In nuclear physics, when you break eg. a nucleus of Uranium, some neutrons are liberated, and the original atom degrades to a lighter element. The energy that was used to keep these subatomic particles together is liberated (strong interaction), and a small part of the mass is converted to energy (E=MC^2) so you get a lot of energy with a small amount of atoms.  So why nuclear fusion (the opposite operation) could even liberate MUCH more energy? I would naively expect it to take a lot of energy, not liberate it.  I'm not a physician nor a student, just interested in physics, and this has always been a mystery to me.   EDIT: actually, both of your answers are great and cristal-clear. That makes perfect sense and is exactly what I was looking for. I wish I could accept both answers, but the one with the graphic was a bit more complete. But thank you two!

Lighter nuclei liberate energy when undergoing fusion, heavier nuclei when undergoing fission.  What is it about the nucleus of an Iron atom that makes it so stable?  Alternatively: Iron has the greatest nuclear binding energy - but why?


A little while ago I came across this question on Yahoo! Answers and it had no replies. I have spent a while trying to figure this out as well, but I couldn't find an algebraic proof. The assumptions are $a+b=1$ and $a,b\in(0,1)$. We want to show that $a^ab^b+a^bb^a\leq 1$. Or equivalently that $a^a(1-a)^{1-a}+a^{1-a}(1-a)^a\leq 1\;\forall a\in(0,1)$.  I tried see if $a$ and $b$ are an upper bound for $a^ab^b$ and $a^bb^a$ respectively, for their sum would be $1$ by assumption. However, I was unable to do so. I managed to show the following, assuming without loss of generality that $b\leq a$ (i.e. $a\geq\frac{1}{2}$): $$\begin{aligned} a^ab^b&=a^{1-b}b^b\\ &=a\left(\frac{b}{a}\right)^b\\ &\leq a. \end{aligned}$$ However, applying this same trick to $a^bb^a$ does not yield the necessary inequality: $$\begin{aligned} a^bb^a&=a^bb^{1-b}\\ &=b\left(\frac{a}{b}\right)^b\\ &\geq b. \end{aligned}$$ Which means that the upperbound $a^bb^b\leq a$ is too weak. By similar means one can find the following bounds: $$b\leq a^ab^b\leq a,$$ $$b\leq a^bb^a\leq a.$$  Now define $f(a):=a^a(1-a)^{1-a}=a^ab^b$. One can show that the following holds for the derivative of $f$: $$f'(a)=f(a)\log\left(\frac{a}{1-a}\right).$$ Since $f(a)=\exp(a\log(a)+(1-a)\log(1-a))>0$, we have that $f'(a)=0\iff a=\frac{1}{2}$. Since $f'(a)>0$ for all $a>\frac{1}{2}$ we have that $f$ has a minimum at $a=\frac{1}{2}$. Also note that $\lim_{a\to 1}f(a)=1$, hence by symmetry $\frac{1}{2}\leq f(a)\leq 1$ for all $a\in(0,1)$.  We'll define $g(a):=a^{1-a}(1-a)^a=a^bb^a$. One finds that $$g'(a)=g(a)\left(\frac{1-2a}{a(1-a)}+\log\left(\frac{1-a}{a}\right)\right).$$ Now $g'(a)=0\iff\log\left(\frac{1-a}{a}\right)=\frac{2a-1}{a(1-a)}$. It's clear that $a=\frac{1}{2}$ is a solution. Note that $\frac{d}{dx}\log\left(\frac{1-a}{a}\right)=\frac{1}{a(a-1)}<0$. Since $\frac{2a-1}{a(1-a)}$ is increasing, $a=\frac{1}{2}$ is the only solution. As $\lim_{a\to 1}g(a)=0$, we have that this is a maximum. Hence $0\leq g(a)\leq\frac{1}{2}$ for all $a\in(0,1)$.  Combining these results with what we already had yields the following inequalities: $$\frac{1}{2}\leq a^ab^b\leq a,$$ $$b\leq a^bb^a\leq\frac{1}{2}.$$ This gives us $$b+\frac{1}{2}\leq a^ab^b+a^bb^a\leq a+\frac{1}{2}.$$ Which is still not all too helpful... So then I got another idea. I construct two parabolas $\phi,\psi$ both with their top at $\left(\frac{1}{2},\frac{1}{2}\right)$. One of them, $\phi$, through $(0,0)$ and $(1,0)$ and $\psi$ through $(0,1)$ and $(1,1)$. Then I'd show that $a^a(1-a)^{1-a}\leq\psi(a)$ and $a^{(1-a)}(1-a)^a\leq\phi$. Then, since $\phi(a)+\psi(a)=1\;\forall a$ the statement is proven. However, this created some difficulty that I am now stuck on. I'll first define $\phi$ and $\psi$: $$\phi:(0,1)\to\left(0,\frac{1}{2}\right]$$ $$x\mapsto-2x^2+2x$$ Then $\psi(x)=1-\phi(x)$. We have $$\begin{aligned} \phi(a)&\geq a^{1-a}(1-a)^a\\ \iff-2a^2+a&\geq a^{1-a}(1-a)^a\\ \iff2a&\geq\left(\frac{a}{1-a}\right)^{1-a}. \end{aligned}$$ I cannot figure out how to prove this inequality though; let alone the other one. Hence now I am stuck!  I'm hoping someone can help figure this out. I realize parts of what I've done may be superfluous, but this is just everything that I have tried put together.

Given $a + b = 1$, Prove that $a^ab^b + a^bb^a \le 1$; $a$ and $b$ are positive real numbers.


Hi i want to add the box description into my flowchart like . (Sorry for bad drawing). first, i don't know how to make that box and shapes inside it and aligned in equal symbol. The box should be about 5x6cm (flexible). And i want the flowchart have caption just below the flowchart. This is my attempt for the flowchart only: \documentclass{book} \usepackage[utf8]{inputenc} \usepackage{tikz} \usetikzlibrary{shapes.geometric, arrows} \usepackage{caption}  \tikzstyle{startstop} = [rectangle, rounded corners, minimum width=3cm, minimum height=1cm, text centered, draw=black, fill=red!30] \tikzstyle{io}=[trapezium, trapezium left angle=70, trapezium right angle=110, minimum width=3cm, minimum height=1cm, text centered, draw=black, fill=blue!30] \tikzstyle{process} = [rectangle, minimum width=3cm, minimum height=1cm, text centered, draw=black, fill=orange!30] \tikzstyle{decision} = [diamond, minimum width=3cm, minimum height=1cm, text centered, draw=black, fill=green!30] \tikzstyle{arrow} = [thick,->,>=stealth] \tikzstyle{line} = [draw, very thick, color=black!50, -latex']  \begin{document}     \begin{tikzpicture}[node distance=2cm]         \node(start)[startstop]{Start};         \node(in1)[io, below of=start]{Input};         \node(pro1)[process, below of=in1]{Process 1};         \node(dec1)[decision, below of=pro1, yshift=-0.5cm]{Decision 1};         \node(pro2a)[process, below of=dec1, yshift=-0.5cm]{Process 2a};         \node(pro2b)[process, right of=dec1, xshift=2cm]{Process 2b};         \node(out1)[io, below of=pro2a]{Output};         \node(stop)[startstop, below of=out1]{Stop};                  \draw[arrow](start)--(in1);         \draw[arrow](in1)--(pro1);         \draw[arrow](pro1)--(dec1);         \draw[arrow](dec1)--node[anchor=east]{yes}(pro2a);         \draw[arrow](dec1)--node[anchor=south]{no}(pro2b);         \path[line, dashed](pro2b) |- (pro1);         \draw[arrow](pro2a)--(out1);         \path[line](out1)--(stop);      \end{tikzpicture}    \end{document}  Please help me. Thanks in advance.

could not figure out an answer to this one.  I have a tikz diagram using tikz/matrix, and I want to have a legend next to it, also using tikz/matrix. But I do not know how to refer to one particular matrix in the style definitons.  Let me visually show the problem:  \documentclass{article}  \usepackage{tikz} \usetikzlibrary{     positioning,     shapes.geometric,     matrix}  \begin{document}  \begin{tikzpicture} \tikzset{     status/.style={rectangle, draw=black, text centered, anchor=north, text=black, minimum width=2em, minimum height=2em},     mstyle/.style={column sep=10em, row sep=3em,nodes={status},font=\bfseries},     line/.style={draw,thick,-latex},     row 1 column 1/.style={nodes={fill=green}},     row 1 column 2/.style={nodes={fill=red}},     row 1 column 3/.style={nodes={fill=yellow}},     row 2 column 3/.style={nodes={fill=gray}},     row 2 column 2/.style={nodes={diamond}}, } \matrix(m)[matrix of nodes,ampersand replacement=\&,mstyle]{     H \& A1 \& A2 \\       \& {} \& D \\ }; \tikzset{     status/.style={rectangle, draw=black, text centered, anchor=north, text=black, minimum width=2em, minimum height=2em},     m2style/.style={column sep=1em, row sep=1em,nodes={status},font=\bfseries}, } \matrix(m2)[right=of m, draw=black, matrix of nodes,ampersand replacement=\&,m2style]{     H  \& Healthy \\     A1 \& Infected  \\     A2 \& Infected  \\     D \& Death  \\ }; \draw[line] (m-1-1) -- (m-1-2) node[pos=0.5,above,align=left] {$\#\mathrm{A1} \geq 1$ or\\$\#\mathrm{A2} \geq R$}; \draw[line] (m-1-2) -- (m-1-3) node[pos=0.5,above] {wait $\tau$ ticks}; \draw[line] (m-1-3) -- (m-2-3) node[pos=0.5,right] {wait 1 tick}; \draw[line] (m-2-3) -- (m-2-2) node[pos=0.5,above] {$p_\mathrm{replenished}$}; \draw[line] (m-2-2) -| (m-1-1) node[pos=0.2,above] {$1-p_\mathrm{infected}$}; \draw[line] (m-2-2) -- (m-1-2) node[pos=0.5,left] {$p_\mathrm{infected}$}; \end{tikzpicture}  \end{document}   I want tikzset to be limited to the scope of the first matrix m ....


I am trying to make a simple irc server and client, and I am trying to find a way to have the server respond differently to certain messages sent to it. I am using the kryonet api for networking. Here is my listener:          public void received(Connection c, Object o){              if (o instanceof TransferMessage){                  TransferMessage msg = (TransferMessage)o;                  if (msg.text == "Hello"){                      System.out.println("User said Hello to me!");                  }                 System.out.println(tag+msg.user+": "+msg.text);                 server.sendToAllTCP(msg);              }          }   I have created a client program which sends the message "Hello" upon connecting. However, the server fails to recognize that msg.text does in fact equal "Hello", or some other anomaly is happening. There is no error produced, but the server ignores the conditional statement. I have also verified that the string that the client sends is what it is supposed to be.

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?


