I'm trying to install the most stable, boring version of GIMP available, without adding special repositories. I'm assuming the correct way to do this is sudo apt install gimp. Unfortunately, I'm getting the following warning:  WARNING: The following packages cannot be authenticated!   libbabl-0.1-0 libsdl1.2debian libamd2.4.1 libcamd2.4.1 libccolamd2.9.1   libcholmod3.0.6 libumfpack5.7.1 libgegl-0.3-0   Should this be happening? I tried running  sudo apt-key update sudo apt-get update   which didn't fix the issue. What should I do?  Edit: Here is the full output of sudo apt install gimp:  Reading package lists... Done Building dependency tree        Reading state information... Done The following packages were automatically installed and are no longer required:   libmircommon5 linux-headers-4.4.0-78 linux-headers-4.4.0-78-generic   linux-headers-4.4.0-79 linux-headers-4.4.0-79-generic   linux-image-4.4.0-78-generic linux-image-4.4.0-79-generic   linux-image-extra-4.4.0-78-generic linux-image-extra-4.4.0-79-generic   linux-signed-image-4.4.0-78-generic linux-signed-image-4.4.0-79-generic   linux-tools-4.4.0-78 linux-tools-4.4.0-78-generic linux-tools-4.4.0-79   linux-tools-4.4.0-79-generic Use 'sudo apt autoremove' to remove them. The following additional packages will be installed:   gimp-data libamd2.4.1 libbabl-0.1-0 libcamd2.4.1 libccolamd2.9.1   libcholmod3.0.6 libgegl-0.3-0 libgimp2.0 libsdl1.2debian libumfpack5.7.1 Suggested packages:   gimp-help-en | gimp-help gimp-data-extras The following NEW packages will be installed:   gimp gimp-data libamd2.4.1 libbabl-0.1-0 libcamd2.4.1 libccolamd2.9.1   libcholmod3.0.6 libgegl-0.3-0 libgimp2.0 libsdl1.2debian libumfpack5.7.1 0 upgraded, 11 newly installed, 0 to remove and 0 not upgraded. Need to get 14.0 MB of archives. After this operation, 79.8 MB of additional disk space will be used. Do you want to continue? [Y/n] y WARNING: The following packages cannot be authenticated!   libbabl-0.1-0 libsdl1.2debian libamd2.4.1 libcamd2.4.1 libccolamd2.9.1   libcholmod3.0.6 libumfpack5.7.1 libgegl-0.3-0 Install these packages without verification? [y/N] n E: Some packages could not be authenticated   EDIT: I don't believe this question is a duplicate of any of the questions suggested by reviewers:    None of the answers are the one that worked for me.  Does not have the answer that worked for me.  This is not the error I got initially.

I'm experimenting with VPNs, so I wanted to install pptpd.  john@desktop:~$ sudo apt-get install pptpd [sudo] password for john:  <snip> The following extra packages will be installed:   bcrelay The following NEW packages will be installed:   bcrelay pptpd 0 upgraded, 2 newly installed, 0 to remove and 11 not upgraded. Need to get 90.5 kB of archives. After this operation, 442 kB of additional disk space will be used. Do you want to continue [Y/n]? y   And everything runs fine until I get down to here:  WARNING: The following packages cannot be authenticated!   bcrelay pptpd Install these packages without verification [y/N]? n E: Some packages could not be authenticated   No authentication?  I'm no security expert, but I'm fairly sure that's a bad thing.  So I .  john@desktop:~$ apt-cache policy pptpd pptpd:   Installed: (none)   Candidate: 1.3.4-3ubuntu1   Version table:      1.3.4-3ubuntu1 0         500 http://us.archive.ubuntu.com/ubuntu/ natty/main i386 Packages john@desktop:~$ apt-cache policy bcrelay bcrelay:   Installed: (none)   Candidate: 1.3.4-3ubuntu1   Version table:      1.3.4-3ubuntu1 0         500 http://us.archive.ubuntu.com/ubuntu/ natty/main i386 Packages   They're coming from Ubuntu.com and not being properly authenticated?  What's going on here?


So I want to build a map where the blocks in spawn can't be broken but if you place a block you can break that block but you still can't break the default blocks kind of like BedWars on Hypixel. I thought this was a duplicate of the existing ones but it seems like those would make it so that you either cant break blocks you build or even build them in the first place.

I am creating a map on Minecraft, and I want the players to break certain blocks like trees, but I don't want them to break other blocks like stone bricks. Is there any command that can help me?


Reduce the quadratic form to a canonical form and find the rank index and signature. Write down the corresponding equations of transformation:  $$ 4x_1^2 + 3x_2^2 - x_3^2 + 2x_2 x_3 - 4x_3 x_1 + 4x_1 x_2 $$  The given solution says: $(w1)^2 + (w2)^2 - (w3)^2 ;$ $r=3, p=2, s=1;$ $x_1 = (1/2)(w_1) - (\sqrt 2/4)(w_2) + (1/2)(w_3),$ $x_2 = (\sqrt 2/2)(w_2) - (1/2)(w_3),$ $x_3 = (1/2)(w_3)$    Can anyone provide with detailed ans,how they come up with this?

This is just an eg. to understand how to solve it.  There is a quadratic form as below we need to bring to canonical form using Lagrange's method and find the coordinate transformation.   Then find the positive & negative indexes, rank of the quadratic form. Finnally, to investigate sign-definiteness of the quadratic form by two ways: according to it's canonical form and the criterion of Sylvester: $4x_{1}^2+2x_{2}^2+10x_{3}^2-4x_{1}x_{2}-12x_{1}x_{3}-8x_{2}x_{3}$  Any clarification on how to do it are very welcomed.


Minecraft won't maximize by the maximize square because I accidentally hit F11 and went into full screen. I have tried to hit F11 again but it wont work. I tried to quickly click on the maximize button, but it appears grey and wont even let me use it when I disable F11. My screen is either too big or too small. I dont know how to restart it.

I was playing in Minecraft in full screen for a while (toggling using F11) and then after a while, I went back to windowed mode (I don't like playing in full screen, I was just testing the functionality of it for... reasons).  However, when I went to maximize the window (not full-screen) I realized the maximize box in the corner had become disabled. I fiddled with full screen a bit more, toggling it in the menu options as well as toggling it with F11, but this had no effect. Restarting Minecraft (closing it completely and re-opening, not just quitting to menu) did fix it however.  Does anyone know what causes this issue, and if restarting Minecraft is the only way to fix it?


My computer won't shut down. If you try to shut down the LED on the Power button will continue to be on and you can clearly hear all the fans inside running. It's stuck in Limbo. Only after pressing the power button for a view seconds it properly shuts down.  I've spent weeks looking for a solution, now I'm just really disappointed by Windows and myself for not getting this fixed. I really hope you guys can help me.  This is my system:  Intel Core i5 2500k CPU Asus P8P67 Rev 3. Mainboard Kingston 16GB HyperX DDR3 RAM Samsung 850 Evo 500GB SSD 2nd 500GB hard disk Corsair 500W CM9 PSU   This is what I've tried so far:   Disabled the Windows 10 hibernate / fast startup feature Tried to downgrade the Intel Management Engine drivers to a 9.xx Version Tried multiple versions of Intel Rapid Storage (now uninstalled it) Tried multiple Intel Chipset drivers Tried to disable every non-windows service to see if it's a software issue. Replaced PSU   I tried to find a solution with Windows and Asus but both keep saying this is not a common issue.  I would really appreciate some help.

I have visited tons of websites and watched dozens of Youtube videos, yet I can't find a solution to my problem.  I recently upgraded my laptop to Windows 10, and now it won't shutdown properly.  I have turned off fast startup and sleep mode from power options.  Here is what happens:   Start menu > Power > Shut down It exits windows and apparently turns off (black screen) The power lights are still on and my computer apparently doesn't shutdown.


What is a real example of Request scope usage in ColdFusion?  From :     Request   The Request scope is used for the current request. These   variables are non-persistent global variables.   Is it like the request object in JEE?  Thanks in advance.

Been going over my predecessor's code and see usage of the "request" scope frequently. What is the appropriate usage of this scope?


Is there a way to preserve the current section counter to continue its counting, independent from the parts separating them? \part{Part 1} \section{Section 1} \section{Section 2} \section{Section 3}  \part{Part 1} \section{Section 4, not 1}

Some document elements (e.g., figures in the book class) are numbered per chapter (figure 1.1, 1.2, 2.1, ...). How can I achieve continuous numbering (figure 1, 2, 3, ...)?  And vice versa: Some document elements (e.g., figures in the article class) are numbered continuously. How can I achieve per-section numbering?  \documentclass{book}% for "vice versa" variant, replace `book` with `article`  \begin{document}  \chapter{foo}% for "vice versa" variant, replace `\chapter` with `\section`  \begin{figure} \centering \rule{1cm}{1cm}% placeholder for graphic \caption{A figure} \end{figure}  \end{document}   Bonus question: Is it possible to adjust the numbering of the sectioning headings themselves? Can I, e.g., switch from per-chapter to continuous numbering of sections in the book class?


I am working for a telco client and I have an interesting use case where I need some help. We have a person account enabled in Salesforce which will be used to send a communication on the closure of a case. Now the customer has a primary email address but he can also provide a temporary email of his son if he wants the case rep to contact his son as he is out of town. now we want to send the email via SFMC to his son for that particular case without creating a new person account and changing the customer preferred email address. Does any one have any suggestion on this? How I can just send a email to an additional email address that I have capture without impacting the primary email address of the customer and indeed all subscriber list should be updated with son email address?

Can someone help clarify use case where we should have Subscriber key related to multiple email addresses?   We generate our own subscriber key but we want to give a Subscriber opportunity to be communicated to different email addresses (what he wants to receive on what email) but we still want to track him through 1 Subscriber key.   How should we approach to this scenario? Possible solutions?


My output:   What it's supposed to look like:   My code:   My output:  Jersey number: 84, Rating: 7 Jersey number: 4, Rating: 5 Jersey number: 66, Rating: 9   Asking for:  Jersey number: 4, Rating: 5 Jersey number: 66, Rating: 9 Jersey number: 84, Rating: 7   I am not sure how to get the jersey numbers in the desired order, I've tried a variety of ways and I'm still unsure. The specific order is where it's getting me. Any ideas? Thank you for reading!

I have a list of dictionaries and want each item to be sorted by a specific value. Take into consideration the list: [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]  When sorted by name, it should become: [{'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}]


To include an external JavaScript file in HTML page you would use this:  <script type="text/javascript" src="images/jquery-2.2.2.js"></script>   The body of this tag is empty, yet I have to type the string script twice. Why isn't XML-style implicit closing tag in HTML, like this:  <script type="text/javascript" src="images/jquery-2.2.2.js" />   If it is valid (at least in HTML5), then how to make sure every browser supporting HTML5 (or XHTML5) can pick up on this usage?  EDIT  I'm gonna take this question to HTML WG. Deprecate the src attribute on <script> as it is the source of this entire confusion, with one of the two options:   <link> tag instead of an empty <script> tag:  <link type="text/javascript" href="images/jquery-2.2.2.js" />   Or optionally with a rel attribute:  <link rel="script" type="text/javascript" href="images/jquery-2.2.2.js" />  PHP-style include() in JavaScript itself: (thus force some content into the tag at all times, and allow multiple scripts to be included in one script tag)  <script type="text/javascript">include("images/jquery-2.2.2.js");</script>

What is the reason browsers do not correctly recognize:  <script src="foobar.js" /> <!-- self-closing script element -->   Only this is recognized:  <script src="foobar.js"></script>   Does this break the concept of XHTML support?  Note: This statement is correct at least for all IE (6-8 beta 2).


