Two days ago, after putting my computer for less than one hour in sleep mode, I noticed weird lines of different colors around all windows as you can see on this screenshot:    Lines are both horizontal and vertical, from different colors, they look like barcodes and on the edges horizontal and vertical lines are scrambled. Those lines appear whether the window is active or not, in the foreground or not.  The lines also appear on the "X" "-" "|_|" icons on the top left hand corner of each window.  The bug also happens in search results in Chrome for instance, actually in any input field with autosuggestion on an HTML webpage:    It also happens when a context menu is open. For instance after a mouse right click there are lines all around the menu.  I use Ubuntu 16.04. I upgraded from Ubuntu 14.04 last week. When I upgraded to Ubuntu 16.04 I had a graphic issue (Computer in low graphics mode, solved with:  ), but after solving the initial issue everything was fine and I don't think I installed anything special after.  If I reboot my computer, the problem is solved. But it reappears after sleep mode.  Specs:   Asus N56VM BIOS: N56VM.206 01:00.0 VGA compatible controller: NVIDIA Corporation GF108M [GeForce GT 630M] (rev a1)   Kernel driver in use: nvidia Kernel modules: nvidiafb, nouveau, nvidia_375_drm, nvidia_375  00:02.0 VGA compatible controller: Intel Corporation 3rd Gen Core processor Graphics Controller (rev 09)   Kernel driver in use: i915 Kernel modules: i915
After some gorgeous updates I witness this:
Attributing contributions to academic work that occur in Stack Exchange Intuition behind the Definition of Conditional Probability (for 2 Events) Where does the Mac store account pictures? I took a picture to add to my account on Mac.   Now I want to use it on messenger, but I can't find it. On the account config, it's too small. I need to find the original.   Where are the accout pictures stored on a Mac? I would be happy to get some hints on the following integral: $$ \int_0^1\frac{x^{19}-1}{\ln x}\,dx $$ I want to copy the data folders from a previous ubuntu(14.04) partition My hard drive is a mess and I want to reformat it and do a fresh install of 16.04. My problem is I only have access to the HDD via 16.04 on a USB stick and although I can navigate to the various folders, nothing I do allows me to copy to an external hard drive. The folders are not encrypted. I also have a small swap partition - I would like to preserve the data on that too if possible.  Here's a listing of the home directories on the drive:  ubuntu@ubuntu:/$ ls -l /media/*/*/home /media/ubuntu/0e228f65-dac7-435f-828b-1eb1a6c3f29a/home: total 4 drwxr-xr-x 46 1000 1000 4096 Dec 21  2015 simon  /media/ubuntu/db79a6d1-7c6c-4d8c-822b-ba6ccd1ddecd/home: total 4 drwxr-xr-x 27 1000 1000 4096 Apr 17  2014 simon  /media/ubuntu/ebe2a6ba-9cb5-4279-8352-5082b8dd4d5e/home: total 4 drwxr-xr-x 39 1000 1000 4096 Aug  3 14:21 simon Logical volume is not mounted but is in use. Please close all applications using this device (eg iscsi) How do I resize a LVM partition using system-config-lvm?  As per   I clicked on the partition and tried to resize, but got a pop-up error from system-config-lvm.     Logical volume is not mounted but is in use. Please close all applications using this device (eg iscsi).     What's the problem and how can I fix it?    See also . MonoTouch.Dialog: How to set limit of number of characters for EntryElement Exact probability of random graph being connected The problem: I'm trying to find the probability of a random undirected graph being connected. I'm using the model $G(n,p)$, where there are at most $n(n-1) \over 2$ edges (no self-loops or duplicate edges) and each edge has a probability $p$ of existing. I found a simple formula online  where $f(n)$ is the probability of $G(n,p)$ being connected. But apparently it's too trivial for the writer to explain the formula (it was just stated briefly).  