Is this really a paradox? God, at the point in time when the question is posed, is Omnipotent - so he must be able to. But in creating something that He cannot later move, does this power simply come at the cost of his own Omnipotence?   Does denying God the power to die also deny his Omnipotence?
The easiest way to explain this question is with a thought experiment:  Consider God, the ultimate of everything, who is wholly omnipotent (all-powerful) and omniscient (all-knowing). Let's just say, hypothetically, that he wanted  (this is a totally random example I just happened to be reading about; I could have just as easily said He wanted Bob to score a free-throw in basketball 3 times in a row tomorrow, or Sally to knit a blue sweater on Sunday, November 4th, 2012).  But anyways, one way He could do this (get X group of people to save a man from a burning car) is: At the beginning of time, God could instantly plan out the future and set in motion a massive chain of events, from the creation of light and stars, the creation of Earth and the creatures, mankind, then civilization with all the conflicts and progress and glory that comes with it, such that one day, a few people would come from out of nowhere—complete strangers to each other—and save this man (also a stranger) from under a burning car. God need not intervene at all this whole time, He merely "ignited" the chain of events from day 1 and watched it unfold exactly as He planned. Let's call this Universe A.  He could also, being all-powerful, just freeze time for a bit and skip a few steps, creating the world and universe instantly, with all the civilizations and animals and things in place, putting all the strangers there together next to a car which is already on fire and the man trapped under it. Then press play, and presto, you have the exact same scenario, without billions of years of buildup. Let's refer to this as Universe B.  God is obviously capable of doing both. My question thus is whether there is an  (really any difference at all) between Universe A and Universe B. Personally, I don't think this kind of power is outside of God's capability; he should, after all, have no problem with creating identical universes which match up exactly once they reach "biker under the car" moment where He unfroze time in Universe B. At that moment, Universe A and Universe B are exactly the same, right? Or is there somethinge special about Universe A, with all it's buildup, that separates it intrinsically from Universe B? In other words, is it outside of God's ability to make them exactly equivalent?
How to disable autorefresh in snap Is it possible to disable autorefresh in snap? This  says that Snaps are updated automatically in the background to the latest version, every day. Some snaps (e.g. vlc) are built daily, and it makes no sense to install them every day. Meaning of $\int\mathop{}\!\mathrm{d}^4x$ What the following formula mean?  $$\int\mathop{}\!\mathrm{d}^4x$$  I know that this $\int f(x)\mathop{}\!\mathrm{d}x$ is the integral of the function $f$ over the $x$ variable, but the following $\int\mathop{}\!\mathrm{d}^4x$ leave empty the argument and also have the $\mathrm{d}$ elevated to the fourth power. What that mean?  Update.  In the Einstein-Hilbert action we have (note that I have understand that the other parts of the integral are relevant only after your answers, thus sorry.):  $$S=\frac{c^4}{16\pi G}\int\mathop{}\!\mathrm{d}^4x \, \ R \sqrt{-g}$$ What is wrong with this fake proof $e^i = 1$? Using sObject.get() to fetch a field value from a related parent object I have a contact in a generic sObject in apex and while I'm able to easily get detail field values using   sObject.get('MyContactField__c')  I'm not able to get fields off the related parent object, and I'm sure the related field is in memory, based on the debug logs. This returns a runtime error:  sObject.get('Account.MyAccountField__c')  with the error     Invalid field Account.MyAccountField__c' for Contact   How do I get a field value off of a parent on a generic sObject? What is the difference between old style and new style classes in Python?  