problem in installing mysql
Permission denied, are you root?
How does one add boarding passes to Apple Wallet? What is the difference between I play and I do play?  For example:  If someone were to ask to me, do you play soccer? Printing an SE page in Firefox shows only the first page I used to print Q&amp;A pages to keep a base of PDFs. I used either the 'print edit' addon or simply the print function of Firefox, which both allow to print a webpage in a file.  Usually the printed file for a webpage extends on 2 or 3 pages. But I'm not able to print more than one page on this site expressly anymore. No matter the method. I just get the top of each discussion. Why it's true? $\arcsin(x) +\arccos(x) = \frac{\pi}{2}$ The following identity is true for any given $x \in [-1,1]$: $$\arcsin(x) + \arccos(x) = \frac{\pi}{2}$$  But I don't know how to explain it. I understand that the derivative of the equation is a truth clause, but why would the following be true, intuitively?  $$\int^{x}_{C1}\frac{1\cdot dx}{\sqrt{1-x^{2}}} + \int^{x}_{C2}\frac{-1 \cdot dx}{\sqrt{1-x^{2}}} =\\  \arcsin(x) - \arcsin(C1) + \arccos(x) - \arccos(C2) = 0 \\  \text{while } \arcsin(C1) + \arccos(C2) = \frac{\pi}{2}$$  I can't find the right words to explain why this is true?    Edit #1 (25 Jan, 20:10 UTC): The following is a truth clause: $$ \begin{array}{ll} \frac{d}{dx}(\arcsin(x) + \arccos(x)) = \frac{d}{dx}\frac{\pi}{2} \\ \\ \frac{1}{\sqrt{1-x^{2}}} + \frac{-1}{\sqrt{1-x^{2}}} = 0 \end{array} $$  By integrating the last equation, using the limits $k$ (a constant) and $x$ (variable), I get the following:  $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = m  \text{ (m is a constant)}\\ \\ \arcsin(x) + \arccos(x) = m + \arcsin(k) + \arccos(k) \\ \\ \text{Assuming that } A = m + \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ Using Calculus, why is that true for every $x \in [-1,1]$?    Edit #2:  A big mistake of mine was to think that $\int^x_k0 = m \text{ (m is const.)}$, but that isn't true for definite integrals.  Thus the equations from "Edit #1" should be as follows: $$ \begin{array}{ll} \int^x_k\frac{1}{\sqrt{1-x^{2}}}dx + \int^x_k\frac{-1}{\sqrt{1-x^{2}}}dx = \int^x_k0 \\ \\ \arcsin(x) - \arcsin(k) + \arccos(x) - \arccos(k) = 0\\ \\ \arcsin(x) + \arccos(x) = \arcsin(k) + \arccos(k) \\ \\ A = \arcsin(k) + \arccos(k) = \frac{\pi}{2} \text{ ,for } x \in [-1,1] \end{array} $$ How can mixed content compromise an entire HTTPS session? elementary ring proof with composition of functions and addition of functions as operations Consider the set $\mathcal{F}=\lbrace f\mid \mathcal{f}:\Bbb{R}\to\Bbb{R}\rbrace$, in which an additive group is defined by addition of functions and a second operation defined as composition of functions. The question asks to verify that the resulting structure does not satisfy the ring properties.     This is the question 24.10 from Modern Algebra, by Durbin, 4th edition.  So what I have so far is that all of the properties for the additive group (with addition of functions as the operation here) hold, associativity of the composition of functions hold, and that the failure must come from the distributive laws.  My proof of my claim, so far;  $$1\,\,\,\,\,[(\mathcal{f}\circ\mathcal{g})+\mathcal{h}](\mathcal{x})=\cdots=\mathcal{f}(\mathcal{g}(\mathcal{x}))+\mathcal{f}(\mathcal{h}(\mathcal{x})).$$ ...but,  $$2\,\,[(\mathcal{f}\circ\mathcal{g})+\mathcal{h}](\mathcal{x})=$$ $$3\,\,(\mathcal{f}\circ\mathcal{g})(\mathcal{x})+\mathcal{h}(\mathcal{x})=$$ $$4\,\,\,\mathcal{f}(\mathcal{g}(\mathcal{x}))+\mathcal{h}(\mathcal{x}).$$ ...which brings me to my contradiction of the properties of rings that;  $$5\,\,\,[\mathcal{f}\circ(\mathcal{g}+\mathcal{h})](\mathcal{x})\neq[(\mathcal{f}\circ\mathcal{g})+\mathcal{h}](\mathcal{x}).$$  So my question is, is this the correct path to showing the distributive laws are not obeyed when composition is taken as an operation on $\mathcal{F}$? particularly my proceeding from 2 to 3 to 4 is what i feel uncomfortable about, like i may be missing a step in there.  