How to sketch the following set $|z|=arg(z)$
What points in the complex plane satisfy $|z|=arg(z)$.
What are the basics needed to learn quantum computing? I was very inspired by Michio Kaku's  on the possibilities of quantum computing and also listening to Talia Gershon's  on it.   As I come from a business &amp; analytics background, what materials can I begin exploring to prepare to learn more about the field of quantum computing? Eg. what would I need to study to understand Peter Shor's algorithm,  on physics, etc. Range of $(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$ is If $a_{1},a_{2},a_{3},a_{4}\in \mathbb{R}$ and $a_{1}+a_{2}+a_{3}+a_{4}=0$ and $a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4}=1.$ Then Range of $$E =(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ is  Try: From $$(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+(a_{3}-a_{4})^2+(a_{4}-a_{1})^2$$ $$=2(a^2_{1}+a^2_{2}+a^2_{3}+a^2_{4})+2(a_{1}a_{2}+a_{2}a_{3}+a_{3}a_{4}+a_{4}a_{1})$$ $$=2-2(a_{1}+a_{3})(a_{2}+a_{4})=2+2(a_{1}+a_{3})^2\geq 2$$ and equality hold when $\displaystyle a_{1}=-a_{3}$ and $a_{2}=-a_{4}$ Could some help me how to find its upper bound, Thanks I can't seem to find a few of the Stones, so, is there a list available with the location of all the stones? Generalized Second Borel-Cantelli lemma I have a text file abc.txt and its contents are:     /lag/cnn/org/one.txt    /lag/cnn/org/two.txt    /lag/cnn/org/three.txt   If I use:  $ tar -cvf allfiles.tar -T abc.txt   I'm getting the tar of files in the list. Similarly is it possible to copy those files in abc.txt to a folder?  I tried this:              $ cp --files-from test1.txt ./Folder   But it is not working. I'm reading through a proof in my lecture notes:     Every convergent sequence of a metric space has a unique limit.   Suppose $\lim_{n \to \infty} x_n = l$ and $\lim_{n \to \infty} x_n = m.$ Then for every $\epsilon &gt; 0,$ there exists $N$ such that for every $n \geqslant N,$  $$d(l,m) \leqslant d(l,x_n) + d(x_n,m) &lt; \epsilon + \epsilon = 2 \epsilon. $$  So $(\star) \hspace{1mm} 0 \leqslant \frac{d(l,m)}{2} &lt; \epsilon \Longrightarrow \hspace{1mm} (\star^\prime) \hspace{1mm} \frac{d(l,m)}{2} =0 \Longrightarrow l = m.$  I can follow the proof up until the jump from $(\star)$ to $(\star^\prime)$. Why does $\frac{d(l,m)}{2}=0$? Prove that $a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c) \leq 3abc.$ Prove that there is a bijection between the set of all subsets of $X$, $P(X)$, and the set of functions from $X$ to $\{0,1\}$. Given any set $X$, let $P(X)$ be the set of all subsets of $X$, and let $\{0,1\}^X$ be the set of all functions $X \rightarrow \{0,1\}$. Construct a bijection (and its inverse) between P(X) and $\{0,1\}^X$.    Let us define $f: P(X) \rightarrow \{0,1\}^X$ by $f(\alpha) = g_{\alpha}$ where $\alpha \in P(X)$ and $g_{\alpha} : X \rightarrow \{0,1\}$ is defined by  $$g_{\alpha}(x) =\begin{cases}        1 &amp; x \in \alpha \\       0 &amp; x \in X-\alpha     \end{cases}$$  We need to check if this function is well-defined. Let $\alpha, \beta \in P(X)$. If $\alpha=\beta$, then $\alpha$ and $\beta$ correspond to the same set, and so correspond to the same function since both $g_{\alpha}$ and $g_{\beta}$ send that set to 1 and the ($X-$that set) to 0.   If we have a function $g_{\alpha} \in \{0,1\}$, then $\alpha$ is a set in $X$ that is sent to 1. But since $P(X)$ is the set of all sets of $X$, we must have $\alpha \in P(X)$. So the function is surjective.  Finally, if $g_{\alpha}=g_{\beta}$ then both $g_{\alpha}$ and $g_{\beta}$ send the set $\alpha$ or $\beta$ to 1. So $\alpha$ and $\beta$ must be same set. In other words, the function is injective; and hence bijective.   