Of course, a while back when searching for something unrelated, I came across a question almost identical to mine, but can't find it now.  Since turning off and restarting my computer a few days ago, I haven't been able to boot into Windows.  My suspicion is that it's related to an update Windows had a couple weeks ago; after the update, I hadn't booted into Ubuntu until just the other day, and upon restarting and trying to start Windows, GRUB just gave an error message(I don't recall the wording, but something about Windows Boot Manager not being found/available) and prompted me to try again.  Tried this a few times with same results.  From Ubuntu, I tried installing and running boot-repair, but all it seemed to do was cause my computer to skip the GRUB screen entirely and boot straight into Ubuntu.  Here is the pastebin link given by boot-repair:    Any help is greatly appreciated.
I installed Windows on my computer, followed by an installation of Ubuntu. However, now I'm unable to boot into my Windows install.  What can I do to fix this?
onConfigurationChanged not getting called Why Was It 'Essential' That Voldemort Kill Harry Potter? ‘So the boy ... the boy must die?’ asked Snape, quite calmly.   ‘And Voldemort himself must do it, Severus. That is essential.’   Deathly Hallows - page 551 - UK Hardcover - chapter 33, The Prince's Tale   Dumbledore tells Snape it is 'essential' that Voldemort be the one to kill Harry, I'm assuming in order for the piece of Voldemort's soul in Harry to be properly killed as well.  But why?   Hermione destroyed the Hufflepuff cup; Ron destroyed the Slytherin locket; Neville killed Nagini; Harry destroyed the diadem and Tom Riddle's diary. So the Horcruxes were not immune to destruction at the hand of someone other than Voldemort. And as it ended up, Harry himself wasn't even a Horcrux, but rather just an unknowing host to a parasitic bit of Voldemort's soul.   Why was it 'essential' that Voldemort be the one to kill Harry in order for the piece of Voldemort's soul to die? split file lines based on first field I have file with content like below and want to convert my output like below   Input  1,a,b,c 2,b,c 3,e,f 4,l  Required output  1,a 1,b 1,c 2,b 2,c 3,e 3,f 4,l    Values on first field is unique and no duplicate lines for 1st field in the input.  I am new to scripting and not sure how can we do this. Studying for an exam, a review question...  Given $G=\langle x,y|x^4=y^4=e,xyxy^{-1}=e\rangle$.    Show $|G|\leq16$.   For this, I want to consider that $x^3=x^{-1}$ and $y^3=y^{-1}$ based on our assumptions. I am a little lost as to how to put the second part, $xyxy^{-1}=e$ to show there are no more than 16 elements.   If $|G|=16$, find the center of the group and find a group that is isomorphic to $G/\langle y^2\rangle$.   I'm pretty sure the group is centerless. Random plot based on a specific random seed Is the latest version of Ubuntu 100% FREE Using 'very' with a noun Compare integer in bash, unary operator expected How to have full bleed between two facing pages Indesign When my receiver that I have hooked up to my pc turns on it causes my monitors to turn black is there a way to stop this? Proving the measure of an increasing sequence of measurable sets is the limit of the measures Show that if $A_1\subseteq A_2\subseteq A_3\subseteq\cdots$ is an increasing sequence of measurable sets (so $A_j\subseteq A_{j+1}$ for every positive integer $j$), then we have $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty}m(A_j)$$  Here is my proof: According to the $\sigma$-algebra property, $\bigcup_{j=1}^{\infty}A_j$ is a measurable set, so it makes sense to talk about $m(\bigcup_{j=1}^{\infty}A_j)$. Firstly, I prove that $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. This is because for any given positive integer $N$, $A_N\subseteq \bigcup_{j=1}^{\infty}A_i$, according to monotonicity, we have $m(A_N)\leq m(\bigcup_{j=1}^{\infty}A_i)$. Take the limit,we will have $\lim_{j\to\infty}m(A_j)\leq m(\bigcup_{j=1}^{\infty}A_j)$. Secondly, I prove that $m(\bigcup_{j=1}^{\infty}A_j)\leq \lim_{j\to\infty}m(A_j)$. For any given positive integer $N$, $\bigcup_{j=1}^N A_j = A_N$. According to monotonicity,we have $m\left(\bigcup_{j=1}^N A_j\right)=m(A_N)\leq \lim_{j\to\infty}m(A_j)$. Take the limit, we will have $m\left(\bigcup_{j=1}^\infty A_j\right) \leq \lim_{j\to\infty} m(A_j)$. Combine the above two arguments, we will see that $$m\left(\bigcup_{j=1}^\infty A_j\right)=\lim_{j\to\infty} m(A_j)$$$\Box$  The above is my proof, unlike many books, my proof does not use the property of countable additivity. So I doubt my proof is correct. Who can point out where are my mistakes? How can I get my usual interface when opening projects created by other people? I've bougth cup models from internet.   