thanks I have to travel to Germany and Switzerland for business purpose. I have already been here 2 years ago. I have applied for business visa to the German consulate twice and both times the application got rejected. They have put a green stamp on the last page twice. The reason is the purpose and information submitted regarding intended stay is not reliable. I have submitted the documents to Swiss consulate and they have accepted and granted me a visa. Now I have to first travel to Germany and then to Switzerland.  So my question is, with a Schengen business visa granted by Switzerland, may I enter in Germany? Does the rejection stamp has any impact for entering at German airport ? Does it mean I cannot travel to Germany but to other Schengen countries ? Why are the number of false positives independent of sample size, if we use p-values to compare two independent datasets? If I run the R code below, it will generate two independent vectors then test them to see if they are related in some way (i.e. p-value &lt; 0.05).  If I repeat this 1,000 times, then 50 of these (5%) will have a false positive, with a p-value &lt; 0.05. This is a type 1 error.  If we increase sampleSize to 1000, or even 100,000, the results are the same (5% false positives).  I am struggling to understand this, as I would have thought that if we collected enough samples, the chance of a false positive would drop towards 0 (just like correlation).  So, I guess my question is: "How is it possible that the number of false positives, based on p-values generated from comparing two independent datasets, be independent of sample size?".  # R code to demonstrate that with a large dataset, we can still  # get significant p-values, purely by chance.  # Change this to 1000, and we still get the same number of false positives (50, or 5%). sampleSize = 20  cat("Sample size:",sampleSize,"\n")  set.seed(1010093) n=1000 pValues &lt;- rep(NA,n) for(i in 1:n){     y &lt;- rnorm(sampleSize)     x &lt;- rnorm(sampleSize)     pValues[i] &lt;- summary(lm(y ~ x))$coeff[2,4] } # Controls false positive rate fp = sum(pValues &lt; 0.05) cat("Out of ",n," tests, ",fp," had a p-value &lt; 0.05, purely by by chance.\n",sep="")  ----output---- Running "true positives - none.R" ... Sample size: 1000  Out of 1000 tests, 52 had a p-value &lt; 0.05, purely by by chance. Is exFAT now the best choice for an external hard drive that will be used for Windows, Mac, and Linux? I am having trouble getting onto a server in Minecraft PE on iOS. When I tap on the person's server it just says invalid name:    I have tried turning off and restarting the iPad, going to home screen and logging back into the server but I have had no luck and no change. Is there a reason why this is so? What are good methodologies for creating sprocs that reduce the pain of debugging? And what tools are out there for debugging stored procedures?  Perhaps most importantly, what are indications to look out for that errors are happening in a sproc and not in the code? I hope I'm not all over the board too terribly bad here. Votes for answers to any of the above. Thanks.  For what it's worth, I work in a .NET environment, SQL servers. Need verification - Prove a Hermitian matrix $(\textbf{A}^\ast = \textbf{A})$ has only real eigenvalues Proof: Let eigenvalue $\lambda \neq 0$ such as  $$\textbf{A}\vec{v}  = \lambda\vec{v}$$ $$\Rightarrow (\textbf{A}\vec{v})^\ast = (\lambda\vec{v})^\ast$$ $$\Rightarrow (\vec{v}^\ast\textbf{A}^\ast)=(\lambda^\ast\vec{v}^\ast)$$ Right-multiply both sides by $\color{orangered}{\vec{v}}$$$\Rightarrow (\vec{v}^\ast\textbf{A}^\ast \color{orangered}{\vec{v}} )=(\lambda^\ast\vec{v}^\ast \color{orangered}{\vec{v}} )$$ $$\textbf{A}^\ast=\textbf{A}$$ $$\Rightarrow(\vec{v}^\ast\textbf{A}\vec{v})=(\lambda^\ast\vec{v}^\ast\vec{v})$$ $$\Rightarrow(\vec{v}^\ast\lambda\vec{v}) = (\lambda^\ast\vec{v}^\ast\vec{v})$$ $$\Rightarrow(\lambda\vec{v}^\ast\vec{v}) =  (\lambda^\ast\vec{v}^\ast\vec{v})$$ $$\Rightarrow \lambda = \lambda^\ast$$ $$\Rightarrow \lambda\in\mathbb{R}$$ I am looking for a short proof that $$\int_0^\infty \left(\frac{\sin x}{x}\right)^2 \mathrm dx=\frac{\pi}{2}.