Say I have a , and my opponent and I each control other creatures. Magus of the Arena's ability is     {3}, {T}: Tap target creature you control and target creature of an opponent's choice he or she controls. Those creatures fight each other.   Is the following sequence of actions on my part legal?   I declare that I am activating the Magus's ability. I choose one of my creatures as a target. My opponent chooses one of their creatures as a target. Specifically, it is a creature I did not want to fight. I say "I decline to tap my lands for mana" since step 601.2f, only says that I have a chance to activate mana abilies, and not that I must do so. Then I say "I do not have enough mana to pay for the spell" (in step 601.2g). Now, according, to step 717, we back out of activating the ability, and do something else.   This seems to follow the letter of the rules, but it also seems to violate the spirit of the card in question.

Ok, I was playing an EDH game with my friend and I had 8 mana available to me. In order for me to win the game, I had to destroy one of his permanents with a  (7 mana). Now I planned to play a  (1 mana and produces 3 mana) so that I'd have enough mana for a  (3 mana) to protect my Spine of Ish Sah.  How I played the turn out is I try to cast Spine of Ish Sah, but I forgot to play Mana Vault first. By the time I've realized that I forgot a very important part in my plan, I've already tapped for 5 mana. At this point I stop tapping for mana and I am unable to pay for the Spine of Ish Sah and it should return to my hand (where I cast it from) and my lands should untap (rule 717). My opponent said that's not the game works. He says that since I started paying for the spell, I have to finish paying for the rest of it. Now, is my friend correct, that if you pay for part of a spell, if you are able to pay for all of the spell, then you must finish paying its cost?  Answer must have rules referencing this.


I need to show that if $X$ is a covering space of $Y$ with the covering map $p$ and $Y$ is connected, then $p^{-1}(y)$ have the same cardinality for every $y\in Y$.   I have this hint: A function $f:W\to Z$ is locally constant iff each $w\in W$ has an open neighbourhood $U$ such that $f\restriction_U$ is constant. Show that any locally constant function is continuous.  I've also found a solution which I didn't understand: Let $k$ be a cardinality. Denote those $y$'s by $B_k$ for which $|p^{-1}(y)|=k$. By the "local triviality" (now I absolutely don't know what does that mean, perhaps it has something to do with the hint I found somewhere else), every $B_k$ is open. But they are also closed at the same time because they are all complements of unions of open sets. Since $Y$ is connected, there is a $k$ such that $Y=B_k$.  If you can explain the meaning of "local triviality" and its connection with the hint and also the hint itself I would be very thankful.

In "Knots and Primes: An Introduction to Arithmetic Topology", the author uses the following proposition   Let $h: Y \to X$ be a covering. For any path $\gamma : [0,1] \to X$ and any $y \in h^{-1}(x) (x = \gamma(0))$, there exists a unique lift $\hat{\gamma} : [0,1] \to Y$ of $\gamma$ with $\hat{\gamma}(0) = y$. Furthermore, for any homotopy $\gamma_t (t \in [0,1])$ of $\gamma$ with $\gamma_t = \gamma(0)$ and $\gamma_t(1) = \gamma(1)$, there exists a unique lift of $\hat{\gamma_t}$ such that $\hat{\gamma_t}$ is the homotopy of $\hat{\gamma}$ with $\hat{\gamma_t}(0) = \hat{\gamma}(0)$ and $\hat{\gamma_t}(1) = \hat{\gamma}(1)$.  The author then follows with; "In the following, we assume that any covering space is connected. By the preceding proposition, the cardinality of the fiber $h^{-1}(x)$ is independent of $x \in X$."   I am not sure why this results is true. This is my attempt at explaining it to myself. Take two different $x_1, x_2$ and their fibers $h^{-1}(x_1), h^{-1}(x_2)$ such that $y_1 \in h^{-1}(x_1)$ and $y_2 \in h^{-1}(x_2)$. Take two paths $\gamma_1, \gamma_2$ such that $\gamma_1(0) = x_1$ and $\gamma_2(0) = x_2$. Then we get two lifts $\hat{\gamma_1}, \hat{\gamma_2}$ with $\hat{\gamma_1}(0) = y_1$ and $\hat{\gamma_2}(0) = y_2$. Now, since our covering space is connected we can continuously deform $\hat{\gamma_1}$ into $\hat{\gamma_2}$ and conclude that every elements in $h^{-1}(x_1)$ is also in $h^{-1}(x_2)$ and vice versa. I feel that this is wrong but cannot figure out the right way to see this. Any help would be appreciated.


$$\gcd(\text {lcm}(x, y), \text {lcm} (x, z)) = \text {lcm}(x,\gcd(y, z))$$ where $x,y,Z$ are three integers. I came across this property on GCD but was not able to prove this.Can anyone suggest me a method.

I honestly have no idea how to approach this question. I have done easier questions eg. show $gcd(a,b)lcm(a,b)=ab$ by letting $a=Alcm(a,b)$ and $b=Blcm(a,b)$ and then using the fact that $A$ and $B$ are coprime. However this approach only works in two variables. Any help with the question in the title would be greatly appreciated, as well as the thought process behind the answer so that I can approach similar problems.


Trying to make the footer in my html css website stick down but nothing works. I've tried changing the position to absolute and fixed and setting bottom: 0 and doing everything but nothing works. Also, is there a better way to make my logo aligned in the middle? Heres my css: .footer{     background-color: #d62929;     clear: both;     width: 100%vw;     display:block;     overflow: hidden;     padding-top:10px;     padding-bottom: 10px;     min-height: 100%vw;     } .contact{     margin-left: 30px;     margin: 0 auto;     display:block;     float: left;     padding-right: 50px;     } .info{     margin-left: 30px;     margin: 0 auto;     padding-left: 30px;     display:block;     float: left;     padding-right: 50px;     } .account{     margin-left: 30px;     margin: 0 auto;     padding-left: 30px;     display:block;     float: left;     padding-right: 50px;      } a{     text-decoration:none;     color: black;     font-family: times new roman;     font-size: 18px;     text-align: center;     } ul{     list-style: none;     text-align: left;     } .logo_footer{     float: left;     padding: 40px 0;     margin-left: 20px;     margin-right: 40px;     } h1{     color: white;     font-size: 24;     } li{ padding: 5px;     }  Heres my html for the footer:

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


The phone has some bugs in this new update I have tried doing a factory reset but it still has the android 8.

Mostly, android devices get updates through OTA to higher versions. I had upgraded to Kitkat(MIUI6) version, on my Redmi 1S using OTA. I have downloaded complete ROM of Jelly bean(MIUI5). But, I cannot install directly, it shows error and failed. That means an upgraded device cannot downgrade by direct installation. What should be done? (Unrooted device)


How to install ubuntu 14.10 on a 16gb usb flash drive. I want to install it, as if it were the hard drive, not live cd, because my hard drive is broken.

If you are hurrying to reply, System → Administration → StartUp Disk Creator -- no, that's not what I'm talking about.   I want to try Ubuntu 11.04's Unity without touching my existing Ubuntu install.  To do this, I need to install the nVidia drivers first (sigh).  To do this, I need changes to persist a reboot.  To do this, I need to really install Ubuntu on a USB key.  How do you do that?    What I tried   I tried to make a USB key from , then boot from it, then choose "Install Ubuntu." The installer refused to install to the installation media itself. I tried, from my installed copy of Ubuntu:  sudo kvm /dev/sdb --cdrom .cache/testdrive/iso/ubuntu_natty-desktop-i386.iso   ...but the installer didn't detect the disk properly.


Here is the solution:   First $f$ is monotone and integrable on $(0,\infty)$, wolg we can assume that $f>0$ and approaches $0$ as $x$ goes to infinity. Observe that  $$xf(2x) \leq \int_x^{2x} f(t)$$ since when $t\in [x,2x]$ we have $f(t) \geq f(2x)$ because $f$ is decreasing. And since $f$ is integrable, we have $$\lim_{x\rightarrow \infty} \int_x^{2x} f(t) = 0,$$ thus $$\lim_{x\rightarrow \infty} 2xf(2x) = 0.$$

Let $f:\mathbb{R}_+ \to \mathbb{R}_+$ be a monotone decreasing function defined on the positive real numbers with $$\int_0^\infty f(x)dx <\infty.$$ Show that $$\lim_{x\to\infty} xf(x)=0.$$   This is my proof: Suppose not. Then there is $\varepsilon$  such that for any $M>0$ there exists $x\geq M$ such that $xf(x)\geq \varepsilon$. So we can construct a sequence $(x_n)$ such that $x_n \to \infty $ and $x_n f(x_n ) \geq \varepsilon$. So $$\frac{\varepsilon}{x_n}\leq f(x_n) \implies \sum_{n\in\mathbb{N}}\frac{\varepsilon}{x_n} \leq  \sum_{n\in\mathbb{N}} f(x_n) \leq \int_0^1 f(x)dx.$$ So we get a contradiction.  I feal like I have the correct idea but some details are wrong.  Any help would be appreciated.


I have problem using combination keys like Ctrl or Alt or Shift. For example when I want to copy selected text I press Ctrl + C,but it isn't working. Or when I want to select all text, I press Ctrl+A, but this also doesn't work. In android studio all combination keys, doesn't work. When I open settings> keyboard> shortcuts , I can't edit shortcuts , because when I enter Ctrl + C for shortcut to copy, ctrl never gets entered , just C.

I use Android Studio in Ubuntu and everything was working fine. Since two weeks that shortcuts stopped working, Ctrl+C, Ctrl+V, Ctrl+Z. I tried to change Keymap in Settings but without success.  What could be the possible problem?


What is the font used  in this image please?  Best Regards,

Is it possible to identify the font used in a specific document/picture?  Answers to this question should identify:   Possible methods to do this (perhaps one answer per method) and adequately describe how to use it (as opposed to merely stating it); Ways of finding the identified fonts, if possible (free or not); and Any prerequisites associated with the method used, if required (for example, "In order to use method X, your document has to be in format Y").   This question is meant as an FAQ, based on an . Its aim is to facilitate the community with the general procedures involved in font identification. Similar cases are solved on a per-usage basis on 's  tag.


I am writing a remote terminal application and i use pipes to take data from a child process' stdout/stderr to then send it to a client. Specifically, I am running Linux, and I don't really care about portability (if the solution is Linux-only, it's OK).  I need to know how many bytes have currently been written to, but not yet read from, the pipe. I was Googling for a long time and couldn't find an answer. I need to read as many bytes from the pipe as possible without blocking.  Is this possible? Thank you for help.

