Wireless not working on Dell XPS 13 9365 Ubuntu 16.04
How to get Intel Dual Band Wireless-AC 3168 work on Ubuntu 16.04?
What is the difference between new/delete and malloc/free? So the vending machine tore my $5 in pieces. What now? The vending machine swallowed half of my $5 bill, and the other half remained in my hand. Can I get my $5 back, or is it all lost forever? What is the minimum $ \sigma$-algebra that contains open intervals with rational endpoints What are the minimum $\sigma$-ring and $\sigma$-algebra on $\mathbb R$ which contain the open intervals with rational endpoints?  Is there a relation between this $\sigma$-algebra and Borels? I have a problem with the interpretation of a test result in which the p-value is 8.6e-28. How should it be interpreted? What is the e for?  Specifically, I used White's Test for the homoscedasticity assumption of a linear regression:  White's general test statistic :  150.0535  Chi-sq( 9)  P-value =  8.6e-28 What are the differences between 32-bit and 64-bit, and which should I choose? What is the difference between 32-bit and 64-bit Ubuntu? I've heard the 64-bit platform performs better and can detect more than 4GB of RAM. Also, while some apps haven't ported to 64-bit yet, ia32-libs lets a 64-bit machine run them.  If so, why not promote 64-bit over 32-bit? What does NOEXEC flag mean when mounting directories on RHEL? I am trying to understand the NOEXEC flag when mounting.  I am having an execution issue within the /tmp directory on someone elses machine that I cannot access atm where the /tmp directory is mounted onto a different drive than '/' and NOEXEC is present.  I wanted to try and recreate this scenario on my machine, but I do not have a second hard drive.  I tried doing the following command:  mount --bind /test1 /test2   I then removed the bind flag and added NOEXEC in /etc/fstab.  Then, I created a file in /test2 called test.sh where it just echos 'hello world'.  I try and run it and it said 'permission denied'.  I then ran chmod 777 test.sh and was able to execute the file just fine.  I thought that the NOEXEC flag should not allow me to execute anything?  Is mount --bind /test1 /test2 not the same as mounting from a completely different physical drive? As in /test1 and /test2 are on different drives? Why upload to academic preprint sites like arXiv? The recent question  made me realize that I don't have any understanding of why one would upload to a preprint site.  (I don't work in a field where they seem to be common.)  Why would one be concerned about getting a manuscript on arXiv, if it's already under review with a journal?   Is it because of the long publishing timeline of journals?  (That could explain why computer science &mdash; which publishes much more in conferences &mdash; doesn't seem to do much with them.) My linear algebra professor proved that      Diagonalizable matrices with complex values are dense in set of $n \times n$ complex matrices.   He defined a metric (I believe) that was somehow related to the usual metric on $\mathbb{R}^{n^2}$.  Then somehow proved that diagonalizable matrices were dense because for any matrix $A$ if $\det(A - \lambda I) = 0$ on an open subset, then $\det(A - \lambda I)$ was the zero polynomial.  I Googled around a bit and found some stuff talking about the Zariski topology, and I am not sure this is what I want. Most of what I have found on this topology is much more general than what he was doing.  Does anyone know any resources that would explain how to go about proving this statement?  Edit: I found out how to prove this the way my professor did.  We define the norm of a matrix by $$ |A| = \max\{|Ax| : |x| = 1 \}.$$ Now we have a distance $d(A, B) &lt; \epsilon$. You can prove that if $(A - B)_{ij} &lt; \epsilon/n$, then $|A - B| &lt; \epsilon$.  Let $p(x)$ be the characteristic polynomial of $A$, an $n \times n$ matrix.  $$p(x) = \prod_1^n (x - x_i) = x^n + (-1)\sigma_1 x^{n - 1} + \cdots + (-1)^n\sigma_n$$ where $\sigma_1, \ldots, \sigma_n$ are the elementary symmetric polynomials of $x_1, \ldots, x_n$ which are the eigenvalues.  The discriminant of the characteristic polynomial is a symmetric polynomial, therefore it can be written in terms of the elementary symmetric polynomials, which in turn can be written in terms of the entries of the matrix.  