Elementary particles such as the electron are quantum mechanical fuzzy objects described by quantum states or wavefunctions. They are not classical billiard balls with some fixed radius. Can we assign a notion of a natural &quot;size'' of the electron irrespective of the state it is in? The only length associated with the electron that I can think of is the Compton wavelength, $\frac{\hbar}{m_ec}$. But I am not sure if it will be proper to refer that as the size of the electron. I don't have a definite answer. Thanks.
According to :     The electron has no known substructure. Hence, it is defined or assumed to be a point particle with a point charge and no spatial extent.   Does point particle mean the particle should not have a shape, surface area or volume?  But when I searched Google for the "electron shape" I got many results (like  and ) that says electrons are round in shape.
Best practices for catching and re-throwing .NET exceptions No Dialect mapping for JDBC type: -9 I am wondering the following question: Let $X$ be a separable Banach space which is linearly isomorphic to a dual Banach space $Y^*$. Is there a Banach space $Z$ such that $X$ is lineraly isometric to the dual of $Z$: $X=Z^*$.  I think that the answer is no, but I do not have a counterexample. Since $L_1$ is not isometric to any dual Banach space, maybe one can find a dual Banach space which is isomorphic to $L_1$...  To finish, do that change anythink if I suppose $X$ to be almost linearly isometric to $Y^*$ ? By almost linearly isometric I mean that for every $\varepsilon &gt;0$ there exist $T$: $X \to Y$ a linear isomorphism satisfying $\|T\|  \|T^{-1}\| \leq 1+\varepsilon$. Why doesn't Cantor's diagonal argument also apply to natural numbers? Uniformly Continuous Function sending Bounded Set to Unbounded One Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and let $f: X \to Y$ be a uniformly continuous function. If $A \subset X$ is bounded, must $f(A) \subset Y$ be bounded?  It is clear to me that in metric spaces that satisfy the Heine-Borel property, such as $\mathbb{R}^{n}$, the answer to this question is yes. However, I can see no reason why this should hold for arbitrary metric spaces. Any ideas? Thanks! I need some help with the exercise I was assigned in math class. We were given a 3x3 square made of unit squares, and we wanted to paint all the squares. We had 3 sets of paint, red, blue, and green. Each paint was distinct from each other, but two unit squares with the same color were the same. Additionally, there's only enough paint of each color to color 3 unit squares, and all the paint has to be used. The goal of the exercise was to find the total amount of distinct possible ways such that no two unit squares that share an edge (that is to say they are next to each other vertically or horizontally, having same color across a diagonal is fine). I've started by trying to take a look at each possible arrangement of the bottom 3 unit squares, but I can't seem to find a pattern between it. Is there a better way than continuing by brute force? Any ideas on how I should proceed from here? Let $G$ be a group and $c\in G$. Define $*$ on $G$ by: $x*y=xdy$ for all $x,y\in G$, where $d$ is the inverse of $c$. Prove $(G,*)$ is a group. Let $G$ be a group and $c$ be a fixed element of $G$. Define a new operation $*$ on G by:   $$x*y=xdy$$   for all $x$ and $y$ in $G$, where $d$ is the inverse of $c$. Prove that $G$ is a group under the operation $*$.    If you want to check, this problem is in the book “A Course In Group Theory”, by John F. Humphreys (problem $3$, page $28$)   I know what are the axioms that $(G,*)$ must obey in order to be a group and I did show that all the axioms where obeyed except the inverse axiom.  My problem is that, whenever I try to verify what would the inverse be for some element $x$, I get an equation where the $x$ is cancelled in both sides.  Can please someone guide me?  Thank you!  (P.S.: This is NOT a homework assignment, I’m studying Group Theory by this book and got problems solving this question) Abelian groups of order n. Is there a number $n$ such that there are exactly 1 million abelian groups of order $n$?   Can anyone please explain. I would yes because numbers are infinitive, and so any number n can be expressed as a direct product of cyclic groups of order n.  