Are atoms getting weaker?
Why does space expansion not expand matter?
How do you read CSS rule values with JavaScript? I would like to return a string with all of the contents of a CSS rule, like the format you'd see in an inline style. I'd like to be able to do this without knowing what is contained in a particular rule, so I can't just pull them out by style name (like .style.width etc.)   The CSS:  .test {     width:80px;     height:50px;     background-color:#808080; }   The code so far:  function getStyle(className) {     var classes = document.styleSheets[0].rules || document.styleSheets[0].cssRules     for(var x=0;x&lt;classes.length;x++) {         if(classes[x].selectorText==className) {             //this is where I can collect the style information, but how?         }     } } getStyle('.test') Assigning values to divergent series I have been looking at divergent series on wikipedia and other sources and it seems people give finite "values" to specific ones. I understand that these values sometimes reflect the algebraic properties of the series in question, but do not actually represent what the series converges to, which is infinity. Why is it usefull to assign values to divergent series?  The only theory I could come up with, is this:  Say you have 2 divergent series, series' A and B, and you assign each a value,   Series ($A= \sum_{n=0}^\infty a_n$), which I assigned the value Q  and series ($B= \sum_{n=0}^\infty b_n$ ), which I assigned the value P  But it just so happens that series $C=A-B= \sum_{n=0}^\infty (a_n-b_n)$ converges. Could that imply that the actual value of series $C$ is the difference of the two assigned values to $A$ and $B$, that is $\sum_{n=0}^\infty (a_n-b_n)=Q-P$ ?  If so, then that would make some sense to me, as to why people sometimes assign values to divergent series. I put 4 different layers into my ArcMap document, but only one of them will show up on the map! I should be seeing the other 3 layers as well. When I do a Zoom to layer, I can see one layer, but I cannot figure out how to make the other 3 show up at once on top of each other.  Before I opened ArcMap, I used ArcCatalog where I right-clicked 3/4 of the shapefiles and  clicked Properties &rarr; Coordinate system &rarr; Import and then clicked on my Trees layer, which has the coordinate system I want the other 3 to have. But when I opened ArcMap the 4 layers did not overlap, nor could I move them above or below each other. MySQL search in comma list Does the closedness of a subgroup $H$ of $G$ imply that for $h\in H, g\in G \setminus H : hg \not \in H$? Does the closedness of a subgroup $H$ of $G$ imply that for $h\in H, g\in G \setminus H: hg \not \in H$?  I'm new to abstract algebra and this would appear to be a useful result. However, by definition, a subgroup is nearly closed, and I don't think this directly implies that for any element $g$ from $G$ not in $H$, the composition $gh \not \in H$.  Is this in fact a true statement? I cannot easily figure out which one is more appropriate to use:     It's important to me.        It's important for me.   Are they the same? If not, what's the difference? Significance contradiction in linear regression: significant t-test for a coefficient vs non-significant overall F-statistic I'm fitting a multiple linear regression model between 4 categorical variables (with 4 levels each) and a numerical output. My dataset has 43 observations.  Regression gives me the following $p$-values from the $t$-test for every slope coefficient: $.15, .67, .27, .02$. Thus, the coefficient for the 4th predictor is significant at $\alpha = .05$ confidence level.  On the other hand, the regression gives me a $p$-value from an overall $F$-test of the null hypothesis that all my slope coefficients are equal to zero. For my dataset, this $p$-value is $.11$.  My question: how should I interpret these results? Which $p$-value should I use and why? Is the coefficient for the 4th variable significantly different from $0$ at the $\alpha = .05$ confidence level?  