I'm a recent college graduate, and will begin working 40 hour weeks soon. I'm not good at cooking, and I prefer to cook all of my food (fish, chicken, turkey, vegetables) by steaming them in a small rice cooker. This currently works out great, and is quick and simple enough to clean up without a mess.  However, I was wondering if it's possible to use one of these "multi-cooker" steamers, such as this  model for steaming and then slow cooking (i.e. "keep warm") chicken breasts for 4-5 hours while being away.  I was thinking about coming home for lunch, starting the steamer, then letting it run for 4-5 hours until I get home from work.   My current rice cooker isn't built for very long durations of "keep warm", and if I use the "keep warm" setting for more than an hour, it makes it more difficult to clean up.    Question:  What would you recommend for an optimal cooking tool given this situation. I'd rather start the cooking mid-day (12pm), then have it finished and ready by the return from work (5pm).

I just bought all the ingredients to make a soup in the crock pot. I was planning on having it cook all day, but I just realized the recipe only calls for it to cook on low for 3-4 hours. If I cook it on low for 3-4 hours, and then put it on warm before I leave for work, will it be a bacteria minefield or will it be warm enough to keep the food safe from bacteria?


Prove that if $X$ is a non-empty topological Hausdorff compact space and $f : X\to X$ is continuous, there exists a non-empty closed set $C \subset X$ such that $f(C)=C$.   My thoughts are to show that we can travel from a starting point $x$ through $f(x),f(f(x)),...,f^n(x)$ such that $f^n(x)=x$. that's we found a closed set. I can't seem to find the use of $f$ being continuous. Thanks!

As far as I know, there are fixed-point-like results for continuous functions from a convex compact subset $K$ of an Euclidean space to itself. I have one question in mind: Does there exist a set $A\subseteq X$ for which $f(A)=A$? Let's say $X$ is a compact metric space and $f$ is continuous.


1) A Web server  2) A Server running MySQL  Please list any technical limitations, if you know them (e.g. max memory).  Thank you

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


I have used a Box-Behnken experimental design. I have a full quadratic model.   However, I had to transform the response, $Y$ for the model to fit; I did this using a Box-Cox transformation with $\lambda=0.5$.   For example, one of the regressions is like this:  $$Y = 1.28 - 0.008X_1-0.025X_2-0.05X_3-0.13X_1^2-0.01X_2^2-0.006X_3^2\\ +0.02X_1X_2-0.05X_1X_3+0.09X_2X_3$$  How do I interpret the terms?

I'm using fourth root (1/4) power transformation on my response variable, as a result of heteroscedasticity. But now I'm not sure how to interpret my regression coefficients.   I assume that I would need to take the coefficients to the fourth power when I back-transform (see below regression output). All of the variables are in units of dollar in millions, but I would like to know the change in dollar in billions.    While holding the other independent variable constant, a billion dollar change in fees, on average, leads to a change of 32 (or 32,000 dollars) in collections. I take 0.000075223 * 1000 (to get to billions) ^ 4 = 0.000032. Now do I multiply this number by 1 million or 1 billion (the original unit of the dependent variable is in millions)?  lm(formula = (Collections^(1/4)) ~ Fees + DIR)                   Estimate      Std. Error  t value            Pr(>|t|) (Intercept)   2.094573355     0.112292375   18.653  0.0000000000000151 Fees        **0.000075223   **0.000008411    8.943  0.0000000131878713 DIR           0.000022279     0.000004107    5.425  0.0000221138881913


I want to figure out what's the difference between buying stocks and entering Ponzi scheme under certain situations. This is more of a theoretical/hypothetical question.  Suppose the company P makes 0 revenue, contributes 0 to the society and there is 0 regulations.  If investor A bought stocks of P, and then sold it at a higher price to investor B. That is, the later comer is paying the earlier comer. Then this is how Ponzi scheme works. Hence, buying stocks of P = entering Ponzi scheme?  Does the reasoning make sense? Do I miss something important?

Sorry if the below seems like a conspiracy theory. It probably is, but I probably don't understand something. (This has bugged me to no end as long as I can remember.)  Okay, so here is what I understand about the stock market?   The main way money enters the stock market is through investors investing and taking money out. The only other cash flow is in through dividends and out when businesses go public.   In a Ponzi scheme, there is usually some small source of income to hide the fact that it's a Ponzi scheme. Dividends are pretty insignificant. For the purposes of this question, we will only consider non-dividend stocks.  When you buy stock, it is claimed that you own a small portion of the company. This statement has no backing, as you cannot exchange your stock for the company's assets.   For example, if I bought $10 of Apple Stock early on, but it later went up to $399, I can't go to Apple and say "I own $399 of you, here you go it back, give me an iPhone." The only way to redeem this is to sell the stock to another investor (like a Ponzi Scheme.)  The stock market goes up only when more people invest in it. Although the stock market keeps tabs on Businesses, the profits of Businesses do not actually flow into the Stock Market.   In particular, if no one puts money in the stock market, it doesn't matter how good the businesses do.    Based on the above, it seems like the Stock Market has little to do with investing businesses to create growth, and primarily grows in a similar way to a Ponzi Scheme. (Some differences include that it is Stochastic (so when it collapses, it can be blamed on luck or the economy) and it has no top operator (similar to a Pyramid Scheme)).  Am I missing something? Does the stock market create value for investors?


"The number $\frac 16 \pi^2$ turns up surprisingly often and frequently in unexpected places." - Julian Havil, Gamma: Exploring Euler's Constant.   It is well-known, especially in 'pop math,' that $$\zeta(2)=\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\cdots = \frac{\pi^2}{6}.$$  is nice. I would like to know where else this constant appears non-trivially. This is a bit broad, so here are the specifics of my question:  We can fiddle with the zeta function at arbitrary even integer values to eek out a $\zeta(2)$. I would consider these 'appearances' of $\frac 16 \pi^2$ to be redundant and ask that they not be mentioned unless you have some wickedly compelling reason to include it. By 'non-trivially,' I mean that I do not want converging series, integrals, etc. where it is obvious that $c\pi$ or $c\pi^2$ with $c \in \mathbb{Q}$ can simply be 'factored out' in some way such that it looks like $c\pi^2$ was included after-the-fact so that said series, integral, etc. would equal $\frac 16 \pi^2$. For instance, $\sum \frac{\pi^2}{6\cdot2^n} = \frac 16 \pi^2$, but clearly the appearance of $\frac 16\pi^2$ here is contrived. (But, if you have an answer that seems very interesting but you're unsure if it fits the 'non-trivial' bill, keep in mind that nobody will actually stop you from posting it.)  I hope this is specific enough. This was my attempt at formally saying 'I want to see all the interesting ways we can make $\frac 16 \pi^2$.' With all that being said, I will give my favorite example as an answer below! :$)$  There used to be a chunk of text explaining why this question should be reopened here. It was reopened, so I removed it.

$\text{What are some interesting cases of $\pi$ appearing in situations that do not seem geometric?}$  Ever since I saw the identity $$\displaystyle \sum_{n = 1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$  and the generalization of $\zeta (2k)$, my perception of $\pi$ has changed. I used to think of it as rather obscure and purely geometric (applying to circles and such), but it seems that is not the case since it pops up in things like this which have no known geometric connection as far as I know. What are some other cases of $\pi$ popping up in unexpected places, and is there an underlying geometric explanation for its appearance?  In other words, what are some examples of $\pi$ popping up in places we wouldn't expect?


Part in my jsp page where I am passing date after submit of this form to the servlet:

                   Date	                   <input type=date step=7 name="day" required="required">

   In my servlet I wrote:  Date day=new Date();  String startDateStr = request.getParameter("day");     System.out.println("stsartdate Str"+startDateStr);     SimpleDateFormat sdf = new SimpleDateFormat("yyyy-MM-dd");     try {         date=sdf.parse(request.getParameter("startDateStr"));         System.out.println("date entered is="+date);     } catch (ParseException e) {         e.printStackTrace();     }   It is prompting me a null pointer exception.  Mar 31, 2016 3:21:20 PM org.apache.catalina.core.StandardWrapperValve invoke SEVERE: Servlet.service() for servlet [com.controller.TravelServlet] in     context with path [/LetsRide] threw exception java.lang.NullPointerException at java.text.SimpleDateFormat.parse(Unknown Source) at java.text.DateFormat.parse(Unknown Source) at com.controller.TravelServlet.doPost(TravelServlet.java:76) at javax.servlet.http.HttpServlet.service(HttpServlet.java:647) at javax.servlet.http.HttpServlet.service(HttpServlet.java:728) at     org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:305) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210) at org.apache.tomcat.websocket.server.WsFilter.doFilter(WsFilter.java:51) at org.apache.catalina.core.ApplicationFilterChain.internalDoFilter(ApplicationFilterChain.java:243) at org.apache.catalina.core.ApplicationFilterChain.doFilter(ApplicationFilterChain.java:210) at org.apache.catalina.core.StandardWrapperValve.invoke(StandardWrapperValve.java:222) at org.apache.catalina.core.StandardContextValve.invoke(StandardContextValve.java:123) at org.apache.catalina.authenticator.AuthenticatorBase.invoke(AuthenticatorBase.java:502) at org.apache.catalina.core.StandardHostValve.invoke(StandardHostValve.java:171) at org.apache.catalina.valves.ErrorReportValve.invoke(ErrorReportValve.java:100) at org.apache.catalina.valves.AccessLogValve.invoke(AccessLogValve.java:953) at org.apache.catalina.core.StandardEngineValve.invoke(StandardEngineValve.java:118) at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:408) at org.apache.coyote.http11.AbstractHttp11Processor.process(AbstractHttp11Processor.java:1041) at org.apache.coyote.AbstractProtocol$AbstractConnectionHandler.process(AbstractProtocol.java:603) at org.apache.tomcat.util.net.JIoEndpoint$SocketProcessor.run(JIoEndpoint.java:310) at java.util.concurrent.ThreadPoolExecutor.runWorker(Unknown Source) at java.util.concurrent.ThreadPoolExecutor$Worker.run(Unknown Source) at java.lang.Thread.run(Unknown Source)   However, the print statement where I am printing the date fetched from the jsp page in the form of string, it is printing:   stsartdate Str2016-03-10   It is throwing null-pointer exception while executing try block. Kindly get me out of this. Thanks in advance.

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 was trying to solve for $i^i$.  I got distracted and did the following: $e^{i \theta} = \cos(\theta) + i\sin(\theta)$ $e^{i 2\pi} = cos(2\pi) + i\sin(2\pi) = 1$ ||| Take ln() of two sides of equation: $\ln(e^{i2\pi}) = \ln(1)$ $i2\pi = 0$ Clearly this is false. What did I do wrong?

$$e^{i} = e^{i2\pi/2\pi} = (e^{2\pi i})^{1/(2\pi)} = 1^{1/(2\pi )} = 1$$  Obviously, one of my algebraic manipulations is not valid.


