How to disable NetworkManager on Ubuntu 15.10?
How can I completely remove NetworkManager?
Proof that all abelian simple groups are cyclic groups of prime order Just wanted some feedback to ensure I did not make any mistakes with this proof. Thanks!  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has either infinite order or even order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has odd order, and by Feit-Thompson, $G\cong Z_p$ for some prime $p$.  Edit: Thanks, I see that Feit-Thompson is too much.  Since $G$ is abelian, every subgroup is normal. Since $G$ is simple, the only subgroups of $G$ are $1$ and $G$, and $|G| &gt; 1$, so for some $x\in G$ we have $x\neq 1$ and $\langle x\rangle\leq G$, so $\langle x\rangle = G$. Suppose $x$ has infinite order. Then $\langle x^2\rangle \leq G$ but $\langle x^2\rangle \neq \langle x \rangle$, a contradiction. So $x$, and therefore $G$, has finite order. Suppose $x$ has composite order $n$ so for some $p &gt; 1$ that divides $n$, $\langle x^p \rangle$ is a proper non-trivial subgroup of $G$, so $G$ is not simple. So $G$ is a cyclic group of prime order. What is the $n$ times iterate of $f(x)=\frac{x}{\sqrt{1+x^2}}$? Is it ok to ask to explain a certain piece of code on SO? Is it OK to ask on SO for help in explaining someone else's working piece of code, from Open Source projects or library examples etc.? Or does this belong somewhere else? What is the memory layout of a .NET array?  Take for instance this array:  Int32[] x = new Int32[10];   I understand that the bulk of the array is like this:  0000111122223333444455556666777788889999   Where each character is one byte, and the digits corresponds to indices into the array.  Additionally, I know that there is a type reference, and a syncblock-index for all objects, so the above can be adjusted to this:  ttttssss0000111122223333444455556666777788889999         ^         +- object reference points here   Additionally, the length of the array needs to be stored, so perhaps this is more correct:  ttttssssllll0000111122223333444455556666777788889999         ^         +- object reference points here   Is this complete? Are there more data in an array?  The reason I'm asking is that we're trying to estimate how much memory a couple of different in-memory representations of a rather large data corpus will take and the size of the arrays varies quite a bit, so the overhead might have a large impact in one solution, but perhaps not so much in the other.  So basically, for an array, how much overhead is there, that is basically my question.  And before the arrays are bad squad wakes up, this part of the solution is a static build-once-reference-often type of thing so using growable lists is not necessary here. Solving an Op-amp circuit The image of the circuit is shown below and it required to find \$V_0\$,  My first attempt at solving this problem is by changing the current source into a voltage source with 1-V and 2k\$\Omega\$ resistance. The fact that the inverting and non-inverting terminals aren't grounded make this problem look difficult.To the point which, I don't know how to proceed with this question or where to start. I would appreciate any help. Following some thought and another schematic from a hint suggested by Alfred, I produced a schematic representing our work.  And  my solution for the problem is below, Using node equation at nodes A and B we have, $$\frac{V_A-V_B}{1k}=-0.5 \text{mA}$$ $$\frac{V_B-(2+V_A)}{1k}=-0.5-x$$ where x is the current that is sent in the output of the op-amp. Using KCL, at the bottom node near the current source we see that the same current that goes through the op-amp also goes through the \$2\text{k}\Omega\$ resistor. Hence, we have, $$\frac{-V_A}{2k}=x$$ Replacing this in the second equation, $$\frac{V_B-(2+V_A)}{1k}=-0.5+\frac{V_A}{2k}$$ And and solving the equations yields \$V_A=-2 \text{-V}\$ and \$V_B=-1.5 \text{-V}\$ Converting point layer to raster grid that shows frequency of points per cell uisng QGIS I have a point layer that shows the distribution of a species.   Is it possible in QGIS to convert this point layer to a raster grid in which the value of each cell corresponds with the amount of points within that cell?   Until now, I only managed to attach attribute values to the grid cells. I have already added a column in my attribute table that has the value '1' for each point, hoping that there is a way to use a sum of attributes for each grid cell. I am trying to solve a problem from Stanley's book, it says:  Fix $k,n \in \mathbb{P}$. Show that: \begin{align} \sum a_1 a_2 \cdots a_k = \binom{n+k-1}{2k-1} \end{align} where the sum ranges over all compositions $(a_1 , a_2 , \ldots , a_k)$ of $n$ into $k$ parts.  I am trying to reason like this:  we need to find the coefficient $c_n = \sum a_1 a_2 \cdots a_k$ from this generating function --  \begin{align} \sum_n c_n x^n &amp;= \sum_n \sum a_1 a_2 \cdots a_k x^n \\  &amp;= \sum_n \sum a_1 a_2 \cdots a_k x^{a_1 + a_2 + \cdots + a_k}\\ &amp;= \sum_n \sum a_1x^{a_1} a_2x^{a_2} \cdots a_kx^{a_k} \end{align}  after that, I have no clue, how do I solve this ?   