Chapter 11: test2

https://www.overleaf.com/learn/latex/Commands

\newcommand

\renewcommand to replace existing command

https://www.physicsread.com/latex-newcommand/

font color[10.6.1]

sin(α),sinn(β),sinm(γ)

cos(2θ)sin(2θ)=cos(4θ)

θ=tan1(xy)

tan(α+β)=tan(α)+tan(β)1tan(α)tan(β)

https://www.alanshawn.com/latex3-tutorial/#latex3-regular-expression-xxviii

\regex_replace_all

https://tex.stackexchange.com/questions/422631/expl3-regex-and-missing-semicolon

\documentclass{article}

\usepackage{tikz,xparse}

\ExplSyntaxOn

\tl_new:N \l_bob_func_tl

\NewDocumentCommand \makelinesegment { m }
 {
  \pgfextra
  \tl_set:Nn \l_bob_func_tl { #1 }
  \regex_replace_all:nnN { ([0-9])- } { \1)--( } \l_bob_func_tl
  \regex_replace_all:nnN { : } { , } \l_bob_func_tl
  \exp_last_unbraced:NNV % this is not necessary, but I find it cleaner, correct me if I'm wrong
  \endpgfextra
  (\l_bob_func_tl)
 }

\ExplSyntaxOff

\begin{document}

% \makelinesegment{1:1--1:2-2:2} % do not use outside `\tikz` or `{tikzpicture}`

\tikz\draw\makelinesegment{1:1--1:2-2:2};

\end{document}

https://stackoverflow.com/questions/41655383/r-markdown-similar-feature-to-newcommand-in-latex

Var(X)

Var[Y]=x=3

0=zl(h(zl1)wlzl+λh(zl)wl+1zl+1)

https://tex.stackexchange.com/questions/353114/latex-equations-colour-all-instances-of-symbol

0=\ensuremathzl(h(zl1)wl\ensuremathzl+λh(\ensuremathzl)wl+1zl+1)

0=zl(h(zl1)wlzl+λh(zl)wl+1zl+1)


0ϵ=¯PFd(P,L)=¯PF¯PP=(x,y)(0,yF)(x,y)(x,yL)=(x,yyF)(0,yyL)=x2+(yyF)2(yyL)2ϵ2=x2+(yyF)2(yyL)2=x2+y22yFy+y2Fy22yLy+y2L0=x2+(1ϵ2)y22(yFϵ2yL)y+(y2Fϵ2y2L)ϵ1=x2+(1ϵ2)[y22(yFϵ2yL)1ϵ2y+y2Fϵ2y2L1ϵ2]=x2+(1ϵ2)[y22(yFϵ2yL)1ϵ2y+(yFϵ2yL1ϵ2)2(yFϵ2yL1ϵ2)2+y2Fϵ2y2L1ϵ2]=x2+(1ϵ2)[(yyFϵ2yL1ϵ2)2+(y2Fϵ2y2L)(1ϵ2)(yFϵ2yL)2(1ϵ2)2]=x2+(1ϵ2)(yyFϵ2yL1ϵ2)2+(y2Fϵ2y2L)(1ϵ2)(yFϵ2yL)21ϵ2

(y2Fϵ2y2L)(1ϵ2)(yFϵ2yL)2=(1ϵ2)y2F(ϵ2ϵ4)y2Ly2F+2ϵ2yFyLϵ4y2L=ϵ2y2Fϵ2y2L+2ϵ2yFyL=ϵ2(yFyL)2


0ϵ=¯PFd(P,L)=¯PF¯PP=(x,y)(0,yF)(x,y)(x,yL)=(x,yyF)(0,yyL)=x2+(yyF)2(yyL)2ϵ2=x2+(yyF)2(yyL)2=x2+y22yFy+y2Fy22yLy+y2L0=x2+(1ϵ2)y22(yFϵ2yL)y+(y2Fϵ2y2L)ϵ1=x2+(1ϵ2)[y22(yFϵ2yL)1ϵ2y+y2Fϵ2y2L1ϵ2]=x2+(1ϵ2)[y22(yFϵ2yL)1ϵ2y+(yFϵ2yL1ϵ2)2(yFϵ2yL1ϵ2)2+y2Fϵ2y2L1ϵ2]=x2+(1ϵ2)[(yyFϵ2yL1ϵ2)2+(y2Fϵ2y2L)(1ϵ2)(yFϵ2yL)2(1ϵ2)2]=x2+(1ϵ2)(yyFϵ2yL1ϵ2)2+(y2Fϵ2y2L)(1ϵ2)(yFϵ2yL)21ϵ2

(y2Fϵ2y2L)(1ϵ2)(yFϵ2yL)2=(1ϵ2)y2F(ϵ2ϵ4)y2Ly2F+2ϵ2yFyLϵ4y2L=ϵ2y2Fϵ2y2L+2ϵ2yFyL=ϵ2(yFyL)2