“group have” vs “group has”
Are collective nouns always plural, or are certain ones singular?
How to quickly and easily create a grid with Adobe Photoshop? Suppose I need to draw a grid in Photoshop, each square being the same size, and a certain amount of pixels tall and wide, and the grid lines having a certain pixel width. Like this:     Currently I do this by drawing one square, copy+pasting+moving+merging it a few times to create one row, copy/pasting that a few times to create more rows above each other.  It works, but feels very clumsy, I'm sure there ought to be a more efficient way to do this?  What are some efficient methods for creating a grid of equally sized squares? Error notification on plone 4 Why is '.' a hard link in Unix? Why can we always drop any constant term in a Lagrangian density in quantum field theory?  This issue is somehow related to the constant term being some kind of cosmological constant.  Can you please explain this issue? What CMYK values should I use for rich black, and how should I handle tints/shades? I'm working on a full color piece in Illustrator which has a few large black and grey areas.  I'd like the black to be a rich black. I was thinking c30 m30 y30 k100. What is the best ways to match my gray areas? Proof that a formula $\theta$ is valid if and only if $\forall x \theta$ is valid I need to prove for first-order logic that $\vDash\theta$ if and only if $\vDash\forall x \theta$. The proof for the forward direction seems simple enough:  Suppose $\theta$ is valid. Let $\mathfrak{A}$ be an arbitrary structure and let $s:V\to |\mathfrak{A}|$ be an arbitrary function assigning values of the universe of $\mathfrak{A}$ to variables. Then we know by assumption that $\vDash_\mathfrak{A}\theta[s]$.  But since $s$ is arbitrary, it follows that $\vDash_\mathfrak{A}\theta[s(x|c)]$, for all $c\in|\mathfrak{A}$. But this means that $\vDash_\mathfrak{A}\forall x\theta[s]$. Since $\mathfrak{A}$ and the function assigning elements of the structure to variables are arbitrary, it follows that $\vDash\forall x\theta$.   Now we have to show that if $\vDash\forall x\theta$, then $\vDash\theta$, and this is where I'm stuck. I understand why it makes sense that this should be true, I'm just having a hard time writing it rigorously. Any suggestions would be greatly appreciated. Can I send in for repair a lens that has a shattered front element?     UPDATE: The front element was not shattered, it was the UV filter. Factory diagnosis: Lens assembly had been impacted; cleaned and checked.  Helpful notes by Crowley, David Richerby (x2), Michael Clark (x2), and   mattdm (on element price vs UV filter).      Canon cleaned and checked my lens back to like new quality for $179 labor, no parts, not including shipping.      Jim Garrison answered my question, but Crazy Dino has the best info all around for a lens element shatter.   Original Question:  I recently opened my camera bag to find a shattered front element on a Canon EF 16-35mm f/2.8 L II USM lens.   There are a lot of micro glass shards on the second element.  Does anyone have experience with this sort of thing?  Some of the following are questions I have:   Can I get it repaired? What will the cost be roughly? Will the lens ever work the same and have a good image quality? Should I ship it as is, or attempt cleaning it with air? Where is the best place to cry a little?   Thanks, any good info on this would be appreciated.  I will never rest a lens on another lens, I nearly lost a 24-70mm L that shattered a UV filter also.   Good reason to have a UV always, never thought it could happen to me.  Update: Looking at it more it looks like maybe just the front ring came undone and the UV filter broke?  I can't remember what it looked like exactly before.  I threw away the glass shards (polished sand) in the Nevada desert.      