I am learning about malloc function & I see this difference but I cannot understand it What is the difference between ptr = (*int) malloc(sizeof(int)*N)  and ptr = malloc(sizeof(int)*N)

In , someone suggested in a  that I should not cast the result of malloc. i.e., I should do this: int *sieve = malloc(sizeof(int) * length);  rather than: int *sieve = (int *) malloc(sizeof(int) * length);  Why would this be the case?


My question is, how can I calculate the total volume such that once reached, the surface tension and cohesion are insufficient to keep the water stuck to the glass? I mean when we add objects to the container and the water level rises and is above the edge of the container, then a dome is formed, the question is about the maximum size that can have this dome at standard temperature and pressure.

When you fill a glass with water, water forms a concave  with constant contact angle $\theta$ (typically $\theta=20^\circ$ for tap water):    Once you reach the top of the glass, the water-air interface becomes convex and water rises up to a height $\Delta h$ above the edge of the glass, allowing you to fill the glass beyond the naive capacity $\pi r^2 h$:    So when getting myself a glass of water, I came to wonder exactly how much this increases the capacity of a glass, and what physical constants are involved.  My intuition would be that for a very large glass, $\Delta h$ converges to a constant so that the effective water capacity of the glass grows like $\pi r^2 (h+\Delta h)$ (to make things simple I'm assuming that the glass is very thin: $\Delta r\ll r$). Perhaps such a constant depends on the precise shape of the rim of the glass. But if not, perhaps it is a constant multiple of the capillary length?  So, what can we say about $\Delta h$, the "rim contact angle" $\alpha$, or the shape of the water-air interface when the glass is filled at maximum capacity?


I have /etc/update-manager/release-upgrades set to lts and here it is 2 days after 18.04 has been released and I am still getting "no new release". What gives fellas?  i am running 16.04 32bit server if that matters.

I'm following the , but on launching the upgrade tool I get this response:  Checking for a new ubuntu release No new release found   Am I doing something wrong? Is there a workaround?


I had a Windows 10 installation in my laptop. I got a new SSD for it so I took out the old HDD and put it in a USB enclosure.  The hard disk works fine and great, better than I thought it would on USB (yay for USB 3.0). However, it won't boot from the USB.  Is there any way I can boot my old OS too via USB?

When Win 7 went RTM, I upgraded my laptop from XP.  But to protect myself against the possibility of having data inaccessible, I got a new disk and put it in the laptop, and took the old (XP) disk and put it in a USB enclosure.  I have no trouble accessing the data on the USB drive under Windows 7.  But I would like to be able to plug it in, tell the BIOS to boot from the USB drive, and be back on my old machine.  The laptop is a Dell Precision M90, and it has a boot from USB option in its boot menu.  But when I try to do that, it does read the drive, gets as far as putting up the Windows XP splash screen and starting the boot progress bar, and then reboots.  What do I need to do to the old disk now running on USB to allow the machine to boot from it?


~$ sudo swapon --show gives me this  NAME      TYPE      SIZE USED PRIO /dev/sda5 partition 976M   0B   -2   I want to increase my swap size. Is there a way to increase this or do I have to create a new partition or a file? I have gparted installed and I see that there are only 2 partitions as lsblk shows below -  NAME   MAJ:MIN RM   SIZE RO TYPE MOUNTPOINT sr0     11:0    1  1024M  0 rom   sda      8:0    0 238.5G  0 disk  ├─sda5   8:5    0   976M  0 part [SWAP] └─sda1   8:1    0 237.5G  0 part /   How do I go about increasing the size?

I have, by default, 250MB of swap space in Ubuntu, and I want to expand it to a larger size. I need 800MB, which I think will be enough to open several applications without having to hit the current limit of swap. I hope someone can help me.


I have destiny on ps4 as a disk but I have none of the dlc's. I just bought the legendary edition and PlayStation store is saying cannot add the follwing content to downloads. A disc version of the same aplication has been used on this ps4, you must delete the disc Version. If I delete it and download the legendary edition will I still have my data.

So I have Destiny on PS3. If I were to buy the Legendary Edition (Containing House of Wolves and Dark Below Expansion) will I be able to play with my characters from my original Destiny disc? May sound silly, but I need confirmation.


In Adventurers League, does Magic Initiate let you take a spell that is not in your PHB+1? We know that "Feats and spells gained by advancement are subject to PHB+1", but in this case Magic Initiate is part of the PHB+1 and that in general is possible obtain spell out the PHB+ if out the advancement in level (see wizards)

I’m thinking of trying out a Hexblade Warlock in Adventurers League, which limits players to PHB+1 for character-building.  The problem I have with this is that the Hexblade is from Xanathar's Guide to Everything, but I really want the booming blade cantrip as well from the Sword Coast Adventurer's Guide. Is this legal in Adventurers League play?


I want to repeatedly run a function that retrieves data from an API every 60 seconds. However, I've realized that I can't put it to sleep, or use a while loop, without interrupting the other functions.  My question is, how do I repeatedly retrieve the data from the API/rerun the code that gets the data in the first place, without interrupting the other functions?

I want to repeatedly execute a function in Python every 60 seconds forever (just like an  in Objective C). This code will run as a daemon and is effectively like calling the python script every minute using a cron, but without requiring that to be set up by the user.  In , the solution appears to effectively just  for x seconds. I don't need such advanced functionality so perhaps something like this would work  while True:     # Code executed here     time.sleep(60)   Are there any foreseeable problems with this code?


I found hard (for me at least) example, in fact it just would be nice to prove.  $$\binom{2n}{n} = \sum_{k=0}^{n} \binom{n}{k}\binom{n}{n-k}$$

How can we prove that     $$\sum_{k=0}^{n}\binom{R}{k}\binom{M}{n-k}=\binom{R+M}{n}?$$   (Presumptive) Source: Theoretical Exercise 8, Ch 1, A First Course in Probability, 8th ed by Sheldon Ross.


From this answer  I am using y filter to process data, all fine until I decide to put a legend entry in the plot. I cannot seem to simultaneously use the yfilter and the addlegendenty lines in the following code.  I get this error  ! Illegal parameter number in definition of \pgfplots@curplotlist.   from this code (later try commenting the two lines):  \documentclass{article} \usepackage{pgfplots} \usepackage{tikz} \begin{document} \begin{tikzpicture} \begin{axis} \addplot[     y filter/.code={\pgfmathparse{#1*10.}\pgfmathresult}% try comment ] coordinates {  ( 1., 1. )  ( 2., 2. )  ( 3., 3. ) }; \addlegendentry{Legend Entry};% try comment \end{axis} \end{tikzpicture} \end{document}   What is wrong, is this a bug?  info: pdfTeX 3.1415926-2.5-1.40.14 LuaTeX, Version beta-0.76.0-2013061217  (TeX Live 2013)


When I was looking at my profile, I discovered that the name of my next tag badge was jammed with the bar showing my progress for me. It looks like this happens when the tag name is too long.    This happens even if I am logged out. The display issue is not as severe only because the icon next to it isn't there.

The Next tag badge block overlaps the badge score block.


I am trying to do something which seems like it would be pretty damn simple, yet I can't get to work; access a custom field (added via the UI at /admin/config/people/accounts/fields), for the current logged in user. I have tried  after , but when it comes to accessing the field value itself, nothing works. Loading the user object is pretty straightforward, but when it comes to getting the actual field values, neither get() nor any other option has worked for me. In my case, I added a multi value entity reference field called field_distributors that references a Distributor node type, and I just need to access those fields in a form to set values for a select list. use Drupal\user\Entity\User;  $user = User::load(\Drupal::currentUser()->id());  works to get a loaded user, object, and I can see the fields in both $user->fields and $user->values. How do I access those values?

I have a custom form that needs to display a default value from a custom user field. The field was created under /admin/config/people/accounts/fields and its machine name is field_r_period_length_sec.  //get user $userCurrent = \Drupal::currentUser(); $uid = $userCurrent->id(); $r_period = $userCurrent->field_r_period_length_sec;   However, the value doesn't show and I get this error:     Notice: Undefined property: Drupal\Core\Session\AccountProxy::$field_r_period_length_sec   What am I doing wrong?


I am writing a program that asks the user to input his name, address and phone number. When the data is entered the program shall print the data and ask the user to verify the data by entering YES or NO. This process shall be repeated until the user is satisfied and answers YES to the question.  Now in my case I may have put a if-else statement inside the while loop in an inappropriate way. That's why it's not working as it is expected to be. How can I solve this?  Also I only have tried the promt asking to enter the users name. But if I want to add more prompt with different question in the similiar way then how could I do that?  Code:  package userinfo;     import java.util.Scanner;  public class UserInfo {      public static void main(String[] args) {             String name;         String yes = "YES";         String no = "NO";             Scanner userInput = new Scanner(System.in);          System.out.println("Enter your name:");         name = userInput.next();         System.out.println("Please varify your name by typing YES or NO");          while (true) {             String input = userInput.next();             if (input == yes) {                 System.out.println("Your name is: " + name);             }             if (input == no) {                 System.out.println("Enter your name again");              } else {                 System.out.println("Invalid input! Enter value again:");                 break;             }             }         }     }

I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference?


I recently found out the ability to use virtual machines instead of dual installation Ubuntu and Windows. Thinking of installing just Ubuntu, and through virtualbox have a Windows installation. What are the pros of cons of this? Thank you for any insight!

I would like to install Ubuntu on my Windows 7 64-bit PC to do analysis of astronomy data. The data files are often large (can be few hundred MB, possibly a GB). The data files may be located on an external hard disk. The PC has 6 GB RAM and 670 GB hard disk.   I'm trying to decide whether to do a proper dual-boot, installing Ubuntu in its own partition (I'm nervous of messing around with the hard disk, have never done it before) or to run Ubuntu using VirtualBox in Windows. How much slower is running in Virtualbox likely to be?   If I do go for a traditional dual-boot, would it make sense to divide the hard disk more or less equally between the two OSs?  I would be very grateful for some advice.  Many thanks Kevin


I found this cool design in Dribbble and wonder how these layers of curves are made in Photoshop. Buildings in the background is also blended well. Is there any proper method to blend them into the background layer? Or is it just made by reducing the opacity level? Would appreciate if there are links which explains what kind of design it is or guidelines to build them.   Thanks in Advance :)    My try is here:

I would like to to create the image above. I have tried to make multiples layers of the waves and edit it's opacity but it does not seem to work. I would really appreciate some feedback and guidance. Thank you.


