Do blockades in star wars make sense?
In-universe explanation for single line of planetary entry/departure in Star Wars
Stars and pins are retained when chat messages are moved! I am using App Backup &amp; Restore for creating backup of my apps but with KitKat its using the internal SdCard as Storage.  When I explicitly set the external SdCard path it gives the error that it is unable to create folder,  I have also tried with creating folder under android/data/[packagename]/files/ but it is still not able to do the same.  Is there any solution so that apps can use external storage for writing files.  Note : I don't want to ROOT my Phone. Proof that disjoint cycles commute I'm trying to prove that in a symmetric group two disjoint cycles commute. But I suspect that something is not right about my proof (a sense of vagueness). Some hints would be appreciated.  Here's my proof:  Let $\sigma=(s_1 s_2 ... s_n)$, $\tau=(t_1 t_2 ... t_m)$ for some integers $m, n$. That is, $s_i \neq t_j$ for any $i \in [1, n]$, $j \in [1, m]$. Now, $\sigma\tau=(s_1 ... s_n)(t_1 ... t_m)$ is a cycle of disjoint permutations, which cannot be represented by any other disjoint permutations. Thus $\sigma\tau=\tau\sigma$. How to evaluate the limit of this expression?  $$\lim_{x\to\infty} \frac{\sin^2\left( \sqrt{x+1}-\sqrt{x}\right)}{1-\cos^2\frac{1}{x}}$$  I managed to simplify the denominator into a sinus form by the Pythagorean formula, and also modified the argument of the sinus in the numerator by dividing with a conjugate which I think is necessary, into this final form:  $$\lim_{x\to\infty} \frac{\sin^2\left(\frac{1}{\sqrt{x+1}+\sqrt{x}}\right)}{\sin^2\frac{1}{x}}$$  And that's where I got stuck. So far I've tried to come up with some solution to eliminate the sinus from the denominator, without success. I'm sure the solution is pretty obvious but I don't quite know how to modify the trigonometric functions in this fashion, so any help would be most appreciated. Grandi's Series; tends to $1/2$, but why is this considered a valid sum? Grandi's series,  $$1+1-1+1-1+1-1+...$$  can be expressed as the below:  $$\sum_{n=0}^\infty(-1)^n$$  Two valid sums that make sense to me are $1$, and $0$, depending on how you approach the series. $(1+1)-(1+1)-(1+1)-...=0$, and $1+(1-1)+(1-1)+(1-1)+...=1$.  There is consensus, however, that the actual sum is $\frac{1}{2}$. Why? I understand the approach of finding partial means of the series, and they do indeed tend to $\frac{1}{2}$, but it seems unintuitive to assert that the sum is neither $1$ or $0$.  A more convincing method I found was assuming the series is $S$, then shifting it such that $S-1 = S$, then through algebra finding $S = \frac{1}{2}$, but again, it seems more intuitive answer is either $0$ or $1$. I say this strictly because adding and subtracting integers should equal an integer, never a fraction.  Is this a characteristic of infinite series, which is not specific to Grandi's series? Setting environment variables for latex and bibtex in .bashrc Possible Duplicate:          I recently started using TeX to write journal articles. I use TeX Live version 2009-10 on a laptop running Ubuntu 10.10 (64-bit). I edit and compile the tex files using vim and vim-latexsuite.   Currently for each journal article I have all associated tex, bib, eps, etc. files in individual folders. I would like to simplify my life a bit and have a dedicated folder containing a "master" bib file - And maybe another folder for my custom style files.   My problem is that can't figure out how I should define the PATH, and other environment variables in my .bashrc file to find the necessary support files (e.g., the .bib file). Which variables do I need to set? PATH? MANPATH? TEXINPUTS? BIBINPUTS? Is $f(x)=10$ a periodic function? I am not getting satisficatory explanation for this. Clearly $f(x+T) = f(x)$ for all values of $T$.  