I dont know the concept or how to answer the question.     Suppose there are two sets, $X=\{1,2\}$ and $Y=\{a,b\}$.         What are all the ordered pairs in the form of (element of $X$, element of $Y$)?   What are all the ordered pairs in the form of (element of $Y$, element of $X$)?

Just reading about the cross product. It says here: $$A\times B= \{(x,y)|x\in A, y\in B\}$$ Does this mean that the result of $A\times B$ is a new set of ordered pairs? If so, what happens if both A and B share a common member? What would the result be, since an ordered pair cannot include duplicates?    Edit: Quoting the free textbook I'm looking at online,     If we are given that A is a set and no other information about A, then there is no   ordering to the elements of A. Thus, we cannot speak of “the second element of the set A”   unless we have specified an ordering of the elements of A. If we wish to regard A as ordered in some way, then we specify this fact explicitly: “The elements of A are ordered a, b, c,” or “A = (a, b, c).” The latter notation replaces the braces with parentheses and designates that A is ordered, left to right, as indicated. We call this an ordered set. An ordered set is also called a linear order. Various other names are also used: list, vector, string, word — all with no repeated elements. Of course, you’ve seen repeated elements in vectors, for example the point in the plane at the coordinates (1,1). That’s fine, it’s just not an ordered set.       Why is it okay to specify a set S = {a, b, c, a} where the element a has been repeated,   but it is not okay to have repeated elements in an ordering of S?   When we say S  =   {a, b, c, a}, we know that S contains just the three elements a, b and c.  If we were to talk   about the ordered set (a, b, c, a) it would not make sense because it would say that the   element a is in two places at once: the first position and the last position.


Post Title : Top 10 WordPress Custom Post Type Tutorials  I am going to link 10 WordPress custom post type tutorials blog posts which I found from the internet in to the above post.  Should I make those links no follow or do follow?

Forgive me for asking such a simple question, I understand why  you would want to rel nofollow untrusted links (forum posts, blog comments etc), but why would you not want to do it on EVERY link on your website?  I know it's greedy and perhaps unethical, but if I write a blog post and link to an interesting website within that post, am I losing out in any way by not rel nofollowing it?  Is my juice leaking?


I've had a few friends, predominantly female, downvoted heavily when first using Stack Overflow, and they haven't come back.  To be fair, they were asking bad questions, or asking questions badly; the downvotes aren't wrong.  But the lack of reason for the downvotes is off-putting to people just getting started in computer science.  It's not training people to ask better questions; it's training them to look elsewhere, and not growing the service in the way it could.  Suggestion  When downvoting a user with fewer than 25 rep, pop up a message to the downvoter similar to:       You're downvoting a new user; can you add a friendly comment   explaining why, so they can avoid bad questions in the future?   This is explicitly not a duplicate of ; I'm suggesting special messages when the user who made the initial post is new to the site, not just a generalized canned message when downvoting.    The canned message is already there; when the poster is new at this, give slightly different advice in the canned message.

Note: If you're looking for a simple explanation as to why comments aren't required on downvotes, see .   I used to get "upset" (though that is too strong a term) when I got downvoted without comment. If my answer isn't good enough then I'd like to know why. Not only does it improve the answer for the OP but it improves my knowledge too.  Where the down-vote has been explained I've found it useful & it has improved my answer, or forced me to delete the answer if it was totally wrong.  So is there any way we can encourage people to leave a comment? Perhaps they don't lose rep if they explain their down-vote?  I must admit though that I haven't always explained my down-vote either so you could call me a hypocrite. I've also grown a thicker skin over the months of using SO (it seems to have come with the higher rep score ;-)), so I'm less bothered about this now.


I know that if a Lie group $G$ is not compact so the map $\exp: \mathfrak{g} \to G$  is not necessarily surjective. But I am not absolutely sure if $\exp$ must be surjective if $G$ is connected.   I'm looking for an answer, maybe an example of connected Lie group for which the $\exp$ map wouldn't be surjective.  Many thanks, in advance.

A  made me realize I didn't know any proof that exponential of a Lie algebra $\mathfrak g$ of a compact connected Lie group $G$ is surjective. After a bit of thinking I've come up with two proofs. First is based on relation between exponential curves and geodesics. This is rather technical but also gives us other useful information. This is not what I want to talk about here though.  The second proof (which I find slicker) is based only on topology and goes like this: Since $\exp$ is a local homeomorphism, it is both open and closed. Therefore $\exp(\mathfrak g)$ is clopen and so equal to $G$.  The trouble with this "proof" is that it also proves the statement for $G$ non-compact (which is false). So I wonder (and this is my question) what precisely went wrong.     Can the above mentioned "proof" be made into a real proof?   My thoughts on this are that $\exp$ is closed and open only when $G$ is compact because then we can pick a bounded open subset $C \subset \mathfrak g$ such that $\exp(C) = G$  and we can use the relation $\exp(A+\epsilon B) \approx \exp(A)\exp(\epsilon B)$ to conclude that $\exp$ is a local homeomorphism everywhere in $C$ (not just around $0$). This implies that $\exp$ is open (since it is locally open) in $C$. Also, since any closed subset of $C$ is compact, it's image is also compact and therefore closed in $G$.     Where exactly does this argument break down when $G$ is not compact.


I have fitted an ARIMA model to forecast GDP using the auto.arima function in R:   Series: tGDPdev   ARIMA(2,0,2)(1,1,0)[12] with drift    Coefficients:           ar1      ar2      ma1     ma2     sar1        1.7038  -0.7521  -1.3467  0.5857  -0.4793  s.e.  0.2523   0.2290   0.2180  0.0917   0.0842         drift        0.5334  s.e.  0.0470   I would need to know the equation thath represents this model, though I'm not a mathematician. Any help would be greatly appreciated!

I would appreciate if someone could help me write the mathematical equation for the seasonal ARIMA (2,1,0) x (0,2,2) period 12. I'm a little confused with how to go about this. I would prefer an equation involving $Y_t , e_t, θ$ and $Θ$.


it is given that f is Continious. and $\forall q\in\mathbb{Q}$ $f(q)=0$    how do I prove that: for all $x\in\mathbb{R}$ :  $f(x) = 0$

The question is: Prove that if $f:\mathbb{R}\to\mathbb{R}$ is continuous and $f(x) = 0$ whenever $x$ is rational, then $f(x) = 0$ for all $x\in\mathbb{R}$.  My proof:  Let $ x\in\mathbb{R} $. If $ x\in\mathbb{Q}$, then $ f(x)=0 $.  Otherwise $ x\in\mathbb{R\backslash Q} $. Then there exists a sequence of rational numbers $ (q_n)_{n\in\mathbb{N}}$ (for example $q_n= \frac{\lfloor{nx\rfloor}}{n} ,\forall n\in\mathbb{N}$) such that $q_n\to x$ as $n\to \infty$. By the properties of $f$, we have $f(q_n)=0$ for all $n\in\mathbb{N}$. So $f(q_n)\to 0$ as $n\to \infty$. Since $f$ is continuous, $q_n\to x$ implies that $f(q_n)\to f(x)=0$. So $f(x) = 0$ for all $x\in\mathbb{R}$.   Thank you for any help.


I am confused since the definition of order says that $(ab)^n=e$, therefore does just showing that ab=ba solves both parts or is it two different solutions?  Assume $(ab)^n=e$, following the definition of order since n is a finite number.  $(ab)^n=\underbrace{(ab)(ab)(ab)\cdots(ab)}_{n~\text{copies}}$  $=a\underbrace{(ba)(ba)(ba)\cdots(ba)}_{n-1~\text{copies}}b$  $=a(ba)^{n-1} b$  From this, we get that $(ba)^{n-1}=a^{-1} b^{-1}=(ba)^{-1}$, therefore we have $(ba)^n=e$. Also proving that $ab=ba$.  Would this answer both of the questions above since then it show that both $ba^n=e$ and that the order of $ba$ would be $n$?

Attempted proof based on  and .  Given: $a,b$ elements of $G$, and $ab$ has finite order $n$.   Hence $\color{magenta}{|ab| = n} \iff (ab)^n = e$.   I must prove that $n$ is the smallest positive integer        such that $(ba)^n = e.$ Notice:  $\begin{align}(\color{darkorange}{a}\color{darkcyan}{b})^n &= e \implies \color{darkorange}{a}\color{darkcyan}{b}(ab)^{n-1} = e \implies  \color{darkcyan}{b}(ab)^{n-1}\color{darkorange}{a} = e \implies (ba)^n = e \end{align} $ $\implies |ba| \le n$.  Now I must prove that there's no positive integer $m$ such that $m<n$ and $(ba)^m = e.$ For a proof by contradiction, suppose $|ba|<\color{magenta}{n}$. Then we could apply the same reasoning to find that $ |ab| ≤ |ba| < \color{magenta}{|ab|}$,  which is absurd. So $|ba| = |ab| = n. \quad \blacksquare$  Questions   Whence does $ |ab| ≤ |ba| $ arise, in the second last line?   What's the intuition of the claim proved?    How do you prognosticate that $|ba|<\color{magenta}{n}$ is impossible, and that you should prove it by contradiction?


I am unable to see the icons in search. I tried sudo apt update, sudo apt upgrade. Logged off, Logged back in, rebooted the system. Still I am unable to see icons when I am searching. How can we resolve this? Thank you very much. :) When I go in the application section, I am able to see the icons.  But when I search with name, nothing shows up.

I did an upgrade today to the latest version of ubuntu. Everything looked good and fine only for me to discover that I cannot find my applications after clicking the 'Show Applications' icon. The same applies when I search for an application in the search bar, I do not see it. It shows me an endless search.  These are the screenshot images of the problem:      In the meantime, I access my applications from the menu at the top bar and also use the terminal to run applications with CLI.


This is a high swing cascode current mirror. My question is what is the significance of the connection from the drain of M3 to the gate of M1 and M2? What would happen if both M3 and M1 were diode connected MOSFETs?

I am currently dealing with current mirrors and came across the circuit shown below, a High-Swing Cascode Current Mirror. I read that this implementation, as the name suggests, has the advantage of a high voltage headroom, since the minimum potential at the drain terminal of M4 is: $$ V_{D4\_min}=V_{DS4}+V_{DS2}=(V_{GS4}-V_{TH})+(V_{GS2}-V_{TH}) $$ So the drain potential at M4 only needs to be twice the overdrive voltage, in case of same transistor dimensions, in order to hold M2 and M4 in saturation. But cannot this be said for many other cascode current mirrors, since that behaviour simply occurs by "stacking" together two transistors (M2 and M4)? Additionally, why is there a connection from the gate of M1 to the drain of M3? I have a hard time understanding whats going on in this circuit, help is greatly appreciated.    EDIT: A bias voltage source is setting the potential at the gate of M3/4. Sorry its not shown in the circuit.  EDIT 2: I just read that the mentioned bias voltage is implemented by a MOSFET in diode configuration, . For further discussion lets call it M5.


Many articles are structured in the following way: 1. [Title of section 1] 1.1.[title of subsection 1.1].text of subsection 1.1 1.2.[title of subsection 1.2].text of subsection 1.2. 2. [Title os section 2] 2.1.[title of subsection 2.1].text of subsection 2.1 2.2.[title of subsection 2.2].text of subsection 2.2. How can I do that (in a proper way)? Thanks.