Does my proof hold up to prove that $0$ is the only eigenvalue of $A$ if $A^2 = 0$?     Let $A$ be an $n \times n$ matrix. $A^2 = A\times A$ because of matrix multiplication. If $A = k$, where $k \neq 0$, then $A^2 \neq 0$. Therefore if $A^2 = 0$, $A$ is the zero matrix.  If $A$ is the zero matrix, det($A^2) = 0$. $\lambda \neq 0$ if $A$ is non-singular. $A$ is a non-singular matrix IFF $\det(A)$ $\neq 0$. The charcteristic equation is false when $\lambda = 0$ for a non-singular matrix so it can only be true when the matrix is singular, making $\det(A) = 0$. $\lambda = 0$, $\rightarrow$ $\begin{vmatrix} 0I - 0 \end{vmatrix} = 0$

I need to prove that:      if a linear operator $\phi : V \rightarrow V$ on a vector space is nilpotent, then its only eigenvalue is $0$.   I know how to prove that this for a nilpotent matrix, but I'm not sure in the case of an operator. How would I be able to relate $\phi$ to a matrix?  Note: A nilpotent operator $\phi$ has been defined as an operator that satisfies $\phi^{n} = 0$ for some $n \geq 1$.


Given a transformation T:V-W that is injective, if {v1,v2,...,vn} is a linearly independent subset of V, I need to prove that the set {T(v1), T(v2),...,T(vn)} will be a linearly independent set of W. To my understanding no values in {v1,v2,...,vn} are equal since they are independent and because there is a one to one mapping from V to W the set {T(v1), T(v2),...,T(vn)} should also have no equal values. This somewhat proves that {T(v1), T(v2),...,T(vn)} is independent. I am struggling to understand how to prove that the set {T(v1), T(v2),...,T(vn)} doesn't contain values that are linear combinations of other values in the set.

Problem 4.3.32 in Linear Algebra, Lay:  Let $V$ and $W$ be vector spaces, let $T:V\to W$ be a linear transformation, and let $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ be a subset of $V$. Suppose that $T$ is a one-to-one transformation [...]. Show that if the set of images $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ is linearly dependent. This fact shows that a one-to-one linear transformation maps a linearly independent set onto a linearly independent set.  All I can think of doing: If $\{T(\mathbf{v}_1), \dots ,T(\mathbf{v}_p)\}$ is linearly dependent, then there exist weights $c_1, \dots, c_p$, not all zero, so that $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = \mathbf{0}$$ Since $T$ is linear: $$c_1 T(\mathbf{v}_1) + \cdots + c_p T(\mathbf{v}_p) = T(c_1\mathbf{v}_1) + \cdots + T(c_p \mathbf{v}_p)=T(c_1 \mathbf{v}_1 +\cdots + c_p \mathbf{v}_p)=\mathbf{0}$$ So the set $\{\mathbf{v}_1, \dots , \mathbf{v}_p \}$ must be linearly dependent. Am I missing something? And why must $T$ be on-to-one for this to be true? An example showing that it is not true if $T$ is not one-to-one would be great!


I have a set of all bounded sequences of real numbers with the metric $d$ defined as:  d(a, b) = sup$_n|a_n - b_n|$.   I want to prove this is a complete metric space, i.e. every Cauchy sequence is convergent.   My attempt at Solution:   Consider a Cauchy sequence $a_n$.  For every $\epsilon > 0$, there exists an $N \in \mathbb{N}$ such that for all $m, n > N$, we have sup$_{n, m} |a_n - a_m| < \epsilon$. $\implies $ $|a_n - a_m| < \epsilon$ for all $n, m > N$.   Every bounded sequence in reals has a convergent subsequence - call this subsequence $a_{n_k}$. Hence, for every $\epsilon > 0$, there exists an $M \in \mathbb{N}$ such that for all $k > M$, we have $|a_{n_k} - L| < \epsilon$ $\implies $ sup$_k |a_{n_k} - L| < \epsilon$.   Let $N^* = \max \{N, n_M \}$ so that for all $n, m > N^*$ and $n_k > N^*$ for some $k > M^*$, we have:   sup$_k|a_{n_k} - L| < \epsilon$ and  sup$_{n, m}|a_n - a_m | < \epsilon$.   Then,   $|a_n - L | = |a_n - a_{n_k} + a_{n_k} - L| \leq |a_n - a_{n_k}| + |a_{n_k} - L| $.   Now if $k > M^*$ such that $n_k > N^* \geq N$ and also $n > N^*$, we have   $|a_n - L| \leq 2 \epsilon$  Hence, sup$_n |a_n - L| \leq 2 \epsilon$.

Let $X$ be the set of all bounded sequences $x=(x_n)$ of real numbers and let $$d(x,y)=\sup{|x_n-y_n|}.$$ I need to show that $X$ is a complete metric space.  I need to show that all Cauchy sequences are convergent. I appreciate your help.


Dark matter supposed to be affected only by gravity and no other force. At the same time, however, dark matter forms halos around galaxies, that become visible examining how galaxies and clusters bend the light. The question is now, what does prevent these huge clouds of dark matter from collapsing? A cloud of "normal" matter has an inner pressure from its temperature. The force working here would be the electromagnetic force because the atoms and molecules of "normal" matter repel each other (compare  Veritasium - Can We Really Touch Anything?) When it collapses as a cold cloud to form stars, the inner pressure is still working, it's heating up until it forms a star hot enough to start fusion fire. When the star collapses because the pressure decreases after the core cools down after fusion slows down at the end of its life, quantum mechanical forces (Pauli principle) keep it from getting down any further until the mass is strong enough to overcome the Pauli forces and create a black hole. Now why doesn't the same happen to dark matter if there's no other force that's affecting dark matter particles (whatever they might be) except for gravity? Inner pressure can't happen because electromagnetic forces don't work here. Pauli principle can't happen because for that it's not dense enough and would again need other forces to be exchanged between the particles. Does that mean, that these DM particles have to be Bosons since they can go through each other and thus never lose any energy to each other do collapse further? However, gravity could produce some kind of inner friction too when one particle loses gravitational energy to another one passing by. PS: I am not a physicist just interested in this stuff, thus I can't do the calculations for myself.

I'm a complete layperson. As I understand, dark matter theoretically only interacts with the gravitational force, and doesn't interact with the other three fundamental forces: weak nuclear force, strong nuclear force, and electromagnetism.  Those are my understandings going in. If I'm wrong, please correct me. I've done some googling, and I haven't found anything confirming or denying that dark matter is affected by either of the fundamental nuclear forces.  So since dark matter only interacts with gravity, what causes any dark matter particle to be repelled from another? If they can pass freely through each other, and they are gravitationaly attracted to each other, why don't such particles clump together in a single 'point' in space?   It seems to me that particles occupying a single 'space' are philosophically not distinct particles, but I don't know how actual physics would play into this.  Edit , author's credentials unknown, but implicitly claims to be a physicist or astronomer, says "...[P]hysicists generally take all dark matter to be composed of a single type of particle that essentially interacts only through gravity."  Edit 2 The author is this , "Professor of Science on the physics faculty of Harvard University."


I'm trying to create a lightbulb using a tutorial on youtube, and I am completely new to Blender.  I'm trying to tighten the bottom of the lightbulb by scaling it down, but it doesn't scale proportionally. Instead it wavers off to the side.  What I have so far:    Expected result (screenshot from the tutorial):

Hello. I'm a beginner.  In the model shown, when I select all of the top vertices and then scale them to the center, the center of scaling is not the center of the 3d-model.   How do I move the center of scaling? How do I move the manipulator widget where I want it after selecting?


Is it (or will it be) possible to import data already collected by other apps into the new health app in iOS 8?

I have a few apps that may not be updated to support iOS 8 and HealthKit, but they do offer CSV export of data they have stored.  Is there any way to import historical data into HealthKit and Health.app?


This question is from Book of Proof by Richard H. Hammack. This is exercise number 13 of chapter 3 section 6. This is exercise I think is related to Pascal identity, but I'm struggling to understand it.

I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$   Thanks in advance for any help.


I have a laptop, which comes with an SMPS charger. The laptop charger is a very common type. I know there are some earthing issues in my home. I need to fix that. But SMPS isolate mains from the supply, but when I touch the body of the laptop, it shocks. Another day, I was charging my phone through USB from the laptop and plugged the laptop charger. I own a metal earphone, and that shocked me. I know there's some problem in my home earthing part, but still aren't SMPS meant to be isolated from mains. Then why I'm getting that shock. Why does electricity want to flow through my body if it's floating? Someone, please explain the incident, with reasoning and possible ways to avoid it.

So, I got some laptop charger from my neighbor and I was thinking about to do something with it. Then one day I feel tiny shock from the output of the charger. I thought ac to dc filtering mechanism maybe shorted. Then I had tested other remaining charger to see if there was any AC current going on. And they all do the same. Is it normal to have flow some AC through DC ? Can I do something about it ? make it safer for work ? It does not feel like comfortable.


I was wondering if there is a limit to $_SERVER['REQUEST_URI'] itself since there seems to be no limit on storing string in a variable (php7).   So, if a user can somehow request using a very very large url (I know there are browser limits), will php store all of the request URI in $_SERVER['REQUEST_URI']?  EDIT: I am sorry this is not duplicate of . The said question addresses browsers' limit. What if someone uses ?

What is the maximum length of a URL in different browsers? Does it differ among browsers? Is a maximum URL length part of the HTTP specification?


I spent an hour today trying to figure out why  return abs(val-desired) <= 0.1   was occasionally returning False, despite val and desired having an absolute difference of <=0.1. After some debugging, I found out that -13.2 + 13.3 = 0.10000000000000142. Now I understand that CPUs cannot easily represent most real numbers, but this is an exception, because you can subtract 0.00000000000000142 and get 0.1, so it can be represented in Python.  I am running Python 2.7 on Intel Core architecture CPUs (this is all I have been able to test it on). I'm curious to know how I can store a value of 0.1 despite not being able to apply arithmetic to particular floating point values. val and desired are float values.

My code:  a = '2.3'   I wanted to display a as a floating point value.  Since a is a string, I tried:  float(a)   The result I got was :  2.2999999999999998   I want a solution for this problem. Please, kindly help me.  I was following .


I have an audio mixer that requires an AC 16 volt 1 amp external power adapter. I need to know if an 18 volt 1 amp AC external power adapter would work without damaging the audio mixer?

Power supplies are available in a wide range of voltage and current ratings.  If I have a device that has specific voltage and current ratings, how do those relate to the power ratings I need to specify?  What if I don't know the device's specs, but am replacing a previous power supply with particular ratings?  Is it OK to go lower voltage, or should it always be higher?  What about current?  I don't want a 10 A supply to damage my 1 A device.


