I've always seen spacetime shown with a ball on top of some cloth that curves around it, and I don't really understand it, since it really only has 2 dimensions and there's some sort of external gravity causing gravity? I was wondering what it would look like if that ball had more balls inside of it (like Earth contains humans, gases, etc.). Is there a way to visualize that? Does the fabric of spacetime go through everything or does it wrap around individual elementary particles? Is it even actually a fabric?  Apologies if this question is too broad.
I've seen many science popularisation documentaries and read few books (obviously not being scientist myself). I am able to process and understand basic ideas behind most of these. However for general relativity there is this one illustration, which is being used over and over (image from Wikipedia):    I always thought that general relativity gives another way how you can describe gravity. However for this illustration to work, there needs to be another force, pulling the object down (referring to a direction in the attached image). If I put two non-moving objects in the image, what force will pull them together?  So where is my understanding incorrect? Or is general relativity not about explaining gravity and just describes how heavy objects bends spacetime (in that case the analogy is being used not correctly in my opinion)?    UPDATE Thank you for the answers and comments. Namely the XKCD comics is a spot on. I understand that the analogy with bent sheet of fabric pretty bad, but it seems that it can be fixed if you don't bent the fabric, but just distort the drawn grid.   Would you be so kind and answer the second part of the question as well - whether general relativity is explaining gravitational force. To me it seems that it is not (bending of spacetime simply can not affect two non-moving objects). However most of the time it is being presented that it does.
How was the Cloak of Invisibility Created? Pattern in Isomorphisms between $\mathbb Z_m \times \mathbb Z_n$ and $\mathbb Z_{mn}$ Different results for integration using Mathematica and MATLAB Sum of indepedent random variables and a constant I am writing an article in English, with some pages in Russian.  \documentclass[12pt]{article}  \usepackage[T1,T2A]{fontenc}  \usepackage[utf8]{inputenc}  \usepackage[russian,english]{babel}  \usepackage{newtxmath} \usepackage{newtxtext} \begin{document}  hello \textbf{hello}  \selectlanguage{russian} привет \textbf{привет}  \end{document}   Output:  hello hello привет привет  I am using \selectlanguage{} to switch between languages when needed.  However, I noticed that when I try to use \textbf{} when in Russian language mode, it does not work because of newtxtext, which I am using for math formulas.  Does anyone know how to solve this problem?  Thank you. Prove that $\det(M-I)=0$ if $\det(M)=1$ and $MM^T=I$ $M$ is a $3 \times 3$ matrix such that $\det(M)=1$ and $MM^T=I$, where $I$ is the identity matrix. Prove that $\det(M-I)=0$  I tried to take $M$ $=$ $$ \begin{pmatrix}         a  &amp;b &amp; c \\         d &amp; e &amp; f \\         g &amp; h &amp; i         \end{pmatrix} $$   but its expansion is too long. Will ESM also become available for Ubuntu 14.04? ESM is available for 12.04, which enables Advantage customers to keep receiving kernel updates and updates for many user space packages. In our case it's very important to know if we'll be able tot use Ubuntu 14.04 safely after April '19, because it is not possible to upgrade our application and migrate it to say, 16.04. Motivating the ansatz for the infinitesimal translation operator Finitely generated monoids- axiomatization- compactness theorem. Restore Missing "Settings" (gnome-control-center) Icon in Ubuntu 18.04 I have been running Ubuntu 18.04 for a couple of weeks. I upgraded from 16.04 and all went smoothly.   I am not sure what I changed, but this morning I noticed that the "Settings" Icon that launches gnome-control-center is missing from "Show Applications" and the applications menu as well. When I searched from "Show Applications" for gnome-control-center or just settings, nothing showed up.   So, I figured I must have accidentally removed it and figured a reinstall of the package would resolve the issue. I have searched and found several postings where this worked. I tried:  sudo apt install gnome-control-center   And then rebooted - which reinstalled the gnome-control-center and I could then launch it from the command line. However, there was still no Settings Icon in any of the menus.   I also tried many other suggestions including:  sudo apt-get install --reinstall gnome-control-center sudo apt-get install ubuntu-desktop sudo apt install ubuntu-desktop --reinstall sudo apt-get install unity-control-center   However, none of these has been successful in restoring the icon into "Show Applications" or any applications menu.   