Unexpected output in dictionary
Why is python ordering my dictionary like so?
Application of Kolmogorov's Zero-One Law Most of the books I read, they only state some examples of tail events. One of them is  $[\sum_n X_n \ converges]$. My main problem is to show that this is indeed a tail event. What is inertia of a body? The definition of  is  "Inertia is the resistance offered by the body whenever its state of motion is changed."   What is inertia of a body? Is inertia actually a force exerted by the body?  If so, then why is there no application of inertia in numerical problems involving application of force on a body? What is the actual application of inertia in theory? can any one tell me how to write long equation in multline i tried with coding below but some errors of \right and \left appear \begin{multline*} \displaystyle \left \| f  \right \|_{M\dot{K}^{\alpha(\cdot),q}_{p(\cdot), \lambda}(\mathbb{R}^{n})} \approx \max \left\{ \sup_{L\leqslant0,L\in  \mathbb{Z}}2^{-L\lambda}\left ( \sum_{k=-\infty}^{L}2^{k\alpha(0)q}\left \|  f\chi_{k} \right \|^{q}_{L^{p(\cdot)}} \right )^{1/q},\\\sup_{L&gt;0,L\in \mathbb{Z}}\left[ 2^{-L\lambda}\left ( \sum_{k=-\infty}^{-1}2^{k\alpha(0)q} \left \| f\chi_{k} \right \|^{q}_{L^{p(\cdot)}} \right )^{1/q}\\ +2^{-L\lambda}\left ( \sum_{k=0}^{L}2^{k\alpha(\infty)q}\left \|  f\chi_{k} \right \|^{q}_{L^{p(\cdot)}} \right )^{1/q}\right] \right\}  \end{multline*} Evaluate forecasting ability of GARCH models with RMSE and MAE How to show that the set of points of continuity is a $G_{\delta}$ Why is variance squared? I know that new arXiv papers usually appear in Google Scholar within a few days or weeks. My question is: if someone submits a replacement of his/her arXiv paper, does the new version of the paper also appear in Google Scholar? Why Didn't People Think That Voldemort Died After His Attack on Baby Harry? After the Avada Kedavra curse rebounded from Harry onto Voldemort, Voldemort disappeared. I'm curious as to why so many key players believed Voldemort was still "alive" somewhere, his powers decimated, but, under the right circumstances, able to return to full power. It seems the witches and wizards in chapter one of Philosopher's Stone, The Boy Who Lived, were happy to believe that Voldemort was gone forever. Why didn't everyone believe Voldemort had died, rather than just disappeared? Was it the lack of a body? Or were there other reasons?  Here's a couple of quotes from Philosopher's Stone if anyone wants to review them. Otherwise, just disregard.     ‘A fine thing it would be if, on the very day You- Know-Who seems to have disappeared at last, the Muggles found out about us all. I suppose he really has gone, Dumbledore?’   ‘It certainly seems so,’ said Dumbledore. ‘We have much to be thankful for.’   Philosopher's Stone - page 13 - Bloomsbury - chapter 1, The Boy Who Lived   ‘But what happened to Vol– sorry – I mean, You-Know-Who?’   ‘Good question, Harry. Disappeared. Vanished. Same night he tried ter kill you. Makes yeh even more famous. That’s the biggest myst’ry, see ... he was gettin’ more an’ more powerful – why’d he go?   ‘Some say he died. Codswallop, in my opinion. Dunno if he had enough human left in him to die. Some say he’s still out there, bidin’ his time, like, but I don’ believe it. People who was on his side came back ter ours. Some of ’em came outta kinda trances. Don’ reckon they could’ve done if he was comin’ back.   ‘Most of us reckon he’s still out there somewhere but lost his powers. Too weak to carry on. ’Cause somethin’ about you finished him, Harry. There was somethin’ goin’ on that night he   hadn’t counted on – I dunno what it was, no one does – but somethin’ about you stumped him, all right.’   Philosopher's Stone - pages 46-47 - Bloomsbury - chapter 4, The Keeper of the Keys How do I make a floor plan with specific dimensions? Why is "rm" so much slower than deleting files via the File Manager? Colors and legend in groupplots/barplot In order to compare three items I use the groupplot command. For each item I have n different criteria to compare.    Is it possible to set the color for each criteria itself? Meaning the first bar in every group goes gray, the second pink etc.. The legend than should contain the n compared criteria.  And... how do I manage to let the whole tikz-picture have a width of 0.9\textwith  \documentclass{standalone} \usepackage{pgfplots,pgfplotstable} \usetikzlibrary{pgfplots.groupplots}  \pgfplotstableread{ nr  fii faa foo a   10  15  12 b   12  18  14 c   13  19  15 d   9   13  10 e   7   11  8 f   10  15  12 }\datatable  \begin{document} \begin{tikzpicture} \begin{groupplot}     [group style={columns=3},         ybar,         ymin=0,         ymax=20,         xtick=\empty,         tickpos=both]  \nextgroupplot[xlabel=fii] \addplot [fill=blue] table [x expr=\coordindex, y=fii] {\datatable}; \nextgroupplot[xlabel=faa] \addplot [fill=green] table [x expr=\coordindex, y=faa] {\datatable}; \nextgroupplot[xlabel=foo] \addplot [fill=red] table [x expr=\coordindex, y=foo] {\datatable};  \end{groupplot} \end{tikzpicture} \end{document} I've just started to learn about the tensor product and I want to show: $$(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z}) \cong \mathbb{Z}/ \gcd(m,n)\mathbb{Z}.