Can anyone please help  me understand.  Thank you. Toggle Switch to swap the connections of two buttons How to find real world Selenium projects Where can I find sample or real Selenium projects. All the material I can find on github and on internet are basic, and never intermediate or advanced.  How does a real Selenium project look in real life? How can I find sample code? How do I import a dxf? I am a noob with blender. Please be as specific as possible. College students have been doing this for me, but I am unable to understand them. How best to sum up lots of floating point numbers? What program can I use to decompile a class file? Will I actually get Java code, or is it just JVM assembly code?  On Java performance questions on this site I often see responses from people who have "decompiled" the Java class file to see how the compiler optimizes certain things. Category-theoretic limit related to topological limit? I need to generate random numbers within a specified interval, [max;min].  Also, the random numbers should be uniformly distributed over the interval, not located to a particular point.  Currenly I am generating as:  for(int i=0; i&lt;6; i++) {     DWORD random = rand()%(max-min+1) + min; }   From my tests, random numbers are generated around one point only.  Example min = 3604607; max = 7654607;   Random numbers generated:  3631594 3609293 3630000 3628441 3636376 3621404   From answers below: OK, RAND_MAX is 32767. I am on C++ Windows platform. Is there any other method to generate random numbers with a uniform distribution? My objective here is really simple -- I'm trying to set an NSString to some test data, then return the class, which should be NSString. Here's my code:  NSString* stringer = [NSString stringWithFormat: @"Test"]; NSLog(@"%@", [stringer class]);   The log says that the class is NSCFString, not NSString. What's going on here? Let $E$ be a metric $\mathbb R$-vector space, $I$ be a nonempty set and $(\mu_t)_{t\in I},(\nu_t)_{t\in I}$ be families of nonnegative finite measures on $\mathcal B(E)$.  that if $(\mu_t)_{t\in I}$ and $(\nu_t)_{t\in I}$ are tight$^1$ and bounded in total variation norm, then the family of convolutions $(\mu_t\ast\nu_t)_{t\in I}$ is tight and bounded in total variation norm as well.  Now I would like to show the converse: If $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, then $(\nu_t)_{t\in I}$ is tight. (Most probably we need to impose a boundedness assumption on $(\mu_t)_{t\in I}$ and on $(\mu_t\ast\nu_t)_{t\in I}$ or even a priori on $(\nu_t)_{t\in I}$.)  For the moment, I'm only able to prove the claim assuming that the $\mu_t,\nu_t$ are probability measures. Under this assumption, the proof goes as follows: Let $\varepsilon&gt;0$. Since $(\mu_t)_{t\in I}$ and $(\mu_t\ast\nu_t)_{t\in I}$ are tight, there is a compact $K\subseteq E$ with $$\sup\left(\sup_{t\in I}\mu_t(K^c),\sup_{t\in I}(\mu_t\ast\nu_t)(K^c)\right)&lt;\varepsilon\tag1.$$ Now $K-K$ is compact and \begin{equation}\begin{split}1-\varepsilon&lt;(\mu_t\ast\nu_t)(K)&amp;=\underbrace{\int_K\mu_t({\rm d}x)\nu_t(\underbrace{K-x}_{\subseteq\:K-K})}_{\le\:\nu_t(K-K)}+\underbrace{\int_{K^c}\mu_t({\rm d}x)\underbrace{\nu_t(K-x)}_{\le\:1}}_{&lt;\:\varepsilon}\\&amp;&lt;\nu_t(K-K)+\varepsilon\end{split}\tag2\end{equation} for all $t\in I$, which immediately yields $$\nu_t(K-K)&gt;1-2\varepsilon\;\;\;\text{for all }t\in I\tag3.$$  Can we drop the probability measure assumption?  EDIT I'm not able to obtain the claim, even with the help of the answer below. If $\varepsilon&gt;0$, the goal is to show that $$\nu_t(E)-\nu_t(K-K)=\nu_t((K-K)^c)&lt;\varepsilon\tag4.$$ If I'm choosing all the constants as generic as possible, we obtain $$\mu_t(E)\le c_1\tag5$$ $$\mu_t(K^c)\le k_1(\varepsilon)\tag6$$ $$(\mu_t\ast\nu_t)(K)&lt;c_1\nu_t(K-K)+c_2k_1(\varepsilon)\tag6$$ $$\mu_t(E)\ge\alpha_1\tag7$$ $$(\mu_t\ast\nu_t)(E)\le c_3\tag8$$ $$\nu_t(E)\le\frac{c_3}{\alpha_1}=:c_2\tag9$$ $$(\mu_t\ast\nu_t)(E)\ge\alpha_3\tag{10}$$ $$\nu_t(E)\ge\frac{\alpha_3}{c_1}\tag{11}$$ $$(\mu_t\ast\nu_t)(K^c)&lt;k_3(\varepsilon)$$ The assumptions that I'm imposing is that $(\mu_t)$ and $(\mu_t\ast\nu_t)$ are bounded above (by $c_1$ and $c_3$) and below (by $\alpha_1$ and $\alpha_3$) in total variation norm and are tight (so that we can find $K$ for arbitrary $k_1(\varepsilon)$ and $k_3(\varepsilon)$). In order to obtain the claim we need to choose $k_1(\varepsilon)$ and $k_3(\varepsilon)$ such that $(4)$ holds. From everything above, we obtain $$\nu_t(E)-\nu_t(K-K)&lt;\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)\tag{12}$$ Even if we simplify this by assuming $\alpha_1=\alpha_3$, $c_1=c_3$ and $k_1(\varepsilon)=k_3(\varepsilon)$, the equation $$\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}+\frac{c_3}{\alpha_1c_1}k_1(\varepsilon)+\frac1{c_1}k_3.