Using the article document class, how can one remove the newline at the end of a subsubsection title?  I am using \documentclass{article}; I do not currently use titlesec or any of the alternatives; I do use fancyhdr.


What i had so far was this: claim: How to proof that for all n $$ \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^3 $$  Proof  To proof:  How to proof that for all n $$ \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^3 $$  basis $n=1$:  $$ \left(\sum_{i=1}^n i\right)^2 = \sum_{i=1}^n i^3\\ (1)^2 = (1^3)\\ 1 = 1$$  Induction $n = k$: $$ \left(\sum_{i=1}^k i\right)^2 = \sum_{i=1}^k i^3 $$ My assumption is that this is true  $n = k+1$: $$ \left(\sum_{i=1}^k+1 i\right)^2 = \sum_{i=1}^k+1 i^3\\ \left(\sum_{i=1}^k i +(k+1)\right)^2 = \sum_{i=1}^k i^3 +(k+1)^3 $$  I got stuck here and i dont know how to continue.

How can I prove that     $$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$    for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.  Thanks


In the SIMPLS formulation of partial least squares (PLS) regression, the weights are constrained to have length of 1,  $$r_a^Tr_a = 1,$$ where $a$ represents a latent component (from $1$ to $A$). This is from the original definition by de Jong (1993).  As the plsgenomics package says in the help for the pls.regression function:     In the original definition of SIMPLS by de Jong (1993), the weight       vectors have length 1. If the weight vectors are standardized to have       length 1, they satisfy a simple optimality criterion (de Jong, 1993).   I read de Jong's paper and he says:     For the development of the theory and algorithm of SIMPLS it was       convenient to choose normalized weight vectors $r_a$. This choice,        however, is in no way essential.   My question is this. Can anyone give me an intuition for what effect constraining the length of the weight vector has on the algorithm?  References:   de Jong, S. . Chemom. Intell. Lab. Syst. 18, 251–263 (1993).

In deriving the eigenvectors for PCA, the vector is subject to the condition that it should be of unit length. Why is this so?


I read this book probably 12-13 years ago, in my high school library. Here's the specifics I remember.   I'm reasonably sure the title had the word "Crystal". The story follows a girl. At the beginning of the story she visits a library, when she leaves it she meets a stranger. She thinks the stranger is going to shoot her but instead they pull out a handkerchief. She meets a boy and falls in love. She is crew on a spaceship leaving earth. He ships out to surprise her but dies in cryosleep, along with one other passenger. I'm unclear on the next bit, but the girl ends up alone on the ship. She journeys to the edge of the universe and watches it collapse, and a new Big Bang. She travels through time to her own era, and becomes the stranger her past self meets in front of the library. She becomes a scientist and clones herself. She becomes the second passenger who dies on the way to the ship. Her clones end up saving the world, I really don't recall much about the end of the book.

I remember quite a bit, but can't seem to find the title or (female) author.   A girl accompanies her father as he captains a mission to travel at near-light speed in order to experience the effects of time dilation and return to Earth 100 years into its future. She leaves behind her boyfriend (Tal or Tel or something), who promises he'll find a way to stay with her despite the time travel that will render him, well, dead before her return. She's accompanied by a close friend who is deaf but lip-reads and speaks so well as to occasionally fool others into believing she is hearing. They spend a lot of time in the ship's garden. I believe he ends up either getting on a different ship to survive into the future via time dilation, or else having himself frozen. Either way, if I recall correctly, Earth isn't doing well at all by the time the mission with the girl returns. There's been a war or some other disaster. There's an extended period of vague, ambiguous, existential crap in which the girl (now a woman) eventually ends up transferring her consciousness into her own daughter, presumably in order to go on living. Everything wraps up somehow and the girl and her boyfriend walk together hand in hand on a desolate planet into the sunset near the end of time.   It was all really rather trashy and silly, I remember, and took itself entirely way too seriously, but I'd like to at least know what it was so I could read it again. Either it'll be worth a laugh, or it'll turn out to be surprisingly good.


I am trying to /testfor 10 wheat anywhere in the players inventory.  I have tried this command:  testfor @a {SelectedItemSlot:0,Inventory:[{Count:10b,id:minecraft:wheat}]}  But it gives me this same error all the time:  The entity UUID provided is in an invalid format  How can I do this?

This is the command I entered in the command block: testfor @p {Inventory:[{id:minecraft:diamond,Count:10}]}  But it doesn't seem to work. What am I doing wrong?


I am trying to use the  reaction environment from chemmacro. How to get the numbers with chapter number like for equation numbers (e.g 1.1, 2.1 and so on)  \chapter{The First Chapter} \begin{reaction}     a + b = c \end{reaction}  \begin{equation}  a + b =2 \end{equation}

Some document elements (e.g., figures in the book class) are numbered per chapter (figure 1.1, 1.2, 2.1, ...). How can I achieve continuous numbering (figure 1, 2, 3, ...)?  And vice versa: Some document elements (e.g., figures in the article class) are numbered continuously. How can I achieve per-section numbering?  \documentclass{book}% for "vice versa" variant, replace `book` with `article`  \begin{document}  \chapter{foo}% for "vice versa" variant, replace `\chapter` with `\section`  \begin{figure} \centering \rule{1cm}{1cm}% placeholder for graphic \caption{A figure} \end{figure}  \end{document}   Bonus question: Is it possible to adjust the numbering of the sectioning headings themselves? Can I, e.g., switch from per-chapter to continuous numbering of sections in the book class?


I have the following versions of the forest and etextools packages:  Package: forest 2015/07/15 v1.0.10 Drawing (linguistic) trees Package: etextools 2010/12/07 v3.1415926 e-TeX more useful tools for LaTeX pack age writers   When I try to compile the following MWE:  \documentclass{standalone}  \usepackage{forest} \usepackage{etextools}  \begin{document}   \begin{forest}     [a [b] [c]]    \end{forest} \end{document}   I get the following error:  ! Argument of \ettl@forloop has an extra }. <inserted text>                  \par  l.9   \end{forest}   The MWE works fine if I remove the etextools include. How can I use both at the same time?

The following code uses the etoolbox package to convert a list a,b,c into an internal list (called \mylist) and then typesets that with colons: a:b:c:.  However, when I load the etextools package (by uncommenting the second line), I get no list output.   \documentclass{minimal} \usepackage{etoolbox} %\usepackage{etextools} \begin{document} A list: \def\mylist{} \forcsvlist{\listadd\mylist}{a,b,c} \def\do#1{#1:} \dolistloop{\mylist} \end{document}   I suspect it is is a bug since etextools claims to depend on etoolbox.  Is this a bug or am I doing something wrong?  In either case, how can I use \dolistloop with  etextools (which I want for goodies like \expandnext)?  Notes   \show\mylist produces the same output a|b|c| in both cases, so the problem is with \dolistloop which expands to \forlistloop\do The \forlistloop command becomes substantially more complicated when etextools is used.  might be relevant (but not obviously).


Simply do not need these. Is there a way to remove them from my machine? sudo rm -rf QuickTime\ Player.app/ didnt work :/  MacOS version 10.14.5 (18F132)  MacBook Pro (Retina, 13-inch, Early 2015)  Edit:  Error on running sudo rm -rf QuickTime\ Player.app/ (trimmed down)  rm: QuickTime Player.app//Contents/XPCServices: Operation not permitted rm: QuickTime Player.app//Contents/Info.plist: Operation not permitted rm: QuickTime Player.app//Contents/PkgInfo: Operation not permitted rm: QuickTime Player.app//Contents/version.plist: Operation not permitted rm: QuickTime Player.app//Contents: Operation not permitted rm: QuickTime Player.app/: Operation not permitted

How can I delete Apple OS X application such as Photos, iPhoto, and others that I never use?


We know the nice properties and consistency results of the MLE estimator don't hold when the true value is on the boundary of the parameter space, for eg. if the parameter space is $[0, \inf)$, so while we can still find  $\hat{\theta}_{MLE}$, how do you go about finding $\hat{\theta}_{MLE} -\theta$ if $\theta$ might be zero?  The usual Taylor expansion of first derivative around true value is no longer equal to zero, so how does one proceed instead?    : For some context.

I'm interested in a good reference for results concerning asymptotic properties of maximum likelihood estimators. Consider a model $\{f_n(\cdot \mid \theta): \theta \in \Theta, n \in \mathbb N\}$ where $f_n(\mathbf x \mid \theta)$ is an $n$-dimensional density, and $\hat \theta_n$ is the MLE based on a sample $X_1, \ldots, X_n$ from $f_n(\cdot \mid \theta_0)$ where $\theta_0$ is the "true" value of $\theta$. There are two irregularities I'm interested in.    The data $X_1, \ldots, X_n$ are not iid and, as a result, the Fisher information about $\theta$ accrues at a rate slower than $n$.  $\Theta$ is a bounded set, and with positive probability $\hat \theta_n$ lies on the boundary. The boundary corresponds to a "simpler" model, and so there is particular interest in whether or not $\theta_0$ lies on the boundary.    My particular questions are   Letting $J_n(\theta)$ denote the observed Fisher information corresponding to $\theta$, and suppose $\theta_0$ lies in the interior of $\Theta$. Under what conditions is $$\left[J_n(\hat \theta_n)\right]^{1/2}(\hat \theta_n - \theta_0)$$ asymptotically normal as $n \to \infty$? In particular, are the regularity conditions similar to the usual ones, with the relevant modification being $J_n(\hat \theta_n) \to \infty$ in some sense? Suppose instead that $\theta_0$ is on the boundary, and again recall that $\hat \theta_n = \theta_0$ happens with positive probability - for concreteness, in a mixed effects model $Y_{ij} = \mu + \beta_i + \epsilon_{ij}$ we can have $\hat \sigma_{\beta}^2 = 0$. Under what conditions does $\hat \theta_n \to \theta_0$ (almost surely or in probability) and under what conditions will $\hat \theta_n = \theta_0$ eventually (this probably fails for the mixed effects model, but corresponds to "oracle" properties for the LASSO and related estimators, so perhaps it is too much to ask for general results)?   Again, just a pointer to a text with results at this level of generality would be greatly appreciated.


Say a circle is rotating about its centre at a relativistic speed.  If it were not rotating in your reference frame, its circumference divided by its diameter would be $\pi$.    Now, because it is rotating, let us say we have been using little sticks all along to measure circumference and diameter.  We would need a ratio of $\pi:1$ sticks to build the static circle.  However, these sticks change when the circle is spinning because of length contraction.   So, the sticks along the diameter become thicker, but not shorter.  The sticks along the circumference, however, become smaller, and not thicker.  Therefore, if a circle is rotating in a reference frame, $\frac{circumference}{diameter}$ is not $\pi$.  This is an example of relativistic geometry.  My question is: how does area change?  I believe the area becomes larger.  Here is my reasoning:  The rotating circle has a larger circumference than before, therefore the 'new $\pi$' is greater than the 'old $\pi$'.  And, because this circle is spinning around its centre, it is still a circle to you even when its spinning.  Therefore, area is still $\pi r^2$; but this time, $\pi$ is larger, therefore the new area is larger.  I can also think of a counterexample:  Imagine using a bunch of smaller circles within the larger circle to estimate its area.  When the circle begins spinning, these circles shrink because of length contraction.  Therefore, these circles become smaller, and the overall area is smaller.  However, this may also be interpreted as changing the metric of the space by requiring more sticks to create the diameter of the circles, contradicting the argument that the circle gets smaller and actually agreeing that it gets larger.  Which is correct?  Why?  EDIT: I chose to distinguish my question from others because the reasoning I offer is not something explored in-depth in other questions/answers and is a part of the question that I believe enhances it.

