it clickable, but since you dont know the number of elements and for loops dont work, I can't add a function upon creation. Not all elements follow the same format also, so simply modifying the existing li isnt an option.  Is there a simple way that I don't know about that does what I want it to do..?

I have a bit of code where I am looping through all the select boxes on a page and binding a .hover event to them to do a bit of twiddling with their width on mouse on/off.  This happens on page ready and works just fine.  The problem I have is that any select boxes I add via Ajax or DOM after the initial loop won't have the event bound.  I have found this plugin (), but before I add another 5k to my pages with a plugin, I want to see if anyone knows a way to do this, either with jQuery directly or by another option.


We are considering expanding our user provisioning to cover a large portion of the workforce that does not currently exist in AD.  Our understanding is that any named user in AD requires a User CAL.  Is that correct even if the employee will not be authenticating or have an email address assigned to them?

This is a  about Licensing.    Questions on licensing are off-topic on Server Fault.  If your question has been closed as a duplicate of this question, then this is because we want to help you understand why licence questions are off topic rather than just telling you "it just is". In all likelihood, this question will not address your question directly, it was not meant to.    I have a question regarding software licensing. Can the Server Fault community please help with the following:   How many licenses do I need? Is this licensing configuration valid? What CALs do I need to be properly licensed? Can I run this product in a virtual environment? Can I downgrade this product to an earlier version? Am I entitled to feature X with license Y?


I've been working on an Add On project for a while to generate buildings/terrain procedurally out of "primitive" Blender objects as elements. As you might expect, one of the most common operations the Add On performs is to duplicate an object (and its children, as 'primitive' is interpreted somewhat loosely here, to allow for multiple objects related through parenting to serve as a single modular building block). Recently, I've been trying to optimize the Add On's run-time a bit, and noticed that it seems to spend almost all of its time -- if I am understanding this cProfiler () output correctly -- updating the view layer:    Ordered by: cumulative time     ncalls  tottime  percall  cumtime  percall filename:lineno(function)         1    4.202    4.202  116.476  116.476 make_building.py:32(makeBuilding)       260    0.078    0.000  107.959    0.415 util.py:68(blDuplicate)       260    0.005    0.000  106.569    0.410 ops.py:184(__call__)       520  105.985    0.204  105.985    0.204 ops.py:155(_view_layer_update)        1    0.002    0.002   56.751   56.751 make_building.py:132(generateFloorsWallsHallways)        1    0.006    0.006   43.853   43.853 make_building.py:287(addCeilingTiles)  (pstats output truncated, but sorted on cumtime so I think this is sufficient to get the idea) To be a bit more confident about this, I also setup this minimal example, run from inside a Blender Console inside a running Blender session: >>> def dup(): ...     bpy.ops.object.duplicate() >>> import cProfile >>> p = cProfile.Profile() >>> p.runcall(dup, *[], **{}) >>> p.dump_stats(bpy.path.abspath("//") + "turkey_baster.dump")  This, similarly, seems to indicate that view layer updating is taking most of the time: >>> ps.strip_dirs().sort_stats('cumulative').print_stats() Sun Feb  7 12:00:42 2021    <filepath removed + "turkey_baster.dump">           13 function calls in 0.233 seconds     Ordered by: cumulative time     ncalls  tottime  percall  cumtime  percall filename:lineno(function)         1    0.000    0.000    0.233    0.233 <blender_console>:1(dup)         1    0.000    0.000    0.233    0.233 ops.py:184(__call__)         2    0.232    0.116    0.232    0.116 ops.py:155(_view_layer_update)         1    0.002    0.002    0.002    0.002 {built-in method _bpy.ops.call}         1    0.000    0.000    0.000    0.000 ops.py:40(__getattr__)         1    0.000    0.000    0.000    0.000 ops.py:82(__getattr__)         1    0.000    0.000    0.000    0.000 ops.py:180(idname_py)         2    0.000    0.000    0.000    0.000 {method 'startswith' of 'str' objects}         1    0.000    0.000    0.000    0.000 ops.py:79(__init__)         1    0.000    0.000    0.000    0.000 ops.py:168(__init__)         1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}  I haven't yet tried to look up ops.py in the Blender source to try to read what _view_layer_update is doing, but in short am wondering if there's much room from the Blender Python API to optimize this operation, or if I just have to accept this cost. I don't really have any idea what view_layer_update does right now, so I'm not sure whether there's much opportunity to reduce its cost. (EDIT: in case it is pertinent, note that I'm working in Blender v2.83.2; I've been upgrading versions only slowly to avoid having to accommodate any assumptions I've made that the version upgrade would break) EDIT2: I notice the layer update percall time is very small (0.001) on a brand-new scene, and I seem to be able to cut it down from 0.116 to 0.07 or so by disabling every single collection (except the Master Collection) from the scene. Maybe I could cut that down further by disabling or deleting still other data structures in the file (e.g., Materials)? I am not sure how practical that is as solutions go, but it seems like relevant evidence, at least.

In my script, I have for loop over many cube objects (~1000) and the treatment is very slow. Looking more in details, I notice that in the same amount of loops:   if I use python operation or simple blender operation like  obj = bpy.data.objects[obj_name]   or   obj.select = True   it takes less than 0.08s (for all 1000 objects) but as soon as I start using blender operators like:  bpy.ops.object.select_pattern() # or bpy.ops.object.duplicate() # or bpy.ops.object.location_clear() # or bpy.ops.object.transform_apply() # etc.   then, the performance drops down deeply with more than 6s for the same amount of objects.   And, a last information, if I reduce the amount of objects from 1000 to 50, the same set of operations with bpy.ops is 0.03s - so extrapolating, it would be 0.6s for the 1000 objects and not 6s. It is as is with 1000 objects, we loose a factor 10 in speed in comparison than with 50 objects.  I tried reducing the complexity of my mesh changing the cubes in planes but it had no effect at whole on performance.  Is there something particular to know to improve these performance or a way to use the bpy.ops methods with better performances? Obviously, I am missing something important probably.


prove a symmetric matrix is diagonalizable without doing any calculations of eigenvalues or eigenvectors.   I know its orthogonally diagonalizable but is that the same as diagonalizable?