If we assume it is periodic, does this mean period =  $0$? How to properly denoise renderings? Since the Denoise feature was added to Blender I only use the default setting with it and it really does the work. But sometimes the denoise is noticable on some of my renders. How can I overcome the blurred effect? Denoising for me is a big help because my system is quite aged. If $M$ isn't Noetherian, $M$ has a submodule maximal with respect to being not finitely generated. I'm answering this question:     Let $M$  be a module. Show that if  $M$ is not Noetherian then $M$ has a submodule $N$ such that $N$ is not finitely generated but $A$ is finitely generated whenever $N &lt; A ≤ M$.   The solution starts as follows (I'm following a given solution): Let $Y$ be the set of non-finitely-generated submodules of $M$. If $(N_α)$ is a chain in $Y$ then $Q= \bigcup_αN_α $ is in $Y$, else $Q$ is f.g., which implies $Q=N_α$ for some $α$, whence $N_α$  is f.g., contradiction.  I don't understand why if $Q$ is f.g., then $Q=N_α$ for some $α$? Am I missing some fact?  Thanks! Logistic Regression - Adding interactions makes Independent variable statistically insignificant My name is Abhi &amp; I am trying to better understand logistic regression by solving a few practice problem. I am using R and RStudio as the development environment  Problem Statement  Given the age, sex and class(first,second,third) for each passenger can you predict if he survived or died when the titanic sank   Attempt1  Simple Logistic Regression. Use age,sex and passenger class directly. The formula (in R)  Survived~Pclass+Sex+Age  This gives fairly decent results - accuracy of 79% and all of independent variables are statistically significant   Attempt2  Add interactions between age,sex and passenger class. Accuracy has improved to 80% but age is no longer significant. Also none of the new terms(age-sex,age-class,sex-class,age-sex-class) are statistically significant. The formula (in R) Survived~Pclass*Sex*Age   Can some one explain why this is happening? I can accept that the new terms may not be significant but why is age no longer significant?   Any help would be much appreciated Disk brakes severely reduced in power after replacing bearings Let $\mathbb{F}$ be a finite field.  Let $a_1,\dots,a_n \in \mathbb{F}$ be given.  I want to know whether there exists $x_1,\dots,x_n \in \mathbb{F}$ such that  $$a_1 x_1^2 + a_2 x_2^2 + \dots + a_n x_n^2 = 1.$$  My question: Can we characterize when this equation has a solution, i.e., by some criterion on $a_1,\dots,a_n$?  Is there a clean characterization, in terms of $a_1,\dots,a_n$ and the properties of $\mathbb{F}$?    Here's what I can see immediately.  First, if $n=1$, there is an easy answer (the equation has a solution iff $a_1$ is a square in $\mathbb{F}$).  Second, if any $a_i$ is a square in $\mathbb{F}$, the equation has a solution (set all other $x_j$'s to $0$), so we can assume that all the $a_i$'s are non-squares and that $n \ge 2$.  Third, if $-1$ is a square in $\mathbb{F}$, then the equation has a solution: we can take $x_2 =  \sqrt{-a_1/a_2} \cdot x_1$.  But what about the case where $-1$ is a non-square in $\mathbb{F}$?  Is it always guaranteed to have a solution in this case? Permission Denied To Delete File First of all, I'm not looking for an answer on how to remove the files. I am looking for a reason why this doesn't work - and a suggestion on how to remove the files without sudo.  