i'm new in Blender. Can you guyz help me. I'm putting UVmapping to Lego character face. But the red and white color not appearing. Only black color just Like this. Can you guyz tell me how to show in color UVmapping ? Thank you

I'm designing this animal-type creature, and want to add a transparent .png image to its face with the face features (object has a subsurf). On the left is the image with the features, on the right is the part of the object where I want it to be.    I tried to import the image as a plane, but matching the shape of the plane with the face didn't work, and the lines of the face left shadows on the head, and could be seen as a different object from some angles (although it's the closest I've gotten yet). It looked like this:    So , I tried selecting the faces of the head that the features should go over, and create a new material for it. Unfortunately, since the .png is transparent, the result produced was this:    (where it's impossible to tell the transparency from a black background default color, and where I wouldn't know where to start to adjust size and position).  I also tried UV mapping, but then the whole object became black with the features repeated all over.   Is it something I have to go into Nodes for? I'm kind of stuck here. Creating a new material and applying it is simple, but applying a material that is an image is a bit more complex it seems... Ideally, I would reproduce the same technique for the adhesive bandage on its arm (seen on the right of image 1).


I'm trying to write a single regular expression to validate phone numbers. Below numbers should be   1111111111 or 111-111-1111 or 111.111.1111 - should fail and at the same time 767-789-9876 or 7677899876 should pass  I have come up with ^(?!([0-9])\1{2}[^d]\1{3}[^d]\1{4}$), but it only helps validate repetitive digits, alphanumeric characters and all succeed. Can anyone help me improve my regex?

I'm trying to put together a comprehensive regex to validate phone numbers. Ideally it would handle international formats, but it must handle US formats, including the following:   1-234-567-8901 1-234-567-8901 x1234 1-234-567-8901 ext1234 1 (234) 567-8901 1.234.567.8901 1/234/567/8901 12345678901   I'll answer with my current attempt, but I'm hoping somebody has something better and/or more elegant.


into a Java application I have the following situation:  I have these 2 lists:  List<Twp1007Progetto> progettiDaValidare = progettoService.getListaProgettiConsultazione(consultazioneFilter);  List<Twp1007Progetto> progettiInValidazione = progettoService.getListaProgettiConsultazione(consultazioneFilter);   and I have to concatenate these 2 lists (respetivelly named progettiDaValidare and progettiInValidazione and put the result into a new list named listaProgetti.  What is the best way to do it in Java?

Conditions: do not modify the original lists; JDK only, no external libraries. Bonus points for a one-liner or a JDK 1.3 version. Is there a simpler way than: List<String> newList = new ArrayList<String>(); newList.addAll(listOne); newList.addAll(listTwo);


It does not seem that multicols environment can cross across pages. Is there an alternative to create multiple columns that are filled automatically (i.e., we just insert the items and the environment fills columns using its own algorithms) with the contents but can break frames if necessary?    \documentclass[aspectratio=169]{beamer}  \usepackage{blindtext} \usepackage{multicol}  \begin{document}  \begin{frame}[allowframebreaks=0.7]{Long Lists}      \begin{multicols}{2}          \blindlistlist[3]{enumerate}          \blindlistlist[3]{enumerate}          \blindlistlist[3]{enumerate}      \end{multicols}  \end{frame}   \end{document}

I want beamer to automatically split my text over multiple frames (as many as is required) while retaining the format of the specified number of multicolumns (in this case 3). I do not want to have to specify the column width that is why i have used "multicols." I found the question entitled  but this question is unique to the multicol environment.  \documentclass[gray]{beamer} \usetheme{default} \begin{document} \begin{frame}[allowframebreaks]{Example Frame} \begin{multicols}{3} \small\raggedright Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. It has survived not only five centuries, but also the leap into electronic typesetting, remaining essentially unchanged. It was popularised in the 1960s with the release of Letraset sheets containing Lorem Ipsum passages, and more recently with desktop publishing software like Aldus PageMaker including versions of Lorem Ipsum. Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. It has survived not only five centuries, but also the leap into electronic typesetting, remaining essentially unchanged. It was popularised in the 1960s with the release of Letraset sheets containing Lorem Ipsum passages, and more recently with desktop publishing software like Aldus PageMaker including versions of Lorem Ipsum. Lorem Ipsum is simply dummy text of the printing and typesetting industry. Lorem Ipsum has been the industry's standard dummy text ever since the 1500s, when an unknown printer took a galley of type and scrambled it to make a type specimen book. It has survived not only five centuries, but also the leap into electronic typesetting, remaining essentially unchanged. It was popularised in the 1960s with the release of Letraset sheets containing Lorem Ipsum passages, and more recently with desktop publishing software like Aldus PageMaker including versions of Lorem Ipsum. \end{multicols} \end{frame} \end{document}


I'm wondering why ubuntu one changes the date of the synced files. Instead of leaving the actual date and time of modification it updates the time and date to the sync time. So, e.g., if I change and save a file today at july 26 2 pm and I go home and sync my home-pc with u1 tomorrow at 10 the 'Date modified' of that file will reflect the syncdate which might be July 27, 10am.  I don't like that behavior and I don't understand if this is a bug or if that is actually intended?

I would like to keep the original time stamp of a file when synchronizing with Ubuntu One.  Now Ubuntu One is messing with my time stamps, and I no longer know when I last worked on the file.  Ubuntu One also seem to mess up time stamps on the original source host.  This really is a show stopper, and we are just about to implement Ubuntu One across the many users in the organization.


I am new to blender and using Blender in 2D. I was just in edit mode moving things around, and then it would only let me move things at certain intervals. To clarify, I could move things, but it would only let me move it by a certain amount, no more, no less  From here To here

My cursor is moving by every selection by large increments!  Normally when I use Blender, I am able to select and move objects to any coordinate. Now, all of the sudden, when I select an object or mesh the selection moves by an inch or so in each direction.  How do I reset it to normal?


I just installed Ubuntu 16.04 LTS. I followed close instruction on how to install the nvidia drivers. I did sudo apt-get purge nvidia-* then I did   sudo add-apt-repository ppa:graphics-drivers/ppa   Then I did sudo apt-get update. and then sudo apt-get install nvidia-current  I rebooted and everything worked. I should also mention that my grub file has "nomodeset quiet splash' written in it. I started up Nvidia server and made sure Nvidia Graphics was selected. It was also selected in settings/updates and software/additional drivers (364).   Now when I restart, I got the black screen. I then took out nomodeset to see if that would help. I get to the login screen but of course, it keeps coming back to that over and over. Could someone please tell what the working solution for this is? I've gone through 3 installs of Ubuntu: 14.04,15.10 and now 16.04 to see if it would work and for each I get the same problem.   I'm using a Dell i7559 laptop with GTX960M graphics card.

I installed or upgraded to Ubuntu 16.04, but I'm having some problems with my graphics card.   The display is blank The display flickers Items on the screen do not display correctly Certain programs complain of graphics issues The computer boots to low-graphics mode Your specific issue may not be here, but try the solutions described in the answers anyway.   This question is very broad, but each one of these issues likely has one solution.    If you are experiencing a login loop, check the last part of my answer.    If you experience one of the listed issues and have your own solution (mine didn't work, or there's a faster way), feel free to post another POLITE answer, describing your problem and how you fixed it.    Those attempting to install Ubuntu, please refer to the section titled: If you are unable to enter a TTY or are trying to install Ubuntu.


AppStore requests a demo video for my app, but at the moment all I have is an iPad 1 with iOS 5 on it. They don't accept a video recorded in a simulator.  Buying a newer iPad is one obvious option, unless there are others.  Any ideas? Thanks.

I need to record a demo on my iPad's screen where I narrate what's going on but I also want the sounds from the iPad to be in the video. It seems that choosing the Mac Built-in Mic option causes only my voice to be in the video while choosing the iPad Mic option cause only the sounds from the iPad to be in the video. What can I do to put both in the video?


When an object is rubbed against another object, it transfers electrons, that much I know.  But what determines the way it's going? Against the object that's being rubbed against, or the object that's rubbing?  I tried searching on the internet for a while, but nothing popped up.

We all have studied that on rubbing a glass rod with silk or human hair, the two get charged with opposite polarities.  What confuses me is that since both glass and silk(or hair) are insulators, they contain no free electrons. So, how do the electrons get transferred?


I am trying to get everything between startStr and endStr for case bbb. I understand how I can get all occurrences between startStr and endStr using sed. I do not see how I would constraint it just to the one instance where bbb occurs.  Sample input:  fff startStr aaa bbb ccc endStr xxx yyy startStr ddd endStr ddd bbb   Required output:  startStr aaa bbb ccc endStr   This is what I have:  $ sed -n -e '/startStr/,/endStr/ p' sample.txt startStr aaa bbb ccc endStr startStr ddd endStr

I have a multi line log entry format that I need to process.   The log looks something like this:   --START-- Device=B Data=asdfasdf Lorem=Ipsum --END-- --START-- Device=A Data=asdfasdf Lorem=Ipsum --END-- --START-- Device=B Data=asdfasdf --END-- --START-- Device=A Data=asdfasdf --END-- --START-- Device=B Data=asdfasdf --END-- --START-- Device=C Data=asdfasdf Lorem=Ipsum --END--   I want to print everything between --START-- and --END-- if a particular pattern is matched.   e.g:   Print all entries where Device=A  --START-- Device=A Data=asdfasdf Lorem=Ipsum --END-- --START-- Device=A Data=asdfasdf --END--   All I've been able to do so far is write:   sed -e -n '/--START--/,/--END--/p' < input   Which effectively prints the input but I think I need to add {} to filter with N and then print if that condition matches.   I also think I'm completely lost.  Any idea on how to print multiple lines if a single line matches a condition?


I am a real estate agent and have a canon 400 D with a wide angle lens EFS 10-18mm and a tripod. I know nothing about photography and the homes I photograph have all beautiful window views being a coastal town.Please help me with taking the photos as well as editing in photoshop CS 5 extended. I have no plugin programs and no external light source as well as no money to spend on expensive gear (need to sell a few houses first) I need to have camera settings for normal lighting as well. I have been through lots of tutorials but cannot get anything right.

I sometimes find I want to photograph a room showing the view outside the window. This is tricky — either  you adjust the shutter speed so that the the window is so bright  you can see no detail, or the room looks very dark and gloomy.    Are there any techniques I can use to get the best photo showing as much as possible of both room and view?   My camera is a Sony Nex3, and for this sort of thing I usually use it with an elderly Pentax-M 1:2.8 / 28mm manual focus lens with an adaptor, although I would be prepared to buy another (cheap manual!) lens.


How could I prove that the following is greater than or equal to $0$? $$\sum_{n=2}^\infty \frac{(-x)^n}{n!}$$  And is this series equal to:$$\frac{e^{-x}}{2}x^2$$ if yes, why?  Question related to:

The Taylor expansion of $\exp(-x)$ is $$e^{−𝑥} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!}  ....$$ What I want to prove is that $$ \frac{x^2}{2!} - \frac{x^3}{3!}  .... $$  is greater than or equal to $0$.  I would really appreciate some help.


How can I scale a polygon about his center? I find only transform function, but the scale function here can only move the polygon with the scale factor and don't enlarge it.

