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\begin{document}

% slide 1
%slide 1 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf 
Multi-scale methods in numerical analysis  
}
\end{center}

\vfill

\begin{center}
{\bf 
W. N. Polyzou
}
\end{center}


\begin{center}
{\bf 
Other contributors: Gerald Payne, Brian Kessler, Fatih Bulut, Tracie Michlin
}
\end{center}

\vfill

\begin{center}
{\bf
The University of Iowa 
}
\end{center}

\begin{figure}
%\begin{minipage}[t]{.3\linewidth}
\centering
\includegraphics[angle=000,scale=.3]{doe.pdf}
%\caption{Full matrix, $\lambda$=0, mass=1}
%\label{fig:8}
%\end{minipage}
%\begin{minipage}[t]{.4\linewidth}
%\centering
%\includegraphics[angle=000,scale=.5]{doe.pdf}
%\caption{Full matrix, $\lambda$=0.2, mass=1}
%\label{fig:9}
%\end{minipage}
\end{figure}


%\begin{figure}
%\centering
%\includegraphics[scale=.3]{mumbai.pdf}
% \caption{V(x)}
%\label{figure 1}
%\end{figure}  

\end{frame}

%slide 2 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Outline
}
\end{center}
\vfill
\begin{itemize}
{\bf 
\item[$\bullet$] Background and motivation.
\vfill
\item[$\bullet$] Fractals self-similarity and the renormalization group. 
\vfill
\item[$\bullet$] Numerical analysis using the renormalization group.
\vfill
\item[$\bullet$] Applications.
}
\end{itemize}
\end{frame}

%slide 3 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Standard numerical methods
}
\end{center}
\vfill

\begin{itemize}
{\bf 
\item[$\bullet$] Designed to work with functions that with enough
magnification locally look like straight lines.
\vfill
\item[$\bullet$] Global basis functions (orthogonal polynomials, Laguerre functions, Hermite functions) are not efficient for representing local structures, or structures on multiple scales.
\vfill
%\item[$\bullet$] Global basis functions $\{\phi_n(x) \}$
%are often associated with a scale.  They may have to be
%rescaled $\{\sqrt{a} \phi_n(ax) \}$ to efficiently represent
%structures on a given scale.
\item[$\bullet$] Even local functions (for example splines) can still result
in large systems of equations.

\vfill

}
\end{itemize}
\end{frame}

%slide 4 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Wavelets are basis functions that can overcome these difficulties
}
\end{center}

\begin{itemize}
{\bf 
\item[$\bullet$] Used for data compression in
digital photography (JPEG).
\vfill
\item[$\bullet$] Efficient at treating images with many different
scales and structures.
\vfill
\item[$\bullet$] A digital photograph is just a matrix of numbers.
Could this same data compression method be used to efficiently solve problems
in linear algebra?
\vfill
\item[$\bullet$] Wavelet bases result in sparse matrices.
Faster algorithms can be used and they require less storage. 
}
\end{itemize}
\end{frame}

%slide 5 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Properties
}
\end{center}
\begin{itemize}
  {\bf
\item [$\bullet$] Fractal valued 

\vfill
\item[$\bullet$] Orthonormal basis
\vfill
\item[$\bullet$] Compact support
\vfill
\item[$\bullet$] Some smoothness
\vfill
\item[$\bullet$] Infinite number with support in any open set.
\vfill
\item[$\bullet$] Pointwise represent low degree polynomials  
\vfill
\item[$\bullet$] Locally finite partitions of unity at different  resolutions.  
}
\end{itemize}

\end{frame}

%slide 5 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


\begin{frame}
\begin{center}
{\bf
Challenges/questions
}
\end{center}

\begin{itemize}
{\bf 
\item[$\bullet$] How do you evaluate fractal functions $\{\phi_n(x)\}$?
\[
f(x) = \sum f_n \phi_n(x) 
\]
\item[$\bullet$] How do you calculate integrals involving fractal functions?
\[
f_n = \int \phi_n^*(x) f(x) dx
\]
\item[$\bullet$] How do you calculate derivatives of fractal functions?
\[
f''(x) = \sum f_n \phi''_n(x) 
\]
}
\end{itemize}
\end{frame}

%slide 6 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
What do we mean by a fractal valued function
}
\end{center}

\begin{itemize}
{\bf 
\item[$\bullet$] Looks like a copy of itself on smaller scales.
\vfill
\item[$\bullet$] How do we change scales mathematically?
\[
Df(x) = \sqrt{2} f(2x)  
\]  
\item[$\bullet$] Shrinks the support of the function by a factor of 2,
preserving the Hilbert space norm of the function.
\vfill
\item[$\bullet$] $D$ is called a scaling or dilatation operator

