Could someone explain how/why the square root of $x$ equals $x$ to the one half power? I know by definition it does, but is there any mathematical process we can go through to get from one to the other?  In addition, I know the $x$ to the 2ed power is equal to $x*x$, but what is $x^{1/2}$ equal to? Is it one half of $x$ times one half of $x$? No!  I know all this stuff by definition, but I want to know why. Thanks
Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x $?   I'm currently studying indices/exponents, and this is a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this.  Thank you.
Young adult book about a boy(?) who gets sent through a portal to another world which is really the distant future When is the limit of a sum equal to the sum of limits? I was trying to solve a problem and got stuck at the following step:  Suppose ${n \to \infty}$ .  $$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = 1$$  Let us rewrite $n^3=n \cdot n^2$ as $n^2 + n^2 + n^2 + n^2 \dots +n^2$,$\space$ n times.  Now we have  $$\lim \limits_{n \to \infty} \frac{n^3}{n^3} = \frac {n^2 + n^2 + n^2 + n^2 + n^2 \dots +n^2}{n^3} $$  As far as I understand, we can always rewrite the limit of a sum as the sum of limits ...  $$\dots = \lim \limits_{n \to \infty} \left(\frac{n^2}{n^3} + \frac{n^2}{n^3} + \dots + \frac{n^2}{n^3}\right)$$  ...but we can only let ${n \to \infty}$ and calculate the limit if all of the individual limits are of defined form (is this correct?). That would be the case here, so we have:  $= \dots \lim \limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n} + \dots + \frac{1}{n}\right) =$[ letting ${n \to \infty}]$ $= 0 + 0 + \dots + 0 = 0$  and the results we get are not the same.  Where did I go wrong? When I analyze my variables in two separate (univariate) logistic regression models, I get the following:  Predictor 1:    B= 1.049,    SE=.352,    Exp(B)=2.85,    95% CI=(1.43, 5.69),    p=.003    Constant:    B=-0.434,    SE=.217,    Exp(B)=0.65,                            p=.046  Predictor 2:    B= 1.379,    SE=.386,    Exp(B)=3.97,    95% CI=(1.86, 8.47),    p&lt;.001    Constant:    B=-0.447,    SE=.205,    Exp(B)=0.64,                            p=.029   but when I enter them into a single multiple logistic regression model, I get:  Predictor 1:    B= 0.556,    SE=.406,    Exp(B)=1.74,    95% CI=(0.79, 3.86),    p=.171 Predictor 2:    B= 1.094,    SE=.436,    Exp(B)=2.99,    95% CI=(1.27, 7.02),    p=.012    Constant:    B=-0.574,    SE=.227,    Exp(B)=0.56,                            p=.012   Both predictors are dichotomous (categorical). I have checked for multicollinearity.  I am not sure if I have given enough info, but I cannot understand why predictor 1 has gone from being significant to non-significant and why the odds ratios are so different in the multiple regression model. Can anyone provide a basic explanation of what is going on? Product Notation for Multiplication in Reverse Order Is there a standard notation for multiplication in reverse order? For example consider the problem  $$x_{k+1} = A_k x_k$$  where $x_i \in \mathbb{R}^n$ and $A_i \in M_n(\mathbb{R})$, ($i=0,1,2,\dots$) without any further assumptions on $A_i$. The solution to this problem is  $$x_k = A_{k-1} \dots A_1 A_0 x_0$$  Obviously, writing  $$x_k = \prod_{i=0}^{k-1} A_i x_0$$  is wrong because of the multiplication order. But I'm also unconfortable to use  $$x_k = \prod_{i=k-1}^{0} A_i x_0$$  as the notation suggests that we need to increment $i$, not decrement it. Also, one may need to write $\prod_{i=k}^{m} A_i$. Then, we need to explicitly state $m &lt; k$ to imply ordering.  I can always write the open form but as similar products frequently occur in calculations in my work I want to express them in a more compact way. The question is, is there a standard and nice notation to do so? There are some suggestions in , but I think none of them are "nice enough" to work with.  I came up with the notation  $$x_k = \coprod_{i=k-1}^{0} A_i x_0$$  to imply decrementing $i$ rather than increment it, but I don't know if it is used in somewhere else or there is a standard notation. Geiger and Marsden's experiment led Rutherford to believe that the positive charge and most of the mass of the atom was concentrated in a small region. I understand what led him to conclude the way the positive charge is positioned in the atom. But how did he conclude that most of the mass was in a small region (the nucleus)?  How did the distribution of the mass matter after all? Given that the electric force is greater than the gravitational force by many magnitudes, the force between the positice charge and the electrons was predominantly electric.   So how did Rutherford conclude that most of the mass is in the nucleus? Proving that without the axiom of choice there is a set with an accumulation point that isn'tthe limit of a convergeant sequence I've been reading the simple parts of Thomas Jech's book on the Axiom of Choice and came across the following proof on page 141. The proof assumes a mathematical model without the axiom of choice.     Lemma. $X$ is D-finite if and only if $X$ does not have a countable subset.       Theorem. There is a set $S$ of real numbers and a real number $a \not \in S$ such that a is in the closure of $S$ but there is no sequence $(x_n)^{\infty}_{n=0} \subset S$ such that $(x_n)^{\infty}_{n=0} \longrightarrow a$.       