UPDATE: Primary reason for thinking more than just the UV was broken,   UV filter split into two rings, depositing the locking collar into the   lens barrel. chkrootkit shows "tcpd" as INFECTED. Is it a false positive? Alternative methods to prove that $Z_m$ is a field under addition and multiplication $\bmod\ m$ iff $m$ is a prime I am looking for ways to prove that $\mathbb{Z}_m=\{0,1,2,\dots,m-1\}$ is a field under addition and multiplication $\bmod\ m$ iff $m$ is a prime.   I tried it this way: If $m$ is a prime $p$,it is obvious that most importantly, we need to prove that every non-zero element in $\mathbb{Z_p}$ has an inverse in $\mathbb{Z_p}$.We note that if $a,b\in \mathbb{Z_p}$, then $a\cdot b\in \mathbb{Z_p} \dots \boxed{*}$. Next, $a^p\equiv a \bmod p$ for prime $p$(the proof follows by induction on $a$).As  $\displaystyle 0\leq a\leq p-1$,$a\not =p$. Thus $$a^{p-1}\equiv 1 \bmod p$$ for prime $p$ which,together with $\boxed{*}$ guarantees the existence of an inverse in $\mathbb{Z_p}$.Proving the other properties is simple.  Conversely,if $\mathbb{Z_m}$ is a field,then every non-zero element of $\mathbb{Z_m}$ has an inverse.Note that inverse of an element is unique.For if $a\cdot b\equiv a\cdot c\equiv 1 \bmod m$,then $(a\cdot b)\cdot c\equiv c\equiv (a\cdot c)\cdot b\equiv b \bmod m$ i.e $m|(b-c)\implies b-c=0(0\leq a\leq m-1,a\in \mathbb{Z_m})$.So, inverses must be unique and every element is either an inverse of itself or has in inverse in the field.That allows us to conclude that $[(m-1)!]^2\equiv 1 \bmod m$ which can only mean that either  $(m-1)!\equiv 1 \bmod m$ or $(m-1)!\equiv -1 \bmod m$. If $m$ is not a prime, it has a prime factor, say $p$. $ 0\leq p\leq m-1\implies p|(m-1)!\implies p|1$or $-1$ which is absurd, a contradiction.Thus $m$ is prime.  I feel that this method is too involved and complicated. Can anyone please suggest some other method?(Some context: This is a problem from Finite Dimensional Vector Spaces by Paul Halmos. I am learning linear algebra for the first time.It seems some of the comments have been posted on the assumption that I know rings and fields; I don't. ) Why is the wild card character * so different between the zip and rm commands? I put together a script to do some file operations for me. I am using the wild card operator * to apply functions to all files of a type, but there is one thing I don't get. I can unzip all files in a folder like this  unzip "*".zip   However, to remove all zip files afterward, I need to do   rm *.zip   That is, it does not want the quotation marks. The unzip, on the other hand, does not work if I just give it the * (gives me a warning that "files were not matched").   Why is this different? To me, this seems like the exact same operation. Or am I using the wild card incorrectly?   in Unix do not really go into this, and I could not locate anything in the rm or zip docs.  I am using the terminal on a Mac (Yosemite). Here is my candidate for  the most elementary proof that $\lim_{n \to \infty}(1+1/n)^n $ exists. I would be interested in seeing others.  $***$ Added after some comments: I prove here by very elementary means that the limit exists. Calling the limit "$e$" names it. $***$  It only needs Bernoulli's inequality (BI) in the form $(1+x)^n \ge 1+nx$ for $x &gt; -1$ and $n$ a positive integer, with equality only if $x = 0$ or $n = 1$. This is easily proved by induction: It is true for $n=1$, and $(1+x)^{n+1} = (1+x)(1+x)^n \ge (1+x)(1+nx) = 1+(n+1)x+nx^2 \ge 1+(n+1)x $. (If $-1 &lt; x &lt; 0$, if $1+nx \ge 0$, the above proof goes through, and if $1+nx &lt; 0$, $1+mx &lt; 0$ for all $m \ge n$ so certainly $(1+x)^m &gt; 1+mx$.)  This proof originally appeared in N.S Mendelsohn, An application of a famous inequality, Amer. Math. Monthly 58 (1951), 563 and uses the arithmetic-geometric mean inequality (AGMI) in the form $\big(\sum_{i=1}^n v_i/n\big)^n \ge \prod_{i=1}^n v_i$  (all $v_i$ positive) with equality if and only if all the $v_i$ are equal.  