The inverse function is $f^{-1}: \{0,1\}^X \rightarrow P(X)$ where $f^{-1}(g_{\alpha}) = \alpha$.     Another way to prove this is to first construct the inverse function, show that it is well defined, and then show that $ff^{-1}$ and $f^{-1}f$ are the identity functions, right?  So if we let $g_{\alpha}=g_{\beta}$, then $\alpha=\beta$, because otherwise $g_{\alpha} (\beta)=0$ and $g_{\beta} (\beta)=1$; a contradiction.   Now we have $f \circ f^{-1} (g_{\alpha}) = f(\alpha) = g_{\alpha}$ and $f^{-1} \circ f(\alpha) = f^{-1}(g_{\alpha}) = \alpha$ , and so $f$ and $f^{-1}$ must be bijective. Theorem:   If $x$ is a positive integer and  $x^3$ is even, then $x$ is even.   My Proof by Contrapositive:  I. Assuming that $x$ is odd, then I will show that $x^3$ is odd.  II. $x$ is odd, so $x$ can be expressed as $2k+1$  III. $(2k+1)^3$  IV. $2k(4k^2 + 6k + 3) +1$  V. Since $2k(4k^2 + 6k + 3) + 1$ is an integer, we can conclude $x^3$ is odd.  Did I go about this correctly? Should step V say "$2$ multiplied by an integer and added by $1$ is odd therefore we can conclude $x^3$ is odd"? .cache/upstart/Unity7.log growing to consume the entire free disk space. Python cypto suspected A weak converse of $AB=BA\implies e^Ae^B=e^Be^A$ from "Topics in Matrix Analysis" for matrices of algebraic numbers. It is a well known fact that if $A,B\in M_{n\times n}(\mathbb C)$ and $AB=BA$, then $e^Ae^B=e^Be^A.$  The converse does not hold. Horn and Johnson give the following example in their Topics in Matrix Analysis (page 435). Let $$A=\begin{pmatrix}0&amp;0\\0&amp;2\pi i\end{pmatrix},\qquad B=\begin{pmatrix}0&amp;1\\0&amp;2\pi i\end{pmatrix}.$$ Then $$AB=\begin{pmatrix}0&amp;0\\0&amp;-4\pi^2\end{pmatrix}\neq\begin{pmatrix}0&amp;2\pi i\\0&amp;-4\pi^2\end{pmatrix}=BA.$$ We have $$e^A=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0&amp;0\\0&amp;2\pi i\end{pmatrix}^k=\sum_{k=0}^{\infty}\frac 1{k!}\begin{pmatrix}0^k&amp;0\\0&amp;(2\pi i)^k\end{pmatrix}=\begin{pmatrix}e^0&amp;0\\0&amp;e^{2\pi i}\end{pmatrix}=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}.$$   For $S=\begin{pmatrix}1&amp;-\frac i{2\pi}\\0&amp;1 \end{pmatrix},$ we have $$e^B=e^{SAS^{-1}}=Se^AS^{-1}=S\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}S^{-1}=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}.$$   Therefore, $A,B$ are such non-commuting matrices that $e^Ae^B=\begin{pmatrix}1&amp;0\\0&amp;1\end{pmatrix}=e^Be^A.$  It is clear that $\pi$ is important in this particular example. In fact, the authors say what follows.      It is known that if all entries of $A,B\in M_n$ are algebraic numbers and $n\geq 2,$ then $e^A\cdot e^B=e^B\cdot e^A$ if and only if $AB=BA.$   No proof is given. How does one go about proving that? Yesterday, Asaf Karagila's  sparked an extensive discussion on ways of proving that all countable ordinals are embeddable in $\Bbb Q$, and whether particular solutions to this use the Axiom of Choice.  Since the arguments used to prove this can be shady on their use of Choice when formulated tersely, fully fleshed out answers are appreciated.  To be concrete, this question asks for proofs of the following fact:     All countable ordinals $\alpha &lt; \omega_1$ are order-embeddable in the rationals $\Bbb Q$.   Especially answers avoiding the Axiom of Choice are appreciated.    Of relevance and related interest is  (but it's not a duplicate, since that question asks for explicit embeddings). First, when I never had installed sl:  diogosaraiva@Ubuntu:~$ sl O programa 'sl' não está instalado actualmente. Pode instalá-lo escrevendo: sudo apt-get install sl   traslaction: "The program 'sl' is not currently installed. You can install it by typing:"  But when I'm tired of it and I remove it by typing sudo apt-get remove sl per example...  