In QGIS (1.8.0-Lisboa) on Windows (OSGeo4W Install), how would you resize a feature proportionally?  The only way I've found to do this is by hand using either the Freehand Editing (0.2.6) plugin or the Reshape Features button. Neither of these provides a means to uniformly scale (as in resize) the selected feature up or down.  For example, I want to increase the features size by 130% so that you retain the shape of the feature but expand the area of the shape itself proportionally in all directions.


We should be less concerned with students' 'cognitive impatience', however, than by what may underlie it: the potential inability of large numbers of students to read with a level of critical analysis sufficient to comprehend the complexity of thought and argument found in more demanding texts.  I want to know if there is anything omitted before the word sufficient? For instance, that is or which is? If so, is it an adjective clause? I think this way of writing is quite common but I just realised that I don't understand what it is "technically." Apart from that, I would appreciate if someone can help me identify the subject, object and all the other elements in this sentence since it's a terrible long one for me, I did spend some time reading it.

I heard phrases like "place nice" and "place safe" several times during watching series and movies and each time I was wondering why an adjective stands after a noun while they teach that adjectives go strictly before nouns. Unfortunately, all what I gained with simple googling is that it could be a set phrase like "Princess royal". And I already checked a  which doesn't answer my question because in this case I see no difference between "a place safe" and "a safe place". Examples:   We always have to go to, you know, someplace nice.     Is there any place safe for her?


There is a series which most people will be very familiar with.  $$S=1-1+1-1+1-1+\cdots$$  We all know that it is divergent hence $S=\lim\limits_{n\to\infty}S_n=\lim\limits_{n\to\infty}\sum\limits_{i=0}^{n}(-1)^n$ has no limit value.  But on the other hand there is a method to make $S$ has a value. Looks like that: $$\begin{equation}S+S=(1-1+1-1+\cdots)+1-1+1-1+\cdots\\ =1+(1-1)+(1-1)+(1-1)+(1-1)+\cdots\\=1\end{equation}$$  Now we have $S=\frac{1}{2}$.    I am wondering that Does the result imply that a divergent series can also have an exact sum? Does the sum of the alternative series contradict the divergence?

Today, a friend gave me a "proof" of $1=2$ and challenged me to find the fallacy.     $1 = 1$      $1 = 1 + 0 + 0 + 0 ...$      $1 = 1 + 1 - 1 + 1 - 1 + 1 - 1 ...$      $1 = 2 - 1 + 1 - 1 + 1 - 1 ...$      $1 = 2 + 0 + 0 ...$      $1 = 2$   My answer was that once you turn the initial $1 + 1$ into a 2, everything is offset so a $-1$ is always left at the end no matter how many times it is repeated. This negative one balances out the $2$ at the beginning so $1=1$ still holds true. I.e.  $$1 = 1 + 0 + 0 + 0 ... = 1 + (1 - 1) + (1 - 1) + (1 - 1) = 2 + (-1 + 1) + (-1 + 1) - 1$$  However, my friend claimed that my answer only applies if the $+ 1 - 1$ repeats for a finite number of times. He argues that because the sequence repeats infinitely and things work differently when working with infinity, my answer is not valid.  Can anyone enlighten me to the true fallacy in this proof?


First have a look at my .  I started with 1 basic reputation point. then I got 100 association bonus. So the total reputation became 1 + 100 = 101.  Then I  one post and earned 2 more reputation points. So, the reputation became 101 + 2 = 103.  Then, on Oct 26, 2 reputation points were deducted with reason "User was removed". But the  is still as it is and the writer of the post also still exists.  My question: why was my account affected because of that unknown user removed?  Sorry, but I do not understand what I am doing something wrong in my calculation.

Most of us have seen on the profile page a loss of reputation citing as a reason "User was removed". This reason comes with the useful link:  which explains (emphasis mine):     This message means that a user who voted for one of your posts had their account deleted (either by request or due to violating the network's terms of service). As a result, all of their votes were removed, and the reputation you gained or lost from them was undone. The resultant reputation change could be any amount; it could even be a reputation gain if enough of the removed votes were downvotes. All the reputation changes from a single user's deletion are rolled into a single event in the reputation page labelled "User was removed".      This removal occurs whenever a user is deleted, unless that user had a very high reputation score. Because high-reputation users have usually cast a great many votes, removing all of them could be that much more disruptive to other users. In such cases, the staff use a special deletion that preserves the votes, resulting in no reputation change for those who had been voted on by that user.   However, I have got recently such a message on one of beta sites, where I have no posts. Here is  on that site. As you can see, I have only got points from association bonus,  and the last event in my reputation history is the reputation loss with the message 'User was removed'. (Notice that the post, where I made suggested edit, was not deleted. And the users who approved the posts are Community♦ and one another user, who did not delete their account.)  You can also see from  on that particular site that I have posted no questions/answers there so far. Which means there are no votes on my posts which could have been removed.     Why this did happen? How can I lose reputation points in this way if I have never made a post on that site?   I am not sure if this is a . (Feel free to remove the tag if it does not fit the question.) (EDIT: The answer and the comments posted so far indicate, that it is not a bug, so I have removed the tag from the question.) But if there is some reason why reputation posts can be lost even without having some posts and actual votes on the posts, then the description of "User was removed" seems to be a bit imprecise. (Of course, if it is a very rare edge case, it is probably not that important to have it included in the explanation.)    Just a short disclaimer: I do not care that much for the 2 points I lost. (Obviously, it would be very silly to make a fuss just for 2 points, and even more on a site which I do not frequent that much.) But if this could be indication of some hidden bug, it might be interesting to know more about this.    I have also noticed two related questions       In both cases in some answers or in comments as a possible explanation was given that probably a post, where a suggested edit was made, has been deleted. As you can see from the links I, this is not the case here.


As far as I understand the postulates of quantum mechanics use only properties of abstract Hilbert space. So could we use any other Hilbert space for calculations instead of $L_2$? What could it be?  I can only suggest infinite-dimensional vector space with Euclidean norm, but it seems to be isomorphic to $L_2$.

I asked this question for many people/professors without getting a sufficient answer, why in QM Lebesgue spaces of second degree are assumed to be the one that corresponds to the Hilbert vector space of state functions, from where this arises? and why 2-order space that assumes the following inner product:  $\langle\phi|\psi\rangle =\int\phi^{*}\psi\,dx$  While there is many ways to define the inner product.  In Physics books, this always assumed as given, never explains it, also I tried to read some abstract math books on this things, and found some concepts like "Metric weight" that will be minimized in such spaces, even so I don't really understand what is behind that, so why $L_2$? what special about them? Who and how physicists understood that those are the one we need to use?


When I print as PDF a work done with a shape point and OpenLayers plugin I notice that the point are shifted  In the dual screen image you can see the PDF exported image (left) and the QGIS working one (right). You can notice that the point are a little shifted in the exported image and I am not able to understand the reason.    In the image you can see the same yellow street and the point on the street (right image) and shifted in the exported image  I use QGIS 2.6.1

I'm having trouble with the OpenLayers Plugin of Qgis and the map composer:  If I create an OSM-background layer and if I want to export this, the OSM-Layer looks perfectly all right in the normal qgis program window. But in the map composer and after export the layer has shifted relatively to my other shape layers (EPSG:32633 - WGS 84 / UTM zone 33N).  The second thing is that the output resolution of the exported osm-layer is very, very poor.   A really bad way of getting around this whole trouble would be to increase the screen resolution and make a screenshot of the map composition window of qgis. But I don't think this would be very professional. It also would cause a lot of pain :)   I'm using Qgis 1.8.0-Lisboa under Linux. The openlayers plugin is version 0.92.


Long story is I'm trying to draw some of the trigonometric graphs, but I'm incrementing a value by 0.1 to get a precise turning point, however when incrementing it's incrementing by 0.100000000006 or some value similar to that, so I'm unable to check if it's equal to another value  I ran a test on a different file to check if the problem was due to some of the sine conversions, incase of a memory leak or something (I'm not sure) and I had the same problem in another file which was just a for loop.  import math xNum = 0 for x in range(180):     print(xNum)     xNum = 0.1 + xNum   And when printing xNum I'm getting values like 0.40000000006 that I'm not sure why it isn't incrementing xnum by 0.1. i used xNum += 0.1 but there seems to be no difference in output

Consider the following code:   0.1 + 0.2 == 0.3  ->  false   0.1 + 0.2         ->  0.30000000000000004   Why do these inaccuracies happen?


How can I easily differentiate between when to use "relationship" and "relation". More often than not,  I tend to use "relationship", but would really like to know how these are to be used in writing. Speaking from a general viewpoint, the definition for both of these is almost similar. (this is something I've gathered from a simple Google Search) When discussing two entities, relationship seems to be an appropriate choice. But there are times when these could be interchangeable. For e.g., the relation between writing and reality  OR the relationship between writing and reality.  So are there are rules or pointers that are to be kept in mind to avoid using these incorrectly?

What is the difference between relation and relationship? Some say that relationship often refers to social connections. For instance,  She has a close relationship with her daughter.  How about the following?  the relationship between poor housing and health problems (Longman Dictionary) the special relationship between Britain and the US (Longman Dictionary)


I have the following method:  public class RangeFilter<T> : Filter<T> { }   Is there a way to restrict T to the following types:  bool  byte, sbyte  char, string decimal, double, float int, uint, long, ulong short, ushort

Can anyone tell me if there is a way with generics to limit a generic type argument T to only:   Int16 Int32 Int64 UInt16 UInt32 UInt64   I'm aware of the where keyword, but can't find an interface for only these types,  Something like:  static bool IntegerFunction<T>(T value) where T : INumeric


If the Samba Icon is not visible using the DASH resource search icon does it mean that the installation failed?  I cannot get the file-sharing function working and I believe it is due to an incomplete installation of Samba.

After a fresh installation of ubuntu 14.04 (my previous installation has seen to be broken somewhere reading .wmv file with VLC), I've installed Samba GUI tool from the software center.  It briefly appears in the launchpad then disappears. From this time I cannot the see in in the dash.  I've found this in askUbuntu   but the command sudo apt-get install samba samba-common gives me nothing to install.  Lecture des listes de paquets... Fait Construction de l'arbre des dépendances        Lecture des informations d'état... Fait samba est déjà la plus récente version disponible. samba passé en « installé manuellement ». samba-common est déjà la plus récente version disponible. Les paquets suivants ont été installés automatiquement et ne sont plus nécessaires :   kde-l10n-engb kde-l10n-fr Veuillez utiliser « apt-get autoremove » pour les supprimer. 0 mis à jour, 0 nouvellement installés, 0 à enlever et 8 non mis à jour.   The following guide  has no result either.  Any idea ?  Laurent06000


what does $dx$ mean in: $$\int f(x)dx$$ and is it true that, if $\frac{df(x)}{dx} = g(x)$, then: $$\int g(x)dx = \int df(x)$$

When I was at school and learning integration in maths class at A Level my teacher wrote things like this on the board.  $$\int f(x)\, dx$$  When he came to explain the meaning of the $dx$, he told us "think of it as a full stop".  For whatever reason I did not raise my hand and question him about it.  But I have always shaken my head at such a poor explanation for putting a $dx$ at the end of integration equations such as these.  To this day I do not know the purpose of the $dx$.  Can someone explain this to me without resorting to grammatical metaphors?


