Do I need UK transit visa?
Do I need a UK or Ireland transit visa when flying from the US to a Schengen country?
`apt update` failed on Debian Linux Attempting to upgrade testing/stretch Debian Linux:  $ sudo apt update  Ign:1 http://dl.google.com/linux/talkplugin/deb stable InRelease Hit:2 http://dl.google.com/linux/talkplugin/deb stable Release                                                                    ... Get:25 http://mirror.vorboss.net/debian unstable/main i386 Packages 2016-03-25-1456.28.pdiff [15.0 kB]              Fetched 686 kB in 7s (89.5 kB/s)                                                                                                                                                         Reading package lists... Done W: gpgv:/var/lib/apt/lists/dl.google.com_linux_talkplugin_deb_dists_stable_Release.gpg: The repository is insufficiently signed by key 4CCA1EAF950CEE4AB83976DCA040830F7FAC5991 (weak digest) W: gpgv:/var/lib/apt/lists/deb.opera.com_opera_dists_stable_InRelease: The repository is insufficiently signed by key 419D0ACF314E8E993F7F92E563F7D4AFF6D61D45 (weak digest) W: Failed to fetch http://dl.google.com/linux/talkplugin/deb/dists/stable/Release  No Hash entry in Release file /var/lib/apt/lists/dl.google.com_linux_talkplugin_deb_dists_stable_Release, which is considered strong enough for security purposes E: Some index files failed to download. They have been ignored, or old ones used instead.  What's wrong? How to fix it? How was Jon Snow able to kill the White Walker in S05E08? When I click the bounty info icon on the iOS app. The added bounty description and reason is not shown.  For example: Reading a binary file and using Response.BinaryWrite() How can I find all files I can not write to?  Would be good if it takes standard permissions and acls into account.  Is there an "easy" way or do I have to parse the permissions myself? Does Time Machine backup programs like Photoshop and iWork? Django: AppRegistryNotReady() Can we set privacy for "last seen" status? What's the right way to hard boil eggs? Unable to stylize raster data as pseudocolor render type on MAC OSX In a raster layer's Style Properties, I can choose pseudocolor render type, and choose a color ramp and classify the raster.      However, when I click the 'Apply' button, or click 'OK', the visualization of the raster never changes.  I was able to output a 'colorized' version of the same raster using GRASS plugin.  I also reinstalled QGIS 2.8, and was able to adjust and see a raster stylings using the steps I outlined above.  So, can anyone confirm this as an issue, or confirm that there's a mistake somewhere in my process?  I should note that I'm on a Mac, 10.10.4, and I'm using the QGIS provided by , 2.10.1-1 .  Update 1: Here's a screenshot of the rest of my settings.      I have Grayscale set to off in the Style setting.  Also, I don't believe grayscale/color-blind Preview is available on the Mac version of QGIS.  They're does't appear to be the option where it , in the Menu Bar under View / Preview / Photocopy (grayscale) etc.  Either way-- I added a shapefile to show color.  Update 2: I've experimented with setting color interpolation to discrete, linear, and exact, but still am not able stylize using Singleband pseudocolor.  I've also confirmed that the 'Render' box on the the main map view is checked.  Also, I am able to adjust Color Rendering settings, like Hue, Blending Mode, etc., and see the results when clicking 'Apply' or 'OK'.  Update 3: As other Mac users are confirming this, I think it may be a bug. I've created a bug report at the official . How to prove $\,x^a-1 \mid x^b-1 \iff a\mid b$ How to prove $x^a-1\mid x^b-1 \iff a \mid b$, where $x \ge 2$ and $a,b,x \in \Bbb Z$.  I've tried the following in attempting to solve this:  $$a\mid b \Rightarrow aq=b \Rightarrow x^{aq}=x^b \Rightarrow x^ax^q=x^b$$  Because $x^q \in \Bbb Z$, it follows that $x^a\mid x^b$.  This is as far as I have gotten; any help getting further is appreciated.  Note: It may be that this identity is not true at all? A problem with my reasoning in a problem about combinations proving how many permutations exist. Show that the number of permutations of n numbers, for n ≥ 2, with two cycles is at most (n−1)!logn. Unsure as to were to start any help would be appreciated. Calculating sequential numbers into sorted table using ArcGIS Desktop? Possible to bake texture to new UV map? Why does Ubuntu 12.10 not detect my Win7 partition? Showing that $\int_0^1 \log(\sin \pi x)dx=-\log2$ I need help with a textbook exercise (Stein's Complex Analysis, Chapter 3, Exercises 9). This exercise requires me to show that $$\int_0^1 \log(\sin \pi x)dx=-\log2$$ A hint is given as "Use the contour shown in Figure 9."    Since this is an exercise from Chapter 3, I think I should use the residue formula or something like that. But the function $f(x)=\log(\sin \pi x)$ becomes singular on $x=0$ and $x=1$, which makes the contour illegal for the residue theorem. Can anyone give me a further hint on this problem? Many thanks in advance!  P.S. This is my first time on Math Stack Exchange. If you find my post ambiguous, let me know. Give an example of a sequence of continuous functions $f_n:\mathbb{R} \to \mathbb{R}$ such that the pointwise limit $f:\mathbb{R} \to \mathbb{R}$ exists but is discontinuous. My idea is $f_n(x)=\displaystyle\frac{x}{n}.$ Then the pointwise limit is $0$ but $f$ is discontinuous on $\mathbb{R}.$ Is that correct? I'm having a hard time understanding the definition of pointwise convergence. I have this definition: $\displaystyle\lim_{n\to \infty} f_n=f$ pointwise iff $\displaystyle\lim_{n\to \infty} f_n(x)=f(x)$ But then what is $f(x)$? Where does the $n$ go? In the example I gave, for instance, $\displaystyle\lim_{n\to \infty} f_n(x)=\displaystyle\lim_{n\to \infty} \frac{x}{n}=0.$ But does this equal $f(x)$? Is $f(x)$=0 a discontinuous limit? As writing matrices in LaTeX is an extremely time-consuming activity, I was wondering if a software or a site exists that automatically gave the transposed form of a matrix, that is to say, given a matrix:  \begin{bmatrix}    0 &amp; -1 &amp; -1 \\    -1 &amp; -1 &amp; 0 \\    -1 &amp; 0 &amp; 1 \\    1 &amp; 0 &amp; 0 \\    0 &amp; 0 &amp; -1 \\    -1 &amp; 2 &amp; 1  \end{bmatrix}   it will produce a transposed matrix, with the columns in place of the lines and conversely...  It would be particularly time-saving! Allowing a group Read-Write Access to a directory