The Stress-Energy-Momentum Tensor is the source of gravity in General Relativity. Momentum, as well as mass, pressure and shear stress are components of that tensor, which represents the flow of energy in every possible direction.   Thereby, is momentum a source of gravity? Meaning that a massive object, travelling close to the speed of light, would have a greater gravitational pull than another object of same mass but no momentum?  Would it be possible, for instance, to accelerate a massive object so close to the speed of light that it becomes a black hole?

At a very basic level I know that gravity isn't generated by mass but rather the  and when I wave my hands a lot it seems like that implies that energy in $E^2 = (pc)^2 + (mc^2)^2$ is the source of gravity.  If the total energy of a particle contributes to the generation of a gravitational field, that seems to imply some strange things.  For example, you could accelerate a particle like an electron to such a high velocity that it has enough momentum / kinetic energy to collapse into a black hole.  An observer traveling with the electron would see an electron and stationary observers would see an arbitrarily massive black hole flying by.  These observations seem to be contradictory.  The same would be true of an extremely high energy photon.  Another consequence is that a stationary particle could appear to have so much mass that it could attract / deflect the trajectory of a rapidly moving observer.  It seems odd that at low velocities you wouldn't even notice a stationary electron as you pass it by but at high velocities the gravitational field you observe could be significant enough to alter your path.  Am I correct that non-mass-energy generates a gravitational field?  If it does, how are these strange observations like a particle looking like a black hole in one reference frame but not another reconciled?


I want to buy a server to accommodate both the Primevera and Sage applications? Would a HP Proliant 350e Gen8 do? Are there any concerns about this platform?

This is a  about  Related:      I have a question regarding capacity planning. Can the Server Fault community please help with the following:   What kind of server do I need to handle some number of users? How many users can a server with some specifications handle? Will some server configuration be fast enough for my use case? I'm building a social networking site: what kind of hardware do I need? How much bandwidth do I need for some project? How much bandwidth will some number of users use in some application?


I'm trying to create a label that will show the following: MP 23.6 The "23.6" is a double type column in my table, the "MP" is the text I want to add in front of it of the number (milepost).  I am new to QGIS.

Is there a way to concatenate fields in field calculator of QGIS? (e.g. hectares + ' ha')


I read a sci-fi story in a collection of shorts about 10 years ago, I believe they were stories from 70s possibly and have been trying to track down the title.  My recollection is that a number of ships encountered problems with a black hole or similar timey-wimey space object, and the ship belonging to the protagonist's girlfriend/love interest sacrifices itself to save the others. The 'black hole's' properties being what they are, the man goes on to live his life on Earth where the ship is visible in space, stuck in the moment of their sacrifice forever.

I am looking for a novel that I'd love to re-read it with my wife (my 13-year old self must be proud of me I guess). Key Plot points The plot was about humanity finding an alien station on an orbit of Earth. (I'm not sure if it wasn't Moon itself). On this station there were thousands of ships ready to travel so there rose a company that paid people to get on those ships and explore the galaxy. The problem was humans didn't understand how to steer those ships so they had to rely on preset targets. Of course there was no telling where particular ship was to go and fair amount of them never returned. Chronology and other relevant information I read it as an early teenager back in early 2000s. It was Polish translation of a foreign (possibly English) story. I think (not sure) that the book wasn't new at the time. It was just one of those books you dig out in a library and take to see if it's good. I don't recall cover either and even if I did it'd be of the Polish version. Additional information could be that the narration was led along the psychoanalytics of the main protagonists where he relives his memories from inside the station and his journeys.


Finding the value of the limit  Let  $ x_{n} $ =$ \sqrt{1+\sqrt{2+\sqrt{3+....+\sqrt{n}}}}$  then  $\lim_{n\to\infty}x_{n}=$?

It is an interesting task to try finding the limit of nested square root expressions.      $$\lim_{n \to \infty}\left( 1 + \sqrt{2 + \sqrt{3+ ... + \sqrt {n + \sqrt{n+1}}}}\right)$$   How to solve this one?


So might be a dumb question and actually quite simple, but I managed to confuse myself, and I don't really want to be learning the wrong thing.  So $a≡b\;(\bmod n)$ can be defined by $a-b=ln,$ $l\in\mathbb{Z}$ (modular equivalence), and if it exists, the modular inverse $p^{-1}p≡1\;(\bmod q)$.   Bezout's identity states that for some $a,b$ there always exists $m,n$ such that $am + bn = \gcd(a, b)$  How should I show the inverse mod as a modular equivalence? I just kind of know how to do them but not how to work them if that makes sense and I'm confusing myself.  How would I then use that with Bezout's Identity to find the gcd?

Isn't finding the inverse of $a$, that is, $a'$ in $aa'\equiv1\pmod{m}$ equivalent to solving the diophantine equation $aa'-mb=1$, where the unknowns are $a'$ and $b$? I have seem some answers on this site (where the extended Euclidean Algorithm is mentioned mainly) as well as looked up some books but there is no mention of this. Am I going wrong somewhere or is this a correct method of finding modular inverses? Also can't we find the Bézout's coefficients by solving the corresponding diophantine equation instead of using the extended Euclidean Algorithm?


Is the Kaiser rule (choosing only PCs with eigenvalues >1) a suitable method to select the best principal components? What are its pros and cons?

Is the Kaiser rule (choosing only PCs with eigenvalues >1, ) a reasonable method to select the number of principal components to retain? What are its pros and cons?


As per , Android apps using FLAG_SECURE in their activities are protected from screenshots being taken by the system.  You get a blank file when using screencap and some other screenshot apps, and this error when using the hardware buttons:     Can't take screenshot due to limited storage space, or it isn't allowed by the app or your organization.   However, this is not a hard guarantee.  In Android Studio, I can still take screenshots of such "secure" apps with the button pointed out in .  I took a look through the Android Studio source code, and the screenshot is acquired through a framebuffer: request to ADB, which in turn uses screencap.  Before I go looking through source for more info on how this works for adb but not for a root terminal and a possible "developer" solution, I thought I should ask: Has anyone already figured out what's happening behind the scenes here and can an end-user make use of it?  Ideally, without needing a PC.  My device is rooted, so I'm open to apps/commands/etc. that require root (or having USB Debugging enabled).  I've tried a few screenshot apps from pre-Android-2.3 that use root, but none of them still work (presumably due to how the framebuffer is now protected).

Some apps don't allow screenshots so how can I still take a screenshot?  It is my phone and I should be able to do whatever I want with it so why is there even a feature to disallow screenshots?  Android 5.0.1


I understand that if you have a dummy variable with $m$ categories that you should include $m-1$ categories in order to avoid perfect collinearity between regressors. However I don't understand why removing the constant term prevents the issue of perfect collinearity, because all of the categories will still sum to 1. For example: Suppose we have a dummy variable 'gender', with the categories being male and female. If we include both categories in the model without an intercept, the model is given by $y = \beta_1male + \beta_2female$. Don't we still have the issue of perfect collinearity because $male + female = 1$? We could rewrite this as $male = 1 - female$ so the regressors are still perfectly collinear because there is an exact linear relationship between them. How has removing the intercept helped us? (I am yet to learn about matrix algebra, so please provide answers that do not require knowledge of matrix algebra if possible).

I am running a binomial logistic regression with a logit link function in R. My response is factorial [0/1] and I have two multilevel factorial predictors - let's call them $a$ and $b$ where $a$ has 4 factor levels  $(a_1,a_2,a_3,a_4)$ and $b$ has 9 factor levels $(b_1,b_2,\dotsc,b_9)$. Therefore:  mod <- glm(y ~ a + b, family = binomial(logit), data=pretend) summary(mod)   The model output would then show all the information about the model as well as the coefficients.  There is a factor level for both a and b missing (a1 and b1) from the summary output. I understand that it is constrained in the "intercept" of the model. I have read that if I want to remove the intercept term and see the estimates for those factor levels I can just add -1 to the model formula, i.e.:  mod2 <- glm(y ~ a + b - 1, family=binomial(logit), data=pretend) summary(mod2)   In the new model (mod2) the intercept term is then gone and variable a's factor-level a1 is given amongst the list of coefficients. But, variable b's factor-level b1 is still missing and given that there is no intercept term anymore, how can I interpret the odds-ratio for that factor level then?  Could someone please explain to me how to get the coefficient for b1 too and why this is happening?


I'm not sure how to show it's not holomorphic. I can find a general equation for $f(z)$ but can only seem to show it's not one to one, and am not sure how to show it is not holomorphic.

There doesnt exist a holomorphic $f: \mathbb C\setminus0 \rightarrow \mathbb C\setminus 0 $, so that $e^{f(z)} = z $ for all $z \in \mathbb C \setminus 0 $  I tried to work with the derivative of f(z) but i didnt manage to get anything out of it.


Is it possible to make use of ASP.NET routing if not using ASP.NET MVC? It seems like a better solution than URL rewriting.

Now I'm trying to work with System.Web.Routing. All is just fine, but I can't understand how to make form authentication  work with url routing (return url, redirection, etc). Google says nothing. Help! :)  UPD: I forgot - I don't use MVC. That's the problem. How to use rounig and form authentication  without MVC  UPD2: more about my problem  What I want to get: urls such “mysite.com/content/123”, “mysite.com/login/”, etc using Routes. It’s important to make login page works like “regular” ASP.NET login form (redirects to login from secure area when not login on, and redirect back to secure area when loggined). That’s what I’m doing. In global.asax on Application_Start, register routes like this:  routes.Add("LoginPageRoute", new Route("login/", new CustomRouteHandler("~/login.aspx"))); routes.Add("ContentRoute", new Route("content/{id}", new ContentRoute("~/content.aspx")) {     Constraints = new RouteValueDictionary {{ "id", @"\d+" }} });   Where CustomRouteHandler and ContentRoute – simple IRouteHandler classes, just like: ...  public IHttpHandler GetHttpHandler(RequestContext requestContext) {     var page = BuildManager.CreateInstanceFromVirtualPath(VirtualPath, typeof(Page)) as IHttpHandler;     return page; }   ...  All seems to be perfect: I’m getting content.aspx when go to “/content/10” and login.aspx when go to “/login/”. But… When I make content secured (in web.config, with deny=”?”), login form doesn’t work like expected.  Now I can’t reach the “/content/10” page:  0. I’m typing “/content/10” in my browser. 1. Site redirects to “/login/?ReturnUrl=%2fcontent%2f10”. (Hm… seems like all problems starts here, right? :) 2. I’m trying to log in. No matter what credentials I’m entered… 3. …site redirects me to “login?ReturnUrl=%2fContent%2f10” (yellow screen of error - Access is denied. Description: An error occurred while accessing the resources required to serve this request. The server may not be configured for access to the requested URL.) So, the problem is how to get ASP.NET understand real ReturnUrl and provide redirection after login.