Let us say you are in an inertial reference frame with a circular planar disk. If you take your meter measuring rods (or perhaps tape measure) you can find the diameter and circumference of the disk. If you divide the circumference by the diameter, you will get exactly $\pi$. Now you start rotating the disk. A book I have claims that now the ratio will be different (lets call it $\pi_\circ$ to avoid ambiguity.) This has caused to reevaluate my understanding of special relativity (this is a set up to general relativity, but the problem itself only requires the special theory.) My problem is that, sure, the measuring rods (or tape) would contract on the circumference, but wouldn't the circumference contract a similar amount, therefore giving a measurement of $\pi$ in every reference frame according to your own measuring rods. Of course, this reasoning goes on to make me question how I understood all of special relativity up this point.


I have this a bit of nightmarish set of equations: $$x \equiv 10\pmod {19271}$$ $$x \equiv 4\pmod {7343}$$ $$x \equiv 8\pmod {9973}$$ For which I am supposed to use Chinese Remainder Theorem to find a solution or to show that one does not exist. Now I have started solving it in a way that would give me $X$ through equation: $$X = r_{1}M_{1}x_{1} + r_{2}M_{2}x_{2} + r_{3}M_{3}x_{3}$$ but then have noticed that $19271$ and $7343$ share a common divisor in 7. The literature I have on my hands doesn't explain that well how to tackle this case, other than having a single remark which states that the theorem used states nothing in the case of moduli not being pairwise relatively prime, and the stuff I went through via online search doesn't really explain it that well due to lack of practical examples. I was wondering if someone knows how to go about this thing and if it would be possible to explain the steps in a relatively simple way as I had trouble understanding what I encountered so far.

We want to find the solution $x$ to the congruence system  $$\begin{align} x &\equiv r_1 \!\!\!\pmod{\!m_1}\\ x &\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \  \gcd(m_1, m_2) = 1$$  These can be rewritten as  $$\begin{align} x &= r_1 + m_1 j\\ x &= r_2 - m_2k\end{align}$$  for unknown integers $j, k$. So we set those two equations as equal and rearrange them to  $$m_1j + m_2k = r_2 - r_1$$  Now assume we perform $\text{egcd}(m_1, m_2)$, which gives us the solution to  $$m_1u + m_2v = \gcd(m_1, m_2)$$  Let the integer $\,h\,$ make $\,h \gcd(m_1, m_2) = r_2 - r_1$.  Then $\qquad hm_1u + hm_2v = h\gcd(m_1, m_2) =  r_2 - r_1 = m_1j + m_2k$  At this point I am lost because I don't think we can just assume  $hm_1u = m_1j$ and extract $j = hu$.   How do I get the values of $j$ or $k$ so I can get the value of $x$?


I know that and is used to join two sentences or phrases. There are some places I've read that have And is used at the beginning of a sentence. What are the occasions when this is done?

Since I first learned English, I have been holding this understanding that "and", as a conj. but unlike "but",  can only connect two clauses, not two sentences ended with periods.  But recently, I have seen so many prints, either in entertainment or in academia, where "And" is popularly used in the beginning of a sentence. It seems like the author is trying to connect the sentence just right before and the sentence following "And" in some intended meaning which I don't quite get.  I was wondering if I have been wrong all the time, or if there is a new trend that I fail to understand and accept? How to understand such usage of "and" in rigorous English grammar? If I am right, why it is used differently from "but" in terms of what I mentioned at the beginning of this post?


Let $p \in \mathbb Z [x]$ be a monic quadratic polynomial. Show that for $n \in \mathbb Z$ there exists $k \in \mathbb Z$ such that $$p(n)p(n+1)=p(k)$$   I came across this problem in an olympiad book. I have tried this question by many different approaches like simply putting $n , (n+1)$ into the quadratic equation and and try to factorise or do something to get the result but it doesn't help. Secondly, I know that  $$p(x)-p(n)p(n+1)=0$$  has a root equal to $k$ and then there is only one thing left to do that is to prove k integer which we can prove by proving discriminant to be a perfect square and other coefficients as integer but the problem arises how to prove discriminant to be a perfect square. Please tell if either of these approaches will take us to the prove or if not please tell the correct steps of proving   I am very confused with problem (not a homework problem). Please help to prove it with proper steps and methods.

Let $f(x) = x^2 + ax + b$ be a quadratic polynomial in which $a$ and $b$ are integers. If for a given $n$ , $f(n)f(n+1) = f(m)$ for some integer $m$ then the value of $m$ is ?  Do we have to substitute $x$ by $n$ and $n+1$ and then multiply it? or is there another method to get the solution because the mentioned process will give an extremely large equation.  Answer is $n(n+1)+an+b$


May question may be duplicate, but what I'm looking here is a clear answers and explanation. So, yeah. Due to high shortage and demands on graphic cards, I plan to build my custom pc completely in US(newegg.com), and would ask my relatives to bring it down to India(Chennai). So here what's my doubt is, the total cost of unit around 900$, which includes processor, motherboard, GPU, HDD, SSD, PSU, cooling system etc. I know there is lot of global and trusted shipping companies are there. but shipping itself cost so much.  need to pay any extra taxes?  in Airport, do they ask any documents like P.O or any other permission?   Kindly, add your valuable advice or alternative approaches. Kindly help me to understand

I am planning to travel to India (COK airport, Kerala) from United States. I want to take my laptop as well as my tablet (iPad). Am I allowed to take both in my backpack with my documents related my work. I think I am allowed to take only one computer as per the rule. Will the customs consider my iPad as a computer?  What all I can take?   One of my friends took 2 laptops (1 work and 1 personal) and had trouble at the airport.  Also I am planning to take some wristwatches (7 of them, each avg price is 45 $). I read in a website that I can take stuff up to max 35000 INR. Is that true? In that case, how will I manage my stuff?


The distance I had to travel with my buddy Pokemon Lapras was originally 3 kilometers after traveling about a total of 40 kilometers with it it dropped to 1.25 kilometers needed to find a candy has anyone else experienced this?

I noticed that the walking distance has changed from 3km for 1 candy to 0.75km for 1 candy. Is this a new update? Is it a glitch? Is it because I've walked an x amount of km's already?


I am setting the min of checkOut as the value of checkIn. My problem comes that i need to add one day to firstdate. (Should not be able to check out on or before the check in day.)  <script> function updatedate() { var firstdate = document.getElementById("checkIn").value; document.getElementById("checkOut").value = ""; document.getElementById("checkOut").setAttribute("min",firstdate); }   </script>  Check In <input type="date" id="checkIn" onchange="updatedate();" name="checkin">  Check out <input type="date" id="checkOut" min="" name="checkout">

How to add days to current Date using JavaScript. Does JavaScript have a built in function like .Net's AddDay?


I have a notebook with Windows 8.1 and Firefox 66.0.3; all the add-ons got suddenly disabled. So I uninstalled them and tried to reinstall them from Mozilla web page but I get a red label that says something about a download failed, check your connection.  Also tried by deleting the user profile with same results.  What can be wrong?

All of my Firefox extensions were spontaneously and simultaneously disabled (converted to Legacy status) a few moments ago.  How can I re-enable them? Firefox 66.0.3 (64-bit)


I've just finished reading up about IP addresses and subnets through this . In the examples provided by the article, IP addresses seem pretty arbitrary. This got me thinking, why are the default IP addresses for most home networks 192.168.0.1/24? Is there a reason (historic or otherwise) for the apparent convention of "192.168.0"? Additionally, are there any implications if I decided I wanted to change my home network's IP address to something arbitrary and easier to type like 1.1.1.1/24?

Why did the  (IETF) choose 192.168/16 to be a private IP address classand not something else?  Why specifically 192.168/16 and 10/8 and 172.16/12 and not 145.243/16 for example?  Is there a reason those IP addresses were chosen to be the standard for private IP addresses over all the other possibilities?


I am working with an Azure SQL Server database in SSMS 2016. I have Intellisense enabled in Tools -> Options -> Text Editor -> Transact-SQL -> Intellisense, however I have never had Intellisense working on this computer. Any advice on how to resolve this would be appreciated.

This problem is really proving to be a tricky one (and quite annoying).  In SQL Server Management Studio 2008, up until a few days ago my IntelliSense was working great.  Then all of a sudden it stopped.  The icon as it as enabled on the toolbar menu, and under Tools -> Options -> Text Editor -> T-SQL -> IntelliSense it says it is enabled there.  I have tried refeshing the IntelliSense cache with Ctrl-Shft-R but that doesn't work either.  Any ideas what happened to my IntelliSense and what I need to do to get it back?


I'm writing in Node.js, ES6. I have an object adaptors passed from a request. I would like to check if the String IOS is in adaptors. And I tried the following commands, which makes me super confused: const adaptorList = Object.values(adaptors); console.log("adaptorList: "+ adaptorList); // adaptorList: IOS,ANDROID console.log("typeof adaptorList: " + typeof(adaptorList));//typeof adaptorList: object console.log(adaptorList[0] == "IOS"); // true  console.log("IOS" in adaptorList); // false  I'm so confused why adaptorList[0] == "IOS" gives true while "IOS" in adaptorList gives false.

What is the most concise and efficient way to find out if a JavaScript array contains a value? This is the only way I know to do it: function contains(a, obj) {     for (var i = 0; i < a.length; i++) {         if (a[i] === obj) {             return true;         }     }     return false; }  Is there a better and more concise way to accomplish this?


I logged into Ubuntu and my menu bar and launch bar are gone.  I cannot use Ubuntu as there as the only this I see is my wallpaper. I have restarted and it has not done anything.  I can open the terminal, and I can run commands, but that is all I can do.    As you can see I don't even have the ability to close, maximise or minimise windows (I can close by using alt+f4 though.)  I have tried various command like alt+f2 and other commands, but these don't appear to work.  Is there a way in the terminal to open the home folder as a window? If I can have a folder open, then I can at least use Ubuntu to an extent.  Thanks in advance.

When I login, nothing happens.   I am presented with my desktop wallpaper.    No Dash, no Launcher, nothing.


Is that proper to get messages about your work (almost critique no constructive feedback) at the early dawn, I freaked out from my sleepings not the first time to expose that besides the high pressure to get results as soon as possible. I feel my body is exhausted and sick and sometimes crying, is this behavior normal in academia?

I am talking with a professor to apply to a postdoc program. I sent a cold email earlier this week and we have exchanged a few emails. In his last email he sent me some papers to read. The thing is, I might end up sending my answer on Saturday, after I read the papers. Is this a problem? Should I just send it and wait until Monday for his answer? Maybe send it but include an apology for writing during the weekend? Or should I just wait until Monday to send it? I usually text everyone during the weekend including my advisor, but I'm not sure if it might be considered rude, and as I do not know the person yet I want to be as careful as possible with etiquette. Thank you.


I've checked the webpages of some their providers, but franly, just like in the case of those in my own language, I literally cannot find my way around them. So if you wouldn't mind sharing a word of two or advice.  I'm looking for a good solution for mobile internet surfing in northern Italy, that can be connected to a laptop (you know, that usb stick with a SIM card inside kind of thing). 15 Gb or so at some decent speed. Can you perhaps give a recommendation or two of what to get?    Edit: Just to define what I'm looking for exactly. A data plan of 15 Gb with an internet key (usb stick into which a SMS card goes) that you plug into a laptop. Preferably non-contractual, so I can cancel when I leave. I am not interested in surfing on my mobile phone since due to size I still prefer the smaller Nokia-like models for calls (smaller to carry) - no internet on them.

