\documentclass[11 pt,xcolor={dvipsnames,svgnames,x11names,table}]{beamer}

\usepackage[english]{babel} 
\usepackage{polski}         
\usepackage[skins,theorems]{tcolorbox}
\tcbset{highlight math style={enhanced,
  colframe=red,colback=white,arc=0pt,boxrule=1pt}}

\usetheme[
	bullet=circle,		% Other option: square
	bigpagenumber,		% circled page number on lower right
	topline=true,			% colored bar at the top of the frame 
	shadow=false,			% Shading for beamer blocks
	watermark=BG_lower,	% png file for the watermark
	]{Flip}

%\logo{\kern+1.em\includegraphics[height=1cm]{SHiP-3_LightCharcoal}}
                            

\usepackage[lf]{berenis}
\usepackage[LY1]{fontenc}
\usepackage[utf8]{inputenc}

\usepackage{emerald}
\usefonttheme{professionalfonts}
\usepackage[no-math]{fontspec}	
\usepackage{listings}
\defaultfontfeatures{Mapping=tex-text}	% This seems to be important for mapping glyphs properly

\setmainfont{Gillius ADF}			% Beamer ignores "main font" in favor of sans font
\setsansfont{Gillius ADF}			% This is the font that beamer will use by default
% \setmainfont{Gill Sans Light}		% Prettier, but harder to read

\setbeamerfont{title}{family=\fontspec{Gillius ADF}}

\input t1augie.fd

%\newcommand{\handwriting}{\fontspec{augie}} % From Emerald City, free font
%\newcommand{\handwriting}{\usefont{T1}{fau}{m}{n}} % From Emerald City, free font
% \newcommand{\handwriting}{}	% If you prefer no special handwriting font or don't have augie

%% Gill Sans doesn't look very nice when boldfaced
%% This is a hack to use Helvetica instead
%% Usage: \textbf{\forbold some stuff}
%\newcommand{\forbold}{\fontspec{Arial}}

\usepackage{graphicx}
\usepackage[export]{adjustbox}

\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{bm}
\usepackage{colortbl}
\usepackage{mathrsfs} 			% For Weinberg-esque letters
\usepackage{cancel}				% For "SUSY-breaking" symbol
\usepackage{slashed}            % for slashed characters in math mode
\usepackage{bbm}                % for \mathbbm{1} (unit matrix)
\usepackage{amsthm}				% For theorem environment
\usepackage{multirow}			% For multi row cells in table
\usepackage{arydshln} 			% For dashed lines in arrays and tables
\usepackage{siunitx}
\usepackage{xhfill}
\usepackage{grffile}
\usepackage{textpos}
\usepackage{subfigure}
\usepackage{tikz}
\usepackage{hyperref}
%\usepackage{hepparticles}    
\usepackage[italic]{hepparticles}     

\usepackage{hepnicenames} 

% Drawing a line
\tikzstyle{lw} = [line width=20pt]
\newcommand{\topline}{%
  \tikz[remember picture,overlay] {%
    \draw[crimsonred] ([yshift=-23.5pt]current page.north west)
             -- ([yshift=-23.5pt,xshift=\paperwidth]current page.north west);}}



% % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % % %
\usepackage{tikzfeynman}		% For Feynman diagrams
\usetikzlibrary{arrows,shapes}
\usetikzlibrary{trees}
\usetikzlibrary{matrix,arrows} 				% For commutative diagram
% http://www.felixl.de/commu.pdf
\usetikzlibrary{positioning}				% For "above of=" commands
\usetikzlibrary{calc,through}				% For coordinates
\usetikzlibrary{decorations.pathreplacing}  % For curly braces
% http://www.math.ucla.edu/~getreuer/tikz.html
\usepackage{pgffor}							% For repeating patterns