Can someone point me to a paper, or show here, why symmetric matrices have orthogonal eigenvectors? In particular, I'd like to see proof that for a symmetric matrix $A$ there exists decomposition $A = Q\Lambda Q^{-1} = Q\Lambda Q^{T}$ where $\Lambda$ is diagonal.


We have a compact set $X$ and a closed set $E \subset X$, how to prove that $E$ is compact?

If S is a compact subset of R and T is a closed subset of S,then T is compact. (a) Prove this using definition of compactness. (b) Prove this using the Heine-Borel theorem.  My solution: firstly I should suppose a open cover of T, and I still need to think of the set S-T. But if S-T is open in R,it can be done because the open cover of T and S-T is a open cover of R. The reality is S-T is not necessarily a open set in R. My question is that How we can find a open cover which covers S-T but misses T! I don't know how to do this thing! in terms of part (b), I know it is bounded, but How to prove it is closed for T.


It is known that every proper ideal of a ring must be contained in a maximal ideal, by the Zorn's lemma. Is this true in general for proper homogeneous ideals in a graded ring?

Let $I$ be a homogeneous ideal of a graded ring $S$, $I\ne S$. I want to show that there exists a homogeneous prime ideal which contains $I$.    I proved the following:  Let $T$ be the set of all homogeneous ideals which contain $I$. Then, by Zorn's lemma, there exists a maximal element of $T$, say $P$. I will claim that $P$ is prime. Suppose that for homogeneous elements $a,b \in S$, $ab \in P$ but, $a\notin P$. Then $\langle a \rangle + P$ is a homogeneous ideal which contains $I$. It contradicts by maximality of $P$ so, $a\in P$.  Is it right???


If $f$ is real-valued and continuously differentiable on $\mathbb{R}$, do we have the following inequality:$$\left(\int |f|^2\,dx\right)^2 \le 4\left(\int |xf(x)|^2\,dx\right)\left(\int |f'|^2\,dx\right)?$$

If $f$ is real-valued and continuously differentiable on $\mathbb{R}$, prove that $$  \left(\int|f|^2dx\right)^2\le 4\left(\int|xf(x)|^2dx\right)\left(\int|f'|^2dx\right) $$  Attempt: I tried the Cauchy-Schwarz inequality as well as the Plancherel theorem, but none of them seems to work.


Contrary to popular belief, and numerous Hollywood movies, you could survive unprotected in space for over a minute - provided you could get back to medical care immediately after.  Moreover, in movies we see astronauts and people in spaceships opening doors and closing them again with no suits on.  Are these things correct or just myths? Is there any way to estimate or predict the time for the above?

How does "outer space" affect the human body? Some movies show it as the body exploding, imploding or even freezing solid.   I know space is essentially a vacuum with 0 pressure and the dispersion of energy makes it very cold. So are the predictions above accurate?


I have a Java Application running on Linux machine, it is generating a file  in which each record is separated by new line,  This file will be used by an application running on Unix SunOS system and  my code below does not generate proper file . I mean following code generates a file in which each line end with ^M charater when I open it with vi editor in Unix system  bw = new BufferedWriter(new FileWriter(file)); bw.write("\n");   Which character I have to write in order to see the file properly in Unix system.

How do I get a platform-dependent newline in Java? I can’t use "\n" everywhere.


I can't understand why these two 2D list is different: list1 = [[' ' for i in range(n)] for j in range(n)] and list2 = [[' '] * n] * n. n = 3 list1 = [[' ' for i in range(n)] for j in range(n)] list1[1][1] = ' ' for i in range(n):     for j in range(n):         print(list1[i][j], end='')     print('')  This code let me get result like this: *** * * *** This is the result as I expected. But in list2, n = 3 list2 = [[' '] * n] * n list2[1][1] = ' ' for i in range(n):     for j in range(n):         print(list2[i][j], end='')     print('')  Result of this code is like this: * * * * * * I think this is so strange. Why this happens?

I needed to create a list of lists in Python, so I typed the following: my_list = [[1] * 4] * 3  The list looked like this: [[1, 1, 1, 1], [1, 1, 1, 1], [1, 1, 1, 1]]    Then I changed one of the innermost values: my_list[0][0] = 5  Now my list looks like this: [[5, 1, 1, 1], [5, 1, 1, 1], [5, 1, 1, 1]]    which is not what I wanted or expected. Can someone please explain what's going on, and how to get around it?


i installed Ubuntu as only operating system, i used the default partitioning scheme using the complete hard drive and now i want to add another partition for data? i used gparted, after input data, but there is no option to resize.

Previously, I have installed Windows 7 on my 320 GB laptop with three partitions 173, 84 and 63 GB each. The 63 GB partition was where the Windows was installed. The rest were for file containers.  Now I changed my OS to Ubuntu 12.04 LTS. I installed Ubuntu by replacing the entire Windows 7 on the 63 GB partition. The rest of the partitions remain as an NTFS Windows partition and I can still access them both (the 173 and 84 GB partitions).  Now I want to change the two partitions of Windows into an Ubuntu format partitions plus most importantly, I want to extend the 63 GB partition to more than a 100 GB because at the moment I am running out of disk space. Whenever I try to install any application, especially using wine, it always complains for a shortage of disk space.  How do I do the extending activity before I entirely format my laptop again and lose all the important files on my partitions?


The way i understand it, a key encoding should return a sequence of bytes in some specific encoding such as UTF-8 for example. However, logging the following:  KeyPairGenerator kpg = KeyPairGenerator.getInstance("EC");  kpg.initialize(256);  KeyPair kp = kpg.generateKeyPair();   Log("TEST : ${kp.public.encoded}")  Log("Test : ${kp.public.encoded}")  Equivalently, in Java: kp.getPublic().getEncoded();  Is giving me 2 different byte arrays! What am i missing here? It seems that every time i call the encoded method, different pairs are generated.