I have a directory in my Desktop that has a few inner directories (don't judge the naming - I just extract what I need from files I'm given) and after getting the files I try and remove the files with rm -rf Bengal\ Tiger\ Creative/, I am surprised to see the following warning for each image.     rm: cannot remove ‘Bengal Tiger Creative/Bengal Tiger Creative/images/customer-services-landscape.jpg’: Permission denied   Below is some code outputs.  tree  nathan@nathan-pc:~/Desktop$ tree . ├── Bengal Tiger Creative │   └── Bengal Tiger Creative │       └── images │           ├── awards.jpg │           ├── blank.gif │           ├── bullet1-light-green.gif │           ├── bullet2-light-green.gif │           ├── bullet3-light-green.gif │           ├── bullet4-light-green.gif │           ├── customer-services-landscape.jpg │           ├── desktop.ini │           ├── header-img1.jpg │           ├── header-img2.jpg │           ├── header-img4.jpg │           ├── header-img5.jpg │           ├── softhr.gif │           ├── Thumbs.db │           ├── tiger_main-img1.jpg │           ├── tiger_main-img2.jpg │           ├── tiger_main-img3.jpg │           ├── tiger_main-img4.jpg │           ├── Vanquis-benefits-icons_iphone-green.gif │           └── white-arrow-green.gif   ls -laR  nathan@nathan-pc:~/Desktop$ ls -laR Bengal* Bengal Tiger Creative: total 12 drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:30 . drwxr-xr-x 6 nathan nathan 4096 Aug 10 11:34 .. drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:34 Bengal Tiger Creative  Bengal Tiger Creative/Bengal Tiger Creative: total 12 drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:34 . drwxrwxr-x 3 nathan nathan 4096 Aug 10 11:30 .. dr-xr-xr-x 2 nathan nathan 4096 Aug 10 11:36 images  Bengal Tiger Creative/Bengal Tiger Creative/images: total 356 dr-xr-xr-x 2 nathan nathan   4096 Aug 10 11:36 . drwxrwxr-x 3 nathan nathan   4096 Aug 10 11:34 .. -rwxrwxrwx 1 nathan nathan  14910 Mar 18 12:29 awards.jpg -rw-rw-r-- 1 nathan nathan   1093 Mar 21  2014 blank.gif -rw-rw-r-- 1 nathan nathan   1978 May  8  2014 bullet1-light-green.gif -rw-rw-r-- 1 nathan nathan   1969 May  8  2014 bullet2-light-green.gif -rw-rw-r-- 1 nathan nathan   2416 May  8  2014 bullet3-light-green.gif -rw-rw-r-- 1 nathan nathan   2449 May  8  2014 bullet4-light-green.gif -rw-rw-r-- 1 nathan nathan 113138 Apr 14  2014 customer-services-landscape.jpg -r--r--r-- 1 nathan nathan    142 Mar 25 11:58 desktop.ini -rw-rw-r-- 1 nathan nathan  14312 Mar 23 11:39 header-img1.jpg -rw-rw-r-- 1 nathan nathan   1480 Mar 23 11:41 header-img2.jpg -rw-rw-r-- 1 nathan nathan   1278 Mar 23 11:41 header-img4.jpg -rw-rw-r-- 1 nathan nathan   1168 Mar 23 11:41 header-img5.jpg -rw-rw-r-- 1 nathan nathan   1703 Mar 21  2014 softhr.gif -rw-rw-r-- 1 nathan nathan  88064 Aug  4 10:39 Thumbs.db -rw-rw-r-- 1 nathan nathan  23607 Mar 23 11:56 tiger_main-img1.jpg -rw-rw-r-- 1 nathan nathan  23400 Mar 23 11:56 tiger_main-img2.jpg -rw-rw-r-- 1 nathan nathan   6970 Mar 23 11:56 tiger_main-img3.jpg -rw-rw-r-- 1 nathan nathan   8996 Mar 23 11:57 tiger_main-img4.jpg -rw-rw-r-- 1 nathan nathan    275 Jul 31  2014 Vanquis-benefits-icons_iphone-green.gif -rw-rw-r-- 1 nathan nathan   1821 Jul  9  2014 white-arrow-green.gif   Doing chmod 777 and then delete does not work.  Doing rm -rf * inside the directory does not work.  Doing sudo rm file does work.   I can easily remove these files using sudo, however, I cannot see why they cannot be removed without sudo. As a relatively new Ubuntu user, I would like to learn why.  Thanks in advance, Nathan Mean and variance of Squared Gaussian: $Y=X^2$ where: $X\sim\mathcal{N}(0,\sigma^2)$? What is the mean and variance of Squared Gaussian: $Y=X^2$ where: $X\sim\mathcal{N}(0,\sigma^2)$?  It is interesting to note that Gaussian R.V here is zero-mean and non-central Chi-square Distribution doesn't work.  