No protocol specified Unable to init server: Could not connect: Connection refused  (nautilus:3041): Gtk-WARNING **: cannot open display: :0
I installed Ubuntu 17.10. Now I am having trouble with gksu:  $ gksu -dg synaptic No ask_pass set, using default! xauth: /tmp/libgksu-HgUjgQ/.Xauthority STARTUP_ID: gksu/synaptic/8760-0-alex-XPS-15-9530_TIME4974977 cmd[0]: /usr/bin/sudo cmd[1]: -H cmd[2]: -S cmd[3]: -p cmd[4]: GNOME_SUDO_PASS cmd[5]: -u cmd[6]: root cmd[7]: -- cmd[8]: synaptic buffer: -GNOME_SUDO_PASS- brute force GNOME_SUDO_PASS ended... Yeah, we're in... Unable to init server: Could not connect: Connection refused (synaptic:8767): Gtk-WARNING **: cannot open display: :1 xauth: /tmp/libgksu-HgUjgQ/.Xauthority xauth_env: (null) dir: /tmp/libgksu-HgUjgQ   If I don't use -g, the password dialog is disabled. So looks like a problem with creating a tty for root.  Any advice?
Windows 10: Pin an item to left side of Start Menu? I noticed that I have three warnings when I submitted my Sitemap. They all say this:      When we tested a sample of URLs from your Sitemap, we found that some   URLs redirect to other locations. We recommend that your Sitemap   contain URLs that point to the final destination (the redirect target)   instead of redirecting to another URL.   I'd like to try and resolve this issues but I'm not sure what to do.  Sitemap url for reference: Let $G$ be a finite group and suppose there exists $f\in\text{Aut}(G)$ such that $f^2=\text{id}_G$, i.e., $f$ is its own inverse, and such that $f$ has no fixed points other than the identity $e$ of $G$, i.e., $f(x)=x\Rightarrow x=e$. Show that $G$ is necessarily abelian.   While trying to do this exercise I noticed two facts.  First, $g$ and $f(g)$ have the same order because $o(f(g))|o(g)$ and, applying $f$ again and using $f^2=\text{id}_G$, $o(f(f(g)))=o(g)|o(f(g))$ and, once the order of any element is $\geq 1$, it follows that $o(g)=o(f(g))$. Also, it's easy to see that $g$ and $f(g)$ commute.  Second, there cannot exist such an automorphism if the order of $G$ is even, because $f(e)=e$ and we can form pairs like $\{g,f(g)\}$ with $f(g)\neq g$ that are invariant under $f$, i.e., $f(\{g,f(g)\})=\{g,f(g)\}$. But once the order of $G$ is even, proceeding with the construction of the pairs, we'll end up with just one element $\neq e$, so we must have some $\gamma\in G\setminus{e}$ with $f(\gamma)=\gamma$, contradicting the hypothesis.  Let $n$ be the order of $G$ and let's fix an enumeration of the elements of $G$, say $G=\{g_1,\ldots,g_n\}$. My approach was the following. For each $\sigma\in S_n$, let $x_{\sigma}=\prod_{i=1}^ng_{\sigma(i)}$. Then we have that $f(x_{\sigma})=x_{\sigma^{\prime}}$. If it's shown that $f(x_{\sigma})=e\;\forall\;\sigma\in S_n$, then we'd have $x_{\sigma}=e\;\forall\;\sigma\in S_n$, which implies that $G$ is abelian. The problem is that $S_n$ has $n!$ elements while $G$ has $n$ elements, so there repetitions among the $x_{\sigma}$, and we cannot apply directly the reasoning of the last paragraph.  Is this the right way, or is there an easier manner to solve this? I (think) that I've configured a static IP address in /etc/network/interfaces:  # The primary network interface  auto eth0  iface eth0 inet static     address 10.1.1.2     netmask 255.255.255.0     network 10.1.1.0     broadcast 10.1.1.255     gateway 10.1.1.1   And I change the file /etc/resolv.conf to include the desired nameservers.   Then I restart networking sudo /etc/init.d/networking restart  But after some hours resolv.conf always reverts back to the DHCP nameserver.   I should also say that this is Ubuntu Server with only l0 and eth0 interfaces. I found the below in  web-site:     We use much in questions and negative clauses to talk about degrees of something.      We don’t use much in affirmative clauses.   Also, it the same web-site, I found the following:     She runs much faster than he does.    also I noticed that some web-sites use much with affirmative clauses:     May we all be having this much fun and getting this much exercise when we're his age!    Is it Okey to use much in affirmative clauses? If yes, when can we use it? and what is the rules?  