I know that it should be easy, but I'm new in Ubuntu, so please help:  sudo dpkg -i virtualbox-5.1_5.1.10-112026-Ubuntu-yakkety_amd64.deb  Wybieranie wcześniej niewybranego pakietu virtualbox-5.1. (Odczytywanie bazy danych ... 301555 plików i katalogów obecnie zainstalowanych.) Przygotowywanie do rozpakowania pakietu virtualbox-5.1_5.1.10-112026-Ubuntu-yakkety_amd64.deb ... Rozpakowywanie pakietu virtualbox-5.1 (5.1.10-112026~Ubuntu~yakkety) ... dpkg: problemy z zależnościami uniemożliwiają skonfigurowanie pakietu virtualbox-5.1:  virtualbox-5.1 zależy od libqt5core5a (>= 5.6.0~beta); jednakże:   Wersją libqt5core5a:amd64 w systemie jest 5.5.1+dfsg-16ubuntu7.2.  virtualbox-5.1 zależy od libqt5widgets5 (>= 5.6.0~beta); jednakże:   Wersją libqt5widgets5:amd64 w systemie jest 5.5.1+dfsg-16ubuntu7.2.  virtualbox-5.1 zależy od libqt5x11extras5 (>= 5.6.0); jednakże:   Wersją libqt5x11extras5:amd64 w systemie jest 5.5.1-3build1.  dpkg: błąd przetwarzania pakietu virtualbox-5.1 (--install):  problemy z zależnościami - pozostawianie nieskonfigurowanego Przetwarzanie wyzwalaczy pakietu systemd (229-4ubuntu12)... Przetwarzanie wyzwalaczy pakietu ureadahead (0.100.0-19)... ureadahead will be reprofiled on next reboot Przetwarzanie wyzwalaczy pakietu gnome-menus (3.13.3-6ubuntu3.1)... Przetwarzanie wyzwalaczy pakietu bamfdaemon (0.5.3~bzr0+16.04.20160824-0ubuntu1)... Rebuilding /usr/share/applications/bamf-2.index... Przetwarzanie wyzwalaczy pakietu desktop-file-utils (0.22-1ubuntu5)... Przetwarzanie wyzwalaczy pakietu mime-support (3.59ubuntu1)... Przetwarzanie wyzwalaczy pakietu hicolor-icon-theme (0.15-0ubuntu1)... Przetwarzanie wyzwalaczy pakietu shared-mime-info (1.5-2ubuntu0.1)... Wystąpiły błędy podczas przetwarzania:  virtualbox-5.1   Ubuntu 16.04

I downloaded VirtualBox OSE from the Ubuntu Software Center but it seems that that version is limited. I prefer to have the complete version and don't really care if it's closed source or not.  Can I install it via a terminal or do I have to download something from a website?  I don't know how to install software I download from the web. So far I've only used the Software Center.


I need to split a string base on two delimiters: one or more space and quote    public static void main( String[] args ) {     String[] aStr = {"Header  \"2018-02-23 picture test\"", "Average Value \"A B \"  3.14"};      for(String str: aStr) {         for(String str1: str.split("\"|\\s"))             System.out.println(str1);     } }   My desired output:  Header 2018-02-18 picture test Average Value A B 3.14

I'm new to regular expressions and would appreciate your help.  I'm trying to put together an expression that will split the example string using all spaces that are not surrounded by single or double quotes. My last attempt looks like this: (?!") and isn't quite working. It's splitting on the space before the quote.  Example input:   This is a string that "will be" highlighted when your 'regular expression' matches something.   Desired output:  This is a string that will be highlighted when your regular expression matches something.   Note that "will be" and 'regular expression' retain the space between the words.


So in my application, a user can select a Widget, and select tags that describe that widget (Fast, Slow, Expensive,i.e).  As of now, the plan was to use a dual-list with a bag of "unchosen" descriptors, and a bag of "chosen" descriptors.  Essentially, my UI currently looks like the .    My issue now is that the user has to be able to select a group of widgets (n > 1) and then apply a descriptor to all of them ("Describle all these widgets as slow").  Clearly, this breaks the UI as the dual list now has to somehow handle Add/Delete operations for (n > 1) widgets.  My next progression was going to be a 3 part list where they would have a first list with   "All"  "Widget 1"  "Widget 2"   But that felt wrong.  Does anyone know of a UI pattern that can allow the user to perform Add/Delete changes on 1 or more Widget items or am I thinking about this incorrectly.  Any help is greatly appreciated.

So a while back, I asked a question about a dual list UI.  I didn't explain my question well enough, and the answers reflected that.  My question now is, "Is there a better way to allow the user to accomplish their goals than with the UI I've created".    The goals are as follows.   User must be able to select Super-Categories to cut down on the Available List size User must be able to Filter Categories to cut down on the Available List Size. User should be able to use a general Add/Remove to move Categories from the Available to the Selected List.   Here is my current setup, pay no mind to the colors, they were used for sizing and positioning.  The thing I'm noticing is how this UI setup seems to always feel like it needs more and more space.  It also feels VERY unbalanced because of the Filter and Super-Category selection being on one side, and nothing being on the other.  I could move the Filter so that it was on the same line as "Sections"/"Super-Categories" and then add a Separator for a mental break.  But I feel that leads the user to think that the filter works on the Selected List when it only works on the AvailableList.  Any help is GREATLY appreciated.  UI Atrocity Below...


How do I install "Double Commander" or "Midnight Commander" file managers in Ubuntu 18.04?  I tried to install Midnight Commander with sudo apt-get install mc but got   Reading package lists... Done Building dependency tree Reading state information... Done Package mc is not available, but is referred to by another package.  This may mean that the package is missing, has been obsoleted, or is only available from another source E: Package 'mc' has no installation candidate   Is there some working file managers for Ubuntu 18.04?

What does the above Error mean? What is an installation candidate?  I was trying to do  sudo apt-get install munin-memcached     and I get this message:  Reading package lists... Done Building dependency tree        Reading state information... Done Package munin-memcached is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source E: Package munin-memcached has no installation candidate   I googled for the problem and someone said to do a apt-get upgrade but it still did not solve my problem.


I'm using 14.04 and would like to move the launcher to the bottom of the screen.  Is there a way to do this ?  Thank's

I can’t get used to the new Unity system. I’d like to move the left Unity panel to the bottom of screen (like Windows 7 – more similar is better).  But I can't find a way. Is it possible?


I'm a semi-frequent traveler (3-5x/year), and I'm debating whether to join the Spirit $9 Fare Club.  My question is, how frequently do these club fares pop up? How does Spirit schedule these things?  For example, do they pop up on Tuesdays, or on a set schedule?  Do they happen during the holidays?  I'm guessing it's a probability game, on the chance there will be a reduced fare on the route I'm taking, at the time I'm booking. But I'm wondering how often during the year and in general these club fares pop up. And if people have tips on the best time to book to grab these reduced fares.

Has anyone tried  membership ($60/year)? What do you get with your membership, and what makes it worth it?


Say we have the earth and a person on it and an astronaut travelling away from the earth in a spaceship in a straight line and then coming back in a straight line at the same speed.   The person on the earth measures the time between when the astronaut leaves and when he comes back as 1 year. This is a proper time since the observer has not moved in his frame of reference. Thus according to the relation $t = \gamma t_o$ the astronaut must have experienced a longer time.   The person on the spaceship measures the time from takeoff to landing as 1 year. Since he's standing at the same point in his frame of reference this is a proper time. Thus according to the relation $t = \gamma t_o$ the person on earth must have experienced a longer time.   The above is contradictory. Either the astronaut experiences a longer time or the person on earth does. The issue is that either the astronaut is thought to be moving relative to the earth or the earth is thought to be moving relative to the astronaut. What's the explanation? Can anyone help me understand?  Okay. I’ve read through the below answers, as well as watched a few videos and the Wikipedia page on it, however, i don’t see any of them explaining it. They claim that the acceleration solves the issue, however, don’t we arrive at the same problem? In the earth point of view the astronaut is rotating, but you might as well see the earth as rotating in the astronauts view. In other words, same problem.

I read a lot about the classical twin paradox recently. What confuses me is that some authors claim that it can be resolved within , others say that you need . Now, what is true (and why)?


Can someone please verify my proof or offer suggestions for improvement?  I am aware that there is a similar question elsewhere, but I want help with my proof in particular.     Let $X$ be a metric space with metric $\operatorname{d}$. Show that $\operatorname{d}:X \times X \longrightarrow \mathbb{R}$ is continuous.   Let $U \subseteq \mathbb{R}$ be open. Assume $\operatorname{d}^{-1}(U) \neq \varnothing$.  Let $(x, y) \in \operatorname{d}^{-1}(U)$ and set $\epsilon_1 = \frac{1}{2}(b-\operatorname{d}(x, y))$  Clearly, if $(\alpha, \beta) \in B_d(x, \epsilon_1) \times B_d(y, \epsilon_1)$, then \begin{eqnarray} \operatorname{d}(\alpha, \beta) &\leq& \operatorname{d}(x, \alpha) + \operatorname{d}(x, y) + \operatorname{d}(y, \beta)\\ &<& \operatorname{d}(x, y) + 2\left(\frac{b-\operatorname{d}(x, y)}{2}\right) \\ \end{eqnarray}  That is,  $$\operatorname{d}(\alpha, \beta) < b$$  Similarly, set $\epsilon_2 = \frac{1}{2}(\operatorname{d}(x, y) - a)$.  Let $(\alpha, \beta) \in B_d(x, \epsilon_2) \times B_d(y, \epsilon_2)$. Assume, for the sake of contradiction, that $\operatorname{d}(\alpha, \beta) \leq a$.  Then,  \begin{eqnarray} \operatorname{d}(x, y) &\leq& \operatorname{d}(x, \alpha) + \operatorname{d}(\alpha, \beta) + \operatorname{d}(\beta, y) \\ \operatorname{d}(x, y) &<& \operatorname{d}(x, y)-a + \operatorname{d}(\alpha, \beta) \\ &<&  \operatorname{d}(x, y) \end{eqnarray}  Therefore, it must be the case that $$\operatorname{d}(\alpha, \beta) > a$$  Set $\epsilon = \operatorname{min}\{\epsilon_1, \epsilon_2\}$.  Then, if $(\alpha, \beta) \in B_d(x, \epsilon) \times B_d(y, \epsilon)$,  $$a < \operatorname{d}(\alpha, \beta) < b$$  Since $\operatorname{d}^{-1}(U)$ can be written as the union of open sets in the metric topology, it follows that $\operatorname{d}^{-1}(U)$ is open. Therefore, $d$ is continuous.

