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What Happens to electrons after Alpha Decay and Nuclear Fission? Where do the electrons go? In alpha decay do 2 electrons follow the alpha particle and make stable Helium or does the larger daughter nucleus become an anion?  Also what do the electrons do in the mixture of fission and alpha decay? With Beryllium-8 it decays into two alpha particles and 4 lonely electrons or do you get two Helium atoms?If someone could add an entry to Holocron it would be greatly appreciated, thanks. What is the number of automorphisms (including identity) for permutation group $S_3$ on 3 letters? How to Self-Study Mathematical Methods? Edit:  Ok, user  made comment that truly helps narrow the focus of my question. Basically, I'm asking for a self-study course of Mathematical Methods.  Thanks to his recommendation I identified a book that is very close to the answer:  Mathematical Methods in the Physical Sciences by Mary L. Boas  Supposedly, the book covers most of the Math a person needs to learn in order to understand the simulations I presented.  Does this book suit my needs?    Hello,  I have always been interested on the idea to study Math at an Undergraduate level to be able to implement simulations like the followings:  Finite Difference Simulation of the Wave Equation    Heat Equation in Two Dimensions    Besides my precollege experience in Math, I have only been able to do very simple simulations due to my lack of knowledge of Advanced Math. Every time I ask about the inner workings of the simulations I get redirected to papers that look like foreign language to me and, unfortunately, I depend on easy-to-read tutorials to understand them, and while I have found   I'm more interested in reaching the solution myself.  I checked some of the questions related to book recommendations for self-study and while they are very helpful I'm interested in learning by example (There is a book called: "Accelerated C++ Practical Programming by Example" that fits very well my study habits)  Thanks in advance. What is the benefit/downside to using a switch statement vs. an if/else in C#.  I can't imagine there being that big of a difference, other than maybe the look of your code. Is there any reason why the resulting IL or associated runtime performance would be radically different? Related: Reading about  I understood that black holes lose energy over time - which is logical in some way (otherwise they would be there forever and heat death would never technically happen)  But - how exactly does it "evaporate"? What happens when it no longer contains enough mass within it's  ? Does it explode somehow? Transforms into "regular matter"? Simply vanishes over time? Or? Is $\{\frac{m}{10^n}\mid m,n\in\mathbb Z,\ n\geq 0\}$ dense in $\mathbb R$? The set $S$ of real numbers of the form $m/10^n$, $m,n$ integers and $n$ greater than or equal to $0$, is a dense subset of $\mathbb R$ or not??   I know dense means closure of $S$ in $\mathbb R$ is $\mathbb R$ then it will be dense. How to prove or disprove I have no idea. "Questions that may already have your answer" box overlaps onto the footer What is LookDev/Material Preview mode for in 2.8? Suppose I am given a morphism of schemes $f: X \longrightarrow Y$ and suppose that for any point $y \in Y$, $\kappa(y)$ denotes the residue field. I would like to show that the fibered product $$X \times_{Y} \text{Spec}(\kappa(y))$$ is homeomorphic to the preimage $f^{-1}(y)$ in the induced topology.  The following is my attempt so far:  The question is obviously affine on the target, so without loss of generality we reduce to the case of a morphism $f: X \longrightarrow \text{Spec }A$. Then we have  $$ f^{-1}(\text{Spec }A) = \bigcup_{i \in I} \text{Spec }B_{i} $$ via morphisms $$ \phi_{i}: A \longrightarrow B_{i}. $$ We can then write, \begin{align*} f^{-1}(\text{Spec } A) \times_{\text{Spec } A}\text{Spec } \kappa(y) &amp;= \bigcup_{i \in I} \left( \text{Spec } B_{i} \times_{\text{Spec }A} \text{Spec }\kappa(y) \right) \\  &amp;= \bigcup_{i \in I} \text{Spec }  \left( B_{i} \otimes_{A} \kappa(y)  \right). \end{align*} Now let $\mathfrak{p}$ be the prime ideal of $A$ corresponding to the point $y \in \text{Spec } A$. Then $\kappa(y) = (A/ \mathfrak{p})_{\mathfrak{p}}$. Then for any of the $B$ (dropping the index for brevity), we have \begin{align*} B\otimes_{A} (A/ \mathfrak{p})_{\mathfrak{p}} &amp;\simeq B \otimes_{A} A / \mathfrak{p} \otimes_{A} A_{\mathfrak{p}} \\ &amp;\simeq B_{\mathfrak{p}} \otimes_{A} A / \mathfrak{p} \\ &amp; \simeq B_{\mathfrak{p}} / \mathfrak{p}B_{\mathfrak{p}} \end{align*}  Now the prime ideals in $B_{\mathfrak{p}} / \mathfrak{p}B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B_{\mathfrak{p}}$ containing $\mathfrak{p}B_{\mathfrak{p}}$. Further, the prime ideals of $B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B$ contained in $\phi(\mathfrak{p})B$. That is, we have \begin{align} \text{Spec }B_{\mathfrak{p}} &amp;= \{  \mathfrak{q} \subseteq B :  \phi^{-1}(\mathfrak{q})  \subseteq \mathfrak{p}  \} \\ &amp;=  \{ \mathfrak{q} \subseteq B : f(\mathfrak{\mathfrak{q}}) \subseteq \mathfrak{p}      \}. \end{align} Note that in the very last line we have abused notation slightly: There is not really any notion of being "inside" $\mathfrak{p}$ when it is being treated as a point in the scheme. So really we mean that $f(\mathfrak{q})$ is a point in the scheme whose corresponding prime ideal is a subset of $\mathfrak{p}$. Now as we said above, the prime ideals of $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ are the following $$ \text{Spec }(B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}) = \{ \mathfrak{q} \subseteq B : f(\mathfrak{\mathfrak{q}}) \subseteq \mathfrak{p}      \text{ and } ???  \} $$ This is the point at which I get stuck. Ideally I would like to make some argument from the statement "Now the prime ideals in $B_{\mathfrak{p}}/\mathfrak{p}B_{\mathfrak{p}}$ correspond precisely to the prime ideals of $B_{\mathfrak{p}}$ containing $\mathfrak{p}B_{\mathfrak{p}}$" to conclude that the question marks are the condition $\mathfrak{p} \subseteq f(\mathfrak{q})$. Could someone point me in the right direction as to how to get this condition?   Is the proof up until this point correct? Different sources online seem to do this many different ways, so it's difficult for me to check. Exact relationship between electric field and intensity I often see $I(t)\propto |E(t)|^2$. What is the exact form ? Which constants are missing to make this an equality and why are they so often omitted ? How widespread is the practice of delegating LoR writing to the applicant? At least two questions (, ) on this site indicate that in more than a few isolated cases at least, letters of recommendation are in fact written by the applicant and only signed off (or perhaps rephrased somewhat) by the recommender, who would actually be responsible for writing the letter.  Is there any evidence about the prevalence of this practice that goes beyond anecdotes? I realize that cross-country (cultural) difference may play a role. Akhil showed that the  is the same as the continuum, using as a step the observation that continuous functions that agree at rational points must agree everywhere, since the rationals are dense in the reals.  This isn't an obvious step, so why is it true? Background on color perception:  Due to the nature of the human eye and visual processing by the brain, there's an optical illusion that the same color will look different depending on the background.   It is known as the , illustrated below — center rectangles are identical:   Source:   An illustration  makes the effect of the background's color even more obvious:     Source:   (Note how the two upper-most colors in the second image appear the same, but are actually completely different.)    What about web design?  My question pertains to web design, specifically in this case, the color of link text on different backgrounds.  For a given color of text, how can one find the complementary color to make it appear the same on a certain background?    Let's take a look at an example for this question:  Consider my example below, with yellow text #FFF000 on the following backgrounds:    white #FFFFFF, black #000000, and grey #555555.          