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I am running QGIS 2.18.14 (QGIS code revision: ff83b9a479) on a windows 7 machine, it comes with GRASS 7.2.2 (Code Revision r71496), but it does not come with r.in.lidar. I have noticed QGIS 2.18.16.1 has the same problem and it shares the same GRASS 7.2.2 (Code Revision r71496).  I know there were some issues with it and later solved as I read , I also certailny used it short ago in 2017, so I wonder which version should I go for before starting the always risky attempt of a new installation. Number Theory: Find all solutions of $\phi(n)=16$ and $\phi(n)=24$ Does there exist a bijective $f:\mathbb{N} \to \mathbb{N}$ such that $\sum f(n)/n^2$ converges? We know that $\displaystyle\zeta(2)=\sum\limits_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$ and it converges.        Does there exists a bijective map $f:\mathbb{N} \to \mathbb{N}$ such that the sum $$\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^2}$$ converges.      If our $s=2$ was not fixed, then can we have a function such that $\displaystyle \zeta(s)=\sum\limits_{n=1}^{\infty} \frac{f(n)}{n^s}$ converges If a creature with Multiattack is affected by Haste, can it make two Multiattacks? It's a fairly straightforward question.   A creature has a Multiattack option, and it also can cast the Haste spell. While under the effect of Haste, can it use the second Action to take another Multiattack Action?  My main concern is the clause of the Haste spell that limits the number of attacks made from taking the Attack Action. This is a different Action, though.   Then there is the concern of being able to use Multiattack at all, because Haste says what kinds of Actions you can take. Review audit ignores closereason filter Replace all occurrence of character except first one Does anyone know where I can find cheat codes for Goldeneye 007 Wii? I know that the game hasn't even been out for a week, but does anyone know if any cheat codes have been released by Activision yet?  I know that they must exist since there is a cheat codes menu within the game. I have just updated Windows 7 to Windows 10.  My touchpad scolling (right-side) is not working anymore.    I have no idea how to fix this in Windows 10.  I have hit the start button and typed "synaptic", but have found nothing useful.  I used to have an icon on the taskbar that showed where I was touching the touchpad.  I used to be able to right-click that icon to change settings, but Windows 10 has not brought that icon over.  Also, the asus website has no synaptic touchpad drivers available for Windows 10.  I have tried the enaltech drivers, but upon the installation atempt I am told that my hardware is not correct. How to solder with one hand? I am a beginner at electronics.  I don't have many tools and stuff, so I am forced to solder by hand.   I don't have a pcb to solder on, so I usually just solder the wires together, but it's really difficult without a holder.  Is there an easier way to do this? If not, is there a substitute for a wire holder?  Any help would be appreciated. In the simple harmonic oscillator model, we talk about the angular frequency and I understand that this tells something about the instantaneous angular velocity of the object. For example, a simple pendulum. But what does this mean when we apply this concept to an object attached to a spring that behaves as a simple harmonic oscillator? There is no instantaneous circular motion for the object, so how can we relate the angular frequency to a system like this? Unexpected Results From Clicking The Unanswered Button Are the floor functions of $0.999\cdots$ and 1 equal?  It is true that $0.999\cdots=1$ but how does one justifies the integer part of $0.999\cdots$ being 1 , where it is not, or alternatively without using $0.999\cdots=1$ how can we show that  $\lfloor0.999\cdots\rfloor = \lfloor 1 \rfloor$ ? Is this statement "Every element in the empty set is greater than itself." true or its negation is true? $p$ : Every element in the empty set is greater than itself.  $\neg p$ : Some element in the empty set is smaller than or equals to itself.      I do not have the answers to this exercise, but it came out in my exams. I simply do not understand. There is nothing in an empty set! How can nothing be greater/equals/smaller/ to itself? If I had to choose an answer, it has to be $\neg p$ since nothing is equals to nothing...   