Let $ n \geq 3 $. By factorising $ n $ or $n + 1 $ (as appropriate), show that $ \mathbb{Z}[\sqrt{-n}] $ is not a UFD.   My thoughts so far:  Define $ N(a + b \sqrt{-n}) = a^2 + n b^2 $.  Suppose $ n $ is odd. Then $ n + 1 $ is even, say $ n + 1 = 2k $. Now $ N(2) = 4 $, and the norm of an element in this ring can never be 2, so we have that 2 is irreducible. Now note that $ 1 + n = (1 + \sqrt{-n})(1 - \sqrt{-n}) $. Is $ 1 + \sqrt{-n} $ irreducible? Well, if $ 1 + \sqrt{-n} = z_1 z_2 $, then $ N(z_1)N(z_2) = 1 + n $. So $ N(z_i) \leq \frac{n+1}{2} < n $. But this means both $ z_i$ must be purely real, which clearly can't be the case. Similarly, $ 1 - \sqrt{-n} $ is irreducible. Neither of these factors are equal to 2, and so 2 appears in one factorisation but not another. Hence for $ n $ odd, we don't have a UFD.  What about $n$ even? How can I factorise $ n $ other than as $ 2k $ for some $k $?  Thanks  EDIT: I overlooked $ n = \sqrt{-n} ( -\sqrt{n}) $!

I'm considering the ring $\mathbb{Z}[\sqrt{-n}]$, where $n\ge 3$ and square free. I want to see why it's not a UFD.   I defined a norm for the ring by $|a+b\sqrt{-n}|=a^2+nb^2$. Using this I was able to show that $2$, $\sqrt{-n}$ and $1+\sqrt{-n}$ are all irreducible. Is there someway to conclude that $\mathbb{Z}[\sqrt{-n}]$ is not a UFD based on this? Thanks.


When we have trouble with starting a car we say "Oh no, the car won't start."  Why do we use future tense here?

You're standing in front of a ticket vending machine and it says "The ticket is printing". Is that correct or should it be "The ticket is being printed"?  EDIT:  If both are correct, which one should be preferred or is more common?


I am triying to obtain a good calification on Page Speed test, now I've obtained 85 points but I am still on the problem of "Leverage browser caching".  This is the Page Speed test message about this problem:  Leverage browser caching Setting an expiry date or a maximum age in the HTTP headers for static resources instructs the browser to load previously downloaded resources from local disk rather than over the network. Leverage browser caching for the following cacheable resources: https://cdn.syndication.twimg.com/…ollowing&lang=es&screen_names=lomasguapa (10 minutes) http://platform.twitter.com/widgets.js (30 minutes) https://apis.google.com/js/api.js (30 minutes) https://apis.google.com/js/platform.js (30 minutes) http://pagead2.googlesyndication.com/pagead/js/adsbygoogle.js (60 minutes) http://pagead2.googlesyndication.com/pagead/osd.js (60 minutes) https://oauth.googleusercontent.com/…e:rpc:shindig.random:shindig.sha1.js?c=2 (60 minutes) http://stats.g.doubleclick.net/dc.js (2 hours) http://www.google-analytics.com/ga.js (2 hours)   However I have seen in many source on Internet that this solution works for this case, the solution is a modification of .htaccess file. There is the first block.  ## EXPIRES CACHING ## **<IfModule mod_expires.c> ExpiresActive On ExpiresByType image/jpg "access 1 year" ExpiresByType image/jpeg "access 1 year" ExpiresByType image/gif "access 1 year" ExpiresByType image/png "access 1 year" ExpiresByType text/css "access 1 month" ExpiresByType text/html "access 1 month" ExpiresByType application/pdf "access 1 month" ExpiresByType text/x-javascript "access 1 month" ExpiresByType application/x-shockwave-flash "access 1 month" ExpiresByType image/x-icon "access 1 year" ExpiresDefault "access 1 month" </IfModule>** ## EXPIRES CACHING ##  # REST OF MY .HTACCESS # BEGIN Better WP Security Options -Indexes  Order Allow,Deny Deny from env=DenyAccess Allow from all SetEnvIF REMOTE_ADDR "^1\.234\.83\.77$" DenyAccess SetEnvIF X-FORWARDED-FOR "^1\.234\.83\.77$" DenyAccess SetEnvIF X-CLUSTER-CLIENT-IP "^1\.234\.83\.77$" DenyAccess SetEnvIF REMOTE_ADDR "^211\.110\.140\.70$" DenyAccess SetEnvIF X-FORWARDED-FOR "^211\.110\.140\.70$" DenyAccess SetEnvIF X-CLUSTER-CLIENT-IP "^211\.110\.140\.70$" DenyAccess SetEnvIF REMOTE_ADDR "^91\.200\.13\.45$" DenyAccess SetEnvIF X-FORWARDED-FOR "^91\.200\.13\.45$" DenyAccess SetEnvIF X-CLUSTER-CLIENT-IP "^91\.200\.13\.45$" DenyAccess SetEnvIF REMOTE_ADDR "^200\.27\.231\.114$" DenyAccess SetEnvIF X-FORWARDED-FOR "^200\.27\.231\.114$" DenyAccess SetEnvIF X-CLUSTER-CLIENT-IP "^200\.27\.231\.114$" DenyAccess # END Better WP Security  # BEGIN WordPress <IfModule mod_rewrite.c> RewriteEngine On RewriteBase / RewriteRule ^index\.php$ - [L] RewriteCond %{REQUEST_FILENAME} !-f RewriteCond %{REQUEST_FILENAME} !-d RewriteRule . /index.php [L] </IfModule>  # END WordPress  <IfModule mod_deflate.c>     # compress text, HTML, JavaScript, CSS, and XML (for more page speed)     AddOutputFilterByType DEFLATE text/plain     AddOutputFilterByType DEFLATE text/html     AddOutputFilterByType DEFLATE text/xml     AddOutputFilterByType DEFLATE text/css     AddOutputFilterByType DEFLATE application/xml     AddOutputFilterByType DEFLATE application/xhtml+xml     AddOutputFilterByType DEFLATE application/rss+xml     AddOutputFilterByType DEFLATE application/javascript     AddOutputFilterByType DEFLATE application/x-javascript </IfModule>  # remove browser bugs (for more page speed) BrowserMatch ^Mozilla/4 gzip-only-text/html BrowserMatch ^Mozilla/4\.0[678] no-gzip BrowserMatch \bMSIE !no-gzip !gzip-only-text/html Header append Vary User-Agent # END    The case is that solution does not work and the page speed is showing again the caching problem. I put all the .htaccess in order to give more information, may be some lines cause any king of problem.

I tested my website with Google PageSpeed Insights.  I did cache optimization in my .htaccess, but it didn't affect resource files that are from Facebook and Google+  Leverage browser caching      http://connect.facebook.net/fr_FR/all.js (20 minutes)      https://apis.google.com/js/plusone.js (30 minutes)      https://oauth.googleusercontent.com/…e:rpc:shindig.random:shindig.sha1.js?c=2 (60 minutes)   What can I do to solve this?


During the Delta missions, you run into Deoxys but don't have a chance to save beforehand. I accidentally fainted Deoxys and failed to catch him.  Soon afterwards I went to the Mossdeep Space Center where the scientists told me Deoxys had been tracked somewhere around route 131.   I searched route 131, and even sky pillar, but I couldn't find him. My question is, where is he now?

I've heard that legendary Pokemon such as Kyogre/Groudon respawn after you defeat the Elite Four (if you didn't catch them). The only information that I can find seems to relate to the original version of the game.  Can players faint legendary Pokemon in order to progress the story and come back later to capture them? And are there any exceptions?  The reasoning for this is some players, myself included, wish to finish the game in order to unlock post-game features before investing time/effort soft resetting for ideal natures in legendary Pokemon.  Please avoid spoilers where possible.


I have this function:  def median(numbers):     middle = int(len(numbers)/2)     return numbers[middle]  print(median([4, 5, 6, 7, 8]))   The writing style of this line is a bit confusing for me:  return numbers[middle]   If I am not mistaken it returns the input numbers' order in a list. Is there a different way to read this line?

I need a good explanation (references are a plus) on Python's slice notation.   To me, this notation needs a bit of picking up.   It looks extremely powerful, but I haven't quite got my head around it.


For some reason, I can't add my strongest Pokémon to a friendly gym? Will someone tell me why this is happening for me?

So I defeated a Team Valor gym earlier today using my highest cp Pidgeot. It turned grey, so I went to claim it with that Pidgeot, except it wouldn't let me. What determines which pokemon you are allowed to place at a gym? Is it just those who have participated in a battle can't? Is there a time period for it? What about training at a friendly gym? Does this impact anything?


i have bought an laptop with Ubuntu version 12.04 and i want to remove it completely from laptop and install windows but i can't and i don't know how to uninstall it , So could anyone tell me please how to remove it ?

I have absolutely no experience with Linux, and I desperately need to get my computer back up and running again with Windows.  How do I remove Ubuntu and reinstall Windows?    Editor's note: many of the answers are about removing Ubuntu from dual-boot but keeping Windows (which is a bit complicated), while other answers are about removing Ubuntu from single-boot (which is easy: basically just format the disk while installing Windows). The question as written is ambiguous between dual-boot or single-boot.


I am trying to generate a list of all multisets of length $k$ in a set with $n$ symbols. For example, if I had the set   $S = {A, B, C}$  I would expect the following output for $k = 2$ and $n = 3$:  $O = {(A,A),(A,B),(A,C),(B,B),(B,C),(C,C)}$  What is the proper way to go about generating this list?

Let's say I have numbers each taken in a set $A$ of $n$ consecutive naturals, I ask myself : how can I found what are all the unique multisets, which could be created with $k$ elements of this set $A$?  For example I've got $A=[1,2,\dots,499]$. If I wanted to create unique multiset of 3 elements, I would search all the multisets $\{a,b,c\}$ such as $a\leq b\leq c$ and then have all the unique multisets. As such for each elements $a$ there is $499-(a-1)=500-a$ possibilities for $b$ and $500-b$ possibilities for $c$. Unfortunately I'm stumped here and can't find the number of possible combination, as I didn't do any maths for years. I know that I should have some kind of product, but I don't know how to find the product anymore.  So first, I would like to know if I am right in my original assumption, or if I am looking in the wrong direction. Second I would like to approach that as if I were doing an homework, as I would like to understand the logic behind it: What would be the formula to give the numbers of multisets of $k$ elements from a set $A$ of $n$ consecutive naturals?  P.S. I'm totally new to math.SE, as such, I'm not sure I tagged the post appropriately.


I recently watched a  that used two images simultaneously for front and side views. The images don't disappear when the speaker moves around the 3d-viewport. It's a little difficult for me to explain it in words, please watch the video to look at the image orientation.  I would love to know how that kind of image reference would be possible, because when I load any images they're only visible in one view and disappear when I switch the view.  Please let me know if any more information is required.

How do I put a backdrop behind the model that I am creating so that I can compare the picture to the model while editing?


I was curious about the independence of justice and moderation with those that embody them on the forums: the moderators.   I wasn't asking on the way the community, its users, such as me, can control moderators.  This is only an example but for instance if a moderator becomes rude on a forum and is flagged, who will moderate? Would he take part to his moderation case? Does it implicate his moderator qualification?