What is the best option for a prepaid data plan for a cell phone for a short stay in Italy (Sardinia)? I am looking for a Micro-Sim Card. I don't need to call and I need approximatively 300-500 MB.  Ideally, I would like to buy the Sim card in Italy. Where can I buy those cards?  Alternatively, are there other options for such a situation? I heard the EU has passed a law that lowers the roaming fees (I have a German carrier) but it still seems pretty expensive.


Hello. When I log into Minecraft it says that there is an "error connecting to servers". When I see the error log, it says:  MagicLauncher 1.0.0 java.home: C:\Program Files\Java\jre7 java.runtime.name: Java(TM) SE Runtime Environment java.runtime.version: 1.7.0_09-b05 os.name: Windows 7 os.version: 6.1 os.arch: amd64 sun.arch.data.model: 64 java.net.SocketException: Connection reset     at java.net.SocketInputStream.read(Unknown Source)     at java.net.SocketInputStream.read(Unknown Source)     at sun.security.ssl.InputRecord.readFully(Unknown Source)     at sun.security.ssl.InputRecord.read(Unknown Source)     at sun.security.ssl.SSLSocketImpl.readRecord(Unknown Source)     at sun.security.ssl.SSLSocketImpl.performInitialHandshake(Unknown Source)     at sun.security.ssl.SSLSocketImpl.startHandshake(Unknown Source)     at sun.security.ssl.SSLSocketImpl.startHandshake(Unknown Source)     at sun.net.www.protocol.https.HttpsClient.afterConnect(Unknown Source)     at sun.net.www.protocol.https.AbstractDelegateHttpsURLConnection.connect(Unknown Source)     at sun.net.www.protocol.https.HttpsURLConnectionImpl.connect(Unknown Source)     at magic.launcher.ak.a(Unknown Source)     at magic.launcher.z.run(Unknown Source)     at java.lang.Thread.run(Unknown Source) Error loading: .http://www.minecraftforum.net/rss/writ/1-news/: Connection reset Error loading: .http://search.twitter.com/search.atom?q=from%3Ajeb_: Connection reset Error loading: .http://search.twitter.com/search.atom?q=from%3ADinnerbone: Connection reset Error checking for new version: Connection reset Error loading: .http://reddit.com/r/minecraft.rss: Connection reset Error loading: .http://search.twitter.com/search.atom?q=from%3Anotch: Connection reset Error loading: .http://notch.tumblr.com/rss: Connection reset

Similar to  on the bugs in Minecraft, I would like a complete list of what error codes exist in Minecraft and how they occur, as well as how they can be resolved.


Sometimes comments are answers to the question. What should I do if somebody used the copy & paste method to write an answer which is built from comments' content? Should I mark that answer as plagiarism for the moderators' attention?

There are a lot of questions which have (often enough) complex and good answers in comments, which should really be posted as answers.  A good example is here: . Virtually all of the questions in "assembly" listed when using "unanswered" have answers provided via comments.  Now I was reading  and notice this comment:     More than that... in case you ask them to post as answer and they don't do that within a day or so, just post your own answer and mark accepted. Case solved. :) –  Sha Wiz Dow Ard   And I'm unsure if this was just a joke or meant seriously.  On the one hand, SO should provide answers, and it would be helpful, if the answer is provided in comments, if it were summarized in a real answer. On the other hand, I would feel a bit like a rip off (is this the proper term?) if I would do this from somebody else's comments and possibly even get the rep for it.


I did a silly thing. I changed the permissions of my /etc folder using  sudo chmod -R ugo+rw /etc   After that I did some damage control using the command  pkexec chmod 755 /etc   Then I was able to the sudo commands but every time I use a sudo command a warning comes  sudo: /etc/sudoers.d is world writable   Is there a way to completely reset default permissions for /etc ?

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?


Two and a half men features Charlie Sheen as a hedonistic jingle-writer.  He has a beach house in Malibu and spends a lot of money while he is rarely shown working.  So, how much does he really earn and how? Royalties?

This is quite tricky to figure out, and I am not looking for a specific number, but for a approximation.  As you know, Charlie (nearly) always had a lot of money to spend, but a few hints suggest that he actually did not have money at all (talking about absolute, not cash flow)   he went bancrupt after a few months without income he told Alan that there is no money, and everything just a house of cards he worries about Bertha giving away a few clothes he has three mortgages on his house when he died, there was no inheritage to Alan or anyone, and they had to sell the house ...   However, considering the lifestyle and spendings, Charlie must have had a substantial amount available.   driving the latest foreign expensive cars owning a malibu beachhouse paying the property tax for said house discussing retirement at 45 spending endless money on booze, betting and women ...   So all things considered, did he have money, or was he just a scam?


TypeScript Math: How does 1 - (80/100) = 0.19999999999999996? Why Doesn't it equal .20? I'm confused how this is happening with such a straightforward math equation. CalculateDiscount.tsx: // Helper Function: Calculate Discount (Percent) export const calculateDiscountPercent = (price: number, salePrice: number) => {   return 1 - (salePrice / price); };  Repl.it:

Consider the following code:   0.1 + 0.2 == 0.3  ->  false   0.1 + 0.2         ->  0.30000000000000004   Why do these inaccuracies happen?


I'm new to Linux.  When I try to use my built-in bluetooth chipset in Huawei Matebook, the adapter seems recognized well but it could not find any bluetooth device.  I also try to use bluetooth with my dongle type adapter and then it is working  here is my hciconfig -a  ~$ hciconfig -a hci0:   Type: BR/EDR  Bus: USB BD Address: FC:3F:7C:63:81:35  ACL MTU: 1021:8  SCO MTU: 64:1 UP RUNNING  RX bytes:2068 acl:0 sco:0 events:116 errors:0 TX bytes:3865 acl:0 sco:0 commands:113 errors:0 Features: 0xbf 0xfe 0xcf 0xfe 0xdb 0xff 0x7b 0x87 Packet type: DM1 DM3 DM5 DH1 DH3 DH5 HV1 HV2 HV3  Link policy: RSWITCH SNIFF  Link mode: SLAVE ACCEPT  Name: 'yeomMate' Class: 0x0c010c Service Classes: Rendering, Capturing Device Class: Computer, Laptop HCI Version: 4.1 (0x7)  Revision: 0x2000 LMP Version: 4.1 (0x7)  Subversion: 0x230f Manufacturer: Broadcom Corporation (15)   I wonder why it say that Bus type is USB.  and dmesg |grep [bB]luetooth is here  $ dmesg |grep [bB]luetooth [    3.276981] Bluetooth: Core ver 2.21 [    3.277036] Bluetooth: HCI device and connection manager initialized [    3.277040] Bluetooth: HCI socket layer initialized [    3.277042] Bluetooth: L2CAP socket layer initialized [    3.277063] Bluetooth: SCO socket layer initialized [    3.300552] Bluetooth: hci0: BCM: chip id 101 [    3.316606] Bluetooth: hci0: BCM4354A2 [    3.317586] Bluetooth: hci0: BCM4354A2 (001.003.015) build 0000 [    3.317608] bluetooth hci0: Direct firmware load for brcm/BCM4354A2-0a5c-6419.hcd failed with error -2 [    3.317611] Bluetooth: hci0: BCM: Patch brcm/BCM4354A2-0a5c-6419.hcd not found [    4.329100] Bluetooth: BNEP (Ethernet Emulation) ver 1.3 [    4.329104] Bluetooth: BNEP filters: protocol multicast [    4.329109] Bluetooth: BNEP socket layer initialized [    8.713237] Bluetooth: RFCOMM TTY layer initialized [    8.713246] Bluetooth: RFCOMM socket layer initialized [    8.713254] Bluetooth: RFCOMM ver 1.11   my working usb-type bluetooth also fail to Direct firmware load. but it works.  nothing is blocked in rfkill list  if you want any other info, I will be glad to append

Bluetooth is on but can't find or be found by any other devices.  Output of sudo service bluetooth status:  ● bluetooth.service - Bluetooth service    Loaded: loaded (/lib/systemd/system/bluetooth.service; enabled; vendor preset: enabled)    Active: active (running) since ხუთ 2015-06-04 22:33:18 GET; 13min ago  Main PID: 26678 (bluetoothd)    CGroup: /system.slice/bluetooth.service            └─26678 /usr/sbin/bluetoothd -n  ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPAG ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPHS ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSource ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSink ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPAG ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/HFPHS ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSource ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Endpoint registered: sender=:1.63 path=/MediaEndpoint/BlueZ4/A2DPSink ივნ 04 22:39:14 Rangoo bluetoothd[26678]: Adapter /org/bluez/26678/hci0 has been enabled ივნ 04 22:39:14 Rangoo bluetoothd[26678]: bluetoothd[26678]: Adapter /org/bluez/26678/hci0 has been enabled   Output of lsusb:  Bus 003 Device 005: ID 0a5c:21d7 Broadcom Corp. BCM43142 Bluetooth 4.0   How can I fix it?


In the task manager of Windows 10, I see this:    What does 100 % disk usage mean in Windows 10? How is the disk usage percentage calculated?  I am particularily confused by the first graph showing a constant percentage at 100 %, while the second graph shows read and write rates that vary.

After startup for 5 minutes my system is very slow. The Task Manager shows that the disk usage is at 100%. (Later this number is around 2-4 %, but sometimes for 1-2 minutes go up again at 100%.)  I searched a lot on Google. For example I read , and also other pages. I opened two topic on Microsoft answer.  and one in Hungarian.  But I didn't found the solution yet. So I would like to ask Community's help. :)  Here the steps I already done:   I disabled everything at startups (including services) and the issue is the same.  I disabled Prefetch and Superfetch, the Baground Intellingent Transmission, Windows Search and Windows Defender but the issue is still persists. I made a health check on HDD, I made defrag and file system error check, but there are no errors. Also I tried to reinstall the HDD's driver. I made a full hardware check up, with no errors. The Windows and all the drivers are up-to-date. I tried a clean start. The issue is the same. I deleted Dropbox and all antivirus. I turned of the MSI mode.      I using also Linux, where this error is not occurs. Good to know, that Windows had 4 partitions, but I deleted 2 of them for the Linux. I don't know what partitions was these but both was only 10 - 20 MB.  I using a Dell Inspiron 15R SE 7520 with Windows 10 Home.  Personally I think it can be the Intel Rapid Store Technology "driver", because the problem started around when I installed it. (However I'm not sure.) And my hardware not supported yet by Intel, so I have no the newest driver. However I deleted Intel's software and driver and the issue is the same.    Thank you in advance for your assistance!