I need a function called SizeOfPipe() which should return the size of a pipe - I only want to know how much data is in the pipe and not actually read data off the pipe itself.   I thought the following code would work:  fseek (pPipe, 0 , SEEK_END); *pBytes = ftell (pPipe); rewind (pPipe);   but fseek() doesn't work on file descriptors. Another option would be to read the pipe then write the data back but would like to avoid this if possible. Any suggestions?


i'm using volley library to receive data from json file I have this problem i don't know the reason for this error     Attempt to invoke virtual method 'android.content.Context.getResources()' on a null object reference     private void fetchRemoteData(final DataStatus callback){     StringRequest stringRequest = new StringRequest(Request.Method.GET,             URL_DATA,             new Response.Listener<String>() {                 @Override                 public void onResponse(String s) {                      List<Post> listItems = new ArrayList<>();                     try {                         JSONObject jsonObject = new JSONObject(s);                         JSONArray array = jsonObject.getJSONArray("bgs");                         for (int i = 0; i < array.length(); i++){                             JSONObject o = array.getJSONObject(i);                             Post item = new Post(                                     o.getString("img")                             );                             listItems.add(item);                         }                         callback.onSuccess(listItems);                     } catch (JSONException e) {                     }                  }             },             new Response.ErrorListener() {                 @Override                 public void onErrorResponse(VolleyError error) {                     callback.onError(error);                 }             });     RequestQueue requestQueue = Volley.newRequestQueue(getActivity());     requestQueue.add(stringRequest); }   error exist here :  @Override                 public void onErrorResponse(VolleyError error) {                     callback.onError(error);                 }

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


I have been trying to tabulate some data using latex but every time i put in a large sentence it doesn't go the next line automatically, but instead ti just extends the table as wide as the sentence is, in addition to that the font sizes in my table isn't the same as the rest of my document.  Here is my LaTeX table code:      \begin{table}[] \centering \caption{Input parameters for the rational method formula, Equation \ref{discharge}.} \label{imput} \begin{adjustbox}{width=1\textwidth} \begin{tabular}{|l|l|l} \cline{1-2} \textbf{Input Parameters} & \textbf{Description}  \\ \cline{1-2} Coefficient of runoff & The coefficient of runoff accounts for the hydraulic behavior of the runoff as it flows over different types of surfaces in the drainage area. The coefficient for different surfaces can be obtained from Table \ref{c}   \\ \cline{1-2} Rainfall Intensity & This parameter represents the amount of rainfall per a specified time. This can be obtained from rainfall intensity vs duration curves. However, In South Africa the rainfall intensity can be obtained from the software discussed in Topic [REF A CHAPTER]   \\ \cline{1-2} \multirow{3}{*}{Time of Concentration} & This parameter represents the amount of time it would take the runoff from any point to get to a drainage outlet. Moreover, this parameter can go further in-detail as to consider the time of surface flow and time of flow in the drainage structure.  \\ \cline{2-2} & The key factors that can affect this system is that the slope of the land and characteristics of the surface that the runoff will be flowing over.  \\ \cline{2-2} & It is recommended that Equation \ref{tcs} should be used for the estimation of surface flow time. Moreover, Equation \ref{tcd} can be used to obtain the drainage flow time. \\ \cline{1-2} \end{tabular} \end{adjustbox} \end{table}

I am trying to create linebreaks in some of the cells of my table, since I have long sentences and the table doesn't fit on an A4 page. How to let LaTeX choose the linebreaks?  Here is my code   \begin{table}[hbt] \centering \begin{tabular}{lllllll}     \toprule     &  Raumgewicht $\gamma$'$_m$ and [$\frac{kN}{m^3}$] &Reibungswinkel $\phi$'$_m$[]& Kohäsion c'[] & M$_E$-Wert []& N$_SPT$ [] & k-Wert []  \\     \midrule     Auffüllungen/Deckschicht& 18& 27 & 0 & 10* & 17 & -\\     Rückstausedimente nicht vorbelastes& 21 & 28 & 10 & 10 & 15 & $5\times10^-8$ - $1\times10^-6$ \\     Rückstausedimente vorbelastet & 22 & 32 & 10 & >50 & 91 & $5\times10^-8$ - $1\times10^-6$ \\     Moräne & 22 & 36 & 5 & >50 & >100 & $1\times10^-8$ - $5\times10^-6$\\     \bottomrule \end{tabular} \caption{Bodenkennwerte} \label{tab:Bodenkennwerte} \end{table}   My preamble is  \documentclass[12pt,a4paper]{scrartcl} \usepackage[utf8]{inputenc} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{graphicx} \usepackage{booktabs} \usepackage{stackengine} \usepackage[top=3cm, bottom=2.5cm, left=2.5cm, right=2.5cm, footskip = 1cm]{geometry} \usepackage{hyperref}


In the context of a tutorial, I need to create this experience using a light spot on cycle engine. We can replace persons with other objects.  I tried to do it using this answer  but the mirror does not reflect the light. here is my .blend . Any help?

Working in Cycles.  Say I have a "disco ball" with a glossy shader. I'm pointing a spot light to it, and I'm trying to have the reflections of that light illuminate the diffuse material on the walls with the patterns from the ball.    I get this:    but I'd like to get something like:    For the illustration I placed a point light inside the ball an removed faces. But what I'd like to do is move the light around the ball so that the light reflected from it changes like it would do in real life.   Is that possible at this point?  EDIT:  Just for the record here's the image rendered in LuxRender (yet another program to learn...)


On Serverfault:   If you try to edit a message in a review queue (Review First Posts for example) and you have too many pending edits, the error message   "You have too many pending edits [...]"    won't show up, nothing happens (tested with Chrome/Edge/Firefox)

While reviewing First Posts on Super User, I noticed that I couldn't edit this question (see below). Clicking the "edit" link simply does nothing. Editing is not possible for me right now because I have already five suggested edits pending. It used to be the case that the link would've been grayed out (and the reason shown as a popover). , and while I'm not sure it's an improvement, the current implementation needs some fixing at least.  This bug appears regardless of the actual reason that the edit isn't possible; for instance,  is another user experiencing the same bug who couldn't suggest an edit because of an already pending one. Apparently, I'm far from the only one who experiences this issue. As of last Sunday morning,  have been posted networkwide (and are still alive) and  is #16.


I have tried the approach that I have used before on earlier releases of Ubuntu but on 15.04 I have found that the following commands are not working.  sudo xhost +SI:localuser:lightdm  sudo su lightdm -s /bin/bash  gsettings set com.canonical.unity-greeter draw-grid false   When I run the last command I get the following:  (process:5354): dconf-WARNING **: failed to commit changes to dconf: Error spawning command line 'dbus-launch --autolaunch=1af246dec6324e1ca014e1c3c06da71c --binary-syntax --close-stderr': Child process exited with code 1   Can anybody suggest what I'm doing wrong, or ways that I could diagnose what is failing?  Cheers

I can change the background image that Unity Greeter displays, but regardless of the image I choose, it's overlaid with a grid of dots. The dots look OK on the default background, but they look horrible on the background I want to use. How can I get rid of them?


have one question, i trapped in a prob..The problem is like that --->I have to manipulate the css of inner html of iframe but the content of this iframe is not coming from same domain so how can it makes possible live eg.-u can take ("tweets feed in our website ,it comes from another domain tweeter.com/bla....blaa but render on our web page encapsulate with iframe,so how can i style tweet section in our website") i have tried by jQuery but its not working can u plz help me out in that ? Thanks in advance Smile | :)

I have a simple page that has some iframe sections (to display RSS links). How can I apply the same CSS format from the main page to the page displayed in the iframe?


Being $f:\mathbb{N} \rightarrow \mathbb{N}$ such that $f(1)=2011$ and $f(n)=  \frac{1}{n^2 -1}(f(1)+f(2)+...+f(n-1))$ for $n \ge 2$. Calculate $f(2011)$  When calculating $f(n)$ I need to consider the sum of the previous elements, from $2011$ to $f(n-1)$. But when considering each of these, I need again the the sum of the previous $f(n)$. I've never met a problem like this. It seems like a sequence  by recursion: are there any rules when approaching problems like this?

Given that $$f(1)=2005$$ and $$f(1)+f(2)+...f(n) = n^{2}f(n)$$ for all $n>1$. Determine the value of $f(2004)$.  My progress:  I first substituted $n-1$ into the equation to get $f(1)+f(2)+...+f(n-1) = (n-1)^{2}f(n-1)$ and then I subtracted it from the first equation to get $$f(n) = n^{2}f(n)-(n-1)^{2}f(n-1)$$. I got a little stuck so I checked the answers for a hint and it said that: $$f(n) = \frac{(n-1)^{2}}{n^{2}-1}f(n-1)$$. How is this so? I don't understand the last part.


This command does not work:   $ sudo echo 3 > /proc/sys/vm/drop_caches  bash: /proc/sys/vm/drop_caches: Brak dostępu   But when I gain root priviliges separately everything works.  root@pc:~# echo 3 > /proc/sys/vm/drop_caches  root@pc:~#    Does this have something to do with > operator redirecting sudo, not echo?

When using sudo to allow edits to files, I regularly get 'permission denied'.  For example, my mouse is jittery and sluggish, so I want to disable polling:  sudo echo "options drm_kms_helper poll=N">/etc/modprobe.d/local.conf   I'm prompted for a password, and then get:  bash: /etc/modprobe.d/local.conf: Permission denied   So I tried to do a temporary change to disable polling by using:  sudo echo N> /sys/module/drm_kms_helper/parameters/poll   Yet again the system responded with:  bash: /sys/module/drm_kms_helper/parameters/poll: Permission denied   Any ideas?


There is a slightly difference between when comparing those commands.

I just mistyped ls as la. Without noticing, I ran it:  tim@Hairy:~$ la detection.sh   output-2.jpg  output-4.jpg  output-6.jpg  output-8.jpg Test1.pdf output-1.jpg   output-3.jpg  output-5.jpg  output-7.jpg  output-9.jpg          tim@Hairy:~$ ls detection.sh   output-2.jpg  output-4.jpg  output-6.jpg  output-8.jpg Test1.pdf output-1.jpg   output-3.jpg  output-5.jpg  output-7.jpg  output-9.jpg    What's the difference between ls and la?


I'm new to Tex and want to install it on my Mac which has OSX Yosemite.   I want to know know the Tex editor which works best with my OS? I saw on it on some forums that MacTeX doesn't work properly on the latest OSX.  Thanks in advance! :)

I am a noob when it comes to TeX, however, I do use LaTeX for typing mathematical and engineering formulas, etc.   Recently I have taken it upon myself to try and write some technical papers to try and get published, however I do not know what... 'type'(?) of TeX software to use best, so that I can hit the ground running with it.   If you look , this can give you context as to the forums that I would be publishing to, (IEEE), and they even have some templates there. However I cam getting easily confused between BibTex, TeX, etc, and how they all 'fit' together, in my goal of writing a paper. I would like guidance on this.   What might be the easiest 'headache less' software to start writing a paper in TeX for a noob such as myself? Am I on the right track?  Thanks!


I have a vague recollection of a cartoon I saw sometime in the late 90s. As Scandinavia was a bit behind in releases back then I believe it was broadcasted in early 90s or late 80s in its original run. The animation style is similar to that of Spider-man the Animated Series from the early 90s and Highlander the Animated Series. The premise of the show is similar to Highlander the Animated Series series, but with a more gritty tone. The world has been plunged into a radioactive wasteland after a nuclear war, I think, or some other catastrophe. A group of kids/young adults traverse wildlands looking for "sages" that has preserved the old world knowledge. The one scene I remember vividly is when the group encounters one of the "sages", the one concerned with atomic/nuclear power. The sage removes his cowl to show a face scarred by radiation, this is never shown as his face is never seen but it is implied. After he has given his knowledge he sends the group away and activates some kind of self-destruct sequence, an atomic explosion follows. The purpose of the group I cant remember but it was similar to that of Highlander, overthrowing a tyrant. Does any of this sound familiar or have I gotten it mixed up with Highlander? It was 25 years ago so it is not unlikely.

