Yes, I know, in stars, fusion occurs up to Iron(-56); but, I want to know if fusion past this nucleus can happen at all.  If so, the daughter element would move to the right of the peak and its parents (away from the point of Iron-56), on the following graph (meaning additional energy is required, as opposed to energy being emitted), right?    Would fusion of heavy elements still be considered a reaction?  This is not a duplicate of . There are multiple questions, including the one about the graph, that are definitely not encountered in the other post. My main question is also different; the question and answers do not speak about fusion of heavy elements, but their rapid capture.  My question itself has nothing to do with stars, the first statements purpose was to dismiss comments telling me that stars only fuse up to Iron-56!
In all the discussions about how the heavy elements in the universe are forged in the guts of stars and especially during a star's death, I usually hear that once the star begins fusing lighter atoms to produce iron (Fe) that's the end of the star's life and the whole system collapses onto itself; and based on how massive the star was initially, it has different outcome - like a white dwarf, a neutron star or a black hole. I have rarely heard a detailed explanation of how the elements heavier than iron are produced. I would appreciate a convincing explanation of this process.
Extend x axis on graph? I'd like to extend my x axis a little longer. Is it possible to do that?  \centering     \begin{tikzpicture}[scale=1.5]      \begin{axis}[         xlabel=Reynolds Number,ylabel=Coefficient Discharge]     \addplot [color=blue,mark=*] coordinates {         (6485.41491,0.408143397)         (7927.942858,0.407370795)         (8663.115504,0.38550864)         (9219.462371,0.366953133)         (9537.374867,0.346531912)         (10491.11235,0.330115983)         (11087.19828,0.312041115)         (12219.76155,0.313951177)     };         \addlegendentry{1/4}     \addplot [color=red,mark=triangle] coordinates {         (6556.945221,   0.412644978)         (7868.334265,   0.404307857)         (8663.115504,   0.38550864)         (9187.671122,   0.365687778)         (9120.114717,   0.331371141)         (10443.42548,   0.328615455)         (11683.28421,   0.328817519)         (12875.45607,   0.330797337)     };         \addlegendentry{1/2 Lift}     \addplot [color=black,mark=*] coordinates {         (6843.066467,   0.430651305)         (8297.516134,   0.426361012)         (9203.566747,   0.40955872)         (8615.42863,    0.342911376)         (9298.940495,   0.337868614)         (10824.92047,   0.340619673)         (11683.28421,   0.328817519)         (12899.29951,   0.331409925)     };         \addlegendentry{3/4 Lift}     \addplot [color=yellow,mark=*] coordinates {         (6854.988186,   0.431401568)         (8398.850742,   0.431568008)         (9267.149246,   0.412388141)         (8567.741755,   0.341013343)         (9346.62737 ,   0.339601273)         (10753.39016,   0.338368882)         (11897.87515,   0.334857025)         (13030.43841,   0.334779157)     };         \addlegendentry{Full Lift}     \end{axis}% \end{tikzpicture}% C++ template typedef Showing that $ \sum_{n=1}^{\infty} \arctan \left( \frac{2}{n^2} \right) =\frac{3\pi}{4}$ I performed principal component analysis (PCA) with R using two different functions (prcomp and princomp) and observed that the PCA scores differed in sign. How can it be?  Consider this:  set.seed(999) prcomp(data.frame(1:10,rnorm(10)))$x              PC1        PC2  [1,] -4.508620 -0.2567655  [2,] -3.373772 -1.1369417  [3,] -2.679669  1.0903445  [4,] -1.615837  0.7108631  [5,] -0.548879  0.3093389  [6,]  0.481756  0.1639112  [7,]  1.656178 -0.9952875  [8,]  2.560345 -0.2490548  [9,]  3.508442  0.1874520 [10,]  4.520055  0.1761397  set.seed(999) princomp(data.frame(1:10,rnorm(10)))$scores          Comp.1     Comp.2  [1,]  4.508620  0.2567655  [2,]  3.373772  1.1369417  [3,]  2.679669 -1.0903445  [4,]  1.615837 -0.7108631  [5,]  0.548879 -0.3093389  [6,] -0.481756 -0.1639112  [7,] -1.656178  0.9952875  [8,] -2.560345  0.2490548  [9,] -3.508442 -0.1874520 [10,] -4.520055 -0.1761397   Why do the signs (+/-) differ for the two analyses? If I was then using principal components PC1 and PC2 as predictors in a regression, i.e. lm(y ~ PC1 + PC2), this would completely change my understanding of the effect of the two variables on y depending on which method I used! How could I then say that PC1 has e.g. a positive effect on y and PC2 has e.g. a negative effect on y?    