Does anyone know how to fix this weird screen output?     ~$ sudo lshw -c video   *-display                       description: VGA compatible controller        product: GF119M [GeForce 610M]        vendor: NVIDIA Corporation        physical id: 0        bus info: pci@0000:01:00.0        version: a1        width: 64 bits        clock: 33MHz        capabilities: pm msi pciexpress vga_controller bus_master cap_list rom        configuration: driver=nvidia latency=0        resources: irq:32 memory:f6000000-f6ffffff memory:e0000000-e7ffffff memory:e8000000-e9ffffff ioport:e000(size=128) memory:f7000000-f707ffff   *-display        description: VGA compatible controller        product: 3rd Gen Core processor Graphics Controller        vendor: Intel Corporation        physical id: 2        bus info: pci@0000:00:02.0        version: 09        width: 64 bits        clock: 33MHz        capabilities: msi pm vga_controller bus_master cap_list rom        configuration: driver=i915 latency=0        resources: irq:29 memory:f7400000-f77fffff memory:d0000000-dfffffff ioport:f000(size=64)

After some gorgeous updates I witness this:


I am reading a book on QFT which is stating the following.  For a massless scalar field $\phi$ the simplest possible Lagrangian is given by   $$ \mathcal{L}(x) = \frac{1}{2} \partial^\mu\phi \partial_\mu\phi  $$ with $\partial_\mu\phi\equiv\partial\phi(x)/ \partial\phi^\mu $. This can be expanded to $\mathcal{L}(x) = \frac{1}{2} (\partial_t\phi)^2 -\frac{1}{2}\nabla\phi\cdot\nabla\phi$. Which I easily see by using the definitions of $\partial^\mu$ and $\partial_\mu$ and having the mixted termes cancelling out each other.  But now the book also states that  $$ \mathcal{L}(x) = \frac{1}{2} \partial^\mu\phi \partial_\mu\phi=\frac{1}{2}(\partial_\mu\phi)^2,  $$ but I totaly fail to see this relation. To my understanding the expantion of the right part should look like  $$ \frac{1}{2}(\partial_\mu\phi)^2= \frac{1}{2} (\partial_t\phi)^2 -\frac{1}{2}\nabla\phi\cdot\nabla\phi+ \partial_t\phi\nabla\phi  $$ which is not equal to the given expantion above.  So what is my error?  [edit] Thanks for the comments that $$ (\partial_\mu\phi)^2 \equiv  (\partial_{\mu}\phi)g^{\mu\nu}(\partial_{\nu}\phi)  $$ And that I should not take the square "seriously". But If I don't take it seriously, how can I later on see that   $$ \partial_\mu \left(  \frac{\partial\mathcal{L}}{\partial(\partial_\mu\phi)} \right)= \partial_\mu\partial^\mu\phi  $$ when using the Euler-Lagrange equation?

The Lagrangian density for a single real scalar field theory is \begin{equation}\mathcal{L}=\frac{1}{2}(\partial_{\mu}\phi)^{2}-V(\phi)\end{equation} I have often seen this written \begin{equation}\mathcal{L}=\frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-V(\phi)\end{equation} which appears to differ in the kinetic term.  Specifically \begin{equation}(\partial_{\mu}\phi)^{2}=(\partial_{\mu}\phi)(\partial^{\mu}\phi)\neq\partial_{\mu}\phi\partial^{\mu}\phi=(\partial_{\mu}\phi)(\partial^{\mu}\phi)+\phi\partial_{\mu}\partial^{\mu}\phi\end{equation} where the product rule has been used in the last equality. Is this just sloppy notation (which is acceptable because only first derivatives in the field are considered anyway), or is my maths very bad?


m@samsung:~$ sudo rm /var/lib/apt/lists/* -vf [sudo] password for m:  rm: cannot remove `/var/lib/apt/lists/partial': Is a directory m@samsung:~$ sudo apt-get update N: Ignoring file 'gnomebaker.lis' in directory '/etc/apt/sources.list.d/' as it has an invalid filename extension E: Malformed line 1 in source list /etc/apt/sources.list.d/gnomebaker.list (URI parse) E: The list of sources could not be read. m@samsung:~$

I have unistalled and reinstalled the Ubuntu Software Center as per info I found in a similar thread and I got the same response about line 91 or something like that.  I just tried to upload a screen shot but since I'm new it won't allow me to.  I also can not figure out how to cut and paste anything so I have to hand type what the error screen says, both when I attempt to open the software center and nothing happens, when I try to enter commands into the terminal to uninstall, reinstall, whatever I get the same following:  COULD NOT INTITIALIZE THE PACKAGE INFORMATION An unresolvable problem occured while initializing the package information Please report t:his bug against the 'update-manager' package and include the following error message: 'E: Malformed line 91 in source list/etc/apt/sources.list (dist parse)  E: The list of sources could not be read.,  E: The package list of status file could not be parsed or opened.   How do I report bugs?  What can be done about this.  I have searched and everything everyone says to do leads me back to the same line error message.  So, I don't know how to get to line 91 in the source list; to tell you what it says. Sorry, I'm really new to this. That is what I need is to find out how to get there and fix what it says. I would really like to NOT have to re partition my hard drive and start from scratch, so I'm really looking forward to getting this problem solved. I need to be able to install new software.


When I use the following command in Bash, it works fine: sed -n '/##/{p; :loop n; /##/q; p; b loop}' FILENAME.txt  It outputs something like this: ##  * Some line   Command substitution However, when I try to use the same command in a command substitution, it suddenly fails completely: echo $(sed -n '/##/{p; :loop n; /##/q; p; b loop}' FILENAME.txt)  This instead prints out all of the files in my working directory in a row and then Some line at the end, which I have no clue how that would even happen:  FILENAME.txt OTHER_FILE.exe dir Some line   How can I make this work? It is crucial for me that the output of sed is echoed with some prefix.

Or, an introductory guide to robust filename handling and other string passing in shell scripts.  I wrote a shell script which works well most of the time. But it chokes on some inputs (e.g. on some file names).  I encountered a problem such as the following:   I have a file name containing a space hello world, and it was treated as two separate files hello and world. I have an input line with two consecutive spaces and they shrank to one in the input. Leading and trailing whitespace disappears from input lines. Sometimes, when the input contains one of the characters \[*?, they are replaced by some text which is is actually the name of files. There is an apostrophe ' (or a double quote ") in the input and things got weird after that point. There is a backslash in the input (or: I am using Cygwin and some of my file names have Windows-style \ separators).   What is going on and how do I fix it?


My wife and I will be travelling to the UK together. My wife depends on me. In my application I completed the income part.  In my wife application:  In the part "planning to spend money" in the UK I will put 40% of my total cost of the trip. For example, I put GBP 3,000 in my trip cost. Now I will put for my wife GBP 1,200. Is this ok?  Secondly, in my wife's application I don't know what to put in "Spend money in each month" because she is dependent and I spend in my home monthly. I just give her pocket money or for purchasing some food items. Now what should I put in this field?

I am an Indian citizen residing in the United States. My wife and I are planning to travel to the UK for a week. We plan to apply for the UK standard visitor visa from the US. I was filling out the application and I have a few questions.   Travel History - The application asks for travel history to the United States, Canada, Australia and New Zealand. Considering that we're resident aliens in the United States, should we consider all the times we've traveled back to the US from India? Travel History - Should we also list out the times we've traveled to India from the US? Cost of the trip - Hypothetically if we plan to spend a total of ~4,000 USD on this trip, should I split up the cost and write down 2,000 USD on my application and 2,000 USD on my wife's application? Income - From what I've read on the forum, my wife must say that "someone else" i.e. me - is paying for her costs and mention my GWF number in the remarks on her application. Would that be sufficient considering that I'll be sending all income documents with my application? Savings - We have joint savings accounts. Should I mention the same amount on both our applications? Posting the documents - Is the up-to-date advice to send all documents in one packet? Or would sending them in separate packets at the same time be better? My thought was to annotate my wife's packet with my GWF number if we were to send them separately. Bank statements - We get e-statements. Would printing those out be enough?


I wish to uninstall Software Center and keep the native GNOME app instead. However, I can't find it neither through Terminal nor Software. Currently it doesn't replay properly the apps on the Explore tab.

I've recently noticed an issue with Ubuntu Software Center which I have tried to resolve following the answers provided in  without result and  which partially worked but didn't persist. The Issue After installing 20.04 I opened Ubuntu Software Center and installed a number of apps without any problem - one of which was Stacer which I will use as an example.  In Show Applications the app is clearly present and right clicking on the icon selecting Show Details provides the image above and it works perfectly well. However, if I open Ubuntu Software Center and type in the search bar stacer it reports that no application is found:  Nor is it listed when selecting the Installed tab. This is the setup for for Software and Updates.  There are other applications I have previously installed which are also now reported as No Application found such as gLabels, gscan2pdf and others. Has something changed since I initially installed 20.04.1 and more importantly, how can I restore good order to the Ubuntu Software Center?


I have calculated the antiderivative of $sin^n(x)$ from $n=1$ to $n=20$ by software. Now I want to know if there are some laws about them？

I am working on trying to solve this problem:     Prove: $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$   Here are the steps that I follow in the example that I am reading:     $u = \sin^{n - 1}{x}$      $du = (n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx$      $v = -\cos{x}$      $dv = \sin{x} \ dx$            $\int \sin^n{x} \ dx =  \sin^{n - 1}{x} \cdot \sin{x} \ dx$      $\int \sin^n{x} \ dx = \underbrace{\sin^{n - 1}{x}}_{u} \cdot \underbrace{-\cos{x}}_{v} - \int \underbrace{-\cos{x}}_{v} \cdot \underbrace{(n - 1) \cdot \sin^{n - 2}{x} \cdot \cos{x} \ dx}_{du}$      $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \cos^{2}{x} \ dx$      $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$   Here is where I get lost. How did we go from $\int \sin^{n - 2}{x} \cdot \left(1 - \sin^{2}{x}\right) \ dx$ to $\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$? Even more specifically, where did $\sin^{n}{x}$ come from?     $\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx - (n - 1) \int \sin^{n}{x} \ dx$   I get this part.     $n\int \sin^n{x} \ dx = -\cos{x} \cdot \sin^{n - 1}{x} + (n - 1)\int \sin^{n - 2}{x} \ dx$      $\int \sin^n{x} \ dx = -\frac{1}{n} \cos{x} \cdot x \ \sin^{n - 1}{x} + \frac{n - 1}{n} \int \sin^{n - 2}{x} \ dx$   Could someone please explain what I am missing?  Thank you for your time.


I've been reading through some posts on how to install Python 2.5.4 in Ubuntu but I can not manage to find a solution to my problem. The version of Ubuntu I have is the 11.10 and the one of Python I intend to install is 2.5.4. It must be that one and not any of the updates since I am following a course which  uses specifically 2.5.4. I have already tried executing in the terminal commands proposed in other posts but I always get a mistake.   I'd be really thankful I some could explain detailed and concisely how I can do it.  As said, thank you in advance,  Dionisio.

I have installed Ubuntu 12.10 on my laptop, now i want to use Python 2.5.4 for educational purpose. Can any one help me with step by step instructions... I also need IDLE for 2.5.4.