One or more of the following things happened to me on a Stack Exchange site:   My post(s) were wrongly deleted or locked by a moderator A moderator posted a wrong comment on one of my posts I was suspended, but did not do anything wrong I asked a meta question, but the question was closed and/or deleted A moderator discredited me in public ...   I think a moderator has abused their power, but I feel powerless against them.  What recourse do I have now? What can I do when I feel that a moderator has abused their power, and has been repeatedly unfair in their dealings with me?


Jacob Rees-Mogg MP, speaking by the invitation of the Speaker of the House of Commons in a , quoted a Financial Times article and, in passing, 'corrected' its grammar.  Quoting the article, Rees-Mogg said of entries on a Wikipedia list (of entries in a table) that 'none are' from Europe and then corrected the article saying that it should be 'none is'.  'None' may refer to 'not one' in which case I would say it is singular. But 'none' may also refer to 'not any' in which case I would suggest it is plural.  Thus I think that Mr Rees-Mogg is not correct in correcting the FT article.  'None of the entries' (that is, not any of the entries) are from Europe' seems quite correct to me.  'Not one of the entries is from Europe' is also correct, I would say.  Is Mr Rees-Mogg correct ?     entry for 'none' :     a. Not any (one) of a number of people or things. Also: neither of two persons or things (now regional).   ======================================  EDIT : I have read the suggested duplicates and they are inconclusive. I was hoping for a definitive (and thus an academic), decisive answer to the question.  =======================================

In my grammar book (English Grammar, HarperCollins Publishers), I read that none is occasionally treated as plural, but it is usually regarded as singular. Can you give me an example of sentence where none is used as plural pronoun?


Im trying to upgrade my system but when I execute the command this happens:  plex@plex-server:~$ sudo apt-get upgrade [sudo] password for plex: Reading package lists... Building dependency tree          Reading state information... Done   You may want to run 'apt --fix-broken install' to fix it. The following packages have unmet dependencies:    libgail-3-0 : Depends: libgtk-3-0 (=3.22.30-1ubuntu4) but 3.22.30-1ubuntu1 is installed   libgtk-3-bin : Depends: libgtk-3-0 (>= 3.22.30-1ubuntu4) but 3.22.30-1ubuntu1 is installed E: Unmet dependencies. Run 'apt --fix-broken install' without packages (or specify a solution).   I have done update before and I've tried autoclean, autoremove, --fix-missing but it still doesn't work.  EDIT This is what I have in my sources.list file:  #deb cdrom:[Ubuntu 18.04 LTS _Bionic Beaver_ - Release amd64 (20180426)]/  bionic main restricted  # See http://help.ubuntu.com/community/UpgradeNotes for how to upgrade to # newer versions of the distribution. deb http://es.archive.ubuntu.com/ubuntu/ bionic main restricted # deb-src http://es.archive.ubuntu.com/ubuntu/ bionic main restricted  ## Major bug fix updates produced after the final release of the ## distribution. deb http://es.archive.ubuntu.com/ubuntu/ bionic-updates main restricted # deb-src http://es.archive.ubuntu.com/ubuntu/ bionic-updates main  restricted  ## N.B. software from this repository is ENTIRELY UNSUPPORTED by the  Ubuntu                                             ## team. Also, please  note that software in universe WILL NOT receive any ## review or updates from the Ubuntu security team. deb http://es.archive.ubuntu.com/ubuntu/ bionic universe # deb-src http://es.archive.ubuntu.com/ubuntu/ bionic universe deb http://es.archive.ubuntu.com/ubuntu/ bionic-updates universe # deb-src http://es.archive.ubuntu.com/ubuntu/ bionic-updates universe  ## N.B. software from this repository is ENTIRELY UNSUPPORTED by the  Ubuntu ## team, and may not be under a free licence. Please satisfy yourself as  to ## your rights to use the software. Also, please note that software in ## multiverse WILL NOT receive any review or updates from the Ubuntu ## security team. deb http://es.archive.ubuntu.com/ubuntu/ bionic multiverse # deb-src http://es.archive.ubuntu.com/ubuntu/ bionic multiverse deb http://es.archive.ubuntu.com/ubuntu/ bionic-updates multiverse # deb-src http://es.archive.ubuntu.com/ubuntu/ bionic-updates multiverse  ## N.B. software from this repository may not have been tested as ## extensively as that contained in the main release, although it  includes ## newer versions of some applications which may provide useful features. ## Also, please note that software in backports WILL NOT receive any  review ## or updates from the Ubuntu security team. deb http://es.archive.ubuntu.com/ubuntu/ bionic-backports main restricted  universe multiverse # deb-src http://es.archive.ubuntu.com/ubuntu/ bionic-backports main  restricted universe multiverse  ## Uncomment the following two lines to add software from Canonical's ## 'partner' repository. ## This software is not part of Ubuntu, but is offered by Canonical and  the ## respective vendors as a service to Ubuntu users. # deb http://archive.canonical.com/ubuntu bionic partner # deb-src http://archive.canonical.com/ubuntu bionic partner  deb http://security.ubuntu.com/ubuntu bionic-security main restricted # deb-src http://security.ubuntu.com/ubuntu bionic-security main  restricted deb http://security.ubuntu.com/ubuntu bionic-security universe # deb-src http://security.ubuntu.com/ubuntu bionic-security universe deb http://security.ubuntu.com/ubuntu bionic-security multiverse # deb-src http://security.ubuntu.com/ubuntu bionic-security multiverse

There are now errors when updating and I cannot install most software due to a corrupted  /etc/apt/sources.list   file. Is there a copy I can download to replace it?  The file would be for Ubuntu 12.04 (Final Beta) in the United States.


As in my earlier environment I was running Drush 6 and Drush 8 together.  I need help with the Drush 8 and Drush 9 installation. Is there any possible way to achieve it?  I tried to do porting command following , without success.  Any help would be appreciated.

So in my environment, I have some websites on Drupal 7 and some on Drupal 8, and since the latest Drush 9 works with Drupal 8.4+ I was wondering if there's a way to be able to have both Drush 9 and (put another version here) to work on both major Drupal versions.  I'd like to be able to just use drush as my main alias and not define different aliases for different Drupal versions.


I cannot express enough how much i love math and physics.Whatever i have learnt till now in my syllabus i have trued visualising and finding new ways to solve same problems.I want to contribute something to this field as well as computer science and physics sometime in the future. Please guide me in what to learn further which might serve as a good next step for my goal. Also:3blue1brown has inspired me a lot in exploring the depths of what was part of my syllabus and a bit more(jee preparation)

I'm a first year econometrics student with a great interest in mathematics. I very much enjoy my study, but still I am interested to learn about more topics in mathematics which are not part of my study. Some of the main topics which I am already familar with or which will soon be covered in my study are: calculus, linear algebra, optimization, statistics/probability/combinatorics.  Which topics in mathematics would you advise me to study still (that are not part of this list)? I have a general interest in mathematics, so any advice on interesting and or essential topics in mathematics that are worth studying is appreciated.   If possible, could you also give me advice on books/references which I should study from?  Thanks in advance.  Edit: To be more specific, I am looking for topics on which the general consensus is that they are essential to know of for any mathematician or very interesting to study.


I am trying to find the odd indices in a list and return them into a new list:   def odd_indices(last):     lst = lst[1::2]     return last     print(odd_indices([4, 3, 7, 10, 11, -2]))   So my question is... why does creating a split between the slice syntax make this work? [1:]  Would start it at the first odd index. But why does [1::2]   work? I don't understand the fundamental concept of adding another ":" and how it now gets all the odd indices? Thanks!

I need a good explanation (references are a plus) on Python's slice notation.   To me, this notation needs a bit of picking up.   It looks extremely powerful, but I haven't quite got my head around it.


Does anyone have a detailed plan that shows the current availability of Ubuntu Phone BQ Aquaris to European countries?

Canonical has recently announced , which is Ubuntu with a specific interface for mobile devices.  Where can I get phones with this version of Ubuntu pre-installed? If no devices are available yet, where can I pre-order one?


It has been said that when you try to jump you are exerting your force on the ground and the ground pushes you up.  I was wondering if this was an over-simplification or not.  If so, what's the full explanation?  Otherwise, I would like to know the derivation of this proof.  For clarification: If I jump and the ground is "pushing me up" (from a normal force), then assuming that this is not a simplification and is truly what happens, why is this true?  My current hypothesis is that it has something to deal with electromagnetism (pushing two repelling magnets together; how atoms swerve to avoid collisions, for example).

There is this famous example about the order difference between gravitational force and EM force. All the gravitational force of Earth is just countered by the electromagnetic force between the electrons on the surface of my feet and the ground.  But also wave-function of electrons(fermions) do not overlap each other due to Pauli exclusion principle.  So which one is the true reason of me not flopping inside the earth forever? Is it only one of them(my guess is Pauli exclusion) or is it both?


I am using ArcMap 10.2 and I want to find duplicates in 2 columns.  For example,  Street_Name, HSN and ID are my columns.  Street_Name:Pyxis--HSN:24--ID:2356  Street_Name:Pyxis--HSN:24--ID:4234   As you can see in example I want to find same HSN on Same street.  How can I do that? Any ideas?

So I found this python script on another post here and was trying to adjust it for my needs.  I'm a very novice python user so I'm struggling with how to modify the script.  I have a Feature Class stored in a feature Dataset that I want to search a field for duplicate values and populate a new field with Y for duplicate or N for none.  The below script is what I found that looks like it will work once I find a way to drill down into my file geodatabase.          from arcpy import *          inShapefile = pointsShapefile          checkField = "xyCombine"         updateField = "dplicate"         #List of values found once         occursOnce = []        #list of values found twice         occursTwice = []       cursor = da.SearchCursor (inShapefile, [checkField])           for row in cursor:           #Check value is not null        if row[0]:     #If not already found to occur twice, proceed     if not row[0] in occursTwice:         #If hasn't occured once yet         if not row[0] in occursOnce:             #Add to occurs once list             occursOnce.append (row[0])         #If value has already been found once         else:             #Add to occurs twice list (duplicates)             occursTwice.append (row[0])     del cursor       cursor = da.UpdateCursor (inShapefile, [checkField, updateField])      for row in cursor:       #Check value is not null         if row[0]:            #check if value in occursTwice list (i.e. is duplicate)     if row[0] in occursTwice:         row[1] = "Y"     else:         row[1] = "N"     cursor.updateRow(row)    del cursor


Tried to google for this basic query related to SSH key generation, but didn't get expected answer  ,   explains to create key without email  Other links explain to create with email     What's the good practice ? Should I use email or avoid it ?

Say I have a public RSA SSH key that looks like:  ssh-rsa ijvoienvejfvmiejnci3onrvejviewlonvcljnrijcnreionv me@machine   If I’m copying this key to share with another machine, do I include the comment at the end (me@machine)? Do I include the header (ssh-rsa)?


I know that with 1gbit ethernet 100mbit ethernet cables won't work. You have to use cables designed for 1gbit.  What about with USB3? If I have a ten year old USB extension cable (ie. one that was made well before USB3 was) will it work at USB3 speeds? Or do I need to buy a new extension cable made specifically for USB3?