I've seen a related question, , but there was an opposite situation: high $t$-test $p$-values and low $F$-test $p$-value. Honestly, I don't quite understand why we would need an $F$-test in addition to a $t$-test to see if linear regression coefficients are significantly different from zero. Prove that if the identity is written as the product of $r$ transpositions, then $r$ is an even number Theorem. If the identity is written as the product of $r$ transpositions, $id=τ_1τ_2\dots τ_r$, then $r$ is an even number. Proof. We will employ induction on $r$. A transposition cannot be the identity; hence, $r &gt; 1$. If $r = 2$, then we are done. Suppose that $r &gt; 2$. In this case the product of the last two transpositions, $τ_{r−1}τ_r$, must be one of the following cases: $(ab)(ab) = id\\ (bc)(ab) = (ac)(bc)\\ (cd)(ab) = (ab)(cd)\\ (ac)(ab) = (ab)(bc),$ where $a$, $b$, $c$, and $d$ are distinct. The first equation simply says that a transposition is its own inverse. If this case occurs, delete $τ_{r−1} τ_r$ from the product to obtain $id=τ_1τ_2···τ_{r−3}τ_{r−2}$. By induction $r − 2$ is even; hence, $r$ must be even. In each of the other three cases, we can replace $τ_{r−1} τ_r$ with the right-hand side of the corresponding equation to obtain a new product of $r$ transpositions for the identity. In this new product the last occurrence of $a$ will be in the next-to-the-last transposition. We can continue this process with $τ_{r−2} τ_{r−1}$ to obtain either a product of $r − 2$ transpositions or a new product of $r$ transpositions where the last occurrence of $a$ is in $τ_{r−2}$. If the identity is the product of $r − 2$ transpositions, then again we are done, by our induction hypothesis; otherwise, we will repeat the procedure with $τ_{r−3}τ_{r−2}$. At some point either we will have two adjacent, identical transpositions canceling each other out or $a$ will be shuffled so that it will appear only in the first transposition. However, the latter case cannot occur, because the identity would not fix $a$ in this instance. Therefore, the identity permutation must be the product of $r − 2$ transpositions and, again by our induction hypothesis, we are done. Question: That is the proof written in the book, but unfortunately after more than $10$ times reading I don't understand the last two paragraph (i.e., from &quot;In each of the other three cases&quot;). Is there another simpler proof or could someone please clarify what is going on? Thank you. Can one do the maths of physics without using $\sqrt{-1}$? The use of imaginary and complex values comes up in many physics and engineering derivations. I have a question about that: Is the use of complex numbers simply to make the process of derivation easier, or is it an essential ingredient, without which it would be impossible to derive some results?   I can identify two different settings where the answer may be different:   It doesn't look like it is mandatory for results in Newtonian mechanics, General Relativity and classical electrodynamics. Can we say the same thing about quantum mechanics either way for sure?   Could this be a difference in quantum mechanics over the classical picture? I am thinking of buying a camera to learn photography. Most of the time I'll be shooting portraits and landscapes. But I am not sure where I should start. Most of my friends tell me that I should get an entry level DSLR, but I am not sure if I have the persistence and enthusiasm to carry it all day in a backpack. But the camera on my cellphone doesn't seem to be enough to begin with.  So what I want is some expert advice (preferably with professional experience). Can most compact fixed-lens or mirrorless cameras fit the situations I mentioned? Or do I really need to go to a DSLR to get a tool to learn photography? Is there a way to display the number of messages for any label in the Gmail label list? The Gmail sidebar displays the list of our labels (or some of them, according to settings). Next to every label name there is a number showing how many unread messages we have for that label.  Is there a way to display also the total number of messages for each label?  Example:  [to answer] (0/36) [to do] (0/55) [waiting answer] (0/7) cooking (2/48) family (6/352) friends (1/1828) website (2/412) work (12/12801) How long was Luke Skywalker's training with Yoda? From the answer to the question  :     We also don't know how much time Luke spent training.   I tried to check some references like  and  related section on their Wookieepedia articles, but there is not even a clue of the amount of time he spent there.  