I have a test csv file and I have written a code via  to show the PCA for that. I also use another tool in Excel () to compare the results. The XLSTAT automatically calculates the number of features, however, based on my understanding, I have to specify how many components are needed using the scikit package. For example, while XLSTAT shows 5 features: Factor scores: F1 F2 F3 F4 F5 A1 -1.293 -0.663 -0.462 -0.713 0.010 A2 -0.297 0.293 -1.429 0.397 0.056 A3 2.328 0.069 0.987 -0.108 0.062 A4 -0.556 -2.273 0.538 0.344 -0.032 A5 1.823 0.775 -0.597 -0.052 -0.085 A6 -2.005 1.799 0.963 0.133 -0.011  In the following code, I specified 2 components: x = StandardScaler().fit_transform(x) pca = PCA(n_components=2) principalComponents = pca.fit_transform(x) print( principalComponents )  [[-1.29292842 0.66325508] [-0.29706395 -0.29346337] [ 2.32751305 -0.06850045] [-0.5558091 2.27288988] [ 1.82312052 -0.77527304] [-2.0048321 -1.7989081 ]]  As you can see, the first column in XLSTAT and scikit are the same. However, the second columns are negated. For example, considering F1 and F2, we see XLSTAT => -1.293 -0.663 scikit =>  [-1.29292842 0.66325508]   Considering the F1 and F2 as a XY scatter point, I want to know why the value of Y in XLSTAT and scikit are opposite?

I performed principal component analysis (PCA) with R using two different functions (prcomp and princomp) and observed that the PCA scores differed in sign. How can it be?  Consider this:  set.seed(999) prcomp(data.frame(1:10,rnorm(10)))$x              PC1        PC2  [1,] -4.508620 -0.2567655  [2,] -3.373772 -1.1369417  [3,] -2.679669  1.0903445  [4,] -1.615837  0.7108631  [5,] -0.548879  0.3093389  [6,]  0.481756  0.1639112  [7,]  1.656178 -0.9952875  [8,]  2.560345 -0.2490548  [9,]  3.508442  0.1874520 [10,]  4.520055  0.1761397  set.seed(999) princomp(data.frame(1:10,rnorm(10)))$scores          Comp.1     Comp.2  [1,]  4.508620  0.2567655  [2,]  3.373772  1.1369417  [3,]  2.679669 -1.0903445  [4,]  1.615837 -0.7108631  [5,]  0.548879 -0.3093389  [6,] -0.481756 -0.1639112  [7,] -1.656178  0.9952875  [8,] -2.560345  0.2490548  [9,] -3.508442 -0.1874520 [10,] -4.520055 -0.1761397   Why do the signs (+/-) differ for the two analyses? If I was then using principal components PC1 and PC2 as predictors in a regression, i.e. lm(y ~ PC1 + PC2), this would completely change my understanding of the effect of the two variables on y depending on which method I used! How could I then say that PC1 has e.g. a positive effect on y and PC2 has e.g. a negative effect on y?    In addition: If the sign of PCA components is meaningless, is this true for factor analysis (FA) as well? Is it acceptable to flip (reverse) the sign of individual PCA/FA component scores (or of loadings, as a column of loading matrix)?


$$\sum_{n=1}^{\infty} \frac{n}{4^{n+1}}$$  Can someone help me find the sum of the series? I keep getting the sum to be  $ {\frac{1}{12}}$ but the answer is $ {\frac{1}{9}}$.  Very detailed explanation will be greatly appreciated.

How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to , but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.  In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$


I have a string that will be different every time you run the code. I want to make an if statement that compares one char, to every individual character of the string. For example, if I were to compare 'c' to cat, it will return that there is a 'c'. BUT If I compared it to orange, it would return that there is not any 'c'. I can't figure out how to do this because the string will change length every time so I cannot simply do c = a && b && c because there could be a d.

In Java is there a way to check the condition:  "Does this single character appear at all in string x"  without using a loop?


I tried induction but got stuck. For $n=1$ it is true. Suppose it holds for $n$, i.e, the order of 3 is $\phi(7^n)$. Now I should prove it for $\phi(7^{n+1})=6\cdot 7^n$.  $7^n\mid3^{6\cdot 7^n}-1$. How do I derive that $7^{n+1}\mid 3^{6\cdot 7^{n+1}} - 1$. I would appreciate your help.

so I am trying to find out how to prove that 3 is a primitive root of $7^k$ for all $k \ge 1$. I am trying to prove this via induction.  Thanks.


I need to extract the numner of pages of a pdf in c#. Actually I trying a bit with  but there only a method where i can extract the page size (the rectangle) but nothing where i can see the number of pages. Any idea?      static void Main(string[] args)     {         Console.WriteLine("starting");         String filename = "d:\\tmp\\t1.pdf";          if (File.Exists(filename)){             byte[] pdfFile = File.ReadAllBytes(filename);             PdfReader reader = new PdfReader(pdfFile);          }       }

I need to determine the number of pages in a specified PDF file using C# code (.NET 2.0).  The PDF file will be read from the file system, and not from an URL.  Does anyone have any idea on how this could be done?  Note: Adobe Acrobat Reader is installed on the PC where this check will be carried out.


Probably it's kind of stupid question, but how do you come up with this symbol in latex

I would like to reproduce the following symbol:     The  does not seem to refer to that symbol, I did not find any font corresponding in the  by Michael Sharpe My pdf reader tells me that the embedded fonts are      But that is related to the OCR technique and does probably not reflect the font actually used.    was not really conclusive:        and  were of no help. I did not manage to find it in the  nor in the    Does someone have an idea? Do I miss something?  EDIT: in context,   EDIT 2: when I copy / paste that symbol from my pdf viewer, it yields a 2, could that be a 2? It really looks like an L!  That also is probably related to some error of the OCR.


I am new to Android Programming and i am following some tutorial and the tutorial is 1 year ago. i am stuck with this error i think the problem is related to the database but i dont know anything yet about android. PID: 23165     java.lang.NullPointerException: Attempt to invoke virtual method 'com.google.firebase.database.DatabaseReference com.google.firebase.database.DatabaseReference.child(java.lang.String)' on a null object reference         at com.example.fooddeliver.MainActivity.checkUserFromFirebase(MainActivity.java:101)         at com.example.fooddeliver.MainActivity.lambda$init$0$MainActivity(MainActivity.java:83)         at com.example.fooddeliver.-$$Lambda$MainActivity$9ehXip3aW7uFlecdkbtLT68AicY.onAuthStateChanged(Unknown Source:2)         at com.google.firebase.auth.zzj.run(com.google.firebase:firebase-auth@@19.3.1:2)         at android.os.Handler.handleCallback(Handler.java:938)         at android.os.Handler.dispatchMessage(Handler.java:99)         at com.google.android.gms.internal.firebase_auth.zzj.dispatchMessage(com.google.firebase:firebase-auth@@19.3.1:6)         at android.os.Looper.loop(Looper.java:223)         at android.app.ActivityThread.main(ActivityThread.java:7656)         at java.lang.reflect.Method.invoke(Native Method)         at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:592)         at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:947) I/Process: Sending signal. PID: 23165 SIG: 9  my code is this. Update i put my init because i dont know how to initialize userRef sorry for the newbie post     private void init() {         providers = Arrays.asList(new AuthUI.IdpConfig.PhoneBuilder().build());          firebaseAuth = FirebaseAuth.getInstance();          dialog = new SpotsDialog.Builder().setCancelable(false).setContext(this).build();         cloudFunctions = RetrofitCloudClient.getInstance().create(ICloudFunctions.class);         listener = firebaseAuth -> {             FirebaseUser user = firebaseAuth.getCurrentUser();             if(user != null)             {                 checkUserFromFirebase(user);             }             else             {                 phoneLogin();             }         };     }  private void checkUserFromFirebase(FirebaseUser user) {         dialog.show();         userRef.child(user.getUid())                 .addListenerForSingleValueEvent(new ValueEventListener() {                     @Override                     public void onDataChange(@NonNull DataSnapshot snapshot) {                         if(snapshot.exists())                         {                             Toast.makeText(MainActivity.this, "You are Already Registered", Toast.LENGTH_SHORT).show();                              UserModel userModel = snapshot.getValue(UserModel.class);                              goToHomeActivity(userModel);                         }                         else                         {                             showRegisterDialog(user);                         }                          dialog.dismiss();                     }                      @Override                     public void onCancelled(@NonNull DatabaseError error) {                         dialog.dismiss();                         Toast.makeText(MainActivity.this, ""+error.getMessage(), Toast.LENGTH_SHORT).show();                     }                 });     }

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 want to search for unanswered questions having two or more tags.   When searched with /unanswered/tagged/java+and+json, I get list of questions that already have answers.

I've searched for this specific answer with no luck. I want to be able to filter unanswered questions with [tag1] AND [tag2], but it doesn't seem to work for unanswered.  When I use a link like this:   Even though it has a "+", it still treats it as an OR. I want unanswered questions containing both tags. Is this possible? Unclear if this is a possible bug/feature/ or user error.


Just out of curiosity I was wondering, can a person log-in or visit TSE using the  Android app and still it would count as a day visited for these badges?  Or does one have to explicitly visit the webpage only?

If I do any of the below, which would normally constitute "activity" on any Stack Exchange site (including mobile):   View my profile Vote on any question or comment Visit one or more questions   ...then Stack Exchange considers me to have "visited" the site.  If I do any of the below on the mobile app:   Vote on any question or comment Visit one or more questions   ...then Stack Exchange does not consider me to have visited the site.  If one wants the app to have feature parity with the mobile site, then shouldn't my participation within the app also count towards the Enthusiast/Fanatic badge?


In Game of Thrones we see evidence that various religions are able to perform magic.  The followers of the Red God are able to give birth to living shadows and bring people back from the dead.  The religion of the Old Gods is founded on magical "Children" and their all-seeing Godswood trees.  Is there any evidence that the Faith of the Seven was founded on supernatural events?  Or is there any evidence that followers have any magical abilities themselves?

In Game of Thrones, the priests of the Red God and the Old Gods of the Forest exhibit magic powers. Do the devotees of the Seven exhibit similar power?


