Convert an int to bool with Json.Net
JSonNet boolean serialization
How to prove or disprove following statement :     There are infinitely many primes of the form : $\lfloor \sqrt {3} \cdot n \rfloor $    Note: This is a problem I made myself.  There is a theorem that states :  $\lfloor nx \rfloor = \begin{cases}  n\lfloor x \rfloor, &amp; \text{if } 0 \leq \{x\} &lt; \frac{1}{n} \\  n\lfloor x \rfloor +1, &amp; \text{if } \frac{1}{n} \leq \{x\} &lt; \frac{2}{n} \\  n\lfloor x \rfloor +2, &amp; \text{if } \frac{2}{n} \leq \{x\} &lt; \frac{3}{n} \\  \vdots \\  n\lfloor x \rfloor +n-1, &amp; \text{if } \frac{n-1}{n} \leq \{x\} &lt; 1 \\ \end{cases}$  where $\{x\}$ is a non-integer part of $x$ .  Hence :  $\lfloor \sqrt{3}\cdot n \rfloor = \begin{cases}  n, &amp; \text{if } 0 \leq \sqrt{3}-1 &lt; \frac{1}{n} \\  n +1, &amp; \text{if } \frac{1}{n} \leq \sqrt{3}-1 &lt; \frac{2}{n} \\  n +2, &amp; \text{if } \frac{2}{n} \leq \sqrt{3}-1 &lt; \frac{3}{n} \\  \vdots \\  2n-1, &amp; \text{if } \frac{n-1}{n} \leq \sqrt{3}-1 &lt; 1 \\ \end{cases}$  How can I proceed from here ? Answers no longer sorted by score/votes For me at least,  has answers that are not sorted by score:    Is this normal?  I've tried refreshing the page a few times, but the order doesn't change... Is there a way to calculate the improper integral $\int_0^\infty \big(\frac{\ln x}{x - 1}\big)^2 dx$?  What have I tried:  $$\int_0^\infty \Big(\frac{\ln x}{x - 1}\Big)^2 dx = \int_0^1 \Big(\frac{\ln x}{x - 1}\Big)^2 dx + \int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2 dx = \int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2dx + \int_1^\infty\Big(\frac{\ln t}{\big(\frac{1}{t} - 1\big)t}\Big)^2dt = 2\int_1^\infty \Big(\frac{\ln x}{x - 1}\Big)^2dx$$  From the above statement we can conclude, that the integral does indeed converge, as $0 \leq (\frac{\ln x}{x - 1})^2 \leq \frac{C}{(x - 1)^{\frac{3}{2}}}$, for some constant $C$. However, this new form of the integral is not much helpful in finding its exact value.  I also tried substituting $z = \ln x$ , then $dz = \frac{dx}{x}$. Then the integral becomes reduced to $2\int_{0}^{\infty} \frac{z^2}{e^z(e^z - 1)^2}dz$, which, unfortunately, also does not seem to be any easier. install ubuntu on thinkpad T440p Is there anybody, who is able to install and fully configure Ubuntu 12.04 or 14.04 on Thinkpad T440p (ver 20AN)? If yes, could you post here your procedure step by step?  Please does not include your experience with any other ThinkPad models.  Edit: Situation is not so simple, see: Why use Chef/Puppet over shell scripts? New to Puppet and Chef tools. Seems like the job that they are doing can be done with shell scripting. Maybe it was done in shell scripts until these came along.  I would agree they are more readable. But, are there any other advantages over shell scripts besides just being readable? Can you explain briefly the main concepts and command line tools used to manage file permissions? How do you debug PHP scripts? How do you debug PHP scripts?  I am aware of basic debugging such as using the Error Reporting. The breakpoint debugging in PHPEclipse is also quite useful.  What is the best (in terms of fast and easy) way to debug in phpStorm or any other IDE? Is it correct to hyphenate with compound premodifiers? If so, where is the hyphen placed? Can non-magical persons see the Hogwarts Express during its journey to and from Hogwarts? Autosegmental Representation in tikz-qtree This is what I want to achieve    and this:    But all I've been able to make with tikz-qtree is the following:  \documentclass[12pt,a4]{article} \usepackage[utf8]{inputenc} \usepackage{tipa,tikz,tikz-qtree} \begin{document}  \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ p ] ] [.R [.$\mu$ a ] [.$\mu$ : ] ] ] \end{tikzpicture} \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ p ] ] [.R [.$\mu$ a ]  ] ] \end{tikzpicture}   And this:  \begin{tikzpicture}[baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ l ] ] [.R [.$\mu$ a ] [.$\mu$ l ] ] ] \end{tikzpicture} \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ : ] ] [.R [.$\mu$ a ] [ l ] ] ] \end{tikzpicture} \end{document}   Which renders:    and    The important part here is the linking from the leaves to more than one parent (it would be nice if I had the option of making the line dashed too). Now, pst-asr doesn't quite achieve what I want it to. It's important that I have the moras ('mu's). (Also it's a pstricks package, which means I must typeset it in DVIPSPDF.) I have tried a bit of the forest-package (especially forest-GP1), but I can't make that work as I want either. Lastly, some have recommended xyling, but I find that package very difficult to use.  Any recommendations?  Thanks. I want to add a check constraint on a column if entered value for column is greater than 3 than it should be saved as 1 else entered value. For that scenario I am using below query but on insertion of data it shows an error message.  