I have a couple of new USB 3.0 devices which connect to my USB 3.0 ports via the new "Blue" USB 3.0 cables that end in what looks like a USB 2.0 Type A connector.  Indeed, the USB 3.0 devices will also work in the USB 2.0 ports.  So, I have some USB 2.0 extender cables.  While I obviously can run the USB 3.0 devices over the USB 2.0 cables, will I be able to get the USB 3.0 speeds or do I need to get the specific USB 3.0 cables - to get the maximum transfer speeds?  It's not the cost - you can get the cables cheap.  It's just understanding how the end-to-end system works (and the "in case of emergency - break glass" effect).


prove that if $a\in \mathbb{R}$ such that $0 \leq a \leq \epsilon,$ for every $\epsilon > 0,$ then $a=0.$ What is the difference between proving this and the following theorem in Bartle, ,  p 28?

I am taking a real analysis course. I have the following statement:     Prove that if  $a$ belongs to $\mathbb R$,  such that  $0\leq a <  \epsilon$, for all $\epsilon > 0$, then $a = 0$   I know how to prove it (using contradiction). Now, I was wondering, how I prove almost the same thing:     Prove that if  $a$ belongs to $\mathbb R$,  such that  $0\leq a \leq \epsilon$ for all $\epsilon > 0$, then $a = 0$   Can someone show me how to do it? I am stuck with the equality to show a contradiction.  Note: this is not homework, just for my own knowledge.


Here is a problem from the book Analysis by Terence Tao:  (Vol 1, Exercise 5.5.2)  Let $E$ be a non-empty subset of $R$, let $n \geq 1$ be an integer, and let $L < K$ be integers. Suppose that $K/n$ is an upper bound for $E$, but that $L/n$ is not an upper bound for $E$. Without using the theorem EoLUB (stated below), show that there exists an integer $L < m \leq K$ such that $m/n$ is an upper bound for E, but that $(m-1)/n$ is not an upper bound for E. (Hint: prove by contradiction, and use induction. It may also help to draw a picture of the situation.)  Theorem EoLUB: Let $E$ be a non-empty subset of $R$. If $E$ has an upper bound, then it must have exactly one least upper bound.  Any idea how to solve this?  Moreover, is it possible to give a constructive proof of this, other than one using contradiction? Because I think it is this exercise that illustrates the existence of least upper bound, I will be very happy to see a constructive proof.  Thanks.

Terence Tao, Analysis I, 3e, Exercise 5.5.2:     Let $E$ be a non-empty subset of $\mathbb{R}$, let $n \ge 1$ be an   integer, and let $L < K$ be integers. Suppose that $K/n$ is an upper   bound for $E$, but that $L/n$ is not an upper bound for $E$. Without   using Theorem 5.5.9, show that there exists an integer $L < m \le K$   such that $m/n$ is an upper bound for $E$, but that $(m-1)/n$ is not   an upper bound for $E$. (Hint: prove by contradiction, and use   induction. It may also help to draw a picture of the situation.)    does not do the trick for me, since I'm looking for the proof by contradiction, equipped only with the set of propositions and theorems covered so far.  What I'm actually looking for is an entry point to the proof, since I find it quite hard to negate the assumption.  My attempt is to induct over $n$. Assume I have been able to prove the statement for $n = 1$, is the following a valid negation of the assumption?  For all integers in the range $(L, K]$, at least one of the following statements holds:   $(m-1)/(n+1)$ is not an upper bound and $m/(n+1)$ is not an upper bound. $(m-1)/(n+1)$ is an upper bound and $m/(n+1)$ is not an upper bound. $(m-1)/(n+1)$ is an upper bound and $m/(n+1)$ is an upper bound.   If I were able to prove that there is an integer in the given range such that none of the above holds, am I done?       Theorem 5.5.9 (Existence of least upper bound). Let $E$ be a non-empty subset of $\mathbb{R}$. If $E$ has an upper bound, (i.e.,   $E$ has some upper bound $M$), then it must have exactly one least   upper bound.     I went through another loop considering  of Steve Kass, but I ended up wondering why the proof needs induction at all:  Assume for all integers $m$ where $L < m \le K$ we have that $m/n$ and $(m-1)/n$ are upper bounds of $E$. Then $(m-1)/n = L/n$ is also an upper bound. But $L/n$ is no upper bound, by assumption. Assume $m/n$ and $(m-1)/n$ are no upper bounds. Then $m/n = K/n$ is also no upper bound. But $K/n$ is an upper bound, by assumption.   Hence, there is an integer in the range $(L, K]$ s.t. $m/n$ is an upper bound for $E$, and $(m-1)/n$ is not.


I have signed up to Blender Cookie and got an download link for the above mentioned file. But when I try to install it with the addon function in Blender it won't install. Does anybody know this file and can tell me how to use it?  Regards, Pingponguin

I am new to Blender. How can I apply some material included in a ?


I have two objects with the same structure and I want to concat them together using Javascript. Is there an easy way to do this?  var num = { '0': 0, '1': 0, '2': 0, '3': 0 }; var let = { 'A':'1', 'B':'2', 'C':'3' }

I have two JSON objects with the same structure and I want to concat them together using Javascript. Is there an easy way to do this?


Is there any link, info or documentation how should be every entry in the \bibitem mentioned? For example  \bibitem{Misiak} A. Misiak. \ \textit{$n$-inner product spaces.} Math.Nachr. {\bf 140} (1989), 299--319.  \bibitem{WKG} R.A. Wibawa-Kusumah and H. Gunawan. \textit{Two equivalent $n$-norms on the space of $p$-summable sequences}. To appear in {\it Period. Math. Hungar}.  \bibitem{ksw} P. Kostyrko, T. \v{S}al\'{a}t and W. Wilczy\'nski, $\mathcal{I}$-convergence, \emph{Real Anal. Exchange} \textbf{26(2)} (2000-2001), 669--686.   This bibitems can contains many entries (author, authortype, editor, editortype, editorX, editorXtype, holder, bookauthor ...). From this three examples I can extract without problem maybe numeric parts and assume they belong to pages or section entry. But when I found many different methods of storing info about for example: authors, I'm quite undecided how should I write bibliography cutter to entries efficiently:  jean de la fontaine Jean de la fontaine Jean {de} la fontaine jean {de} {la} fontaine Jean {de} {la} fontaine Jean De La Fontaine jean De la Fontaine Jean de La Fontaine   This is the case when \bibitem contains only one author, but there is also cases that it could contains even more. How could I logical extract bibliography entries? Is there anywhere complex structure mentioned?

I have an old document written in LaTeX which I need to edit/re-write. The document has hundreds of citations without using BibTex but just old style thebibliography environment and \bibitem's. Now I would like to re-use the old bibliographic references and add new ones but using BibTex. Is there a way I can re-use my old references? The thing that comes to mind is a script or something which translates \bibitem's entry and put them into BibTex format, but any solution is welcome. By the way, a (admittedly superficial) search on google and stack exchange did not provide any solution.


I'm using CSV to sort the players highest sore. I have the name and scores in a list like this:  ['4', 'Score One'] ['8', 'Score Two'] ['6', 'Score Three']   Which has been created from looping through the file, finding the maximum score and making a new list with this score and the player name.   for row in reader:         highscores = ([row[-1], row[0]])   When I print (sorted(highscores)) the same list is returned to me. What I want it to sort as is:  ['8', 'Score Two'] ['6', 'Score Three'] ['4', 'Score One']   If someone could explain to me why sorted doesn't work and what I could do to fix it, that'd be great.

I've got a list of Python objects that I'd like to sort by an attribute of the objects themselves.  The list looks like:  >>> ut [<Tag: 128>, <Tag: 2008>, <Tag: <>, <Tag: actionscript>, <Tag: addresses>,  <Tag: aes>, <Tag: ajax> ...]   Each object has a count:  >>> ut[1].count 1L   I need to sort the list by number of counts descending.  I've seen several methods for this, but I'm looking for best practice in Python.


To illustrate, assume that I have two tables as follows:  VehicleID Name 1         Chuck 2         Larry  LocationID VehicleID City 1          1         New York 2          1         Seattle 3          1         Vancouver 4          2         Los Angeles 5          2         Houston   I want to write a query to return the following results:  VehicleID Name    Locations 1         Chuck   New York, Seattle, Vancouver 2         Larry   Los Angeles, Houston   I know that this can be done using server side cursors, ie:  DECLARE @VehicleID int DECLARE @VehicleName varchar(100) DECLARE @LocationCity varchar(100) DECLARE @Locations varchar(4000) DECLARE @Results TABLE (   VehicleID int   Name varchar(100)   Locations varchar(4000) )  DECLARE VehiclesCursor CURSOR FOR SELECT   [VehicleID] , [Name] FROM [Vehicles]  OPEN VehiclesCursor  FETCH NEXT FROM VehiclesCursor INTO   @VehicleID , @VehicleName WHILE @@FETCH_STATUS = 0 BEGIN    SET @Locations = ''    DECLARE LocationsCursor CURSOR FOR   SELECT     [City]   FROM [Locations]   WHERE [VehicleID] = @VehicleID    OPEN LocationsCursor    FETCH NEXT FROM LocationsCursor INTO     @LocationCity   WHILE @@FETCH_STATUS = 0   BEGIN     SET @Locations = @Locations + @LocationCity      FETCH NEXT FROM LocationsCursor INTO       @LocationCity   END   CLOSE LocationsCursor   DEALLOCATE LocationsCursor    INSERT INTO @Results (VehicleID, Name, Locations) SELECT @VehicleID, @Name, @Locations  END      CLOSE VehiclesCursor DEALLOCATE VehiclesCursor  SELECT * FROM @Results   However, as you can see, this requires a great deal of code.  What I would like is a generic function that would allow me to do something like this:  SELECT VehicleID      , Name      , JOIN(SELECT City FROM Locations WHERE VehicleID = Vehicles.VehicleID, ', ') AS Locations FROM Vehicles   Is this possible?  Or something similar?

Consider a database table holding names, with three rows:  Peter Paul Mary   Is there an easy way to turn this into a single string of Peter, Paul, Mary?


My math teacher recently asked my class to find if there are any functions that comply to the rule $$(f\cdot g)^\prime = f^\prime \cdot g^\prime$$ I have searched the web for an answer but I couldn't find it, could anyone help me with that? or at least point me to the right direction?

In general, $\frac{d}{dx}(f(x) \cdot g(x)) \neq  \frac{d}{dx}f(x) \cdot \frac{d}{dx}g(x)$   When does this result hold true? My first try is to use product rule on left side and compare the two sides, but this hasn't helped at all. Any suggestions?


I have learnt that the formula for calculating the magnetic field at the centre of a current-carrying coil of $N$ turns is:- $$ B = \frac {\mu N I}{2r}$$ (where $r=$ radius of the loop, $I=$ current in the coil) And, the magnetic field at the centre of a current-carrying solenoid of $N$ turns is:- $$ B = \frac{\mu N I}{L}$$ (where $L$ & $I$ are the length and the current in the solenoid respectively and $\mu=\mu_0\mu_r$ is the magnetic permeability).  As we can see, both these formulas are different. But I can't figure out why that is.  (Since from what I have read about solenoids, they are just a number of coils wound closely together).  So my question is- why are there two different formulas for magnetic field at the the centre of a coil of $N$ turns and of a solenoid?  (Is a solenoid somehow different from a coil having many turns?)

