Collision between stationary and moving vehicle
Car accident, put in park or neutral?
The title word filter is one of the worst ideas ever implemented on SO There are numerous (, , ) questions concerning this problem. I realize this is a borderline duplicate, but I'll re-iterate the futility of this particular filter and why it should be outright removed from Stack Overflow.  Do you know how effective this is in preventing users who just don't care from posting questions with bad titles? 100% ineffective. What do these users do?   Replace it with a synonym: "issue," "trouble," "predicament," "botheration" or what have you. Or worse, they simply work around the filter by deliberately misspelling the word or inserting space or punctuation:    () ?    and the coup de grâce:      I mean, even Jeff . This adds zero value to the title. In fact, it looks worse. Why, oh why are people  to scrap a bad feature? All this does is:   Prevent people who have a  to use the word "problem" in the title from posting questions with an appropriate title. Break the search feature because of workarounds or the need to use less descriptive titles than the error itself   This is an extraordinarily useless (and harmful) feature. Please, for the love of Cthulhu,  remove, eliminate, abolish, eradicate, extirpate this "feature" from Stack Overflow.    Bonus:     So I can't mention the Halting Problem in the title? – Keith Thompson Nov 1 at 19:05      @KeithThompson Maybe it's a "Halting Issue". Or "Need help with Halting". Or the "Halting thing in Computability Theory". – slhck Nov 1 at 19:14 Nesting HTML- anchor tags Show that the series $$\sum\limits_{n=2}^{\infty} \frac {(n^3+1)^{1/3}-n}{\log n}$$ converges.   I showed it using Abel's theorem and limit comparison test. Any other simpler method? breaking of output into new lines I created a script ./abc.  I want to break the ouput into new lines.  mycode:  #!/bin/bash dir="$1"  echo -n "size of file:" du -hba $dir | sort -n -r | tail -1   my output is:   size of files: 20kb /path/to/sport   I want the output as:  size of files: 20kb          /path/to/sport I am running on a dual core AMD E450 CPU. When I tried to run a 64-bit OS that requires hardware virtualization using VirtualBox, it showed me an error message that "AMD-V is not enabled".    My AMD processor is supposed provide AMD-V support, but I can find no option for enabling AMD-V in BIOS.    How can I solve this problem? How can I enable AMD-V for my CPU?  Thanks in advance  lscpu :-  Architecture:          x86_64 CPU op-mode(s):        32-bit, 64-bit Byte Order:            Little Endian CPU(s):                2 On-line CPU(s) list:   0,1 Thread(s) per core:    1 Core(s) per socket:    2 Socket(s):             1 NUMA node(s):          1 Vendor ID:             AuthenticAMD CPU family:            20 Model:                 2 Stepping:              0 CPU MHz:               1650.000 BogoMIPS:              3291.72 Virtualization:        AMD-V L1d cache:             32K L1i cache:             32K L2 cache:              512K NUMA node0 CPU(s):     0,1   EDITED:-  Error message from VirtualBox:-  Failed to open a session for the virtual machine XXX.  AMD-V is disabled in the BIOS. (VERR_SVM_DISABLED).  Result Code: NS_ERROR_FAILURE (0x80004005) Component: Console Interface: IConsole {1968b7d3-e3bf-4ceb-99e0-cb7c913317bb} Screen brightness isn't taking effect on a Lenovo Z570 My Lenovo Z570 has two GPU's, nVidia and Intell. Brightness control has no effect what so ever over the screen brightness, and because I've had some major issues in the past when I was trying to fiddle with nVidia drivers and settings, and I wanna be cautious now. I haven't find similar questions referring 14.04 LTS with similar configuration as mine.   I've read that it could be fixed by adding/changing content in xorg.conf but when I run  sudo gedit /etc/X11/xorg.conf   the file comes out to be absolutely empty. What do I need to do, what do I need to add to it and not get my laptop screwed with a black screen again? Can a probability distribution value exceeding 1 be OK? On the , there is this line:      $p(\mathrm{height}|\mathrm{male}) = 1.5789$ (A probability distribution over 1 is OK. It is the area under the bell curve that is equal to 1.)    