When should I use one or the other? Are functions of independent variables also independent? The email for getting notifications can be changed by clicking the "preferences" tab in your profile. After entering a new email address, the "Save email settings" button becomes clickable:    Clicking the button turns it grey:    So all looks well. But if you don't confirm the email by clicking on the link sent to your email inbox, the next time you go back to the preferences, you will see your old email address with no indication that it changed.  You can subscribe to the weekly newsletter from the "edit" tab of your profile. If you click the option and save, the next time you visit the edit page before confirming your email you get this message:    Once you confirm the email, the resend link disappears so you know the subscription is set up:    Ideally, the two systems would be unified so that all email subscriptions happen the same way. But short of that, it would be very helpful to indicate that an email address is unconfirmed for notification emails just like newsletters. Determinant of a rank $1$ update of a scalar matrix, or characteristic polynomial of a rank $1$ matrix Alpha and Omega limit sets (dynamical systems) What are all the possible $α$- and $ω$-limit sets of points for the linear system of equations $x'=Ax$ given by the following matrices $A$:  I guess I don't really know how to get started, Could I have some guidance? So I was told to calculate the eigenvalues (fine) and eigenvectors (I kind of forgot how to do that...)  I know that the $ω$-limit sets of points are the set of points that the system of equation approach as time goes to infinity, and the $α$-limit sets of points are the points approached as t goes to negative infinity.   $    A=   \left[ {\begin{array}{cc}    -4 &amp; -2 \\    3 &amp; 11 \\   \end{array} } \right] $   $λ_1=-10$ $λ_2=-5$  Since both eigenvalue are negative, this is a stable node, and the $ω$-limit set=0.  for $λ_1=-10$:  eigenvector $    =   \left[ {\begin{array}{cc}   1 \\    3 \\   \end{array} } \right] $  for $λ_2=-5$:  eigenvector $    =   \left[ {\begin{array}{cc}   2 \\    1 \\   \end{array} } \right] $   $    A=   \left[ {\begin{array}{cc}    1 &amp; 1 \\    3 &amp; 1 \\   \end{array} } \right] $   $λ_1=4$ $λ_2=-2$  Since one eigenvalue is positive and one is negative, it is not stable, so I can't use the same reasoning that I used above to find the $ω$-limit set... so I am not sure how to continue  for $λ_1=4$:  eigenvector $    =   \left[ {\begin{array}{cc}   1 \\    1 \\   \end{array} } \right] $  for $λ_2=-2$:  eigenvector $    =   \left[ {\begin{array}{cc}   -1 \\    1 \\   \end{array} } \right] $   $    A=   \left[ {\begin{array}{cc}    4 &amp; 5 \\    -5 &amp; -2 \\   \end{array} } \right] $   $λ_1=1+4i$ $λ_2=1-4i$  for $λ_1=1+4i$:  eigenvector $    =   \left[ {\begin{array}{cc}   -3/5-(4i)/5 \\    1 \\   \end{array} } \right] $  for $λ_1=1-4i$:  eigenvector $    =   \left[ {\begin{array}{cc}   -3/5+(4i)/5 \\    1 \\   \end{array} } \right]$ How do I check the series $\sum_{n=2}^{\infty} \frac{(-1)^n }{n+(-1)^n}$ for absolute convergence/conditional convergence ?  How can I solve this problem? $(a,b) \mathbin\# (c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$. Multiplicative inverse? Let $r\in \mathbb{R}$ and let $0\neq s \in \mathbb{R}$. Define operations $\#$ and $\&amp;$ on $\mathbb{R}$ x $\mathbb{R}$ by $(a,b) \mathbin\#(c,d)=(a+c,b+d)$ and $(a,b) \mathbin\&amp;(c,d)=(ac-bd(r^2+s^2), ad+bc+2rbd)$ as addition and multiplication, respectively. Do these operations form a field?  I have shown that every axiom of the field is satisfied except for the "existence of multiplicative inverses" axiom. This is where I need help. We either need to show that this axiom fails and so we don't have a field or that the axiom $aa^{-1}=1$ is satisfied for every $a\neq 0 \in \mathbb{R}$  The multiplicative identity in this problem is $(1,0)$.  