In the B field of the loop, if R is the radius and z is the distance along the axis perpendicular to the center of the loop, let z go to zero, and multiply by N loops. Starting with the B field of the loop axis:  $$B=\frac{\mu_0 i R^2}{2(R^2+z^2)^{3/2}}$$ Becomes: $B_{\text{N loops}}=\frac{\mu_0 i N}{2R}$ and not $B_{\text {solenoid}}=\mu_0 i n$. What is the difference?


I did some changes in user preferences, but I forgot that changes and where I did them. How to I get back the initial blender settings?

I accidentally opened too many menus (or windows) and I want to reset them to default.  It is really annoying, and effecting my animation.  Thanks


I'm trying to display the login on the navigation drawer header , it worked fine when I used this piece if code       Bundle b = getIntent().getExtras(); name.setText(b.getCharSequence("username"));   But when I tried going back to the navigation drawer from another activity using the back arrow my app crashes my logcat says null pointer exception on this line   name.setText(b.getCharSequence("username"));   P.S the username is name of the variable on my php code..  help

What are Null Pointer Exceptions (java.lang.NullPointerException) and what causes them?  What methods/tools can be used to determine the cause so that you stop the exception from causing the program to terminate prematurely?


I have a Lenovo A5500-H tablet with a sim card slot and i want to know if i can make calls on it.  I hope you can help me fix this.

I have an Android tablet without telephony capabilities. I can put in a SIM card. When somebody calls my SIM card number (mobile number), can I receive this call? What application do I need?


I am using natbib and amsthm. If I use \citep[Thm. 4.5]{key} in the options of my theorem environment I get the error:  Argument of \NAT@@citetp has an extra }. \begin{thm}[\citep[Thm]  Why can't I put a citep with options into the options of a theorem environment? Below I have my code sample.  Thank you, for your help  \documentclass[a4paper]{scrartcl} \renewcommand\sf{\sffamily}  \usepackage{scrhack}  \usepackage[numbers]{natbib}  \usepackage{amsmath, amsthm, amssymb, dsfont} \usepackage{amsfonts} \usepackage{amsopn} \usepackage{bbold} \usepackage{bbm}  \newtheoremstyle{mystyle}%                % Name   {}%                                     % Space above   {}%                                     % Space below   {\normalfont}%                          % Body font   {}%                                     % Indent amount   {\bfseries}%                            % Theorem head font   {}%                                    % Punctuation after theorem head   { }%                                    % Space after theorem head, ' ', or \newline   {}%                                     % Theorem head spec (can be left empty, meaning `normal')  \theoremstyle{mystyle} \newtheorem{Theorem}{Theorem}%  \begin{document}   \begin{Theorem}[\citep[Thm]{cs_Foucart13}] test \end{Theorem}  \bibliographystyle{plainnat-d} \bibliography{literatur}

LaTeX seems to be having problems handling nested optional arguments. I've included a minimal code example that generates the error:  \documentclass[12pt,oneside]{book} \begin{document}  \newcommand{\termbad}[2][]{#2#1} \newcommand{\termgood}[2][]{#2#1}  \newcommand{\up}[1]{#1} \newcommand{\dd}[1][]{#1}  \termgood{\up{Tart}}{Pop} \termbad[\up{Tart}]{Pop}      \termgood{\dd[Tart]}{Pop} \termbad[{\dd[Tart]}]{Pop} %\termbad[\dd[Tart]]{Pop}   \end{document}   If you run this code, it works. If you remove the % in the second to last line, it fails with:  Runaway argument?  Tart{Pop} \par \par \end {document}  !File ended while scanning use of \dd.   After further tinkering, I have discovered the problem: LaTeX is reading the ] for the inner optional argument as the end of the outer optional argument and then passing a malformed argument in. So it is passing in \dd[Tart as the optional argument to \termbad instead of \dd[Tart].  This still feels like a bug to me. Nesting braces in arguments works, brackets should as well. The following are workarounds:   Enclose the inner optional argument in braces (as in the line before %). Redefine your inner function to require its argument instead of using optional arguments. Redefine your outer function to require its argument instead of using optional arguments.   Any one of the three works around the issue.  (Reposted here; question was originally posted to stackexchange and then closed instead of being migrated.)


How can I strip the $ and , characters from the following value, the code i am using is not working  var asset_value =   second_col.find("input[type=text]").val().replace('$', '');   second_col.find("input[type=text]").val() looks like      $1,080.00   Update:  I am not sure why i am getting voted down! the duplicate solution does not solve my question, nor does any answer below, except for this, very strange!  second_col.find("input[type=text]").val(function(i, val) {     return val.replace(/\$|,/g, ''); });  var asset_value =   second_col.find("input[type=text]").val();

I have a text box that will have a currency string in it that I then need to convert that string to a double to perform some operations on it.   "$1,100.00" → 1100.00  This needs to occur all client side. I have no choice but to leave the currency string as a currency string as input but need to cast/convert it to a double to allow some mathematical operations.


How to install FreeMind (free mind mapping software) in Ubuntu 16.4 with command line ?

Back in the dark days when I used to run windows I used to use MindJet Mindmanager and I found it to be very good.  Since I have been using Ubuntu as my main operating system I have been trying to find a replacement mind map software to use but have not had any luck finding a good replacment.  So far I have tried VYM, kdisset and Semantik. I have either found them not to be very good or have a nasty habit of crashing.   So I was wondering could anyone recommend good mind mapping software to use under Ubuntu?


So I was surprised I didn't find anything regarding this.  I have a python script which is testing a C++ program. It needs to format a float in the same way std::setprecision does. That is a float like 1.265 should be rounded UP to 1.27 (2 dp).  Now I have the following code: "{:.2f}".format(myFloat)  The issue is that numbers like 1.265 are rounded to 1.26 and my tests fail. setprecision rounds 1.265 to 1.27. What is the best way to fix this issue?

I want a to be rounded to 13.95.  >>> a 13.949999999999999 >>> round(a, 2) 13.949999999999999   The  function does not work the way I expected.


I am using this code to split a string at \.*[\s$], but preserving the split characters as an item in the list:  import re  markup = r"\{caption Figure 1: Leaf shapes\} \image:leaf_shapes.tiff" tokens = re.split(r'(\.*[\s$])', markup)   I expected it to return this list:  ['\\caption ', 'Figure 1: Leaf shapes', '\\} ', '\\image:leaf_shapes.tiff']   But instead I get this:  ['\\{caption', ' ', 'Figure', ' ', '1:', ' ', 'Leaf', ' ', 'shapes\\}',  ' ', '\\image:leaf_shapes.tiff']   Why are white spaces returned as separate items in the list?  Edit  I was supposed to escape the backslash, not the period, so I've changed my code to this:  markup = r"\{caption Figure 1: Leaf shapes\} \image:leaf_shapes.tiff" tokens = re.split(r'(\\.*[\s$])', markup)   But now I get this result:  ['', '\\{caption Figure 1: Leaf shapes\\} ', '\\image:leaf_shapes.tiff']   Please look at my expected result so you know what I'm trying to achieve.

I have this gigantic ugly string: J0000000: Transaction A0001401 started on 8/22/2008 9:49:29 AM J0000010: Project name: E:\foo.pf J0000011: Job name: MBiek Direct Mail Test J0000020: Document 1 - Completed successfully  I'm trying to extract pieces from it using regex. In this case, I want to grab everything after Project Name up to the part where it says J0000011: (the 11 is going to be a different number every time). Here's the regex I've been playing with: Project name:\s+(.*)\s+J[0-9]{7}:  The problem is that it doesn't stop until it hits the J0000020: at the end. How do I make the regex stop at the first occurrence of J[0-9]{7}?


I met a girl from Russia and she wants to see Denmark. I told her to meet first in Moscow to know each other better - anyway she asked me for 500 Euro for an international passport (she never had one she says) and a visa for Denmark. All that could be finished within 14 days. I won't send the money anyway but is 14 days realistic for a NEW passport?

On Travel.SE we often see questions about a person (usually a love interest) asking for money to pay for airfare, visas, permits and other expenses. For example        What general advice can be given in such situations to the person asking the question? Is it all a scam or can there be legitimate situations in which an online friend could be asking for money?


I am having a problem to verify on why CRONTAB is not executing a command set   0 8 * * * cp /directoryA/fileA /directoryB/fileB >/dev/null 2>&1 --> working fine  0 9 * * * cp /directoryA/$(date '+%m%d.csv') /directoryB/fileC --> NOT working  e.g.    on June 8 the file 0608.csv exists, but CRONTAB to not copying the file as above.   using cp /directoryA/$(date '+%m%d.csv') /directoryB/fileC via command line --> OK    How can I verify the CRONTAB log (and where) if the command as in CRONTAB was executed and why if it was not working (Error log?)  Thanks in advance Regards, ET

The following crontab does not work:  # TEST LINE DOES DOT RUN */1 * * * * /bin/echo 'test '`/bin/date +%Y-%m-%d` >> /tmp/test   I also tried starting it with:  SHELL=/bin/bash   Update: I thought the backtick characters ` were the villains, but as the answer below clarifies the percentage % was the culprit!


Let $R$ be a ring and $I$ and $J$ be ideals in $R$ such that $I + J = R$. Show that there exists a ring isomorphism $R/(I\cap J) \cong R/I \times R/J$. I've already proved that $IJ= I\cap J$. Feels like I am pretty closed to the answer, but trapped. Thanks!

Let $I$ and $J$ be two ideals of a commutative ring $R$ with $1.$ Give a necessary and sufficient condition so that    $$R/IJ\cong R/I\times R/J.$$   Prove your claim. Then decide whether the following ring isomorphisms are true or not:   $$\mathbb{Q}[x]/\langle x^2-1\rangle\cong \mathbb{Q}[x]/\langle x-1\rangle\times \mathbb{Q}[x]/\langle x+1\rangle,$$   $$\mathbb{Z}[x]/\langle x^2-1\rangle\cong \mathbb{Z}[x]/\langle x-1\rangle\times \mathbb{Z}[x]/\langle x+1\rangle.$$   It has been a while since I took Abstract Algebra and I am preparing for the prelims. I am not sure how to tackle this one. Any help/suggestion/hint will be much obliged. This is what I have so far:     This is a simplified version of the chinese remainder theorem. The map   $R\to R/I\times R/J$ defined by $r\mapsto (r+I,r+J)$ is a ring homomorphism with kernel $I\cap J.$ If the ideals $I$ and $J$ are comaximal, then this map is surjective and $I\cap J=I\cdot J$, so    $$R/(I\cdot J)=R/(I\cap J)\cong R/I\times R/J.$$   My question is about the second part of the question, do I have to check whether: $\langle x-1\rangle$ and $\langle x+1\rangle$ are comaximal and $\mathbb{Q}[x]$ and/or $\mathbb{Z}[x]$ are ring homomorphisms? I believe that these ideals are comaximal in both cases but forget how to prove it.


I was playing on a multiplayer server with the owner and had been given OP privileges.  I had turned on Creative mode and activated the 'God' command, but before the owner left, he de-opped me. Now I can not get out of Creative or God mode.   So I've tried:    Killing myself & falling into the void  Going to the middle of the nether with coordinates.  Making a portal but I can not light up the portal (Multiplayer server rules).    None of which worked.  Finally I wait for another player to come online, but the player is also not an op so I cannot teleport to him.   Please help me! The only thing I can do is to wait for the Owner of the server!

