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%\usepackage{opensans}          % Can't make this work so far. Shame. Would be lovely.
\usepackage[normalem]{ulem}     % For underlining links
% Most of the following are not required for the majority
% of cheat sheets but are needed for some symbol support.
\usepackage{amsmath}            % Symbols
\usepackage{MnSymbol}           % Symbols
\usepackage{wasysym}            % Symbols
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% Document Info
\author{Dois}
\pdfinfo{
  /Title (discrete-math.pdf)
  /Creator (Cheatography)
  /Author (Dois)
  /Subject (Discrete Math Cheat Sheet)
}

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% Header and Footer
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\fancyhead[L]{
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{C}
    \SetRowColor{DarkBackground}
    \vspace{-7pt}
    {\parbox{\dimexpr\textwidth-2\fboxsep\relax}{\noindent
        \hspace*{-6pt}\includegraphics[width=5.8cm]{/web/www.cheatography.com/public/images/cheatography_logo.pdf}}
    }
\end{tabulary}
\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Discrete Math Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Dois} via \textcolor{DarkBackground}{\uline{cheatography.com/11428/cs/1340/}}}
\end{tabulary}
\end{multicols}}

\fancyfoot[L]{ \footnotesize
\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Dois \\
  \uline{cheatography.com/dois} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Published 30th October, 2013.\\
   Updated 13th May, 2016.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Sponsor}}  \\
  \SetRowColor{white}
  \vspace{-5pt}
  %\includegraphics[width=48px,height=48px]{dave.jpeg}
  Measure your website readability!\\
  www.readability-score.com
\end{tabulary}
\end{multicols}}




\begin{document}
\raggedright
\raggedcolumns

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% www.emerson.emory.edu/services/latex/latex_169.html
\footnotesize % Small font.

\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{x{2.63781 cm} x{2.33919 cm} }
\SetRowColor{DarkBackground}
\mymulticolumn{2}{x{5.377cm}}{\bf\textcolor{white}{Complex Numbers}}  \tn
% Row 0
\SetRowColor{LightBackground}
j\textasciicircum{}2\textasciicircum{} = -1 & j\textasciicircum{}3\textasciicircum{} = -j \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
j\textasciicircum{}4\textasciicircum{} = 1 & z = a + bj \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
z = r(sin θ + jsinθ) & z = re\textasciicircum{}jθ\textasciicircum{} \tn 
% Row Count 4 (+ 2)
% Row 3
\SetRowColor{white}
tan\textasciicircum{}-1\textasciicircum{} b/a =  θ & cos\textasciicircum{}-1\textasciicircum{} a/r =  θ \tn 
% Row Count 5 (+ 1)
% Row 4
\SetRowColor{LightBackground}
sin\textasciicircum{}-1\textasciicircum{} b/r =  θ & (a + bj)* = a - bj \tn 
% Row Count 6 (+ 1)
% Row 5
\SetRowColor{white}
|z|=r=sqrt(a\textasciicircum{}2\textasciicircum{} + b\textasciicircum{}2\textasciicircum{}) & |z|\textasciicircum{}x\textasciicircum{} = |z\textasciicircum{}x\textasciicircum{}| \tn 
% Row Count 8 (+ 2)
% Row 6
\SetRowColor{LightBackground}
arg(z)\textasciicircum{}x\textasciicircum{} = x arg (z) & arg(z) =  θ + 2kπ \tn 
% Row Count 10 (+ 2)
% Row 7
\SetRowColor{white}
(cos θ + jsin θ)\textasciicircum{}k\textasciicircum{} & = cos kθ + jsin kθ \tn 
% Row Count 12 (+ 2)
% Row 8
\SetRowColor{LightBackground}
= (e\textasciicircum{}jθ\textasciicircum{})\textasciicircum{}k\textasciicircum{} = e\textasciicircum{}jkθ\textasciicircum{} & \textless{} DeMoivre's Theorum \tn 
% Row Count 14 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\SetRowColor{LightBackground}
\mymulticolumn{2}{x{5.377cm}}{* means conjugate \newline j = i = sqrt(-1) = imaginary unit \newline Find roots example: \newline z\textasciicircum{}2\textasciicircum{} = -4j \newline Convert to exponential form first: \newline z\textasciicircum{}2\textasciicircum{} = 4e\textasciicircum{}-jπ/2\textasciicircum{} \newline  \newline |z\textasciicircum{}2\textasciicircum{}| = r\textasciicircum{}2\textasciicircum{} = sqrt(0\textasciicircum{}2\textasciicircum{} + 4\textasciicircum{}2\textasciicircum{}) = 4 \newline |z| = r = 2 \newline k = (0, 1 ...n where n = expon' of z) = 0, 1 \newline arg(z\textasciicircum{}2\textasciicircum{}) = 2 arg(z) = -π/2 + 2kπ \newline arg(z) = -π/4 + kπ \newline  \newline Substitute values of k (0, 1) for z = |z|e\textasciicircum{}jarg(z)\textasciicircum{} = 2e\textasciicircum{}-jπ/4\textasciicircum{}, 2e\textasciicircum{}j3π/4\textasciicircum{}}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}--}
\end{tabularx}
\par\addvspace{1.3em}