Thanks. I have upgraded 14.04 to 16.04 . Now I am trying to open the gedit from terminal by passing the command:    sudo gedit /some/where/file.txt   but it is throwing an error stating that:    (gedit:2090): Gtk-WARNING : Calling Inhibit failed: GDBus.Error:org.freedesktop.DBus.Error.ServiceUnknown:   The name org.gnome.SessionManager was not provided by any .service files**   However, it is opening the file. When I try to save any modification then it is showing this error after saving the file:    (gedit:2090): WARNING : Set document metadata failed: Setting attribute metadata::gedit-position not supported   I am not getting why this is happening in 16.04 only; it was not on 14.04. How do I make a Theorem n followed by a Theorem n' ? $ p(x)\in\mathbb{R[X]} $ is a polynomial of degree $n$ with no real   roots. Show that: $$\int\limits_{-\infty}^{+\infty}\dfrac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n^{3/2}\pi.$$   It's easy to see that the degree of $ p$ has to be even. For $n=2$ this integral is at most $2\pi$. For $n&gt;2$ the maximum value of this integral is obtained when all the imaginary parts of the roots of $p(x)$ tend to $0$, but I couldn't go further.    Any help would be appreciated, thanks.    Edits by David Speyer: It seems very likely now that the optimum bound is $n \pi$, not $n^{3/2} \pi$. As pointed out in the comments below, and further in , this is the value we get in the limit where $p$ has double roots on the real axis.   It seems likely that  moving the roots of $p$ away from the real axis can only make the integral less. Write the roots of $p$ as $a_j \pm i b_j$, so $p(x) = \prod ((x-a_j)^2 + b_j^2)$ and  $$\frac{p'(x)}{p(x)} = \sum \frac{2 (x-a_j)}{(x-a_j)^2 + b_j^2}.\quad (\ast)$$  So making the $b_j$ larger tends to make $p'(x)/p(x)$ smaller, which makes the integral smaller. But this argument is not rigorous, because the terms of $(\ast)$ can have both positive and negative sign, so it could be that making the individual terms closer to $0$ makes the absolute value of $(\ast)$ larger.  I don't see how to beat this issue easily.  Thus, I'm putting up a bounty for proving or disproving   $$ \int_{-\infty}^{+\infty}\dfrac{(p'(x))^2}{(p'(x))^2+(p(x))^2}\,dx \leq n\pi.$$ By $A\oplus B$ we denote the symmectric difference of two sets. The definition is $A\oplus B =(A\setminus B) \cup (B\setminus A)$. Now I hope to show that $A\oplus (B\oplus C) = (A\oplus B)\oplus C$. I remember that there's an elegant proof, but I forget its detail. (The first step is to show $A\oplus (A\oplus B) = B$, and maybe applying this formula, we can obtain associativity.) Show that the boundary of a set equals the boundary of its complement $\newcommand{\bdy}{\operatorname{bdy}}$  I'm trying to show that $\bdy(A) = \bdy(A^c)$.  I know that $\bdy(A) = \operatorname{closure} A \setminus \operatorname{int}(A) = (\operatorname{int}(A^c))^c \setminus \operatorname{int}(A)$, but I don't know where to go from there.  Any help or hints would be very much appreciated. A case of optional “that”: “check the” vs. “check that the” Consider the following use case:   Please check the username and password are correct. Please check that the username and password are correct.   In this case, I would say that that is required because it feels more natural to me and so the flow is better. However, is the lack of that wrong in this case?  The question boils down to: when is that optional? See . A quick summary:   That is optional if the pronoun is the object. That is required if the pronoun is the subject.   I think that we fall into the second case: that is required because the relative pronoun is the subject. But I'm not sure if I'm interpreting these rules correctly. Can someone please advise?