The desired formula: $f(n) = 1-\sum\limits_{i=1}^{n-1}f(i){n-1 \choose i-1}(1-p)^{i(n-i)}$  My method is: Consider any vertex $v$. Then there's a probability ${n-1 \choose i}p^i(1-p)^{n-1-i}$ that it will have $i$ neighbours. After connecting $v$ to these $i$ neighbours, we contract them ($v$ and its neighbours) into a single connected component, so we are left with the problem of $n-i$ vertices (the connected component plus $n-i-1$ other "normal" vertices.   Except that now the probability of the vertex representing connected component being connected to any other vertex is $1-(1-p)^i$. So I introduced another parameter $s$ into the formula, giving us:  $g(n,s)=\sum\limits_{i=1}^{n-1}g(n-i,i){n-1 \choose i}q^i(1-q)^{n-1-i}$,  where $q=1-(1-p)^s$. Then $f(n)=g(n,1)$. But this is nowhere as simple as the mentioned formula, as it has an additional parameter...  Can someone explain how the formula $f(n)$ is obtained? Thanks! Does a car use friction to move? I'm using a docker image as a base for my own development that adds the jessie backports repository in its Dockerfile and uses that to install a dependency. This image uses the following command to add the repository:  echo "deb http://ftp.debian.org/debian jessie-backports main" &gt;&gt; /etc/apt/sources.list   The problem is that fetching packages from the backports repository now  fails with the following error (this used to work previously):  W: Failed to fetch http://ftp.debian.org/debian/dists/jessie-backports/main/binary-amd64/Packages 404  Not Found  W: Failed to fetch http://deb.debian.org/debian/dists/jessie-updates/main/binary-amd64/Packages  404  Not Found   I looked on that server, and those paths are indeed not present there.  I tried to figure out on the Debian backports site whether this particular repository should still be available, and I didn't find any indication that this was deprecated or something like that.  Is this a temporary issue with the repository, or is the jessie-backports repository not available anymore? And if this is not a temporary issue, what options do I have to use this or an equivalent repository without upgrading to the newer Debian stable version? Normal + Connected -> Uncountable Here's a solution (to the exercise "prove that if $X$ is a space with more than one element, and is normal and connected then $X$ is uncountable"):   by Urysohn's lemma, given $A$ and $B$, closed and disjoint in $X$, there exists a continuous function from $X$ into $[0,1]$ such that $f(A)={0}, f(B)={1}$.   If $X$ were countable, so it would be $f(X)\subset [0,1]$; choose $r\in (0,1)\setminus f(X)$, then $X=f^{-1}([0,r))\cup f^{-1}((r,1])$, so $X$ is disconnected.  But the full power of the lemma wasn't used, all that was used is that there exists a continuous function $X\longrightarrow [0,1]$, so the condition of being normal seems too much.  So my question is, what are the weakest conditions on $X$ for the existence of such continuous functions into $[0,1]$ (or $\mathbb{R}$, for that matter)? Prove that $A \smallsetminus (A \smallsetminus B) = A \cap B$ $A$ and $B$ are any sets, prove that $A \smallsetminus (A \smallsetminus B) = A \cap B.$ This formula makes sense when represented on a Venn diagram, but I am having trouble with proving it mathematically.  I have tried letting $x$ be an element of $A$ and continue from there, but it doesn't seem to work out as a valid proof anyways.   Could anyone please point me in the right direction?  Many thanks. Prove $\gcd(a+b, a-b) = 1$ or $2\,$ if $\,\gcd(a,b) = 1$ I want to show that for $\gcd(a,b) = 1$ $a,b \in Z$   $\gcd(a+b, a-b) = 1$ or $\gcd(a+b, a-b) = 2$ holds.      I think the first step should look something like this:      $d = \gcd(a+b, a-b) = \gcd(2a, a-b)$   From here I tried to proceed with two cases. 1: $a-b$ is even, which leads to $\gcd(a+b, a-b) = 2$ 2: $a-b$ is odd, which leads to $\gcd(a+b, a-b) = 1$  My main problem I think is, that I do not know how I should include $\gcd(a,b) = 1$ in the proof.  Any help is appreciated. Thx in advance. Cherio Woltan Proving $\mathbb{F}_p/\langle f(x)\rangle$ with $f(x)$ irreducible of degree $n$ is a field with $p^n$ elements I am attempting to prove what the title says, that $\mathbb{F}_p/\langle f(x)\rangle$ with $f(x)$ irreducible of degree $n$ is a field with $p^n$ elements.  