Update:  Here is another example on how often the word "much" is used  in informal style:     After much consideration I will NOT be performing at the Inauguration with my improv group Capitol Hill-arious When I wake the laptop from the sleep I need to get the VPN state as I left it (turned on). Can't find the setting of autoconnect in VPN settings... Need some easy way to ask the system to autoconnect the VPN when the internet is available.  Edit based on comments:  Ubuntu 18.04 has no option to Always connect to VPN when using using this connection. reason for exec in wrapper scripts I have seen wrapper script examples which in a nutshell are following:  #!/bin/bash  myprog=sleep echo "This is the wrapper script, it will exec "$myprog""  exec "$myprog" "$@"   As seen above, they use exec to replace the newly created shell almost immediately with the $myprog. One could achieve the same without exec:  #!/bin/bash  myprog=sleep echo "This is the wrapper script, it will exec "$myprog""  "$myprog" "$@"   In this last example, a new bash instance is started and then $myprog is started as a child process of the bash instance.  What are the benefits of the first approach? How does Java's serialization work and when it should be used instead of some other persistence technique? I've been lately trying to learn more and generally test Java's serialization for both work and personal projects and I must say that the more I know about it, the less I like it. This may be caused by misinformation though so that's why I'm asking these two things from you all:  1: On byte level, how does serialization know how to match serialized values with some class?  One of my problems right here is that I made a small test with ArrayList containing values "one", "two", "three". After serialization the byte array took 78 bytes which seems awfully lot for such low amount of information(19+3+3+4 bytes). Granted there's bound to be some overhead but this leads to my second question:  2: Can serialization be considered a good method for persisting objects at all? Now obviously if I'd use some homemade XML format the persistence data would be something like this  &lt;object&gt;     &lt;class="java.util.ArrayList"&gt;     &lt;!-- Object array inside Arraylist is called elementData --&gt;     &lt;field name="elementData"&gt;         &lt;value&gt;One&lt;/value&gt;         &lt;value&gt;Two&lt;/value&gt;         &lt;value&gt;Three&lt;/value&gt;     &lt;/field&gt; &lt;/object&gt;   which, like XML in general, is a bit bloated and takes 138 bytes(without whitespaces, that is). The same in JSON could be   {     "java.util.ArrayList": {         "elementData": [             "one",             "two",             "three"         ]     } }   which is 75 bytes so already slightly  smaller than Java's serialization. With these text-based formats it's of course obvious that there has to be a way to represent your basic data as text, numbers or any combination of both.  So to recap, how does serialization work on byte/bit level, when it should be used and when it shouldn't be used and what are real benefits of serialization besides that it comes standard in Java? Can we make it more plain to recruiters that we're not open to relocation? The preference for being open to relocation is a checkbox, and if you don't check it, nothing about relocation shows up in your profile.  Evidently recruiters can't understand that a lack of "I am open to relocation" is the same thing as "I am not open to relocation", and so every single contact I've gotten through Careers is for a job that I would have to relocate for.  Can we make it more obvious to the recruiters (especially ones from Amazon) that we have not checked that box on purpose? CAML query daterange not working What do I say to make a new line when using voice recognition? I've found I can make punctuation happen by saying "comma", or "period", but I can't find the word for new line.  I tried "paragraph", and "line break", but it just prints those words. Can I Turn Off or Remove Map Layers in QGIS Print Composer The title pretty much says it all.  I have a base map in QGIS that I am using with several Print Composers.  In one of the Print Composers, I don't want one of the layers to show up.  