I am looking for an animated TV series which was probably made in the '80s or '90s.  The main character had red hair.  The cartoon is set in a post apocalyptic environment and the character lives with his mother and an old man —his mentor I guess— with a nomadic group.  In one early episode, one of the group is wounded in an ambush, he is taken back to the camp where the mentor performs surgery which is revealed he learned during the war. The wound is caused by a bullet, then the mentor is surprised saying that guns were lost and forgotten.  In another episode the main character is captured and sent to a prison. Then after he escapes with other prisoners, they look outside the gate and see grass and soil which they either didn't see in their lives or were in the prison for a long time.  The main character later finds a sword like Excalibur which can cause some sort of electrical disruptions.  I think his mother was kidnapped in the beginning.


I used the prcomp() function to perform a PCA (principal component analysis) in R. However, there's a bug in that function such that the na.action parameter does not work. ; two users there offered two different ways of dealing with NA values. However, the problem with both solutions is that when there is an NA value, that row is dropped and not considered in the PCA analysis. My real data set is a matrix of 100 x 100 and I do not want to lose a whole row just because it contains a single NA value.  The following example shows that the prcomp() function does not return any principal components for row 5 as it contains a NA value.  d       <- data.frame(V1 = sample(1:100, 10), V2 = sample(1:100, 10),                        V3 = sample(1:100, 10)) result  <- prcomp(d, center = TRUE, scale = TRUE, na.action = na.omit) result$x                                # $ d$V1[5] <- NA                           # $ result  <- prcomp(~V1+V2, data=d, center = TRUE, scale = TRUE, na.action = na.omit) result$x   I was wondering if I can set the NA values to a specific numerical value when center and scale are set to TRUE so that the prcomp() function works and does not remove rows containing NA's, but also does not influence the outcome of the PCA analysis.  I thought about replacing NA values with the median value across a single column, or with a value very close to 0. However, I am not sure how that influences the PCA analysis.  Can anybody think of a good way of solving that problem?


The warlock's  "Armor of Shadows" says:     You can cast mage armor on yourself at will, without expending a spell slot or material components.   Does this mean I know the spell? That is, can I cast it on someone else by expending a spell slot and material component as normal?

My friends and I are going to be starting another campaign in the near future. I've going to be playing as a Warlock and I was hoping to get some clarification on the use of the Eldritch Invocation feature of the class.  Some of the invocations allow you cast spells. One instance is Armor of Shadows, whose text reads     You can cast mage armor on yourself at will, without expanding a spell slot or material components.   Does this indicate that Mage Armor has to be one of the spells you know (in other words, count against the spells known for your level?) Or is this considered something separate from normal spellcasting for Warlocks?


Enviroment: Plesk v12, latest Windows Server 2012 SmarterMail  My web hosting provider told me that someone hacked into my server a couple of weeks ago and was hitting various mail servers (Gmail, Yahoo, etc).  I worked with the company to cure the problem, basically: * Change all passwords * Update server and all software to latest version * Turn on Windows firewall, was somehow disabled  Doing these tasks and waiting a couple of weeks finally took me off all the blacklists and restored my MXToolbox rating.  I found out today that Microsoft blacklisted my IP (not reported by the free version of MxToolbox). They said that a few days ago (8/28 and 8/29) that there was still Namespace Mining going on, but offered no information on how they know, how to prevent, or anything else that I asked. They wrote me back to my questions a few minutes ago that upon further investigation that they will lift the block.  Questions:   How does Namespace mining occur? Is that the result of a hacker finding out or hijacking an email account? Aside from inspecting the logs, how can I determine if there is Namespace mining occuring? How do I prevent Namespace mining beyond what I did? Would that be by installing a professional security software (e.g. third party Antivirus / Firewall / Email security software)?   I resolved the current problem, but I would like to know how to find out if I have the problem. In this case, I tried to send an email to a Microsoft account of mine and received a failure report. There should be a better way to go about a check then accidentally discovering the problem.

This is a  about Server Security - Responding to Breach Events (Hacking)   See Also:                  Canonical Version I suspect that one or more of my servers is compromised by a hacker, virus, or other mechanism:   What are my first steps? When I arrive on site should I disconnect the server, preserve "evidence", are there other initial considerations? How do I go about getting services back online? How do I prevent the same thing from happening immediately again? Are there best practices or methodologies for learning from this incident? If I wanted to put a Incident Response Plan together, where would I start? Should this be part of my Disaster Recovery or Business Continuity Planning?   Original Version       2011.01.02 - I'm on my way into work at 9.30 p.m. on a Sunday because our server has been compromised somehow and was resulting in a    attack on our provider. The servers access to the Internet   has been shut down which means over 5-600 of our clients sites are now   down. Now this could be an FTP hack, or some weakness in code   somewhere. I'm not sure till I get there.      How can I track this down quickly? We're in for a whole lot of   litigation if I don't get the server back up ASAP. Any help is   appreciated. We are running Open SUSE 11.0.            2011.01.03 - Thanks to everyone for your help. Luckily I WASN'T the only person responsible for this server, just the nearest. We managed   to resolve this problem, although it may not apply to many others in a   different situation. I'll detail what we did.      We unplugged the server from the net. It was performing (attempting to   perform) a Denial Of Service attack on another server in Indonesia,   and the guilty party was also based there.      We firstly tried to identify where on the server this was coming from,   considering we have over 500 sites on the server, we expected to be   moonlighting for some time. However, with SSH access still, we ran a   command to find all files edited or created in the time the attacks   started. Luckily, the offending file was created over the winter   holidays which meant that not many other files were created on the   server at that time.      We were then able to identify the offending file which was inside the   uploaded images folder within a  website.      After a short cigarette break we concluded that, due to the files   location, it must have been uploaded via a file upload facility that   was inadequetly secured. After some googling, we found that there was   a security vulnerability that allowed files to be uploaded, within the   ZenCart admin panel, for a picture for a record company. (The section   that it never really even used), posting this form just uploaded any   file, it did not check the extension of the file, and didn't even   check to see if the user was logged in.      This meant that any files could be uploaded, including a PHP file for   the attack. We secured the vulnerability with ZenCart on the infected   site, and removed the offending files.      The job was done, and I was home for 2 a.m.            The Moral   - Always apply security patches for ZenCart, or any other CMS system for that matter. As when security updates are released, the whole   world is made aware of the vulnerability.   - Always do backups, and backup your backups.   - Employ or arrange for someone that will be there in times like these. To prevent anyone from relying on a panicy post on Server   Fault.


I bought an Hp computer a few weeks ago second hand and it has ubuntu 12.04 installed onto it. The guy i bought it off gave me the installation cd he used to put it on the computer but i really dont like it, i want to get windows back on the computer but have no idea how an earth to do that, i have a product key for windows 8 but no installation disk for it. HOW DO I REINSTALL WINDOWS AND UNINSTALL UBUNTU?????

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


My question is why is the formula for finding the standard deviation of a given data (either grouped or non grouped) the way it is?  so let me start from the definition of a standard deviation with my own words, which i believe is "what is the average dispersion between the mean and each data."   Logically from what the definition says i believe the equation should have been sum of all the differences between the mean and each value divided by the total frequency. (am sorry i don't now how to right equations in this site so i would appreciate any editing).  Well there is a catch here since by definition mean is the data that is by average at equal distance from each data point, the standard deviation calculated using the above formual (idea) will be zero which doesn't describe what we are looking for, and when we analyse where this problem may have come we can see that the definition of the standard deviation shouldn't depend on the direction of the data points relative to the mean, by that i mean smaller or greater in quantity we just need to now by how much they vary so i believe to alleviate this problem mathematician has came with the idea of variance which is squaring the difference, and then latter taking the square root in order to find the standard deviation, which is were i depart, i don't really get why we square the deviation between the mean and each data, i believe what we should do is take the absolute value of each difference and then divide the sum by the total frequency. so in general what i am saying is Shouldn't the formula of a standard deviation be the what i said when we come at the problem from the definition.

In the definition of standard deviation, why do we have to square the difference from the mean to get the mean (E) and take the square root back at the end? Can't we just simply take the absolute value of the difference instead and get the expected value (mean) of those, and wouldn't that also show the variation of the data? The number is going to be different from square method (the absolute-value method will be smaller), but it should still show the spread of data. Anybody know why we take this square approach as a standard?  The definition of standard deviation:  $\sigma = \sqrt{E\left[\left(X - \mu\right)^2\right]}.$   Can't we just take the absolute value instead and still be a good measurement?  $\sigma = E\left[|X - \mu|\right]$


I have a long list of named Pandas data frames. My goal is to sort this list alphabetically by the names of the data frames. Here is a MWE:  import pandas as pd     df_1 = pd.DataFrame(None) df_1.name = 'a' df_2 = pd.DataFrame(None) df_2.name = 'b' df_3 = pd.DataFrame(None) df_3.name = 'c' df_list = [df_2, df_1, df_3]   Now, my goal is to sort df_list such that I get a new list [df_1, df_2, df_3] reflecting the alphabetical order of the names of the data frames.

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


It's these type of questions that I don't know what approach to even take. Which is stressful in an exam condition. My first thought was, well, why not assume there are at least two different topologies that are Hausdorff on $X$?  Say $\tau_1,\tau_2$ are distinct topologies on $X$ with the Hausdorff property. Say we pick some distinct $x,y \in X$ such that $\exists U_1,V_1 \in \tau_1$ that are open and disjoint, by Hausdorffness. We pick $x,y$ so that $U_1$ or $V_1$ is not open in $X$ with $\tau_2$. We can find such $x,y$ since we assume $\tau_1,\tau_2$ are different topologies therefore at least one subset in $X$ considered open in one is not in the other. By Hausdorffness of $\tau_2$ we can also find open disjoint sets $U_2,V_2 \in \tau_2$ such that $x \in U_2$ and $y \in V_2$.  Here's where I start running out of ideas  Perhaps take the intersection? $U_1 \cap U_2 \neq \phi$...$V_1 \cap V_2 \neq \phi$.  Then it stops. No idea what and how these will help me. At this point I am not sure if my approach was utterly pointless in the first place assuming two distinct Hausdorff topologies...or I'm kind of in the right direction but not seeing what properties to look at? I am trying to look for some contradiction but well, so far, nothing seems to be out of control, everything seems fine to have at least two different Hausdorff topologies on a finite set...so, why not? Can someone prove this...?

Theorem: $X$ is a finite Hausdorff. Show that the topology is discrete.   My attempt: $X$ is Hausdorff then $T_2 \implies T_1$ Thus for any $x \in X$ we have $\{x\}$ is closed. Thus $X \setminus \{x\}$ is open. Now for any $y\in X \setminus \{x\}$ and $x$ using Hausdorff property, we get $\{x\}$ is open.  Am I right till here? And how to proceed further?


