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=tan−1(xy)
tan(α+β)=tan(α)+tan(β)1−tan(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(zl−1)⋅wl−zl‖+λ‖h(zl)⋅wl+1−zl+1‖)
https://tex.stackexchange.com/questions/353114/latex-equations-colour-all-instances-of-symbol
0=∂∂\ensuremathzl(‖h(zl−1)⋅wl−\ensuremathzl‖+λ‖h(\ensuremathzl)⋅wl+1−zl+1‖)
0=∂∂zl(‖h(zl−1)⋅wl−zl‖+λ‖h(zl)⋅wl+1−zl+1‖)
0≤ϵ=¯PFd(P,L)=¯PF¯PP′=‖(x,y)−(0,yF)‖‖(x,y)−(x,yL)‖=‖(x,y−yF)‖‖(0,y−yL)‖=√x2+(y−yF)2√(y−yL)2ϵ2=x2+(y−yF)2(y−yL)2=x2+y2−2yFy+y2Fy2−2yLy+y2L0=x2+(1−ϵ2)y2−2(yF−ϵ2yL)y+(y2F−ϵ2y2L)ϵ≠1=x2+(1−ϵ2)[y2−2(yF−ϵ2yL)1−ϵ2y+y2F−ϵ2y2L1−ϵ2]=x2+(1−ϵ2)[y2−2(yF−ϵ2yL)1−ϵ2y+(yF−ϵ2yL1−ϵ2)2−(yF−ϵ2yL1−ϵ2)2+y2F−ϵ2y2L1−ϵ2]=x2+(1−ϵ2)[(y−yF−ϵ2yL1−ϵ2)2+(y2F−ϵ2y2L)(1−ϵ2)−(yF−ϵ2yL)2(1−ϵ2)2]=x2+(1−ϵ2)(y−yF−ϵ2yL1−ϵ2)2+(y2F−ϵ2y2L)(1−ϵ2)−(yF−ϵ2yL)21−ϵ2
(y2F−ϵ2y2L)(1−ϵ2)−(yF−ϵ2yL)2=(1−ϵ2)y2F−(ϵ2−ϵ4)y2L−y2F+2ϵ2yFyL−ϵ4y2L=−ϵ2y2F−ϵ2y2L+2ϵ2yFyL=−ϵ2(yF−yL)2
0≤ϵ=¯PFd(P,L)=¯PF¯PP′=‖(x,y)−(0,yF)‖‖(x,y)−(x,yL)‖=‖(x,y−yF)‖‖(0,y−yL)‖=√x2+(y−yF)2√(y−yL)2ϵ2=x2+(y−yF)2(y−yL)2=x2+y2−2yFy+y2Fy2−2yLy+y2L0=x2+(1−ϵ2)y2−2(yF−ϵ2yL)y+(y2F−ϵ2y2L)ϵ≠1=x2+(1−ϵ2)[y2−2(yF−ϵ2yL)1−ϵ2y+y2F−ϵ2y2L1−ϵ2]=x2+(1−ϵ2)[y2−2(yF−ϵ2yL)1−ϵ2y+(yF−ϵ2yL1−ϵ2)2−(yF−ϵ2yL1−ϵ2)2+y2F−ϵ2y2L1−ϵ2]=x2+(1−ϵ2)[(y−yF−ϵ2yL1−ϵ2)2+(y2F−ϵ2y2L)(1−ϵ2)−(yF−ϵ2yL)2(1−ϵ2)2]=x2+(1−ϵ2)(y−yF−ϵ2yL1−ϵ2)2+(y2F−ϵ2y2L)(1−ϵ2)−(yF−ϵ2yL)21−ϵ2
(y2F−ϵ2y2L)(1−ϵ2)−(yF−ϵ2yL)2=(1−ϵ2)y2F−(ϵ2−ϵ4)y2L−y2F+2ϵ2yFyL−ϵ4y2L=−ϵ2y2F−ϵ2y2L+2ϵ2yFyL=−ϵ2(yF−yL)2