Any solutions or hints are greatly appreciated. Google Chrome PPA upgrade invalid signature I tried to export my Geogebra graphs into tikz-code. In General it works fine, but plot x^1/3 doesn't work:  \documentclass[10pt]{article} \usepackage{pgf,tikz} \usetikzlibrary{arrows} \pagestyle{empty} \begin{document} \definecolor{ccqqqq}{rgb}{0.8,0,0} \definecolor{qqttcc}{rgb}{0,0.2,0.8} \definecolor{cqcqcq}{rgb}{0.75,0.75,0.75} \begin{tikzpicture}[line cap=round,line join=round,&gt;=triangle 45,x=1.0cm,y=1.0cm] \draw [color=cqcqcq,dash pattern=on 1pt off 1pt, xstep=1.0cm,ystep=1.0cm] (-4,-4) grid (4,4); \draw[-&gt;,color=black] (-4,0) -- (4,0); \foreach \x in {-4,-3,-2,-1,1,2,3} \draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize $\x$}; \draw[-&gt;,color=black] (0,-4) -- (0,4); \foreach \y in {-4,-3,-2,-1,1,2,3} \draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize $\y$}; \draw[color=black] (0pt,-10pt) node[right] {\footnotesize $0$}; \clip(-4,-4) rectangle (4,4); \draw[color=qqttcc, smooth,samples=100,domain=-4.0:4.0] plot(\x,{(\x)*(\x)*(\x)}); \draw[color=ccqqqq, smooth,samples=100,domain=-4.0:4.0] plot(\x,{((\x))^(1/(3))}); \draw [color=qqttcc](2.06,4.18) node[anchor=north west] {$f(x)=x^3$}; \draw [color=ccqqqq](2.06,1.18) node[anchor=north west] {$f(x)=\sqrt[3]{x}$}; \begin{scriptsize} \draw[color=qqttcc] (-1.74,-7.16) node {$f$}; \end{scriptsize} \end{tikzpicture} \end{document}      Error line 21: !Missing number, treated as zero.   So plot has problems with the negative x I suppose, but I don't know. Any help would be great! I googled it for nearly 1 hour - but no real answers.  As suggestet it rotated and mirrored the graph - works great!:  \documentclass[10pt]{article} \usepackage[utf8]{inputenc} \usepackage{pgf,tikz} \usetikzlibrary{arrows} \pagestyle{empty} \begin{document} \definecolor{ccqqqq}{rgb}{0.8,0,0} \definecolor{qqttcc}{rgb}{0,0.2,0.8} \definecolor{cqcqcq}{rgb}{0.75,0.75,0.75} \begin{tikzpicture}[line cap=round,line join=round,&gt;=triangle 45,x=1.0cm,y=1.0cm] \draw [color=cqcqcq,dash pattern=on 1pt off 1pt, xstep=1.0cm,ystep=1.0cm] (-4,-4) grid (4,4); \draw[-&gt;,color=black] (-4,0) -- (4,0); \foreach \x in {-4,-3,-2,-1,1,2,3} \draw[shift={(\x,0)},color=black] (0pt,2pt) -- (0pt,-2pt) node[below] {\footnotesize \x}; \draw[-&gt;,color=black] (0,-4) -- (0,4); \foreach \y in {-4,-3,-2,-1,1,2,3} \draw[shift={(0,\y)},color=black] (2pt,0pt) -- (-2pt,0pt) node[left] {\footnotesize \y}; \draw[color=black] (0pt,-10pt) node[right] {\footnotesize 0}; \clip(-4,-4) rectangle (4,4); \draw[color=qqttcc, smooth,samples=100,domain=-4.0:4.0] plot(\x,{(\x)*(\x)*(\x)}); \begin{scope}[yscale=-1,xscale=1] \draw[rotate=90, color=ccqqqq, smooth,samples=100,domain=-4.0:4.0] plot(\x,{(\x)*(\x)*(\x)}); \end{scope} \draw [color=qqttcc](2.06,4.00) node[anchor=north west] {$f(x)=x^3$}; \draw [color=ccqqqq](2.06,1.00) node[anchor=north west] {$f(x)=\sqrt[3]{x}$}; \begin{scriptsize} \draw[color=qqttcc] (-1.74,-7.16) node {f}; \end{scriptsize} \end{tikzpicture} \end{document} Calibre No Longer Working Calibre stopped working recently. Tried uninstalling and reinstalling. When I open a terminal and call $calibre I get the following error message: Traceback (most recent call last):   File &quot;/usr/bin/calibre&quot;, line 20, in &lt;module&gt;     sys.exit(calibre())   File &quot;/usr/lib/calibre/calibre/gui_launch.py&quot;, line 73, in calibre     main(args)   File &quot;/usr/lib/calibre/calibre/gui2/main.py&quot;, line 543, in main     listener = create_listener()   File &quot;/usr/lib/calibre/calibre/gui2/main.py&quot;, line 514, in create_listener     return Listener(address=gui_socket_address())   File &quot;/usr/lib/calibre/calibre/utils/ipc/server.py&quot;, line 110, in __init__     self._listener._unlink.cancel() AttributeError: 'NoneType' object has no attribute 'cancel'  Do I need to install some dependency that the sudo apt install didn't get? I tried installing via command line and the software installer/updater GUI. I'm on version Ubuntu 20.04.1 LTS. I made sure that my software in general is up to date also. Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots. Prove $f=1+x+x^2+x^3+\cdots+x^n$ has no multiple roots.  My attempt:  Consider the polynomial $g=(x-1)(1+x+x^2+x^3+\cdots+x^n)$  As $f\mid g, g$ all the roots of $f$ are roots of $g$.  This means I have to prove the statement:  If $c$ is a multiple root of $f \Longrightarrow c$ is multiple a root of $g$.  Equivalently:  If $c$ is not a multiple root of $g \Longrightarrow c$ is not a multiple root of $f$.  I'll try to prove this last statement:  Suppose $\exists c$ such that $c$ is a multiple root of $g$, then   $$g(c)=0 \implies g'(c)=0$$  $$ g=(x-1)(1+x+\cdots+x^n) \implies  g'=nx^n $$ $$ g'(c)=nc^n=0 \implies c=0 $$ But  $$ g(0)\neq 0 $$ And we've found the contradiction: $\not \exists c: g(c)=g'(c)=0$. Then $f$ has no multiple roots.  Is this correct?   Also: Assuming this is correct, is it possible to prove $\sum_{k=0}^n \frac {x^n}{n!} $ has no multiple roots in a similar manner? Adding new group layer with ArcPy? finite elements-exercice We consider in $\mathbb{R}^2$ the set of points $$\{M_1(-1,1),M_2(0,1), M_3(2,1),M_4(-1,0),M_5(1,0),M_6(2,0)\}$$  Let $\Omega$ a rectangular structure consisting of the heads $\{M_4(-1,0),M_6(2,0), M_3(2,1),M_1(-1,1)\}$  Let the two polygons $\sum_1 = \{M_4,M_5,M_2,M_1\}$ and $\sum_2 = \{M_5,M_6,M_3,M_2\}$   let the mesh $\{(e_1,\sum_1,Q_1),(e_2,\sum_2,Q_2)\}$ such $$Q_1=\{f : \mathbb{R}^2 \rightarrow \mathbb{R} , f(x,y) = a + bx + cy + d xy , a , b, c , d \in \mathbb{R}\}$$  1- Prove that $(e_1,\sum_1,Q_1)$ and $(e_2,\sum_2,Q_1)$ are two finite elements of Lagrange.  2- Let $V$ the finite-dimensional space that corresponds to the mesh. Prove that $V \nsubseteqq H^1(\Omega)$  My problem is to prove 2. How to debug EntityMalformedException? I've got fatal error EntityMalformedException: Missing bundle property on entity of type node. in entity_extract_ids() (line 7700 of .\includes\common.inc) when trying to access user/xyz.  I tried to retrieve info about the malformed node at line 7700, where the error message is built, thinks like :  if (!isset($entity-&gt;{$info['entity keys']['bundle']}) || $entity-&gt;{$info['entity keys']['bundle']} === '') {  dpm($info);// or dpm($entity);  throw new EntityMalformedException(t('Missing bundle property on entity of type @entity_type.', array('@entity_type' =&gt; $entity_type))); }   dpm($entity) returns an unexpected user object, and $info a huge qty of things.   Could somebody put me on the right way ?  I already read whatever I could find about Missing bundle property errors, but none helped solving.  dpm($entity) returns  uid (String, 2 characters ) 70 name (String, 9 characters ) John pass (String, 55 characters ) $S$DUwPuOuDPiDL4nRTYXqc7a5uOfMKey7pyhOFUEKka1XM... mail (String, 30 characters ) john@example.com theme (String, 0 characters ) signature (String, 0 characters ) signature_format (String, 13 characters ) filtered_html created (String, 10 characters ) 1396286331 access (String, 10 characters ) 1397146661 login (String, 10 characters ) 1396513460 status (String, 1 characters ) 1 timezone (NULL) language (String, 2 characters ) fr picture (NULL) init (String, 30 characters ) john@example.com data (Array, 5 elements) roles (Array, 1 element) og_user_node (Array, 0 elements) message_subscribe_email (Array, 1 element) field_bio (Array, 0 elements) field_name_first (Array, 1 element) field_name_last (Array, 1 element) field_facebook_url (Array, 0 elements) field_linkedin_url (Array, 0 elements) field_twitter_url (Array, 0 elements) user_trusted_contacts (Array, 1 element) group_group (Array, 1 element) group_access (Array, 1 element) metatags (Array, 0 elements) rdf_mapping (Array, 3 elements) realname (String, 13 characters ) John Doe content (Array, 13 elements) entity_view_prepared (Boolean) TRUE privatemsg_disabled (Boolean) FALSE Calculate sum of series $\sum \frac{n^2}{n!}$ I have to calculate sum of series $\sum \frac{n^2}{n!}$. I know that $\sum \frac{1}{n!}=e$ but I dont know how can I use that fact here.. How to Update Multiple Array Elements in mongodb