$$  Can you tell me if my proof is right:  $\mathbb{Z}/m\mathbb{Z}$ and $\mathbb{Z} / n \mathbb{Z}$ are both finite free $\mathbb{Z}$-modules with the basis consisting of one single element $\{ 1 \}$. So $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$ has the basis $\{ 1 \otimes 1 \}$.  Therefore, any element in $(\mathbb{Z}/m\mathbb{Z}) \otimes_\mathbb{Z} (\mathbb{Z} / n \mathbb{Z})$  is of the form $(ab) 1 \otimes 1$ and any element in $\mathbb{Z}/ \gcd(m,n)\mathbb{Z}$ is of the form $k 1 = k$ where $k \in \{ 0, \dots , \gcd(n,m) \}$.  I would like to construct an isomorphism that maps $ab$ to some $k$. Let this map be $ab (1 \otimes 1) \mapsto ab \bmod \gcd(n,m)$.  This is a homomorphism between modules: it maps $0$ to $0$ because it maps the empty sum to the empty sum. It also fulfills $f(a + b) = f(a) + f(b)$ because there is only one element, $a = 1$.   It is surjective. So all I need to show is that it is injective. But that is clear too because if $ab \equiv 0 \bmod \gcd(m,n)$  then both $a \equiv 0 \bmod n$ and $b \equiv 0 \bmod m$ so the kernel is trivial.  Many thanks for your help!! Do you have to pass delete the same pointer that was returned by new, or can you pass it a pointer to one of the classes base types?  For example:  class Base { public:     virtual ~Base();     ... };  class IFoo { public:     virtual ~IFoo() {}     virtual void DoSomething() = 0; };  class Bar : public Base, public IFoo { public:     virtual ~Bar();     void DoSomething();     ... };  Bar * pBar = new Bar; IFoo * pFoo = pBar; delete pFoo;   Of course this is greatly simplified.  What I really want to do is create a container full of boost::shared_ptr and pass it to some code that will remove it from the container when it is finished. This code will know nothing of the implementation of Bar or Base, and will rely on the implied delete operator in the shared_ptr destructor to do the right thing.  Can this possibly work? My intuition says no, since the pointers will not have the same address.  On the other hand, a dynamic_cast&lt;Bar*> should work, so somewhere the compiler is storing enough information to figure it out.  Thanks for the help, everybody who answered and commented. I already knew the importance of virtual destructors, as shown in my example; after seeing the answer I gave it a little thought, and realized the whole reason for a virtual destructor is this exact scenario. Thus it had to work. I was thrown by the absence of a visible means of converting the pointer back to the original. A little more thinking led me to believe there was an invisible means, and I theorized that the destructor was returning the true pointer for delete to release. Investigating the compiled code from Microsoft VC++ confirmed my suspicion when I saw this line in ~Base:  mov eax, DWORD PTR _this$[ebp]   Tracing the assembler revealed that this was the pointer being passed to the delete function. Mystery solved.  I've fixed the example to add the virtual destructor to IFoo, it was a simple oversight. Thanks again to everyone who pointed it out. prove $\sqrt{a_n b_n}$ and $\frac{1}{2}(a_n+b_n)$ have same limit I have some edges that are not on the x, y or z axis. What is a simple way to straighten them? For example, I can scale and then use Shift+Z to straighten on the x and y axis.  Any shortcuts other than temporarily rotating the entire object to put the edges on an axis? How to remove symbolic link I was trying to create this symbolic link:  sudo ln -s /usr/share/phpmyadmin /var/www/phpmyadmin   but I accidentally typed:  sudo ln -s /usr/share/php,yad,in /var/www/phpmyadmin   So now I want to correct it but it says symbolic link already exist. Let $f(z)$ be an entire function whose real part is bounded by a polynomial in $|z|$. Does it follow that $f(z)$ is a polynomial?  Or, without loss of generality and more suggestively $$(f(z)=\sum_{k=0}^\infty a_k z^k \quad \land \quad \operatorname{Re}(f(z))\leq |z|^n)\quad \forall z\in \mathbb{C} \qquad \Rightarrow \qquad f(z)=\sum_{k=0}^n a_k z^k \quad\forall z\in \mathbb{C}$$ I'm curious of how illusions would affect spells or abilities that require a target to be looking at something. A good example of this would be Medusa's Petrifying Gaze which has the following effect:  Petrifying Gaze: When a creature that can see the medusa's eyes starts its turn within 30 feet of the medusa, the medusa can force it to make a DC 14 Constitution saving throw if the medusa isn't incapacitated and can see the creature. If the saving throw fails by 5 or more, the creature is instantly petrified. [...]  So if someone were to cast Minor Illusion to put a bag over Medusa's head, would other player/creatures who do not realize that the bag is an illusion be affected by the gaze if they were looking in the direction of it? Does the illusion block the effect? $f^{-1}([-\infty,r))$ measurable for all rational numbers $r$ If $X$ is measurable space, $f : X\to [-\infty,+\infty]$, function such that $f^{-1}([-\infty,r))$ measurable for all rational numbers $r$,prove than $f$ is measurable  I know that $f$ is measurable iff set $\{x\in X : f(x)&lt;a\}$ is measurable for all $a\in \mathbb{R}$ but for $a\in\mathbb{Q}$ I don't know how to proceed. How to replace a character by a newline in Vim I'm trying to replace each , in the current file by a new line:  :%s/,/\n/g    But it inserts what looks like a ^@ instead of an actual newline. The file is not in DOS mode or anything.  What should I do?  If you are curious, like me, check the question  as well.