But since the discriminant is a polynomial, it only has finitely many roots. Therefore for $\epsilon &gt; 0$, by changing the entries of the matrix less than $\epsilon / n$ we can find a new matrix $B$ such that $|B - A| &lt; \epsilon$ and the discriminant is not zero.   The discriminant being not zero means $B$ has distinct eigenvalues, thus has a basis of eigenvectors. Therefore, it is diagonalizable.  Thus the set of diagonalizable matrices is dense in the set of matrices with respect to that metric. Remove border from IFrame How would I remove the border from an iframe embedded in my web app? An example of the iframe is:  &lt;iframe src="myURL" width="300" height="300"&gt;Browser not compatible.&lt;/iframe&gt;   I would like the transition from the content on my page to the contents of the iframe to be seamless, assuming the background colors are consistent. The target browser is IE6 only and unfortunately solutions for others will not help. Extract data from a file and place in different files based on one column value We will generate a csv file with below values  yp1234,577,1,3 yp5678,577,3,5 yp9012,132,8,9   I need to extract data and create files based on second column. If it's 577 then the whole line has to be extracted and placed in a separate file. I mean I need a file having lines with second column as 577 alone and another file with second column as 132 alone  I tried using IF but didn't work How can you programmatically tell an HTML SELECT to drop down (for example, due to mouseover)? How can you programmatically tell an HTML select to drop down (for example, due to mouseover)? I came across this problem when I was at a book store inside of a book made to prepare Berkeley graduates to pass a mandatory exam. I wanted to buy the book, but, alas, I didn't have the money (forty bucks is a lot of money when you don't have a job). So I took my phone out and started taking as many pictures as I could. Unfortunately, I didn't take any pictures of the solutions!  Enough backstory. Time for math. The problem is as follows:  Prove that $\cos^p( \theta) \le \cos(p \theta)$ if $0\le\theta\le\frac\pi2$ and $0\le p\le 1$.  I tried using the series expansion for cosine, but that seemed to be a dead end. Then I tried using Euler's theorem, but I got stuck. Then I got distracted and started to think about other cosine identities. For example, $$\cos\left(\frac\theta2\right)=\pm\sqrt{\frac{1+\cos(\theta)}{2}}.$$ Then what is $\cos(\frac\theta3)$ equal to? I tried to figure it out then I realized that I needed to find the root of a cubic polynomial. Then I realized that I'm better at staying focused than finding the roots of cubic polynomials.  Anyways, a proof would be nice here. I really appreciate any hints or answers. I apologize for my digressions! Let $X$ be an infinite cofinite space and let $A \subset X$ be infinite. Then $\overline A = X$  I am having a tough time in trying to understand this. Why is every infinite set in an infinite cofinite space dense ? "If I would have lost you" vs "If I had lost you" I watched a (Hollywood?) film the other day where a character visiting his just-hospitalised wife (who it seems will actually survive) says "If I would have lost you [I don't know what I'd do]".  I'd always thought this was an error made by non-native speakers (I myself would only expect "If I had lost you"), but there was nothing to suggest that in the movie. Is it a regionalism?  EDIT: Actually, that wasn't true about only expecting "If I had lost you". I'd have been perfectly happy with "If I'd have lost you", so long as I didn't think too hard about what was being elided. I'd usually elide "have" there to just a schwa (neutral vowel) anyway, and think no more about it. I am trying to balance underbraces in the equation below, how can I ensure that both underbraces at the same depth?  \begin{equation} y = \underbrace{ \left [ x^2 + r^2 (a^2+\frac{a^{0.3}}{b}) \right]}_\text{growth rate} \times  \underbrace{ C_0}_\text{initial} \end{equation} Every open set in $\mathbb{R}$ is a countable union of closed sets Was the Wookiee language fully designed or is it just random emotionally-relevant sounds? We know that some of the other Star Wars languages were fairly well developed by the movie's creators, in terms of vocabulary, grammar, etc.  