How could I create a new environment to automated the following code?  \noindent\fcolorbox{white}{lightgray}{ \begin{minipage}{0.96\textwidth}  some text in here \end{minipage} }

I have a problem, I'm fairly new to LaTeX but I think I got the basics.  My problem is that wants to create a environment that can create a minipage with a background color and some text in it.  I can make this by doing it by hand, but I have a lot of boxes, and to write mostly the same over and over again is to my a bit redundant.  \documentclass[12pt,a4paper]{book}   % Book settings + layout  \usepackage{xcolor}         % Extended colors \usepackage{color}         % Color extended names  %DEFINE ENVIRONMENT BLOCK % Riddle \newenvironment{colbox}[3]{        % Riddle environment   \begin{center}                   % Centering minipage     \colorbox[HTML]{#1} {          % Set's the color of minipage       \begin{minipage}[b]{380px}   % Starts minipage    \textbf{#2}\\ \textit{#3}       % Set's title and starts italic for text   \end{minipage}}                  % End minipage }{\end{center}}                    % End Riddle environment   \begin{document}  \begin{center}     \colorbox[HTML]{F8E0E0}{      \begin{minipage}[c]{380px}       \textbf{Riddle: }\\ \textit{some text here}      \end{minipage}} \end{center}  Some other text  \begin{colbox}{F8E0E0}{Riddle:}   some text here \end{colbox}  Some more text  \begin{colbox}{F8E0E0}{Riddle:}   {some text here} \end{colbox}  \end{document}     As I can see it, the problem is that in my environment colorbox is not ended by the end parameter, and I need to move \end{minipage} the same place, but I can't move it before the \end{colorbox} is moved, is there a way to do this right?  %DEFINE ENVIRONMENT BLOCK % Riddle \newenvironment{colbox}[3]{        % Riddle environment   \begin{center}                   % Centering minipage     \colorbox[HTML]{#1} {          % Set's the color of minipage       \begin{minipage}[b]{380px}   % Starts minipage    \textbf{#2}\\ \textit{#3}       % Set's title and starts italic for text   \end{minipage} }                 % End minipage }{\end{center}}   I can "fix" the out-of-box problem by putting {} on the text like this  \begin{colbox}{F8E0E0}{Riddle:}   {some text here} \end{colbox}   But that's not really what I want.  If it's possible I would like to keep everything in what's default in LaTeX cause I'm not the only one maintaining this document and special packages is not well liked,


I have two GPUs: Intel HD 5500 and AMD Radeon R5 M330.  I want to know which video card is being used when I'm watching videos on YouTube or just simply navigating through websites on Chrome.  When does Windows switch between them and how can I find out? I'm not playing games.

I use:   HP Pavilion dm4-3001tx Entertaiment Notebook PC Windows 7 Ultimate 64bit   In  it has two graphic cards.     But in  it only has 1 graphic card.    And in several programs, it tells me that I need to configure my graphic card for better performance. Let's take nox for instance    So, I'm guessing my AMD drivers is not installed properly.  I did reinstall my OS but I'm using the same windows as before (which is windows 7 64bit) and a driver from HP official site. I've also installed HP Support Assistant to check my drivers but everything is seems to be updated (according to HP Software).  Any solution to show my AMD graphic into dxdiag, or remove nox graphic suggestion (as it means no problem with my graphic cards anymore)?


I want to calculate the sum of this series:  $$S = 1\cdot1! + 2\cdot2! + 3\cdot3! + 4\cdot4! +\dots+ n\cdot n!$$  Is their any formula for finding this sum?

I discovered that the summation $\displaystyle\sum_{k=1}^n k\cdot k!$ equals $(n+1)!-1$.  But I want a proof. Could anyone give me one please? Don't worry if it uses very advanced math, I can just check it out on the internet. :)


I have a div tag with width set to 800 pixels. When the browser width is greater than 800 pixels, it shouldn't stretch the div, but it should bring it to the middle of the page.

How can I horizontally center a
 within another
 using CSS?

Foo foo


I would like to ask if this is possible:  I have a class with a private constructor which of course throws a fatal error if I try to instantiate that class from an invalid context (outside the class), can I catch this fatal error and the name of the class and then throw an exception like:  // catch the private constructor from invalid context fatal error // and store somehow the name of the class inside $class // and then: throw new UninstantiableClassException("The class " . $class . " cannot be instantiated");   Is that possible, or basing on this post I have just found it's not? ->   If it is not possible, should I make the constructor public and then throw and exception inside of it like this?:  public function __construct() {    throw new UninstantiableClassException("The class " . get_class($this) . " cannot be instantiated"); }

I can use set_error_handler() to catch most PHP errors, but it doesn't work for fatal (E_ERROR) errors, such as calling a function that doesn't exist. Is there another way to catch these errors?  I am trying to call mail() for all errors and am running PHP 5.2.3.


So I've watched FMA: Brotherhood and Star of Milos. I want to watch conqueror of Shamballa but it seems like it continues from the  original series.  I don't want to have watch the original series since I know that the earlier portions of the story are the same as Brotherhood . Is there an episode in the original series I can start watching from that's starts to have a different story path from Brotherhood? Or should I just read about the originals ending?

I just finished watching Fullmetal Alchemist 2003 series. I would like to read/watch the manga/Brotherhood anime now. Does anyone know at what chapter the story of FMA2003 and the story of the manga/Brotherhood anime go their separate ways?


Here is my code  bash-3.2$ read num 100 bash-3.2$ for k in {1..$num} > do >    echo Printed $k times > done {1..100} bash-3.2$ echo $num 100 bash-3.2$ echo $k {1..100}   Does anyone have any idea why it is doing this? It isn't the variable num that's wrong because I checked that with my echo at the end.

How do I iterate over a range of numbers in Bash when the range is given by a variable?  I know I can do this (called "sequence expression" in the Bash ):   for i in {1..5}; do echo $i; done   Which gives:     1    2    3    4    5   Yet, how can I replace either of the range endpoints with a variable? This doesn't work:  END=5 for i in {1..$END}; do echo $i; done   Which prints:     {1..5}


I am trying to make a gold ore material on my object but I cannot seem to make it look right with different nodes. Can anybody help me?!! I am just learning about nodes so I am fairly new to how they work and it needs to be explained in detail what to do. This is what I have that is slightly close to what I am looking for...

Brand new to blender, and I'm trying to make a metallic object that is reflective.  There are lots of tutorials on youtube, however none of them are compatible with blender 2.71 with the cycles render.   Here are the tutorials I have tried:        All of these tutorials have a 'transparency' and a 'mirror' option in the outliner menu that I just don't seem to have.   How can I make this object reflective? Am I missing plugins or some other menu?


Show that if $G$ is 2-connected, so each vertex $u$ has vertex $v  \neq\ u$ and circle $C$ that contains $u, v$ and all the neighbors of $v$.

Show that if a graph $G$ is $2-$connected, then for any two vertices $u$ and $v$ there exists a cycle $C$ such that $u, v \in V (C)$  I tried to use the fact that a graph $G$ with at least $3$ vertices is $2-$connected, if and only if for any $u, v \in V$, with $u \neq v$, there are at least $2 (u, v)-$ trajectories that do not have internal vertices in common.   Any suggestions would be great!


I'm a fairly new user with ubuntu. But What's the best/easiest way to get wine to work. It never seems to work for me.  thanks

I am looking for a step by step instruction to guide me in the ways of installing, configuring and debugging many of the issues new users have when dealing with Wine in Ubuntu.  What is the recommended way of installing Wine, some of the key issues that it has and how to solve them?  If I have a specific problem with an App in Wine, how do I go about solving it? What are the recommended steps to ensure that I can fix a problem related to an app I wish to use?


To borrow a phrase, every editor who is not too stupid or too full of himself to notice what is going on knows that propagating the myth of "objective journalism" is indefensible. A newspaper or radio program may try to hide or obscure the fact that the people responsible for its content have opinions, convictions, and biases. But it is impossible to function as a journalist without making subjective judgment calls about newsworthiness, relevance and emphasis, or covering issues about which you have an opinion. Pretending otherwise requires willfully misleading the public.   What does To borrow a phrase mean in the above paragraph?

How do you say to someone that you will reuse a sentence (or a joke) you've just heard from them, as-is, because you liked it a lot ?   In Italian we say "Questa me la rivendo", that translated is "I'm gonna resell that one" (ri-vendo / re-sell is the idiomatic part, the other words could be recombined / replaced).  I can't find anything in the web, so... what is the English version ?


As we know that oblivious transfer is done with probability of 1 out of 2. But how to implement it with probability of 1 out of 3 or 1 out of 4??

I'm a student and I don't know how to solve the following problem :   Alice is proposing a 10 questions list to Bob. Bob choose only 1 question and wants to know the answer without Alice knowing the question.  Alice wants that Bob can only have 1 correct answer.   Is it possible ? If yes how ?    In my mind it should begin like this :   Bob & Alice knows the 10 questions Alice encrypts each answer with a different key  Alice sends the 10 encrypted answers to Bob Bob chose a question (ex : question 2)   Here comes the intricate part in my mind :   Bob should crypt the question (How?)  Bob sends the encrypted question to Alice Alice answer to bob with the encrypted decrypting key (How?) Bob can decrypt his answer, Alice doesn't know what key she gave   Thanks :)


Prove or disprove: If $\mu$ is an additive (i.e., 2-additive) set function on a semi-ring $S$, then $\mu$ is finitely additive (i.e., n-additive for every finite $n$) on $S$.   If $S$ is a ring, this is true by induction. For a semi-ring, the naive induction proof fails because if $A_1,\ldots,A_n$ are disjoint sets in $S$ and if $\bigcup_{i=1}^{n} A_i$ is in $S$, it does NOT generally follow that $\bigcup_{i=1}^{n-1} A_i$ is in $S$.  Every measure theory book I've looked at works with finitely additive sets functions on semi-rings, never just additive set functions on semi-rings. So maybe additive doesn't imply finite additive on a semi-ring.

A semiring $\Pi$ on a set $X$ is a non-empty family of subsets of $X$ with the following properties.     1) $P \cap Q \in \Pi$ whenever $P\in \Pi$ and $Q\in\Pi$.      2) $P - Q$ is a finite disjoint union of members of $\Pi$ whenever $P\in \Pi$ and $Q\in\Pi$.   Let $\lambda: \Pi\rightarrow [0, \infty]$ be a non-negative extended real-valued set function such that $\lambda(\emptyset) = 0$. Suppose $\lambda$ is additive, i.e. if $P, Q_1, Q_2$ are members of $\Pi$ and $P = Q_1\cup Q_2$ is a disjoint union, then $\lambda(P) = \lambda(Q_1) + \lambda(Q_2)$.  Then $\lambda$ is finitely additive? In other words, if $P, Q_1, \cdots, Q_n$ are members of $\Pi$ and $P = \bigcup_{i=1}^n Q_i$ is a disjoint union, then $\lambda(P) = \sum_{i=1}^n \lambda(Q_i)$?  I came across this problem when trying to solve this question:   I also found this is an exercise in Halmos's Measure Theory, Ch. II, sec. 7. It has some hints, but I was unable to solve it.


Consider the hyperpower equation  $$x^{x^{x^{x...}}}=2$$  We will use the method that  Let $y=x^{x^{x^{x...}}}$ so $x^y=x^{x^{x^{x...}}}=2$ and $x^2=2$ to give the solution $x=\sqrt2$  However consider another equation such $x^{x^{x^{x...}}}=4$  Use the similar method so we have $y=x^{x^{x^{x...}}}$ to give $x^y=x^4=4$ , this also makes $x=\sqrt2$ as our solution  But the $\sqrt2^{\sqrt2^{\sqrt2^{\sqrt2...}}}$ have to equal to something, not 2 and 4 at the same time.  If there is only one solution $x=\sqrt2$, then what is wrong with the second equation?