It is said to be a good practice to write the following block of code in your python modules:  if __name__ == '__main__':   # execute only if run as a script   main()   This prevents executing statements when you want to import this module into another program.  However, the curious thing over here is that main() function. We also need to put our code into an actual main method as follows:  def main()     # All code goes here     print("Inside main")   I feel that this is a roundabout way. Are there any benefits to creating a separate main() function and then calling it inside the if-clause?   Instead what is the harm in writing the code directly inside the if-clause?  if __name__ == '__main__':     # All code goes here     print("Inside main")

Given the following code, what does the if __name__ == "__main__": do?  # Threading example import time, thread  def myfunction(string, sleeptime, lock, *args):     while True:         lock.acquire()         time.sleep(sleeptime)         lock.release()         time.sleep(sleeptime)  if __name__ == "__main__":     lock = thread.allocate_lock()     thread.start_new_thread(myfunction, ("Thread #: 1", 2, lock))     thread.start_new_thread(myfunction, ("Thread #: 2", 2, lock))


Edit: My question is different because I am not asking how permanent magnets come about. Rather, I am asking why (taking permanent magnets as a given) they attract or repel each other. I.e. What equation describes the force that causes a permanent magnet to stick to my fridge.  I have been searching online for some qualitative explanation of why permanent magnets attract each other (for example fridge magnets) and can't find anything.  The closest I can find is attraction/repulsion of two parallel current carrying wires.  Here's using the same steps to two permanent magnets situated on top of one another:  Lorentz force law says force is proportional to $\vec{v} \times \vec{B}$. Each magnet is composed of many atoms all oriented the same way. All together the electrons orbiting the individual nuclei sum to create a stronger magnetic field (let's say pointing up to the other magnet).  This other magnet is also composed of individual, similarly oriented atoms.  Since I know it will attract when the "south" pole of the top magnet is above the "north" pole of the bottom magnet, let's try that situation.  Trying to apply the Lorentz force law, $v$ is going around in a circle on the top magnet. $B$ is coming straight up (from the first magnet). This seems to imply that the force would be perpendicular to the $B$ field, but spinning around in time (as the electron spins around the nucleus).    Where does the force that attracts the two permanent magnets towards one another come from?

Are they working with motion of electrons?


I could see ~/Library through Unix command in Mac, yet I cannot see it in a Finder window. I need to go as Finder -> Go to folder -> ~/Library, then I can find it. Does anyone know why is it like that? Thanks! It's macOS Big Sur.

I just installed Lion, and I can't find the Library folder in my user folder.  Does the folder no longer exist? If it doesn't, where are my preferences? If it does, how do I get to it?


$(document).ready(function(){  var nam="";  $("#sayHello").click(function(){   nam = $("#textBox1").val();   if(nam == ""){    alert("Please let know what your name is.");   }   else if(/  /i.test(nam)){    alert("Try to make sure that your not adding an extra space. It can be any name make one up if that helps.");   }   else{    $("#welcome").html("
Hello "+ nam +",
 How was your day?
" +     "<button id='good'>GOOD</button> <button id='okay'>okay</button>" +    " <button id='bad'>Bad</button>
").show();    $("#inputBox").hide();   }   });   $("#good").click(function(){    $("#goodHit").html("
I'm so happy for you!
" +     "
Here is a quote to help you along your way.
" +     "<blochquote>The good old days are now.</blochquote>" +     "
Tom Clancy
").show();    $("#hi").hide();     });   }); <!doctype html>       <title></title>

   Here is my first text box.      <input id="textBox1" type="text" value=""/>      <button id="sayHello">Submit</button>

                      In my first .click(function(){..}); I make some buttons and put them into a div. I then hide my div that gave me that ability and show the buttons but when I want to use those buttons nothing is happening. I am new to javascript and jquery please help. here is my code.  $(document).ready(function(){ var nam=""; $("#sayHello").click(function(){     nam = $("#textBox1").val();     //some checks are here but unimportant   Here is where the buttons are made and put into my html.          $("#welcome").html("
Hello "+ nam +",
 How was your day?
" +          "<button id='good'>GOOD</button> <button id='okay'>okay</button>" +         " <button id='bad'>Bad</button>
").show();         $("#inputBox").hide();     }     });    I then try and use one for what I want next to happen.  $("#good").click(function(){         $("#goodHit").html("
I'm so happy for you!
" +          "
Here is a quote to help you along your way.
" +          "<blochquote>The good old days are now.</blochquote>" +          "
Tom Clancy
").show();         $("#hi").hide();  });

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


hi all I have a quick question, to get my +n in the skills column for proficient skills do i add the multiplier from the attributes list ie my charisma is +3 do i then add my proficiency bonus of +5 to get +8 in my skills list and then when I get expertise in a skill do I double the entire thing or just the base score of +3 then add the +5 bonus? sorry it's just very unclear in the PHB

Bards get Expertise at level 3. How does this work?  For example, if I have a standard proficiency bonus of +2, and if I have proficiency in Perception with a Wisdom ability modifier of +2, then my Perception is +4. If I choose expertise in Perception, does it double the +4 to Perception +8, or does it double the proficiency bonus only to +4 for a total of Perception +6?


Which of the following example sentences is correct? (one uses who and the other uses whom)     This has caused a problem for students who, up until recently, were in good shape.   or     This has caused a problem for students whom, up until recently, were in good shape.   I am aware I should use whom when it's the object of the preposition, but I'm confused about how to analyze the grammar of these sentences.

I can never figure out whether I should use who and whom. Most people use who for both colloquially, but some people say this is not correct.  What’s the rule for using who and whom correctly?


Let $(x_n)$ be a real-valued sequence with partial sums $s_n = x_1 + x_2 + ... +x_n$. We define $\sigma_n = \frac{1}{n}(s_1 + s_2 + ... + s_n)$. Now, supposing $\Sigma\,x_n$ is convergent, I need to show that the sequence $(\sigma_n)$ is convergent and that $lim\,\sigma_n = \Sigma\,x_n$.  My intuition is to simply say that since $\Sigma\,x_n$ converges, then $lim\,sup\,s_n = lim\,s_n = L$, some real number, and then try to find a way to use a convergence test.   But I don't have any information about the $x$-values, so I can't say they're non-negative, which means I can't use comparison, and the ratio and root test both come out inconclusive since $lim\,\frac{\sigma_{n+1}}{\sigma_n}=1$.  Should I be finding a way to use the Cauchy criterion here? Or am I missing a different technique altogether?   EDIT: This isn't actually a duplicate of the , because there is only one series of partial sums in the arithmetic mean. This problem takes the sum of each of those sums from 1 to n and presents a new problem with that.  EDIT 2: Upon further review, although not identical, the problem turns out to essentially be a re-skin of the arithmetic mean. See answer below.

Let $(a_n)_{n \in \mathbb{N}}$ be a convergent sequence with limit $a \in \mathbb{R}$. Show that the arithmetic mean given by: $$s_n:= \frac{1}{n}\sum_{i=1}^n a_i \tag{A.M.} $$ also converges to $a$.  I have read:  but unfortunately the answers there don't help me much because I don't understand their substitutions and most of it all, why their substitutions seem to work.   What I know from the problem is that since $a_n$ is convergent and $\epsilon >0$ is given, I can say that:   $$\exists N_1 \in \mathbb{N}, \forall n \geq N_1: |a_n-a|<\epsilon_1 $$ I also know that since $a_n$ is convergent, it is bound, so $(a_n) < M, \ \forall n \in \mathbb{N}$   I need to show that: $$\exists N_2 \in \mathbb{N}, \forall n \geq N_2: |s_n-a|< \epsilon_2$$ I started as follows: $$ \left|s_n-a \right| = \left|\frac{1}{n}\sum_{i=1}^na_i -a\right|= \left|\frac{1}{n}\sum_{i=1}^m(a_i-a)+\frac{1}{n}\sum_{i=m+1}^n(a_i-a) \right| \\ \leq \frac{1}{n} \sum_{i=1}^m|(a_i-a)|+\frac{1}{n}\sum_{i=m+1}^n|a_i-a|$$ I believe to understand that the left sum after the $\leq$ is finite, bound and doesn't depend on the values that $n$ takes on. However, I don't understand where all the substitutions come from and make this proof so seemingly easy to complete.   Is there a general idea I can follow to complete such proofs? Because I know that the last step is to show that the given sum is smaller than $\epsilon$. I also know that I should bring the condition $a_n < M$ into place somewhere, but I don't know where.  If I choose an $n \geq N: |a_n-a|< \epsilon'$ what does that tell me about $|a_i-a|$? I know that they are seemingly the same just with a different index.


I want to show that $Z =\{(u^3, u^2v, uv^2, v^3) \mid  u,v \in \mathbb{C}\}$ is an algebraic set. So I need functions in $\mathbb{C}[x_1, x_2, x_3, x_4]$ that vanish for all $(u^3, u^2v, uv^2, v^3)$ where $u,v \in \mathbb{C}$.  I already found out that the function $f_{uv}(x_1, x_2, x_3, x_4) = -x_1 - 3x_2 - 3x_3 - x_4 + (u+v)^3$ vanishes at $(u^3, u^2v, uv^2, v^3)$. But how can I find one that vanishes for all complex numbers $u$, $v$?

How do I go about proving the subset $V = \{(s^3, s^2t, st^2, t^3)\mid s, t \in k\}$ is an affine algebraic subset of $\mathbb{A}_k^4$ and find $\mathbb{I}(V) \subset k[x_0, x_1, x_2, x_3]$?


By a commented line I mean either:   #!/usr/bin/perl -w   and the comments in /boot/grub/menu.lst file as described in .  It already has a comment, i.e the line begins with a #, but still is being interpreted. Why is that so?

I have seen many files which have this line as the first line in them. What exactly is this?  #!/usr/bin/env python   What does it mean? Why is it necessary?


If k and n are relative primes, do I get all the remainders (0, ..., n-1) from these numbers (mod n): 0, k, 2k, 3k, ..., (n-1)*k? If so, what theorem is about this topic? Also, what if they are not relative primes?

I have this problem assigned for homework and I'm struggling with the proof of it:  If $a_1,a_2,\dotsc,a_n$ is a complete set of residues modulo $n$ and $\gcd(a,n)=1$, prove that $aa_1,aa_2,\dotsc,aa_n$ is also a complete set of residues modulo $n$.  (Hint: It suffices to show that the numbers in question are incongruent modulo $n$.)  I'm in elementary number theory so I'm not allowed to use an high-level theorems to prove this, I pretty much have to use the basics of modulo.  Thanks!


I am new to Ubuntu and I have tried using LibreOffice, but I face a lot of problem using LibreOffice Calc instead of MS Excel.  Please help me get MS Excel back so I can do my work smoothly.  Thanks in advance.

I want to know how to install Microsoft Office (specifically, Microsoft Office 2010) in Ubuntu?   I am a die hard fan of MS Office and I want to know if there is a way I can use it in Ubuntu.


public class AutomaticProperty    {        public integer MyReadOnlyProp { get; }        public string MyWriteOnlyProp { set; }     }   could you please give any examples related to this concept.can we use any where only get and only set?  Thanks in advanced!!!