\usetikzlibrary{decorations.pathmorphing}	% For Feynman Diagrams
\usetikzlibrary{decorations.markings}
\tikzset{
	% >=stealth', %%  Uncomment for more conventional arrows
	vector/.style={decorate, decoration={snake}, draw},
	provector/.style={decorate, decoration={snake,amplitude=2.5pt}, draw},
	antivector/.style={decorate, decoration={snake,amplitude=-2.5pt}, draw},
	fermion/.style={draw=gray, postaction={decorate},
		decoration={markings,mark=at position .55 with {\arrow[draw=gray]{>}}}},
	fermionbar/.style={draw=gray, postaction={decorate},
		decoration={markings,mark=at position .55 with {\arrow[draw=gray]{<}}}},
	fermionnoarrow/.style={draw=gray},
	gluon/.style={decorate, draw=black,
		decoration={coil,amplitude=4pt, segment length=5pt}},
	scalar/.style={dashed,draw=black, postaction={decorate},
		decoration={markings,mark=at position .55 with {\arrow[draw=black]{>}}}},
	scalarbar/.style={dashed,draw=black, postaction={decorate},
		decoration={markings,mark=at position .55 with {\arrow[draw=black]{<}}}},
	scalarnoarrow/.style={dashed,draw=black},
	electron/.style={draw=black, postaction={decorate},
		decoration={markings,mark=at position .55 with {\arrow[draw=black]{>}}}},
	bigvector/.style={decorate, decoration={snake,amplitude=4pt}, draw},
}

% TIKZ - for block diagrams, 
% from http://www.texample.net/tikz/examples/control-system-principles/
% \usetikzlibrary{shapes,arrows}
\tikzstyle{block} = [draw, rectangle, 
minimum height=3em, minimum width=6em]




\usetikzlibrary{backgrounds}
\usetikzlibrary{mindmap,trees}	% For mind map
\newcommand{\degree}{\ensuremath{^\circ}}
\newcommand{\E}{\mathrm{E}}
\newcommand{\Var}{\mathrm{Var}}
\newcommand{\Cov}{\mathrm{Cov}}
\newcommand\Ts{\rule{0pt}{2.6ex}}       % Top strut
\newcommand\Bs{\rule[-1.2ex]{0pt}{0pt}} % Bottom strut

\graphicspath{{images/}}	% Put all images in this directory. Avoids clutter.

% SOME COMMANDS THAT I FIND HANDY
% \renewcommand{\tilde}{\widetilde} % dinky tildes look silly, dosn't work with fontspec
%\newcommand{\comment}[1]{\textcolor{comment}{\footnotesize{#1}\normalsize}} % comment mild
%\newcommand{\Comment}[1]{\textcolor{Comment}{\footnotesize{#1}\normalsize}} % comment bold
%\newcommand{\COMMENT}[1]{\textcolor{COMMENT}{\footnotesize{#1}\normalsize}} % comment crazy bold
\newcommand{\Alert}[1]{\textcolor{Alert}{#1}} % louder alert
\newcommand{\ALERT}[1]{\textcolor{ALERT}{#1}} % loudest alert
%% "\alert" is already a beamer pre-defined
\newcommand*{\Scale}[2][4]{\scalebox{#1}{$#2$}}%

\def\Put(#1,#2)#3{\leavevmode\makebox(0,0){\put(#1,#2){#3}}}

\usepackage{gmp}
\usepackage[final]{feynmp-auto}

\usepackage[backend=bibtex,style=numeric-comp,firstinits=true]{biblatex}
\bibliography{bib}
\setbeamertemplate{bibliography item}[text]

\makeatletter\let\frametextheight\beamer@frametextheight\makeatother

% suppress frame numbering for backup slides
% you always need the appendix for this!
\newcommand{\backupbegin}{
	\newcounter{framenumberappendix}
	\setcounter{framenumberappendix}{\value{framenumber}}
}
\newcommand{\backupend}{
	\addtocounter{framenumberappendix}{-\value{framenumber}}
	\addtocounter{framenumber}{\value{framenumberappendix}} 
}


\definecolor{links}{HTML}{2A1B81}
%\hypersetup{colorlinks,linkcolor=,urlcolor=links}

% For shapo's formulas:

% For shapo's formulas:
\def\lsi{\raise0.3ex\hbox{$<$\kern-0.75em\raise-1.1ex\hbox{$\sim$}}}
\def\gsi{\raise0.3ex\hbox{$>$\kern-0.75em\raise-1.1ex\hbox{$\sim$}}}
\newcommand{\lsim}{\mathop{\lsi}}
\newcommand{\gsim}{\mathop{\gsi}}
\newcommand{\wt}{\widetilde}
%\newcommand{\ol}{\overline}
\newcommand{\Tr}{\rm{Tr}}
\newcommand{\tr}{\rm{tr}}
\newcommand{\eqn}[1]{&\hspace{-0.7em}#1\hspace{-0.7em}&}
\newcommand{\vev}[1]{\rm{$\langle #1 \rangle$}}
\newcommand{\abs}[1]{\rm{$\left| #1 \right|$}}
\newcommand{\eV}{\rm{eV}}
\newcommand{\keV}{\rm{keV}}
\newcommand{\GeV}{\rm{GeV}}
\newcommand{\im}{\rm{Im}}
\newcommand{\disp}{\displaystyle}
\def\be{\begin{equation}}
\def\ee{\end{equation}}
\def\ba{\begin{eqnarray}}
\def\ea{\end{eqnarray}}
\def\d{\partial}
\def\l{\left(}
\def\r{\right)}
\def\la{\langle}
\def\ra{\rangle}
\def\e{{\rm e}}
\def\Br{{\rm Br}}
\def\fixme{{\color{red} FIXME!}}
\def\mc{{\color{Magenta}{MC}}}
\def\pdf{{\rm p.d.f.}}
\def\cdf{{\rm c.d.f.}}
\def\ARROW{{\color{JungleGreen}{$\Rrightarrow$}}\xspace}  
\def\ARROWR{{\color{WildStrawberry}{$\Rrightarrow$}}\xspace}  

\author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)}
\institute{UZH}
\title[Specific \pdf~generation]{Specific \pdf~generation}
\date{\fixme}
\newcommand*{\QEDA}{\hfill\ensuremath{\blacksquare}}%
\newcommand*{\QEDB}{\hfill\ensuremath{\square}}%

\author{ {\fontspec{Trebuchet MS}Marcin Chrz\k{a}szcz} (Universit\"{a}t Z\"{u}rich)}
\institute{UZH}
\title[Partial Differential Equation Solving]{Partial Differential Equation Solving}
\date{\fixme}


\begin{document}
\tikzstyle{every picture}+=[remember picture]

{
\setbeamertemplate{sidebar right}{\llap{\includegraphics[width=\paperwidth,height=\paperheight]{bubble2}}}
\begin{frame}[c]%{\phantom{title page}} 
\begin{center}
\begin{center}
	\begin{columns}
		\begin{column}{0.9\textwidth}
			\flushright\fontspec{Trebuchet MS}\bfseries \Huge {Partial Differential Equation Solving, vol 2.}
		\end{column}
		\begin{column}{0.2\textwidth}
		  %\includegraphics[width=\textwidth]{SHiP-2}
		\end{column}
	\end{columns}
\end{center}
	\quad
	\vspace{3em}
\begin{columns}
\begin{column}{0.44\textwidth}
\flushright \vspace{-1.8em} {\fontspec{Trebuchet MS} \Large Marcin Chrząszcz\\\vspace{-0.1em}\small \href{mailto:mchrzasz@cern.ch}{mchrzasz@cern.ch}}

\end{column}
\begin{column}{0.53\textwidth}
\includegraphics[height=1.3cm]{uzh-transp}
\end{column}
\end{columns}

\vspace{1em}
%		\footnotesize\textcolor{gray}{With N. Serra, B. Storaci\\Thanks to the theory support from M. Shaposhnikov, D. Gorbunov}\normalsize\\
\vspace{0.5em}
	\textcolor{normal text.fg!50!Comment}{Monte Carlo methods, \\ 12 May, 2016}
\end{center}
\end{frame}
}

\begin{frame}\frametitle{Announcement}

\begin{Large}
There will be no lectures and class on 19$^{th}$ of May
\end{Large}