In Java, arrays don't override toString(), so if you try to print one directly, you get the className + '@' + the hex of the  of the array, as defined by Object.toString():  int[] intArray = new int[] {1, 2, 3, 4, 5}; System.out.println(intArray);     // prints something like '[I@3343c8b3'   But usually, we'd actually want something more like [1, 2, 3, 4, 5]. What's the simplest way of doing that? Here are some example inputs and outputs:  // Array of primitives: int[] intArray = new int[] {1, 2, 3, 4, 5}; //output: [1, 2, 3, 4, 5]  // Array of object references: String[] strArray = new String[] {"John", "Mary", "Bob"}; //output: [John, Mary, Bob]


A colleague told me that it is not recommended to use USB hard disks due to the lack of inherent integrity checking of copied files. I am looking for details on that. In what respect is copying to a USB device different from copying to a hard disk connected via e.g. SATA? If anybody has further sources/information on that, I would appreciate sharing it with me.  edit: As explained below, I know that in many cases, the used hard disk is effectively nothing else but a SATA disk in a casing. My question is whether the USB <=> SATA interface introduces data security issues. My colleague referred to a case where a broken USB controller led to data loss. Data could be copied to the hard disk without any problems, but the data written to the disk was corrupted. Replacement of the USB controller resolved the issue in this case.


Using the Python add-in wizard I need to restart ArcMap every time I want to test something.  For example, when I've got syntax errors or something little that I need to troubleshoot.    What approach would you recommend to negate the need to start ArcMap repeatedly?  It has been noted in comments that this question has already been asked and vaguely answered; however it's totally unclear to me how you integrate this into your script.    Can someone give me an example?  I've tried to include the following script as noted in the other question linked here.  import os  esri = "C:\Program Files\Common Files\ArcGIS\bin\ESRIRegAddIn.exe" cwd = os.getcwd()  mapdoc = "C:\Data\Shaun Pilot Area\PilotAreaMXD.mxd"  try: os.system("TASKKILL /F /IM ArcMap.exe") except: pass  os.system(os.path.join(cwd, "makeaddin.py"))  os.system('"{0}" {1} /s'.format(esri, os.path.split(cwd)[-1] + ".esriaddin")  os.system(mapdoc)   Here's the output I get  ERROR: The process "ArcMap.exe" not found. 'C:\Data\Shaun' is not recognized as an internal or external command, operable program or batch file. 'C:\Program' is not recognized as an internal or external command, operable program or batch file. 'C:\Data\Shaun' is not recognized as an internal or external command, operable program or batch file.

I have been developing a few python addins and I'm just wondering what is the best way to re-compile the addin anytime I edit the script.  Currently, if I make an edit to the script, it seems like I have to   Close ArcMap Go to My addin folder and double click the 'makeaddin' python file Double click the 'Python Addin File Restart ArcMap for the changes to come into effect   If I don't do these steps, then the edits aren't recognized. This gets pretty annoying and time consuming when I want to place maybe a simple print line somewhere or I do a small edit just to test a result.  Is there a better way of doing this?


So this is my question and I also have to do it so it has 8 solutions, my problem is that I don't really understand how to approach the problem at all.  Any type of help would be appreciated thanks

Consider the polynomial $x^2+x=0$ over $\mathbb Z/n\mathbb Z$   a)Find an n such that the equation has at least 4 solutions  b)Find an n such that the equation has at least 8 solutions  My idea is to check individual n, and I found the answer for a is n=6 at x={0,2,3,5} f(x)={0, 6, 12, 30)  Is there any shorter way?


In Middle-earth, Elves are bound to the world and will live as long as it lasts. Men are of course mortal, as are Dwarves, Hobbits, Eagles, Trolls, Orcs, Ainur (so it would seem), and, most importantly, foxes. In the early stages of the Ringbearer's Quest, one fox comes across the sleeping Hobbits far from where Hobbits tend to sleep. He finds it rather odd, and we get a peek inside his head for a short monologue.  My question is, do Foxes count as sentient beings of Middle-earth? Are they something more akin to the Eagles, being a representative of a Vala in the wide world?

When Frodo, Pippin, and Sam were traveling to Rivendell, they slept under a tree the first night. Tolkien writes about a fox that wonders to himself why three hobbits were sleeping under a tree. Why can the fox think for himself? This isn't Narnia where animals can talk, but this is the only time I can think of where an animal thinks for itself in Lord of the Rings. Here is an excerpt from the book:     A fox passing through the wood on business of his own stopped several minutes and sniffed. "Hobbits!" he thought. "Well, what next? I have heard of strange doings in this land, but I have seldom heard of a hobbit sleeping out of doors under a tree. Three of them! There's something might queer behind this." He was quite right, but he never found out more about this.   -Fellowship of the Ring   What other strange thing could the fox have heard of and who or what did he hear them from?  Could the Great Eagles be considered a true animal according to this? They have no equivalent in the real world of the same size, so I don't count them.


Options are:  A. $20$ B. $12$ C. $12.5$ D. $13.3$  And steps to  calculate would be appreciated.  Edit: Apologies for having repeated the same question here.

If $a, b$ are positive real numbers and $a+b = 1$, prove that :  $$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$  I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.  Thank you.


Let $(A, +, \cdot)$ be a ring with $1$.  An element $a\in A$ is nilpotent if there exists $n\in \mathbb{N}$ so that $a^n=0$.      Show that if $a$ is nilpotent then $1+a$ and $1-a$ are invertible.

If $ua = au$, where $u$ is a unit and $a$ is a nilpotent, show that $u+a$ is a unit.   I've been working on this problem for an hour that I tried to construct an element $x \in R$ such that $x(u+a) = 1 = (u+a)x$. After tried several elements and manipulated $ua = au$, I still couldn't find any clue. Can anybody give me a hint?