The Force.com Apex Code Developer's Guide documentation for  contains the following:     Properties do not require additional code in their get or set accessor   code blocks. Instead, you can leave get and set accessor code blocks   empty to define an automatic property. Automatic properties allow you   to write more compact code that is easier to debug and maintain. They   can be declared as read-only, read-write, or write-only. The following   example creates three automatic properties:   public class AutomaticProperty {    public integer MyReadOnlyProp { get; }    public double MyReadWriteProp { get; set; }    public string MyWriteOnlyProp { set; } }   This had me scratching my head a bit. MyReadWriteProp makes perfect sense. The usefulness of MyReadOnlyProp and MyWriteOnlyProp eludes me.  Why would you want a read-only or write-only automatic property?   As far as I can tell you could never actually do anything useful with them. Either you could read a value that you could never set or write a value that you could never read.  Am I missing something or is this just a poor example in the documentation? Maybe there is some way to access the automatically generated backing property to set/read the value?


is it just for me or is the review option missing from stack overflow? Has it been removed or is it a real bug because i searched everywhere but i couldn't find the review option anymore.! I used to edit and review "first posts" and "Late Answers".

Is it just for me, or has the privilege to review new posts and late answers changed for everyone?  I recently started being more active in reviewing, but the rep requirement just shot up and I can not review anymore? The rep requirement has shot up to 500. Is this just for me, as a 'process", or has it been changed site wide?


Summation of answers and comments: Use "an" before a word that begins with a vowel sound, otherwise, use "a". Still not clear on which to use before acronyms that can be read as letter-by-letter or as the words they represent. Side note: In the future, I'll only use acronyms where there is no need for "a" or "an" before them.  If an "a" or an "an" is needed, I'll write the words of the acronym. I'd use: "Nightmare of an operator of N.Y.C.'s subway/Subway..." Simpler sentence: Instead of "An NYU student" or "A New York University student", "A student of NYU" or "A student of New York University".  Original Question: I'm pretty sure I use the correct "a"/"an". Something like: if the next word (expanded from acronym, if exists) starts with "a"/"e"/"i"/"o"/"u"(/"y"?) => use "an", otherwise => use "a". A (bad?) example from : "An N.Y.C. Subway Operator's..." What are the "official" rules?

Which indefinite article should precede hour — a or an?        an hour     a hour      Does the usage of an vs a depend on the pronunciation — a history, a hobby, but an hour, an honor?


So I've actually come across this kind of sentences quite a lot, when we're describing the capacity in which the subject does something. For example, sometimes we say:  The vendor's solicitors will receive the deposit as agent for the vendor  I always thought, strictly grammatically speaking, the correct formulation was:  The vendor's solicitors will receive the deposit as an agent for the vendor  Are both correct? There seems to be a lot of inconsistency, as I've come across both formulations quite often.

I'm used to always hearing or seeing a  before certain nouns.  Yet on certain occasions the article is totally omitted, and it bothers me.  I'm wondering what the justification for omitting the article is, especially when the definite article would seem to be required, or if it is just bad English either on the part of the individual in question or more likely by tradition.  The most frequent place I hear this is at the doctor or dentist's office. In every one I have ever visited, the receptionist or assistant always says things like "Doctor will see you shortly" instead of "The doctor will see you shortly."  The wikipedia article linked above says this: "A definite article indicates that its noun is a particular one (or ones) identifiable to the listener".  In the case of a clinical setting, I know who the doctor is, especially if this is not my first visit.  The people in the office know the doctor even better than I, so it would seem to be inappropriate to omit the article. Even the indefinite article would be incorrect. It further says, under "Zero article":  "In languages having a definite article, the lack of an article specifically indicates that the noun is indefinite."  Since the noun is not indefinite, why are they omitting the article?  Another place I see the article omitted is with the word "bar".  Bartenders apparently don't say "I tend the bar" they say "I tend bar".  Unless a bartender works in multiple bars, it would seem like an article would be required.  Furthermore, "the bar" can mean any bar a person goes to, so even in a case like , it seems like the word "the" should be in the sentence.  The only thing I can think of in this case is maybe the speaker doesn't want to confuse the listener into thinking they work in the legal field.  Is this just a bad habit, a historical corruption or tradition, or is there really a valid reason (or perhaps more than one) for these omissions?


I have a technical question:  I want to know why/where/when the exactness of floating point modulus (fmod / remainder) matters.  It's not about programming, I know how to program it. It's more about marketing, because I'd like to replace naive implementation found in most Smalltalk dialects by a standard conforming one. Adhering to standards and providing state of the art floating point in Smalltalk is my first goal. However, I anticipate that adhering to standard will not be enough to convince my peers to change things, so explaining in which circumstances that would really matter would help a lot. I could not find a good example so far.  Is there a site where I could ask such question?

There seems to be more than a few computer science/programming Stack Exchange networks (is that the correct term?). Stack Overflow, being the first, has by far the most users, questions, and answers.  What is the reasoning for creating the others, and are there clear guidelines for which kinds of questions should be posted? I can see a large amount of potential overflow, many cases of people not getting a good answer to their question, because the person who has the answer isn't browsing that particular network at the moment.  I understand that they were probably created for organizational purposes, but wouldn't it almost make more sense to just have them as categories under Stack Overflow, keep them separated but still connected, instead of making people have to create multiple 'account's, one for each network?  I am sure there was a very good reason to break them up, but as someone that is new to SE, it can be somewhat intimidating to decide which one to post in to ensure you get a good answer. For example, if I am a computer science student, my first instinct might to be to post in the computer science network, until I see that it literally has 1% of the users as the Stack Overflow network, which still seems to be for programming/computer science related questions.  So my second instinct would then to be to post it into the Stack Overflow site on the basis that I have a significantly higher chance of my question being seen and getting a good answer. Then I notice that there is also a 'programmers' network, and I don't even begin to know where that fits in. I, and I am assuming most people, will probably just post on Stack Overflow to be safe.  My question is, other than the short little description blurb of each one, is there a clear cut set of guidelines which what each network is intended for, what kinds of questions should go to each one, and is anything being done to encourage people to post in these newer, smaller (more specialized?) networks as opposed to just posting in big daddy Stack Overflow?  Now again, this is for computer science/programming/"why isn't this code doing what I want?" related questions, I am not saying that if I had a question about Linux or WordPress or something I would have the same confusion.


What are the special cases when we can't use "the" before superlative degree? I'd like the answers with proper explanations. Some of the examples:    He is most likely to succeed. It is fairest of all methods. I have read many books. "Against all odds" is best of all.

For example:     I could not face being alone again and losing the person dearest to me.   I wonder why there is no "the" in front of "dearest".


I'll often highlight a bunch of files in Finder to find their collective size with Cmd+I. Sometimes the Get Info window shows the pooled size in a single window, but other times it'll open 40 separate Get Info windows. What determines when this happens?  (I've since learned that  will always show the highlighted items' collective size, but when does Cmd+I alone act one way instead of the aforementioned other?)

Is there some plist I can edit to decrease the number of files that Finder's Get Info function will open separate info windows for?  The number of Get Info windows that will open simultaneously is . The windows slowly appear one at a time, and once they finally finish appearing I have to close them all one at a time. Pure madness, and whoever chose the default value for this feature should go outside and think about what they've done.   If I get info on thirty-six files, it finally automatically consolidates them into a  I would like to change this threshold of 36 to something reasonable, like 4.  I know about the  to open the Inspector and Summary Info windows, however sometimes I accidentally forget to hold Option or Control, and I would like to eliminate the possibility of this happening by accident.   A satisfactory alternate solution would be if there were a way to make Get Info always open a Summary Info window instead of opening separate file info windows.  MacBook Pro (Retina, Mid 2012) running 10.10.5.


Is there a tool that will help me decide? I think there should be one. There two many interrelated stack exchange sites, and it is in the best interest of the network that post be well categorized (unless some sort of grand unification is planned, which might not be a bad idea--aren't these subsites just another type of tag after all?). At the same time, humans shouldn't be expected to be deeply familiarized with the intricacies of what makes a post more suitable for "Stack Overflow", "Ask Ubuntu", "Web Applications", "Server Fault", "Unix and Linux", "Webmasters", "Super User", "Programmers", etc. when you well may have a question that touches each label.  I think those intricacies are well embedable in a piece of software that could direct/redirect me to particular subsite based on the text content of my question, its title, and possibly my answers to a couple of questions, if I choose to answer those.


Suppose I create an object in javascript.  var myObject = {};   What is the difference between...  myObject.someFunc = function(){...};    and   myObject.prototype.someFunc = function(){...}    in javascript?  I am having trouble understanding the difference or if there is a difference and how and when to use either of these syntaxes.  It seems when I code something like this that there really is no difference.  I am looking for both a client side (browser) and server side (like node.js) answers.  I want to code properly and accurately and this is really bothering me.

What's the difference between  var A = function () {     this.x = function () {         //do something     }; };   and  var A = function () { }; A.prototype.x = function () {     //do something };


I have a virus file that gets re-created by some other infected file/process. Antivirus catches it, but the file still gets created once in a while.  Is there a way to figure out/catch which process created the file? I'm open to any solutions. If it requires a 3-rd party app, so be it. This is all happening on a Win7 32-bit MacBook Pro. Thanks!

I am looking for an app that can monitor my C: (system) drive and find which program is creating a folder, with the rather generic name TEMP1\, on that drive.  I wouldn't mind if it could hook the creation of files too – as there a couple of files created as well (quite probably from another process). Google does not help me, as the names of those files are a bit generic too (not as TEMP1 though – dvmaccounts.ini and dvmexp.idx).   The problem is that every time I delete those, they are recreated on startup, so the utility I'm searching must start up on boot and also start fast. It should also be really clever, as my C: drive is huge.


Question: Let $X$ be a set. Let $\tau$ consists of $\varnothing$ , and any set $U \subseteq X$ such that $X\setminus U$ is either finite or countable. Show that $\tau$ is a topology on $X$.  Looking only for $\mathbf{hints}$. Let $\tau=\left \{ \varnothing,U \right \}$. Obviously, $X \subseteq X$ so $X \in \tau$ Now, $\tau=\left \{ \varnothing, \left \{ U_{i} \right \}_{i \in I} \right \}$ with $X \in \left \{ U_{i} \right \}_{i \in I}: \exists i \in I$ By hypothesis, $X\setminus U$ is countable so it is expected that $\exists$ a bijection $f:\mathbb{Z}^{+}\rightarrow X\setminus U$ I would like to take this further. My intuition says that if I can show that the elements in $U$ are open sets, then the elements in $U$ are elements of $\tau$ which then satisfies one of the axiom for topological space. However, I am unable to do so. Any Hint is appreciated. Thanks in advance.