I've gone into the Nether and seemed to have lost my portal out of it, I also can't die in lava or by mobs attacking me, I'm appear to be invincible and can't die.  This is Single Player by the way.  Why is this happening?


I've seen websites like google and facebook redirect from "facebook.com" or "www.facebook.com" to "" versions of their website. How can I redirect like that when people first visit my website?  This is my code so far:  var link = HttpContext.Current.Request.Url.AbsoluteUri.ToLower(); var path = HttpContext.Current.Request.Url.AbsolutePath; if (link.StartsWith("http://www") == false) { Response.Redirect("http://www.arcadeLegend.com" + path); }

I am using ASP.NET 2.0 C#. I want to redirect all request for my web app with "www" to without "www"  www.example.com to example.com  Or  example.com to www.example.com  Stackoverflow.com is already doing this, I know there is a premade mechanism in PHP (.htaccess) file. But how to do it in asp.net ?  Thanks


I want to copy move some files from my PC to my USB but, when I run it it says:  WindowsError: [Error 5] Access denied: 'F:\\'  How can I use admin rights in python to avoid this error. Also, here is the code:  import os import shutil from os import walk  ruta='C:\\' final='F:\\' for (path,ficheros,archivos) in walk(ruta):     for i in archivos:         shutil.copytree(ruta,final)     for i in ficheros:         shutil.copytree(ruta,final)

I want my Python script to copy files on Vista. When I run it from a normal cmd.exe window, no errors are generated, yet the files are NOT copied. If I run cmd.exe "as administator" and then run my script, it works fine.  This makes sense since User Account Control (UAC) normally prevents many file system actions.  Is there a way I can, from within a Python script, invoke a UAC elevation request (those dialogs that say something like "such and such app needs admin access, is this OK?")  If that's not possible, is there a way my script can at least detect that it is not elevated so it can fail gracefully?


I have an old drive where I put files which I do not care much about. I mounted it as /mnt/crap (well chosen name!).  There I create a directory with my name: /mnt/crap/alexis.  I create a link to that folder from my home directory:  /home/alexis/crap -> /mnt/crap/alexis   I can go there with Nautilus, add files, etc. since I setup the proper permissions on that alexis directory it all works as expected.  However, when I go in Chromium and I click on the crap folder, it tells me:     Permissions denied.   and it shows nothing. If I go to /, the try to click on mnt, same error.  Is there a way to give chromium permission to access files outside of my home folder?

I just upgraded to Ubuntu 19.10 and to my horror I discovered that the default browser Chromium can no longer access other partitions than /home.  This means that my partitions containing music and videos (/multimedia) for instance are no longer accessible e.g. to upload into Google Drive. This includes USB partitions.   I found another description of this phenomenon here:  where it says  "Also, while we acknowledge that most users will only need to access their home folders, it’s worthy to mention that you no longer can access system files & folders from inside Chromium (You can’t access anything except your home folder). This means that if you are to upload a file or an image for example into the Internet using an uploading website, then this file/image must be in your home folder, as you can not access directories such as /usr, /etc or other directories located in your root partition. In fact, you no longer can access any other partition at all, this includes USB sticks and other booted partitions:"  I wonder why this behaviour been introduced? it violates the whole idea of multiple disks, partitions and symlinks. Btw, this still works for Firefox.  My questions are   Are there any plans to fix this? I expect one reason for this behaviour is about security. If so, is it possible to explain why and how that makes for better security? The files on other partitions are legitimately accessible by the user anyway e.g. through symlinks. But with this change, I'll have to move them to /home before uploading so still accessing them although more slowly and cumbersome.  If this behaviour is now permanent, the question is how should we incorporate multiple disks in our filesystem?  Alternatively, can someone tell me how to find and install the debian package. I haven't been able to locate it?   This is not a duplicate of  because that is about a specific error during installation that I didn't get. My question is about the access chromium has in Ubuntu 19.10.  Anyway, I solved the problem via the following command by using the original debian package from Google:  sudo apt-get install google-chrome-stable   After that I can now access my other partitions again. I know this is hacky in the sense that the package isn't native to Ubuntu so I would really like to get some advice on how to incorporate multiple disks and partitions with this new restricted behaviour of Chromium.  Thank you very much


Let $S(n)$ be the sum of digits of $n$,where $n$ is an natural number. If $n=p$ be a prime number, then my questions are   Are there infinitely many primes $p$ such that, $S(p)=q$ and $q$ is also a prime? If there are only finitely many such prime numbers, then what is the largest such prime?

Here's a possibly interesting prime puzzle. Call a prime $p$ flirtatious if the sum of its digits is also prime. Are there finitely many flirtatious primes, or infinitely many?


I believe every ordinal numbers $\alpha$ can be written as $\alpha=\lambda+n$ for a limit ordinal $\lambda$ and a natural number $n$. Is this true? Is it easy to see? Is there some terminology for the $\lambda$ and the $n$, something like limit part and natural part? (in analogy to integer and fractional part of reals)

I've been browsing through Jech's and Levy's texts on set theory, and the ideas of ordinals come up fairly quickly. The idea of a limit ordinal is introduced, which is an ordinal with no maximum element. My question is, can any infinite ordinal be written as the sum of a limit ordinal and a finite ordinal, possibly unique?  My thinking was, if $\alpha$ is an infinite ordinal with no maximum, it is a limit ordinal, so $\alpha=\alpha+0$. Otherwise, suppose $\alpha$ has some order type $\{a_0,a_1,\ldots, b\}$, so $\alpha=\omega+1$. Similarly, if $\alpha$ has order type $\{a_0,a_1,\ldots,b,c\}$, we could write it as $\omega+2$.     (Thanks to Arturo Magidin, for pointing out that the following example I gave is not an ordinal.) But what about an order type like $\{a_0,a_1,a_2,\ldots, b_2,b_1,b_0\}$, this has order type $\omega+\omega^*$, would it still be possible to write is a sum of a limiting ordinal and a finite ordinal? Thanks.


I am booting Ubuntu 20.04 from a USB drive. Using commands such as iwconfig does not show any sign of a wireless device being connected to this PC. Using USB tethering from my phone to my PC does allow me to access the internet, but using an external internet adapter puts less strain on my phone. My external internet adapter is the D-Link AC600 (the version with the antenna) Link:   Could this be a driver issue not supporting my internet adapter? Not sure if this information is relevant, but my internet speed using USB tethering is around 10mbps when I usually get around 80mbps. Any help regarding this issue is appreciated.  lspci command results:  00:00.0 Host bridge: Intel Corporation Xeon E3-1200 v6/7th Gen Core Processor Host Bridge/DRAM Registers (rev 05) 00:01.0 PCI bridge: Intel Corporation Xeon E3-1200 v5/E3-1500 v5/6th Gen Core Processor PCIe Controller (x16) (rev 05) 00:14.0 USB controller: Intel Corporation 200 Series/Z370 Chipset Family USB 3.0 xHCI Controller 00:16.0 Communication controller: Intel Corporation 200 Series PCH CSME HECI #1 00:17.0 SATA controller: Intel Corporation 200 Series PCH SATA controller [AHCI mode] 00:1c.0 PCI bridge: Intel Corporation 200 Series PCH PCI Express Root Port #5 (rev f0) 00:1c.7 PCI bridge: Intel Corporation 200 Series PCH PCI Express Root Port #8 (rev f0) 00:1d.0 PCI bridge: Intel Corporation 200 Series PCH PCI Express Root Port #9 (rev f0) 00:1f.0 ISA bridge: Intel Corporation 200 Series PCH LPC Controller (B250) 00:1f.2 Memory controller: Intel Corporation 200 Series/Z370 Chipset Family Power Management Controller 00:1f.3 Audio device: Intel Corporation 200 Series PCH HD Audio 00:1f.4 SMBus: Intel Corporation 200 Series/Z370 Chipset Family SMBus Controller 01:00.0 VGA compatible controller: NVIDIA Corporation GP107 [GeForce GTX 1050 Ti] (rev a1) 01:00.1 Audio device: NVIDIA Corporation GP107GL High Definition Audio Controller (rev a1) 03:00.0 Ethernet controller: Realtek Semiconductor Co., Ltd. RTL8111/8168/8411 PCI Express Gigabit Ethernet Controller (rev 15)   lsusb results:  Bus 002 Device 002: ID 0781:5581 SanDisk Corp. Ultra Bus 002 Device 001: ID 1d6b:0003 Linux Foundation 3.0 root hub Bus 001 Device 005: ID 046d:c084 Logitech, Inc. G203 Gaming Mouse Bus 001 Device 004: ID 258a:0016 BY Tech Usb Gaming Keyboard Bus 001 Device 003: ID 2001:3318 D-Link Corp. 11ac adapter Bus 001 Device 002: ID 0d8c:0005 C-Media Electronics, Inc. Blue Snowball Bus 001 Device 007: ID 2717:ff80 Xiaomi Inc. Mi/Redmi series (RNDIS) Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub

On Ubuntu 19.04 I had current rtl8812au (AC1200) and rtl8814au (AC1750) dkms drivers for my USB wireless dongles installed and running fine. They both broke after I upgraded to Ubuntu 19.10. I've been able to find a replacement/newer rtl8812au driver that works with kernel 5.3 on Ubuntu 19.10. I haven't been able to find a replacement/newer rtl8814au driver that works with kernel 5.3 on Ubuntu 19.10. Github.com has a number of rtl8814au drivers available, even some that proport to compile with kernel 5.2. I believe that I've tried about 6 different rtl8814au drivers from github, but none of them work, even though some of them do compile with kernel 5.3. Does anybody know of a working rtl8814au driver that works with kernel 5.3 on Ubuntu 19.10?