Anyone have any suggestions on something I haven't tried? Would be much appreciated.  Or, is there a way to just manually recreate the icon?   Output of locate gnome-control-center.desktop:  /usr/share/applications/gnome-control-center.desktop  /home/dccrens/.local/share/applications/gnome-control-center.desktop /home/dccrens/.gnome/apps/gnome-control-center.desktop   I am running Ubuntu 18.04 desktop on kernel 4.15.0-45-generic. How do I trim leading/trailing whitespace in a standard way? Is there a clean, preferably standard method of trimming leading and trailing whitespace from a string in C? I'd roll my own, but I would think this is a common problem with an equally common solution. Why can't a single ReLU learn a ReLU? As a follow-up to  I simplified even more and tried to train a single ReLU (with random weight) to a single ReLU.  This is the simplest network there is, and yet half the time it fails to converge.  If the initial guess is in the same orientation as the target, it learns quickly and converges to the correct weight of 1:      If the initial guess is "backwards", it gets stuck at a weight of zero and never goes through it to the region of lower loss:        I don't understand why.  Shouldn't gradient descent easily follow the loss curve to the global minima?  Example code:  from tensorflow.keras.models import Sequential from tensorflow.keras.layers import Dense, ReLU from tensorflow import keras import numpy as np import matplotlib.pyplot as plt  batch = 1000   def tests():     while True:         test = np.random.randn(batch)          # Generate ReLU test case         X = test         Y = test.copy()         Y[Y &lt; 0] = 0          yield X, Y   model = Sequential([Dense(1, input_dim=1, activation=None, use_bias=False)]) model.add(ReLU()) model.set_weights([[[-10]]])  model.compile(loss='mean_squared_error', optimizer='sgd')   class LossHistory(keras.callbacks.Callback):     def on_train_begin(self, logs={}):         self.losses = []         self.weights = []         self.n = 0         self.n += 1      def on_epoch_end(self, batch, logs={}):         self.losses.append(logs.get('loss'))         w = model.get_weights()         self.weights.append([x.flatten()[0] for x in w])         self.n += 1   history = LossHistory()  model.fit_generator(tests(), steps_per_epoch=100, epochs=20,                     callbacks=[history])  fig, (ax1, ax2) = plt.subplots(2, 1, True, num='Learning')  ax1.set_title('ReLU learning ReLU') ax1.semilogy(history.losses) ax1.set_ylabel('Loss') ax1.grid(True, which="both") ax1.margins(0, 0.05)  ax2.plot(history.weights) ax2.set_ylabel('Weight') ax2.set_xlabel('Epoch') ax2.grid(True, which="both") ax2.margins(0, 0.05)  plt.tight_layout() plt.show()     Similar things happen if I add bias: 2D loss function is smooth and simple, but if the relu starts upside down, it circles around and gets stuck (red starting points), and doesn't follow the gradient down to the minimum (like it does for blue starting points):    Similar things happen if I add output weight and bias, too.  (It will flip left-to-right, or down-to-up, but not both.) EXTRA PROBLEM I have NEI on minecraft with some other mods and NEI is supposed to have a saves thing on the left, and items on the right. The only thing on the left is the options. I want to see the saves and time changing options too.  Cheat Mode When I click on a item to get it it keeps showing me the recipe. Though in the options menu from NEI it is on cheatmode. I saw another post saying I need to change something to this lockmode=-1 or cheatmode=2 but it is already that way. That didn't work how do i get it to give me the items and also recipe when I press R. I'm working in Lee's book on topological manifolds and have gotten stumped on the first question in chapter 5, the chapter on cell complexes.  The problem is:   Let $D$ and $D'$ be two closed cells not necessarily of the same dimension.     Show that any continuous map $f:\partial D \to \partial D'$ can be extended to a continuous map $F:D \to D'$ such that $F(Int \; D) \subseteq Int \; D'$. Given points $p \in Int \; D$ and $p' \in Int \; D'$, show that $F$ can be chosen so $F(p) = p'$. Show that if $f$ is a homeomorphism then $F$ can be chosen to be a homeomorphism.   I proved the first part roughly as follows: First suppose $D$ and $D'$ are convex and each contain $0$ in their interior (in their respective ambient spaces).  Then every element other than $0$ in $D$ can be expressed uniquely in the form $\lambda q$ where $q \in \partial D$ and $\lambda\in (0,1]$ (an equivalence relation on the chords connecting $0$ to boundary points partitions $D- \{0\}$).  