I have struggle with AWS EC2 ubuntu server ssh connection refuse. After I wrote mount path on /etc/fstab and reboot, I can`t access the instance.  I found your comment at   Can you help me about this.  Thank you

I've replicated this two or three times, so I'm guessing there's something wrong with what I'm doing.   Here are my steps:   Launch new instance via EC2 Management console using: Ubuntu Server 13.10 - ami-ace67f9c (64-bit) Launch with defaults (using my existing key pair) The instance starts. I can SSH to it using Putty or the Mac terminal. Success! I reboot the instance 10 minutes later, when the instance should be back up and running, my terminal connection shows:  stead:~ stead$ ssh -v -i Dropbox/SteadCloud3.pem ubuntu@54.201.200.208 OpenSSH_5.6p1, Op`enSSL 0.9.8y 5 Feb 2013 debug1: Reading configuration data /etc/ssh_config debug1: Applying options for * debug1: Connecting to 54.201.200.208 [54.201.200.208] port 22. debug1: connect to address 54.201.200.208 port 22: Connection refused ssh: connect to host 54.201.200.208 port 22: Connection refused stead:~ stead$    Fine, I understand that the public IP address can change, so checking the EC2 management console, I verify that it is the same. Weird. Just for fun, I try connecting with public DNS hostname: ec2-54-201-200-208.us-west-2.compute.amazonaws.com. No dice, same result.  Even using the Connect via Java SSH client built into the EC2 console, I get Connection Refused.  I checked the security groups. This instance is in group launch-wizard-4. Looking at the inbound configuration for this group, Port 22 is allowed in from 0.0.0.0/0, so that should be anywhere. I know that I'm hitting my instance and this is the right security group, because I can't ping the instance. If I enable ICMP for this security group, all of a sudden my pings go through.  I've found a few other posts around the internet with similar error messages, but most seem to be easily resolved by tweaking the firewall settings. I've tried a few of these, with no luck.  I'm guessing there's a simple EC2 step I'm missing. Thanks for any help you can give, and I'm happy to provide more information or test further!  Update - Here are my system logs from the Amazon EC2 console:


I have a problem with cross-referencing a table in my appendix. Somehow, the table is appearing as "Table I" when it is created, but when I try to cross-reference it later in the appendix, it is being referenced as "Table A" (see code and resulting pdf below). Has anyone else experienced this problem before? Any suggestions would be appreciated. \documentclass[prx,aps,epsfig,twocolumn]{revtex4}  \usepackage[colorlinks, citecolor=blue, linkcolor=blue, linktocpage]{hyperref} \usepackage{graphicx}  % needed for figures \usepackage{dcolumn}   % needed for some tables  \begin{document}  \begin{appendix}  \section{Section Title}  \begin{table}[tbp!] \label{tab:example} \caption{caption here} \begin{tabular}{|c|c|c|} \hline  \textbf{column 1} &  \textbf{column 2} & \textbf{column 3} \\ \hline a & b & c \\ \hline d & e & f \\ \hline g & h & i \\ \hline \end{tabular} \end{table}  Let's try to reference Table \ref{tab:example}. Why does it appear as "Table A" instead of "Table I"?????  \end{appendix}  \end{document}

\documentclass{article} \usepackage{algorithm}  \begin{document} \setcounter{section}{54} \section{A section with no label} \begin{algorithm} \label{myalg} \caption{An algorithm with the label `myalg'.} \end{algorithm} Reference to myalg: \ref{myalg}. \end{document}   With this code, the reference givens 55 (the section number); if I move the label after the caption command, I get the correct reference (1).  Why should this be the case? I find it confusing: when thinking about an algorithm as an environment, you expect to be able to label it right after you began it. At the very least I would expect some kind of error or "??" to appear, rather than the previous environment or reference-able entity being used.


I am trying to write a Python program to verify the law of logarithms which states that log(ab)=loga+logb.When I run the program it runs with certain set of numbers but it does not work always.I have enclosed my code below.Please help me. import math a=float(input("Enter first number:")) b=float(input("Enter second number:")) c=float(input("Enter a base:")) d=round((math.log(a,c)),1) print (d) e=round((math.log(b,c)),1) print(e) f=round((math.log((a*b),c)),1) print(f) if (d+e)==f:     print("We have verified the first property of logarithms")

Consider the following code:   0.1 + 0.2 == 0.3  ->  false   0.1 + 0.2         ->  0.30000000000000004   Why do these inaccuracies happen?


How does one use the value of networking bits of a class A address to find the number of subnet bits and host bits? For example, with a subnet mask of /22, what would be the process of finding the host bits and subnet bits?  *I know a class C address would be networking bits - 24 = subnet bits and 8-subnet bits = host bits. This method doesn't work directly for class A addresses.

This is a  about IPv4 Subnets.      Related:               How does Subnetting Work, and How do you do it by hand or in your head?  Can someone explain both conceptually and with several examples? Server Fault gets lots of subnetting homework questions, so we could use an answer to point them to on Server Fault itself.   If I have a network, how do I figure out how to split it up? If I am given a netmask, how do I know what the network Range is for it? Sometimes there is a slash followed by a number, what is that number? Sometimes there is a subnet mask, but also a wildcard mask, they seem like the same thing but they are different? Someone mentioned something about knowing binary for this?


\documentclass[12pt,twoside]{report} \usepackage[square,sort&compress,comma,numbers]{natbib} \usepackage[sectionbib]{chapterbib} \usepackage{booktabs} \usepackage{amsmath,amssymb} \usepackage{caption} \usepackage{rotating} \usepackage{url} \usepackage{graphics} \usepackage{amsfonts} \usepackage{epsfig} \usepackage{array} \usepackage{multirow} \usepackage{titletoc} \usepackage{float,here} \usepackage{color}  %\usepackage{lipsum} %\usepackage{microtype} %\usepackage[none]{hyphenat}   \setlength{\oddsidemargin}{0.1 cm} \setlength{\evensidemargin}{-0.28cm} \setlength{\topmargin}{-0.65 cm } \setlength{\textwidth}{6.4in} \setlength{\textheight}{9.4in}  \input{rgb} \pagestyle{myheadings}   \def\fps@figure{htbp} \def\fps@table{htbp}  \renewcommand{\textfraction}{0.5} \renewcommand{\floatpagefraction}{0.4} \newcommand{\FH}{{\textsc{\tiny FH}}}  \floatsep 36pt \textfloatsep 36pt \intextsep 36pt   \def\vitalsecdepth#1{\setcounter{secnumdepth}{#1}    \setcounter{tocdepth}{#1}} \vitalsecdepth{3} % 5/13/92 kelliher  \sloppy   \begin{document}   %\include{first-page} %\include{statem} %\include{certi1} %\include{acknow2}  \clearpage \baselineskip=20pt %\include{TOC}  \pagenumbering{roman}  \renewcommand{\contentsname}{Table of Contents}  \newcommand{\setupname}[1][\chaptername]{ \titlecontents{chapter}[0pt]{\vspace{1ex}}{\bfseries#1~\thecontentslabel:\quad}{\bfseries}{\bfseries\hfill\contentspage}[] }  \tableofcontents \listoffigures \listoftables  \pagenumbering{roman}  \clearpage \pagenumbering{arabic}  %\chapter{Introduction} \include{chap1} \include{chap2} \include{chap3} \include{chap4} \include{chap5} \include{chap6} \include{chap7} \clearpage \baselineskip=20pt \pagenumbering{roman} %\include{reference}   \end{document}

At here  the word Chapter added in \documentclass{book}. Now I want to use them in \documentclass{report}. How can I do that?


I created a dictionary of objects. I'm trying to update the class variable (numbers) by removing an element from it. The problem is that it updates every class variable of the remaining objects. If I update the variable with a single integer it does fine. Does anyone have an idea why it happens and how could I fix it?  class Matrix():     numbers = set(range(1,10))     def __init__(self):         pass  def gen():     entries = set(range(1,10))     integer = random.randint(1,9)      dict = {}     for i in range(0,10):         dict['A{0}'.format(i)] = Matrix()      dict['A0'].numbers.remove(integer)  gen()

Is there any meaningful distinction between:  class A(object):     foo = 5   # some default value   vs.  class B(object):     def __init__(self, foo=5):         self.foo = foo   If you're creating a lot of instances, is there any difference in performance or space requirements for the two styles? When you read the code, do you consider the meaning of the two styles to be significantly different?


There is a problem when I set up my server system ,IP would be different when I use Wired network connection and wireless network connection，how can I set unchanging IP when I connect to the system? Thanks for your help.

I am a new with Linux, having years experience with Windows servers/desktops and am having issues setting a static IP. I am using a method used for previous versions of Ubuntu, which doesn't seem to work with 16.04  I have used the command sudo nano /etc/network/interface and added the following  iface enp0s25 inet static address 10.10.8.2 netmask 255.255.0.0 gateway 10.10.1.1 dns-nameservers 8.8.8.8 8.8.4.4   I have rebooted the system and the Ethernet is pretty much dead, ping doesn't work at all. I have tried to modify /etc/NetworkManager/NetworkManager.conf and made the following changes  #dns=dnsmasq (comment out the dnsmasq) [ifupdown] managed=true (changed from false)   With this I can get Ethernet to work sporadically, however it eventually fails.  I have tried this configuration on two other machines plus a virtual machine as well and all have the same results. I can confirm these settings work fine when I install Windows on any of these machines. As well when I let DHCP auto configure, everything works fine no issues.  I figure I am missing something here, setting up a static IP should not be difficult at all.


I am trying to delete some elements from an std::map. In particular the elements that have      value = 100   should be deleted. However after running the following piece of code a bug occurs. An element with value 100 appears in the remaining map, after the erase(it) method. Where is the bug?   Code:  #include <iostream> #include <map>  int main() {     std::map<int, int> outcomes;     outcomes.clear();      if ( !outcomes.insert( std::make_pair( 10, 1 ) ).second )          return 0;     if ( !outcomes.insert( std::make_pair( 11, 100 ) ).second )          return 0;     if ( !outcomes.insert( std::make_pair( 12, 100 ) ).second )          return 0;     if ( !outcomes.insert( std::make_pair( 13, 100 ) ).second )          return 0;     if ( !outcomes.insert( std::make_pair( 14, 100 ) ).second )          return 0;     if ( !outcomes.insert( std::make_pair( 15, 101 ) ).second )          return 0;     if ( !outcomes.insert( std::make_pair( 16, 101 ) ).second )         return 0;     if ( !outcomes.insert( std::make_pair( 17, 100 ) ).second )         return 0;      std::map<int, int>::iterator it = outcomes.begin();     while(it != outcomes.end())     {         std::cout<<it->first<<" :: "<<it->second<<std::endl;         it++;     }      it = outcomes.begin();     while(it != outcomes.end())     {         if( it->second == 100 )             outcomes.erase(it);         it++;     }     std::cout << "after delete" << std::endl;     it = outcomes.begin();     while(it != outcomes.end())     {         std::cout<<it->first<<" :: "<<it->second<<std::endl;         it++;     }     return 0; }     Result is the following:  10 :: 1 11 :: 100 12 :: 100 13 :: 100 14 :: 100 15 :: 101 16 :: 101 17 :: 100 after delete 10 :: 1 14 :: 100 15 :: 101 16 :: 101

In the following code I loop through a map and test if an element needs to be erased.  Is it safe to erase the element and keep iterating or do I need to collect the keys in another container and do a second loop to call the erase()?  map<string, SerialdMsg::SerialFunction_t>::iterator pm_it; for (pm_it = port_map.begin(); pm_it != port_map.end(); pm_it++) {     if (pm_it->second == delete_this_id) {         port_map.erase(pm_it->first);     } }   UPDATE: Of course, I then  which I didn't think would be related but answers my question.


I am new to Python and I am struggling a bit with regular expressions. If I have an input like this:      text = <tag>xyz</tag>\n<tag>abc</tag>   Is it possible to get an output list with elements like:      matches = ['<tag>xyz</tag>','<tag>abc</tag>]   Right now I am using the following regex      matches = re.findall(r"<tag>[\w\W]*</tag>", text)   But instead of a list with two elements I am getting only one element with the whole input string like:      matches = ['<tag>xyz</tag>\n<tag>abc</tag>']   Could someone please guide me? Thank you.

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 remember seeing red messages like these when I look at my list of questions:  Have you considered accepting an answer or starting a bounty for this question?  But I have not seen these (annoying) reminders recently. Did I gain a new privilege that would explain why I no longer see these red reminders when I look at my list of questions?

Sometimes, I notice the following message on the Questions tab in my profile on Stack Exchange site:  Have you considered accepting an answer or starting a bounty for this question?  When does this message get displayed? Is it based on time of non-availability of answers or inactivity period?


question clearly explains about how to find all the commands which are provided by a particular installed package.  My question is, how to find(or lists) all the available commands which are provided by a particular package before installing that package via terminal ?