(\varepsilon)=\varepsilon\tag{13}$$ doesn't seem to admit a solution $(k_1(\varepsilon),k_3(\varepsilon))$ with positive $k_1(\varepsilon)$ and $k_3(\varepsilon)$; unless $\frac{c_3}{\alpha_1}-\frac{\alpha_3}{c_1}\ge0$; which clearly doesn't need to be the case (it is clearly the case for probability measures though ...)  $^1$ Remember that a family $\mathcal F$ of finite signed measures on $\mathcal B(E)$ is called tight if for all $\varepsilon&gt;0$, there is a compact $K\subseteq E$ such that $$\sup_{\mu\in\mathcal F}|\mu|(K^c)&lt;\varepsilon.$$ How long does it take for DNS records to propagate? Update:   Once upon a time there was a feature called . Its goal was to help our most active users find the questions they can answer. &quot;New nav&quot; never made it out of beta and that beta only ever made it to Stack Overflow users. Several months ago we gave new nav a sad, but overdue burial. As we promised, we've been working on a replacement for new nav, and now it's time for us to start rolling it out. We're calling it custom question lists and we are really excited about its potential to help all users get at the questions that are most interesting to them. While we hope what we've built hits the mark for many of you who have requested a feature like this (e.g., , , , , , ), we know how difficult it is to make something complex simple. So, we are rolling this out in beta to gather feedback from you and iterate on it. (Don't worry, we promise to actually release it this time.) When it's live later you'll be able to opt-in to the beta by turning on a new account preference. This will enable the feature for Stack Overflow, Meta Stack Overflow, and Meta Stack Exchange. It will be available for all network sites when the beta period is over. Schedule for future improvements This feature will roll out in four phases:  Phase I: Filter and sort controls to let you modify a question list view (Julyish) Phase II: Saved custom lists (In progress, ETA November 2018) Phase III: Notifications on custom lists (December 2018) Phase IV: Final fixes and official release (January 2019)  Sneak peak Phase I: Filter and sorts We'll add filter and sort controls to question lists. This does two things:  Clarifies the definition of built-in views - Naming is hard and short names can't always describe what a view is showing you. By exposing the filter and sort controls it is more obvious. Introduces the ability to create new custom views - This won't be super useful until we add the ability to save the views to the UI (phase II). Until then you can bookmark the URL and get back to your views.  Question list with new controls   Views drop down   Filter   Sort   Phase II: Saved custom question lists In this phase we will add the ability to save your custom questions lists. Create custom list In our current design, a dialog will be pre-populated based on the filters and sorts you're currently using. Give it a name, confirm your options, and click save.  Accessing your saved lists When you save your custom lists they will show up in the &quot;More&quot; drop down for question list views. Additionally, you can add them to the left navigation bar for easy access (see image below).  Phase III: Notifications For phase III we'll enable indicators for when new items are available in a custom question list. When you pin a question list to the left nav you can see a dot on any list with new items.  Feedback welcome As usual, we're interested in your constructive feedback. This is a work in progress and, as mentioned, will be rolled out in phases. You can provide feedback on the concepts now based on the information in this post and provide feedback based on the working elements as they are released. As a(n) noun vs as only