Some way to integrate $\sin(x^2)$?
Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods?
"High response time" warnings in Google Webmaster Tools In my Google Webmaster Tools account, there are warnings about my site.     Some URLs listed in this sitemap have a high response time. This may indicate a problem with your server or with the content of the page.   What is the solution? If a>0 and b>0,  $ \lim_{x\rightarrow +\infty } \sqrt[x]{a^x+b^x} = ? $  What I was trying to do: Suppose a>b. Then, for sufficiently large values of x, $ a^x &gt;&gt; b^x $; so $\sqrt[x]{a^x+b^x} \rightarrow \sqrt[x]{a^x} \rightarrow a $ when $x \rightarrow +\infty$.   Is that idea correct? How can I formalize it? How can I use an array of function pointers? How should I use array of function pointers in C?  How can I initialize them? I am trying to proof the theorem that states that:  "A map to a smaller dimensional space is not injective"  So I first suppose that V and W are finite-dimensional vector spaces such that dimV > dimW and I let T be the linear transformation from V to W.  Given the above, you have that:  dim(null(T)) = dim(V) - dim(range(T)) >= dim(V) - dim(W) > 0  So I understand the first and second equalities. However, I don't understand why dim(V) - dim(W) is strictly greater than 0. I've browsed the forums and found a number of posts that have addressed this issue, but none of the solutions seem to work for me. I have the following script that I just copied from the bibtex home page to get familiar with it. Instead of the citation number I get a question mark. I compile using Latex+Bibtex+Latex+Latex+PDFLatex+ViewPDF just as has been previously suggested and the problem persists.   \documentclass[11pt]{article} \usepackage{cite}  \begin{document}  \title{My Article} \author{Nobody Jr.} \date{Today} \maketitle  Blablabla said Nobody ~\cite{Nobody06}.  \bibliography{mybib} \bibliographystyle{plain} \end{document}   My bibliography (Bib.bbl)  @misc{ Nobody06, author = "Nobody Jr", title = "My Article", year = "2006" }    Looking at previous posts one thing that is concerning is that my .bbl looks empty as shown below. Further, I don't have a .blg  \begin{thebibliography}{}  \end{thebibliography} Finding out the area of a triangle if the coordinates of the three vertices are given What is the simplest way to find out the area of a triangle if the coordinates of the three vertices are given in $x$-$y$ plane?   One approach is to find the length of each side from the coordinates given and then apply . Is this the best way possible?   Is it possible to compare the area of triangles with their coordinates provided without actually calculating side lengths? Is there a new auto-login feature? Generalizing a sequence. Consider a sequence of the form:  $x[n]=-1,-1...,1,1,...,-1,-1...,1,1... \quad n&gt;0$  One can think of this as a square wave ($\pm1$) with a 50 % duty cycle (coming from EE).  For the simplest case of such a series i.e. $-1,+1,-1,+1,-1...$, a general equation is quite trivial ($x[n]=-1^n$).  Can we have e general equation for the other cases i.e. when there are multiple $-1$'s followed by $+1$'s ? Would someone explain wheel sizes.  If I am not mistaken, the wheels on my Trek 520 touring bike are 700C. There are no markings, and of course, no documentation of any kind. I assume 700 means 700mm, so 700 / 25.4 = 27.56 inches.  If I measure the rim however, its diameter is approximately 25.1 inches.  If I buy Continental tires, the size is 28 inch, but I have to buy 27 inch Bontrager tubes.  To make matters worse, per this link, tires come in sizes of 26, 27, and 28 inch, and size 700.     How can I determine my wheel size, and are these various sizes interchangable. Also, is the 'C' suffix important, and what does it mean?  Of course the tires are marked, but I am under the impression (per the dealers comments) that Continental marks their tires differently than everyone else. Also, the tire markings don't help for ordering a new wheel.   