I want to create a math operator that is an integral with a square (\square) stack onto it such that the integral is still formated the way that the operator \int is. How do I do this?
Good Afternoon all,  I am trying to typeset a symbol that I have been using to represent the integral of a regulated function. This symbol is the standard integral symbol, with the letter "R" superimposed on it.  I have tried using \mathllap, \mathclap and \mathrlap (as suggested ), however none of these have managed to give the desired effect. When I tried doing it without the limits of integration, it got close to working, however then I was unable to have the limits to the right hand side of the symbol:  \mathop{\mathrlap{\int}R}_a^b f   Any help would be greatly appreciated, thank you in advance!  Minimal Example:  \documentclass{article}  \usepackage{amsmath,mathtools}  \begin{document}  Take the regulated function $f$, we want to compare the Riemann integral, $\int_a^b f$ to the regulated integral $\mathop{\mathrlap{\int}R}_a^b f$ by taking sequences of step functions.  We will prove that $$ \mathop{\mathrlap{\int}R}_a^b f = \int_a^b f $$  \end{document}
Use Stokes's Theorem to show $\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2$ I am a little stuck on the following problem:  Use Stokes's Theorem to show that  $$\oint_{C} y ~dx + z ~dy + x ~dz = \sqrt{3} \pi a^2,$$  where $C$ is the suitably oriented intersection of the surfaces $x^2 + y^2 + z^2 = a^2$ and $x + y + z = 0$.  OK, so Stokes's Theorem tells me that:  $$\oint_{C}\vec{F} \cdot d\vec{r} = \iint_{S}\operatorname{curl} \vec{F} \cdot \vec{N} ~dS$$  I have calculated:  $$\operatorname{curl} \vec{F} = -\vec{i} - \vec{j} - \vec{k}.$$  I then figured that on the surface $S$ we must have:  $$\vec{N}dS = \vec{i} + \vec{j} + \vec{k}dxdy$$  since this follows from the equation of the given plane.  However this will then give me:  $$\operatorname{curl} \vec{F} \cdot \vec{N} = -1 -1 -1 = -3$$  And thus I would get, if project this onto the $xy$-plane:  $$\iint_{S} \operatorname{curl} \vec{F} \cdot \vec{N} ~dS = -3 \iint_{A} dA = -3 \pi a^2$$  which is obviously not correct.  I would greatly appreciate it if someone could help me with this.  I actually had multivariable calculus a few years ago, and I know that I knew this stuff then.  However, now that I need it again I notice that I've become quite rusty.  Thanks in advance :) SQL server Read Only permission automatically for new databases Unable to Change Apple ID and password I have tried repeatedly to change my Apple ID and password without success. I have followed all of the directions trying both my emails and asked for new passwords----no luck whatsoever. From a link from the question  to the HP Wikia, I started mulling over something I read about Voldemort and the "regeneration" potion (not a canon potion, but I'll use it for lack of a better way of describing it. . .) and would really like to know:  How did Voldemort drink the regeneration potion to turn his body into the grotesque childlike form we see in Goblet of Fire while he was still in a "mangled spectral state"?  Spectres (i.e. spirits or ghosts) are non-corporeal; they obviously can't consume food or drink liquids. So, how did Voldemort overcome this difficulty and manage to drink the regeneration potion? Which is stronger, a rope without knots or a rope with knots? Suppose I have two identical ropes, one of which I manually tie some knots in. If I use them to hang clothes, which one is more durable?      Personally I think that rope with knots will be more durable, but I can't come up with a satisfactory reason. Could someone please explain? I couldn't find anything on the internet, everything talks about how to go about it in some way, but nothing says exactly what it is.   Also, what is a fully trusted assembly and how do they differ from one another?  I have a MS certification exam and this is the only topic that I just don't understand.  EDIT: Thanks guys. Now I have a better understanding of security in .NET. I was able to pass my certification exam. Compute: $\lim_{x\to 0}\frac{\sqrt[m]{\cos(\alpha x)}-\sqrt[n]{\cos{(\beta x)}}}{x^2},\;m,n\in\mathbb N,\alpha,\beta\in\mathbb R$ Compute: $$\lim_{x\to 0}\frac{\sqrt[m]{\cos(\alpha  x)}-\sqrt[n]{\cos{(\beta x)}}}{x^2},\;m,n\in\mathbb N,\alpha,\beta\in\mathbb R$$   My attempt:  $$L=\lim_{x\to 0}\frac{\sqrt[m]{\cos(\alpha x)}-\sqrt[n]{\cos{(\beta x)}}}{x^2}=\lim_{x\to 0}\frac{\sqrt[m]{\cos(\alpha x)}-1+1-\sqrt[n]{\cos(\beta x)}}{x^2}\\\displaystyle=\lim_{x\to 0}\frac{\left(1+(\cos(\alpha x)-1)\right)^{\frac{1}{m}}-1-\left(\left(1+(\cos(\beta x)-1)\right)^{\frac{1}{n}}-1\right)}{x^2}$$  Transformed expression: $$-\frac{(1+(\cos(\alpha x)-1))^{\frac{1}{m}}-1}{\cos(\alpha x)-1}\cdot\frac{1-\cos(\alpha x)}{\alpha^2x^2}\alpha^2+\frac{(1+(\cos(\beta x)-1))^{\frac{1}{n}}-1}{\cos(\beta x)-1}\cdot\frac{1-\cos(\beta x)}{\beta^2x^2}\beta^2$$  I applied the standard limits