I would like to know what some specific package installs, for example, when installing ncurses, I have found that TAB key expands:  sudo apt-get install ncurses-   to show:  ncurses-base      ncurses-doc       ncurses-hexedit ncurses-bin       ncurses-examples  ncurses-term   How could I know what, say, ncurses-term installs? I am interested mainly in programs, but libraries and any other filetypes could be useful too.  Command-line method, if possible, would be preferred (any others accepted too).


I bought Garry's Mod, and it's currently a 2 gb download. I have poor wifi, so I want to know if I can transfer Gmod from my friend's account to mine. He has Gmod on Steam as well. Is it possible with the backup and restore method and a pen drive?

I use (and have it much bigger) drive D: to store all my user data and games, but I have installed the Steam client on drive C:. When I was installing a first game (Warhammer 40000), I've chosen to place my library folder on drive D: and the game went there.  But now as I've got installed the second game (Portal), I can see that it was put on drive C: despite to the choice I've done previously. When I open Steam - Settings - Downloads + Cloud - Steam library folders, I can see both installation places in the list. Can I merge them into one?  I would even agree to use the default drive C: library (as I've replaced it with a symlink already). When I press to remove one library it says it can only remove empty ones.


This might be a very simple question. I read one previous post    I know that in Cartesian coordinates, the kinetic energy $T=\frac{1}{2}mv^2$. And $T$ is not an explicit function of position. So $\frac{\partial T}{\partial x}=0$, where we suppose $x$ is a coordinate.  But I got confused by one example, we have a ball move vertically from the origin O, with a velocity of $\vec V_0$.    then when the ball reach $y_1$ in the positive $y$ axis, we have  $$mgy=\frac{1}{2}mv_0^2-\frac{1}{2}mv_1^2$$ so the velocity at y is  $$v_1=\sqrt {\frac{1}{2}mv_0^2-mgy}$$ Does it means that velocity and kinetic energy both are explicit function of position y in this case? I know this is a special case, but the statement that $\frac{\partial T}{\partial x}=0$ in Cartesian coordinates seems to be quite general. So where have I missed so far?  Thanks guys!

,I got one confusion when reading Goldstein's Classical Mechanics (page 20, third edition). After getting the equation $$ \sum \left\{\left[\frac{\mathrm{d}}{\mathrm{d}t}{\left(\frac{\partial T}{\partial \dot{q_{j}}} \right)} - \frac{\partial T}{\partial q_{j} } \right] - Q_{j} \right\} \ \delta q_{j} = 0 $$ then it says that     Note that in a system of Cartesian coordinates the partial derivative of $T$ with respect to $q_{j}$ vanishes. Thus, speaking in the language of differential geometry, this term arises from the curvature of the coordinates $q_{j}$. In polar coordinates, e.g., it is in the partial derivative of $T$ with respect to an angle coordinate that the centripetal acceleration term appears.   My question is:  Is the above statement general, i.e., that the kinetic energy $T$ does not depend on the position. I wonder why velocity can't depend on the particle's position vector. I mean, why couldn't we have cases where $\vec{v} = \vec{v}(\vec r ,t)$, so that the kinetic energy depends on $q_{j}$ or $\vec{r}$?    I will use the example of the second answer in this post   the second answer,   suppose we have a velocity distribution of gas whose function is $\vec {v}=\vec {v}(x,y,z,t)$  then the partial derivative with respect to $x$ is  $$\frac{\partial T} {\partial x}=mv\frac{\partial v} {\partial x}$$  since $v$ depends on $x$, I think in this case. $\frac{\partial T}{\partial q_{j} }$ does not vanish.


Does there exist a continuous function $f$ such that $f(x^2)=2f(x)$ and $f(0)=0$? How do you solve this? I understand that this is nothing like a normal equation, because you can't solve for $x$, or $f(x)$, because of that $f(x^2)

I ran into a problem recently where I obtained the following constraint on a function. $$f(x^2) = 2f(x) \,\,\,\, \forall x \geq 0$$ and the function $f(x)$ is continuous. Can we conclude that $f(x)$ is of following form? $$f(x) = \begin{cases} a\log(x) & \text{for } x \geq 1\\ 0 & \text{for  }x \in [0,1]\end{cases} \tag{$\star$}$$ where $a \in \mathbb{R}$. It is easy to conclude for $x \in [0,1)$ it should be zero, since $f(0) = 0$ and $f(x) = \dfrac{f(x^{2^n})}{2^n}$. Now letting $n \to \infty$, thanks to continuity, we can conclude that $f(x) = 0$ for $x \in [0,1)$.  But how do I show $(\star)$ for $x \geq 1$? If not under what further constraints, will I be able to conclude ($\star$)? This is similar to the Cauchy functional equation $f(xy) = f(x) + f(y)$, but with $y=x$.


Where may I get an example of blender python script? Currently, I'm working on doing a whole python script from scratch to import, operate, export models. And I'm stuck in some issues.

Since asking for tutorials, videos, resources, links to other content or downloads is considered off topic here, this is a specifically created Community Wiki which gathers resources for Blender and it has been approved by the .  Just write in the appropriate answer/section. If you have concerns, questions, post a  question, so we don't clutter the comments, but you can link your meta question from the comments.  Questions regarding such resources are not allowed anymore, except for very specific and on topic requests (ask on  if you're unsure about your question). Follow the instructions made in the question about how to post, what can be posted, etc.  Learning             Assets    (also see: )   Reference Images/Blueprints (also see: )    Tools    (also see: )      Other   Special Interest ()     Also see the .


I read the following script in a TOEIC Speaking book:  “Hi, Mohammed. This is Sam. I just got your voicemail about the printing job. I would rather go with the original printer. Maybe you can ask them to print twenty copies first and make sure they deliver them on time, before you have a meeting with a client, and then they can finish up the rest of the copies later… after you have a meeting.”  I think the bold-faced sentence can be analyzed into the following:   You can ask them to print twenty copies first and (you can ask them to) make sure they deliver them on time. You can ask them to print twenty copies first and (you can) make sure they deliver them on time.   Which one do you think is correct? #1? #2? Thanks in advance.

I'm wondering the subject of the second clause in a sentence like   You should tell him to get up and get back to work   The subject of get back to work is ambiguous to me. It could be interpreted as either  You should tell him to get up and [you] get back to work   or  You should tell him to get up and [him] get back to work   How should I form the sentence so that either option is explicit and unambiguous?


I have some captions created with the command \DeclareCaptionType[name=Gráfico]{grafico}  and I want wrapfigure to use this caption. How can this be done?

I which to have a wrapped figure in a custom float environment: the caption should be on top and carry the name "Infographic". Unfortunately I don't know how to do this.   This is what I came up with  \documentclass{article}  \usepackage{float} \usepackage{wrapfig}  \floatstyle{plain} \newfloat{infographic}{H}{info} \floatname{infographic}{Infographic}  \begin{document}  \begin{wrapfigure}{r}{.4\textwidth} \begin{infographic} \centering \includegraphics[width=.4\textwidth]{test} \end{infographic} \end{wrapfigure}  \end{document}   but the newfloat completely messes up the wrapfigure.


Were know that three of the Nazgul are of Numenorean origin, Khamul was an Easterling, the origin of the other 5 are unknown.  Considering the black Numenoreans rebelled against the faithful, is it more likely that the three Nazgul of Numenorean origin are probably black Numeanoreans from the southern regions of middle earth?

I recall reading somewhere of the history of the nine riders, but I cannot remember where it was. (It seems perhaps they were from northern kingdoms?)  The leader of the Nine was the Witch King of Angmar, but what were the names of the others, and where did they come from?


Up graded from 13.10 to 14.04 using pop up upgrade recommendation. No problems during upgrade. After restarting and logging in, none of the enu bars are visible. Displays desktop image and mouse can be seen, but unable to open anything or even shut it down properly. Any idea what might have happened and/or how to fix it?

When I login, nothing happens.   I am presented with my desktop wallpaper.    No Dash, no Launcher, nothing.


I am stuck with the following recurrence relations (actually asked in a programming question).Given the following relationships, is it possible to find the index of the occurrence of any given number which is in the series?  $f(0) = 1$ $f(1) = 1$ $f(2) = 2$ $f(2n) = f(n) + f(n + 1) + n$          (for $n > 1$) $f(2n + 1) = f(n - 1) + f(n) + 1$      (for $n >= 1$)  So the series is like:(Index: 0,1,2,3...n) $1,1,2,3,7,4,13,6,15,11,22,12,25,18,28,20,34...$    Example: Given: $f(n) = 8074$ Result: $n = 2441$    Also, some other condition given are: A number may occur more than once - the resultant index should be the last time it occurs. An example: $f(14812) = 65389$ Also, $f(16623) = 65389$ So, the result is $n = 16623$  By programming in python, I was able to observe that any number that repeats occurs utmost twice. Also, since the input can be pretty large - simple recursion (with hashing) takes too long and mostly causes memory errors. So, I tried to reduce the first recurrence formula by substituting the second in it, but still I was not able to find any proper relation between $f(n)$ and $f(2n)$ or $f(n-1)$ since only $f(n)$ is given as input. Is it possible to solve this using any other relations? (Edit: I've also posted on StackOverflow )

For the recurrence relation:  $f(0)=1$  $f(1)=1$  $f(2)=2$  $f(2n)=f(n)+f(n+1)+n\ \ \ (\forall n>1)$  $f(2n+1)=f(n)+f(n-1)+1\ \ \ (\forall n\ge 1)$  (the first numbers of the sequence are: 1, 1, 2, 3, 7, 4, 13, 6, 15, 11, 22, 12, 25, 18, 28)  Given some number $S$, I would like to find $n$ such that $f(n) = S$, using an approach that runs in $O(\log n)$ time. Any ideas or tips how to solve it? I tried substitution so far but didn't get far.


Since the neutral wire and ground wire are connected in the main fuse box then why doesn't some of the returning electricity flow through the ground wire to earth?

If neutral is connected to ground and current desires ground then I would think it would go to ground.  But my understanding is that it is kept off of ground until it's "upstream" of the breaker box, or unless a malfunctioning device shorts hot to ground.  So what "prevents" the electricity from going to the ground that the neutral is connected to?