Out of boredom, I decided to recall the following equation:   $$x^{x^{x\cdots}} = 2.$$  Which, I simply rewrote like this:   $x^2 = 2$, and therefore $x = \sqrt{2}$.  Then I took a look at the more general form:  $$x^{x^{x\cdots}} = p.$$  I then concluded that $x = (p)^{\frac{1}{p}}$, and this was the solution set for all $p$. Then I thought for a while and determined that there are multiple values where this function equals the square root of $2$, so its not injective. However, this led to an awkward statemend based on the fact that $\sqrt{2} = (4)^{\frac{1}{4}}$. We deduce the following:  The equation $x^{x^{x\cdots}} = 2$ and the equation $x^{x^{x\cdots}} = 4$ have the same solution, namely $x = \sqrt{2}$. So we conclude that $2 = 4$.   I showed this to friend of mine, and he conjectured that there are no valid solutions for all $p > e$. To prove this, I tried writing the following:   I tried differentiating $f(x) = (x)^{(1/x)}$ and setting it equal to $0$ to find the maximum. I determined that there was a maximum at the point where $x = e$. Does this mean that the maximum possible solution $x$ for the infinite power is $e^{(1/e)}$ therefore showing that the maximum value for $p$ is indeed $p = e$?


Question in title. When I say n prime factors, I don't mean n distinct prime factors.

If $\pi_k(n)$ is the cardinality of numbers with k prime factors (repetitions included) less than or equal n, the generalized Prime Number Theorem (GPNT) is:  $$\pi_k(n)\sim \frac{n}{\ln n} \frac{(\ln \ln n)^{k-1}}{(k-1)!}.$$  The qualitative appearance of the actual distribution of $\pi_k(n)$ for k = 1,2,3,..., agrees very well with the GPNT, for numbers $n$ within reach of my laptop. But I noticed that as $n$ and $k$ get large, "most" of the numbers less than $n$ seem to have relatively few factors.     Writing $n = 2^m$ and replacing $k$ by $x$ we can graph   $$f(x) =\frac{2^m (\ln\ln 2^m)^{x-1}}{\ln 2^m (x-1)!}$$    from $x = 1$ to $m$ (since no number will have more than m factors) and see that for relatively small fixed $m$, most of the area under the curve f is is contained in a steep bell-shaped curve on the far left of the image.   I take this to suggest that as we consider very large sets, $S_m = \{ 1,2,3,...,2^m\},$ almost all elements of these sets have a "very small" number of factors (including repetitions).  Can this idea be (or has it been) quantified? The phrase "very small" is frustrating, and I think we might be able to say something more concrete about, for example, concentration of the proportion of area as a function of x and m...?  Thanks for any suggestions.   Edit: the answer Eric Naslund gave below is splendid and I won't neglect to accept it. In response to the answer, I wonder if there is any reason not to be able to get something like that answer from the expression $f(x)$?   After all, $f(x)$ appears to be a Poisson-like curve with a mean near the average number of prime factors. If I let m = 100 and then 500 (i.e., we're using $2^{100},2^{500}$), $f'(x) = 0$ at $x \approx 4.73, 6.34$, respectively, while $\ln\ln 2^m$ is respectively 4.23, 5.84. If f is a valid expression for the asymptotic behavior of $\pi_k(n)$, wouldn't we expect it to give us this additional information? Can we not prove it?


I am researching issues regarding oracle. I'm creating stored procedures and boot the following errors I also show them in the picture, I hope to help me resolve the error.  []

The UPSERT operation either updates or inserts a row in a table, depending if the table already has a row that matches the data:  if table t has a row exists that has key X:     update t set mystuff... where mykey=X else     insert into t mystuff...   Since Oracle doesn't have a specific UPSERT statement, what's the best way to do this?


One example could be when you meet a person (Peter Parker)  and remember only that his name starts with P or may Par.

Is there a word for the feeling/situation where we almost remember a thing/word/scene,etc. but are not able to exactly recall it. For example, the other day I was trying to recall the author of the book Contact (i.e. Carl Sagan). I wasn't able to exactly recall 'Carl Sagan' but was in that situation where I could feel I almost remember it. I could recall there was the letter 'C' in his name. Finally after some effort I could recall the exact name 'Carl Sagan'. Is there any word for such a feeling/situation? EDIT:  The feeling is not necessarily of trying to remember a single word(as in from vocabulary). It could be trying to remember a movie or an experience that you were a part of, etc. Eg: If say I was with friends and we were having a conversation. After few days I was able to recall a statement said by someone, but wasn't able to remember by whom; i.e. I had a clouded recollection and I could almost remember but not really.


I've coded a little comment function on my website by PHP. I have noticed that there is a huge security risk. Users can write some HTML/PHP/MySQL and insert it into the table. Is there a easy way to prevent this?  This is my PHP insert script:  <?php     if ( $_SERVER["REQUEST_METHOD"] == 'POST' ) {     $kommentar_neu = $_POST["kommentar"];      if (!empty($kommentar_neu)) {          $order = "INSERT INTO $db_kommentare         (kommentar, teilnehmer, kommentardatum, projektnummer)         VALUES         ('$kommentar_neu', '$teilnehmer_name', '$zeiteingabe',           '$projektnummer_hw_order')";     }     } ?>

If user input is inserted without modification into an SQL query, then the application becomes vulnerable to , like in the following example:  $unsafe_variable = $_POST['user_input'];   mysql_query("INSERT INTO `table` (`column`) VALUES ('$unsafe_variable')");   That's because the user can input something like value'); DROP TABLE table;--, and the query becomes:  INSERT INTO `table` (`column`) VALUES('value'); DROP TABLE table;--')   What can be done to prevent this from happening?


Hi anybody know how to get ATMEL unique ID from Atmega microcontroller?  I read  question but the code in answer not work in Atmel Stadio.  Edit:  Because we produce about 1000 board per week I want to a way that find our board in a project with broadcast a message to them and they replay a Uniq code that we could identify them.

I'd like a couple of Atmega328s to run the same program, but they're going to interface with each other and need unique IDs. Is there any such ID or serial number hard-coded on the chip?


How do I simplify the following:   $a^{b} \ \pmod {b}$ where $b$ is very large.  For example:  $2^{499} \pmod {999}$   How do I find the result without computing all?  I don't want to use such a quantity of memory for storing the value of $2^{499}$.  Is there any sort of property to find the result in a faster way?

How do I efficiently compute $a^b\bmod c$:  When $b$ is huge, for instance $5^{844325}\bmod 21$? When $b$ is less than $c$ but it would still be a lot of work to multiply $a$ by itself $b$ times, for instance $5^{69}\bmod 101$? When $(a,c)\ne1$, for instance $6^{103}\bmod 14$?  Are there any other tricks for evaluating exponents in modular arithmetic?  This is being asked in an effort to cut down on duplicates, see  and .


Suppose I have a vector v1 with values in the set {-1,0,1}  > v1 0  0 -1  1  0  0  0  0  0  0  0  0  0  0 -1  1  0  0  0  0  0  0  0  0  0  0  1 -1  0  0  0  0  0  0  0  0  0  0 -1  1  0  0  0  0   How to generate a vector v2 (with values in the set {-1,0,1}) that has zero correlation (or close to zero) with v1?

For a simulation study I have to generate random variables that show a predefined (population) correlation to an existing variable $Y$.  I looked into the R packages copula and CDVine which can produce random multivariate distributions with a given dependency structure. It is, however, not possible to fix one of the resulting variables to an existing variable.  Any ideas and links to existing functions are appreciated!     Conclusion: Two valid answers came up, with different solutions:   An R  by caracal, which calculates a random variable with an exact (sample) correlation to a predefined variable An R  I found myself, which calculates a random variable with a defined population correlation to a predefined variable     [@ttnphns' addition: I took the liberty to expand the question title from single fixed variable case to arbitrary number of fixed variables; i.e. how to generate a variable having predefined corretation(s) with some fixed, existing variable(s)]


I need to chmod something into drwxrwx---. What is drwxrwx--- in number form?

I need to set the same chmod, how to get number for -rw-r--r-- ?


I was surfing the Internet when I saw this sentence:  From there, you can see the beautiful scene where the sunset's reflecting over the ocean.  As far as I've known, "the" is used when you already have something written above, and you're referring it; or when you want to describe "only" things (e.g. the Moon, the Sun). Is it correct to write the sentence like so? If not, will replacing the "the" in "the beautiful scene" with "a" fix the sentence? Thanks for your answers!

Why do people say "I met him on the street" instead of "on a street", even though they're talking about a street for the first time and another person doesn't know what exact street they mean?


Why it is used division of sizeof(arr1[0]) to calculate length of array? It can be calulated simply without dividing it with sizeof(arr1[0]). Kindly explain this if anyone knows the reason     #include <stdio.h>    int main()   {       //Initialize array        int arr1[] = {1, 2, 3, 4, 5};        //Calculate length of array arr1       int length = sizeof(arr1)/sizeof(arr1[0]);        //Create another array arr2 with the size of arr1.        int arr2[length];        //Copying all elements of one array into another       for (int i = 0; i < length; i++) {            arr2[i] = arr1[i];        }          //Displaying elements of array arr1        printf("Elements of original array: \n");       for (int i = 0; i < length; i++) {            printf("%d ", arr1[i]);       }        printf("\n");        //Displaying elements of array arr2        printf("Elements of new array: \n");       for (int i = 0; i < length; i++) {            printf("%d ", arr2[i]);       }       return 0;   }

How do I determine the size of my array in C?   That is, the number of elements the array can hold?


Every time I run apt-get update, it spits out this. And when I try downloading things from apt-get install or Ubuntu software center, it doesn't work. Please help. Thanks.      Reading package lists... Done W: GPG error: http://us.archive.ubuntu.com/ubuntu yakkety-updates Release: The following signatures were invalid: BADSIG 40976EAF437D05B5 Ubuntu Archive Automatic Signing Key <ftpmaster@ubuntu.com> E: The repository 'http://us.archive.ubuntu.com/ubuntu yakkety-updates Release' is no longer signed. N: Updating from such a repository can't be done securely, and is therefore disabled by default. N: See apt-secure(8) manpage for repository creation and user configuration details. W: The repository 'http://ppa.launchpad.net/fkrull/deadsnakes/ubuntu yakkety Release' does not have a Release file. N: Data from such a repository can't be authenticated and is therefore potentially dangerous to use. N: See apt-secure(8) manpage for repository creation and user configuration details. W: The repository 'http://ppa.launchpad.net/gwendal-lebihan-dev/hexchat-stable/ubuntu yakkety Release' does not have a Release file. N: Data from such a repository can't be authenticated and is therefore potentially dangerous to use. N: See apt-secure(8) manpage for repository creation and user configuration details. W: The repository 'http://ppa.launchpad.net/ubuntuhandbook1/audacity/ubuntu yakkety Release' does not have a Release file. N: Data from such a repository can't be authenticated and is therefore potentially dangerous to use. N: See apt-secure(8) manpage for repository creation and user configuration details.

