Is "she don't" sometimes considered correct form? Recently I was exposed to a lot of uses of "She don't + infinitive" ( + don't), instead of "she doesn't + infinitive" ( + doesn't). I'm not sure if it is a mistake or just accepted usage sometimes.   I found it in a very famous songs such as:   "" by Bon Jovi   Also in  (in 3:04)     But she don't know you like I know you Slim, no one does   She don't know what it was like for people like us growin' up, you gotta call me man   I'll be the biggest fan you'll ever lose   Sincerely yours, Stan, P.S. we should be together too   You can find a lot more by searching in you tube "he don't" or "she don't".
She don't know that she's a lucky girl Here's from an old song (1963) "" by Lesley Gore.       She don't know that she's a lucky girl   Here's another example, "Silence is Golden" (1964) by The Tremeloes [YouTube ]      Oh, don't it hurt deep inside to see someone do something to her.   I wonder how common it is to say "She don't" or "don't it". Do native English speakers often use these?  By the way, Lesley Gore is American, and the Tremeloes is British.
In other words, why do we need to connect the battery positive to the negative to get electron flow? As far as I know, voltage difference is what drives current flow.  From what I understand, there's a surplus of negative charge (electrons) in the positive end of a battery (weird I know, but I guess they do it for mathematical reasons). Between the positive and negative ends of the battery is some kind of wall that prevents the electrons from diffusing, so they have to go the long way (through a wire to the other end of the battery) to diffuse and reach the receiver atoms/molecules.  So why not just have the surplus of electrons on one end go to ground to equalize (e.g. 5 volts to 0 volts should produce current)? Why does it have to be the other end of the battery? I'm sure there's atoms/molecules in the ground that could accept the extra electrons (receivers atoms/molecules for the donors in the battery). is there a way to add hyperlinks in a Illustrator document? I am creating a CV in Illustrator and exporting it to a PDF. I just read an article which stated that it is a good idea that we use hyperlinks in the CV so that the employer can just click and not have to copy the address. I like this idea, but I don't know how to add a hyperlink to text in Illustrator. Is it possible? What are the differences between "lay" and "lie"? Proving that a Hölder space is a Banach space I am trying to show that the Hölder space $C^{k,\gamma}(\bar{U})$ is a Banach space. To do this, I successfully proved that the mapping $\| \quad \| : C^{k,\gamma}(\bar{U}) \to [0,\infty)$ is a norm, by proving its properties.   But how do I show that the sequence $\{u_k\}_{k=1}^\infty \subset C^{k,\gamma}(\bar{U})$ converges to $u \in C^{k,\gamma}(\bar{U})$, that is, how do I show that $$\lim_{k \to \infty}\|u_k-u\|=0,$$ which would mean the normed linear space is complete, and hence a Banach space?  Here are the following taken from PDE Evans, 2nd edition, page 255:     Definition. The Hölder space $$C^{k,\gamma}(\bar{U})$$ consists of all functions $u \in C^k(\bar{U})$ for which the norm $$\|u\|_{C^{k,y}(\bar{U})}:= \sum_{|\alpha|\le k} \|D^\alpha u \|_{C(\bar{U})}+\sum_{|\alpha|=k} [D^\alpha u]_{C^{0,\gamma}(\bar{U})}$$ is finite.   Also from page 254,     Definitions. (i) If $u : U \to \mathbb{R}$ is bounded and continuous, we write $$\|u\|_{C(\bar{U})}:=\sup_{x\in U}|u(x)|.$$      (ii) The $\gamma$th-Hölder seminorm of $u : U \to \mathbb{R}$ is $$[u]_{C^{0,\gamma}(\bar{U})}:=\sup_{\substack{x,y\in U \\ x \neq y}} \left\{\frac{|u(x)-u(y)|}{|x-y|^\gamma} \right\},$$ and the $\gamma$th-Hölder norm is $$\|u\|_{C^{0,\gamma}(\bar{U})}:=\|u\|_{C(\bar{U})}+[u]_{C^{0,\gamma}(\bar{U})}.