moreover, what is the range in the inner sum ?    If we consider Mark Riedel's answer, and assume $n=4$, $k=2$; then the sum will be  \begin{align} \sum (z + 2z^2)^2 = z^2 + 4z^3 + 4z^4 \end{align}  On the other hand the compositions will be $(1,3), (2,2), (3,1)$, therefore the above sum will be counted as:  \begin{align} (1.3)z^{1+3} + (2.2)z^{2+2} + (3.1)z^{3+1} &amp;= 1z^1.3z^3 + 2z^2.2z^2 + 3z^3.1z^1\\  &amp;= 3z^4 + 4z^4 + 3z^4 = 10z^4 \end{align}   what's going on? what am I missing? Dictionary application needs to download dictionary "Apple Dictionary" before use Proof of this fairly obscure differentiation trick? Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$  But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:  $$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$  which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.  It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with $$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$ which is the correct expression for $f'(x)$.  This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.  These observations have led me to the following conjecture:     Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).$$   (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)  This formulation also seems to lend itself to following generalisation:     Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x_1,x_2,\ldots,x_N$. For $n=1,2,\ldots,N$ define $f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum_n f'_n(T).$$   Thus my question is:   Is this true? How can it be proven? (Specifically in the case $N=2$ but also in the general case.) Print file content without the first and last lines The Lie algebra of $SL_n(\mathbb C)$ are the matrices where the trace is $0$. But what is the Lie algebra of $SL_n(\mathbb H)$ where $\mathbb H$ is the quaternions? Center image element in parent div How to quickly and safely defrost chicken? What is the safest, quick way to defrost chicken? Microwave? Running water over it? I know that children’s books is correct, but for some reason I want to say kids books, even though I know that it’s a plural possessive noun ending in an s.   A quick search on the web turns up Barnes &amp; Noble, who have a web page title that reads “Children's Books, Kids Books”.   Is the major bookseller wrong here? Or is there some exception I don’t remember?  More simply, is it kids books or kids’ books — and why? Basic Web Development IDE/Editor like Dreamweaver? I've used Windows for many years, using Dreamweaver as my editor as it had a file tree which is based on a project (via Local, FTP or SFTP).  I need a web development IDE with the following functionality:   syntax highlighting for PHP, HTML, CSS, JavaScript  file tree - Local, FTP, SFTP, Project-Based (e.g. via the methods of Local/FTP/SFTP) in other words, like Dreamweaver can open multiple files (tabbed)   Additionally, I wouldn't mind features like debugging, class listings, etc, as long as I can hide them away. $a =\prod_{i = 1}^{r} p_{i}^{ai}$ with $a_{i} > 0$ for each $i$ is the canonical representation of $a$... Using strtok with a std::string I have a string that I would like to tokenize. But the C strtok() function requires my string to be a char*. How can I do this simply?  I tried:  token = strtok(str.c_str(), " ");    which fails because it turns it into a const char*, not a char* If derivative of function is less than or equal to 4, prove the existence of point $c$ so $f(c)=c^2$. I am struggling to construct a simple proof of the following.  Suppose $f(x)$ is differentiable for any real value $x$ and that $f'(x) \le 4$ for all real values of x. Prove there is at most one real value $c&gt;2$ such that $f(c) =c^2$.  I am completely lost on where to start on this proof and any help would be greatly appreciated! Server vs single-player world effects Are there any differences with regards to chunk generation or other settings in single-player vs server mode of equivalent versions?  I have a map I'd like to host on a private server for my friend to play with me. However, when I'm on the road without ready internet, I'd like to play it single-player as well.  Will this cause any issues, especially with chunk generation, drops, items, etc (i.e., getting a different chunk generated in server-mode than single-player would have created), or do all server-client version pairs (I'm using 1.6.2) use the same game code?  The main point is that I don't want gameplay to change at all, between single-player and server mode. Is there a way to determine how many digits a power of 2 will contain? Is there a direct way to determine how many digits a power of 2 will contain without actually performing the multiplication?  An estimation would help as well if there is no absolute solution.  EDIT: In both decimal and binary bases.