Was the Wookiee language designed in this level of detail or did they just try to put together sounds that made sense emotionally? In categorical terms, why is there no canonical isomorphism from a finite dimensional vector space to its dual? I've read in several places that one motivation for category theory was to be able to give precise meaning to statements like, "finite dimensional vector spaces are canonically isomorphic to their double duals; they are isomorphic to their duals as well, but not canonically."  I've finally sat down to work through this, and -  Okay, yes, it is easy to see that the "canonical isomorphism" from $V$ to $V^{**}$ is a functor that has a natural isomorphism (in the sense of category theory) to the identity functor.  Also, I see that there is no way that the functor $V\mapsto V^*$ could have a natural isomorphism to the identity functor, because it is contravariant whereas the identity functor is covariant. My question amounts to:     Is contravariance the whole problem?   To elaborate:   I was initially disappointed by the realization that the definition of natural isomorphism doesn't apply to a pair of functors one of which is covariant and the other contravariant, because I was hoping that the lack of a canonical isomorphism $V\rightarrow V^*$ would feel more like a theorem as opposed to an artifact of the inapplicability of a definition.  Then I tried to create a definition of a natural transformation from a covariant functor $F:\mathscr{A}\rightarrow\mathscr{B}$ to a contravariant functor $G:\mathscr{A}\rightarrow\mathscr{B}$. It seems to me that this definition should be that all objects $A\in\mathscr{A}$ get a morphism $m_A:F(A)\rightarrow G(A)$ such that for all morphisms $f:A\rightarrow A'$ of $\mathscr{A}$, the following diagram (in $\mathscr{B}$) commutes:  $$\require{AMScd}\begin{CD} F(A) @&gt;m_A&gt;&gt; G(A)\\ @VF(f)VV @AAG(f)A\\ F(A') @&gt;&gt;m_{A'}&gt; G(A') \end{CD}$$  This is much more stringent a demand on the $m_A$ than the typical definition of a natural transformation.  Indeed, it is asking that $m_A=G(f)\circ m_{A'}\circ F(f)$, regardless of how $f$ or $A'$ may vary. Taking $\mathscr{A}=\mathscr{B}=\text{f.d.Vec}_k$, $F$ the identity functor and $G$ the dualizing functor, it is clear that this definition can never be satisfied unless $m_V$ is the zero map for all $V\in\text{f.d.Vec}_k$ (because take $f$ to be the zero map). In particular, it cannot be satisfied if $m_V$ is required to be an isomorphism.     Is this the right way to understand (categorically) why there is no natural isomorphism $V\rightarrow V^*$?   As an aside, are there any interesting cases of some kind of analog (the above definition or another) of natural transformations from covariant to contravariant functors?  Note: I have read a number of math.SE answers regarding why $V^*$ is not naturally isomorphic to $V$. None that I have found are addressed to what I'm asking here, which is about how categories make the question and answer precise. ( was closest.) Hence my question here. Can more than one antibody bind the same antigen I'd like to clear something up about antibodies that I'm not sure I've understood in the articles I've read. Looking at concepts such as "affinity maturation", "monovalent antigens" and "polyvalent antigens" it seems as if there are multiple antibodies which can bind to the same antigen. (Although I think I'm right in saying that individual antibodies can only bind to one antigen?) However, when I read sentences like "It has been estimated that humans generate about 10 billion different antibodies, each capable of binding a distinct epitope of an antigen" on the Wikipedia page (for antibody) it does make it sound like it's one antibody per antigen? Is this the case? Or is it one antibody per epitope. Either way, to put this question in a nutshell:  Is it true to say that the same substance could bind with (and, therefore, be recognised by) more than one antibody, or not?  Also, if the answer is no, does that mean that antibodies with different binding affinities are classed as the same antibody or not? When I open a window to air out the room, how does the smell disperse? Let's say I'm in a room with some kind of noxious stink, possibly of flatulent nature.  The quickest way to right the world that comes to mind is to open a window.  When I open a window, how do the stank particles leave the room?