$$   This is all I got so far: \begin{align} \|u_k-u\|_{C^{k,\gamma}(\bar{U})}&amp;=\sum_{|\alpha|\le k} \|D^\alpha u \|_{C(\bar{U})}+\sum_{|\alpha|=k} [D^\alpha u]_{C^{0,\gamma}(\bar{U})} \\ &amp;= \sum_{|\alpha|\le k} \sup_{x\in U} |u_k(x)-u(x)|+ \sum_{|\alpha|=k} \sup_{\substack{x,y\in U \\ x \neq y}} \left\{\frac{|[u_k(x)-u(x)]-[u_k(y)-u(y)]|}{|x-y|^\gamma} \right\}. \end{align} Now, where can I go from here, to show that $\lim_{k \to \infty}\|u_k-u\|_{C^{k,\gamma}(\bar{U})}=0$? The sequence is Cauchy, and I have to use that fact somehow. This question is a result of having too much free time years ago during military service. One of the many pastimes was playing tic-tac-toe in varying grid sizes and dimensions, and it lead me to a conjecture. Now, after several years of mathematical training at a university, I am still unable to settle the conjecture, so I present it to you.  The classical tic-tac-toe game is played on a $3\times3$ grid and two players take turns to put their mark somewhere in the grid. The first one to get three collinear marks wins. Collinear includes horizontal, vertical and diagonal lines. Experience shows that the game always ends in a draw if both players play wisely.  Let us write the grid size $3\times3$ as $3^2$. We can change the edge length by playing on any $a^2$ grid (where each player tries to get $a$ marks in a row on the $a\times a$ grid). We can also change dimension by playing on any $a^d$ grid, for example $3^3=3\times3\times3$. I want to understand something about this game for general $a$ and $d$. Let me repeat: The goal is to make $a$ collinear marks.  I assume both players play in an optimal way. It is quite easy to see that the first player wins on a $2^d$ grid for any $d\geq2$ but the game is a tie on $2^1$. The game is a tie also on $3^1$ and $3^2$, but my experience suggests that the first player wins on $3^3$ but the game ties on $4^d$ for $d\leq3$. It seems quite credible that if there is a winning strategy on $a^d$, there is one also on $a^{d'}$ for any $d'\geq d$, since more dimensions to move in gives more room for winning rows.  to a related question tells that for any $a$ there is $d$ so that there is a winning strategy on $a^d$.  This brings me to the conjecture:     There is a winning strategy for tic-tac-toe on an $a^d$ grid if and only if $d\geq a$. (Refuted by TonyK's answer below.)   Is there a characterization of the cases where a winning strategy exists? It turns out not to be as simple as I thought.  To fix notation, let $$ \delta(a)=\min\{d;\text{first player wins on }a^d\} $$ and $$ \alpha(d)=\max\{a;\text{first player wins on }a^d\}. $$ The main question is:     Is there an explicit expression for either of these functions?   Or decent bounds?   Partial answers are also welcome.   Note that the second player never wins, as was discussed in .    A remark for the algebraically-minded: We can also allow the lines of marks to continue at the opposite face when they exit the grid; this amounts to giving the grid a torus-like structure. Now there are no special points, unlike in the usual case with boundaries. Collinear points on a toric grid of size $a^d$ corresponds to a line (maximal collinear set) in the module $(\mathbb Z/a\mathbb Z)^d$. (If $a$ is odd, then $a$ collinear points in the mentioned module add up to zero, but the converse does not always hold: the nine points in $(\mathbb Z/9\mathbb Z)^3$ with multiples of three as all coordinates add up to zero but are not collinear.) This approach might be more useful when $a$ is a prime and the module becomes a vector space. Anyway, if this version of the game seems more manageable, I'm happy with answers about it as well (although the conjecture as stated is not true in this setting; the first player wins on $3^2$). How to take the gradient of the quadratic form? I wonder  whether there is a function $f\colon\Bbb R\to\Bbb R$ with the folowing characteristic?   