To find product $\cos{\theta}\cos{2\theta}\cos{3\theta}...\cos{999\theta}$
Find the value of : $\cos x \cos 2x...\cos 999x$ given that $x=\frac {2\pi}{1999}$
What should I do when a hacker gives me hacked money? I was peacefully playing GTA Online (yeah, was killing someone coming out of ls customs) then a hacker came and killed me using explosive ammos, I called him a hacker and he teleported to me and gave me 790k. I know that people get banned for receiving this kind of money, what should I do to avoid the ban? Show that $\frac 1{1+x+y^{-1}}+\frac1{1+y+z^{-1}}+\frac1{1+z+x^{-1}}=1$ if $xyz=1$ If $x.y.z=1$ show that $\dfrac 1{1+x+y^{-1}}+\dfrac1{1+y+z^{-1}}+\dfrac1{1+z+x^{-1}}=1$  My attempt - L.H.S$=\dfrac 1{1+x+y^{-1}}+\dfrac1{1+y+z^{-1}}+\dfrac1{1+z+x^{-1}}$  $=\dfrac y{y+xy+1}+\dfrac z{z+yz+1}+\dfrac x{x+zx+1}$  $=\dfrac{yz}{yz+xyz+z}+\dfrac z{z+yz+1}+\dfrac x{x+zx+1}$  What to do next? How to detect true Windows version? In some literature, I have read that a regression with multiple explanatory variables, if in different units, needed to be standardized.  (Standardizing consists in subtracting the mean and dividing by the standard deviation.)  In which other cases do I need to standardize my data?  Are there cases in which I should only center my data (i.e., without dividing by standard deviation)? Why do I get "Exception; must be caught or declared to be thrown" when I try to compile my Java code? In order to animate a character so that it grabs an object an throws it, I added a Child-Of constrained and keyframed its influence. In the F-curve Editor I can't set the curve which controls the influence to interpolation-mode Constant without changing the other curves (Which should be linear or bezier).   How can the interpolation mode of a curve changed independently? Custom environment in wrapfigure While browsing Stack Overflow, I noticed that I lost a lot of reputation:        Is this serial downvoting? Will it be corrected by a rep-recalculation? How often are those done?    Here's a second instance: How should an educational institute woo good faculty? What steps must an educational institution wanting to recruit good faculty take? There is a cycle in the academia:    Good doctorates from top universities invariably join other top universities as faculty members.  Good students aspire to study in top universities under excellent faculty members.   Suppose a new educational institution or department springs up and desires to set up a top-quality group, how does it break into this cycle? I suppose this could take some years, but is money-power the only thing that could be wielded by the institute to achieve its end? How to find the build / version of Windows 10 During the beta of Windows 10 it was hard know what version you were running unless it had it posted to the desktop. Once that wasn't there - how do you tell what version / build you are running?   This will become more of an issue as Microsoft starts releasing more builds with the new update mechanism. When to use Stateful session bean over Stateless session bean? Inverse metric in linearised gravity From what I've read, in the framework of linearised gravity, one perturbs the metric around a Minkowski background, $\eta_{\mu\nu}$, such that  $$g_{\mu\nu}(x)=\eta_{\mu\nu}+h_{\mu\nu}(x)\tag{1}$$  where $h_{\mu\nu}(x)$ is a small perturbation, i.e. $$\big\lvert h_{\mu\nu}\big\rvert&lt;&lt;1.\tag{2}$$  The inverse metric is then found by assuming the following ansatz:  $$g^{\mu\nu}(x)=\eta^{\mu\nu}+\tilde{h}^{\mu\nu}\tag{3}$$  where $\tilde{h}^{\mu\nu}$ is also small (i.e. $\big\lvert \tilde{h}_{\mu\nu}\big\rvert&lt;&lt;1$).   Using this, it is easy to find that  $$g^{\mu\nu}(x)=\eta^{\mu\nu}-h^{\mu\nu} \tag{4}$$  to first order.   My question is, what is the justification for this ansatz? Is it simply that one expects the inverse metric to have a similar form to the metric in order to satisfy $$g^{\mu\alpha}g_{\alpha\nu}=\delta^{\mu}_{\;\nu}~?\tag{5}$$ Maximum power transfer and Thevenin I am trying to find the thevenin resistance value and the Vth for maximum power transfer. I have tried to find Rth by calculating R3 and R2 as parallel resistance and same for R4 and R5 and then adding them with R1 but I am not sure if I have done it the right way. For Vth I have tried using KVL but I think I have calculated wrong current values using current divider rule. I calculated the total current by diving V1 and R1. I hope someone can help. I have attached the picture of the circuit. I was editing a curve for hair, and pressed W to bring the specials menu and Subdivide it, but instead everything but the curve I had active dissapeared, and my cursor turned into a circle like the one from the circle select mode.  I'm guessing they changed the specials menu and I'm inside some unknown to me function, but I'm not sure how to turn everything back to normal. Powershell Parameters $|a_{n+1}-a_n| \leq \frac{1}{2} |a_{n}-a_{n-1}|$. I'm trying to show that $(a_n)$ is convergent by showing that $(a_n)$ is Cauchy.  