Let $a_n = (1+1/n)^n$ and $b_n = (1+1/n)^{n+1}$.  We will prove that $a_n$ is an increasing sequence  and $b_n$ is an decreasing sequence.  Since $a_n &lt; b_n$, this implies,  for any positive integers $n$ and $m$ with $m &lt; n$  that $a_m &lt; a_n &lt; b_n &lt; b_m$.  For $a_n$, consider n values of $1+1/n$ and $1$ value of $1$.  Since their sum is $n+2$ and their product is $(1+1/n)^n$, by the AGMI,  $((n+2)/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $(1+1/(n+1))^{n+1} &gt; (1+1/n)^n$,  or $a_{n+1} &gt; a_n$.  For $b_n$, consider $n$ values of $1-1/n$ and $1$ value of $1$.  Since their sum is $n$ and their product is $(1-1/n)^n$, by the AGMI,  $(n/(n+1))^{n+1} &gt; (1-1/n)^n$  or $(1+1/n)^{n+1} &lt; (1+1/(n-1))^n$,  or $b_n &gt; b_{n+1}$.  Since $b_n-a_n  = (1+1/n)^{n+1} - (1+1/n)^n = (1+1/n)^n(1/n) =a_n/n $ and every $a_n$ is less than any $b_n$ and $b_3 = (1+1/3)^4 = 256/81 &lt; 4$, $b_n-a_n &lt; 4/n$, so $b_n$ and $a_n$ converge to a common limit.  These proofs do not seem to be really elementary,  since they use the AGMI.  However, they use a special form of the AGMI,  where all but one of the values are the same,  and this will now be shown to be implied by BI,  and thus be truly elementary.   Suppose we have $n-1$ values of $u$ and $1$ value of $v$ with $u$ and $v$ positive.  The AGMI for these values is  $(((n-1)u+v)/n)^n \ge u^{n-1}v$  with equality if and only if $u = v$.  We will now show that this is implied by BI: $(((n-1)u+v)/n)^n \ge u^{n-1}v$  is the same as  $(u+(v-u)/n)^n \ge u^n(v/u)$.  Dividing by $u^n$,  this is equivalent to  $(1+(v/u-1)/n)^n \ge v/u$.  By BI,  since $(v/u-1)/n &gt; -1/n &gt; -1$, $(1+(v/u-1)/n)^n \ge 1+n((v/u-1)/n) = v/u$  with equality only if $n=1$ or $v/u-1 = 0$. Thus BI implies this version of the AGMI. Broken URL for data.stackexchange.com help page When using data.stackexchange.com I fail to remember if as [post link] is the correct alias to get the url for the post. Luckily the site has an  link. Unfortunately it is now broken. Can this be fixed? Proof of this fairly obscure differentiation trick? Suppose we're tying to differentiate the function $f(x)=x^x$. Now the textbook method would be to notice that $f(x)=e^{x \log{x}}$ and use the chain rule to find $$f'(x)=(1+\log{x})\ e^{x \log{x}}=(1+\log{x})\ x^x.$$  But suppose that I didn't make this observation and instead tried to apply the following differentiation rules:  $$\frac{d}{dx}x^c=cx^{c-1} \qquad (1)\\ \frac{d}{dx}c^x = \log{c}\ \cdot c^x \quad (2)$$  which are valid for any constant $c$. Obviously neither rules are applicable to the form $x^x$ because in this case neither the base nor the exponent are constant. But if I pretend that the exponent is constant and apply rule $(1)$, I would get $f'(x)\stackrel{?}{=}x\cdot x^{x-1}=x^x.$ Likewise, if I pretend that the base is constant and apply rule $(2)$ I obtain $f'(x)\stackrel{?}{=}\log{x}\cdot x^x$.  It isn't hard to see that neither of the derivatives are correct. But here's where the magic happens: if we sum the two “derivatives” we end up with $$x^x+ \log{x}\cdot x^x=(1+\log{x})\ x^x$$ which is the correct expression for $f'(x)$.  This same trick yields correct results in other contexts as well. In fact, in some cases it turns out to be a more efficient way of taking derivatives. For example, consider $$g(x)=x^2 = \color{blue} x\cdot \color{red} x.$$ If we pretend the blue $\color{blue} x$ is a constant we would get $g'(x)\stackrel{?}{=}\color{blue}x\cdot 1=x$. Now if we pretend the red $\color{red}x$ is constant we get $g'(x)\stackrel{?}{=}1\cdot \color{red} x=x$. Summing both expressions we end up with $2x$ which is of course a correct expression for the derivative.  These observations have led me to the following conjecture:     Let $f(x,y)$ be a differentiable function mapping $\mathbb{R}^2$ to $\mathbb{R}.