Let $X$ be a metric space with metric $d$. Define $d: X \times X \to \mathbb{R}$, show that $d$ is continuous.  I would like to show that the function is continuous the topology way (since it is a course on topology).   Let $(a,b)$ be a basic open set in $\mathbb{R}$, then $d^{-1}(a,b) = \{(x,y): a<d(x,y)<b\}$. Define $A= \bigcup_{x \in X} B_{b} (x)$, where $B_b (x) = \{y \in X| d(x,y)<b$}. Clearly $A$ is open. Now define $C = \{(x,y):d(x,y)>a \}.$ Essentially I hope that $A \cap C$ is open. But I am stuck in showing that $C$ is open.  Any help please? :)


Let $U$ denote the set of all $n\times n$ matrices $A$ with complex entries such that $A$ is unitary. Then $U$, as a topological subspace of $\mathbb{C}^{n^2}$ is,   Compact, but not connected. Connected, but not compact. Connected and compact. Neither connected nor compact.   I know what the terms are but I am really not getting the idea how to solve this. Please help.

Is the set of all unitary matrices in $M_2(\mathbb{C})$ is compact? I can show that as determinant map is continuous so unitary matrices are closed but how to show they are bounded?   Please help.


Pages like this that have many embedded videos, freeze Ubuntu for a long time, then Flash crashes.  Is this just a Ubuntu 64-bit problem? Do I need more than my current 2gb ram?   I'm surprised that it can completely freeze up Firefox, since Firefox uses a separate process for plugins? Should this be reported as a bug, in Firefox, or Ubuntu?

I've noticed that Flash applications tend to be more sluggish under Ubuntu than they do under Windows on the same machine. This is particularly noticeable when watching HD video or playing graphics/physics-heavy games. Are there any ways of improving the performance of Flash under Ubuntu, or is this just an issue with the Linux version that I will have to live with?  Currently I'm just cutting down on the number of tabs open, blocking flash ads, and closing other programs, but I'm looking for ways to affect Flash itself.  Other things I have already been doing include using Youtube's HTML5 feature and playing videos straight from /tmp in VLC. I was wondering if there was some way of streamlining Flash itself though. Perhaps not.  More Specific Question: Is there anything I can do in mms.cfg to boost performance?


I have never faced with such operator... what does '$\setminus$' mean? Does this expression make any sense? $(A \cup B) \setminus C = A \cup (B \setminus C)$

I have a task to calculate a double integral on a region $D=A\setminus B$, where $A=\{(x,y): \text{some   system of inequalities}\}, B= \{(x,y): \text{another system of inequalities}\}$. What does operator "$\setminus$" mean?


I am trying to find all instances of some words, but only if they have a space on either side.  So for example, if I have the text " Word1 Word2 " I want to match both Word1 and Word2. However, the Word1 match includes the space after it, so Word2 is not counting it as a preceding space.   This is my Regex: (?: )(Word1|Word2)(?: )  Is there a way to re-use the non-capture group characters so that both Word1 and Word2 can use the same space in their match? I do not want the space included in the match.

I can't seem to find an answer to this problem, and I'm wondering if one exists. Simplified example:  Consider a string "nnnn", where I want to find all matches of "nn" - but also those that overlap with each other. So the regex would provide the following 3 matches:   nnnn nnnn nnnn   I realize this is not exactly what regexes are meant for, but walking the string and parsing this manually seems like an awful lot of code, considering that in reality the matches would have to be done using a pattern, not a literal string.


I reviewed a couple of low-quality posts, and when I want back to get the question, all I get is the review status, not the post itself:   :   :     Answer posts still show fine. Same results in private browsing.

On Ask Ubuntu, when I try to view my Low Quality Posts review history and click on the action taken on questions, I can only see the actions taken and no preview of the question.   :    The question still exists on the site:   It seems question review history page is missing this HTML class:  <div class="reviewable-post reviewable-post-post_id">   On the other hand, everything is working fine for answers.


I have come across it in the 9th episode of the 10th season of Friends. Here is the contex:  Chandler: Monica, I want a baby too, but this woman is giving away her child. She deserves to know who it's going to. Monica: (realizes Chandler is right. She's almost crying) okay, right. (They hug) Chandler: So, we'll tell the truth and who knows, maybe she'll like us for us.  I think it means to like someone for what they are, but it's the first time I have seen the phrase. If it means exactly the same as like some for what they are, would it sound as natural if Chandler said maybe she'll like us for what we are? If not, what are other ways to paraphrase it?

What does the following mean?  They like me for me.  I have never seen this expression before.


Using either the standard ArcMap labelling engine, or the Maplex engine, it is possible to automatically omit labels from the map when the area is too congested.   Eg in the centre of this image, there are too many towns close together so their labels aren't placed:    Is there a similar method to switch off those towns' symbols as well?  Ideally I'm looking for something automated, which works based on the current viewport (in the same way that the labels are automatically re-positioned if you pan or zoom the map)

Using ArcMap 10.0, I am working on a map that shows many villages. Each village has a name attribute that I want to use for a label, but in some areas there are so many villages that not all can be labeled. I am using the Maplex label engine and have not checked the "Never Remove (allow overlap)" box.  I am quite happy that not all village names are displayed as this would be way too much. And it is ok that Maplex chooses which labels to display, as I have no way to know which village is more important than another one. This also means that I can't base the labeling on a field such as population or something else, even though that would be less random.  However, I would also like the village points that are not labeled to similarly not be displayed. I think it unlikely that someone would want to see the point without knowing the name of the village. If all points are shown, it is also difficult to tell which point a name label goes with in crowded areas. So instead of choosing just labels not to be shown, I also want the computer to chose features not to be shown.  To give you a better idea on what I mean:    Is there a way to make these features/points not draw if their label is not shown?


The nozzle makes the water flow through the upper part of the table tennis ball. If the angle and velocity of the water jet are well controlled, the table tennis ball will rise to the sky. The string is used to hold the table tennis ball and prevent it from being washed away by water. Someone told me that it's because of Bernoulli's law, because the water above table tennis is is moving, and the air below table tennis is is static. But I don't believe that explanation. Because of the relativity of movement, I can also say that the air under table tennis is is moving, while the water is static.

Caution: Apparently this problem is harder than it seems!    There is a well known phenomena, which I first learnt about when I was a 10 years old kid. You can levitate a ping pong(or whatever not-so-heavy) ball in the air by blowing throw an empty pen body or using a hair dryer. An interesting demonstration can be seen .    While this demonstration is usually  (for instance look at  or other citation links from ), it seems that it is explained using the . The Coandă effect is the tendency of a fluid jet to be attracted to a nearby surface.    I am looking for a simple(and humbly realistic) model which is capable of answering the following questions quantitatively: Assume the ball's mass is $m$ and its radius is $r$, the diameter of our pen is $d$, the  of air through the pen body is $Q$ and the gravitational constant is $g$   What is the minimum of $Q$ which makes this demonstration possible? Perhaps using this minimum rate $Q$ and , how can I estimate the longest time that a normal person can levitate a ping pong ball in the air using his breath? I have noticed that the smaller $d$ is the longer I can levitate the ball. For a normal pen body, my personal experience is about 3 seconds. An estimation of the maximum angle that the pen can make with vertical direction? If we deflect the ball by a tiny amount, what will be the frequencies of motion along the vertical and horizontal direction?


I have this code:  \documentclass[tikz,border=5mm]{standalone} \usepackage[edges]{forest}  \definecolor{foldercolor}{RGB}{124,166,198}  \tikzset{pics/folder/.style={code={%     \node[inner sep=0pt, minimum size=#1](-foldericon){};     \node[folder style, inner sep=0pt, minimum width=0.3*#1, minimum height=0.6*#1, above right, xshift=0.05*#1] at (-foldericon.west){};     \node[folder style, inner sep=0pt, minimum size=#1] at (-foldericon.center){};}     },     pics/folder/.default={20pt},     folder style/.style={draw=foldercolor!80!black,top color=foldercolor!40,bottom color=foldercolor} }  \forestset{is file/.style={edge path'/.expanded={%         ([xshift=\forestregister{folder indent}]!u.parent anchor) |- (.child anchor)},         inner sep=1pt},     this folder size/.style={edge path'/.expanded={%         ([xshift=\forestregister{folder indent}]!u.parent anchor) |- (.child anchor) pic[solid]{folder=#1}}, inner xsep=0.6*#1},     folder tree indent/.style={before computing xy={l=#1}},     folder icons/.style={folder, this folder size=#1, folder tree indent=3*#1},     folder icons/.default={12pt}, }  \begin{document}   \begin{forest}     for tree={font=\sffamily, grow'=0,     folder indent=.9em, folder icons,     edge=densely dotted}     [main folder       [images, this folder size=20pt           [wallpapers]           [logo.pdf, is file]]       [tex-files           [chapter1.tex, is file]]       [main.tex, is file]       [main.aux, is file]     ]   \end{forest} \end{document}   Output:    The "main folder" text isn't centered above the line. How can I change it ?   Thank you.

How to draw the directory tree image shown below, indicating folders and files?     To draw the folder tree I used the code from , but there is no obvious option to include files, and I therefore edited the image in Inkscape, as indicated in red.  The replicated code from Gonzalo Medina post () is shown below      \documentclass[border=5pt]{standalone}     \usepackage{forest}      \definecolor{folderbg}{RGB}{124,166,198}     \definecolor{folderborder}{RGB}{110,144,169}      \def\Size{4pt}     \tikzset{       folder/.pic={         \filldraw[draw=folderborder,top color=folderbg!50,bottom color=folderbg]           (-1.05*\Size,0.2\Size+5pt) rectangle ++(.75*\Size,-0.2\Size-5pt);           \filldraw[draw=folderborder,top color=folderbg!50,bottom color=folderbg]           (-1.15*\Size,-\Size) rectangle (1.15*\Size,\Size);       }     }      \begin{document}      \begin{forest}       for tree={         font=\ttfamily,         grow'=0,         child anchor=west,         parent anchor=south,         anchor=west,         calign=first,         inner xsep=7pt,         edge path={           \noexpand\path [draw, \forestoption{edge}]           (!u.south west) +(7.5pt,0) |- (.child anchor) pic {folder} \forestoption{edge label};         },         before typesetting nodes={           if n=1             {insert before={[,phantom]}}             {}         },         fit=band,         before computing xy={l=15pt},       }       [system       [config       ]       [lib         [Access         ]         [Plugin         ]       ]       [templates       ]       [tests       ]     ]     \end{forest}      \end{document}


I've always thought the answer was 'duh, obviously' but I'm hard-pressed to find RAW evidence all of a sudden. Assume you have a hand crossbow, loaded, in one hand, while your other hand is occupied (holding a sword or a very important relic or even missing for that matter). Can you fire that single shot from the crossbow before having to drop it for something more practical? I'm unsure, on account of not being able to clearly show that such a thing as a 'loaded state' exists for a crossbow, by RAW:  Ammunition Drawing the Ammunition from a Quiver, case, or other container is part of the Attack (you need a free hand to load a one-handed weapon).  Emphasis mine. That part might imply that the loading (and thus the hand requirement) is to happen as part of the attack. Then again, when later referring to using a Loading weapon for a melee attack,  A sling must be loaded to deal any damage when used in this way.  This could imply the sling (and by extension other loading weapons) can be pre-loaded. Or that it needs proper loading, ammo etc. even as an improvised melee weapon. Is there a solid RAW answer on this?