I then define the map $F(\lambda q) = \lambda f(q)$ which is continuous since if $\lambda_n q_n \to \lambda q$ in $D$ then $F(\lambda_n q_n) \to F(\lambda q)$.  Lastly $\lambda q \in Int \; D$ implies $\lambda &lt;1$ and so $\lambda f(q)$ is an interior point of $D'$ since $f(q)$ is a boundary point and $D'$ is convex.    Now if we suppose $D$ and $D'$ are arbitrary closed cells there are homeomorphisms $g_1: \overline{\mathbb{B}^n} \to D$ and $g_2: \overline{\mathbb{B}^m} \to D'$ (where possibly $m=n$) then we have that $g_2^{-1}\circ f \circ g_1$ is a continuous map between the boundaries of two closed balls and so by the first part can be extended to a continuous map $F:\overline{\mathbb{B}^n} \to \overline{\mathbb{B}^m}$.  The mapping $g_2 \circ F \circ g_1^{-1}$ is a continuous map that satisfies the desired claim.    My issue is that parts 2 and 3 don't seem (to me at least) to follow that easily from the proof I constructed.  Maybe my proof is wrong and I'm not seeing it, but more likely I think that I'm missing the spirit in which Lee is intending for us to approach this problem. JUnit Testing Exceptions Simplifying adjacent polygons using QGIS Simplify Geometries tool found a 1994 uncashed profit sharing retirement plan check I found an old un-cashed check dated 1994 for $487 from my Xerox Retirement plan with no expiration date. Can I still cash it? limit of $\lim\limits_{x \to \infty}((x+1)^a - x^a)$ Hello everybody, i couldn't figure out how to demonstrate this equation without L'Hopital i wonder if its possible.   $a \in (0,1)$ $\lim\limits_{x \to \infty}((x+1)^a - x^a) = 0$ Over the past few weeks, there have been a number of complaints about users making suggested edits en masse whose quality was at least questionable.  Let an example of a low-quality edit be the addition of the web-frameworks tag to dozens of questions about the PHP web framework Laravel. No one is ever going to use it in a search, but Laravel is a framework, right?  (If you disagree about this being a low-quality edit, please  bear with me anyway - my point here is that the system through which such conflicts could be settled is broken.)   Examples          This is a toxic situation guaranteed to create strife because -   The new user is happy to have found a way to participate and earn some rep; they may be genuinely feeling that they are improving the site. The reviewers greenlighting most of their suggestions confirms them in their belief.   The reviewers usually see only one edit - and it's easy to see how they might go, "meh, this isn't a terribly useful edit, but hey, maybe it helps someone, and it's a new user. No point discouraging them."   The veteran users who check out the editor see dozens of what they perceive as low-quality edits, usually ignoring anything else that should be fixed in the post. To the veterans, the new user is systematically looking for excuses to make their pointless edits, creating a giant mess in order to  gain some rep - a mess they have to clean up, while there are no consequences for the editor, who gets to keep their ill-gotten points. While a bad question or answer can be downvoted - usually leading to a change of behaviour eventually - the edit system looks like it's wide open for gaming.   So tensions are high, and when there's dialog it can be angry. I'm not saying either side is completely right. I'm completely prepared to believe that many of these edits are in good faith, etc. But it's a mess because it's frustrating people. As said, there are no consequences for bad-faith mass editing; apart from commenting on some random post, there is no way to communicate with an editor. The outcome of a moderator flag is very much uncertain.  The absence of a clear set of rules means there's plenty of room for heated arguments.    Something needs to change here.  Either   We need super clear rules on what makes a good tag edit Tag-only edits should be allowed only for 2k+ users, or they should not gain 2 reputation points for &lt;2k users The possible 1000 rep gain for suggested edits needs to be rate-limited (say, 200 per month) (this is my favourite!) The possible 1000 rep gain for suggested edits needs to be lowered, say to 500 or 250 Rollbacks of suggested edits should result in annulment of the 2 points gained (this is terrible) There should be a system to vote on edits (yes, that's been suggested before) There should be a defined and officially sanctioned pathway, and ruleset, for moderator flags cast about these users    ... or something else. But something needs to give. Let $\varphi:G\rightarrow H$ be a group homomorphism from group $G$ to group $H$. Show that, if $H$ is abelian, all subgroups of $G$ that contain $\mathrm{ker} (\varphi)$ are normal in $G$.