I should draw this:    This is what I have done so far.    I do not know how to draw the horizontal arrows. Besides, in order to draw the dashed line from point Q to point P, I wrote  \draw [fill] (0.3,2) circle (1.5pt) node [above] (Q) {Q}; \draw [fill] (1.1,-1.6) circle (1.5pt) node [left] (P) {P}; \draw [thick,densely dashed] (0.3,2) -- (1.1,-1.6);   Is there a faster way, something like  \draw [thick,densely dashed] (Q) -- (P);   ? This is my code:  \documentclass[border=3cm]{standalone} \usepackage{tikz}  \begin{document} \begin{tikzpicture} % \draw [help lines] (0,0) grid (7,7); \draw plot [mark=none, smooth] coordinates {(1.16,-2.4) (1.6,-0.8) (2,1.6) (2.04,2.8)}; \draw [thick,-latex] (0,0) -- (5,0.0) node (xaxis)[below] {$x$}; \draw [thick,-latex] (0,-4) -- (0,4) node (yaxis)[left]  {$y$}; \node at (2.5,2.6) {$U(y)$}; \draw [thick,latex-latex] (3,1) -- (3,-1) node [midway,xshift=0.4cm,yshift=0.3cm]  {$2l_{m}$}; \draw [thick,-latex] (1.3,-0.2) -- (2.2,-0.2) node [below]  {$\tau_{xy}$}; \draw [thick,-latex] (0.5,0.2) -- (-0.5,0.2) node [above]  {$\tau_{xy}$}; \draw [fill] (0.3,2) circle (1.5pt) node [above] (Q) {Q}; \draw [fill] (1.1,-1.6) circle (1.5pt) node [left] (P) {P}; \draw [thick,densely dashed] (0.3,2) -- (1.1,-1.6); \end{tikzpicture} \end{document}   Edit: thanks to  and  I managed to draw the arrows. Here is the code and the output:  \documentclass[border=3cm]{standalone} \usepackage{tikz} \usetikzlibrary{intersections}  \begin{document} \begin{tikzpicture} % \draw [help lines] (0,0) grid (7,7); \draw [thick, name path=curve] (1.16,-2.4) .. controls (1.6, -0.8) and (2, 1.6) .. (2.04, 2.8); \draw [thick,-latex] (0,0) -- (5,0.0) node (xaxis)[below] {$x$}; \draw [thick,-latex] (0,-4) -- (0,4) node (yaxis)[left]  {$y$}; \node at (2.5,2.6) {$U(y)$}; \draw [thick,latex-latex] (3,1) -- (3,-1) node [midway,xshift=0.4cm,yshift=0.3cm]  {$2l_{m}$}; \draw [thick,-latex] (1.3,-0.2) -- (2.2,-0.2) node [below]  {$\tau_{xy}$}; \draw [thick,-latex] (0.5,0.2) -- (-0.5,0.2) node [above]  {$\tau_{xy}$}; \coordinate (Q) at (0.3,2); \coordinate (P) at (1.1,-1.6); \fill (Q) circle (1.5pt) node [above] {Q}; \fill (P) circle (1.5pt) node [below,xshift=-1mm] {P}; \draw [thick,densely dashed] (Q)--(P); % \foreach \y [count=\yi] in {-2,-1.6,...,2}{%     \path[name path=a\yi] (0,\y) --++ (3,0);     \draw[name intersections={of=curve and a\yi},semithick,-latex] (0,\y) -- (intersection-1); } \end{tikzpicture} \end{document}

Is there a better way to draw an arrow until it reaches the curve? I have just hard coded the numbers, but was wondering if there was a way to label the curve then draw until the curve.   \begin{figure}[h!]     \centering     \begin{tikzpicture}         \draw[thick, ->] (-0.5, 0) -- (3.5, 0) node[anchor=west] {$x$};         \draw[thick, ->] (0, -0.5) -- (0, 1) node[anchor=south] {$y$};         \fill (0, 0) circle (0.7mm) node[anchor=north east] {$O$};         \draw[thick] (0, 0) .. controls (1, 1) and (2, 1) .. (3, 0);         \draw[thick, ->] (0.5, 0) -- (0.5, 0.35);         \draw[thick, ->] (1, 0) -- (1, 0.6);         \draw[thick, ->] (1.5, 0) -- (1.5, 0.7);         \draw[thick, ->] (2, 0) -- (2, 0.6);         \draw[thick, ->] (2.5, 0) -- (2.5, 0.35);     \end{tikzpicture} \end{figure}   produces the image


The official ubuntu 18.04 repo stops at 2.8  On the official download page, they suggest using flatpak, although non better specified "limitations".  There is an unofficial PPA provided by a GIMP's developer (Otto Kesselgulasch)  What's your advice about it?

I understand that there is a new version of the GIMP available, although it appears that the only way of doing this is by compiling from source :P  Is there another way of getting it, or if I have to compile can I do that through the terminal or better still; a GUI?


How can my 6 TB hdd display as full while there's 3,4 TB of files on there? It says 0 bytes available for some reason.

When administering Linux systems I often find myself struggling to track down the culprit after a partition goes full. I normally use du / | sort -nr but on a large filesystem this takes a long time before any results are returned.  Also, this is usually successful in highlighting the worst offender but I've often found myself resorting to du without the sort in more subtle cases and then had to trawl through the output.  I'd prefer a command line solution which relies on standard Linux commands since I have to administer quite a few systems and installing new software is a hassle (especially when out of disk space!)


I want to create a bash script or something similar that I can execute in the current shell:   echo '#!/bin/bash    export foo="bar" ' > zoom.sh   but if I want to run this script in the current process:  exec ./zoom.sh   then my terminal will just exit "process completed", even if I use set +e, it still exits.  There's source and eval which will run stuff in the current shell, so maybe eval is my best bet?  A bash function will work - but I have to rely on the user to source it in the first place.

There are several ways to execute a script. The ones I know are: /path/to/script # using the path (absolute or relative) . script        # using the . (dot) source script   # using the `source` command  Are there any other way? What are the differences between them? Are there situations that I must use one and not another?


Does bivariate normal distribution mean the two random variables have normal distributions? is that enough for two random variables to have a bivariate normal distribution or are there some other conditions that must be true?

Somebody asked me this question in a job interview and I replied that their joint distribution is always Gaussian. I thought that I can always write a bivariate Gaussian with their means and variance and covariances. I am wondering if there can be a case for which the joint probability of two Gaussians is not Gaussian?


The question  has awarded the user the  badge.  Yet - it was closed for being too specific - and hence supposedly not generally useful. Somehow that doesn't make sense to me; why are questions closed for this reason counted for badges? At least, for that kind of badge?  (This not a duplicate since I'm asking about a specific reason for closing rather than in general.)

A while ago . Today it hit 1,000 pageviews, awarding me the  out of the blue.   Is this intended behavior; should badges still be awarded for activity on questions after they are closed? It sends a confusing mixed message — "this question is off topic but you get a cookie for it" — although I can see how we might not want to penalize people out of the achievement system just because a question was closed.  I'm not concerned about the closure itself (wasn't getting the answer I needed anyway). I'm just making sure that this is by design and not a bug.


Code long double sumArray = 0; for (int i = 0; i < input; i++) {     sumArray = sumArray + array[i]; }  cout << sumArray;  Output should look like this 5 1000000001 1000000002 1000000003 1000000004 1000000005 5000000015  Instead, it showing this 5 1000000001 1000000002 1000000003 1000000004 1000000005 5e+09  I want to get the sum of all long double variable like this.  5000000015

In my  I was printing a double using cout that got rounded when I wasn't expecting it.  How can I make cout print a double using full precision?


How can I remove this picture:      I've tried  without any luck.

Windows 7 has a large icon above the username on the right-hand pane of the Start menu. By default it is a faceless torso in a blue collarless shirt that I call the "zombie icon". (I don't know the icon's real name: . (Although "Molly Clark" has a zinnia icon instead of a zombie icon.)  The zombie icon fades from one pointless icon to another as you move your mouse over "Documents", "Computer", "Devices and Printers" etc. This icon is a waste of space and the fades are distracting.  How do I make it go away?


Let $A,B$ be square real matrices of the same size such that $A+B=AB$. Show that $AB=BA$.   I have got the result for invertible matrices, but what about non-invertible ones?

I was doing the problem $$ A+B=AB\implies AB=BA. $$  $AB=BA$ means they're invertible, but I can't figure out how to show that $A+B=AB$ implies invertibility.


I was solving an exercise and I came across this one $a_n = \frac{n}{6^n}$ well obviously we can see that this one converges to 0 but could someone explain it to me or even prove it?

I think should be something very simple... $$\lim_{n\to\infty}nq^{n}$$ wher $|q|<1$ i tried to use binomial theorem, but no success... I know that $q^{n}$ is convergent to $0$, but can't deal with this $n$


I recently came across this integral and found the following solution:  I don't understand the "By symmetry we can write" argument. Can you explain this in a little more detail and is this a sufficient argument? I solved it like this, but I'm not completely sure: $$\int_0^1\frac{\ln(x+1)}{x^2+1}\,dx=\text{?}$$ We apply the classical Feynman method, consider the function $I(y)=\displaystyle\int_{0}^{1}\frac{\ln(yx+1)}{x^2+1}\,dx$, our integral is equal to $I(1)$. \begin{align} \frac{d}{dy}I(y)= {} & \int_0^1 \frac{d}{dy} \left( \frac{\ln(yx+1)}{x^2+1} \right)\,dx \\[8pt] = {} & \int_0^1 \frac{x}{(yx+1)(x^2+1)} \,dx \\[8pt] = {} & \frac{1}{y} \int_0^1 \frac{yx+1-1}{(yx+1)(x^2+1)} \,dx \\[8pt] = {} & \frac{1}{y} \int_0^1\frac{dx}{x^2+1} \\[2pt] & {} - \frac{1}{y} \int_0^1 \frac{1}{(yx+1)(x^2+1)} \,dx \\[8pt] = {} & \frac{1}{y} \cdot \frac{\pi}{4}-\frac{1}{y} \int_0^1 \frac{1}{(yx+1)(x^2+1)} \,dx \end{align} $\displaystyle\int_{0}^{1}\frac{1}{(yx+1)(x^2+1)} \,dx=\displaystyle\int_{0}^{1}\frac{-\frac{y}{y^2+1}\cdot x+\frac{1}{y^2+1}}{x^2+1}\,dx+\frac{y^2}{y^2+1} \cdot\displaystyle\int_{0}^{1}\frac{1}{yx+1}\,dx=-\frac{y}{y^2+1}\cdot\displaystyle\int_{0}^{1}\frac{x \cdot dx}{x^2+1}+\frac{1}{y^2+1}$ $\displaystyle\int_{0}^{1}\frac{dx}{x^2+1}+\frac{y}{y^2+1} \cdot \displaystyle\int_{0}^{1}\frac{1}{1+\frac{1}{y}} \,dx=-\frac{\ln{2}}{2}\cdot\frac{y}{y^2+1}+\frac{\pi}{4}\cdot\frac{1}{y^2+1}+\frac{y}{y^2+1}\cdot \ln(y+1)$ $\frac{d}{dy}I(y)=\frac{1}{y}\cdot \frac{\pi}{4}+\frac{\ln{2}}{2}\cdot\frac{1}{y^2+1}-\frac{\pi}{4}\cdot\frac{1}{y(y^2+1)}-\frac{1}{y^2+1}\cdot \ln(y+1)=\frac{1}{y}\cdot \frac{\pi}{4}+\frac{\ln{2}}{2}\cdot\frac{1}{(y^2+1)}-\frac{\pi}{8}\cdot\frac{2y}{y^2(y^2+1)}-\frac{1}{y^2+1}\cdot\ln(y+1)=\displaystyle\int_{0}^{1}\frac{d}{dy}I(y)=I(1)(I(0)=0)=\frac{\pi}{4}\cdot \ln{y}+\ln{2}\cdot\frac{\pi}{8}-\frac{\pi}{8}\cdot(2\ln{y}-\ln(y^2+1))-I(1)=\ln{2}\cdot\frac{\pi}{8}+\frac{\pi}{8}\cdot \ln{2}-I(1)$ $I(1)=\frac{1}{2}\cdot 2 \cdot \frac{\pi}{8} \cdot \ln{2} = \frac{\pi}{8} \cdot \ln{2}$


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What is wrong with the following “proof” by mathematical induction that all cats are black?  Let P(n) denote the statement  “In any group of n cats, if one cat is black, then they are all black.”   Step 1 The statement is clearly true for n = 1.  Step 2 Suppose that P(k) is true. We show that P(k+1) is true.    Suppose we have a group of k+1 cats, one of whom is black; call this cat “Tadpole.” Remove some other cat (call it “Sparky”) from the group. We are left with k cats, one of whom (Tadpole) is black, so by the induction hypothesis, all k of these are black. Now put Sparky back in the group and take out Tadpole. We again have a group of k cats, all of whom—except possibly Sparky—are black. Then by the induction hypothesis, Sparky must be black too. So all k+1 cats in the original group are black. Thus by induction p(n) is true for all n. Since everyone has seen at least one black cat, it follows that all cats are black.