In the ZIP I get the blend file and pressing F12 I can render the project (Do you call it project?)  But I don't see anymore the "Layout" tab, that is what I use the most.   Why and how can I go back to my usual interface?    Like in the Opening Layout:    UPDATE 1  I could open the file with the default layout, however I'm not seeing the objects in there:    UPDATE 2  How to open Collection 1? It's grayed out. Why doesn't 'find' show this file? I have a 16GB flash drive. I'm running Ubuntu 14.04. I can't open my flash drive and I can't see it in GParted.  After running tail -f /var/log/syslog . After inserting my flash:  USB disconnect, device number 10 Jul  3 16:39:55 rihanna-VGN-Z56GG-B kernel: [ 3309.712166] usb 2-3: new high-speed USB device number 11 using ehci-pci Jul  3 16:39:55 rihanna-VGN-Z56GG-B kernel: [ 3309.846591] usb 2-3: New USB device found, idVendor=090c, idProduct=1000 Jul  3 16:39:55 rihanna-VGN-Z56GG-B kernel: [ 3309.846601] usb 2-3:   New USB device strings: Mfr=1, Product=2, SerialNumber=3     Jul  3 16:39:55 rihanna-VGN-Z56GG-B kernel: [ 3309.846609] usb 2-3: Product: Silicon-Power     Jul  3 16:39:55 rihanna-VGN-Z56GG-B kernel: [ 3309.846615] usb 2-3: Manufacturer: UFD 2.0     Jul  3 16:39:55 rihanna-VGN-Z56GG-B kernel: [ 3309.846622] usb 2-3: SerialNumber: 1401208423400131     Jul  3 16:39:55 rihanna-VGN-Z56GG-B kernel: [ 3309.847176] usb-storage 2-3:1.0: USB Mass Storage device detected     Jul  3 16:39:55 rihanna-VGN-Z56GG-B kernel: [ 3309.847309] scsi12 : usb-storage 2-3:1.0     Jul  3 16:39:55 rihanna-VGN-Z56GG-B mtp-probe: checking bus 2, device 11: "/sys/devices/pci0000:00/0000:00:1d.7/usb2/2-3"     Jul  3 16:39:55 rihanna-VGN-Z56GG-B mtp-probe: bus: 2, device: 11 was not an MTP device     Jul  3 16:40:17 rihanna-VGN-Z56GG-B kernel: [ 3332.176107] usb 2-3: reset high-speed USB device number 11 using ehci-pci     Jul  3 16:40:33 rihanna-VGN-Z56GG-B kernel: [ 3347.288080] usb 2-3: device descriptor read/64, error -110     Jul  3 16:40:48 rihanna-VGN-Z56GG-B kernel: [ 3362.504081] usb 2-3: device descriptor read/64, error -110     Jul  3 16:40:48 rihanna-VGN-Z56GG-B kernel: [ 3362.720076] usb 2-3: reset high-speed USB device number 11 using ehci-pci     Jul  3 16:40:52 rihanna-VGN-Z56GG-B kernel: [ 3366.812110] usb 2-3: reset high-speed USB device number 11 using ehci-pci     Jul  3 16:41:07 rihanna-VGN-Z56GG-B kernel: [ 3381.924082] usb 2-3: device descriptor read/64, error -110     Jul  3 16:41:23 rihanna-VGN-Z56GG-B kernel: [ 3397.356093] usb 2-3: reset high-speed USB device number 11 using ehci-pci     Jul  3 16:41:38 rihanna-VGN-Z56GG-B kernel: [ 3412.468057] usb 2-3: device descriptor read/64, error -110     Jul  3 16:41:54 rihanna-VGN-Z56GG-B kernel: [ 3428.664065] usb 2-3: reset high-speed USB device number 11 using ehci-pci     Jul  3 16:42:09 rihanna-VGN-Z56GG-B kernel: [ 3443.776100] usb 2-3: device descriptor read/64, error -110     Jul  3 16:42:24 rihanna-VGN-Z56GG-B kernel: [ 3458.992085] usb 2-3: device descriptor read/64, error -110     Jul  3 16:42:25 rihanna-VGN-Z56GG-B kernel: [ 3459.208063] usb 2-3: reset high-speed USB device number 11 using ehci-pci     Jul  3 16:42:40 rihanna-VGN-Z56GG-B kernel: [ 3474.320039] usb 2-3: device descriptor read/64, error -110     Jul  3 16:42:50 rihanna-VGN-Z56GG-B kernel: [ 3484.516109] usb 2-3: reset high-speed USB device number 11 using ehci-pci     Jul  3 16:42:50 rihanna-VGN-Z56GG-B kernel: [ 3484.768108] usb 2-3: reset high-speed USB device number 11 using ehci-pci Jul  3 16:43:05 rihanna-VGN-Z56GG-B kernel: [ 3499.880081] usb 2-3: device descriptor read/64, error -110 Jul  3 16:43:20 rihanna-VGN-Z56GG-B kernel: [ 3515.096081] usb 2-3: device descriptor read/64, error -110 Jul  3 16:43:21 rihanna-VGN-Z56GG-B kernel: [ 3515.312048] usb 2-3: reset high-speed USB device number 11 using ehci-pci Jul  3 16:43:36 rihanna-VGN-Z56GG-B kernel: [ 3530.424032] usb 2-3: device descriptor read/64, error -110 Jul  3 16:43:46 rihanna-VGN-Z56GG-B kernel: [ 3540.505230] scsi 12:0:0:0: Device offlined - not ready after error recovery The  asks about functions $f \colon \mathbb R \to \mathbb R$ such that $$f(x+y)=f(x)+f(y).