As expected, doesn't look the same at all. ()  How can we make it look the same on all three? Clearly, the text on the white background needs to be a darker yellow, and the one on the grey needs hue/tone adjustments. (The surrounding text will also affect it — but that's likely harder to account for.)  Can this perfect color somehow be calculated, based on the Hex, RGB or another system? That's where the crux of this question lies, as it would be much more efficient than manually approximating. If the order of all nontrivial elements in a group is 2, then the group is Abelian. I know of a proof that is just from calculations (see below). I'm wondering if there is any theory or motivation behind this fact. Perhaps to do with commutators?    Proof: $a \cdot b = (b \cdot b)  \cdot (a \cdot b) \cdot (a \cdot a) = b \cdot (b \cdot a) \cdot (b\cdot a) \cdot a = b \cdot a$. I've recently installed Ubuntu on Oracle's Virtualbox.  Now I want to install the guest additions.  I started up my Ubuntu guest OS, clicked "Devices" and chose "Install Guest Additions". Nothing happened, and I don't see the CD icon on my desktop.  How can I make this work? The primitive types (number, string, etc.) are passed by value, but objects are unknown, because they can be both passed-by-value (in case we consider that a variable holding an object is in fact a reference to the object) and passed-by-reference (when we consider that the variable to the object holds the object itself).  Although it doesn't really matter at the end, I want to know what is the correct way to present the arguments passing conventions. Is there an excerpt from JavaScript specification, which defines what should be the semantics regarding this? Computing the Galois group of polynomials $x^n-a \in \mathbb{Q}[x]$ I have some problems with this exercise. I don't know if it can be done. Consider the polynomial $ x^n - a \in \mathbb{Q} $ Can I compute the Galois group of this over $\mathbb{Q}$? Maybe having a nice "basis.  The splitting field is given by $\mathbb{Q}(\zeta_n,\alpha)$ , where $\zeta_n$ is a primitive root of unity , and $\alpha$ is some number such that $\alpha^n = a $. Well first of all, I want a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different (in this case the computation of the galois group is very simple).  For example one easy case, it's when $a&gt;0$, then $(a)^{\frac{1}{n}} \in \mathbb{R}$ , so clearly the minimal polynomial of $(a)^{\frac{1}{n}} , \zeta_n$ are distincs, and I'm done. If $n$ is odd then , it's also easy, since one root it's also real, for example $x^3-3 $, the real root is $ \root 3 \of { - 3}  =  - \root 3 \of 3  $ , so I can consider the splitting field as $\mathbb{Q}(-\root 3 \of 3 , \zeta_3 )=\mathbb{Q}(\root 3 \of 3 , \zeta_3 )$. The difficult case is when $n$ is even and $a&lt;0$ , for example  $x^8+20$ or $x^4+20$ in some cases as in the second, there are particular cases since there exist algorithms for the Galois group of quartics, but in general. It can be done?  $\underline{Remark}$ I'm searching a $\underline{"good basis"}$ for the splitting field. In the sense that the minimal polynomials, of the adjoined elements, are different since in this case the computation of the galois group is very simple. I know that $S_n$ is generated by a number of things, like all transpositions, all transpositions of the form $(1\,a)$, the transpositions $(1\,2),(2\,3),\dots,(n-1\, n)$, and  just the two elements $(123\dots n),(1\,2)$.      Suppose $n$ is prime. If you just have $(123\dots n)$ and some arbitrary transposition $(a\, b)$, how does this also generate $S_n$?    Can you somehow get to $(1\,2)$ or reduce it to some other previous case? Python Script returns unintended "None" after execution of a function As mentioned on :     For something that can be done so   easily by accident, there should be   the ability to undo mistakes.   I know that I have clicked several times on a comment up-vote without actually wanting to.  I would suggest being able to cancel the up-vote (maybe only for a couple of minutes or even less than a minute, no problem, as long as I can undo a miss-click).