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\author{Boko}
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  /Title (analysis-part-3-4.pdf)
  /Creator (Cheatography)
  /Author (Boko)
  /Subject (Analysis Part 3-4 Cheat Sheet)
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\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}}
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\columnbreak
\begin{tabulary}{11cm}{L}
    \vspace{-2pt}\large{\bf{\textcolor{DarkBackground}{\textrm{Analysis Part 3-4 Cheat Sheet}}}} \\
    \normalsize{by \textcolor{DarkBackground}{Boko} via \textcolor{DarkBackground}{\uline{cheatography.com/55472/cs/15364/}}}
\end{tabulary}
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\noindent
\begin{multicols}{3}
\begin{tabulary}{5.8cm}{LL}
  \SetRowColor{FootBackground}
  \mymulticolumn{2}{p{5.377cm}}{\bf\textcolor{white}{Cheatographer}}  \\
  \vspace{-2pt}Boko \\
  \uline{cheatography.com/boko} \\
  \end{tabulary}
\vfill
\columnbreak
\begin{tabulary}{5.8cm}{L}
  \SetRowColor{FootBackground}
  \mymulticolumn{1}{p{5.377cm}}{\bf\textcolor{white}{Cheat Sheet}}  \\
   \vspace{-2pt}Not Yet Published.\\
   Updated 1st May, 2018.\\
   Page {\thepage} of \pageref{LastPage}.
\end{tabulary}
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  \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}
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\begin{multicols*}{3}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Numerical Integration}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Area under the curve} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{{\emph{Single Integral }}}}} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Trapezoidal Rule}}} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{fit linear  function} \tn 
% Row Count 4 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Single Segment}} A=(f(a)+f(b))*(b-a)/2} \tn 
% Row Count 5 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Multiple Segments}} \seqsplit{A=(f(xo)+2sum(f(xi))+f(xn))*(b-a)/2n}} \tn 
% Row Count 7 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{n is number of intervals- same width} \tn 
% Row Count 8 (+ 1)
% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{increase accuracy.....increase intervals} \tn 
% Row Count 9 (+ 1)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Exception: -linear function -fluctuating function} \tn 
% Row Count 10 (+ 1)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{as dx decreases, come closer to function} \tn 
% Row Count 11 (+ 1)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{* inefficient but no limit on \# of intervals} \tn 
% Row Count 12 (+ 1)
% Row 11
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{if non equal intervals, calculate separately and add} \tn 
% Row Count 14 (+ 2)
% Row 12
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Simpson's 1/3 Rule}}} \tn 
% Row Count 15 (+ 1)
% Row 13
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{fit quadratic function} \tn 
% Row Count 16 (+ 1)
% Row 14
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Single}}  \seqsplit{A=(f(x0)+4f(x1)+f(x2))*(b-a)/6}} \tn 
% Row Count 17 (+ 1)
% Row 15
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{equidistant  x1} \tn 
% Row Count 18 (+ 1)
% Row 16
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Multiple}}  \seqsplit{A=(f(x0)+4sum\_odd(f(xi))+2sum\_even(f(xi))+f(xn))*(b-a)/3n}} \tn 
% Row Count 20 (+ 2)
% Row 17
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\# of intervals is even} \tn 
% Row Count 21 (+ 1)
% Row 18
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{* even \# of intervals} \tn 
% Row Count 22 (+ 1)
% Row 19
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{most popular bec accuracy is not that significant from 3/8 with less computation} \tn 
% Row Count 24 (+ 2)
% Row 20
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Simpson's 3/8  Rule}}} \tn 
% Row Count 25 (+ 1)
% Row 21
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{fit cubic function} \tn 
% Row Count 26 (+ 1)
% Row 22
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{A= \seqsplit{(f(x0)+3f(x1)+3f(x2)+f(x3))*(b-a)/8}} \tn 
% Row Count 27 (+ 1)
% Row 23
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{1/3 rule is most widely used as computational efficiency it provides outweighs the accuracy provided by 3/8 rule} \tn 
% Row Count 30 (+ 3)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Numerical Integration (cont)}}  \tn
% Row 24
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Trapezoidal rule can reach same accuracy of 3/8 rule by increasing number of intervals} \tn 
% Row Count 2 (+ 2)
% Row 25
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{* multiple of 3 \# of intervals} \tn 
% Row Count 3 (+ 1)
% Row 26