I'm teaching my self topology with the aid of a book. I'm trying to prove the following is a topology:     Let X be an infinite set. Show that $\mathscr{T}_2=\{U \subseteq X : U = \emptyset $ or $ X\setminus U $ is countable$ \} $ is a topology.   My book calls this set the ""  Please let me know if my proof is valid.     corollary [A]: The union of countable sets are countable.    proof: I'm going to take as a given that that the union of finite set are finite, and just prove the case of . A set is countable infinite if there is a bijection between the set and the natural numbers. Let $A$ and $B$ be a countable infinite sets. Now let $f$ be the bijection between $A$ and $\mathbb{N}$ and $g$ be $B$'s bijection.  $f:\mathbb{N}  \rightarrow  A$  $g:  \mathbb{N}  \rightarrow B$  There can always exist a bijection $h$ that maps $A \cup B $ to $\mathbb{N}$. To account for non disjoint sets, let $B' = B \setminus A$; (note: $A \cup B = A \cup B'$)  $h:  \mathbb{N}   \rightarrow  A \cup B'$  $ h(x) = \left\{         \begin{array}{ll}             f(\frac{1}{2}x); & \quad x \in 2\mathbb{N} \\             g(\frac{x+1}{2}); & \quad x \in 2\mathbb{N}+1         \end{array}     \right. $   $h^{-1}(x)$ would map $A$ to the even numbers, and $B'$ to the odd numbers.  Therefore, The union of countable sets are countable. (QED)  Now to verify the definition.     (i) X and Ø are elements of $\mathscr{T}$.   Ø is given in the definition of $\mathscr{T}_2$. Also, the empty set is considered finite, with cardinality zero. Thus $X$ is in $\mathscr{T}_1$ because its complement is finite, and finite numbers are countable.      (ii) $\mathscr{T}$ is closed under finite intersections.   Let:$V_k \subseteq X$ be any countable set. (ie: $V_1, V_2...$ are all countable sets)  So Then, $U_k = X \setminus V_k$, would be the subsets of $\mathscr{T}_2$.  $U_k \cap U_j = X \setminus (V_k \cup V_j) $, due to De Morgan's law.   Thus, $U_k \cap U_j \in  \mathscr{T}_2$, since the Union of two countable sets is also countable[A]. And This reasoning can be extended to any finite amount of intersections.       (iii) $\mathscr{T}$ is closed under arbitrary unions.   Let $A=\bigcup_{i \in I} U_i$ be an arbitrary union of sets whose complements are are countable.Then by De Morgan's law, $A=X \setminus (\bigcap_{i \in I}V_i)$, for some countable sets $V_i$. By the definition of intersection, $\bigcap_{i \in I}V_i \subseteq V_i$ for all $i \in I$. Since $V_i$ is countable so is any of its subsets.  Hence the complement of $A$ is countable.


I am working on this problem and got stuck:  Let $G$ be a group with three normal subgroups $N_1, N_2, N_3$. Suppose whenever $i$ and $j$ are distinct in $\{1,2,3\}$, $G = N_i N_j$, and $N_i \cap N_j = \{e\}$.  Show that $N_1 \simeq N_2 \simeq  N_3$.  I have shown that $G$ is in fact abelian, hence $N_i$'s are abelian subgroups. In order to show that they are isomorphic to each other, i tried to define an isomorphism between $N_i$ and $N_j$ but I can't come to the conclusion. Is my approach not feasible? Should I try another way?

The following is a problem from the Berkeley Problems book.      Let $G$ be a group with three normal subgroups $N_1$ , $N_2$, and   $N_3$. Suppose $N_i \cap N_j = \{ e\}$ and $N_iN_j  = G$ for all   $(i,j )$ with $i \ne j$ . Show that $G$ is abelian and $N_i$, is   isomorphic to $N_j$ for all $i, j$ .   Attempt:   I have not got very far. The only thing that comes to my mind is that as $|HK|=\frac{|H||K|}{|H \cap K|}$, $|N_1|=|N_2|=|N_3|=m$, and order of $|G|=m^2$. I have made other futile attempts of trying to show m prime, finding an isomorphism between the $N's$. I think the fact that the equation $N_i N_j=G$, could be seen as, left coset of $N_j$ with the multiplying elements restricted to only $N_i$ could provide me an answer, but I haven't got anywhere with this.   Thanks in advance.


I am trying to prove this $2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})$ if $n \ge 1$ and using this to prove $2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( 2\sqrt{m}-1)$ if $m\ge 2$ and in particular i want to show that $m=10^6$ the last inequality is between 1998 and 1999 i am using prove by induction ,first ,for the first statement but i having trouble establish for n=1 and i am confuse how to prove if n is true by hypothesis then n+1.for the second part i didn't get to that part Edit reading the answers and comments bellow now i find myself trying to prove the second part that is prove $2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( 2\sqrt{m}-1)$. My attempt $2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})\Leftrightarrow \sum^m_{n=1}2( \sqrt{n+1}-\sqrt n )< \sum^m_{n=1}\frac{1}{\sqrt n}<\sum^m_{n=1}2( \sqrt{n}-\sqrt {n-1})\Leftrightarrow $ i assume this to be true $2\sum^m_{n=1}( \sqrt{n+1}-\sqrt n )< \sum^m_{n=1}\frac{1}{\sqrt n}<2\sum^m_{n=1}( \sqrt{n}-\sqrt {n-1})\Leftrightarrow $ now using telescoping that is $\sum^m_{n=1} (a_n - a_{n-1})=(a_m - a_0)$ i can not make sense of it

Apostol's Calculus, exercise number I 4.7 13.     Prove that if $n \geq 1$, then   $$ 2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1}) $$   and use this to prove that if $m \geq 2$, then   $$ 2 \sqrt{m} - 2 < \sum_{n = 1}^m \frac{1}{\sqrt{n}} < 2\sqrt{m} - 1 $$   In particular, when $m = 10^6$ the sum lies between $1998$ and $1999$.   After establishing the trivial case, for $n=1$, I can't think of a way to perform the inductive step:  assuming $2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}} < 2(\sqrt{n} - \sqrt{n-1})$ and using this, deducing the same for $n+1$ is true as well:  $2(\sqrt{n+2} - \sqrt{n+1}) < \frac{1}{\sqrt{n+1}} < 2(\sqrt{n+1} - \sqrt{n})$  Any help is very appreciated, thanks


I can't upgrade anything at all. I can't even open the software updater or the software center. And if I try to upgrade through the terminal I only get messages about error and failing.  I want to upgrade to a newer version of xubuntu and I am at a loss, what to do?

Recently I have installed an older version of Ubuntu on my old machine. Whenever I try to install any software, I get an error saying it couldn't be found:  $ sudo apt-get install vlc Reading package lists... Done                Building dependency tree        Reading state information... Done     E: Couldn't find package vlc


I see this message in my logs:  systemd[1]: foo.service holdoff time over, scheduling restart.   I could not find the term "holdoff time" in here:    Where can I change the holdoff time?

I have installed Puppet Enterprise on CentOS 7. I can see that several Puppet services fail to start saying holdoff time over. What does this mean?  Sep 03 20:50:16 l-pm1 systemd[1]: pe-puppetdb.service holdoff time over, scheduling restart. Sep 03 20:50:16 l-pm1 systemd[1]: Starting pe-puppetdb Service...


When previewing , the links at the very bottom point to something epic:  I get that it's a placeholder link, but the "visiting your subscriptions page" and "unsubscribe from all stackexchange.com emails" links can probably be set straight. That is, of course, unless this is what is actually being emailed out — that'd take it to a whole new level of epic!

The problem: The links at the bottom of the Stack Exchange newsletter link to a YouTube video instead of the right place on the website. I´m not sure if this is an oversight or a clever hack.  Screenshot:


We are displaying a list of items, each with a set of options (or single option in some cases). The options are displayed using radio buttons because only one may be selected. A selection is required - if the user tries to proceed without making a selection they will get an error message.   I'm trying to determine the best treatment for defaulting (or not) the radio button selections. I'm leaning towards treatment "C" but would love to hear other thoughts.

I am working on a re-design for  a eCommerce site and they have a section where they offer a free gift card if the user makes a purchase over $1000. The idea of the gift card is to get users to purchase more on the site and since the products are generally expensive (its a jewelry site) chances are they will spend the gift card plus some more money on the site.  So the question I have is should the radio button group which asks user if they want a free gift card be pre-selected ?         – Wireframes created with   There are two aspects to it which I am trying to consider and I would appreciate it if answers consider those a parameters   The user experience if the radio button is pre-selected as yes and the user is wondering when did he sign up for a gift card even though its free The aspect of conversion from a e-commerce perspective


Fair coin flipping (50% on both sides) $P_1$ and $P_2$ plays a few games of fair coin flipping. Assume player $A$ starts with $x$ coins and player $B$ with $y$ coins. Let $P_n$ denote the probability of player $A$ winning all coins. Find $P_0$ and $P_{x+y}$, then write a difference equation for $P_n$. Solve the difference equation Find the odds of player $A$ winning over player $B$ if $A$ starts with, e.g. half the coins $B$ starts with. How far I've come I've been able to produce a recurrence relation $P_n = \frac{1}{2}P_{n+1} + \frac{1}{2}P_{n-1}$. And I assume $P_0 = 0, P_{x+y}=1$, since when player $A$ has 0 coins, he also has $0\%$ chance of winning, and when player $A$ has $x+y$ coins, he's already won, therefore it would be $100\%$ Correct me if I'm wrong. All help appreciated, thanks a lot.

Edit 3. Fixed question to be more clear and include current solution Problem Two players player 1 and player 2 plays a game of fair coin flipping. Player 1 starts with $A$ coins and Player 2 with $B$ coins. Let $P_n$ defines the probability of player 1 winning all coins in a game when player 1 has $n$ coins. Find the recurrence relation of $P_n$ and then find its closed formula. This formula can then be used to know the probability of player 1 winning if he has, i.e. twice the coins player 2 has. Solution? I know that $P_0 = 0$, because if she has $0$ coins, then she can't win, what about $P_{A+B}?$ $$P_n = 0.5P_{n+1} + 0.5P_{n-1}$$ Which gives the characteristic equation: $$x^2-2x+1 = (x-1)^2=0 \implies x=1\ (double\ root)$$ This can then be put into the general solution for linear homogenous recurrence relations: $P_{hn} = C_1+nC_2$ and here I'm stuck once again. To find these two values I'd need two boundary points


The default PDF reader in Ubuntu doesn't support bookmarking pages, writing notes and underlining text. Because I need this I am looking for another PDF reader.   I was trying to install Adobe Reader but it did not work. Anyone got any good PDF readers to suggest, or a working procedure to get Adobe Reader installed on 17.10?

Currently I need to highlight certain sections in PDFs, or add annotations (comments/notes). These modifications would need to be saved.   What tools are out there to do this on Ubuntu?


I have completed my PhD in chemistry in an area which doesn’t have scope to get job or a postdoc position.   Now I’m at the age of 29–30, I’m not eligible to write CSIR, a public eligibility test for lectureship and PhD admissions in India, but I’m eligible to write GATE-2016, (Graduate Aptitude Test for Engineering) another public test whose score (valid for 2 years) is also considered for PhD admissions in India and some other countries like Singapore. Shall do another PhD with a valued GATE score again, in my selected field of chemistry? The only stipend available for me is from GATE score.

I am studying a combined bachelor of engineering (electrical) and bachelor of mathematics; I just started this year and will graduate in 2018. The reason why I am doing double degrees and not a single degree is because I love both electrical engineering and mathematics and I could not ignore any of them. So with this in mind, I am thinking of doing two PhDs when I graduate (one in electrical engineering and one in mathematics). Is this a good path or I should concentrate on only one of them?