In addition: If the sign of PCA components is meaningless, is this true for factor analysis (FA) as well? Is it acceptable to flip (reverse) the sign of individual PCA/FA component scores (or of loadings, as a column of loading matrix)? What approach should I take to establish this logical proof? I need to design a logical math proof:  Write a detailed structured proof to prove that if m and n are integers, then either 4 divides mn or else 4 does not divide n.   Hint: Think about the form of the statement.   I was thinking about first proving that either 4 should divide m or n in order for the first part to be true, then use proof by cases to prove what happened when m was 4, and another case when n was 4...  But I got stuck, what do you recommend:    This is what I got so far: Claim: (m ∈ Z ∧ n ∈ Z) => (4 | mn ˅ 4 ∤ n)  Negated claim: (m ∈ z ∧ n ∈ z) ∧ ( P(n) ∧ ¬P(mn) )  Proof by contradiction  Assume m , n ∈ z:    Assume P(n) ∧ ¬P(mn):       Then ∃ q ∈ R, n = 4q       Let q0 be such that n = 4q0       Then mn = m(4q0)       Then mn = 4(mq0)       Then ∃ q ∈ R, mn = 4q        Then 4 | mn     Then P(n) ∧ P(mn) Then (m , n ∈ z) ∧ P(n) ∧ P(mn))    I know that is wrong, how can I fix it? Why can array constants only be used in initializers? I am writing pseudo code in the algorithmic environment. It is quite long and doesn't fit on one single page. Instead of it continuing on the next page, it is cut off. I tried inserting it into a figure environment, but the problem persists. Is there a solution to have the code continue on the next page? How do I survive in the Nether? How to let PHP to create subdomain automatically for each user? How do I create subdomain like  ? Do i have to access htaccess somehow? Is it actually simply possible to create it via pure php code or I need to use some external script-server side language?  To those who answered: Well, then, should i ask my hosting if they provide some sort of DNS access?? Normalization of basis vectors with a continuous index? I have an infinite basis which associates with each point, $x$, on the $x$-axis, a basis vector $|x\rangle$ such that the matrix of $|x\rangle$ is full of zeroes and a one by the $x^{\mathrm{th}}$ element. The book on Quantum Mechanics by Shankar says that the inner product between a basis vector and itself is not one, why not? Why can't these basis vectors be normalized to one, only to the ? I realise it's unlikely that posting the file would cause a problem. However, since I know nothing about what a what's in a minidump file I thought I should check. When a ball moves to the right, friction acts to oppose the motion, in other words, to the left. However, when a car travels around a bend, the friction acts in the perpendicular direction to the car's velocity and provides the centripetal force. I just cannot understand why friction would act in that direction. I agree that every page in a site should have a unique TITLE, and that every page should have one and only one H1 element, but should the text of each H1 be completely unique too?  Instinctively one would think so, but there are potentially many cases where (IMHO) mind this doesn't make sense and attempting to do so will impair the user experience &amp; presentation of the page - for example, where pages are part of a closely related cluster, e.g. a multi-step process/tool/wizard.  Surely the process name should go in the H1 (e.g. "Pregnancy Planner"), and the current step you're on should be marked up as a H2 (e.g. "Step 1 - Trying for a baby"), rather than shoehorning both into a single H1 together (e.g. "Pregnancy Planner - Step 1 - Trying for a baby")?  Not only does this make the H1 rather long and potentially impact the presentation of the page, I would argue that in combining both into one you also lose semantic goodness, as the page is surely semantically richer by keeping parent (process) and child (step) as separate, nested headings?  Furthermore, the current proposal I have read is that on the first page the H1 should be the name of the process and the first step an H2, but on subsequent pages the same text should not be a H1 at all (just text styled exactly the same as an H1) and the current step should become the H1 instead. For me this is introducing unnecessary complexity to the page authoring/production process, and inconsistency in the semantic structure of the pages within the stepped process.  So it all comes back to the basic assertion that each H1 must be completely unique, which I question. Can you convince me that this should be so and that the cost/complexity of implementing this (especially in situations such as that above) is worth it?  