The Crossbow Expert feat states that as a bonus action, you can "attack with a loaded hand crossbow you are holding."  However, nowhere in the rules does it specify a "loaded" versus "unloaded" state for any weapon. There is a "Loading" property (PHB p.147) that just says that "because of the time to load this weapon, you can fire only one piece of ammunition from it..." but doesn't specify an amount of time, type of action, or anything to load the weapon. (And, in any case, if you have Crossbow Expert you can ignore the "Loading" property.)  Under "Ammunition" (PHB p.146) it just says that "drawing the ammunition from a quiver, case, or other container is part of the attack" which seems to imply that crossbows are not kept loaded all the time, but are loaded and fired as part of the same attack. So is there some other action that needs to be performed to load it (so you can have it loaded first, before the trigger hits?)  [The specific situation that I have in mind was the case of a character with Crssbow Expert and two hand crossbows, one in each hand: Can he fire both crossbows every turn, using his normal and bonus actions respectively, even though he doesn't have a hand free to load the crossbows? Do you need a hand free to load a crossbow?]


My brother has came up with this equality:  $$ {e}^{i\pi} = -1 \quad\Rightarrow\quad {\left({e}^{i\pi}\right)}^{2} = 1 \quad\Rightarrow\quad \boxed{2\pi i = 0} $$  But that result sounds impossible, so, where is the error?

We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$  Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?


Using the PROC GENMOD of SAS, in "Criteria For Assessing Goodness Of Fit", I got :  Deviance = 0.6841  Scaled Deviance = 1.1002   I know that a model with too large a deviance doesn’t fit the data well, but I dont know which deviance should I take  Some help would be appreciated

I was reading . They are performing a logistic regression (with R). At some point they calculate the probability of getting a residual deviance higher than the one they got on a $\chi^2$ distribution with degrees of freedom equal to the degrees of freedom of the model. Copying-pasting from their website...   > glm( cbind(using,notUsing) ~ age + hiEduc + noMore, family=binomial)  Call:  glm(formula = cbind(using, notUsing) ~ age + hiEduc + noMore,            family = binomial)   Coefficients: (Intercept)     age25-29     age30-39     age40-49       hiEduc       noMore       -1.9662       0.3894       0.9086       1.1892       0.3250       0.8330    Degrees of Freedom: 15 Total (i.e. Null);  10 Residual Null Deviance:      165.8  Residual Deviance: 29.92        AIC: 113.4        The residual deviance of 29.92 on 10 d.f. is highly significant:  > 1-pchisq(29.92,10) [1] 0.0008828339       so we need a better model     Why does it make sense to compute 1-pchisq(29.92,10) and why does a low probability indicate that something is going wrong with their model?


I'm trying to figure out how to find the screen resolutions available for a linux machine. If you set the wrong resolution or set vga=ask then it prompts you for available resolutions. Is there a way I can find that list somewhere after boot?

I'm looking for a method of reporting display resolution. I want to set up scripts to launch rdesktop, and I want to launch it on several machines with different resolutions, so I want a way to dynamically determine it.


Prove that two 2x2 homogenous systems having the same solutions are equivalent.    This is from Hoffman and Kunze where Equivalent Systems are systems where all equalities in one system are linear combinations of equalities in the other.   My attempt-  I split the problem into 3 cases of 3 different ranks: I made an observation that for the null space to be the same (and hence the homogenous systems to have the same solution), the rank of both matrices had to be same. If it wasn't the dimension of the solutions for the two equations wouldn't be the same and hence would contradict the given data.  If the matrices are rank 0, then they are both null matrices and hence are trivially equivalent.  If both matrices are rank 2, then both of them are invertible and hence, the systems are equivalent since the row space of both spans all of $R^2$. Hence, any equality can expressed as a linear combination of the equalities of the other.  If both systems have rank 1, then I am in a confusion. That only implies that all of the four fundamental subspaces have dimension 1. How do I show that the two systems are equivalent from this?

I've just begun to work learn Linear Algebra on my own through Hoffman and Kunze's book and the first problem set already has a question that I can't solve:  Prove that if two homogeneous systems of linear equations in two unknowns have the same solutions, then they are equivalent.  I can't seem to figure out how to prove this without resorting to case work where you account for the cases where one of the coefficients are zero and when both are.  Is there an elegant way to prove in general that when two systems of linear equations have the same solutions, they are equivalent? The converse is obvious enough though.  Definition of equivalence from the text :     Let us say that two systems of linear equations are equivalent if each equation in each system is a linear combination of the equations   in the other system.


Since I upgraded to the latest version of Ubuntu, I cannot install the Java OpenJDK package needed to compile Android custom ROMs. When I run sudo apt-get install openjdk-7-jdk, the following error appears:  Reading package lists... Done Building dependency tree        Reading state information... Done Some packages could not be installed. This may mean that you have requested an impossible situation or if you are using the unstable distribution that some required packages have not yet been created or been moved out of Incoming. The following information may help to resolve the situation:  The following packages have unmet dependencies:  openjdk-7-jdk :  Depends: openjdk-7-jre (= 7u51-2.4.6-1ubuntu4) E: Unable to correct  problems, you have held broken packages.   How do I fix this problem?

I would like to upgrade my main system to 16.04, but I work on projects that require OpenJDK 7.   Apparently it is not available from a trivial apt-get install openjdk-7-jdk. Only versions 8 and 9 are listed in the repository.  Can anyone point me to instructions on how to install it?


Let's say I have one variable that has a known metric (e.g., variable A has a mean of .7 and standard deviation of .3). How do I scale a second variable (variable B) to have the same mean and standard deviation?   I know how to make both variables have a mean of zero and standard deviation of one. (That's easy--translate them to z scores), but I can't think of how to center them around a particular value with a particular sd. Thanks!

I am looking for a method to transform my dataset from its current mean and standard deviation to a target mean and a target standard deviation.  Basically, I want to shrink/expand the dispersion and scale all numbers to a mean. It doesn't work to do two separate linear transformations, one for standard deviation, and then one for mean.  What method should I use?


Ubuntu, as always, is very risky to set up. I believe I selected the option "erase disk" on the USB. It never mentioned Windows, like it did the first time I installed it, earlier tonight. I thought it was referring to the USB itself with a sort of misnomer, and went ahead with that. Now i can't boot Windows 10. Is it really gone? Is there any way I can recover enough memory to boot Windows 10 again?  I tried TestDisk, recovering the disks as best I could, but I still couldn't boot.  I really have to be less careless. I'm not this stupid, just really careless.  Thanks.

What steps can I take to try to recover lost or inaccessible data from any storage device?  Answers:   This applies to any computer storage device, e.g. internal/external hard drives, USB sticks, flash memory. The most important thing is to STOP using it, any type of I/O can ruin your chances of a recovery.     We have separate questions covering common problems with USB flash drives in greater detail:


Show that each natural number can be written as the difference of two natural numbers having the same number of prime factors.  What I Tried: Upon some inspection, I can see that all odd numbers, and all numbers of the form $4k$ can be written as a difference of two squares, because :- $$\rightarrow n= p^2 - q^2 = (p+q)(p-q)$$ Works for all odd $n$ if you take $(p + q) = n$ , $(p - q) = 1$ and similarly works for all $n$ of the form $4k$ if you take both $p,q$ to be even, directly factoring out a $4$. The problem with this, however, is that both the squares $p^2$ and $q^2$ aren't necessarily going to have the same prime factors as I expected, (like $6^2 - 5^2$) but I couldn't get any more ideas. Another thing left to solve here is to show that, all numbers of the form $(4k + 2)$ can be represented as a difference of two numbers having same number of prime factors. I noticed by some experiments that all these numbers can be represented as a difference of two primes, like :-  $1)$ $2 = 5 - 3.$  $2)$ $18 = 29 - 11.$  $3)$ $42 = 47 - 5.$ And so on. But I was not able to get a proof of it.  Regarding all my attempts with this problem, can anyone help me with the solution? Thank You.

Can we prove the following statement ?  Let $\omega(m)$ denote the number of all distinct prime factors of a positive integer $m$. Prove that every  natural  number $n$ can  be  written  as $n=s-t$, where $s$ and $t$ are positive integers with $\omega(s)=\omega(t)$, in other words, as the difference of two positive integers with the same number of distinct prime factors.   If $n=1$ , we can choose $\ s=3\ $ and $\ t=2$.  If $n$ is even , we can choose $\ s=2n\ $ and $\ t=n$.


We are moving a database off of a older physical server and putting it onto a Virtual Machine (SQL 2005 to 2014 I hope).  I need to copy a single database from the physical to virtual with as little downtime as possible. Originally I was just going to do a full backup and then restore on the new machine, but that will take almost an hour on this database.  My question is: If I can get the new virtual machine setup before the downtime, would it be effective to setup a database mirror from the old Physical box to the new Virtual box?  Then when the downtime came I would cut the mirror (database should be in sync) and realias the server to be the same name as the old physical database server?  I have never had to use mirroring so I was not sure if this was a "Smart" way to do it. If it would work the only downtime would be the name change.

I have a production server running SQL Server 2008 Web Edition, and it has to keep running 24x7 with almost no down time, and it's the time for us to move to a more advanced server hardware, also it's a good chance to move to SQL Server 2012 Enterprise.  Now my question, how can I do this migration without taking the database down at all? Like for example setting something between the two servers as replication or so, so I have exactly identical database up-to-second on both servers, then I can just take the main services down for 1 minute, changing the IP address of the database connection from old server to new server and I be sure that the new server has identical copy of the old one.  I can't go with backup/restore approach as it will take time more than acceptable period, that's why I need something which guarantee identical databases exactly as if I am doing backup/restore. 5 minutes could be the max affordable downtime.  The database size is about 100GB.  New server is running 2012 Enterprise, and old server is running 2008 Web  I can exceptionally archive records from the database and minimize its size down to 20GB like.  I have the option to upgrade my SQL Server 2008 Web edition to Enterprise edition if this will help with more solutions, like replication maybe?


Let us consider the set $X$ of the following infinite dimensional vectors:  $$ x = (x_{1}, x_{2}, \dots, x_{k}, 0, \dots,),  $$ such that $x_{i} \in \mathbb{Z}$ and for any $x$ there exist $k < \infty$ such that $x_{j} = 0$ for all $j > k$.   The statement: $X$ is not countable.   Try: I tried to show this using the fact that the set of all subsets of all integers is uncountable

On the one hand concatenation of the finite sequence produces a set that is a subset of the natural numbers. On the other hand, while Cantor's diagonal argument fails to produce a sequence with trailing zeros (at the same time not on the list)--I was wondering if there is a way to salvage the diagonal argument?