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{{\emph{Multiple Integral }}}}} \tn 
% Row Count 4 (+ 1)
% Row 27
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Step 1 at y=0 find A  {\bf{repeat}}} \tn 
% Row Count 5 (+ 1)
% Row 28
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Step 2 find A of A(y)} \tn 
% Row Count 6 (+ 1)
% Row 29
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{Tavg= A(A(y))/area or T=A(A(y))} \tn 
% Row Count 7 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Numerical Differentiation}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Taylor series}}} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{f(xi+1)= f(xi)+f'(xi)h+f''(xi)h\textasciicircum{}2\textasciicircum{}/2!+............+f\textasciicircum{}n\textasciicircum{}(xi)h\textasciicircum{}n\textasciicircum{}/n!} \tn 
% Row Count 3 (+ 2)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Exponential, you need infinite order because f\textasciicircum{}n\textasciicircum{}(xi)=e\textasciicircum{}x\textasciicircum{} which is never 0} \tn 
% Row Count 5 (+ 2)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{First Forward difference }}} \tn 
% Row Count 6 (+ 1)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f'(xi)= (f(xi+1)-f(xi))/h} \tn 
% Row Count 7 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{to increase accuracy, decrease h that will decrease the rest of Taylor series}}} \tn 
% Row Count 9 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{First Backward difference }}} \tn 
% Row Count 10 (+ 1)
% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{f'(xi)= (f(xi)-f(xi-1))/h} \tn 
% Row Count 11 (+ 1)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{First Centered difference }}} \tn 
% Row Count 12 (+ 1)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{f'(xi)= (f(xi+1)-f(xi-1))/2h} \tn 
% Row Count 13 (+ 1)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{Higher Order}}} \tn 
% Row Count 14 (+ 1)
% Row 11
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{First Forward difference }}} \tn 
% Row Count 15 (+ 1)
% Row 12
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f'(xi)= -f(xi+2)+4f(xi+1) -3f(xi) /(2h)} \tn 
% Row Count 16 (+ 1)
% Row 13
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{First Backward difference }}} \tn 
% Row Count 17 (+ 1)
% Row 14
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f'(xi) = 3f(xi) -4f(xi-1) +f(xi-2) /(2h)} \tn 
% Row Count 18 (+ 1)
% Row 15
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{First Centered difference }}} \tn 
% Row Count 19 (+ 1)
% Row 16
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f'(xi)= -f(x+2) \seqsplit{+8f(xi+1)-8f(xi-1)+f(xi-2)} /(12h)} \tn 
% Row Count 20 (+ 1)
% Row 17
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Second Forward difference }}} \tn 
% Row Count 21 (+ 1)
% Row 18
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f''(xi)= (f(xi+2)-2f(xi+1)+f(xi))/h\textasciicircum{}2\textasciicircum{}} \tn 
% Row Count 22 (+ 1)
% Row 19
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Second Backward difference }}} \tn 
% Row Count 23 (+ 1)
% Row 20
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f''(xi)= (f(xi)-2f(xi-1)+f(xi-2))/h\textasciicircum{}2\textasciicircum{}} \tn 
% Row Count 24 (+ 1)
% Row 21
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Second Centered difference }}} \tn 
% Row Count 25 (+ 1)
% Row 22
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f''(xi)= (f(xi+1)-2f(xi)+f(xi-1))/h\textasciicircum{}2\textasciicircum{}} \tn 
% Row Count 26 (+ 1)
% Row 23
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{* more accurate} \tn 
% Row Count 27 (+ 1)
% Row 24
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Lagrange }}} \tn 
% Row Count 28 (+ 1)
% Row 25
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{fit interpolated polynomial then differentiate- function that passes by all points} \tn 
% Row Count 30 (+ 2)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Numerical Differentiation (cont)}}  \tn
% Row 26
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{general method} \tn 
% Row Count 1 (+ 1)
% Row 27
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{f'(x)= pt1+pt2+pt3} \tn 
% Row Count 2 (+ 1)
% Row 28
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{pt1: 2x-xi-(xi+1)/ ((xi-1)-xi) ((xi-1)-(xi+1))  * f(xi-1)} \tn 
% Row Count 4 (+ 2)
% Row 29
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{pt2: 2x-(xi-1)-(xi+1)/ (xi-(xi-1)) (xi-(xi+1))  * f(xi)} \tn 
% Row Count 6 (+ 2)
% Row 30
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{pt3: 2x-(xi-1)-xi/ ((xi+1)-(xi-1)) ((xi+1)-xi)  * f(xi+1)} \tn 
% Row Count 8 (+ 2)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}

\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Ordinary Differential Equations}}  \tn
% Row 0
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Why solve numerically?} \tn 