The 2019 miniseries "Chernobyl" appears to state that the "lava like" substance of melted radioactive matter from the core, the so-called "corium", could eat its way down into the water reservoirs below the reactor core. This could then possibly cause a second, much stronger explosion.  They indicated this could be on the orders of several megaton, I believe 2-4.  This seems to be on the order of a hydrogen bomb, and I'm curious if this is factually accurate. If it is factually accurate, how exactly is such explosive force possible? Was it assumed fusion would take place? How would the required pressure for fusion be achieved, and why would not fizzling (the explosion itself starting and immediately pushing away the material so no very significant explosion would take place) be an issue for such a large explosion?  I see this topic relating to the miniseries turns up on sites such as ArsTechnia and Reddit, but I'm curious what the community here has to say about it.  Any sources that can be cited would be great.

So the  of the HBO series began to cover the  that led to them sending three divers into the water below the reactor to drain the tanks.  This occurred after the initial explosion that destroyed the reactor, and after the fire in the core had been put out. But at this point the decay heat and remaining fission reaction kept the core at more than 1200°C, causing it to melt through the concrete floors below the reactor.  And below the reactor were water tanks which contained 7,000 cubic meters of water (according to the TV show. If anyone has a real figure, I'd love to hear). When the lava of the melted core hit it, it would cause an enormous .  Finally, my question: About how large would this explosion have been? The character in the show says "2-4 megatons" (of TNT equivalent, I assume). I'm pretty sure this is absurd and impossible. But real estimates are hard to come by. Other sources vary wildly, some repeating the "megatons" idea, and others saying it would've "". This still seems crazy.  tl;dr:  I know a lot of it hinges on unknowns and the dynamics of the structures and materials involved, so I can simplify it to a constrained physics question:  Assuming 7,000 cubic meters of water instantly flashes to steam, how much potential energy is momentarily stored in that volume of steam occupying the same volume as the water did?  I don't know what to assume the temperature of the steam is. There were hundreds of tons of core material at temperatures near 1200°C, so worst case scenario you could assume all the steam becomes that temperature as the materials mix. Best case scenario, I guess we could assume normal atmospheric boiling point (100°C)?


We know that a photon can split into an electron-positron pair and that when an electron and positron come into contact they annihilate and produce a photon. But why does this happen?

Pair production is where an energetic photon on its interaction with strong electric field surrounding a nucleus produces an electron-positron pair. Annihilation is its converse where an electron-positron pair produces two photons on interaction with nucleus.  But how do these happen? I mean, why do they require strong electric field and nuclei?  (Is strong nuclear force necessary for this process?)


I have to decice if the following function is Lebesgue-integrable on $[0,1]$:  $$g(x)=\frac{1}x\cos\left(\frac{1}x\right) $$ where $x\in[0,1]$.  $g(x)$ is Lebesgue integrable if and only if the integral of $|g(x)|$ is finite  So, $\int_{[0,1]} |\frac{1}x\cos\left(\frac{1}x\right)| dm < infinite???$   I don't know how to proof that,  I have thought about using the monotone convergence theorem but I don't have any idea of how to define $f_n$ a sequence of measurable functions.


We have a production machine running Window Server 2003. Previously, development took place in a WAMP environment but a few users are moving to IIS.  The issue is, we have a directory on the server called /home/ that when visited, directs all information elsewhere.  For instance  yourwebsite.com/home/   would actually point to  yourwebsite.com/tabular/   without redirecting the page (i.e. the path wouldn't change)? How is this possible in IIS7.5?  Thanks,  Sam

This is a  about  the use of a *AMPP's stack.    I recently had a talk with some experienced people and they suggested to me not to use a WAMP stack, and instead install apache, mysql and php separately.  I don't understand why they have suggested this, though, so can anyone tell me?  Is there a particular disadvantage of WAMP, or a particular advantage to installing all of them separately?  Since a WAMP stack itself is composed of apache, mysql and php, then what's the difference between using the WAMP stack and installing them all separately?


$f(x,y)=2^{3x+8y}$   $x^2+y^2=1$  Polar Coordinates approach:  Let $x=cos(t), y=sin(t), 0\le t \le 2\pi$ (We can see that this solved the circle equation).  Substituting them into my function:  $f = 2^{3\cos(t)+8\sin(t)} \Longrightarrow f=e^{(3\cos(t)+8\sin(t))ln(2)}$  Taking derivative to find critical points:  $f' = \ln(2)(-3\sin(t)+8\cos(t))e^{(3\cos(t)+8\sin(t))ln(2)}=0$  Now $e^x>0$, so I need to find when  $-3\sin(t)+8\cos(t)=0 \Longrightarrow 3 \sin(t)=8\cos(t) \Longrightarrow \tan(t) = \frac{8}{3} \Longrightarrow t=\arctan(\frac{8}{3}), \arctan(\frac{8}{3}) + \pi $ So I get two Points:  $(\cos(\arctan(\frac{8}{3})) ,\sin(\arctan(\frac{8}{3})))$  $(\cos(\arctan(\frac{8}{3}) + \pi), \sin(\arctan(\frac{8}{3}))$  and now I need to find $f_{xx}, f_{yy}, f{xy}$ in order to know which point is minimum and which is maximum:  $f_{x}=8^x3ln(2)$  $f_y = 256^y 8ln(2)$  $f_{xx} = 9\ln^2(2) 8^x$  $f_{yy} = 64ln^2(2) 256^y$  $f_{xy} = 0$  $f_{yx} = 0$.    for first point: $f_{xx} = 8.97$, $f_{yy} = 5530.131$.  So $f_{xx}*f_{yy} - 0 > 0$ , and $f_{xx} > 0 $, which means it is a local minimum point.   for second point: $f_{xx} = 2.03$ , $f_{yy} = 0.17$. and same as above, it is a minimum point.  Would appreciate any help in approving this solution or finding mistakes in it, and if there's any more efficient approaches, thanks in advance.   EDIT: Approach from Parcly Taxel answer:  Checking maximum and minimum for $3x+8y$:  $\phi (x,y,\lambda) = 3x+8y + \lambda(x^2+y^2-1)$  $\phi_x = 3 + 2\lambda x = 0 \Longrightarrow x= \frac{-3}{2\lambda}$  $\phi_y = 8 + 2 \lambda y = 0 \Longrightarrow y= \frac{-4}{\lambda}$   $\phi_{\lambda} = x^2 +y^2 -1 = 0 \Longrightarrow \frac{9}{4\lambda^2} + \frac{16}{\lambda^2}-1=0 \Longrightarrow 9+64=4\lambda^2 \Longrightarrow \lambda^2 = \frac{73}{4} \Longrightarrow \lambda = \pm \frac{\sqrt{73}}{2}$  So: $x= \pm 0.3511 $  $y = \pm 0.9363$  $(0.3511, 0.9363) , (-0.3511, -0.9364)$

