First, I ran:  sudo chmod -R -777 /usr    Now, my /usr dir permissions looks like this:  d---------   So nobody has permissions I can't even get in the dir or run bash commands.
I accidently ran  sudo chmod 755 -R /   instead of  sudo chmod 755 -R ./   I stopped it after few seconds, but now there is some problems such as  sudo: must be setuid root   How can I revert permissions back?
AlwaysOn AG: Current Host Server - Owner Node - Primary Instance Continuity of Integration for a measurable function I am having difficulties with this problem:  Let $f$ be a measurable function on $R$. Show that $\lim\limits_{n \to \infty} \int\limits_{-n}^{n} | f | = \int\limits_{R} | f | $  My idea is to use Continuity of integration: We need an ascending collection of measurable sets. These are $E_n = R \cap [-n, n]$ $R = \bigcup E_n$ Now if we apply Continuity of Integration we have:  $\lim_{R} |f| = \lim\limits_{n \to \infty} \int_{E_n} =\lim\limits \int_{-n}^{n} |f |$. My problem is Continuity requires f to be intergrable and all i know is that f is measurable.  Another way i was thinking to use monotone convergence theorem, but i am not sure how to create the sequences.  Any Ideas? thanks Add Item to Finder Sidebar There's something wrong with the vertical alignment on the :   (click to enlarge)  On the , which uses a similar list but without the Favorite column, the vertical alignment is correct.  (Tested in the latest Firefox, Safari and Chrome on macOS Catalina.) How do I determine whether or not two lines intersect, and if they do, at what x,y point? How can I set desktop defaults for new users? How can I set up desktop defaults, much like most distributions do, for every new user? (ex. default wallpaper, apps, etc) I want to find the interval of convergence of $$\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(2x)^n$$ By the root test, $\sqrt[n]{|(\frac{n}{n+1})^{n^2}(2x)^n|}=|(\frac{n}{n+1})^{n}(2x)|$  And $(\frac{n}{n+1})^n=(1-\frac{1}{n+1})^n\to\frac1e$ as $n\to\infty$  So I've got an open interval(of convergence) s.t. $|\frac1e 2x|&lt;1\implies |x|&lt;\frac e2$.  Now I need to figure out whether the series is convergent or not at the end points, $x=\frac e2, -\frac e2$.  So there are two series I need to deal with, namely, $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}e^n$ and $\sum_{n=0}^\infty (\frac{n}{n+1})^{n^2}(-1)^ne^n$.  Suppose $(\frac{n}{n+1})^{n^2}$ is increasing. Then $$(\frac{n}{n+1})^{n^2}&lt;(\frac{n+1}{n+2})^{(n+1)^2}\iff 1&lt;(\frac{n+1}{n+2})^{(n+1)^2}(\frac{n+1}{n})^{n^2}$$ $$\iff 1&lt;(1-\frac{1}{n+2})^{2n+1}(1+\frac{1}{n^2+2n})^{n^2}$$  But $(1-\frac{1}{n+2})^{2n+1}\to\frac{1}{e^2}$ and $(1+\frac{1}{n^2+2n})^{n^2}&lt;(1+\frac{1}{n^2})^{n^2}\to e$  So I have $1&lt;\frac1e&lt;1$, which is a contradiction. Hence $(\frac{n}{n+1})^{n^2}$ is not increasing. But I don't know if it's decreasing. (is it true that $(1+\frac{1}{n^2+2n})^{n^2}\to e$?)  If $(1+\frac{1}{n^2+2n})^{n^2}\to e$ is true, then $(\frac{n}{n+1})^{n^2}e^n$ converges to some positive number, so $\sum(\frac{n}{n+1})^{n^2}e^n$ diverges. But I don't know if it's decreasing: can't apply the alternating series convergence test(Leibniz's criterion).  So, can you help me with the two end points? Disable all software installation A loss and gain problem This is a very simple but confusing puzzle.  A customer buys goods worth $200$ rupees from a shop. The shopkeeper selling these goods makes zero profit from this purchase.  The lady gives him a $1000$ rupee note.  The shopkeeper has no change, so he goes next door to another shopkeeper to get change for the $1000$ rupee note. He keeps $200$ for himself and returns $800$ to the customer.  Later, the second shopkeeper from next door comes back with the $1000$ rupee note with a stamp on it saying "counterfeit" and takes his money back.  How much loss does the first shopkeeper face? NSInvalidUnarchiveOperationException: Could not instantiate class named NSLayoutConstraint Updating list definition using features I have a feature that deploys content types, a list definition and template that uses these content types and finally a list instance. This feature is already in production.  Now we have a change request to add new views to this existing list. So I make the changes to schema.xml and upgrade the solution. But alas, the new views don't show up. If I create a new site, the list instance is created with the new views. I tried deactivating/reactivating, iisreset but still no success. Is there anything I'm missing? What is model identifiability? I know that with a model that is not identifiable the data can be said to be generated by multiple different assignments to the model parameters. I know that sometimes it's possible to constrain parameters so that all are identifiable, as in the example in Cassella &amp; Berger 2nd ed, section 11.2.  