}
\end{itemize}
\end{frame}

%slide 7 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Renormalization group equation 
%solutions $s(x)$ are fractal!  
}
\end{center}
\[
{\color{red}s(x)} = \underbrace{D
\underbrace{(\sum_{l=0}^{2K-1} h_l T^l {\color{red}s(x)})}_{\mbox{weighted average}}
}_{\mbox{rescale}} .
\]
\begin{center}
{\bf
solutions $s(x)$ are fractal!  
}
\end{center}

\begin{center}
{\bf $T$: unit translation \qquad $D$ scale transformation}
\end{center}
\vfill
\[
Ts(x) = s(x-1) \qquad Ds(x) = \sqrt{2} s(2x) .
\]
\begin{center}
{\bf $h_l$ are numbers}
\end{center}

\vfill
\begin{center}
  {\bf The renormalization group equation is homogeneous \\
    $s(x)$ a solution implies
    $cs(x)$ is a solution\\
    the scale $c$ is fixed by}
\end{center}
\[
\int dx s(x) =1
\]
\end{frame}

%slide 8 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Properties of  $s(x)$  
}
\end{center}
\[
\tilde{s}(k) := {1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i k x} s (x) dx
\qquad
\tilde{h}({k}) :=
\sum_l {h_l \over \sqrt{2}} e^{-i k l }.
\]

\[
\tilde{s} (k) = \tilde{s}\left ({k \over 2}\right )  
\tilde{h}\left ( {k \over 2}\right )  
\] 

\[
\tilde{s} (k) = \lim_{n\to \infty} \tilde{s}({k \over 2^n}) \prod_{m=1}^n  \tilde{h}({k \over 2^m}) 
=\tilde{s}(0) 
\prod_{m=1}^{\infty}  \tilde{h}({k \over 2^m}) .
\]

\[
k=0 \qquad \to \qquad  
1 =  \prod_{l=1}^{\infty}  \tilde{h}({0}) = \tilde{h}({0})  =
\sum_{l=0}^{2K-1} {h_l \over \sqrt{2}} 
\]

\begin{center}
{\bf {\color{red}$\sum_{l=0}^{2K-1} h_l = \sqrt{2}$}: Necessary for the renormalization group
equation to have a solution}
\end{center}

\end{frame}

%slide 9 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf Support of the solution $s(x)$ to the RG equation}
\end{center}
  
\[
s(x) = {\tilde{s}(0)  \over \sqrt{2 \pi}} \int_{-\infty}^{\infty} dk
e^{i k x} \prod_{m=1}^{\infty} 
\tilde{h}({k \over 2^m}) =
\]
\[
\sqrt{2 \pi} \tilde{s}(0)
\prod_{m=1}^{\infty}(
\sum_{l=0}^{N-1} {h_l \over \sqrt{2}}\delta(x-{l\over 2^m}))
\]
\vfill
\begin{center}
{\bf
which vanishes for $x \notin [0,N-1]=[2K-1]$. 
}
\end{center}
\vfill
\begin{center}
{\bf {\color{red}Support of $s(x) \subseteq [0,2K-1]$ where $N=2K$ is the number of $h_l$'s}\\
Support is compact!}
\end{center}

\end{frame}

%slide 10 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf All fractals are not created equal}
\end{center}

\begin{center}
{\bf Additional properties of $s(x)$ depend on the choice of $h_l$}
\end{center}
\[
\sum_{l=1}^{2K-1}h_l = \sqrt{2}
\]
\begin{center}
{\bf Additional conditions: }
\end{center}
\[
\int s(x-n) s(x-m) dx = \delta_{mn}
\]
\[
\int x^m \sum_{l=0}^{2K-1} g_l s(x-l)dx = 0 \qquad m=0,1,\cdots K-1  
\]
\[
g_l:=(-)^l h_{2K-l-1}
\]
\begin{center}
{\bf Conditions define {\color{red}Daubechies} K scaling functions: }
\end{center}

\end{frame}  

%slide 11 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}{\bf
Ingrid Daubechies} 
\end{center}
  
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{2.5in}{!}{
\includegraphics{{thumb_image_7597592.jpg}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
  
\end{frame}  



%slide 11 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf Conditions that fix $h_l$: }
\end{center}

\begin{itemize}
{\bf 
\item[$\bullet$] $s_n (x) := T^n s(x) = s(x-n)$ {\color{red}are orthonormal}.