% Row Count 1 (+ 1)
% Row 1
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{efficiency or cannot solve analytically} \tn 
% Row Count 2 (+ 1)
% Row 2
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\bf{butterfly effect }}-{}-sensitive to minor changes} \tn 
% Row Count 3 (+ 1)
% Row 3
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{chaotic systems}} -{}-system sensitive to initial conditions but with predictable behavior} \tn 
% Row Count 5 (+ 2)
% Row 4
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{ODE}} with respect to 1 independent variable} \tn 
% Row Count 6 (+ 1)
% Row 5
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{PDE}} with respect to more than 1 independent variable} \tn 
% Row Count 8 (+ 2)
% Row 6
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{both can have many dependent variables} \tn 
% Row Count 9 (+ 1)
% Row 7
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Euler's Method}}} \tn 
% Row Count 10 (+ 1)
% Row 8
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{f(xi+1)=f(xi)+ k1h} \tn 
% Row Count 11 (+ 1)
% Row 9
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{to decrase error, either decrease timestep (h) or take more slopes} \tn 
% Row Count 13 (+ 2)
% Row 10
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{* to decrease timestep (h), take into account round off error propagates and computational efficiency} \tn 
% Row Count 16 (+ 3)
% Row 11
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{assume slope constant in interval} \tn 
% Row Count 17 (+ 1)
% Row 12
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Heun's Method}}} \tn 
% Row Count 18 (+ 1)
% Row 13
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{second order} \tn 
% Row Count 19 (+ 1)
% Row 14
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{yi+1=yi +kavg h} \tn 
% Row Count 20 (+ 1)
% Row 15
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{implicit method (yi+1 predictor) -{}-{}-iterative} \tn 
% Row Count 21 (+ 1)
% Row 16
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Midpoint Method}}} \tn 
% Row Count 22 (+ 1)
% Row 17
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{yi+1= yi + k2 h} \tn 
% Row Count 23 (+ 1)
% Row 18
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{Explicit Method} \tn 
% Row Count 24 (+ 1)
% Row 19
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{ymid= yi+ (k1 h)/2} \tn 
% Row Count 25 (+ 1)
% Row 20
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Runge Kutte}}} \tn 
% Row Count 26 (+ 1)
% Row 21
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{generalized formula for methods} \tn 
% Row Count 27 (+ 1)
% Row 22
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{yi+1= yi + phi h} \tn 
% Row Count 28 (+ 1)
% Row 23
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{phi is weighted average of slopes} \tn 
% Row Count 29 (+ 1)
% Row 24
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{number of ks reflects order} \tn 
% Row Count 30 (+ 1)
\end{tabularx}
\par\addvspace{1.3em}

\vfill
\columnbreak
\begin{tabularx}{5.377cm}{X}
\SetRowColor{DarkBackground}
\mymulticolumn{1}{x{5.377cm}}{\bf\textcolor{white}{Ordinary Differential Equations (cont)}}  \tn
% Row 25
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{there are infinite methods} \tn 
% Row Count 1 (+ 1)
% Row 26
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{fourth order Runge Kutte}}} \tn 
% Row Count 2 (+ 1)
% Row 27
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{yi+1 = yi + h/6 (k1+ 2k2+2k3+k4)} \tn 
% Row Count 3 (+ 1)
% Row 28
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\emph{Ralston's Method}}} \tn 
% Row Count 4 (+ 1)
% Row 29
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{phi= 1/3 K1 + 2/3 k2} \tn 
% Row Count 5 (+ 1)
% Row 30
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{k2 calculated at 3/4th of interval} \tn 
% Row Count 6 (+ 1)
% Row 31
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{{\emph{third order Runge Kutte}}} \tn 
% Row Count 7 (+ 1)
% Row 32
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{phi = (k1 + 4k2+ k3)/6} \tn 
% Row Count 8 (+ 1)
% Row 33
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{k2 is midway and k3 is at the end} \tn 
% Row Count 9 (+ 1)
% Row 34
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{{\bf{systems of odes}}} \tn 
% Row Count 10 (+ 1)
% Row 35
\SetRowColor{LightBackground}
\mymulticolumn{1}{x{5.377cm}}{need to look at each independently but solve simultanously} \tn 
% Row Count 12 (+ 2)
% Row 36
\SetRowColor{white}
\mymulticolumn{1}{x{5.377cm}}{\#intial conditions = \# dependent variables} \tn 
% Row Count 13 (+ 1)
\hhline{>{\arrayrulecolor{DarkBackground}}-}
\end{tabularx}
\par\addvspace{1.3em}


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

\end{document}