Proof. Let $D$ be a D-finite infinite set of reals. We claim that the set $D$ must have an accumulation point. To prove this claim suppose that all $d \in D$ are not accumulation points, then let $\{I_n:n=0,1,...\}$ be a fixed enumeration of open intervals with rational endpoints. Assign the least $n$ such that $I_n \cap D={d}$ to each $d \in D$. This makes $D$ countable and by the above lemma gives us our contradiction. So $D$ has an accumulation point $a$. Let $S=D \setminus\{a\}$ and since $S$ is D-finite, every convergent sequence in $S$ is eventually constant.   Could somebody explain this proof? Why is the last sentence true/ Why does this mean every convergent sequence is eventually constant?  I think the following theorem is important     Theorem. There is a model of ZF which has an infinite set of real numbers without a countable subset What is the difference between SAN, NAS and DAS? What is the difference between SAN, NAS and DAS? XeLaTeX update [20201004] and item ball in Beamer Beamer item balls go missing while compiling in XeLateX since yesterday's TeXlive update, however, the same code works with pdfLaTeX. Any solution, please. Here is my MWE: \documentclass{beamer}  \setbeamertemplate{itemize item}[ball] \setbeamertemplate{enumerate item}[ball]  \begin{document}  \begin{frame} \frametitle{A tested title} text \begin{enumerate}     \item this is an item \end{enumerate} text3 \begin{itemize}     \item this is an item \end{itemize} \end{frame} \end{document} Adjectives and nouns to describe an answer that doesn't answer the question? I've heard of the adjectives 'nonresponsive' or , but are they restricted to law^?     Collins Dictionary:  =       not responsive, not reacting to a stimulus   The definition above differs from the definitions of adjectives like 'equivocate,  evade' and nouns like , but these are negative and accusive; I prefer neutral connotation.  ^Only if you wish, please see: ,  by David Berg,  by Thomas A. Mauet. I was viewing some old posts on Stack Overflow (two-year-old posts) and found that due to "fast technology trend changing," some of them are no longer valid. Stack Overflow (and the Stack Exchange sites generally) should have some sort of criteria to tag questions as Obsolete, or a vote-based system to mark them as such.  I am not in favour of closing obsolete questions. How do I parse a string to a float or int? In Python, how can I parse a numeric string like "545.2222" to its corresponding float value, 545.2222? Or parse the string "31" to an integer, 31?  I just want to know how to parse a float str to a float, and (separately) an int str to an int. I've heard of the idea of bootstrapping a language, that is, writing a compiler/interpreter for the language in itself.  I was wondering how this could be accomplished and looked around a bit, and saw someone say that it could only be done by either   writing an initial compiler in a different language. hand-coding an initial compiler in Assembly, which seems like a special case of the first   To me, neither of these seem to actually be bootstrapping a language in the sense that they both require outside support.  Is there a way to actually write a compiler in its own language? What happened to Rhaegal? In Game of Thrones S08E03 ("The Long Night"), the Night King's dragon tore at the one Jon Snow rides (Rhaegal). He falls off of it and we don't see the dragon anymore, is it dead? Why does this not work with an imaginary exponent? How to install and use Ubuntu from a USB flash drive on Windows 8? When using vagrant Stderr: VBoxManage: error: VT-x is not available (VERR_VMX_NO_VMX) I am trying to install open edx with  ubunty 16.04 and I get this error  There was an error while executing `VBoxManage`, a CLI used by Vagrant for controlling VirtualBox. The command and stderr is shown below.  Command: ["startvm", "c60f0594-d584-4e0b-b863-2801168e641e", "--type", "headless"]  Stderr: VBoxManage: error: VT-x is not available (VERR_VMX_NO_VMX) VBoxManage: error: Details: code NS_ERROR_FAILURE (0x80004005), component ConsoleWrap, interface IConsole   this occurs when using vagrant up How to simulate from a Gaussian copula? Suppose that I have two univariate marginal distributions, say $F$ and $G$, which I can simulate from. Now, construct their joint distribution using a , denoted $C(F,G;\Sigma)$. All the parameters are known.  Is there a non-MCMC method for simulating from this copula? Just freeze an application (suspend the CPU, save its state to disk) and 'defrost' it when I want to use it 4 weeks later. How can I deactivate Bluetooth on system startup? How can I deactivate Bluetooth on system startup? What's the best way to break from nested loops in Javascript?  //Write the links to the page. for (var x = 0; x &lt; Args.length; x++) {    for (var Heading in Navigation.Headings)    {       for (var Item in Navigation.Headings[Heading])       {          if (Args[x] == Navigation.Headings[Heading][Item].Name)          {             document.write("&lt;a href=\""                 + Navigation.Headings[Heading][Item].URL + "\"&gt;"                 + Navigation.Headings[Heading][Item].Name + "&lt;/a&gt; : ");             break; // &lt;---HERE, I need to break out of two loops.          }       }    } }