I have already proven that for any field $K$ and polynomial $f(x)$ in $K[x]$, $K[x]/\langle f(x)\rangle$ is a field if and only if $f(x)$ is irreducible in $K[x]$. So I know that $\mathbb{F}_p/\langle f(x)\rangle$ is a field for sure. Where I'm lost is figuring out how to know for sure that there are $p^n$ elements. It's fairly clear if $n\in\{ 0 ,1\}$, but I don't think induction is going to help.  I know from a proposition in class that since $F_p = \mathbb{Z}/\langle p\rangle$ and $F_p = \mathbb{Z}/\langle p\rangle$ is an integral domain, $F_p[x] = (\mathbb{Z}/\langle p\rangle)[x] = (\mathbb{Z}[x])/(\langle p\rangle[x])$, but I'm not sure whether or not that's helpful information. I am having problems booting my Ubuntu after an attempted hardware fix today. My laptop was working fine but had some minor issues, so we tried a different motherboard, which had a major issue, so we changed it back to the original board. However, now I have problem booting.  The title pretty much says it all. I am using Ubuntu 16.04, but on boot, I get:  [    0:000000] [Firmware Bug]: TSC_DEADLINE disabled due to Errata; please update microcode to version: 0x52 (or later)   and I am now stuck at the BusyBox (initramfs).  Does anyone know how to fix this issue? I am not so familiar with bash or initramfs, so I would appreciate a detailed step-to-step guide if possible. Unique (non-repeating) random numbers in O(1)? I'd like to generate unique random numbers between 0 and 1000 that never repeat (i.e. 6 doesn't show up twice), but that doesn't resort to something like an O(N) search of previous values to do it. Is this possible? Caps lock indicator not synced I have a laptop which I hook up to an external display when I'm at home. I also connected an external keyboard and mouse which makes it more comfortable for me.  I have noticed that pressing caps lock on the external keyboard does not light up the capslock indicator of the laptop keyboard (and vice versa).  When googling, I came across the setleds function, but when I type it into a console all I get is this:  KDGKBLED: Inappropriate ioctl for device Error reading current flags setting. Maybe you are not on the console?   Any idea if I can get this working? I have some questions regarding suspended users:   Why do suspended users have a reputation of one? According to the default reputation of a new user, he/she will have reputation of one and he/she has the privilege of posting questions. When comparing this, this is getting violated when compared to suspended users' reputation. Even though they have one reputation point, they cannot post questions. As suspended users do not have privilege of posting questions, there is no use of making "Ask Question" visible. This is an unnecessary server hit. Am I right? As suspended users do not have the privilege of answering questions, there is no use of making "Answer Question" visible. Why are we using the squared residuals instead of the absolute residuals in OLS estimation?  My idea was that we use the square of the error values, so that residuals below the fitted line (which are then negative), would still have to be able to be added up to the positive errors. Otherwise, we could have an error of 0 simply because a huge positive error could cancel with a huge negative error.  So why do we square it, instead of just taking the absolute value? Is that because of the extra penalty for higher errors (instead of 2 being 2 times the error of 1, it is 4 times the error of 1 when we square it). How to change fonts I'm sure this is a newbie question, but I can't figure this out. I have tried to follow the advice here:  and here:   But I can not seem to be able to generate any font change. I either get no change or error messages.  Here is the output of one failed attempt to change the font to Times using this command:  \rmfamily \renewcommand{\rmdefault}{ptm}     Apparently I either don't have a correct package installed or...  The output says " Metric (TFM) file not found." but I am not sure where to go from there.