for every two real numbers $\alpha,\beta,\alpha\lt\beta$,  $$\{f(x):x\in(\alpha,\beta)\}=\Bbb R$$   I can't  say such a function does not exist, neither can I construct a example  Thanks a lot! Inspired by  I was wondering why the Half Blood Prince left his book as a class copy. One would think that it was precious to him, since he invested so much in it. He made up potions and spells up by his own. I would expect him to keep it to himself. Both, because of the personal value of it, and to prevent others to (as happened at last) use his work (misguided and/or against him).  Is there any canon explanation why his book was left as class copy? Lets say I'm building a base class which will be extended upon by the children class. So a base class is called Base and children can be Child1, Child2, etc.  In the base class's constructor, how can i get the value of Child1/Child2?  This is all using PHP lstlisting in two columns I would like to have two listings, side by side.  They have approximately the same length, therefore I don't care about how it's layed out if one part is longer.  I tried  \begin{minipage}{\textwidth} \begin{parcolumns}{2} \colchunk{\begin{lstlisting}[caption=code 1,frame=tlrb]{Name} void code() {  } \end{lstlisting}}  \colchunk{\begin{lstlisting}[caption=code 2,frame=tlrb]{Name} void code() {  } \end{lstlisting}}  \colplacechunks \end{parcolumns} \end{minipage}   while the border of the listings end up in chunks as follows:   Btw I'm using the texlive-package &amp; pdflatex on Ubuntu and the document class is  \documentclass[ a4paper, 10pt, twoside, openright, notitlepage, parskip=half, ]{scrreprt}   Any idea is appreciated. Invariance of a maximally entangled state under unitary operation $U \otimes U^\dagger$ Apparently the (d-dimensional) maximally entangled state, $|E \rangle = \sum_{i} |ii\rangle /\sqrt{d}$ is invariant under operations of the form $U \otimes U^\dagger$. I want to prove this result, which amounts to showing that  $$\left(\sum_i U | i \rangle \otimes V |i \rangle = \sum_i | ii \rangle \right)\Rightarrow (V = U^\dagger )$$  I have no idea how to even start. I suppose it's some simple linear algebra result, but I don't see it. A hint would be appreciated. "No worthy mechs found" when trying to relay email to Gmail using Postfix I've been trying to configure email to forward to Gmail, using Postfix to relay email to smtp.gmail.com. However, I'm failing to get it to authenticate with smtp.gmail.com, which is a rather vital prerequisite to getting anything working…  The mail logs show only:  Oct 29 15:50:14 gsnedders-1 postfix/master[6596]: daemon started -- version 2.7.1, configuration /etc/postfix Oct 29 15:50:19 gsnedders-1 postfix/pickup[6598]: EBA1F78750: uid=1000 from=&lt;gsnedders&gt; Oct 29 15:50:19 gsnedders-1 postfix/cleanup[6603]: EBA1F78750: message-id=&lt;20111029145019.EBA1F78750@mail.gsnedders.com&gt; Oct 29 15:50:19 gsnedders-1 postfix/qmgr[6599]: EBA1F78750: from=&lt;gsnedders@gsnedders.com&gt;, size=324, nrcpt=1 (queue active) Oct 29 15:50:19 gsnedders-1 postfix/cleanup[6603]: F2D557874F: message-id=&lt;20111029145019.EBA1F78750@mail.gsnedders.com&gt; Oct 29 15:50:19 gsnedders-1 postfix/local[6605]: EBA1F78750: to=&lt;me@gsnedders.com&gt;, orig_to=&lt;me&gt;, relay=local, delay=0.04, delays=0.03/0.02/0/0, dsn=2.0.0, status=sent (forwarded as F2D557874F) Oct 29 15:50:19 gsnedders-1 postfix/qmgr[6599]: F2D557874F: from=&lt;gsnedders@gsnedders.com&gt;, size=454, nrcpt=1 (queue active) Oct 29 15:50:19 gsnedders-1 postfix/qmgr[6599]: EBA1F78750: removed Oct 29 15:50:20 gsnedders-1 postfix/smtp[6606]: warning: SASL authentication failure: No worthy mechs found Oct 29 15:50:20 gsnedders-1 postfix/smtp[6606]: F2D557874F: SASL authentication failed; cannot authenticate to server smtp.gmail.com[74.125.157.108]: no mechanism available   And the postfix config is:  relayhost = [smtp.gmail.com]:587  smtp_use_tls = yes smtp_sasl_auth_enable = yes smtp_sasl_password_maps = hash:/etc/postfix/sasl/passwd smtp_sasl_security_options = noanonymous  smtp_tls_eccert_file = smtp_tls_eckey_file = smtp_tls_security_level = may smtp_tls_CAfile = /etc/postfix/cacert.pem smtpd_tls_received_header = yes tls_random_source = dev:/dev/urandom smtpd_tls_security_level = may I went into Live CD and activated Terminal.  