It is my understanding that the Children of the Forest first created the White Walkers in order to protect themselves from the First Men. However, something must have happened that then the Children of the Forest and the First Men teamed up to defeat, or at least repel, the White Walkers. What is that something? Or am I misinterpreting the show?
After viewing episode 5 of season 6 of Game of Thrones, we have come to know that      the Children of the Forest created the White Walkers.   But it seems that the White Walkers attack the Children of Forest. My question is why they would do so, given their origin.
How could a raw, untrained Force user be so adept at using the Force? Everything that I seemed to know about the Star Wars universe indicated that in order to be successful at being a force user, the user required decades of training and experience.  Which is why I thought that the Jedi trained their pupils from birth to be force users. The lengthy training period was also why the Jedi council said Anakin was too old, and subsequently why Yoda said Luke was too old to begin the training. This can be seen in comparing Luke's raw abilities in the original trilogy compared to that of the Jedi in the prequel trilogy.   However, in The Force Awakens,      Rey not only fends off Kylo Ren, she comes very close to killing him outright, even though she had never even held a lightsaber before.   This happens not long after      she uses her Force persuasion power to escape prison, despite not even believing she was Force sensitive until that point.   It doesn't make any sense that the fight wasn't over in a matter of seconds. What am I missing? Can someone give me an example for this question on reflexivity please. How is this reflexive? Can you show an example with numbers from the set?     Question 6   Let $X = \{0,1,2,3,\dotsc, 9\}$. Define a relation $\mathcal R$ from $X$ by "$x$ is related to $y$ if $x$ and $y$ gives the same remainder on division by $3$".   (a) Show that $\mathcal R$ is an equivalence relation.   Anwser:   (a) $\mathcal R$ is clearly both reflexive and symmetric, for every number has the same (unique) remainder on division by $3$ as itself and if $a$ has the same remainder on division by $3$ as does $b$, then $b$ has the same remainder on division by $3$ as $a$ also. Similarly $\mathcal R$ is transitive, for if $a\mathcal R b$ and $b\mathcal R c$, then $a$, $b$, and $c$ all have the same remainder on division by $3$, so $a\mathcal R c$ also. There are a lot of software in Windows to merge PDF files but how can we do the same in Ubuntu? Suppose $a_n>0$ and $\sum_{n=1}^{\infty}{a_n}$ diverges. Determine convergence of $\sum_{n=1}^{\infty}{\frac{a_n}{s_n^2}}$, where $s_n=\sum^n a_n$. Zombies - Why do they need to feed on living animals/humans? So what I'm looking for here is something like PHP's  function.  This is so I can debug my scripts by seeing what's the state of the object in question. Removing overlapping part of polygons using QGIS? Here is what I'd like to achieve in QGIS.   Is there a processing tool to remove the overlapping part of polygons?    It is also very important that they do not get merged during the process and that they keep their original attributes since I need to work with them afterwards. Prove or find a counterexample: For all real numbers x and y it holds that x + y is irrational if, and only if, both x and y are irrational. The following is a Homework Question that I've been working on and I would like some feedback on my answer: Prove or find a counterexample: For all real numbers $x$ and $y$ it holds that $x + y$ is irrational if, and only if, both $x$ and $y$ are irrational.  So I've got (or at least like to think I've got) a counter example.  Let $x = 1$ and $y = \sqrt{2}$.   Thus $x$ is rational and $y$ is irrational. Adding $x$ and $y$ gives us $1 + \sqrt{2}$ which cannot be simplified any further (right?) and is an irrational number.  Thereby proving that $x + y$ can be irrational without both x and y being irrational.   Is this okay? Any feedback is greatly appreciated thank you! How to install the latest Arduino IDE? I have an Arduino UNO, which I need to get working with Ubuntu.  Can somebody explain to me the steps required to install the IDE on Ubuntu? I've had  on the web since 1999, and I'd like to add a page on why complex numbers are (or should be) considered to be numbers.  I'm frequently asked that question.  This is not a question about the existence or usefulness of complex numbers as they do exist as mathematical entities and they are very useful.  It is a question about what makes them numbers.  It could be broadened to ask what constitutes numbers, but I'm specifically interested in complex numbers.  