Is there any way to accomplish this? What does a binomial coefficient with commas mean? I'm reading through  paper (on Dyck Paths). Near the middle of the second page, the author states the following:     Remark For the set $h_n$ there are ${k\choose t_1, t_2, ... , t_m }$ $n + k \choose{k}$ $=$ $ n + k\choose {n, t_1, t_2, ..., t_m}$ different Dyck paths.   What does ${k\choose t_1, t_2, ... , t_m }$ mean? I know what the binomial coefficient is, but I'm not sure how to interpret it when one of the parameters is a set. Is there a canonical explanation for how Ant-Man's weight works when he shrinks down that's consistent with the various events depicted in the film?    I think the explanation given in the film is roughly "the suit compresses the space in-between atoms, so your size decreases, your density increases, and your total mass stays the same", however it seems to be portrayed in contradictory ways.  For instance:   He can float in water, and even be sent shooting through the air by a relatively gentle flow of water filling a bathtub.  This wouldn't happen if he retained his original mass, as he'd essentially be the densest thing on the planet and should sink like a stone even in a vat of molten lead. Upon landing after being ejected from the bathtub, the ceramic tile underneath him shatters.  That takes a significant amount of force, and similar effects are used at other points in the movie where he falls onto (or straight through) various things. He's able to get sucked into a vacuum-cleaner.  One that didn't even appear to be a high-end, extra powerful vacuum-cleaner, let alone one capable of sucking up a 200 pound piece of dust.   He can run along the barrel of a gun as someone shoots it, apparently without causing them to drop the weapon or even have trouble continuing to aim it.     Does a more consistent explanation exist in other media, that would fit with the above examples?  Or is this simply a case of the power being made to do whatever happened to be most convenient at the time? QThread and QTimer Proof by induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Prove via induction that $\sum_{i=1}^n \frac{i}{(i+1)!}=1- \frac{1}{(n+1)!}$ Having a very difficult time with this proof, have done pages of work but I keep ending up with 1/(k+2).  Not sure when to apply the induction hypothesis and how to get the result $1- \frac{1}{(n+2)!}$.  Please help! thanks guys, youre the greatest! How do I turn off the Minecraft 1.7.2 Super Secret Settings? How do I turn off the Minecraft 1.7.2 super secret settings? I tried F3 and F4, hitting Esc repeatedly, all things mentioned on blogs – nothing is working. Currently, if you access a deleted question you will get a short message informing you that the question has been deleted. However, if you click on a direct link to a deleted answer or comment (which you can get via notifications), nothing is displayed.  This is naturally quite confusing as you will get the message about a new answer or comment via notifications, click on the link and simply see nothing.  Therefore, I propose to display some sort of message to the user, informing her/him that the answer or question has been deleted. Note that this is different from always showing a deleted answer/comment, as I specifically request it in case you access the specific URL.  One example for a comment would be:   One example for a deleted answer is:   This question is related to ,  but requests a feature to circumvent the problem altogether. Applying MVT and IVT (?) to show something about f'''? (edit: actually, Taylor's Theorem using Lagrange remainder) I believe that the correct answer to this question will involve applying both the Mean Value Theorem and the Intermediate Value Theorem, perhaps several times.  However, I can't see how to proceed.  I went down what I thought were the typical paths and just reached dead ends.  Does someone maybe have a hint?  I think that with the appropriate nudge I should be able to solve this question.  If not, I'll edit and ask if someone knows how to solve the whole thing.  Here's the question:  Let $f: \mathbf R \to \mathbf R$ be such that $f$, $f'$, $f''$, and $f'''$ exist and are continuous on $\mathbf R$ and satisfying $f(-3)=-1$, $f(0)=0=f'(0)$, and $f(3)=8$.  Prove that there exists $\xi \in (-3, 3)$ such that $f'''(\xi) \ge 1$. What is a good way to draw images using pygame?