$$ It is a very well-known functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for some advanced questions. Solutions of this equation are often called . Also a few other equations related to this equation are often studied. (Equations which can be easily transformed to Cauchy functional equation or can be solved by using similar methods.) Is there some overview of basic facts about Cauchy equation and related functional equations - preferably available online? What is the benefit of launching a satellite "a few days before the monthly report" I've seen this hint in the game as well as on this site and I've been wondering if there is any specific advantage to launching satellites closer to the monthly report rather than any time during that month.  The way I see it, monthly maintenance costs are taken out during the reports (correct me if i'm wrong), so regardless of when you launch it during the month it would cost the same.   Is the benefit of holding on to them longer simply a safety net in case some countries go into the red during that time? How can I expand a partition into non adjacent free space using GParted? My dual-boot machine has partitions as follows:   sda1 50 GB NTFS sda2 15 GB unallocated sda3 20 GB ext3 sda4 20 GB ext4   I would like to allocate the unused space on sda2 to my Ubuntu partition on sda4, hopefully leaving the other two partitions untouched, leaving this configuration:    sda1 50 GB NTFS sda3 20 GB ext3 sda4 35 GB ext4   Is this possible? If so, how can it be done, preferably using GParted? Convert Shapefiles to GeoJSON? I'm trying to convert Shapefiles from the  website into a GeoJSON file for use with Google Maps API. I'm using QGIS to facilitate but I can't get the projection/coordinates right. So far I've tried Google Mercator (EPSG:900913) and WGS 84 (EPSG:4326) with no luck. I'm using the OpenLayers to verify results and it keeps placing Texas near the  Gulf of Guinea (0 lat, 0 lng). Is it possible to add new alphabets to unicode-math? In ordinary LaTeX, one can easily add new math alphabets by using \DeclareMathAlphabet. One is then able to use, say, cmsy, eus, and rsfs letters side-by-side. Unfortunately, there does not seem to be a similar mechanism in unicode-math, or at the very least \DeclareMathAlphabet doesn't do what it's supposed to do.  Question. How can I set up my own math alphabets in unicode-math?   Attempted workarounds  Since unicode-math has lots of math alphabets, one workaround might be to replace an unused alphabet with the desired one using \setmathfont[range=...-&gt;...]. This doesn't quite work right in my example:  \documentclass[12pt]{article} \usepackage{fontspec,xltxtra,xunicode} \defaultfontfeatures{Mapping=tex-text}  \usepackage{unicode-math}  \scrollmode  \setmathfont{xits-math.otf} \setmathfont[range={\mathscr,\mathbfscr},StylisticSet=0]{xits-math.otf} \setmathfont[range={\mathcal,\mathbfcal},StylisticSet=1]{xits-math.otf} \setmathfont[range={\mathtt-&gt;\mathcal}]{latinmodern-math.otf}  \begin{document} \[ \begin{array}{c} \mathcal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathscr{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathtt{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \end{array} \]   I suspect the problem is that the codepoints of the calligraphic letters in Unicode are not contiguous.   Using \mathversion seems to work better:  \documentclass[12pt]{article} \usepackage{fontspec,xltxtra,xunicode} \defaultfontfeatures{Mapping=tex-text}  \usepackage{unicode-math}  \setmathfont{xits-math.otf}  \setmathfont[version=eucal]{latinmodern-math.otf} \newcommand{\eucal}[1]{\mathchoice %   {\mbox{\mathversion{eucal}$\displaystyle\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\scriptstyle\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\scriptscriptstyle\mathscr{#1}$}}}  \setmathfont[range={\mathcal,\mathbfcal},StylisticSet=1]{xits-math.otf}  \begin{document} \[ \begin{array}{c} \mathcal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathscr{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \eucal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \end{array} \] \end{document}   But this feels very inefficient. Homeomorphism between the unit disc and the unit square I know that the function that describes the homeomorphism is: $f(x,y) =       \begin{cases}     \frac{x^{2}+y^{2}}{\max(|x|,|y|)}(x,y) &amp;\quad\text{if } (x,y)\ne  (0,0) \\       \text{} (0,0) &amp;\quad\text{if } (x,y) =  (0,0) \\       \end{cases}$  But it's impossible to me to show that $f(x,y)$ is an homeomorphism. Mostly i can't see the injectivity.