What is the answer? Incorrect bookmarks and page number in table of contents I have problem with table of contents and with bookmarks. In pdf file   table of contents shows wrong page number of Introduction and Bibliography, Bookmarks: introduction and bibliography don't bring me to the correct page.   Part of my code  \documentclass[12pt,a4paper]{book}  \usepackage[polish]{babel} \usepackage{polski} \usepackage[cp1250]{inputenc} \usepackage[pdftex,pagebackref=false,draft=false,pdfpagelabels=false,pdfstartview=FitH,pdfstartpage=1,bookmarks=true,pdfauthor={LiN},pdftitle={PD},pdfsubject={Titlei},pdfkeywords={words},unicode=true]{hyperref}  \usepackage{tocloft}  \begin{document} \tableofcontents \addcontentsline{toc}{chapter}{Introduction} \chapter*{Introduction} \chapter{Chapter1} \section{Section} \addcontentsline{toc}{chapter}{Bibliography} \begin{thebibliography}{9999999}% \bibitem{1}Author, \emph{Title of book}, Publisher. \end{thebibliography} \end{document}   How can I change that? I am looking for QGIS function to extract (or detect) edge from a raster, which was referred in here,  as answer by @MappaGnosis.  The name could have changed since then, or it may have been integrated into other tool... I cannot find it in my QGIS Processing Toolbox (2.14.10 and 2.18.2).  There are similar tools such as GRASS "i.zc Zero-crossing" and SAGA "Gaussian filter", but I am hoping to find this QGIS geoalgorithm. An example of a sequence of Riemann integrable functions $(f_n)$ that converges pointwise to a function $f$ that is not Riemann integrable. Parentheses around more than one paragraph Does English allow one to put parentheses around more than one paragraph? Example:     Paragraph 1.      (Paragraph 2.      Paragraph 3.)      Paragraph 4.   (Of course, this is just a simple example. In reality, it looks more clean since the paragraphs are more than one line long, and are not separated by an empty line.) Java obfuscators I'm looking for a good Java obfuscator.  I've done initial research into the following Java obfuscators: proguard, yguard, retroguard, dasho, allatori, jshrink, smokescreen, jobfuscate, marvin, jbco, jode, javaguard, jarg, joga, cafebabe, donquixote, mwobfu, bbmug, zelix klassmaster, sandmark, jcloak, thicket, blufuscator, and java code protector.  I tried proguard and it has a really nice GUI, seems really stable, and seems to be the most popular, but it seemed to not like some enumeration on a referenced jar file (not within the code I was trying to obfuscate) which was weird.  Yguard seems to require some interaction with ant, which I didn't know too much about.  What is a good java obfuscator?  It doesn't need to be free, it just needs to work well and be easy to use. I am trying to understand the derivation of the stars and bars formula. I understand how a problem can be converted to the stars and bars format, but I am confused about how the problem of choosing all the different ways to put k stars in n sections can be written as ${n+k-1 \choose k}$. Generally "n choose k" is for counting the number of combinations of k objects we can get given n distinct objects right? That I understand since we're doing permutation divided by the number of orderings for each group since order doesn't matter. I need help figuring out how stars and bars fits into this. Suppose $Y_1, \ldots, Y_n$ is a random sample from a normal distribution with mean $\mu$ and variance $\sigma^2$. Let $S^2$ be the sample variance, which is unbiased for $\sigma^2$.  GOAL: Show that $S = \sqrt{S^2}$ is a biased estimator of $\sigma$.  I realize that $(n-1)S^2/ \sigma ^2$ is chi-square with $(n-1)$ degrees of freedom.  I am given that $$E(S) = E\left[\frac{\sigma}{\sqrt{n-1}} \left[\frac{(n-1)S^2}{\sigma^2}\right]^{1/2}\right] = \frac{\sigma}{\sqrt{n-1}} \int_0^\infty v^{1/2} \frac{1}{\Gamma[(n-1)/2] 2^{(n-1)/2}} v^{(n-1)/2} e^{-v/2} \, dv$$  Can anyone help me understand why they have $v^{1/2}$ and $v^{(n-1)/2}$ (combining them $v^{n/2}$ as I think it should be $v^{1/2}$ and $v^{\frac{(n-1)}{2}-1}$ (combining both terms give $v^{(n-2)/2}$ since the pdf for chi-square distribution is $$ f(v) = \frac{v^{(x/2)-1} e^{-v/2}}{2^{x/2}\Gamma(x/2)},$$ where $x$ is degree of freedom?  Could anyone please tell me which one is the correct answer with explanation?  Thank you