I frequently cross this issue, and always have to google for an answer. Does anyone have a permanent fix for BADSIG errors from apt-get?     W: GPG error:    lucid   Release: The following signatures were   invalid: BADSIG 54422A4B98AB5139   Oracle Corporation (VirtualBox archive   signing key)


I do Know What a NullPointerException means But in my case I'm receiving the exception from Http11Processor.java, Http11InputBuffer.java which I have never accessed it.  Complete Exception:     java.lang.NullPointerException       at org.apache.catalina.connector.CoyoteAdapter.service(CoyoteAdapter.java:389)       at org.apache.coyote.http11.Http11Processor.service(Http11Processor.java:1110)       at org.apache.coyote.AbstractProcessorLight.process(AbstractProcessorLight.java:66)       at org.apache.coyote.AbstractProtocol$ConnectionHandler.process(AbstractProtocol.java:785)       at org.apache.tomcat.util.net.NioEndpoint$SocketProcessor.doRun(NioEndpoint.java:1425)       at org.apache.tomcat.util.net.SocketProcessorBase.run(SocketProcessorBase.java:49)       at java.util.concurrent.ThreadPoolExecutor.runWorker(ThreadPoolExecutor.java:1142)       at java.util.concurrent.ThreadPoolExecutor$Worker.run(ThreadPoolExecutor.java:617)       at org.apache.tomcat.util.threads.TaskThread$WrappingRunnable.run(TaskThread.java:61)       at java.lang.Thread.run(Thread.java:745)   And a huge amount of warning Saying:      WARNING: Unable to add the resource at [/WEB-INF/lib/tomcat-i18n-ja.jar] to the cache because there was insufficient free space available after evicting expired cache entries - consider increasing the maximum size of the cache   Jan 25, 2017 4:46:11 PM org.apache.catalina.webresources.Cache getResource      INFO: The background cache eviction process was unable to free [10] percent of the cache for Context [/Duck] - consider increasing the maximum size of the cache. After eviction approximately [10,212] KB of data remained in the cache.   And my web project is getting really slow and is getting really hard to load if in my browser.  I'm using Apache 9.x and java 8  All replies are much appreciated

What are Null Pointer Exceptions (java.lang.NullPointerException) and what causes them?  What methods/tools can be used to determine the cause so that you stop the exception from causing the program to terminate prematurely?


Let $f(x_1,x_2)=\frac{1}{(x_1+ix_2)^2} \times \chi_{(-1,1) \times (-1,1)}(x_1,x_2)$. Evaluate the $L^p(R^2)$ norm of $f(x_1,x_2) \times \chi_{ \{(x_1,x_2):x_1^2+x_2^2 \leq 1\} }$ for every $1 \leq p \leq \infty$. This last function is =f inside the disk of radius 1 and is =0 outside the disk.  $||f||_{L^p(R^2)}=(\int_{ \{(x_1,x_2):x_1^2+x_2^2 \leq 1 \} } |f(x_1,x_2)|^p dx_1dx_2)^{1/p} =(\int_{ \{(x_1,x_2):x_1^2+x_2^2 \leq 1 \} } \frac{dx_1dx_2}{(x_1+ix_2)^{2p}})^{1/p} =(\int_0^{2 \pi} \int_0^1 \frac{rdrd \theta}{(rcos \theta+irsin \theta)^{2p}})^{1/p} =(\int_0^{2 \pi} \int_0^1 \frac{rdrd \theta}{(rcos \theta+irsin \theta)^{2p}})^{1/p} =(\int_0^{2 \pi} \int_0^1 \frac{drd \theta}{r^{2p-1}e^{i \theta 2p}})^{1/p} =(\int_0^{2 \pi} e^{-i \theta 2p} d \theta \times \int_0^1 r^{1-2p}dr)^{1/p} =(\frac{e^{-i \theta 2p}}{-i2p}]_0^{2 \pi} \times r^{-2p}]_0^1)^{1/p} =[(\frac{e^{-i4 \pi p}}{-i2p}+\frac{1}{i2p} \times (1-?)]^{1/p}$  Where I have the question mark is my problem can anyone help me finish solving the problem?


I have a model similar to the following:  y = a + b + c + d + e;   a,b, and c are binary variables while d is other control variables and e is error term. For my whole sample, each observation has a 1 for either a,b, or c -- each observation must belong to either a,b, or c (no observations can be 0 in all three). To avoid dummy variable trap, I can run my model two ways:  y = a + b + c + d + e (no intercept)  or  y = intercept + b + c + d + e  I've read around, including here, that intercept should never be dropped unless I am sure the regression goes through the origin. That would mean I should use the second model. However, is it possible for me to drop the intercept in this situation and use the first model -- would my estimates be biased if I dropped the intercept for the first model?  Thanks!

I tried a regression in the form ${\rm logit}(Y) = {\rm coefficient}\times X + 0 + e$, where $Y$ is a binomial variable and $X$ is a factor variable with $n$ levels. I noticed that removing the intercept yields higher $p$ values. I'm wondering how to interpret it though.  Since removing the intercept makes it equal to $0$, I believe that the coefficients returned are relative to a $0$ probability of the event $Y$ and that all $X$ factors are in the $0$ state. But I think this is impossible isn't it?  $X$ are mutually exclusive factors, therefore it's impossible to have a case where no factor is $1$, at least in the presented observations. And it cannot be interpreted like the coefficient is relative to hypothetical cases in which really no one of the factors is present, because we have no data like that.  Regarding $Y$ having a $0$ intercept, wouldn't it mean forcing the probability of the event to $0$ when none of the factors is present? Again, this is an impossible case.  Nonetheless this kind of regression would allow me to retrieve pure probability range of the event Y given a factor by transforming the coefficients in the confidence intervals given as $\exp({\rm coefficient})/(1 + \exp({\rm coefficient}))$, and the $p$ values would test whether this probability is not $50\%$. This could also be a valuable result, since it would give independent probabilities for each factor.  Am I wrong?


I own a Canon EF-S 17-55mm f/2.8 IS USM Lens. The other day I was trying to take a photo with the camera pointed directly towards the ground and mounted on a tripod. When I let go of the lens after adjusting the zoom, it would fully extend on its own due to gravity.  I am not certain if it has just now started doing this or if it has always done it and I haven't noticed because I am normally holding the lens while shooting. Is this normal behavior for this lens? If not is there an adjustment I can make to eliminate this behavior?

The Nikkor 18-200mm VR lens creeps until fully extended if pointed vertically down.  This is preventing me from using it on a .  Is there a piece of equipment or a simple (and non-destructive!) hack to hold the zoom in place?  I found  half-overlapping the zoom ring, but that doesn't sound very secure. I'm really after something that will hold it exactly where I put it for a really long time.


I have query like this   update new_keyword set key_count=key_count-1  where  id in     (select b.id  from new_keyword b ,new_news a ,new_mudoal_label c where c.label_id=b.id and a.id=c.id and c.type=1 and c.id=186)    The sub query return id for exmple (1,2,3) but mysql show me error

I have a table story_category in my database with corrupt entries. The next query returns the corrupt entries:  SELECT *  FROM  story_category  WHERE category_id NOT IN (     SELECT DISTINCT category.id      FROM category INNER JOIN         story_category ON category_id=category.id);   I tried to delete them executing:  DELETE FROM story_category  WHERE category_id NOT IN (     SELECT DISTINCT category.id      FROM category        INNER JOIN story_category ON category_id=category.id);   But I get the next error:      #1093 - You can't specify target table 'story_category' for update in FROM clause    How can I overcome this?


I have tried the following code, it displayed the date values for the first two times of the run, but later started throwing a nullpointerexception, any idea why  this is happening?  DBCursor c=test.find(); while(c.hasNext()) {     DBObject dbc=c.next();     Date n=(Date)dbc.get("time");     System.out.println(n);      Calendar cal = new GregorianCalendar();     cal.setTime(n);     System.out.println(cal.get(Calendar.DAY_OF_MONTH));  }


Could I get some recommendations on an IPS solutions that incorporates some form of DDOS protection / what do you currently implement on your network?  Thanks

This is a  about DoS and DDoS mitigation.   I found a massive traffic spike on a website that I host today; I am getting thousands of connections a second and I see I'm using all 100Mbps of my available bandwidth.  Nobody can access my site because all the requests time out, and I can't even log into the server because SSH times out too!  This has happened a couple times before, and each time it's lasted a couple hours and gone away on its own.  Occasionally, my website has another distinct but related problem: my server's load average (which is usually around .25) rockets up to 20 or more and nobody can access my site just the same as the other case.  It also goes away after a few hours.  Restarting my server doesn't help; what can I do to make my site accessible again, and what is happening?  Relatedly, I found once that for a day or two, every time I started my service, it got a connection from a particular IP address and then crashed.  As soon as I started it up again, this happened again and it crashed again.  How is that similar, and what can I do about it?


The outcome of the logistic model is binary (0;1) - intervention that occurs in hospital (3838 interventions in 51096 cases). I am modeling the effect of age groups (60 to 74; 75 to 89 and 90+) and sex (female) on the outcome. My reference age group is male aged 18-59.  My logistic regression coefficients look like this:  intercept  -4.313 female      0.5117 age 60-74   0.4081 age 75-89   1.1761 age 90+     1.5612.   Corresponding odds ratios:  female      1.668 years60_74  1.504 years75_90  3.242 years90     4.764   Now I am trying to obtain a probability of an outcome = 1 for each case. What is the right way to do this?

I have created a Logistic Regression using the following code:  full.model.f = lm(Ft_45 ~ ., LOG_D) base.model.f = lm(Ft_45 ~ IP_util_E2pl_m02_flg) step(base.model.f, scope=list(upper=full.model.f, lower=~1),      direction="forward", trace=FALSE)   I have then used the output to create a final model:  final.model.f = lm(Ft_45 ~ IP_util_E2pl_m02_flg + IP_util_E2_m02_flg +                             AE_NumVisit1_flg + OP_NumVisit1_m01_flg + IP_TotLoS_m02 +                             Ft1_45 + IP_util_E1_m05_flg + IP_TotPrNonElecLoS_m02 +                             IP_util_E2pl_m03_flg + LTC_coding + OP_NumVisit0105_m03_flg +                            OP_NumVisit11pl_m03_flg + AE_ArrAmb_m02_flg)   Then I have predicted the outcomes for a different set of data using the predict function:  log.pred.f.v <- predict(final.model.f, newdata=LOG_V)   I have been able to use establish a pleasing ROC curve and created a table to establish the sensitivity and specificity which gives me responses I would expect.   However What I am trying to do is establish for each row of data what the probability is of Ft_45 being 1. If I look at the output of log.pred.f.v I get, for example,:  1 -0.171739593     2 -0.049905948     3 0.141146419     4 0.11615669     5 0.07342591     6 0.093054334     7 0.957164383     8 0.098415639     . . . 104 0.196368229     105 1.045208447     106 1.05499112   As I only have a tentative grasp on what I am doing I am struggling to understand how to interpret the negative and higher that 1 values as I would expect a probability to be between 0 and 1.  So my question is am I just missing a step where I need to transform the output or have I gone completely wrong. Thank you in advance for any help you are able to offer.