from the table: $$\lim_{x\to 0}\frac{(1+x)^a-1}{x}=a\;\&amp;\;\lim_{x\to 0}\frac{1-\cos x}{x^2}=\frac{1}{2}$$ $$L=-\frac{1}{m}\cdot\frac{1}{2}\alpha^2+\frac{1}{n}\cdot\frac{1}{2}\beta^2=\frac{m\beta^2-n\alpha^2}{2mn}$$ Is this correct? How to disable all div content Why app updates download entire app instead of only "updates"? Python: using a recursive algorithm as a generator Xdotool using "DISPLAY=:0" not works in Crontab If $R_{1}$ and $R_{2}$ are relations on a set S with $R_{1};R_{2}=R_{2};R_{1}$. Then $R_{1}$ and $R_{2}$ are bijective maps For the surjectivity part, I showed that if $(a,b)\in I$ then $(a,c)\in R_{1}$ and $(c,b)\in R_{1}$ and $(a,c)\in R_{2}$ and $(c,b)\in R_{2}$ for some $c\in S$. Now for arbituary $(x,y)\in R_{1}$ we have $(x,z)\in I$ which implies that $(y,z)\in R_{2}$. But also $(y,z)\in R_{1}$, so $R_{1}$ is surjective. Similar argument for $R_{2}$  And for Injectivity, I've tried to show that if $aR_{1};R_{2}c$ and $bR_{2};R_{1}c$, then $a=b$ but I am not entirely sure what $a=b$ means in this context. Do I have to show that there is an equivalence relation containing (a,b)?  edit: $R_{1};R_{2}$ is a composition of two relations and $I=R_{1};R_{2}$ How do I go about proving this?  If $\gcd(a,b)=1$ and $\gcd(a,c)=1$, then $\gcd(a,bc)=1$. I'm very confused with gcd proofs. ETS() function, how to avoid forecast not in line with historical data? I am working on an alogorithm in R to automatize a monthly forecast calculation. I am using, among others, the ets() function from the forecast package to calculate forecast. It is working very well.  Unfortunately, for some specific time series, the result I get is weird.  Please, find below the code i am using :  train_ts&lt;- ts(values, frequency=12) fit2&lt;-ets(train_ts, model="ZZZ", damped=TRUE, alpha=NULL, beta=NULL, gamma=NULL,              phi=NULL, additive.only=FALSE, lambda=TRUE,              lower=c(0.0001,0.0001,0.0001,0.8),upper=c(0.9999,0.9999,0.9999,0.98),              opt.crit=c("lik","amse","mse","sigma","mae"), nmse=3,              bounds=c("both","usual","admissible"), ic=c("aicc","aic","bic"),             restrict=TRUE)   ets &lt;- forecast(fit2,h=forecasthorizon,method ='ets')      Please, you will find below the concerned history data set :   values &lt;- c(27, 27, 7, 24, 39, 40, 24, 45, 36, 37, 31, 47, 16, 24, 6, 21,  35, 36, 21, 40, 32, 33, 27, 42, 14, 21, 5, 19, 31, 32, 19, 36,  29, 29, 24, 42, 15, 24, 21)   Here, on the graph, you will see the historical data (black), the fitted value (green) and the forecast(blue). The forecast is definitely not in lines with the fitted value.  Do you have any idea on how to "bound" the forecat to be "in line" with the historical sales? Summation of Central Binomial Coefficients divided by even powers of $2$ Whilst working out  the following summation emerged: $$\sum_{m=0}^n\frac 1{2^{2m}}\binom {2m}m$$ The is equivalent to  $$\begin{align} \sum_{m=0}^n \frac {(2m-1)!!}{2m!!}&amp;=\frac 12+\frac {1\cdot3}{2\cdot 4}+\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}+\cdots +\frac{1\cdot 3\cdot 5\cdot \cdots \cdot(2n-1)}{2\cdot 4\cdot 6\cdot \cdots \cdot 2n}\\ &amp;=\frac 12\left(1+\frac 34\left(1+\frac 56\left(1+\cdots \left(1+\frac {2n-1}{2n}\right)\right)\right)\right) \end{align}$$  and terms are the same as coefficients in the expansion of $(1-x)^{-1/2}$.  Once the solution  $$ \frac {n+1}{2^{2n+1}}\binom {2n+2}{n+1}$$ is known, the telescoping sum can be easily derived, i.e.    $$\frac 1{2^{2m}}\binom {2m}m=\frac {m+1}{2^{2(m+1)-1}}\binom {2(m+1)}{m+1}-\frac m{2^{2m-1}}\binom {2m}m$$  However, without knowing this a priori, how would we have approached this problem? Let $V$ be a vector space over a field $F$ and $T$ a linear operator on $V$. If $T^2$$=$ $0$, what can you say about the relation of the range of $T$ to the null space of $T$? Background:   Story: A colleague of mine told me about  and it seems good to me. I am always trying to use tools which Mac OS X itself gives me, so I've given the Keychain a try. It seems satisfying, but I don't know how to tell Google Chrome to use my specific keychain. I need it to separate saved stuff for work and for private (syncing private to work, but separated from work data).  Stuff:   Google Chrome 21.0.1180.57 (Fully updated) Mac OS X Lion (Fully updated) OS X Mountain Lion (Fully updated)   Question:  How do I tell to Chrome to use a specific keychain and to save data from/to this keychain? What happens if 2 separate abilities require you to sacrifice a creature? Can I travel to Kuwait on a visa with old passport number? I have a Kuwait visa, obtained from a hotel in Kuwait, with my old passport number. But as the validity of the present passport is expiring within 6 months, I got a new passport with a new passport number. Can I travel with the same visa or do I have to obtain a new visa with my new passport number? Every Artinian ring is isomorphic to a direct product of Artinian local rings