With so many tire sizes available, how can I know if I am using the correct size.  After reading the various answers here, I should explain that my confusion on this matter stemmed from the fact that I used to work as an auto mechanic. In the auto industry, car tires are based on wheel size, 13", 14", 15" etc, and there is no possible way to mount a 15" tire on a 13" rim for example.  So when I saw all these different tire sizes, I assumed they all required a different wheel size, but in fact, there is essentially just one wheel diameter for my type of bike, which they refer to as 700C, which is actually 622mm.   I also initially assumed that the C referred to the wheel's width, but that isn't true either. The C is apparently just antiquated terminology.  So at least in theory, I can mount a 26", 27" 28" or 29" tire on my 700C (622mm) rim. I just need to verify the width of the tire is suitable my wheel width, which is what Sheldon Brown's article explains.  Also, I measured my Continental tires, which are marked as 28 inch. They actually measure 27 inches (fully inflated), which must be why my dealer calls their sizing non standard. I want Ctrl+Alt+B bring up a browser.  Ctrl+Alt+Delete is handy.  Can I set up Ctrl+Alt+B to bring up a browser (either default or specific)? For $V$ vector space of dimension $n$ under $\mathbb C$ and $T: V \to V $ linear transformation , show $V= \ker T^n \oplus $ Im $T^n$ Derivation of the formula for Ordinary Least Squares Linear Regression Black screen on boot after AMDGPU-Pro install (16.04) I have been having some screen freezing issues on my new Ubuntu 16.04 system, and have been in the troubleshooting process for this issue. I thought it was a graphics driver problem, as I am running an AMD R9 380x card which was supposed to be fully supported by the open-source AMDGPU driver preinstalled with 16.04, but after looking around online, many people seem to have issues with. I thought the answer would be to install the recently released AMDGPU-Pro driver , following the instructions from . However, when I reboot I am getting a black screen. Is there an easy way to revert my changes back? My machine was semi-functional with the freezing issue but now with no video at all is useless to me. How to proceed? I've got the new Canon Pixma MG 3150 and so far, I can't seem to find a driver anywhere. I do have the CD for both Windows and Mac, but this doesn't help me, as I have Ubuntu 11.10. I'd prefer to have it running on the wifi.  I'm really new to this, so I need for you to explain this to me as basically as possible. I'm a beginner with blender so this is probably a very basic question.  I have this basic model below. I've ended up with a lot of unwanted geometry that I want to simplify. I wanted to start by merging the two selected faces below but no matter what I try to delete or merge it just destroys the geometry.  How can I achieve what I want to do (which is combining all of the unwanted geometry, either starting with this merge or some other way?) Why do clocks measure arc-length? Apologies in advance for the long question.  My understanding is that in GR, massive observers move along timelike curves $x^\mu(\lambda)$, and if an observer moves from point $x^\mu(\lambda_a)$ to $x^\mu(\lambda_b)$, then his clock will measure that an amount of time $t_{ba}$ given by the curve's arc length; $$   t_{ba} = \int_{\lambda_a}^{\lambda_b}d\lambda \sqrt{-g_{\mu\nu}(x(\lambda))\dot x^\mu(\lambda)\dot x^\nu(\lambda)} $$ will have elapsed where $g_{\mu\nu}$ is a metric on spacetime with signature $(-,+,+,+)$.    Why is this so?     Here is how I would attempt to justify this fact in special relativity with $g_{\mu\nu} = \eta_{\mu\nu}$.  Consider an inertial observer $O$ in $\mathbb R^{3,1}$, and suppose that this observer sees a clock, which I'll call observer $O'$ moving around on a curve $x^\mu(\lambda)$.  If $O'$ were also an inertial observer, then given any event with coordinates $x^\mu$ as measured by $O$, observer $O'$ would measure the coordinates of the event to be $x'^\mu = \Lambda^\mu_{\phantom\mu\nu} x^\nu + x_0^\mu$ for some Lorentz transformation $\Lambda$.  