How can a value $&gt;1$ be OK? I thought all probability values were expressed in the range $0 \leq p \leq 1$. Furthermore, given that it is possible to have such a value, how is that value obtained in the example shown on the page? Arbitrary precision spline interpolation $f,g$ continuous from $X$ to $Y$. if they are agree on a dense set $A$ of $X$ then they agree on $X$ Problem:      Suppose $f$ and $g$ are two continuous functions such that $f: X \to Y  $ and $g : X \to Y $. $Y$ is a a Hausdorff space. Suppose $f(x) = g(x)  $ for all $x \in A \subseteq X $ where $A$ is dense in $X$, then $f(x)  = g(x) $ for all $x \in  X $.   Attempt at a solution:  Put $h(x) = f - g $. Therefore, $h: X \to Y $ is continuous and $Y$ is Hausdorff by hypothesis. Also we know $h(x) = 0 $ for all $x \in A $ such that $A$ is dense in $X$. I want to show that $h(x)$ vanishes everywhere in $X$. We can show $h(x) = 0 $ for all $x \in X \setminus A $. Suppose $h(x) &gt; 0 $ on $X \setminus A$. Pick points $y_1,y_2 \in Y $. Since $Y$ is Hausdorff, can find open set $O_1, O_2 \subseteq Y $ which are disjoint such that $y_1 \in O_1$ and $y_2 \in O_2$. By continuity, $f^{-1}(O_1), f^{-1}(O_2)$ are open in $X$.   I know that if I can show that one of the $f^{-1}(O_i)$ lies in $X \setminus A $, then we would have a contradiction since we have non-empty open set in $X \setminus A$ and this implies $A$ cannot be dense in $X$. But this is the part I am stuck. Any help would greatly be appreciated.  Also, Would be be possible to prove this without using the Hausdorff condition on $Y$? How to find and compare flights carried out by a specific type of plane? Is it possible to interrupt spellcasting? Kinetic energy conservation in collision.  In any type of collision type of collision if we take the system to comprised of both bodies then the net external force is zero. So the work done by external force is also zero. By the relation $W=K_f-K_i$ as $W$ is zero hence kinetic energy is conserved. But this not the case.   Can you tell me what is the flaw in my argument? Some people say that the argument has neglected the fact that some kinetic energy is transformed into heat. But heat is created due to the interactions between bodies during collision which is thus an internal force and thus shouldn't affect kinetic energy. Having Overfull and underfull box warnings. Please help out as there are many matrices and equations Example where Hamiltonian $H \neq T+V=E$, but $E=T+V$ is conserved I'm looking for an example of a Hamiltonian $H$, where $H\neq T+V$, but the total energy in the system, $E=T+V$, is still conserved.   While I'm at it, I might as well add that I'd be most interested in an example from a classical field theory. Furthermore, I am looking for a nontrivial example, e.g. where $H$ doesn't just differ from $E$ by some gradient function that integrates to 0 in the action.   In particular, I am trying to get a better understanding of the physical implications of $H$ being conserved when nontrivial variables are used, i.e. physically, what does $H$ this correspond to. Conversely, can one have a system where $E = H$ in one frame, and is conserved, while in another frame $E'\neq H'$, and $H'$ is conserved while $E'$ is not? Natbib puts reference data one line per item I'm using:  \documentclass[oneside,article,openbib,12pt,titlepage]{memoir} \usepackage[authoryear,round]{natbib} % Better bibliography support \bibpunct{(}{)}{,}{a}{,}{,}   in the preamble of my document and  \backmatter \renewcommand{\bibname}{References} \bibliographystyle{chicago}    \bibliography{listb,mybib}   in the body of the file. But a typical formatted reference is:     Freely, I. (1997).   A small paper.   The journal of small papers -1. to appear.   I've tried other non-natbib styles (e.g., regstud), and the reference flows and wraps as it should. So how does one get natbib to do the same?  