So I am trying to make a wrapper function for generating a hash using Cryptop++ I created this test function:   #include <cryptopp/sha.h> #include <cryptopp/hex.h> #include <cryptopp/cryptlib.h>  #include <vector> #include <cstdint>  #include <string>  #include <iostream>  void test2() {     CryptoPP::SHA1 hash;     CryptoPP::byte digest[CryptoPP::SHA1::DIGESTSIZE];      std::vector<uint8_t> v;     for (uint32_t i = 0; i < 1000; ++i)     {         v.push_back(i % 9);     }      hash.CalculateDigest(digest, v.data(), v.size());      CryptoPP::HexEncoder encoder;      std::string output;      encoder.Attach(new CryptoPP::StringSink(output));     encoder.Put(digest, sizeof(digest));     encoder.MessageEnd();      std::cout << output << std::endl; }   And compiled it with the following clang++ string:  clang++ main2.cpp -lcryptopp. However, when I use it in my project where the function is defined like this:   template<typename Hash> std::string get_hash(std::vector<uint8_t> data) { Hash hash;  // Intialise a byte "array" with space enough for the result CryptoPP::byte digest[Hash::DIGESTSIZE];  // Create hash for the data hash.CalculateDigest(digest, data.data(), data.size());  CryptoPP::HexEncoder encoder; // result will hold the hex representation of the hash std::string result;  // Tell the Hex encoder that result is the destination // for its operations encoder.Attach(new CryptoPP::StringSink(result)); encoder.Put(digest, sizeof(digest)); // As we will not put more in the message we end it encoder.MessageEnd();  return result; }   And call it like this: hash::get_hash<CryptoPP::SHA1>(pair.pivot); with following compiler command: clang++ -std=c++17 -Wall -Werror -Wextra -pthread -pthread -lpqxx -lpq -lcryptopp examples/sql_index_example/sql_index_example.cpp.1.o -o/home/tools/git/alexandria/build/examples/sql_index_example/sql_index_example -Wl-Bstatic -L. -lalexandria -Wl-Bdynamic  I get a ton of undefined references to Crypto++, such as:   examples/sql_index_example/sql_index_example.cpp.1.o: In function `alexandria::data::hash::GetHash[abi:cxx11](std::vector<unsigned char, std::allocator<unsigned char> >)': sql_index_example.cpp:(.text+0x197): undefined reference to `CryptoPP::StringSinkTemplate<std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >::StringSinkTemplate(std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >&)'   and I am quite lost as to what actually happens when the simple test works. So hope someone can help.   EDIT:  Already tried:

Why does the order in which libraries are linked sometimes cause errors in GCC?


When I go into camera lock mode and zoom out on my model. The distant parts grey out and disappear. Why is that?  Kind regards,  Maarten

I have created a simple model here   When rendering it the end of it is cut off   How can I get the camera to render a longer distance?


I currently have a $560k mortgage and took out an additional $105k HELOC piggy back loan to make up the 20% needed for the Jumbo loan last year. I've been saving for the past year and a half and now have $100k in savings. Should I be paying down the HELOC early or investing/saving the money? the HELOC is variable with a current 5.95 APR

I've heard the principle of "Pay yourself first", implying that saving some money each month should be a priority.  However, I've also heard that being debt free is important, and it would be good not to lose income to the interest payment on the loan.  In general, we're trying to get to a savings cushion by setting some money aside each month.  But we are also trying to pay off a HELOC so that we are not throwing away money on interest. If we did have a financial crisis, we could draw more from our home equity line of credit. In prioritizing where the money goes each month, is funding the savings account more important than paying off the HELOC as soon as possible?


My question is: why has the  the form  $$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{c^2}}}$$  and not, for example one of those:  $$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v}{c}}}$$  or  $$\gamma = \dfrac{1}{\sqrt{1 - \dfrac{v^2}{2 c^2}}}$$  or  $$\gamma = \dfrac{1}{\sqrt[3]{1 - \dfrac{v^2}{c^2}}}$$  Or one of the other many possibilities?  I mean: nobody ever told me where did the definition of $\gamma$ came from. Can you explain it to me?

What does the following mean with respect to ?  $$\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$


I bought a Cherry mechanical keyboard yesterday. Cherry is a German keyboard-making company. I had a chance to use a Cherry keyboard before. One of my coworkers used a Cherry keyboard, and he let me use it for some time. It felt very nicer to type on than my membrane keyboard. I hadn't understood why people buy such expensive keyboards, but after using his, I realized why people buy them. My native English-speaking friend told me that I can't say "I hadn't understood" because it sounds like a discrete action and thus sounds very weird to him. I used the past perfect tense to refer to a time earlier than the time when I had a chance to use his keyboard. Could you grammatically explain why it is wrong?

Suppose that in a conversation, our conversational partner just said something completely new to us, and we want to express that we didn't know (or haven't known) about it before. For example,  A: Some dictionaries will tell you whether a noun is countable or not. B1: Oh, I didn't know that. B2: Oh, I haven't known that. B3: Oh, I didn't know that until you mentioned it. B4: Oh, I haven't known that until you mentioned it.  Which responses are possible? (I think all are possible.) Which one is the most likely response by a native speaker?


$AUC$ is the area under the receiver operating characteristic curve. It's said that a loss function set to $1-AUC$ can be minimized by maximizing $AUC$. How does this tie into logit (logistic) regression, if at all, and does logit regression maximize $AUC$ directly/indirectly somehow?  described mathematically for both the discrete-class and probability-based logit regression would be nice. Also, what is a rank loss function?

Given a data set with binary outcomes $y\in\{0,1\}^n$ and some predictor matrix $X\in\mathbb{R}^{n\times p}$, the standard logistic regression model estimates coefficients $\beta_{MLE}$ which maximize the binomial likelihood. When $X$ is full rank $\beta_{MLE}$ is unique; when perfect separation is not present, it is finite.  Does this maximum likelihood model also maximize the ROC AUC (aka $c$-statistic), or does there exist some coefficient estimate $\beta_{AUC} \neq \beta_{MLE}$ which will obtain a higher ROC AUC? If it is true that the MLE does not necessarily maximize ROC AUC, then another way to look at this question is "Is there an alternative to likelihood maximization which will always maximize ROC AUC of a logistic regression?"  I am assuming that models are otherwise the same: we're not adding or removing predictors in $X$, or otherwise changing the model specification, and I'm assuming that the likelihood-maximizing and AUC-maximizing models are using the same link function.


consider the measure space (X,$\mathcal{M}$,$\mu$).imagine the non-negative and Lebesgue measurable function $f:X\rightarrow \mathbb{R}$ is given. Prove that there is a sequence of simple functions  $$\lim_{i \to\infty} f_i(x)=f(x), \forall x \in X$$

This is from Tao Exercise 19.1.3. Let $\Omega \subseteq \mathbb R^n$ measurable and $f: \Omega \rightarrow \mathbb R$ be measurable. Assume $f \geq 0$ on $\Omega$. Then there exists a sequence $\{f_n: \Omega \rightarrow \mathbb R\}_{n=1}^\infty$ s.t. all $f_n$ are simple, non-negative and increasing and $f_n \rightarrow f$ pointwise on $\Omega$. Tao gives the hint to define $$f_n(x) := \sup \left \{\frac j {2^n} : j \in \mathbb Z, \frac j {2^n} \leq \min(f(x),2^n) \right \}$$ Can someone help me to prove why all $f_n$ are measurable ?!


I uploaded many videos on my YouTube account: how can I retrieve them easily?

I have uploaded quite a few videos to YouTube. I'd like to download them all and move them to a different video sharing service. I don't necessarily have the original final files that I uploaded, so I'd like to get them out in the same format I uploaded them in. (Don't want to have to re-edit.)   How can I get my own videos from YouTube?


Hey I have lost my village call as lonelison willa from my Google account n I've sing in with a new village in that sam account so how do I get back my old village in that same account????

I unintentionally and stupidly wiped my Clash of Clans game data. I went to Application Manager on my Android and clicked 'Clear data'. (I used this for fixes before and it never started a new game.)   Now, my level 60 or so base has been wiped as if was to start a new game. Have I doomed myself to starting again or is there a way in which I can retrieve my old game save? Maybe by contacting Supercell? I have a snapshot of my base.


I'm trying to use dpfloat to set two figures on a two-page spread. Because the figures are rather wide, I would like to use pdflscape to set these pages in landscape. However, the following fails when \begin{landscape} is on an odd page:  ... % This fails iff on an odd page. If I add \cleartoevenpage here it works \begin{landscape}   \begin{figure}[p]     \begin{leftfullpage}     ...     \end{leftfullpage}   \end{figure}   \begin{figure}[p]     \begin{fullpage}       ...     \end{fullpage}   \end{figure} \end{landscape} ...   When in the above example \begin{landscape} is on an even page (left page), then dpfloat behaves fine. If it begins on an odd page, dpfloat fails to float to an even page, and it reports  dpfloat moved a leftpage float from page 21.   100 times before crashing at the \end{landscape} line, perhaps because it is blocked by landscape (?). If I manually insert \cleartoevenpage before \begin{landscape} begins, it always works, however in that case I have blank paper before the float.  Here is a MWE that has similar but not identical symptoms. Instead of crashing when the landscape begins on an odd page, it floats to an even page but the page is not set in landscape.  \documentclass{memoir} \usepackage{dpfloat} \usepackage{pdflscape} \usepackage{tikz} \usepackage{kantlipsum}  \begin{document}  \chapter{Analysis}  \kant[3]  \section{Introduction}  \kant[4-7] % \kant[8-12] % Uncomment this line to be on an odd page and make landscape fail  \begin{landscape}   \begin{figure}[p]     \begin{leftfullpage}       \begin{tikzpicture}[scale=6]         \fill[blue] (0,0) rectangle(2, 1.2);         \end{tikzpicture}     \caption{Box plot of number of positions sent per iteration using this scheme}     \end{leftfullpage}   \end{figure}   \begin{figure}[p]     \begin{fullpage}       \begin{tikzpicture}[scale=6]         \fill[blue] (0,0) rectangle(2,1.2);         \end{tikzpicture}       \caption{Detailed plot.       \emph{Top}: overview. \emph{Bottom}: nightly curve.       \label{fig:overview}}     \end{fullpage}   \end{figure} \end{landscape}  \kant[1-3]  \end{document}   How can I set two figures in landscape on a two-page spread?

I have two sideways tables that I want on two facing pages so the reader can turn the entire book and read them both in one turn. For the moment I keep moving them around to keep them on facing pages. Is there a way I can make the first one float until it's on a left page and then place the other one right after? Or perhaps some similar way to automatically get them on the same spread?


Suppose I have a bash script called foo.sh  #!/bin/bash echo "Type your name" read personname if["$personname" == "kevin"]; then     echo "Your name is kevin"     exit 1 fi   Now, I give it global access as cp foo.sh /usr/bin/foo   So, I find it difficult now to call the script and pass the argument at the same time.  I can only do  bash foo john   instead of doing   bash foo john

Say, I have a script that gets called with this line: ./myscript -vfd ./foo/bar/someFile -o /fizz/someOtherFile  or this one: ./myscript -v -f -d -o /fizz/someOtherFile ./foo/bar/someFile   What's the accepted way of parsing this such that in each case (or some combination of the two) $v, $f, and  $d will all be set to true and $outFile will be equal to /fizz/someOtherFile?