Your thoughts please... Is there a way to see my previous performances with respect to winter-bash? May be something like leagues for winter-bash? I understand that hats are different each year, but at least it would be nice to see the number of hats collected. php SimpleXML check if a child exists I am running Ubuntu 18.04. I wanted to install node and npm on my machine. The apt repository version of node is 8.x, whereas I wanted to have version 12.x (Current LTS version).  I followed the guide here:   As per the guide, I downloaded curl -sL https://deb.nodesource.com/setup_12.x -o nodesource_setup.sh  and then ran it with sudo -E ./nodesource_setup.sh    However, my machine fails to run sudo apt update with the following (full) output:  maxim@fujitsu:~/scripts$ sudo apt update Hit:1 http://gb.archive.ubuntu.com/ubuntu bionic InRelease                  Hit:2 http://gb.archive.ubuntu.com/ubuntu bionic-updates InRelease          Get:3 http://security.ubuntu.com/ubuntu bionic-security InRelease [88.7 kB] Hit:4 http://gb.archive.ubuntu.com/ubuntu bionic-backports InRelease               E: Repository 'http://security.ubuntu.com/ubuntu bionic-security InRelease' changed its 'Suite' value from 'bionic-updates' to 'bionic-security' N: This must be accepted explicitly before updates for this repository can be applied. See apt-secure(8) manpage for details. Do you want to accept these changes and continue updating from this repository? [y/N] N Fetched 88.7 kB in 8s (10.5 kB/s)                                                       Reading package lists... Done E: Failed to fetch http://security.ubuntu.com/ubuntu/dists/bionic-security/InRelease   E: Some index files failed to download. They have been ignored, or old ones used instead.    suggests that one should use apt update instead of apt-get update. None the less, I am running apt update.   suggests to answer "Y" to the prompt.  , from the Docker community, makes it clear that doing so is a bad idea.  However, , suggests that sometimes repositories change the names of package sources.  So my question is, is it OK to answer Y here? I don't want to  just blindly accept whatever error the package manager throws at me. This problem has been seen on mid-2012 MacBook Pros on Mavericks and on late-2014 Retina MacBook Pros on Yosemite.  One day while using iMessage on the Mac, all the contacts turned red and messages failed to send, immediately returning a "not delivered" error. It also reported when trying to resend the messages that the contacts were no longer associated with an iMessage account.  Things that didn't fix this behavior:   Deleting the com.apple.ids.madrid Removing the Apple ID information from my phone, turning off iMessage on both of the devices, and placing my iPhone in airplane mode for several minutes before turning it on logging in again.   Could someone help me figure out how to solve this? Solving linear congruences by hand: modular fractions and inverses How to restore Windows 7 MBR without a CD I have been playing with Ubuntu for a few weeks now, and I'd like to revert my computer back to it's original - factory - defaults.   On the computer I have a recovery partition (it's a netbook). I went through the process of recovery and everything seemed fine. However, when I restart the computer I'm presented with   grub rescue &gt;   Now, my understanding is that when I installed Ubuntu "side by side" it replaced the MBR or something like it, with GRUB.   I've read on a slew of forums, that I need to use a Windows Recovery Disk.   Here are my issues:   a) I don't have a recovery disk, I have a recovery partition - it's a netbook. b) I don't have an external cd drive.   What I do have is a USB key that has about 1gb of space on it.   Thanks in advance. Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G has subgroup of order $\operatorname{lcm}(n,m)$. Let $G$ be abelian, $H$ and $K$ subgroups of orders $n$, $m$. Then G   has subgroup of order $\operatorname{lcm}(n,m)$.   This is a statement that my lecturer mentioned in my (beginners') Abstract Algebra class. I'm not sure I understand why it's true.  What I have so far: Use the abelian group structure theorem on $\langle H, K\rangle$ (finite group generated by $H$ and $K$). Then $\langle H,K\rangle=C_{a_1}C_{a_2}\dotsm$ and $n, m |\langle H,K\rangle$. Which means it's also true that $\operatorname{lcm}(n,m)|\langle H,K\rangle$. Can I leverage this to say there's a subgroup of order $\operatorname{lcm}(n,m)$?