Given the following code, is there a way I can call class A's version of method X?  class A {   virtual void X() { Console.WriteLine("x"); } }  class B : A {   override void X() { Console.WriteLine("y"); } }  class Program {   static void Main()   {     A b = new B();     // Call A.X somehow, not B.X...   }

In the code below I tried in two ways to access the parent version of methodTwo, but the result was always 2. Is there any way to get the 1 result from a ChildClass instance without modifying these two classes?  class ParentClass {     public int methodOne()     {         return methodTwo();     }      virtual public int methodTwo()     {         return 1;     } }  class ChildClass : ParentClass {     override public int methodTwo()     {         return 2;     } }  class Program {     static void Main(string[] args)     {         var a = new ChildClass();         Console.WriteLine("a.methodOne(): " + a.methodOne());         Console.WriteLine("a.methodTwo(): " + a.methodTwo());         Console.WriteLine("((ParentClass)a).methodTwo(): "          + ((ParentClass)a).methodTwo());         Console.ReadLine();     } }   Update ChrisW posted this:     From outside the class, I don't know   any easy way; but, perhaps, I don't   know what happens if you try   reflection: use the Type.GetMethod   method to find the MethodInfo   associated with the method in the   ParentClass, and then call   MethodInfo.Invoke   That answer was deleted. I'm wondering if that hack could work, just for curiosity.


"I have fallen down." I thought adjective at first, but it isn't really an adjective right? Because it doesn't really modify me, it modifies how I fell? So it is modifying the verb.

I need to know what down in this specific sentence means. I don't know if it is a preposition or an adverb.


I have a generic method   Foo<T>   I have a Type variable bar   Is it possible to achieve something like Foo<bar>  Visual Studio is expecting a type or namespace at the bar.

What's the best way to call a generic method when the type parameter isn't known at compile time, but instead is obtained dynamically at runtime?  Consider the following sample code - inside the Example() method, what's the most concise way to invoke GenericMethod<T>() using the Type stored in the myType variable?  public class Sample {     public void Example(string typeName)     {         Type myType = FindType(typeName);          // What goes here to call GenericMethod<T>()?         GenericMethod<myType>(); // This doesn't work          // What changes to call StaticMethod<T>()?         Sample.StaticMethod<myType>(); // This also doesn't work     }      public void GenericMethod<T>()     {         // ...     }      public static void StaticMethod<T>()     {         //...     } }


A friend upgraded my device to iOS7. It didn't work properly so I downgraded it a few days later to iOS6.  I have backups both in iCloud and on OS X (iTunes). When I try to restore either I get a message:     To restore this iPhone from this backup, you must first set up the   iPhone as new and restore the software to the latest version.   Without this backup I am very inconvenienced, so I ask is it possible to restore an iPhone from a iOS7 Backup if you downgraded the phone to iOS6?

I'm trying to downgrade my iPhone from the current iOS to an older one, as per .  But every time I try, I always receive this error:     The iPhone could not be restored. This device isn't eligible for the requested build.     I have tried many times with and without DFU mode. Any ideas regarding this?


Often I visit a number of stackexchange sites. There does not seem to be a way to log out of all of the sites with one click. IMHO this would be a nice feature. If there is such a feature I would appreciate learning how to use it.

I propose adding a "log out globally" button on user pop-up to fast log out from all Stack Exchange pages. Why enter a  page and then logout if we can have one button ?


What does it mean when my neural network always gets stuck at the exact number 8.6791 when I use binary-crossentropy loss? Some strange local minimum?  It happens regardless of my learning rate, initialization, activation function, regularizer, or choice of optimizer.   The only thing that changes it is a change of architecture or a new loss function, which are things I don't want to change...

I'm training a neural network but the training loss doesn't decrease. How can I fix this?  I'm not asking about overfitting or regularization. I'm asking about how to solve the problem where my network's performance doesn't improve on the training set.    This question is intentionally general so that other questions about how to train a neural network can be closed as a duplicate of this one, with the attitude that "if you give a man a fish you feed him for a day, but if you teach a man to fish, you can feed him for the rest of his life." See this Meta thread for a discussion:   If your neural network does not generalize well, see:


Consider the following code:  float x = 0.1; if ( x> 0.1){     printf("if"); }   Now when I check equality, it is explained that x is being converted to double by padding 0s at the end, but 0.1 on RHS is stored in double, hence the inequality. But if I go through this logic, the "if" in the above code should have given false, but istead it is true. Why?

Consider the following code:   0.1 + 0.2 == 0.3  ->  false   0.1 + 0.2         ->  0.30000000000000004   Why do these inaccuracies happen?


We have just received this email regarding my partners UK visitor visa application. I know that we should just wait the 3 days for him to collect his passport to see the outcome, but I can’t help but wonder if this email means he has been rejected? I’ve been doing some research and I’ve came across forums where people have said they have received different worded emails based on the outcomes of their visa applications.   I just want to get everybody’s comments or experiences...

I received the following email this morning. Does it mean that my visa has been granted?  UK Visas & Immigration has now assessed your UK visa application and made a decision. Your documents and the decision will be sent back to either the UK Visa Application Centre where you applied, where we will contact you by email over the next few days to collect them, or if you are using the courier return service, will be sent directly back to the address provided. Please note that TLScontact does not know the outcome of the assessment and has played no role in the decision-making process. Kind regards, TLScontact UK VIsas & Immigration


Let $S$ be a unit sphere in $\mathbb{R}^3$ - I am told that $\partial S = \emptyset $, but why does the unit sphere have no boundary?

The unit sphere in $\mathbb{R}^3$ is $\{(x,y,z) : x^2 + y^2 + z^2 = 1 \}$. I always hear people say that this is closed and that it has no boundary. But isn't every point on the sphere a boundary point? Since for every point $x$ on the sphere, any open ball in $\mathbb{R}^3$ (defined by $\{ y : |y-x| < r\}$) contains points on the sphere and outside/inside the sphere. Hence it is a boundary point.  I also sometimes see that people say instead of an open ball in $\mathbb{R}^3$, I should be looking at an open neighbourhood of the sphere but this doesn't make any sense I don't think as the definition of the open set is as I gave above.  Or is it that people mean it's a closed surface (analogous to a closed curve)?


I bought a ipod i tried contacting the owner i tried everything but there apple id and password is on the ipod i cant login or try to set it up for me becuase it says its linked to an apple id but the person i bought it from isnt answering my calls and i cant get in the ipod how do i get in it is there any other way????? any website ideas anything? i payed 200.oo i really need help geeting into the ipod a differnt way they basicly put find my iphone on the device

I bought a used iPhone 4.  The previous owner re-set the phone, but did not remove the device from their iCloud.  It is asking me for their iCloud information when I try to set-up the phone.  I don't have contact with the previous owner anymore.  Is there a way to remove the device from their iCloud so I can use a new one?


I'm a bit confused about an object casting example in Java.  Say you have the following:  Complex cr = new Real(1.2);   Here, Complex is the superclass, Real is the subclass.  If I cast cr to a Complex, say:  (Complex)cr.method();   it acts the same in the program as just regular  cr.method();   Why is it?

I have seen this mentioned a few times and I am not clear on what it means. When and why would you do this?   I know what interfaces do, but the fact I am not clear on this makes me think I am missing out on using them correctly.   Is it just so if you were to do:  IInterface classRef = new ObjectWhatever()   You could use any class that implements IInterface? When would you need to do that? The only thing I can think of is if you have a method and you are unsure of what object will be passed except for it implementing IInterface. I cannot think how often you would need to do that.   Also, how could you write a method that takes in an object that implements an interface? Is that possible?


Let $A$ be a square matrix such that $a_{i,j}=|j-i|$, show that the determinant is $D_n=(-1)^{n-1}(n-1)2^{n-2}$   I have been struggling with this problem for a while now and I was hoping to get the slightest of nudges towards the solution. I would appreciate it if you phrase your hints in forms of questions so I can work through the idea. Please don't post a complete solution unless I asked, and thank you for your time. I wanted to obtain a recurrence relation between $D_n$ and $D_{n-1}$ but I failed. So far, I attempted to use Gaussian elimination on the first and last row and I obtained $(-1,1,\ldots,1)$ and $(1,1,\ldots,-1)$ for the first and last row respectively but I didn't know how to proceed from then on.

A friend of mine asked me to calculate the value of a perticular determinant, but after spending quite some time on it I'm unable to find a solution to it. The given determinant is:  $$         \begin{vmatrix}         0 & 1 & \cdots &n-2&n-1 \\         n-1 & 0 & \cdots &n-3&n-2 \\         \vdots & \vdots & \ddots & \vdots &\vdots\\         1 & 2 & \cdots & n-1 &0 \\         \end{vmatrix} $$  Also my friend is supposed to solve this determinant by more or less elementary methods, like elementary row operations, transpose... and no eigenvalues, nor adjoints.    First of all it seems like the value of the determinant is:  $$         \frac{1}{2n}\begin{vmatrix}         0 & n & \cdots &n&n \\         n & 0 & \cdots &n&n \\         \vdots & \vdots & \ddots & \vdots &\vdots\\         n & n & \cdots & n &0 \\         \end{vmatrix} $$  Anyway I tried adding all the rows to the last one and by taking $\frac{n(n-1)}{2}$ in front to have a row of 1s, but it didn't helped me much.  Then I tried to use the fact that:  $$         \begin{vmatrix}         A & C \\         0 & B \\         \end{vmatrix} = \det(A) \cdot \det(B) $$, where $A,B$ are square matrices. By subtracting the $(\frac{n}{2} + i)$-th column from the $i$-th ($i=0,1,2...$) and then adding the $i$-th row to the $(\frac{n}{2} + i)$-th row, one can obtain the wanted form, but again nothing more. Also this method pretty much fails when $n$ is odd.  My best attepmt was to notice that $A_{ij} + A_{ji} = n$ and using $\det(A^{T}) = \det(A)$, but unfortunatelly  $$\begin{vmatrix}         0 & n & \cdots &n&n \\         n & 0 & \cdots &n&n \\         \vdots & \vdots & \ddots & \vdots &\vdots\\         n & n & \cdots & n &0 \\         \end{vmatrix} \not = \begin{vmatrix}         0 & 1 & \cdots &n-2&n-1 \\         n-1 & 0 & \cdots &n-3&n-2 \\         \vdots & \vdots & \ddots & \vdots &\vdots\\         1 & 2 & \cdots & n-1 &0 \\         \end{vmatrix} + \left( \begin{vmatrix}         0 & 1 & \cdots &n-2&n-1 \\         n-1 & 0 & \cdots &n-3&n-2 \\         \vdots & \vdots & \ddots & \vdots &\vdots\\         1 & 2 & \cdots & n-1 &0 \\         \end{vmatrix}\right)^T$$ I tried to decompose the LHS by using the $n-$linearity of the determinants, by then one gets $2^n$ determinants on the RHS similar to the first one, but things are messy.


Can someone list the fields in history object? Since I wanted them to display in a page block in vf page despite using history related list.

I cannot find anything equivalent to:  <apex:relatedList list="ProcessSteps" />   which works for Approvals.  The only solution I can think of is to build a controller/extension that gets the raw history data (from MyObj__History) and write my own page elements to render it in a table. Is there a neater solution?


Why does TLS as used in https need DHKE? Since the server presents a certificate with a public key, can that not be used for session key establishment? The client (browser) can generate a session key, encrypt it using the server's public key & then send it to the server & that can be used for encrypting all future messages from client to server & from server to client. So why is DHKE required in TLS at all?

Specifically with TLS,   What is RSA used for during key exchange? What is Diffie-Hellman used for? Since RSA already provides authenticity and confidentiality, why is Diffie-Hellman used during the exchange? I've heard DH exchanges aren't secure. Is that something to be worried about in an RSA with Diffie-Hellman exchange? If DH is used for forward secrecy. What's wrong with just generating new RSA key pairs and exchanging keys with that?   The use of Diffie-Hellman along with RSA is the one thing I don't understand about modern Public-Key Cryptography.