create table tblTestCheckConstraint  ( id int, NewColumn int )   alter table tblTestCheckConstraint add constraint chk_tblTestCheckConstraint_NewColumn1 CHECK   (     CASE         WHEN NewColumn &gt;4 THEN 1         ELSE NewColumn     END = 1 )   Insert Statement:   insert into tblTestCheckConstraint values ( 1,5) insert into tblTestCheckConstraint values ( 1,2)   Error Message:   Msg 547, Level 16, State 0, Line 1 The INSERT statement conflicted with the CHECK constraint "chk_tblTestCheckConstraint_NewColumn1". The conflict occurred in database "DBName" , table "dbo.tblTestCheckConstraint", column 'NewColumn'. The statement has been terminated. Selecting through one-to-many relationship I'm trying to make a species distribution project where I can select a species from a non-spatial list of species and the selection would highlight (select) all the counties to which the species has been attributed.  Simplified example data: species, a non-spatial table of species (CSV with header): "species" "Mus borealis" "Mus australis"  species_distributions, a non-spatial relation table (csv with header): "species","county" "Mus borealis","North" "Mus borealis","Central" "Mus australis","South" "Mus australis","Central"  counties, a spatial feature class representing county boundaries (first row = field name): "county" "North" "South" "Central" "West"  So if I select "Mus borealis" from the species table, the "North" and "Central" counties would be highlighted in the map. Seems like a simple scenario, but I have been unable to uncover a solution on my own. The closest I've found is a plugin "Select by relationship" but it doesn't appear to be available for QGIS 3 (I'm using QGIS 3.4.3). I have a PostGIS server available. Are PPAs safe to add to my system and what are some "red flags" to watch out for? Why does camera this tracking fail completely even when the solution value is very low? This is a follow-up question to the discussion about the finite axiom of choice .  Suppose we have a countable collection of non-empty sets $\{A_1, A_2, A_3,\cdots\}$ Reasoning as indicated in that discussion, we can prove the existence of a choice function $c_n$ such that $c_n(i)$ belongs to $A_i$ for $i = 1,\cdots,n$  Again reasoning as suggested and using induction, we can prove the existence of a function $C$ defined over $\{1, 2, \cdots\}$ such that $C(n)$ is a choice function defined for $i=1,\cdots,n$ with $C(n)(i)$ an element of A_i. We can require in addition that $C(n)(i)$ and $C(n-1)(i)$ are equal for $i=1,\cdots,n-1$.  Now it would it seem that we can prove the countable axiom of choice by taking the union of $C(1)$, $C(2)$, $\cdots$  The question is as follows: why is that proof wrong? I suspect the answer may have something to do with the axiom of union. The package babel with the ngerman option (and most other languages as well, but not english) provides the shorthand "|, which prevents a ligature at the given position (see p. 81 of the , e.g. Auf"|lage will result in the word Auflage without an fl-ligature (as is desired because, in German, you don't want ligatures across morpheme borders).  However, when placed in a macro, "| doesn't work anymore. Why is this and how can I circumvent it?  \documentclass{article}  %\usepackage[T1]{fontenc}% doesn't help \usepackage[ngerman]{babel}  \newcommand{\foo}{"|}  \begin{document}  ff  f"|f  f\foo f  \end{document} Can non basic lands be tapped for colorless mana? For example,  has a mana fixing ability, but it doesn't say you can tap it for mana. Other non-basic lands have a mana ability listed on the card. Should I assume that Evolving Winds cannot be tapped for mana?  Basic lands don't have mana tapping abilities listed, but are assumed you can tap them for mana of the color they represent. "is a field and used like a type" error Derivation of survival copula $$ \begin{align}F(x,y) &amp;= P(X\leq x, Y\leq y) \\  &amp;= C(F(x), F(y)) \end{align}$$ The copula of a bivariate distribution is equal to its CDF, $C(u,v) = F(x,y)$. The density of the data's CDF therefore can be expressed as $$f(x,y) = c(u,v) \times f_X(x) f_Y(y)$$ where $c(u,v)$ is the copula density and $f_X(x)$ is a marginal. If $\hat{C}(u,v)$ is the survival copula for determining $$ \begin{align}\bar{F}(x,y) &amp;= P(X&gt; x, Y &gt; y) \\  &amp;= \hat{C}(1-F(x), 1-F(y)) \end{align}$$ how is the first equation of the ordinary copula, $C(u,v)$, linked to the survival copula as $$\hat{C}(u,v) = u + v - 1 + C(1-u,1-v)$$ are $u$ and $v$ merely vectors containing the (ranked) observations of the uniform marginals? A derivation in integrals and probabilities (rather than $F$s) would be good. In the standard PrintDialog there are four values associated with a selected printer: Status, Type, Where, and Comment.  If I know a printer's name, how can I get these values in C# 2.0?