Then, I typed gksu nautilus to gain temporary access to my root directory. Once the Manager window opened, I clicked on File System and navigated to the /etc/default/grub directory. I then double clicked on the Grub file to make changes to the TIMEOUT and TIMEOUT_QUIET.  I saved the file after I made the changes. Then I went back to the terminal mode and typed:  gksu update-grub   The error message I got was:     Can not find a device for / (is /dev mounted?)   I'm not sure what to do at this point. How can I run update-grub to update the changes? Thanks in advance for your help. Please be very detailed and specific with your response as I am totally new to this environment. Is it possible to extend an Indian business visa from India? I'm an Egyptian girl from Cairo. I'm planning a six-month trip to India in December for training purposes, but getting a six-month visa is so hard from my country.  If I can only get 3 months instead of 6 months, is it possible to extend the business visa in India? $a_{n+1}=\log(1+a_n),~a_1>0$. Then find $\lim_{n \rightarrow \infty} n \cdot a_n$ Splitting pieces of a mesh into a new object I've just made a set of armor by using a male body mesh model I downloaded and selecting certain faces from the model and duplicating them to create viable armor pieces.   Since the pieces are just duplicated from the same mesh male model, then it's still part of the same object.   How do I separate the armor piece from the model to make the pieces separate? How to determine $\int x^2 e^{-x^2}\mathrm dx$? I was wondering how to determine $$\int x^2 e^{-x^2}\mathrm dx?$$ I tried to move the exponential into the integrator, and apply integration by parts, but that will make the integrand more complicated. Thank you! We are doing the research about applying remote sensed data on hydro logical model . After delineating, the river basin from DEM data using Argis9.3, We have to map the remote sensed data derived from products like PERSIANN. Our task now is to create the grid with the same resolution with the available remote sensed data( for example 0.25degree x 0.25 degree). and have a certain corner that matches with the obtained data.   Does anyone know how to deal with our problem? Description of varieties in $\mathbb{P}^2\times \mathbb{P}^1$ If $[x:y]$ are coordinates of $\mathbb{P}^1$ and $[X:Y:Z]$ are coordinates of $\mathbb{P}^2$, what do the following varieties look like?   $V(x^2X+y^2Y+xyZ)\subset \mathbb{P}^2\times \mathbb{P}^1$ $V(x^2X^2+y^2Y^2+xyZ^2)\subset \mathbb{P}^2\times \mathbb{P}^1$   As of now, I don't quite know how to visualize these varieties. A detailed geometric picture would be especially welcome. Diffeomorphism group of the unit circle I am given to understand that the group of diffeomorphisms of the unit circle, $\operatorname{Diff}(\mathbb{S}^1)$, has two connected components, $\operatorname{Diff}^+(\mathbb{S}^1)$ and $\operatorname{Diff}^-(\mathbb{S}^1)$, the diffeomorphisms that preserve or reverse the canonical (counterclockwise) orientation, respectively.  Question 1: How does one prove that?  Question 2: Given $\Phi, \Psi \in \operatorname{Diff}^+(\mathbb{S^1})$, can one construct an explicit path joining them?  I'd be satisfied already with a proof that we can join $\Psi, \Phi \in \operatorname{Diff}^+ (\mathbb{S}^1)$, without giving the path explicitly. I tried the obvious path, $t \mapsto \dfrac{t\Phi + (1-t)\Psi}{|t\Phi + (1-t)\Psi|}$, but that doesn't seem to work.  Thanks.