let $|a_2 - a_1 |  =x$.   $|a_3-a_2| \leq \frac{1}{2}x$  $|a_4-a_3| \leq \frac{1}{2}|a_3-a_2|\leq \frac{1}{4}x$  In general, for every $n$:  $|a_{n+1}-a_n| \le \frac {1}{2^{n-1}}x$  Let $\epsilon &gt;0$  for every $m,n &gt; 0 $, say $m&gt;n$ the following holds:  $$|a_m-a_n| = |a_m-a_{m-1} + a_{m-1} - a_{m-2} \dots a_{n+1}- a_n| \le$$  $$|a_m-a_{m-1}| + |a_{m-1} - a_{m-2}| + \dots |a_{n+1}-a_n| \leq \frac{x}{2^{m-2}} + \frac{x}{2^{m-3}} + \dots + \frac{x }{2^{n-1}}$$.  I can say that for big enough $n,m$, $|a_m-a_n|&lt;\epsilon$ and therefore, this sequence is Cauchy? It sounds not good enough to me yet I don't know what else to do here. Syntax of bash alias definition I'm using Cygwin, and have installed all the developer packages as well as most if not all of the other packages. When I provide an alias for gcc, I get the error  -bash: alias: gcc: not found   What could I be doing wrong?  I am typing the following command into the terminal:  alias gcc "gcc -ansi -Wall -g -O0 -Wwrite-strings -Wshadow -pedantic-errors -fstack-protector-all"   FYI, this alias works on other systems. I am using latest version of VirtualBox (VirtualBox 4.3.6 for Windows hosts) and after installing any OS and clicking inside of it mouse disappear. It is not visible until I press right CRTL. Anyone know what can cause this problem. I have also installed extension pack. How do I split a forest tree into two trees with proper vertical alignment? \documentclass[oneside,12pt]{article} \usepackage{geometry} \usepackage{microtype} \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} \usepackage{lmodern}  \usepackage{enumitem} \setlist[enumerate,1]{label=\bfseries\Alph*,align=left,leftmargin=*,   labelsep=1.5em} \setlist[enumerate,2]{label=\arabic*.,labelindent=1em,labelsep=1.5em,   leftmargin=*}  \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsthm} \theoremstyle{definition} \newtheorem*{sol}{Solution}  \usepackage{forest} \forestset{   ass/.append style={     before computing xy={l=\baselineskip},     no edge   },   prooftree/.style={     for tree={       child anchor=north,       parent anchor=south     }   } }  \begin{document} \subsection*{Exercises 29} \begin{enumerate} \item Use QL trees to evaluate the entailment claims (1) to (10) in Exercises   28A.   \begin{enumerate}   \item $\forall x(Fx\supset Gx)\vDash\forall x(Gx\supset Fx)$      \begin{forest}       prooftree       [$\forall x(Fx\supset Gx)$       [$\neg\forall x(Gx\supset Fx)\checkmark$,ass       [$\exists x\neg(Gx\supset Fx)\checkmark$,ass       [$\neg(Ga\supset Fa)\checkmark$,ass       [$Ga$,ass       [$\neg Fa$,ass       [$(Fa\supset Ga)\checkmark$,ass       [$\neg Fa$] [$Ga$]]]]]]]]     \end{forest}      $\neg Fa$ and $Ga$ make the premisses and the negation of conclusion true.     Thus, the entailment is invalid.   \item $\forall x(Fx\supset Gx)\vDash\forall x(\neg Gx\supset\neg Fx)$      \begin{forest}       prooftree       [$\forall x(Fx\supset Gx)$       [$\neg\forall x(\neg Gx\supset\neg Fx)\checkmark$,ass       [$\exists x\neg(\neg Gx\supset\neg Fx)\checkmark$,ass       [$\neg(\neg Ga\supset\neg Fa)\checkmark$,ass       [$\neg Ga$,ass       [$\neg\neg Fa$,ass       [$(Fa\supset Ga)\checkmark$,ass       [$\neg Fa$ [*,ass]] [$Ga$ [*,ass]]]]]]]]]     \end{forest}      Since the tree closes, the entailment is valid.   \item $\forall x\exists yLxy\vDash\forall y\exists xLyx$      \begin{forest}       prooftree       [$\forall x\exists yLxy$       [$\neg\forall y\exists xLyx\checkmark$,ass       [$\exists y\neg\exists xLyx\checkmark$,ass       [$\neg\exists xLax\checkmark$,ass       [$\forall x\neg Lax$,ass       [$\exists yLay\checkmark$,ass       [$Lab$,ass]]]]]]]     \end{forest}      \begin{forest}       prooftree       [$\neg Lab$,ass [*,ass]]     \end{forest}   \item $\forall x((Fx\wedge Gx)\supset Hx)\vDash\forall     x(Fx\supset(Gx\supset Hx))$   \item $(\forall xFx\vee\forall xGx)\vDash\forall(Fx\vee Gx)$   \item $\forall x(Fx\supset Gx),\forall x(\neg Gx\supset Hx)\vDash     \forall x(Fx\supset\neg Hx)$   \item $\exists x(Fx\wedge Gx),\forall x(\neg Hx\supset\neg Gx)\vDash     \exists x(Fx\wedge Hx)$   \item $\forall x\exists y(Fy\supset Gx)\vDash\forall y\exists x(Gx\supset     Fy)$   \item $\forall x\forall y(Lxy\supset Lyx)\vDash\forall xLxx$   \item $\forall x(\exists yLxy\supset\forall zLzx)\vDash\forall x\forall y     (Lxy\supset Lyx)$   \end{enumerate} \end{enumerate} \end{document}   Usually, I have to manually split a forest tree into two to emulate page breaking in a forest tree. However, the second tree doesn't vertically align with the first as in the picture below. How can I vertically align them together? @cfr Appositives with possessiveness? I would like to say something like     The boy, Adam's, favorite toy was a bike.   What is the proper way to say this?