\end{frame}


\begin{frame}\frametitle{Dirichlet conditions:expected number of steps}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW find the function $u(x_1,x_2,...,x_k)$ such that if fulfils the Laplace equation:
\begin{align*}
\dfrac{\partial^2 u }{\partial x_1^2} + \dfrac{\partial^2 u }{\partial x_2^2}+...+\dfrac{\partial^2 u }{\partial x_k^2}=0,~~~(x_1,x_2,...,x_k) \in D \subset \mathbb{R}^k
\end{align*}
In the domain $D$, on the the $\Gamma(D)$ the $u$ function is given by:
\begin{align*}
U(x_1,x_2,...,x_k)=f(x_1,x_2,...,x_k),~~~~(x_1,x_2,...,x_k) \in \Gamma( D )
\end{align*}
\ARROW Now lets assume that the domain $D$ is a hyperball:
\begin{align*}
0 \leq \sum_{i=1}^k x_i^2 \leq r^2,~~~r={\rm const}
\end{align*}
\ARROW Now $\pi_{\nu}(x_1,x_2,...,x_k)$ is a probability that a particle starting from $(x_1,x_2,...,x_k)$ will end up on the edge after $\nu$ steps. The $\kappa(x_1,x_2,...,x_k)$ is the estimated number of steps for this trajectory.
\begin{tiny}

\begin{align}
\pi_0=\begin{cases}
1,~~& (x_1,x_2,...,x_k) \in \Gamma(D)\\
0,~~& (x_1,x_2,...,x_k) \in D
\end{cases}
\label{eq1}
\end{align}
\begin{align}
\pi_{\nu}=\frac{1}{2k} \sum^{ \prime} \pi_{\nu}(x_1\prime,x_2\prime,...,x_k\prime)
\label{eq2}
\end{align}

\end{tiny}



\end{footnotesize}

\end{minipage}

\end{frame}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}\frametitle{Dirichlet conditions:expected number of steps}
\begin{minipage}{\textwidth}
\begin{footnotesize}

\ARROW From Eq.~\ref{eq1} and \ref{eq2} one gets:
\begin{align*}
\kappa(x_1,x_2,...,x_k)=\sum_{\nu=1}^{\infty} \nu \pi_{\nu}(x_1,x_2,...,x_k)
\end{align*}
one gets:
\begin{align*}
\kappa(x_1,x_2,...,x_k)=\frac{1}{2k}\sum_{\nu=1}^{\infty} \left[ \nu \sum^{\prime}\pi_{\nu -1 }(x_1,x_2,...,x_k) \right]\\ = \frac{1}{2k} \sum_{\nu =1 }^{\infty} \left[ (\nu-1)\sum^{\prime} \pi_{\nu-1}(x_1\prime,x_2\prime,...,x_k\prime) \right] + \frac{1}{2k} \sum_{\nu=1}^{\infty} \sum^{\prime} \pi_{\nu-1}x_1\prime,x_2\prime,...,x_k\prime)
\end{align*}
\ARROW From which we get:
\begin{align*}
\kappa(x_1,x_2,...,x_k)=\frac{1}{2k}\sum^{\prime}\kappa(x_1\prime,x_2\prime,...,x_k\prime) +1
\end{align*}

\ARROW Now this is equivalent of the Poisson differential equation:
\begin{align*}
\frac{\partial^2 \kappa}{\partial x_1^2}+\frac{\partial^2 \kappa}{\partial x_2^2}+...+\frac{\partial^2 \kappa}{\partial x_k^2} = -2k,~{\rm b.~con.}~~ \kappa(x_1,x_2,...,x_k)=0,~~(x_1,x_2,...,x_k) \in \Gamma(D)
\end{align*}