Suppose that there is a trigonometric equation of the form $a\sin x + b\cos x = c$, where $a,b,c$ are real and $0 < x < 2\pi$. An example equation would go the following: $\sqrt{3}\sin x + \cos x = 2$ where $0<x<2\pi$.  How do you solve this equation without using the method that moves $b\cos x$ to the right side and squaring left and right sides of the equation?  And how does solving $\sqrt{3}\sin x + \cos x = 2$ equal to solving $\sin (x+ \frac{\pi}{6}) = 1$


My Ubuntu 18.04.2 system is working OK except for the following problem:  I cannot seem to be able to login on any of the virtual terminals (Ctrl+Alt+F1 to F6. Specifically, I can type my username fine, and the keyboard works without a problem, no unusual behaviour, until I press Return (or Enter on the keypad).  At that point, the password prompt appears but immediately it is as if Return is pressed again, so entering a blank password and giving Login incorrect. The login prompt reappears but again it is as if Return is immediately pressed, again giving Login incorrect. The repeats a few more times until the screen clears and the login prompt shows again. Now Return does not keep repeating.  Any ideas? The Return key is working no problem in graphical mode.

I upgraded from 16.04 to 18.04. When I press Ctrl+Alt+F3 I get the terminal which prompts for the username. When I type my username and press Enter, it doesn't wait for my password (its as if I pressed Enter without entering anything). Even if I enter something, the text shown in plaintext and then it keeps on saying my password is wrong. After few times, the screen is cleared and it prompts for my username again and the same loop goes on.


I want to buy a laptop, wipe it and install Ubuntu 14.x.x on it.  I saw this HP Stream with this specs:   Celeron N3060 speed1.6 GHGz Graphics  cardIntel GMA HD Ram 2 GB  SSD  32 GB (eMMC)  Do you think I can Install Ubuntu 14 on this one? Im suspicious of the eMMC disk in particular.  Thank you all.

I'm thinking about installing Ubuntu Desktop, but I don't know what flavor is the best for my system. What are the minimum and recommended hardware requirements? What kind of CPU? How much memory? Should I have Hardware Acceleration? What flavor should I use?  This is an attempt of a canonical answer. My answers have the "official  requirements", the recommended are a mix of official sources and opinion based (along with the answer it's told the source). You can comment or edit if you feel that the information is obsolete or incomplete. It is a good rule of thumb that any system capable of running Windows Vista, 7, 8, x86 (Intel) OS X will almost always be a lot faster with any Ubuntu flavor even if they are lower-spec than described below.


Let me know about this two examples.   1) set of all polynomials with rational coefficients.  2) set of all polynomials whose coefficients belongs to set ${\{1, 2\}}$   I know the first one is countable set. Because we can write first set as union of countable sets $P_n$ which is set of polynomial with rational coefficients having degree at most n"   But, I am stuck at second one, as I see in the first one coefficient belongs to countable set $Q$ so the given set is countable and in second one also, coefficient belongs to set $\{1, 2\}$ which is also countable set and hence set given in 2) is also countable? Is am I right?

Using the definition of enumerability of sets:  A non-empty set B is enumerable iff there is a bijection $f:\mathbb{N}\supset L \rightarrow B$.  So, I have to prove that the set of polynomial of one variable with coefficients on $\mathbb{Q}$ is enumerable. What I thought is that I can write every polynomial $p_0=A_0^0+A_1^0X+\dots +A_n^0X^n$ as  $$p_0=(A_0^0,A_1^0,\dots ,A_n^0,\dots ),$$  where the $A_i's$ are 0 except for a finite number of them. To say that this set is enumerable is the same that to say (denoting this set by B):  $$B=\{p_1,p_2,\dots\}$$  Now define the polynomial   $$ p=(x_0 \neq A_0^0, x_1 \neq A_1^1, \dots), $$  where every $x_i \neq A_i^i$. This polynomial is obviously not in the list B.  My question is: my approach is right to conclude that the set B is non-enumerable?


Well, it's been 4 months. And I am impressed.  I am also a bit disappointed.  Before you read this, take it with a pinch of salt. I'm kinda new, still. I'm still learning the ropes. But I've done a lot of browsing through old questions and new questions, and all the comments in between.  Why am I disappointed, you might ask? Well, I'm disappointed because of the (re)occurence of the yousuck effect. No, I'm not talking about the stern warnings to "read the faq" or the "please post your code". I'm talking about this question. And there are others, if anyone wants to add them.    The first comment on the question was by Mitch Wheat: "It's that time of day again..."  Look, I might have got an easy welcome to Stack Overflow. I posted some dumb stuff. I also posted some good questions, if I say so myself.  But just because it's the time of day where the foreign people (not sure from where because they could be many nationalities) come onto the site and post so-so questions DOES NOT mean that you should treat them like crap.  I know that I wouldn't want to be treated like a useless statistic if I was a foreign person with a question.  And for all those other high-rep people out there, I know what you are like. I know that you are generous. I know that you help me out when I get stuck. But please, please just realize that every new account is, well, a new person. And that is an account that should be treated with utmost respect. And really, the high-rep people should be setting the good examples.  ... not to pick on you, but Mitch, I guess that the comment was deleted for a good reason. I think that it was a good move to delete it.    So in conclusion, just realize that there are people who care. I hate this stuff. I did a "dodgy upvote" because I felt bad for the poor guy. So please, please just do the downvotes and the flags. Maybe you could try to help them out once in a while.  But try to keep the snarky comments to yourself.  Thanks.