If $O'$ is not inertial, then this is no longer true, and there is some more complicated family of transformations, say $T_\lambda$ between events as seen by both observers.  I would argue, however, that if we were to partition the interval $[\lambda_a, \lambda_b]$ into a large number $N$ of intervals $I_1=[\lambda_a, \lambda_i], \dots, I_N=[\lambda_{N-1}, \lambda_b]$ with $\lambda_n = \lambda_a+n\epsilon_N$ and $\epsilon_N=(\lambda_b-\lambda_a)/N$, then on each interval $I_n$, $O'$ is approximately an inertial observer in the sense that $$   T_{\lambda_n} = P_n + \mathcal O(\epsilon_N), \qquad (\star) $$ for some Poincare transformation $P_n$.  Then we would note that since $O'$ is stationary in his own reference frame, he measures his worldline to have the property $\dot x'^\mu(\lambda) = (\dot t(\lambda), \mathbf 0)$ so that $$   I_{ba}=\int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x'^\mu\dot x'^\mu} = \int_{\lambda_a}^{\lambda_b} d\lambda \, \sqrt{\dot t^2} = t(\lambda_b) - t(\lambda_a) = t_{ba} $$ On the other hand the integral on the left can be written as a Riemann sum using the partition above, and we can invoke ($\star$) above to get \begin{align}   I_{ba}  &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x'^\mu(\lambda_n)\dot x'^\nu(\lambda_n)}\right] \notag\\ &amp;= \lim_{N\to\infty}\left[\sum_{n=1}^N \epsilon_N\sqrt{-\eta_{\mu\nu}\dot x^\mu(\lambda_n)\dot x^\nu(\lambda_n)} + \mathcal{O}(\epsilon_N^2)\right] \notag\\ &amp;= \int_{\lambda_a}^{\lambda_b}d\lambda \,\sqrt{-\eta_{\mu\nu}\dot x^\mu\dot x^\mu} \end{align} Combining these two computations gives the desired result.  How do others feel about this argument?  I'm not completely comfortable with it because of the assumption $(\star)$ I made on $T_\lambda$.  I imagine that in GR a similar argument could be made by invoking local flatness of the metric. Prove that $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$ If $\alpha + \beta + \gamma = \pi $ and $\tan(\frac{-\alpha + \beta + \gamma}4)\tan(\frac{\alpha - \beta + \gamma}4)\tan(\frac{\alpha + \beta - \gamma}4) = 1$  Then prove that:     $1 + \cos\alpha + \cos\beta + \cos\gamma = 0$.   I have no idea how to go about this.   Please help. Overview of basic facts about Cauchy functional equation The  asks about functions $f \colon \mathbb R \to \mathbb R$ such that $$f(x+y)=f(x)+f(y).$$ It is a very well-known functional equation, which appears in various areas of mathematics ranging from exercises in freshman classes to constructing useful counterexamples for some advanced questions. Solutions of this equation are often called . Also a few other equations related to this equation are often studied. (Equations which can be easily transformed to Cauchy functional equation or can be solved by using similar methods.) Is there some overview of basic facts about Cauchy equation and related functional equations - preferably available online? Is there anything that is totally random? When I say totally random, I mean absolutely random, not pseudorandom. If I want to say "totally random" numbers such as 1,26,17,4,1 and 27, although I see them to be totally random, they aren't. These are numbers that I think are influenced by my childhood, ideology and everything that I've seen. So, what do you think? Is there anything completely random? Is the set $S=\left\{\left(z_1,z_2\right)\in \mathbb C\times \mathbb C:z_1^2+z_2^2=1\right\}.$ compact? Consider the set $$S=\left\{\left(z_1,z_2\right)\in \mathbb C\times \mathbb C:z_1^2+z_2^2=1\right\}.$$  Is this set compact in $\mathbb C^2$ ?  As $\mathbb C^2$ is a finite dimensional space so a subset of it is compact if and only if it is closed and bounded.  But $z_1^2+z_2^2=1$ gives $(x_1^2+x_2^2)-(y_1^2+y_2^2)=1$ and $x_1y_1+x_2y_2=0$ , where $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$.  But from here how I can show that the set is closed and bounded ?