\end{footnotesize}

\end{minipage}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{frame}\frametitle{Dirichlet conditions:expected number of steps}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW From previous equation: $\kappa(x_1,x_2,...,x_k)=\psi(x_1,x_2,...,x_k)-\sum_{i=1}^k x_i^2$ 
we get the for the $\psi$ function the Laplace equation:
\begin{align*}
\dfrac{\partial^2 \psi }{\partial x_1^2} + \dfrac{\partial^2 \psi }{\partial x_2^2}+...+\dfrac{\partial^2 \psi }{\partial x_k^2}=0
\end{align*}
because on the border ($\Gamma(D)$):
\begin{align*}
\psi(x_1,x_2,...,x_k)=r^2= {\rm const}
\end{align*}
so also inside the $D$:
$\psi(x_1,x_2,...,x_k)=r^2= {\rm const}$
\ARROW From which we can estimate the number steps in the random walk:
\begin{align*}
\kappa(x_1,x_2,...,x_k)=r^2-\sum_{i=1}^k\leq r^2
\end{align*}
\begin{alertblock}{Important conclusion:}
The expected number of steps in the random walk (the time of walk) from the point $(x_1,x_2,...,x_k)$ till the edge od the domain can be estimated by $r$ number (the LINEAR! size). It is completly independent of the $k$!
\end{alertblock}



\end{footnotesize}

\end{minipage}

\end{frame}


\begin{frame}\frametitle{Dirichlet conditions as linear system}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW In the discrete form we can write the Dirichlet conditions as (2-dim case):
\begin{align*}
u(x,y) & =\frac{1}{4} \left[ u(x-1,y)+u(x+1,y)+u(x,y-1)+u(x,y+1) \right],~~(x,y) \in D \\
u(x,y) & =f(x,y),~~~(x,y) \in \Gamma(D)
\end{align*}
\ARROW Now we can order the grid ($(x,y) \in D \cup D$), we can represente the above equations as a linear system:
\begin{align*}
u_i=a_i+\sum_{j=1}^n h_{i j}u_j, ~~~~i=1,2,....,n
\end{align*}

\begin{exampleblock}{The trick:}
So to solve a differential equation with Dirichlet boundary condition we can use all the methods of solving linear equation systems such as Neumann-Ulam or Wassow.

\end{exampleblock}


\end{footnotesize}

\end{minipage}

\end{frame}


\begin{frame}\frametitle{Dirichlet conditions as linear system - example}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\begin{columns}

\column{3in}
\begin{itemize}
\item To do this we act as following: we number separately the points inside the $D$ domain and on the border $\Gamma(D)$. 
\item We write for each point inside the domain the Laplace equation as system of linear equations:
\end{itemize}
\column{2in}
\includegraphics[width=0.99\textwidth]{images/grid1.png}

\end{columns}
\begin{tiny}
\begin{align*}
&u_1 & -u_2/4 & &-u_4/4 & & & & = (f_1+f_{10})/4\\
&-u_1/4 & u_2 & -u_3/4 & & -u_5/4 & & & = (f_2)/4\\
& & -u_2/4  & ~~~~u_3 &  &  & -u_6/4 & & = (f_3+f_4)/4\\
&-u_1/4 &   &  & u_4 & -u_5/4 & & & = (f_8+f_9)/4\\
&-u_1 /4  &  &  & -u_4/4 &~~~~ u_5 & -u_6/4 & -u_7/4 & = 0\\
&  &  & -u_3/4 &  & -u_5/4 & u_6 &  & = (f_5+f_6)/4\\
&    &  &  &  & -u_5/4 &  &~~~~~ u_7 & = (f_5+f_6)/4
\end{align*}
\end{tiny}