There is a distinct decline in the level of civility on all the sites here.  Some of this is due to new users coming in and posting spam and other nonsense, but the off-topic and downvote buttons are doing a pretty good job of keeping this under control. Unfortunately, a lot of this is coming from more experienced users, and the site's built-in moderation system does not (and probably cannot) handle this very well.  Folks are rushing to pound new users down with "this belongs on meta!", "this is off topic", "this is a duplicate!" and "read the Help!". (Which is correct, but should be done nicer) All this, of course, is accompanied by a flurry of downvotes.  This is not very welcoming to new users who don't know about meta, the Help, or what counts as off-topic. Now I am not proposing that we just allow off-topic, meta, or duplicate questions.  However, I think we could be gentler in the way we express these sorts of things.  Explain what meta and the FAQ are and provide useful links.  Just using please and thank-you when asking folks to read the FAQ or post something on meta would be an improvement.  I also think we could rein in the downvoting a bit.  Not that we shouldn't vote stuff down, but unless a new user's post is clearly spam, voting it down to -1 or -2 should be sufficient to send a message without piling on. I like Stack Exchange and I want it to become a resource for everyone, not just an elitist site for high-rep users on the sites.


Reading the basics of $\mathcal{O}_{X}$ modules, where $\left(X,\mathcal{O}_{X}\right)$ is a fixed ringed space, i understood that arbitrary direct products of quasi-coherent $\mathcal{O}_{X}$ modules are quasi-coherent $\mathcal{O}_{X}$ modules. I also know by a result of Gabber that $\mathsf{Qcoh}\left(X\right)$ is a Grothendieck category for an arbitrary scheme $X$ and consequently  it is a complete category. But, i was wondering: is the product of quasi-coherent sheaves coincide with the product in the category of $\mathcal{O}_{X}$ modules? Or there is a counter-example that arbitrary products of quasi-coherent sheaves are not quasi-coherent?

I have a short question:  Does the category of quasi-coherent sheaves on a scheme have arbitrary products? I know that it does if the scheme is affine and I know that they will not be isomorphic to the product as $\mathcal{O}_{X}$-modules, but I haven't been able to find a counterexample.  Any help is appreciated!


I accidentally deleted a few important work related folder from my Ubuntu 14.04. I wanted to recover them. All the options on net gives instruction for recovering files.   I tried scalpel but I don't remember any of the file names so i could not recover any files. Is there a way to list the folders i deleted and recover the one i wanted? I need these files please help me.

The folder is massive (390GB) and contains many different files. The file system is ext4  I was about to copy this file to another hard drive using gksudo nautilus so I didnt have do a massive chmod operation on the destination hard drive, but then I accidentally hit move to trash, which in gksudo nautilus means permanent deletion. I checked the trash but it's not even an operation...  I have seen tools like testdisk and other recovery tools but none that can recover and whole folder because its made up of different files.   I don't think that this would work but if I deleted the partition could I just recover the whole partition at once?  Or would this be wrong because i have already deleted the partition's content? This is my first job in Linux IT.  I don't have a backup... I was trying to make one because the old backup drive died :(


REASONS FOR REFUSAL   You have applied for a visa to visit the UK.   In deciding whether you meet the requirements of Appendix V: of the Immigration Rules for visitors (), I have considered:         your application and any additional relevant information you have   provided with it     your immigration history         The decision   I have refused your application for a visit visa because I am not satisfied that you meet the requirements of paragraph 4.2 of Appendix V because:         You are applying to come to the UK with for 2 weeks. I note that this   will be your first trip to the UK. You state that you are employed,   where you earn £1,257.62 per month, you have savings of £7000.6, and   you intend to spend £1,000.00 on your trip to the UK.     In support of your application, you have provided an employment   letter, payslips and a bank statement, and you have explained the   cash deposits credited to your account. However, the information you   have provided does not satisfactorily demonstrate your monthly   outgoings, or explain how the bank balance has escalated in a short   period of time. I am therefore not satisfied that these funds are   available to you for your personal use.     Given the above, I am not satisfied by your intentions in wishing to   travel to the United Kingdom for a limited period, or that you will   leave the UK at the end of your visit.         Furthermore, I am not satisfied that you have sufficient funds available to cover your costs whilst in the UK without working or accessing public funds. Your application for a visit visa has been refused under paragraph V4.2 (a) (c) and (e) of the Immigration Rules.        In relation to this decision, there is no right of appeal or right to administrative review

Many of the UK visa refusals we see here share a common pattern and the prevailing reasons refer to V 4.2 (a) and (c).    I do understand that, while the applicants may describe very different individual circumstances, there's a consistent pattern and, broadly, fall into specific categories:    Credibility (lifestyle, lack of ties, visit history, lies & omissions) Funding (insufficient funds, provenance of funds, funds parking) Sponsorship (family, friends, employer)   Question: What most commonly triggers a UK visa refusal where V 4.2 (a) and (c) are given as the grounds?  Secondarily: Given that there is a clear pattern, to what extent are these refusals predictable? Does sponsorship make a difference? Is there a set of personal circumstances, however abstract, where a refusal is all but guaranteed? For example, why do those we see here on TSE who sought entry for the PLAB or British Army appear to be refused with a common theme? Is there a uniform shortcoming or is it just discrimination?   PLAB: What is about applications for a visa, for the purpose of sitting the Professional and Linguistic Assessments Board (PLAB) exam, that seem to invite refusals. After all, the test is given so that international medical graduates can show that they qualify to practise medicine in the UK. The first part, PLAB 1, can be taken in centres outside of the UK. However, PLAB 2 can be taken only in the UK. Why can't you can't get a visa just to sit the exam, promising that you'll leave immediately after?    British Army: Since Commonwealth citizens are eligible to apply online to join, even those who don't reside in the UK, why is it so difficult to get a visa just to attend an interview to see whether you are suitable? Even with an invitation from the Army, such visa applications seem to be unsuccessful. Isn't a career in the British Army a valid reason?     Lastly:  After such a refusal, what approaches would increase the chance of a successful application?