Perhaps it's just historical and sociological, but I'm hoping for an answer that there's something intrinsic about complex numbers that makes them fit into the category of number. Why can't I use a type argument in a type parameter with multiple bounds? So, I understand that the following doesn't work, but why doesn't it work?  interface Adapter&lt;E&gt; {}  class Adaptulator&lt;I&gt; {     &lt;E, A extends I &amp; Adapter&lt;E&gt;&gt; void add(Class&lt;E&gt; extl, Class&lt;A&gt; intl) {         addAdapterFactory(new AdapterFactory&lt;E, A&gt;(extl, intl));     } }   The add() method gives me a compile error, "Cannot specify any additional bound Adapter&lt;E&gt; when first bound is a type parameter" (in Eclipse), or "Type parameter cannot be followed by other bounds" (in IDEA), take your pick.  Clearly you're just Not Allowed to use the type parameter I there, before the &amp;, and that's that. (And before you ask, it doesn't work if you switch 'em, because there's no guarantee that I isn't a concrete class.) But why not? I've looked through Angelika Langer's FAQ and can't find an answer.  Generally when some generics limitation seems arbitrary, it's because you've created a situation where the type system can't actually enforce correctness. But I don't see what case would break what I'm trying to do here. I'd say maybe it has something to do with method dispatch after type erasure, but there's only one add() method, so it's not like there's any ambiguity...  Can someone demonstrate the problem for me? Number of total possibilities for an equation I need to find the number of possibilities for which the following equation exists:  $$x_1 + x_2 + x_3 + \cdots + x_{10} \leq 70$$  Each variable is a non-negative integer.  I tried simplifying the question to the point of finding the number of possible solutions to each equation seperatly:  $$x_1 + x_2 + x_3 + \cdots + x_{10} = 70,$$ which is ${79}\choose{10}$.  $$x_1 + x_2 + x_3 \cdots + x_{10} = 69,$$ which is ${78}\choose{10}$.  $$x_1 + x_2 + x_3 + \cdots + x_{10} = 68,$$ which is ${77}\choose{10}$, and so forth.  Then I thought about adding all of these together. This seems a bit too much work considering that the textbook solution is: ${80}\choose{10}$, and I can't seem to figure out the idiomatic way of approaching these kind of questions.   Any help is greatly appreciated. How to set environment variable in systemd service? I have an  and I've created my own service. The configuration service at /etc/systemd/system/myservice.service looks like this:  [Unit] Description=My Daemon  [Service] ExecStart=/bin/myforegroundcmd  [Install] WantedBy=multi-user.target   Now I want to have an environment variable set for the /bin/myforegroundcmd. How do I do that? Is there a way to solve this integral?  $$\int \frac{dx}{\sqrt{-ax^2 + bx +c}}, \;\;\;\;\; \forall\; a,b,c \in (0,+\infty) \;in\;R$$  I have tried it by changing of variable and by parts, but no result. I don't know what else to do.  Thanks in advance. Entire function with vanishing derivatives? Let $f:\mathbb{C}\rightarrow\mathbb{C}$ be an entire function. And assume that at each point, one of it's derivatives vanishes.  What can you say about $f$?  A hint suggests that $f$ must be a polynomial. I'm trying to to execute a script, which uses the command rename.   When I execute it, I get a message rename: command not found. Any ideas?   if [[ -z '$3' ]]; then     shopt -s dotglob     rename "$EXPRESSION" ./* else     rename "$EXPRESSION" "$3" fi Proving square root of a square is the same as absolute value Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:  $\begin{align} f(x) &amp;= \sqrt {x^2} \\ f(-1) &amp;= \sqrt {(-1)^2} \\ f(-1) &amp;= \sqrt {1} \\ f(-1) &amp;= 1 \end{align}$  $\begin{align} g(x) &amp;= x \\ g(-1) &amp;= -1 \end{align}$  thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:  :    :    :    From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$? Can I use the Twig template engine? I do a lot of  development and I like their templating language. , the , looks very much like it.  How can I use Twig in Drupal 7 or even Drupal 6? Let $f$ be a differientable real function such that $$(b-a)f'(\tfrac{a+b}{2}) = f(b)- f(a)$$ for all reals $a,b$.     Is $f$ polynomial of degree $\leq 2$ ? I am a newbie in linux. I wanted to install ubuntu to my desktop. I currently have two partitions, Windows 8 in installed on one, and i have some important files in another partition. While installing ubuntu, if I chose "Replace Windows 8 with Ubuntu", do I lose the other partition too?   And are there other ways to leave the other partition as it is, and install ubuntu in the partition  where windows was installed.