I do Kirchhoff law then I found I1=232uA, I2=158uA, I3= 0,074uA  now I want to calculate voltage Thevenin (voltage between A and B)?  I do like this 0.000232 * 10000=3.23  but I 3.23 not correct   Can you guide me

how to calculate voltage between A and B?


I need to calculate the date of the last Monday that before the latest weekend. Having used what I believe to be the most common Stack Overflow suggestions I have the following code:   const getDayOfTheWeek = () => {   let date = new Date();   let clonedDate = new Date(date.getTime());   console.log(clonedDate);   const dow = clonedDate.getDay();   console.log(dow);   const offset = dow+6;   console.log(offset);   const newDate = new Date(clonedDate.setDate(clonedDate.getDate() - offset));     console.log(newDate);   const newNewDate = new Date(newDate.getTime());   console.log(newNewDate);   const day = newNewDate.getDate();   const month = newNewDate.getMonth();   const year = newNewDate.getYear();    console.log('the year is ',year, 'the month is ', month); } getDayOfTheWeek();   It returns the year as 118 and the month as 5 which are ... not that Monday I need. newNewDate, on the other hand is the last Monday. I was wondering what causes it. I am aware that there are too many reassignments that are not needed. Please help.

Why does this javascript return 108 instead of 2008? it gets the day and month correct but not the year?  myDate = new Date(); year = myDate.getYear();   year = 108?


I have a slope created in percentage of declivity and I class this slope in some ranges (see the image), and I need to know how much % in 0 to 15 in that slope, how much % in 15 to 20 range.

How does one go about obtaining a summary of the counts of cell values within a raster in QGIS? I know QGIS is doing this, in order to provide the summary statistics and histogram in the properties of the layer. I do not need the summary statistics nor a histogram image, but rather counts of individual values.   It looks like there was at one point an attempt to add a special feature for this (Feature #4321, ).   Am I missing something really obvious?


I'm using window.getSelection().anchorOffset and focusOffset to turn selected area into an element. It works fine the first time. But then the parent element divided by new element and getSelection() start count from new element. I don't know how to get current element of getSelection(), this is the only way I can think of to fix this.  var temp = window.getSelection();     if (temp.anchorOffset < temp.focusOffset) {         var start = temp.anchorOffset;         var end = temp.focusOffset;     }else{         var end = temp.anchorOffset;         var start = temp.focusOffset;     }      var mid = text.substring(start,end);      $(".main-element").html(text.substring(0,start)+"<b>"+mid+"</b>"+ text.substring(end)); });

I'm trying to find a way with javascript to highlight the text the user selects when they click some odd highlight button (as in highlighted text). It only has to work with either WebKit or Firefox, but it seems to be well nigh impossible because it has to work in the following cases:
this is text

I eat food
   When the user selects from "is text" through "I eat" in the browser (can't just put a span there).  and this case:  this is textmiddle textthis is text   When the user selects from "is text" to "this is" in the browser (even though you can wrap your highlight spans around each element in the selection, I'd like to see you try to get that middle text highlighted).  This problem doesn't seem to be solved anywhere, frankly I doubt it's possible.  It would be possible if you could get the Range that you get from the selection as a string complete with html which could be parsed and then replaced, but as far as I can tell you can't get the raw html of a Range.. pity.


I recently moved my website to a new domain but seems like the link values are not passed till yet. I researched it a bit and found on Moz documentation that it may take sometime before the link values are passed to the new domain. Any ideas on how much time it can take? Any way to expedite it?

I have a friend who had around 2000 visitors per day to his website before a web designer with no SEO knowledge changed all his URLs. I've used the Wayback Machine and Google Analytics to determine the old URLs and I've done a 301 redirect and checked that it redirects properly.  My question is how long does it take before Google notices the redirects and passes link juice? And will those long-lost backlinks which where pointing to 404 pages count in Google Webmaster Tools - will I see a higher number of backlinks?   And last question, will I get a higher PageRank?


I know the principle of nested intervals can help to prove the question. But I can't complete the proof of the proposition.

Claim: If $f \in {\mathscr R[a,b]}$, then $f$ has infinitely many points of continuity.   1.) I read that it is a corollary of the Lebesgue integrability criterion.  Is it possible to prove the claim without invoking the concept of measure(or using less abstraction) ?    2.) Here is attempt: Given $\epsilon > 0, \exists$ Partition, $P =  \left\{ x_0 =a,...,x_n =b \right\} $ of $[a,b]$ such that $\sum^n_{i=1} (M_i -m_i)\Delta x_i < (b-a)\epsilon$, where  $M_i= \sup \left\{ f(x) : x \in \Delta x_i \right\}$ and $ m_i= \inf \left\{ f(x) : x \in \Delta x_i \right\} $  Let $(M_j -m_j)=\min \left\{ M_i -m_i : i=0,...,n \right\} $.  $ \implies (M_j-m_j)(b-a) \leq \sum^n_{i=1} (M_i -m_i)\Delta x_i <  (b-a)\epsilon$ $ \implies (M_j-m_j)<\epsilon. $  Let $ c\in (x_{j-1},x_j)$ and $\delta$ be any positive number such that $(c-\delta, c+\delta) \subseteq (x_{j-1}, x_j)$.   It follows that $ \left| f(x) - f(c) \right| < \epsilon, $ whenever $ \left| x - c \right| < \delta $.   Since $c$ is arbitrary and $[x_{j-1},x_j]$ is an interval,  there are infinitely many points of continuity.  There is something wrong with my proof since Thomae's function is a counterexample. Could anyone point out the mistakes in my proof? Thank you.


What is the point of using the keyword const? for example when making a game, one of the first things to do is to set the width and height of it. And most of the time you'll use for example:   const int Width   and  const int height   Now I know that you should do it like that because the width and height of the screen will not change throughout the game, but what is the point of doing so ? you can do the same thing without using const and it will work just fine.  That was just an example. so what I'm confused about right now is:  What is the point of using the const keyword anywhere if you won't change the variable anyway?

So why exactly is it that it's always recommended to use const as often as possible?  It seems to me that using const can be more of a pain than a help in C++.  But then again, I'm coming at this from the python perspective:  if you don't want something to be changed, don't change it.  So with that said, here are a few questions:   It seems like every time I mark something as const, I get an error and have to change some other function somewhere to be const too.  Then this causes me to have to change another function somewhere else.  Is this something that just gets easier with experience? Are the benefits of using const really enough to compensate for the trouble?  If you don't intend on changing an object, why not just not write code that doesn't change it?   I should note that at this point in time, I'm most focused on the benefits of using const for correctness and maintainability purposes, although it is also nice to have an idea of the performance implications.


I'm reading a textbook and I see this question:    So there are 200 women, and the DF is 196, implying that the equation for DF is $n - k - 1$. There are 3 variables: bp, age, and type so $k == 3$. What's the intuition behind this?  Also, why is the degrees of freedom for linear regression n - 2?

I'm working through an econometrics textbook and it's proving that   $$ \sigma^2 = E(\hat{\sigma}^2) = \frac{SSR}{n-2} $$  I followed the proof (an example of which is shown on ) until it reaches this step:  \begin{align} E\left[\displaystyle\sum\limits_{i=1}^n \hat{u}_{i}^{2}\right] &= E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})^{2}\right] + (\hat{\beta}_{1}-\beta_{1})^{2} E\left[\displaystyle\sum\limits_{i=1}^n (x_{i} - \bar{x})^2\right] \\ &+ 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})(x_{i}-\bar{x})\right] \\ &= \dots \end{align}  The book says that:   The expected value of the first term should be $(n-1)\sigma^2$ The expected value of the second term should be $\sigma^2$. The expected value of the third term should be $2 \sigma^2$ because the term itself can be written as $2(\hat{\beta}_1-\beta_1)^2 \hat{\sigma}_x^2$.   For the first term, I can work backwards and get that:  \begin{align} \hat{\sigma}^2 &= \frac{1}{n-1} \displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})^{2} \\ &= \sigma^2 \end{align}  but it seems like this only works if we assume that $\hat{\sigma}^2$ is an unbiased estimator.  For the second term:  \begin{align} (\hat{\beta}_{1}-\beta_{1})^{2} E\left[\displaystyle\sum\limits_{i=1}^n (x_{i} - \bar{x})^2\right] &= (\hat{\beta}_{1}-\beta_{1})^{2} E\left[(n-1) \hat{\sigma}_x^2\right] \\  &= (\hat{\beta}_{1}-\beta_{1})^{2} (n - 1) E\left[\hat{\sigma}_x^2\right] \\ &= (\hat{\beta}_{1}-\beta_{1})^{2} (n - 1) \hat{\sigma}_x^2 \end{align}  but this doesn't match what the text shows, so I'm a bit confused as to the final step for this term.   For the third term: \begin{align} 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n (u_{i}-\bar{u})(x_{i}-\bar{x})\right] &= 2 (\hat{\beta}_{1}-\beta _{1}) E\left[\displaystyle\sum\limits_{i=1}^n u_{i}(x_{i}-\bar{x})\right] \\ &= 2 (\hat{\beta}_{1}-\beta_{1}) \displaystyle\sum\limits_{i=1}^n E\left[u_{i}(x_{i}-\bar{x})\right] \\ &= 2 (\hat{\beta}_{1}-\beta_{1}) \displaystyle\sum\limits_{i=1}^n E\left[u_{i}\right]E\left[(x_{i}-\bar{x})\right] \\ \end{align}  I would have thought that because $E[u_i] = 0$, this term would cancel out, so I'm not sure how to derive $2\sigma^2$ from it.  Can someone give me a hint in the right direction for these steps? I'm firmly trying to work through all of the proofs myself with only a little guidance from the book, but I'm not quite there yet.  I noticed , which links to , so maybe with more reading I'll see where to apply that. (If that isn't it, hints are welcome, but I'll still work through those questions to see if it helps)


Okay, there are various questions.  First, "matter and energy bends space-time" does this mean any form of energy can bend space-time?  Does theory of relativity assume that there is no other form of energy other than mass energy (includes kinetic energy) and light energy?  And is it true? I mean, is every form of energy in basic form is mass energy (includes kinetic energy) or light energy? For example, I think that electrostatic potential energy is virtual potential energy caused due to exchange of bosons, which basically is mass energy.  So, Does electrostatic potential or magnetic potential energy bend space-time? Can field theory and relativity co-exist?

It is written everywhere that gravity is curvature of spacetime caused by the mass of the objects or something to the same effect. This raises a question with me: why isn't spacetime curved due to other forces or aspects of bodies?  Why isn't it that there are curvatures related to the charge of a body or the spin of particles or any other characteristics?


