I wanted to try using LaTeX in Scrivener. I loaded the TEX template but don’t really know how I would use this. I tried pasting some LaTeX code in Scrivener, but it has all the markup, and compiles a PDF with the markup listed. Here is a simple tex document that I was trying out, but I can’t figure out a way of editing it in Scrivener:
\documentclass[10pt]{scrartcl}
\usepackage[inner=.5in, outer=1in, top=.75in, bottom=.75in]{geometry}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{lmodern}
\usepackage{fontspec,xltxtra,xunicode}
\defaultfontfeatures{Mapping=tex-text}
\setromanfont[Mapping=tex-text]{DTL Valiance TPRO}
\setsansfont[Scale=MatchLowercase,Mapping=tex-text]{DTL Caspari TOT Medium}
\setmonofont[Scale=MatchLowercase]{Andale Mono}
\usepackage{commath}
\raggedbottom
\title{Proof of Schwartz Inequality}
\date{\vspace{-5ex}}
\begin{document}
\maketitle
Prove $$x_1y_1+x_2y_2\leq\sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}$$\
It is helpful to assume it is true and then work backwards:
\begin{align*}
x_1y_1+x_2y_2 &\overset{?}{\leq}\sqrt{(x_1^2+x_2^2)}\sqrt{y_1^2+y_2^2}\
(x_1y_1+x_2y_2)^2 &\overset{?}{\leq}(x_1^2+x_2^2)(y_1^2+y_2^2)\
(x_1y_1+x_2y_2)^2 &\overset{?}{\leq}x_1^2y_1^2+x_2^2y_2^2+x_1^2y_2^2+x_2^2y_1^2\
x_1^2y_1^2+x_2^2y_2^2+2x_1x_2y_1y_2 &\overset{?}{\leq} x_1^2y_1^2+x_2^2y_2^2+x_1^2y_2^2+x_2^2y_1^2\
2x_1x_2y_1y_2 &\overset{?}{\leq} x_1^2y_2^2+x_2^2y_1^2\
0 &\overset{?}{\leq} x_1^2y_2^2+x_2^2y_1^2-2x_1x_2y_1y_2 \
0 &\overset{?}{\leq} (x_1y_2)^2-2x_1x_2y_1y_2+(x_2y_1)^2 \
0 &\overset{?}{\leq} (x_1y_2-x_2y_1)^2 \
\end{align*}
The last statement is true, since all squares are $\ge$ 0\
So to prove this
\begin{align*}
0 &\leq (x_1y_2-x_2y_1)^2 \
0 &\leq (x_1y_2)^2-2x_1x_2y_1y_2+(x_2y_1)^2 \
0 &\leq x_1^2y_2^2+x_2^2y_1^2-2x_1x_2y_1y_2 \
2x_1x_2y_1y_2 &\leq x_1^2y_2^2+x_2^2y_1^2\
x_1^2y_1^2+x_2^2y_2^2+2x_1x_2y_1y_2 &\leq x_1^2y_1^2+x_2^2y_2^2+x_1^2y_2^2+x_2^2y_1^2\
(x_1y_1+x_2y_2)^2 &\leq x_1^2y_1^2+x_2^2y_2^2+x_1^2y_2^2+x_2^2y_1^2\
(x_1y_1+x_2y_2)^2 &\leq(x_1^2+x_2^2)(y_1^2+y_2^2)\
x_1y_1+x_2y_2 &\leq\sqrt{x_1^2+x_2^2}\sqrt{y_1^2+y_2^2}\
\end{align*}
Proof of general form $\sum\limits_{i=1}^n x_iy_i \leq \sqrt{\sum\limits_{i=1}^n x_i^2} \sqrt{\sum\limits_{i=1}^n y_i^2} $\[2em]
Start with the general form of the first proof and work backwards:
\begin{align*}
\sum\limits_{i=1}^n\sum\limits_{j=1}^n (x_iy_j-x_jy_i)^2 &= \sum\limits_{i=1}^n x_i^2 \sum\limits_{j=1}^n y_j^2+\sum\limits_{i=1}^n y_i^2 \sum\limits_{j=1}^n x_j^2-2\sum\limits_{i=1}^n x_iy_i\sum\limits_{j=1}^n x_j y_j\
\sum\limits_{i=1}^n\sum\limits_{j=1}^n (x_iy_j-x_jy_i)^2 &= 2\biggl(\sum\limits_{i=1}^n x_i^2\biggr)\biggl(\sum\limits_{i=1}^n y_i^2\biggr)-2\biggl(\sum\limits_{i=1}^n x_i y_i\biggr)^2\
0&\leq 2\biggl(\sum\limits_{i=1}^n x_i^2\biggr)\biggl(\sum\limits_{i=1}^n y_i^2\biggr)-2\biggl(\sum\limits_{i=1}^n x_i y_i\biggr)^2\
0&\leq \biggl(\sum\limits_{i=1}^n x_i^2\biggr)\biggl(\sum\limits_{i=1}^n y_i^2\biggr)-\biggl(\sum\limits_{i=1}^n x_i y_i\biggr)^2\
\biggl(\sum\limits_{i=1}^n x_i y_i\biggr)^2&\leq \sum\limits_{i=1}^n x_i^2\sum\limits_{i=1}^n y_i^2\
\sum\limits_{i=1}^n x_iy_i &\leq \sqrt{\sum\limits_{i=1}^n x_i^2} \sqrt{\sum\limits_{i=1}^n y_i^2}
\end{align*}
\end{document}