\columnbreak
\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Discrete Probability \& Sets \& Whatever}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Probability \newline % Row Count 1 (+ 1)
1. P(x) = \textasciicircum{}n\textasciicircum{}Cx . p\textasciicircum{}x\textasciicircum{} . (1-p)\textasciicircum{}n-x\textasciicircum{} \newline % Row Count 2 (+ 1)
2. P(x) = (\textasciicircum{}X\textasciicircum{}Ck)((\textasciicircum{}N-X\textasciicircum{})C(n-k))/\textasciicircum{}N\textasciicircum{}Cn \newline % Row Count 3 (+ 1)
Set Theory \newline % Row Count 4 (+ 1)
A = B when A subset of B \& B subset of A \newline % Row Count 5 (+ 1)
A - B = A n B' \newline % Row Count 6 (+ 1)
A u (A n B) = A \newline % Row Count 7 (+ 1)
A n (A u B) = A \newline % Row Count 8 (+ 1)
A u A' = U \newline % Row Count 9 (+ 1)
A n A' = nullset or \{\} \newline % Row Count 10 (+ 1)
Power set of S is the set of ALL SUBSETS of S e.g. S = \{1,2\} , P(S) = \{ \{\}, \{1\}, \{2\}, \{1,2\}\} \newline % Row Count 12 (+ 2)
|A| = n, |P(A)| = 2\textasciicircum{}n\textasciicircum{} \newline % Row Count 13 (+ 1)
Sets A and B are disjoint iff A n B = \{\} \newline % Row Count 14 (+ 1)
Cardinality of union: |A u B| = |A| + |B| - |A n B| \newline % Row Count 16 (+ 2)
Proof by induction: \newline % Row Count 17 (+ 1)
Show that when p(k) is true, p(k + 1) follows.% Row Count 18 (+ 1)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{1. Binomial Distribution \newline n = trials, x = successes, p = probability of success \newline  \newline 2. Hypergeometric Distribution \newline N = deck size, n = draws, X = copies of card, k = successes}  \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Matrix Manipulations}}  \tn
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{A\textasciicircum{}T\textasciicircum{}: Transpose of A - Switch Rows with Columns (R1 becomes C1, R2 becomes C2 etc.) \newline % Row Count 2 (+ 2)
-A = -1 . A \newline % Row Count 3 (+ 1)
A\textasciicircum{}-1\textasciicircum{}: Inverse of A \newline % Row Count 4 (+ 1)
A\textasciicircum{}-1\textasciicircum{} . I = I = A . I \newline % Row Count 5 (+ 1)
A\textasciicircum{}-1\textasciicircum{}A=I \newline % Row Count 6 (+ 1)
Augment Identity matrix to matrix and perform Guass-Jordon elimination on both to get change Identity matrix to the Inverse. \newline % Row Count 9 (+ 3)
EROs: \newline % Row Count 10 (+ 1)
Switch Rows \newline % Row Count 11 (+ 1)
Scale Row (Multiply entire row) \newline % Row Count 12 (+ 1)
Add multiple of different row to another \newline % Row Count 13 (+ 1)
A matrix A is in row echelon form if \newline % Row Count 14 (+ 1)
1. The nonzero rows in A lie above all zero rows (when there is at least a nonzero row and a zero row). \newline % Row Count 17 (+ 3)
2. The first nonzero entry in a nonzero row (called a pivot) lies to the right of the pivot in the row immediately above it.% Row Count 20 (+ 3)
} \tn 
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


% That's all folks
\end{multicols*}

\end{document}