I search and found there are different way of checking object memory created for Java.  1. One is done by checking Runtime memory available    2. The other one is using the ByteArrayOutputStream to check    So I have written a program using both approach and check if they are the same  import java.io.ByteArrayOutputStream; import java.io.IOException; import java.io.ObjectOutput; import java.io.ObjectOutputStream; import java.io.Serializable;  class LotsOfBooleans implements Serializable {     boolean a0; }  class LotsOfInts implements Serializable {     int a0; }   public class Test {     private static final int SIZE = 1000000;      public static void main(String[] args) throws Exception     {                 LotsOfBooleans[] first = new LotsOfBooleans[SIZE];         LotsOfInts[] second = new LotsOfInts[SIZE];          System.gc();         long startMem = getMemory();          for (int i=0; i < SIZE; i++)         {             first[i] = new LotsOfBooleans();         }          System.gc();         long endMem = getMemory();          System.out.println ("Size for LotsOfBooleans (by checking Runtime) : " + (endMem-startMem));          System.gc();         startMem = getMemory();         for (int i=0; i < SIZE; i++)         {             second[i] = new LotsOfInts();         }         System.gc();         endMem = getMemory();          System.out.println ("Size for LotsOfInts (by checking Runtime memory) : " + (endMem-startMem));          System.out.println ("Size for LotsOfBooleans (using ByteArrayOutputStream) : " + getBytes(first).length);         System.out.println ("Size for LotsOfInts (using ByteArrayOutputStream) : " + getBytes(second).length);          // Make sure nothing gets collected         long total = 0;         for (int i=0; i < SIZE; i++)         {             total += (first[i].a0 ? 1 : 0) + second[i].a0;         }         System.out.println(total);      }      private static long getMemory()     {         Runtime runtime = Runtime.getRuntime();         return runtime.totalMemory() - runtime.freeMemory();     }      private static byte[] getBytes(Object o) {         ByteArrayOutputStream bos = new ByteArrayOutputStream();         ObjectOutput out = null;         try {             out = new ObjectOutputStream(bos);             out.writeObject(o);             return bos.toByteArray();         } catch (IOException e) {             e.printStackTrace();         } finally {             try {                 if (out != null) {                     out.close();                 }             } catch (IOException ex) {                 // ignore close exception             }             try {                 bos.close();             } catch (IOException ex) {                 // ignore close exception             }         }         return null;     } }   I compile it using   The result as below  $javac Test.java $java -Xmx128M -Xms16M Test Size for LotsOfBooleans (by checking Runtime) : 15914240 Size for LotsOfInts (by checking Runtime memory) : 15999984 Size for LotsOfBooleans (using ByteArrayOutputStream) : 7000072 Size for LotsOfInts (using ByteArrayOutputStream) : 10000064 0   From the result, clearly shows they are different. I have the feeling that ByteArrayOutputStream is more accurate. But the checking Runtime approach is by , which is from a renown StackoverFlow contributor.  So which one should I refer to?   Or both of them are correct, but referring to a different size of object (mind explaining how they are different if both are correct?  If both approach is not right, do suggest what is a good way to know the size of my object in Java?

I have an application that reads a CSV file with piles of data rows.  I give the user a summary of the number of rows based on types of data, but I want to make sure that I don't read in too many rows of data and cause OutOfMemoryErrors.  Each row translates into an object.  Is there an easy way to find out the size of that object programmatically?  Is there a reference that defines how large primitive types and object references are for a VM?  Right now, I have code that says read up to 32,000 rows, but I'd also like to have code that says read as many rows as possible until I've used 32MB of memory.  Maybe that is a different question, but I'd still like to know.


Why is $\sqrt{x^2 } = |x|$ ?   Squaring always produces a positive result, and you obviously square the equation before taking the root of it. So where exactly is the problem?

Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:  $\begin{align} f(x) &= \sqrt {x^2} \\ f(-1) &= \sqrt {(-1)^2} \\ f(-1) &= \sqrt {1} \\ f(-1) &= 1 \end{align}$  $\begin{align} g(x) &= x \\ g(-1) &= -1 \end{align}$  thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:  :    :    :    From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$?


After reading this  , I have question on how do we know or "see" the sense on the 4th statement, "Hence it is either prime itself, or divisible by another prime greater than $p_n$ " which meant if P is not a prime, it must be a product of higher prime? I start by considering it cannot be any even number, as it contain constitution of "2" which is divisible by 2 which left only odd number but the odd number must be prime number? Also in most of the proof of infinite prime $P=p_1p_2....p_n+1$ , are the proof actually meant P is not divisible by any prime in the list only(excluded higher prime like $p_{n+1}$),by definition from the list of prime use in above formula and definition of prime, P is a new prime?

I've expressed Euclid's proof on the infinitude of primes on Mathematica:  f[x_] := Product[Prime[n], {n, 1, x}] + 1 TableForm[Table[{f[x], PrimeQ[f[x]]}, {x, 1, 20}]]   Which results in:  $\begin{array}{ll}  3 & \text{True} \\  7 & \text{True} \\  31 & \text{True} \\  211 & \text{True} \\  2311 & \text{True} \\  30031 & \text{False} \\  510511 & \text{False} \\  9699691 & \text{False} \\  223092871 & \text{False} \\  6469693231 & \text{False} \\  200560490131 & \text{True} \\  7420738134811 & \text{False} \\  304250263527211 & \text{False} \\  13082761331670031 & \text{False} \\  614889782588491411 & \text{False} \\  32589158477190044731 & \text{False} \\  1922760350154212639071 & \text{False} \\  117288381359406970983271 & \text{False} \\  7858321551080267055879091 & \text{False} \\  557940830126698960967415391 & \text{False} \\ \end{array}$  The proof flaws for all those values, how is it considered a proof then? I guess that there might be infinite prime numbers according to the proof, but what is the guarantee that at some point it won't fail indefinitely?


Ok, so I want to run a forge file for Minecraft. I double clicked it thinking it would open and all I would have to do is install the client. But it said What do you want to use to open this file. So i clicked Java(TM) Platform SE Binary. But every time i select it then double click the file, it comes up with a Java Control Panel. I jhave no idea what this is or how to use it, and the apply button is greyed out. Can somebody help me out? Thank you!

I'm having a really, really hard time figuring out how to install mods on Minecraft. I've looked everywhere I can think of for help but the information is almost always outdated or unhelpful.  I've seen on several sites about the use of something called Forge. The Mod I want to use requires 1.6.4 Forge and I have both Forge and the Mod folders sitting in my downloads folder.  I haven't installed any kind of mods for any game prior to this and I have little clue to the technical side of Minecraft, so please keep this in mind when you give me an answer please!  Things that might help you figure out how to help me:   I want to play this mod  and am using the 1.6.4 Forge given on a link that website gives


I have a piece of code that prepares renderable array for my images:  if( !empty($row['image_file_id'])) {   $row['thumbnail'] = array(     '#theme' => 'image_style',     '#path' => _mymodule_uri_by_id((integer)$row['image_file_id']),     '#style_name' => 'product_thumbnail_small',     '#attributes' => array(       'alt' => $row['image_alt'],       'title' => $row['image_title'],     ),   ); }   It works OK:    But I want to add else with image provided within a module. I still want this image to go thorough all image style processing etc. I want my theming colleagues to be able to simply not care if they'll ever decide to add some effects or change resolution. But of course this image is in module directory, not in public://. And it means this will not work:  else {   $row['thumbnail'] = array(     '#theme' => 'image_style',     '#path' => drupal_get_path('module', 'mymodule') . '/frontend/noimage.png',     '#style_name' => 'product_thumbnail_small',     '#attributes' => array(       'alt' => 'some universal stub',       'title' => 'some universal stub',     ),   ); }   It gives:    And browser tries to read that image from   This gives nothing (status reports says "access denied"), unless I manually place image in public://sites/default/modules/mymodule/frontend - but I don't want it to be in public file system, I want it to stay in module space. How can I make it work?

Probably coming at this from too much of a Drupal 6 view point but...  How do I theme an image in a given image style when that image lives in a module folder?  This is driving me batty and I know I'm just missing something simple.  Here is some example code showing what I'm trying to do:  $file_uri = 'public://sites/all/modules/my_module/logo.jpg'; $data['logo'] = theme('image_style', array('style_name' => 'medium', 'path' => $file_uri));   Even if I use file_build_uri(), I end up with a path to a file that doesn't exist. If I move the image to the files directory and change the $file_uri to 'public://logo.jpg', it works just fine.  Do I need to copy the image to the files directory as part of the module code?  If so, isn't this ridiculous?  What am I missing?  Is there documentation on how to do something like this?  I've spent over an hour mining the Drupal documentation and came up completely dry.


I'm using mobile version of stackoverflow.com every now and then and have been able to spot a problem that close button on question page is not functioning: close dialog doesn't open.  The problem has been reproduced in the latest versions of Chrome mobile and Opera Mini browsers on iPhone 4G. The problem might be connected with the recent update of close dialog functionality, as it had been working before that update. Switching to 'full site version' yields correct functionality.  Technically, the close button sends a redirect with a hash sign appended to the current URL with no subsequent action undertaken. Probably there might be an error in the JavaScript engine used on SO.  Any clues on such behaviour?

This morning (it's about six AM here) I find that I can no longer submit close votes; the link just jumps to the beginning of the page.  I'm on iOS, using a third-party browser called Grazing, but a brief check with Mobile Safari shows the same problem there.   Here on Meta, where my score is much lower, I only get a "delete" link on my own question, and that works fine. (Oops :-))   Update: As per comments, this affects the "flag" link, too, which doesn't even scroll to the top of the page. Update 2: Latest comment suggests this affects Android, too. Update 3: This also affects desktop computers when viewing mobile (when you click "mobile" in the footer) Update 4:  mentions Windows 8, too; simply deleting platform-specific tags


Suppose there is a population, with goods and bads. The bad rate of the population(=bads/(bads+goods)) is of course unknown.   Now, I have a sample of $N$ from the population and I know the bad rate of this sample as $b$. The question is can I calculate the confidence interval based on $N$ and $b$ ONLY? In other words, can I calculate the confidence interval $x$ such that with, say, 95% confidence the population bad rate falls in $[\text{range}_1,\text{range}_2]$, where $\text{range}_1$ will be $b-x$ and $\text{range}_2$ will be $b+x$.

I have a random sample of Bernoulli random variables $X_1 ... X_N$, where $X_i$ are i.i.d. r.v. and $P(X_i = 1) = p$, and $p$ is an unknown parameter.  Obviously, one can find an estimate for $p$: $\hat{p}:=(X_1+\dots+X_N)/N$.  My question is how can I build a confidence interval for $p$?


I am trying to display the fraction of two integrals as follows:  \begin{equation} E_{sym} = \frac{\iint\displaylimits_{A_{sym}}\sqrt{(u + u_r)^2 + (v + v_r)^2 + (w - w_r)^2}dA}{{\iint\displaylimits_{A_{sym}}\sqrt{u^2 + v^2 + w^2} dA}}, \end{equation}     The problem is that in the frac command the lower limit on the double integrals appear next to the double integral signs, instead of below them. Outside the frac command this does not seem to be a problem:  \begin{equation} E_{sym} = \iint\displaylimits_{A_{sym}}\sqrt{(u + u_r)^2 + (v + v_r)^2 + (w - w_r)^2}dA/{\iint\displaylimits_{A_{sym}}\sqrt{u^2 + v^2 + w^2} dA}, \end{equation}     Any ideas how this could be fixed?  Any advice would be much appreciated.  Thanks very much,  Louis

I am trying to apply a limit law in which the limit is distributed to both the numerator and the denominator. However, the \lim_{x\to \infty} does not render properly when I place it in a fraction (\frac{\lim_{x\to \infty}}{\lim_{x\to \infty}}). The _{x\to \infty} part shows up next to the lim.   Any help would be greatly appreciated.


log4j:WARN No appenders could be found for logger (org.apache.http.client.protocol.RequestAddCookies). log4j:WARN Please initialize the log4j system properly. log4j:WARN    See for

Working with selenium + maven project in eclipse. I am facing below warnings:  log4j:WARN No appenders could be found for logger(org.apache.http.client.protocol.RequestAddCookies). log4j:WARN Please initialize the log4j system properly.  I have tried  Adding log4j.properties Corrected jdk version updated pom.xml file with accurate dependancy Ensure that those files are physically available and link in build with correct + valid path cleaned project in eclipse updated all maven dependancies  Anyone has experience and face, please guide/suggest how those can be fixed?


trying to brute force a specific length of ciphered pad, how do we measure the total number of possible encryptions in One Time Pad

I have a ciphertext encoded with One Time Pad. I do not know the key. How can I find the possible keys which when decoded, will result in a meaningful plain text?