$$ What do you think?  It is kind of amazing that $$\int_0^\infty \frac{\sin x}{x} \mathrm dx$$ is also $\frac{\pi}{2}.$ Many proofs of this latter one are already in this . Proving $\sum\limits_{l=1}^n \sum\limits _{k=1}^{n-1}\tan \frac {lk\pi }{2n+1}\tan \frac {l(k+1)\pi }{2n+1}=0$ Prove that $$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$$     It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:  $\begin{matrix}  \tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}  &amp;  \tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} &amp; \tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} &amp; \tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} &amp; \tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}  \\  \tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}  &amp;  \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} \\ \tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} &amp; \tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} &amp; \tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} \\ \tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} &amp; \tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} &amp; \tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} \\ \tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} &amp; \tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} &amp; \tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} &amp; \tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} &amp; \tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} \\ \tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} &amp; \tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} &amp; \tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} &amp; \tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} &amp; \tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}  \end{matrix}$  one can notice then, that the first column vanish the fourth one :  $\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$  $\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$  and the third column vanish the fifth one :  $\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$  $\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$  $\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$  $\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$  $\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$  $\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$  while the second column is self-vanishing:  $\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$  $\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$  $\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .  So the equality occurs. But how to generalize the proof? Where to download Chrome 32bit? Google has  so on a 32bit Debian machine it's not possible anymore to update the APT repository:  W: Failed to fetch http://dl.google.com/linux/chrome/deb/dists/stable/Release  Unable to find expected entry 'main/binary-i386/Packages' in Release file (Wrong sources.list entry or malformed file)   Also has the .deb file removed from Google servers as it seems.  Is there any alternative .deb download of the last release (46?) available somewhere?  Reason: For an automated Debian build system (debootstrap-based) I urgently need a way to install Google Chrome 32bit v46 (or similar). Switching to another browser (including Chromium) is not a quick enough option for me at the moment.. Under the  tag I have seen many questions on earth-616. Are there any other versions of earth as well? How do they differ from each other? Is $\mathbb{Z}[x]/(x^2+1)$  isomorphic to  $\mathbb{Z}[i]$?   My attempt is that try to define a mapping $g$ from $\mathbb{Z}[x]$ to $\mathbb{Z}[i]$ by $g(f(x))= f(i)$, for $f(x)\in\mathbb{Z}[x]$. If it is possible then $\ker g$ is $(x^2+1)$? Am I on the right track? Please Help. How to run batch file without admin privillege? Which is the recommended way to plot: matplotlib or pylab? How do I view the SQL generated by the Entity Framework?