I'm bugged by the following that's summarized on p. 109 of .     False theorem: All horses are the same color.      Proof by induction:      $\fbox{$P(n)$ is the statement: In every set of horses of size $n$, all $n$ horses are the same color.}$   $\fbox{Base Case or $P(1)$:}$ One horse is the same color as itself. This is true by inspection.   $\fbox{Induction Step:}$ Assume $P(k)$ for some $k \geq 1$.   $\fbox{Proof of $P(k + 1) :$}$   Since $\{H_1, H_2, ... , H_n\}$ is a set of $n$ horses, the induction hypothesis applies to this set. Thus, all the horses in this set are the same color.   Since $\{H_2, H_3, ... , H_{n+1}\}$ is also a set of $n$ horses, the induction step likewise holds for this set. Thus, all the horses in this set are the same color too.         Therefore, all $n +1$ horses in $\{H_1, H_2, H_3, ... , H_n , H_{n+1}\}$ are the same color. QED.   The issue the instructor was trying to point out is clearly valid. For the case $n = 1$, there is only ${H_1}$. So this case says nothing about possible overlapping elements of each set of $(n + 1)$, for instance $H_2$ in the above proof.    But it was proposed in the class discussion that this was the only problem. Had you been able to prove $P(2)$ true, then a proof of the above format would have been fine.    My interpretation is that yes, you could prove all horses are the same color, if you can prove that any set of two horses will be the same color. But this format would not work.  Why not? The problem I see is that the above proof is for the existence of at least one particular pair of sets of horses of sizes $n$ and $n + 1$, such that in each set, all horses are the same color.  Particularly when the set of size $n$ is a subset of the set of size $n + 1$.  In order to prove the induction step, don't you need to prove that sets of sizes $n$ and $n + 1$, do not necessarily contain the same, overlapping elements?  You could prove that any horse can be added to a set of 2 horses.  Take the last two, and they must be the same color, and so on.  Wwhatever color the first two happen to be, all other horses must thus be the same color.    Am I misinterpreting the example, or am I making a logical error? Thanks in advance.


What does actually occur- ATP is hydrolysed and esterification between glucose and HPO4^2- (formed as a result of hydrolysis) take place to form Glucose-6-phosphate or nucleophilic attack of O- of glucose on ATP leads to formation of ADP and Glucose-6-phosphate?

The oxygen anion in the phosphate group is likely to be bonded with a hydrogen cation and thus forming a hydroxyl.  When the phosphate group is condensed with glycerol to make a phospholipid, is the hydroxyl from the glycerol molecule taken or the hydroxyl from the phosphate group taken?  I believe it's related to the difference of electronegativity of oxygen and phosphorus which causes unequal sharing of electrons between atoms, thus making a hydroxyl from one side easier to get taken off than the hydroxyl from the other side is.


Steps to reproduce:   Filter on tags Type a tag Select a tag Choose a question Tap back in menu bar   This takes me back to the tag filtering page instead of the filtered question page.   App Version: 1.0.89 Device Manufacturer: HUAWEI Device Model: HUAWEI P7-L10   OS Version: 5.1.1 (C900B852)

This looks a little bit like a variation on  but specific to the tags view.  Reproduction (app version 1.0.71):   Select any site (say Meta) from the sidebar Switch to the tags view using the top bar Choose a tag, any tag ( is nice)   optionaly refine your search with more terms  Go to any question found, go back   you end on the unfiltered tags view  Go back again   lucky you, your search is here


I tried working from both sides: $| |x| + |y|| = | |x| + |x| - |x| + |y|| = | 2|x| + |y| - |x|| \le |2x| + | |y| - |x| |$ $\Longrightarrow||x| + |y|| = |x| + |y| \le |2x| + | |y| - |x| |$ Then, $|2x| = |x - y + x + y| \le |x-y| + |x+y| $ Can someone help me from this point on?

How would I go about proving that  $$|a|+|b| \leq |a+b|+|a-b|$$  Using the triangle inequality?  I tried squaring both sides, yielding:  $$|a|^2 +|b|^2 +2|a||b| \leq 2|a|^2+2|b|^2+2||a|^2-|b|^2|$$  Is it correct to move the terms to the other side? I tried $$2|a||b| \leq |a|^2+|b|^2+2||a|^2-|b|^2|$$ Then it is obvious that  $$0\leq |a|^2+|b|^2-2|a||b| +2||a|^2-|b|^2|$$ $$0\leq (|a|-|b|)^2 +2||a|^2-|b|^2|$$ $$0\leq ||a|-|b||^2 +2||a|^2-|b|^2|$$  But I have neither used the triangle inequality, nor it looks mathematically rigorous for my real analysis class. Any tips and tricks for solving these problems?


Let $A = 4444^{4444}$;  Then sum of digits of $A = B$; Then sum of digits of $B = C$; Then sum of digits of $C = D$;  Find $D$.  What should be the approach here?


Can I have another function to return the last timestamp. I want to compare with this query to use the best query. My query: CREATE OR REPLACE FUNCTION f_last_update1(nomTable text , OUT updated timestamp)   LANGUAGE plpgsql AS $func$ BEGIN    EXECUTE 'SELECT max(updated) FROM ' || nomTable    INTO updated;    END $func$;

I want to get last timestmap from a table with the column updated in PostgreSQL. And I want a function to do that. So I can pass the table name as parameter and the function returns the latest timestamp in column updated. My table:  Query used: select max(updated) from mytable;


I'm trying to make a simple animation: ice cubes falling into a glass. I modeled my glass, my cube and applied Rigid body behavior to them (Active/Passive). But for some reason (probably me being stupid), the cube is stopped in its fall by a non-existing face at the top of the glass. I tried to switch the collision shape (tried pretty mush all the possibilities) and it didn't solve the issue.  What am I doing wrong here? Thanks in advance !

When i apply rigid body modifier after punching a hole in a cube for inserting a rotatable propeller so that it rotates only when something collides with it.    But when i run the simulation the propeller shaft pops out of the hole and the cube behaves like it had no hole in it, even after after selecting "mesh" for the shape of physics body.


What is a simplified example with xmodmap to have AltGr similar to windows?  clear Mod1 keysym Alt_R = Alt_R add Mod1 = Alt_L Alt_R keycode 56 = x X x X f F   I assumed this would work but it does not.

I want to remap my AltGR+Right_Windows but I can't figure out how. No matter what I try, pressing the right windows key emits the name keysym when right alt is pressed or not.  These are the contents of the xmodmap file I am testing right now. In my keyboard, 38 is the keycode for "a", 48 is the keycode for aportrophe/doublequote and 134 is the keycode for the right windows key.  keycode  38 = 1 2 3 4 5 6 7 8 keycode  48 = 1 2 3 4 5 6 7 8 keycode 134 = 1 2 3 4 5 6 7 8   When I press Right_Alt+apostrophe, it outputs 5 but Right_Alt+a and Right_Alt+Right_Windows give 1 (both when right alt is pressed and when it isn't) which is not what I wanted.  My current keyboard layout is us(alt-intl) and the right alt is apparently set up to act as AltGr / Iso_Level3_Shift. The output of xmodmap -pm is the following:  xmodmap:  up to 4 keys per modifier, (keycodes in parentheses):  shift       Shift_L (0x32),  Shift_R (0x3e) lock        Caps_Lock (0x42) control     Control_L (0x25),  Control_R (0x69) mod1        Alt_L (0x40),  Meta_L (0xcd) mod2        Num_Lock (0x4d) mod3       mod4        Super_L (0x85),  Super_R (0x86),  Super_L (0xce),  Hyper_L (0xcf) mod5        ISO_Level3_Shift (0x5c),  Mode_switch (0xcb)


I do not know of any special meanings for the hash character (#) in Java. Yet, I cannot create a variable with a name containing the hash character. Is there any reason? Is there actually a special meaning?

What characters are valid in a Java class name? What other rules govern Java class names (for instance, Java class names cannot begin with a number)?


I was young when Bioshock 1 came out and did not have a PC strong enough to play it. Over the past years I forgot about it but saw other Bioshock titles coming out but I never had a closer look at it.   Now I'm thinking about playing them but I don't know where to start... Is it necessary to play all the Bioshock games through to understand the story or can I just start with Bioshock Infinite? Is the story even connected in all the games?  Please no spoilers :*

I know that Bioshock 1+2 are considered good games, but do I have to play them in order to understand the story in Bioshock Infinite? Or can I just start with Infinite right away?


I am new to Linux and Ubuntu and need help installing Ubuntu along with Windows 10 in a dual boot setup. My HP laptop with 600GB hard drive came with windows 10 preinstalled and already has 4 partitions. I used shrink partition in windows disk management to shrink the C: partition from 531 GB to 431 GB leaving 100GB of unallocated space to install Ubuntu onto it.I cannot create a new partition on this unallocated space.  When using Ubuntu's disk Utility to create a new partition in the newly created unallocated space I get this error message: "Cannot create a new partition - There are already four primary partitions." I do not know how to proceed at this point. Please Help!

I would like to install Ubuntu alongside Windows 7 on a HP G62 Notebook. Although I have installed  Ubuntu in a dual-boot many times before, I found out that this model has already four partitions.    Partitions:   SYSTEM (NTFS, 199MB, used 66.59MB) Partition without any tag: NTFS (579GM, used 129GB)  RECOVERY (NTFS 16.74 GB, used 2.42GB)  HP_TOOLS (FAT32, 103.34 MB, used 13.23MB)   Since I am not an expert with partitions I would like to get advice on how to do this.  My first idea is this one:   Free some space from /dev/sda2 (I don't know if could also free some other space) delete the HP_Tools partition (I have already created a backup) create an extended partition with the free space in #1 containing three parititions: swap (1gb); / (EXT4, 30GB);  /home (EXT4, 120GB)   Another option is to use wubi instead.  What do you think? Is there any other way to achieve this?  PS: I really think this HP policy of using 4 partitions is not a coincidence  PS: I tried using gparted from the live CD and I got a warning message saying that if I freed some space from /dev/sda2 I could create serious issues in the system


Glow and glowing are different forms of a verb, but does the tense make any difference?  Which one of these (in each pair) is correct and why?   "I saw you dancing" or "I saw you dance". "I will watch you cook" or "I will watch you cooking". "I am watching you sing" or "I am watching you singing".

Which one of the following is the correct one?   I should have seen it glow. I should have seen it glowing.   Or are both correct? Would you parse them please?