Given a particular model, how can I evaluate whether or not it's identifiable? Dirac delta function and volume charge density I just got introduced to the Dirac delta function and one of the questions was to express volume charge density $\rho({\bf r})$ of a point charge $q$ at origin. I saw that the answer is related to Dirac delta function as: $$\rho({\bf r}) = q \delta^3({\bf r})$$ where $\delta^3({\bf r})$ is the 3-dimensional Dirac Delta function. Why is it so? .toArray(new MyClass[0]) or .toArray(new MyClass[myList.size()])? Assuming I have an ArrayList  ArrayList&lt;MyClass&gt; myList;   And I want to call toArray, is there a performance reason to use  MyClass[] arr = myList.toArray(new MyClass[myList.size()]);   over  MyClass[] arr = myList.toArray(new MyClass[0]);   ?  I prefer the second style, since it's less verbose, and I assumed that the compiler will make sure the empty array doesn't really get created, but I've been wondering if that's true.  Of course, in 99% of the cases it doesn't make a difference one way or the other, but I'd like to keep a consistent style between my normal code and my optimized inner loops... Let $(x_n)_{n=1}^{\infty}$ be a sequence such that $|x_{n+1} - x_n| &lt; r^n$ for all $n \geq 1$, for some $0 &lt; r &lt; 1$. Prove that $(x_n)_{n=1}^{\infty}$ is a Cauchy sequence.  I understand that a Cauchy sequence means that for all $\varepsilon &gt; 0$ $\exists N$ so that for $n,m \ge N$ we have $|a_m - a_n| \le \varepsilon$. But this one is really giving me a headache.  I tried doing something like: let $ m &gt; n$. Therefore $x_n - x_m$ =  $(x_n - x_{n-1}) + (x_{n-1} - x_{n-2}) + ... + (x_{m+1} - x_m) $ and then somehow using the triangle inequality to compute some sum such that $x_n - x_m$  &lt; sum which would be epsilon?  any help is appreciated, thank you. Is is possible to convert a .vdi file into a .iso that can be burned to a cd or dvd and make it like an installer. Or Is it possible to convert virtual machines to physical environments? What is the word for act of monotonously demanding for something (many be a toy) repeatedly? Integrating Knowledge into an existing website How do I manage how many revisions snapd keeps of an application? Grant the OP a binding close as duplicate vote Since the  to the close as duplicate mechanism it is a lot more visible to the OP that there is/might be a duplicate to their question. Empirically, I've noticed a lot more people agreeing, in the comments under their question, that their question is a dupe.   For instance, :        I'm aware that allowing  is  (back in the stone age) but now Stack Exchange is so much larger, and Stack Overflow is groaning under the weight of questions, I think this should be partially revisited.  Based on the principle that the simplest way to keep a site clean is to have as many people as possible nibbling away, I propose that the OP gets granted a binding close-as-duplicate vote.  It should simply be an additional line in the banner, which is . It might be something like this:     This question may already have an answer here:   &nbsp; "" 3 answers   &nbsp; Does this fully answer your question?  Yes / No   I'm not wedded to the Yes / No thing I just can't think of a better way at the moment.  Clicking on Yes would take the OP to the close as duplicate dialogue with that question highlighted. If there are multiple duplicate questions you'd need to ask the OP multiple times. A No could be used to track the relative success of the project, i.e.   Whether the question subsequently gets closed. Whether the question gets reopened again. How many times the OP actually clicks. How many times the OP disagrees with the analysis of the community.   Potentially, a No could be used to open a comment dialogue asking the OP to explain why they think their question is not a duplicate thereby stopping them simply putting "this is not a duplicate" in the question and formalising this process somewhat. Not sure about this idea...  I do not think this should be restricted by reputation, anyone should be able to close their own question as a duplicate. It gives people with lower reputation a more positive experience of the closing process and emphasises the importance of duplicate finding.  Most importantly, self-closes should not, initially, count towards the question ban. We can help new users learn how to use a lot more of the site than they might otherwise see and encourage them to search for duplicates more thoroughly. If the OP then goes on to do the same thing 10 times in a row then you can start using it toward the ban as they're obviously using the users as a personalised search service but at least the first 3-4 times this should not have any effect.  This question is, deliberately, only requesting 20% of the linked question (from 2009) and then only based on the new UI. The linked question asks for binding migration votes etc, which I do not want to see and I am not requesting a binding close-as-duplicate for the OP until someone else has already voted to close (though I think if the OP has 3k rep this might be allowable). This, I hope, addresses the concern's raised in the linked question and explains why this is not a duplicate of that question.