I get this when I type sl:  -bash: /usr/games/sl: Ficheiro ou directoria inexistente   translate: "-bash: /usr/games/sl: No such file or directory" instead of:    diogosaraiva@Ubuntu:~$ sl O programa 'sl' não está instalado actualmente. Pode instalá-lo escrevendo: sudo apt-get install sl   This appears with others packages too...  How I get rid of this: -bash: /usr/games/sl: Ficheiro ou directoria inexistente ?  I want see the original:  diogosaraiva@Ubuntu:~$ sl O programa 'sl' não está instalado actualmente. Pode instalá-lo escrevendo: sudo apt-get install sl   Output of type sl  sl está hasheado (/usr/games/sl) In follow up to , it appears that some numbers cannot be represented by floating point at all, and instead are approximated.  How are floating point numbers stored?  Is there a common standard for the different sizes?  What kind of gotchas do I need to watch out for if I use floating point?  Are they cross-language compatible (ie, what conversions do I need to deal with to send a floating point number from a python program to a C program over TCP/IP)? Is there a tool to encrypt a file or directory? What's the most popular way to encrypt individual files or folders? Teenage girl in life-sustaining valley on poisoned Earth visited by another survivor What does the term saturating nonlinearities mean? I was reading the paper  and in section 3 were they explain the architecture of their Convolutional Neural Network they explain how they preferred using:     non-saturating nonlinearity $f(x) = max(0, x). $   because it was faster to train. In that paper they seem to refer to saturating nonlinearities as the more traditional functions used in CNNs, the sigmoid and the hyperbolic tangent functions (i.e. $f(x) = tanh(x)$ and $f(x) = \frac{1}{1 + e^{-x}} = (1 + e^{-x})^{-1}$ as saturating).   Why do they refer to these functions as "saturating" or "non-saturating"? In what sense are these function "saturating" or "non-saturating"? What do those terms mean in the context of convolutional neural networks? Are they used in other areas of machine learning (and statistics)? Seeking QGIS tutorials and web resources? I wish to familiarize myself with the capabilities of QGIS up to and beyond QGIS version 3.0.   Can you recommend some good web resources and tutorials that would show me QGIS' capabilities?   I am interested in both using QGIS through its GUI and in developing applications using QGIS. I was pondering a bit more about  question regarding being able to &quot;omit&quot; the base case in a proof by strong induction due to vacuous truth. The post states:  Strong induction proves a sequence of statements $P(0), P(1), …$ by proving the implication &quot;If $P(m)$ is true for all nonnegative integers $m$ less than $n$, then $P(n)$ is true.&quot; for every nonnegative integer $n$. There is no need for a separate base case, because the $n=0$ instance of the implication is the base case, vacuously.  However, if we consider $n=0$, we would have that the statement is vacuously true, which I would take to mean that the implication is true regardless of the validity of $P(0)$. However, clearly it's necessary for $P(0)$ to hold for an induction proof to be valid. So I'm confused on how, by omitting the base case, $n=0$ isn't just a tautology, making the implication true regardless of whether $P(0)$ actually holds. While deploying code in vsCode from sandbox to production I get the below error      sf:INVALID_OPERATION: INVALID_OPERATION: testLevel of NoTestRun cannot   be used in production organizations   As per the error, I need to change the value of testLevel but I am not able to find where and how can I change it to RunLocalTests  Could someone please help me to change the value of testLevel in vsCode