Let $a$ and $b$ denote two positive constants. If $$\lim_{x\to 0} \Bigg(\frac{\int_0^x \frac{t^2}{\sqrt{a+2t^5}}dt}{bx-e\sin x}\Bigg)= \frac{1}{\pi},$$ find the value of $a$.  What I've tried: I've used L'Hospital's rule since it is a $\frac00$ form, to get $$\frac{\lim_{x\to 0} \Big(\frac{x^2}{\sqrt{a+2x^5}}\Big)}{{\lim_{x\to 0} ( b-e\cos x)}}$$ If  $x$ tends to $0$, the equation will become $\frac{\frac{0}{\sqrt{a}}}{b-e}$, making it impossible to become $\frac{1}{\pi}$. Did I make a mistake somewhere?

I am given the following problem which I have problem to know where to even start:     The question: $\lim_{x\to 0} \frac{\int_0^x\frac{t^2}{\sqrt{a+2t^5}}dt} {bx-esinx}=\frac{1}{\pi}$ The part where I will like to know where to start:  $\int_0^x\frac{t^2}{\sqrt{a+2t^5}}dt$   I appreciate suggestions on how I may get about solving the integration part before I move on to solve the limit question as a whole.


$$\sum_{i=1}^{\infty}i\cdot((1-\frac{1}{N!})^{i-1}\cdot\frac{1}{N!})$$  where N is an integer and $N \geq 2$?  I obtained this series from the following problem:  Given a sequence of $N > 2$ unique numbers (not in sorted order), we randomly shuffle the sequence until it becomes sorted in non-decreasing order. What is the expected number of shuffles needed to sort the $N$ numbers?  Intuitively it should be $N!$ since there are $N!$ permutations, but how does one prove this formally (by evaluating the series above, for instance)?

Why does the following hold:  \begin{equation*} \displaystyle \sum\limits_{n=0}^{\infty} 0.7^n=\frac{1}{1-0.7} = 10/3\quad ? \end{equation*}  Can we generalize the above to      $\displaystyle \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$ ?   Are there some values of $x$ for which the above formula is invalid?  What about if we take only a finite number of terms? Is there a simpler formula?     $\displaystyle \sum_{n=0}^{N} x^n$   Is there a name for such a sequence?    This is being repurposed in an effort to cut down on duplicates, see here: Coping with  duplicate questions.  and here: .


I've searched quite a lot but yet didn't found any solution to burn a bootable usb flash or cd/dvd with Windows from Linux having an iso file with Windows. I've tried dd if= of= ... but when I tried to boot from that usb flash, it just didn't boot.  Is there a reliable solution?

I'd like to create a Windows bootable USB stick, but I don't have a Windows machine with me at the moment to do so. How could I achieve the same using Ubuntu?


I upgraded from Raring to Saucy, and now my bluetooth is giving me problems.  First, even though I have a bluetooth inbuilt in my laptop, which I prefer to keep off unless I am using it, there is no greyed out bluetooth icon in the notification area. This was present in Quantal, but was missing in Raring too. I was hoping to get this fixed in Saucy. Seems otherwise.  Secondly, and more importantly, in saucy when I am turning on my bluetooth, it immediately (or within seconds) turns off automatically, without giving me any time to even make the device visible.

Everytime I turn off my bluetooth, the indicator icon in the system tray disappears, as opposed to just greying out as expected. Next time I need to go to the settings select bluetooth to turn it on, which makes the indicator to reappear.  Is there anyway I can make the indicator icon not disappear but stay greyed out when not in use?


After update to 15.04 I have abiword starting by itself upon login to GNOME or Unity. It does not happen in openbox or blackbox.  How do I prevent this application from starting without removing the application itself ?  I do not have abiword added to Startup Application, nor to .config/autostart, nor it is in /etc/xdg/autostart. The rc.local is clean, as well.  (=^・^=) [2] serg @ ubuntu:/home/xieerqi >_  ls .config/autostart/                                caribou-autostart.desktop chinese-calendar-autostart.desktop chromium-browser.desktop.bak firefox.desktop gnome-do.desktop gnome-terminal.desktop.bak gnome-user-share.desktop indicator-multiload.desktop lxrandr-autostart.desktop nemo-autostart.desktop reconnect-task.desktop scrlayout.desktop telepathy-indicator.desktop ubuntuone-launch.desktop vino-server.desktop wicd-tray.desktop.bak  (=^・^=) [3] serg @ ubuntu:/home/xieerqi >_  grep 'abiword' .config/autostart/*                    (=^・^=) [4] serg @ ubuntu:/home/xieerqi >_  grep 'abiword' /etc/xdg/autostart/*                   (=^・^=) [5] serg @ ubuntu:/home/xieerqi >_     pstree however, reports upstart as the parent of abiword's process.  (=^・^=) [6] serg @ ubuntu:/home/xieerqi >_  pstree | grep abiword         |         |         |-upstart-+-abiword-+-{abiword}   Better look:  |-lightdm-+-kwalletd         |         |         |-upstart-+-abiword-+-{abiword}         |         |         |         |         |-{dconf worker}         |         |         |         |         |-{gdbus}         |         |         |         |         `-{gmain}

Abiword starts automatically each restart, on any user. Already checked out Startup Applications. Remember session is turned off.   ()  Additional info, may be it'll give a clue - I configured VNC server on this machine to run, so sometimes, in additional to regular Ubuntu/Unity sessions I start XFCE4 session. I think this behavior started after VNC4 configuration, but ~/.vnc/xstartup does not contain abiword - just regular xfce startup stuff.  What are additional places where I can look?


Is there some way, when installing Ubuntu-20.04 Desktop, to tell the installer to use a proxy server?  I didn't see any option for that (unlike when installing Ubuntu Server) and trying to google an answer seems impossible because the results are all about setting a proxy after install.  I have a slow, metered internet connection so I want to install though my caching proxy so that any subsequent installs will not have to re-download hundreds of megabytes of packages.