Also, how does one get natbib to include the authors' entire first names? Adjective meaning "argumentative for argument's sake"? Emulate "double" using 2 "float"s Proving $\sum_{k=1}^n\frac1{\sqrt k}<2\sqrt n$ by induction I am just starting out learning mathematical induction and I got this homework question to prove with induction but I am not managing. $$\sum_{k=1}^n\frac1{\sqrt k}&lt;2\sqrt n$$ Please help! Computing the cardinality of a combinatorial set Define two sets: $S_1$ and $S_2$ where we have that $a$ is an odd natural number and $b\in \mathbb{N}$, $S_1 = \Big\{ \frac{2}{a},\frac{2}{a-2},\frac{2}{a-4},...,\frac{2}{a-(a-1)},$ $\frac{2+4}{a+2},\frac{2+4}{a},\frac{2+4}{a-2},...,\frac{2+4}{a-(a-1)},$ $\frac{2+8}{a+4},...,\frac{2+8}{a-2},...,\frac{2+8}{a-(a-1)},$ $\vdots$ $\frac{2+4\lfloor\frac{b}{2}\rfloor}{a+\lfloor\frac{b}{2}\rfloor},\frac{2+4\lfloor\frac{b}{2}\rfloor}{a+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+4\lfloor\frac{b}{2}\rfloor}{a-(a-1)},$ $\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a+\lfloor\frac{b}{2}\rfloor},\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+4(\lfloor\frac{b}{2}\rfloor+1)}{a-(a-1),}$ $\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a+\lfloor\frac{b}{2}\rfloor-2},\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a+\lfloor\frac{b}{2}\rfloor-4},...,\frac{2+4(\lfloor\frac{b}{2}\rfloor+2)}{a-(a-1)}$ $\vdots$ $\frac{2+4b}{a},\frac{2+4b}{a-2},...,\frac{2+4b}{a-(a-1)} \Big\}$ and $S_2 = \Big\{ \frac{2}{\lceil\frac{a}{2}\rceil}, \frac{2}{\lceil\frac{a}{2}\rceil-1},\frac{2}{\lceil\frac{a}{2}\rceil-2},...,\frac{2}{1}$ $\frac{2+2}{\lceil\frac{a}{2}\rceil+1},\frac{2+2}{\lceil\frac{a}{2}\rceil},\frac{2+2}{\lceil\frac{a}{2}\rceil-1},...,\frac{2+2}{1},$ $\vdots$ $\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{\lceil\frac{a}{2}\rceil+(\lfloor\frac{b}{2}\rfloor-1)},\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{\lceil\frac{a}{2}\rceil+(\lfloor\frac{b}{2}\rfloor-2)},...,\frac{2+2(\lfloor\frac{b}{2}\rfloor-1)}{1}$ $\frac{2+2\lfloor\frac{b}{2}\rfloor}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor},...,\frac{2+2\lfloor\frac{b}{2}\rfloor}{1},$ $\frac{2+2(\lfloor\frac{b}{2}\rfloor+1)}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor-1},...,\frac{2+2(\lfloor\frac{b}{2}\rfloor+1)}{1},$ $\frac{2+(\lfloor\frac{b}{2}\rfloor+2)}{\lceil\frac{a}{2}\rceil+\lfloor\frac{b}{2}\rfloor-2},...,\frac{2+(\lfloor\frac{b}{2}\rfloor+2)}{1}$ $\vdots$ $\frac{2+2b}{\lceil\frac{a}{2}\rceil-1},...,\frac{2+2b}{1} \Big\}$ The question is, how do I determine the cardinality of the intersection of these two sets i.e. how would I compute the value $|S_1 \cap S_2|$? Note that duplicates may appear within both sets such as $\frac{1}{2}=\frac{2}{4}$ etc. Therefore we must make sure that for each fraction $\frac{n}{m}$ we have $\text{gcd}(n,m)=1$. I am aware of the existence of , but not entirely sure if that identity would be useful here. Set theory: difference between belong/contained and includes/subset? This is a total noob question.  I am reading , and I'm having difficulty to understand something which seems to be trivial.  In the first chapter he writes:     If $x$ belongs to $A$ ($x$ is an element of $A$, $x$ is contained in $A$), we shall write      $x\in A$   I understand this.  Then, he write:     If $A$ and $B$ are sets and if every element of $A$ is an element of $B$, we say that $A$ is a subset of $B$, or $B$ includes $A$, and we write:      $A \subset B$   I understand this too.  Then he says:     The working of the definition implies that each set must be considered to be included in itself ($A \subset A$); this fact is described by saying that set inclusion is reflexive.   I understand this too.  But then:     Observe that belonging ($\in$) and inclusion ($\subset$) are conceptually very different things indeed. One important difference has already manifested itself above: inclusion is always reflexive, whereas it is not at all clear that belonging is ever reflexive. That is: $A \subset A$ is always true;  is $A\in A$ ever true? It is certainly not true of any reasonable set that anyone has ever seen.   And this is where I don't think I understand anything.  There is not more elaboration on this point in the text.  I tried to skip this but it seems it is quite fundamental for understanding what follows in the book.  Could someone explain what is meant here?