$ Let $f'_1 (x,y)=\frac{\partial}{\partial x} f(x,y)$ and $f'_2 (x,y)=\frac{\partial}{\partial y} f(x,y)$. Then for any $t$ we have: $$\frac{d}{dt}f(t,t)=f'_1 (t,t) + f'_2 (t,t).$$   (I apologise for the somewhat awkward notation which I could not seem to get around without causing undue ambiguity.)  This formulation also seems to lend itself to following generalisation:     Let $f:\mathbb{R}^N \to \mathbb{R}$ be a function differentiable in each of its variables $x_1,x_2,\ldots,x_N$. For $n=1,2,\ldots,N$ define $f'_n(x_1,x_2,\ldots,x_N)=\frac{\partial}{\partial x_n}f(x_1,x_2,\ldots,x_N)$. Let $t$ be any real number and define the $N$-tuple $T=(t,t,\ldots,t)$. Then one has: $$\frac{d}{dt} f(T)=\sum_n f'_n(T).$$   Thus my question is:   Is this true? How can it be proven? (Specifically in the case $N=2$ but also in the general case.) Short circuit evaluation and side effects Uniform convergence with two sequences of functions For, $n\ge 1$ let, $g_n(x)=\sin^2(x+1/n)$ , $x\in [0,\infty)$ and $f_n(x)=\int_0^xg_n(t)\,dt.$  Then, which is(/are) correct ?  (A) $\{f_n(x)\} $ converges pointwise to a function $f$ on $[0,\infty)$ but does not converge uniformly on $[0,\infty)$.  (B) $\{f_n(x)\} $ does not converge pointwise to any function on $[0,\infty)$.  (C) $\{f_n(x)\} $ converges uniformly on $[0,1]$.  (D)  $\{f_n(x)\} $ converges uniformly on $[0,\infty)$.  I found that, $$f_n(x)=\frac{1}{2}\left[x-\frac{1}{2}\sin(x+1/n)+\frac{1}{2}\sin(1/n)\right].$$  So, $f(x)=\lim_{n}f_n(x)=\frac{x}{2}-\frac{1}{4}\sin x$.  Now, $|f_n(x)-f(x)|\le 3/4$. So, by M-test, $f_n(x)$ is not uniformly convergent, but point-wise convergent. So option (A) is correct..Am I right?? How can I delete a symbolic link? I created a symlink using mklink.  Now I need to change it but I can't figure out how to delete it so I can recreate it correctly. Warning or confirmation on new tag creation In retrospect of this , I would like to ask the community how they feel about a warning or confirmation (or just special highlighting) on new tag creation.  Everybody hates stock "are you sure?" confirmation, but a lot of tags get created by mistake (and then used ad nauseam).   For example, I think I might have created the  tag by accident. Assuming the right tag was , there would be two tags for c# 4.0, until they are made synonyms. With a warning, some accidental tags could be prevented. In cricket how does someone become an umpire at international level? Is there some procedure for this? How are they selected and what is the criteria for this? Suppose that we don't know logarithm, then how we would able to calculate $\sqrt x$, where $x$ is a real number? More generally, is there any algorithm to calculate $\sqrt [ n ]{ x } $ without using logarithm? More simple techniques would be nice.  Here is a simple technique used to approximate square roots by :  For example: $\sqrt {78}\approx 8\frac { 14 }{ 17 } $ , where $8$ is the nearest integer root of $78$, $14 = 78 - 8^2$, $17 = 2 \times 8 + 1$.  if $n=2^k$ we can use the method above.   For example, for $k=2$ Lets calculate $\sqrt [ 4 ]{ 136 } $: $$\sqrt [ 4 ]{ 136 } =\sqrt { \sqrt { 136 }  } \approx \sqrt { 11\frac { 136-{ 11 }^{ 2 } }{ 11\times 2+1 }  } =\sqrt { 11\frac { 15 }{ 23 }  } \\ \sqrt { 11\frac { 15 }{ 23 }  } \approx 3\frac { 11\frac { 15 }{ 23 } -{ 3 }^{ 2 } }{ 3\times 2+1 } =\frac { 544 }{ 161 } =3.38\\$$ The exact result is$$ \sqrt [ 4 ]{ 136 } =3.4149\cdots$$ The method approximates well, but it is working for only $n=2^k$ as I know. Recovering a accidently deleted file in Ubuntu I'm using ubuntu 9.10 and accidently deleted a specific file within a directory on my system. I checked in the Trash bin and could not locate the file. I deleted it 10mins ago, are there any tools available that I can recovery this file? It's a really important file.  Any help would be highly appreciated  My partition is ext4