Is there any canon information on the subject? Or any extrapolation of how long Luke's journey on Dagobah was? Was it a day? Several weeks? Some months? The problem is stated as follows:  "Let $R$ be a Noetherian ring and $\theta$ be a ring homomorphism from $R$ to $R$. Show that if $\theta$ is surjective then it is also injective."  Regardless of the right solution, I don't understand why is the following wrong:     We have $\theta: R\to R$. By the isomorphism theorem $R/\ker\theta\cong\operatorname{Im}\theta$. Since $\operatorname{Im}\theta = R$, it follows $\ker\theta =0$, so it's injective.   Harsh criticism will be appreciated. Thanks. How do I find a recursion formula to solve the following question:  In how many sequences of length $n$, with elements from $\{{0,1,2,3,4}\}$, the difference between every 2 adjacent elements is $1$ or $-1$?  I've been trying to solve the problem for the last 5 hours maybe, I'm sure that I'm missing something..   This is a problem in combinatorics, and the solution must be by finding a recursion formula for the problem.   best solution I've come so far is that if we say that $a_n$ is the solution to the question, so let's count how many sequences of size $n-1$ there are that meet the requirement, and we complete the first element to make it a sequence of $n$ elements. so for $n-1$ there $a_{n-1}$ sequences, and for each sequence we have $2$ options to choose the first element (one option is to take a plus $1$ number from the number that start the $n-1$ sequence, and the other option to to take a minus $1$ number from the number that start the $n-1$ sequence), only problem as you figured out is that if the $n-1$ sequnce starts with $0$ or $4$ there is only $1$ option left for the first element in the sequnce.. so $a_n=2a_{n-1}$ won't work.  I think that the solution may involve using $2$ or maybe $3$ recursion formulas.  Hope I was clear enough. Find the ratio of $\frac{\int_{0}^{1} \left(1-x^{50}\right)^{100} dx}{\int_{0}^{1} \left(1-x^{50}\right)^{101} dx}$ $$I_1=\int_{0}^{1} \left(1-x^{50}\right)^{100} dx$$ and $$I_2=\int_{0}^{1} \left(1-x^{50}\right)^{101} dx$$ Then find $\frac{I_1}{I_2}$  I tried by subtracting $I_1$ and $I_2$   $$I_1-I_2=\int_{0}^{1}\left(1-x^{50}\right)^{100}\left(1-(1-x^{50}\right))dx$$ so  $$I_1-I_2=\int_{0}^{1} \left(1-x^{50}\right)^{100} x^{50} dx$$ Now using Integration by Parts we get  $$I_1-I_2= \left(1-x^{50}\right)^{100} \times \frac{x^{51}}{51}\bigg|_{0}^{1} -\int_{0}^{1} 100 \left(1-x^{50}\right)^{99} \times -50 x^{49} \times \frac{x^{51}}{51} dx$$  So $$I_1-I_2=\frac{5050}{51} \times \int_{0}^{1}\left(1-x^{50}\right)^{99} x^{100} dx$$  Now $x^{100}=\left(1-x^{50}\right)^2-(1-x^{50}-x^{50})$ so  $$\frac{51}{5050}(I_1-I_2)=\int_{0}^{1} \left(1-x^{50}\right)^{99} \times \left(\left(1-x^{50}\right)^2-(1-x^{50})+x^{50}\right)dx=I_2-I_1+\int_{0}^{1} \left(1-x^{50}\right)^{99} x^{50} dx$$  Need a hint to proceed further. Given that $xyz=1$, prove that $\frac{x}{1+x^4}+\frac{y}{1+y^4}+\frac{z}{1+z^4}\le \frac{3}{2}$ "Knowledge of" vs. "knowledge on" How is Anti Aliasing Implemented in Ray Tracing? After reading a few articles online I can confidently say I am clueless on how Anti-Aliasing works when using Ray Tracing.  All I understand is that A Single Pixel/Ray is split into 4 sub-pixels and 4 rays rather than 1.  Could somebody please explain how this is done (preferably with code)? Can we make a Vpn server run on Android. Can applications do this? If yes, which protocols are supported and is this a good idea to use such apps? If not, what are some issues that we may face? Find the center of the symmetry group $S_n$. Find the center of the symmetry group $S_n$.   Attempt:  By definition, the center is $Z(S_n) = \{ a \in S_n : ag = ga \forall\  g \in S_n\}$. Then we know the identity $e$ is in $S_n$ since there is always the trivial permutation. Suppose $a$ is in $S_n$, but not equal to identity. Now we can imagine the permutation as bijective function that maps from $\{1,2,\dotsc,n\}$ to $\{1,2,\dotsc,n\}$.  So suppose $p$ is a permutation map. Then $p$ maps from a location $i$ to a location $j$. Take $p(i) = j$ where $i\neq j$. Let $k$ be in $\{1,2,\dotsc,n\}$, where $k$, $i$ and $j$ are all different elements. The cycle $r = (jk)$, then we will see if this commutes.  $rp(i) = rj$    Can someone please help me, I am stuck?  Thank you.