%-u_1/4 & u_2 & -u_3/4 & & -u_5/4 & & & = (f_1+f_10)/4\\


\end{footnotesize}

\end{minipage}

\end{frame}


\begin{frame}\frametitle{Dirichlet conditions as linear system - example}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW The above equation we can transform the above equation into the iterative representation:
\begin{align*}
\overrightarrow{u}=\overrightarrow{a}+\textbf{H}\overrightarrow{u}
\end{align*}
where $\overrightarrow{u}=(u_1,u_2,...,u_7)$ is the vector which represent the values of the function inside the $D$ domain, $\overrightarrow{a}$ is the linear combinations of the boundary values. In our example:
\begin{columns}
\column{2in}
\begin{align*}
\textbf{H}=\begin{pmatrix}
0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & 0 & 0 & 0\\
\dfrac{1}{4} & 0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & 0& 0\\
0 & \dfrac{1}{4} & 0 & 0 & 0 & \dfrac{1}{4} & 0\\
\dfrac{1}{4} & 0 & 0 & 0 & \dfrac{1}{4} & 0 & 0\\
0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & \dfrac{1}{4}\\
0 & 0 & \dfrac{1}{4} & 0 & \dfrac{1}{4} & 0 & 0\\
0 & 0 & 0 & 0 & \dfrac{1}{4} & 0 & 0\\
\end{pmatrix}
\end{align*}
\column{3in}
\ARROW To find the solution to aka $\overline{u}$ one can use the methods we already know: Neumann-Ulam and Wasow, etc.\\
\ARROW There are tricks and tips one can use to make this problem faster as each of the entry is $\frac{1}{4}$. 

\end{columns}


\end{footnotesize}

\end{minipage}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Neumann-Ulam method}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW We put the particle in $(x,y)$. \\
\ARROW We observe the trajectory of the particle until it reaches the boundary. Point $P_k$ is the last point before hitting the boundary.\\
\ARROW For each trajectory we assign a value that the arithmetical mean of the boundary points that are neighbours of the point $P_k$.\\
\ARROW We repeat the above $n$ times and calculate the mean.\\
\ARROW The example solution for $20$ trajectories:
\begin{align*}
u(2,2)=1.0500\pm 0.2756
\end{align*}
\ARROW E 10.1 Solve the above linear system using the Neumann-Ulam method for an assumed boundary conditions.
\end{footnotesize}

\end{minipage}

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Dual Wasow method}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW We choose the boundary conditions with arbitrary chosen probability \pdf~$p(Q)$ the starting point.
\ARROW We choose with equal probability the point inside $D$ where the particle goes.\\
\ARROW With equal probability we choose the next positions and so on until the particle hits the boundary in the point $Q^{\prime}$.\\
\ARROW We count all trajectories $N((x_1,x_2,x_3,...,x_k)$ that that have passed the point $(x_1,x_2,x_3,...,x_k)$.
\ARROW For the point $(x_1,x_2,...,x_k)$ we calculate:
\begin{align*}
w(x_1,x_2,...,x_k)=\frac{1}{2k}N(x_1,x_2,...,x_k)\frac{f(Q)}{p(Q)}
\end{align*}
\ARROW The above steps we repeat $N$ times.\\
\ARROW After that we take the arithmetic mean of $w$.

\end{footnotesize}

\end{minipage}

\end{frame}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{frame}\frametitle{Random walk with different step size}
\begin{minipage}{\textwidth}
\begin{footnotesize}
\ARROW If $u(x,y)$ is a harmonic function that obeys the Laplace equation and $S_r(x,y)$ is a circle in with the middle point $(x,y)$ and radius $r$. Then a theorem states:
\begin{align*}
S_r(x,y)=\frac{1}{2 \pi } \int_0^{2\pi} u(x+r \cos \phi ,y+ r \sin \phi) d \phi
\end{align*}
\ARROW The above is true for in all the dimensions.\\
\ARROW The E.Muller method:
\begin{itemize}
\item At the begging we set the point in the initial point: $(x_1,x_2,...,x_k)$. 
\item We construct a $k$ dimensional sphere with center $(x_1,x_2,...,x_k)$ and radius $r$. The $r$ has to be choosen in a way that the whole is inside the $D$: $S_r(\overrightarrow{x}) \in D$. We choose a random point from $\mathcal{U}(0,2\pi)$ on the sphere which is our new point. 
\item We stop the walk when the point is on $\Gamma(D)$.
\end{itemize}
\ARROW We repeat this $N$ times.\\
\ARROW The final result if the arithmetical mean of all trajectories and is equal of the $u(x_1,x_2,...,x_k)$.



\end{footnotesize}

\end{minipage}

\end{frame}



\backupbegin   

\begin{frame}\frametitle{Backup}


\end{frame}

\backupend			

\end{document}