When I went to school I was specifically told that fields are material (they occupy some region in space, and they "exist" there) and continuous.   Recently, studying quantum physics I came across the notion of gauge particles which carry all kinds of interactions. From this perspective, if I judge it correctly, fields do not exist - they are essentially these particles being exchanged all the time between the interacting objects.   So, what is the current approach to the concept of fields?

There was a reason why I constantly failed physics at school and university, and that reason was, apart from the fact I was immensely lazy, that I mentally refused to "believe" more advanced stuff until I understand the fundamentals (which I, eventually, never did). As such, one of the most fundamental things in physics that I still don't understand (a year after dropping out from the university) is the concept of . No one cared to explain what a field actually is, they just used to throw in a bunch of formulas and everyone was content. The school textbook definition for a field (electromagnetic in this particular case, but they were similar), as I remember it, goes like:  An electromagnetic field is a special kind of substance by which charged moving particles or physical bodies with a magnetic moment interact.  A special kind of substance, are they for real? This sounds like the authors themselves didn't quite understand what a field is so they decided to throw in a bunch of buzzwords to make it sounds right. I'm fine with the second half but a special kind of substance really bugs me, so I'd like to focus on that. Is a field material? Apparently, it isn't. It doesn't consist of particles like my laptop or even the light. If it isn't material, is it real or is it just a concept that helps to explain our observations? While this is prone to speculations, I think we can agree that in scope of this discussion particles actually do exist and laws of physics don't (the latter are nothing but human ideas so I suspect Universe doesn't "know" a thing about them, at least if we're talking raw matter and don't take it on metalevel where human knowledge, being a part of the Universe, makes the Universe contain laws of physics). Any laws are only a product of human thinking while the stars are likely to exist without us homo sapiens messing around. Or am I wrong here too? I hope you already see why I hate physics. Is a field not material but still real? Can something "not touchable" by definition be considered part of our Universe by physicists? I used to imagine that a "snapshot" of our Universe in time would contain information about each particle and its position, and this would've been enough to "de" it but I guess my programmer metaphors are largely off the track. (Oh, and I know that the uncertainty principle makes such (de)serialization impossible — I only mean that I thought the Universe can be "defined" as the set of all material objects in it). Is such assumption false? At this point, if fields indeed are not material but are part of the Universe, I don't really see how they are different from  except for perhaps a more geeky flavor. When I talked about this with the teacher who helped me to prepare for the exams (which I did pass, by the way, it was before I dropped out), she said to me that, if I wanted hardcore definitions,  a field is a function that returns a value for a point in space.  Now this finally makes a hell lot of sense to me but I still don't understand how mathematical functions can be a part of the Universe and shape the reality.


Which one is the right expression? Assuming that this friend is still great today. And for some reason, I regret not telling you earlier about it.  1) I should have told you you are a great friend.  2) I should have told you you were a great friend.  Thank you!

I know the past tense carries the past tense in every dependent clause, but referring specifically to places or to things that are eternal, like the Earth, seems a bit weird and therefore we sometimes (I believe incorrectly) say     He didn't know that New Jersey was actually on the East Coast.   Because it still is. Or     He thought the Earth was round.   So is it square now?  Logically speaking, would you consider the use of past tense here a bit confusing in a day-to-day speech in these examples? Would you instinctively opt for using the present tense?


I wrote a program to run the insertion sorting algorithms. Everything seems good.  But when I execute it there is a strange result.  It seems that the programs uses garbage in memory.  How could I fix it?   #include<stdio.h> #include<stdlib.h> #include<time.h>  void insertion_inc(int Array[], int size) {      int i, j, key;      for (i = 1; i < size; i++)     {         j = i - 1;         key = Array[i];         while (j >=0 && Array[j]>key )         {             Array[j + 1] = Array[j];             j--;         }         Array[j++] = key;     } }   void insertion_dec(int Array[], int size) {      int i, j, key;     for (i = 1; i < size; i++)     {         j = i - 1;         key = Array[i];         while (j >= 0 && Array[j] < key)         {             Array[j + 1] = Array[j];             j--;         }         Array[j++] = key;     } }  void print_array(int arr[], int size) {      for (int i = 0; i < size; i++)         printf(" %d ", arr[i]); }   /////////////////////////////////////////////  int main() {      time_t t;     srand((unsigned)time(&t));     int Arr[10];     int size = sizeof(Arr);     printf("The unsorted Array : \n ");     for (int i = 0; i < size;i++)         Arr[i] = rand()/10;     print_array(Arr, size);     printf("\n\n\n The sorted Array : \n");     insertion_inc(Arr, size);     print_array(Arr, size);     return 0; }   **The output always looks like this : **  The unsorted Array :  12  23  54  87  2  68  29  57  97  12  -858993460  272204808  7601072  7742451  1  15039136  15044096  1  15039136  15044096  7601164  7742023  272204692  7738189  7738189  8462336  0  0  0  0  0  0  0  0  7775604  7775616  0  7601080  0  7601272  The sorted Array :  23  23  23  23  23  23  23  23  23  23  23  23  23  23  23  57  68  7601272  7738189  7738189  7738189  7742023  7775616  8462336  8462336  8462336  15039136  15039136  15039136  15039136  15044096  272204692  272204808  272204808  272204808  272204808  272204808  272204808  272204808  272204808  Even when I ignore the use of srand() and rand(), there is still the problem.

How do I determine the size of my array in C?   That is, the number of elements the array can hold?


3.333 A function cannot be its own argument, because the functional sign already contains the prototype of its own argument and it cannot contain itself. If, for example, we suppose that the function F(fx) could be its own argument, then there would be a proposition “F(F(fx))”, and in this the outer function F and the inner function F must have different meanings; for the inner has the form φ(fx), the outer the form ψ(φ(fx)). Common to both functions is only the letter “F”, which by itself signifies nothing. This is at once clear, if instead of “F(F(u))” we write “(∃φ) : F(φu) . φu = F u”. Herewith Russell’s paradox vanishes.  I fairly understand what the author means here, but it is in the bold part that I'm having difficult understanding what is meant. I don't understand why the function φ should be unique in this case, and how it relates to Russel's paradox.

Can anyone explain to a non-logician how Tractatus 3.333 refutes (or fails to refute) Russell's Paradox? Please explain his use of symbols!


I am working with .jpg images that are all B/W. Here is an example image I am using:    and here is what I am using to put the image into my document:  \begin{figure}[!htb] \centering \includegraphics[width=0.5\textwidth]{1980image} \caption[Sample Caption]{Sample Caption} \end{figure}   However, when I run, the image becomes entirely black. Here is a screenshot from my document:    Has anyone seen this before? Changing the filetype to .png does not fix anything.

I write some of my network notes in TeX. I wanted to make a table, and inside it insert an image:   .   I wrote a document:  \documentclass{article} \usepackage[demo]{graphicx} \usepackage{booktabs} \usepackage{array}  \begin{document} \begin{tabular}{m{3cm}m{3cm}m{5cm}} \toprule Configuration for ... & Commands & Topology\\ \midrule PPP: CHAP, PAP & \#sh int se 0/0/0  & \includegraphics[width=4cm, height=4cm]{chappap.png}\\  & & \\ \bottomrule \end{tabular} \end{document}   But instead of my network topology image, I see only this: .   I don't know what is wrong, cause my image is 100% in the same dir, as my *.tex file.  EDIT:  Changed my code as @Jubobs suggested, now everything is as I wanted:  \documentclass{article}     \usepackage{graphicx}     \usepackage{booktabs}     \usepackage{array}      \begin{document}     \begin{tabular}{m{3cm}m{3cm}m{5cm}}     \toprule     Configuration for ... & Commands & Topology\\     \midrule     PPP: CHAP, PAP & \#sh int se 0/0/0  & \includegraphics[width=4cm, height=4cm]{chappap.png}\\      & & \\     \bottomrule     \end{tabular}     \end{document}


I found some answers to this question, but that was with only two dice, with 6 sides. What if I have 30 dice, with 10 sides each (from 1 to 10). I can't make a matrix and count as was suggested on the other post.  E.g. what is the probability of getting the sum of at least 100 with 30 dice that has 10 sides each?

This is a dice problem  1) I want to calculate the probability to have more than X throwing 3 dice of 6 faces. X = addition of the result of the 3 dice.  2) This is the first step but if you can also provide me a solution to calculate the probability to have more than X with Y dices of Z faces it would be really great.


I have found that there are two $(x,y)$ that fulfill the property that $x^y=y^x$, $x\neq y$:   $(2,4)$ $(4,2)$   From this:   How can I find more, if any? How can I prove that there are no other numbers that fulfill this property, if there aren't any more such numbers?   I am doing two aspects: one with various domains. I am rather confused as to how I might find more answers like this, and am nearly convinced that there aren't any.  Any help is deeply appreciated!  Edit: Not a duplicate  I am also asking if there are solutions in different domains. I would like solutions in $\mathbb{Z,R,}$ and $\mathbb C$, if possible.

We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\neq y$) which satisfies the equality $x^y = y^x$?


On Stackoverflow with it's higher visibility and factual natural, this might not come across often, but on stack exchange for say "interpersonal skills", a lot of non-factual opinions are expressed, so that can cause moderator bias.  For example if you don't like someone's wording on something or believe in certain things, you can remove all opposing comments when they are completely legit. For example, . Mutiple people pointed out that the correct top two answer aren't useful. It might sound nice to "simply to truthful and respectful and ask", but that's not what the OP is asking and he clearly mentions it (that he doesn't want to risk offending the other person). Even the question is posted as "how to check" not "how to politely ask". All such comments were deleted for some random reason... Why?

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


this question is simple about "Open/Run in Terminal"....  Looks Ubunutu doesn't support bash/shell-scripts the way we know from KDE and any other OS:  I tried the obvious:    Right click in file manager   Open with other Application... Show all applications   but there is no Terminal application listed, only "Find Online Applications". You can't even browse for a new application as you would do in any other OS.   Any ideas? I found only hacks like creating a launcher item but that's far from a solution or ideal.   To make it really clear:   I want to run any bash script accessible with the file manager via right-click, "Open with" -> "Terminal" so i can see the output. Its really like double click on a batch file under Windows and it just runs the bloody script, not closing the terminal except I wrote so in the script. So finally this is NOT about double click to run, its about the running the script through context menu as said in the title already!  thanks!  Ubuntu-x64-14.04/Unity/No Mods or tweaks.

How can I make a bash script executable by double clicking just like .exe files in Windows? I tried , but there are two consequences:   the terminal twinkles, disappears, and nothing is done. you must specify to run in terminal in order to work.   I have a script that installs tomcat on an offline PC, including all dependencies of tomcat in the script. I need to make the script work on double clicking like windows since most who use the script will not be familiar with Ubuntu.  Forget the above explanation. I want to make a script that can be run by double-clicking on it, without using the terminal. Anybody knows how?