For some of the Ubuntu installations I use netboot mini.iso from  . This installation method includes the step of configuring proxy. It is very useful for me as I have Squid-Deb-Proxy configured in my network, so I can lower traffic usage by using it.  But I do not see such an option in Ubuntu Desktop graphical installers, for example for Ubuntu MATE. They use Ubiquity as a common. Is it possible to set APT proxy during installation for such installers?


I run the history command in bash but it gives me everything...  I noticed that I'm able to limit the number of previous commands displayed by running history | tail -n <num-of-lines+1>; so is there some way to see only what I've run since I logged in using the number of commands I've run since I logged in in bash?

Maybe I'm overlooking something but is there a way to get your current bash history for the current session you are using like  if i run  ssh host $ pwd $ ls $ cd /tmp   I just want to see those 3 commands and nothing else


I'm new with template in C++ and I'm puzzled on how I could run my program. The code I'm supposed to do is use a template to "create a static stack of any data type." This is the first time I'm working with templates and I only have a very basic knowledge about it. Hoping to get inputs! I'm using codio and the code is broken into three files: Main.cpp, Data.h, and Implementation.cpp. The error is this: /tmp/cckg9fEh.o: In function `main': Main.cpp:(.text+0x24): undefined reference to `IntStack<char>::IntStack(int)' Main.cpp:(.text+0x48): undefined reference to `IntStack<char>::push(char)' Main.cpp:(.text+0x6c): undefined reference to `IntStack<char>::push(char)' Main.cpp:(.text+0x90): undefined reference to `IntStack<char>::push(char)' Main.cpp:(.text+0xb4): undefined reference to `IntStack<char>::push(char)' Main.cpp:(.text+0xd8): undefined reference to `IntStack<char>::push(char)' Main.cpp:(.text+0xfe): undefined reference to `IntStack<char>::pop(char&)' Main.cpp:(.text+0x13b): undefined reference to `IntStack<char>::pop(char&)' Main.cpp:(.text+0x178): undefined reference to `IntStack<char>::pop(char&)' Main.cpp:(.text+0x1b5): undefined reference to `IntStack<char>::pop(char&)' Main.cpp:(.text+0x1f2): undefined reference to `IntStack<char>::pop(char&)' collect2: error: ld returned 1 exit status Stack_Project/test.sh: 6: Stack_Project/test.sh: ./Main.h: not found codio@smile-queen:~/workspace$   My code is as follows: Data.h #ifndef INTSTACK_H #define INTSTACK_H #include <iostream> #pragma once  template <class T> class IntStack {   private:     T *stackArray;     int stackSize;     int top;    public:     IntStack(int size);     void push(T let);     void pop(T &let);     bool isFull();     bool isEmpty(); };  #endif  Implementation.cpp #include <iostream> #include "Data.h" using namespace std;  template <class T> IntStack<T>::IntStack(int size) {    stackArray = new T[size];     stackSize = size;     top = -1; }   template <class T> void IntStack<T>::push(T let) {    if (isFull())    {        cout << "The stack is full.\n";    }    else    {        top++;        stackArray[top] = let;    } }  template <class T> void IntStack<T>::pop(T &let) {    if (isEmpty())    {        cout << "The stack is empty.\n";    }    else    {        let = stackArray[top];        top--;    } }  template <class T> bool IntStack<T>::isFull() {    bool status;     if (top == stackSize - 1)        status = true;    else        status = false;    return status; }  template <class T> bool IntStack<T>::isEmpty() {    bool status;     if (top == -1)        status = true;    else         status = false;    return status; }  Main.cpp #include <iostream> #include "Data.h" using namespace std;  int main() {     IntStack<char> stack(5);     char reclaim;      cout << "Pushing M\n";     stack.push('M');     cout << "Pushing C\n";     stack.push('C');     cout << "Pushing P\n";     stack.push('P');     cout << "Pushing U\n";     stack.push('U');       cout << "Pushing D \n";     stack.push('D');      cout << "Popping...\n";     stack.pop(reclaim);     cout << reclaim << endl;     stack.pop(reclaim);     cout << reclaim << endl;     stack.pop(reclaim);     cout << reclaim << endl;     stack.pop(reclaim);     cout << reclaim << endl;     stack.pop(reclaim);     cout << reclaim << endl; }  Thanks for the help!

Quote from :     The only portable way of using templates at the moment is to implement them in header files by using inline functions.   Why is this?  (Clarification: header files are not the only portable solution. But they are the most convenient portable solution.)


I recently starting learning JS so a beginner level answer would be great! Thanks.   function contains(arr, item) {    // check to see if item is inside of arr    // return true if it is, otherwise return false  }

What is the most concise and efficient way to find out if a JavaScript array contains a value? This is the only way I know to do it: function contains(a, obj) {     for (var i = 0; i < a.length; i++) {         if (a[i] === obj) {             return true;         }     }     return false; }  Is there a better and more concise way to accomplish this?


The ring of polynomials in $X,Y$ all of whose partial derivatives with respect to $X$ vanish for $Y=0$ is Noetherian ?

I have the following two questions on rings of polynomials. They seemed similar enough that I thought I'd go ahead and group them here as opposed to making separate listings for them. The questions are:  ($1$) Determine whether or not the ring of polynomials in $z$ whose first $k$ derivatives vanish at the origin for some fixed positive integer $k$ is Noetherian. ($2$) Determine whether or not the ring of polynomials in $z$ and $w$ whose partial derivatives with respect to $w$ vanish for $z=0$ is Noetherian.  Note that $z$ and $w$ are complex variables, that is, the polynomials are over $\Bbb C$. My work: It seemed like I could view each of these as subrings of $\Bbb C[z]$ and $\Bbb C[z,w]$, and since both of these are Noetherian, I wanted to conclude that each of the above were Noetherian, but I realized it isn't necessarily the case that a subring of a Noetherian ring is Noetherian. Can anyone offer some help for me? Thanks!