\vfill
\item[$\bullet$] $x^m = \sum_n c_n s_n (x)$ {\color{red}pointwise for $m \leq K$}.

\vfill  
\item[$\bullet$] $\sum_{l=1}^{2K-1}h_l = \sqrt{2}$ {\color{red}necessary for existence of solution.}

  
\vfill
\item[$\bullet$] These conditions determine $h_l$ up to
reflection: $h_l \to h_l'= h_{2k-1-l}$.  
}  
\end{itemize}  

%slide 12 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\end{frame}

%slide 12 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
  
\begin{center}
\begin{table} % [hbt]
{\bf Weight coefficients $h_l$ for different $K$ values} \\[1.0ex]
\begin{tabular}{|l|l|l|l|}
\hline
$h_l$ & K=1 & K=2 & K=3   \\
\hline					      		      
$h_0$ &$1 / \sqrt{2}$  & $(1+\sqrt{3})/4\sqrt{2}$ &$(1+\sqrt{10}+\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
$h_1$ &$1 / \sqrt{2}$  & $(3+\sqrt{3})/4\sqrt{2}$ & $(5+\sqrt{10}+3\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
$h_2$ &$0$  & $(3-\sqrt{3})/4\sqrt{2}$ & $(10-2\sqrt{10}+2\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
$h_3$ &$0$  & $(1-\sqrt{3})/4\sqrt{2}$ & $ (10-2\sqrt{10}-2\sqrt{5+2\sqrt{10}})/16\sqrt{2} $ \\
$h_4$ &$0$  & $0$ & $(5+\sqrt{10}-3\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
$h_5$ &$0$  & $0$ & $(1+\sqrt{10}-\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
\hline
\end{tabular}
% \label{tabspec}
\end{table}
\end{center}
\end{frame}

%slide 13 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center} {\bf Basis construction
($L^2(\mathbb{R})$)}
\end{center}
  
\begin{center}
{\bf Rescale and translate fixed point, ${\color{red}s(x)}$}
\end{center}
\[
s^k_n (x) := D^k T^n s(x) = 2^{k/2} {\color{red}s}\left (2^k(x-2^{-k}n)\right ) .
\]
\begin{center}
{\bf Support of  $s^k_n(x)$ is $[{\color{red}2^{-k}}n,{\color{red}2^{-k}}(n+2K-1)]$ }
\end{center}


\begin{center}
  {\bf ${\cal S}_k :=$ resolution $2^{-k}$ subspace
    of $L^2(\mathbb{R})$:
    %$\{s^k_n(x)\}$ basis for ${\cal S}_k$
}
%\quad \sum_{n=-\infty}^\infty \vert c_n \vert^2 < \in
\end{center}

\[
{\cal S}_k := \{ f(x) \vert f(x)= \sum_{n=-\infty}^\infty c_n s^k_n(x),
\quad \sum_{n=-\infty}^\infty \vert c_n \vert^2 < \infty \}.
\]

\[
{\cal S}_k := D^k {\cal S}_0
\]

\begin{center}
{\bf Renormalization group equation implies}
\end{center}

\[
{\cal S}_{k} \subset {\cal S}_{k+n} \qquad n> 0 
\]

\[
  {\cal S}_{k+1} = {\cal S}_{k} \oplus {\cal W}_{k}
  \qquad {\cal W}_k \not= \{\emptyset \}.  
\]

\end{frame}

%slide 14 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center} {\bf Multi-resolution decomposition of
$L^2(\mathbb{R})$}
\end{center}
\[
{\cal S}_{k+1} = {\cal S}_{k} \oplus {\cal W}_{k} 
\]
\begin{center}
{\bf Iterate}
\end{center}
\[{\color{red}
L^2 (\mathbb{R}) = {\cal S}_{k}\oplus {\cal W}_{k} \oplus {\cal W}_{k+1}
\oplus {\cal W}_{k+2} \oplus {\cal W}_{k+3} \oplus  \cdots}  =
\]
\[
\cdots \oplus {\cal W}_{k-2}\oplus {\cal W}_{k-1} \oplus {\cal W}_{k}
\oplus {\cal W}_{k+1} \oplus {\cal W}_{k+2} \oplus  \cdots
\]

\begin{center}
{\bf Wavelets ($\{w^k_n(x)\}$ orthonormal basis for ${\cal W}_k$)}
\end{center}

\[
w(x) :=D \sum_{l=0}^{2K-1} g_l T^l {\color{red}s(x)}
\qquad g_l = (-)^l h_{2K-1-l}
\]

\[
w^k_n (x) := D^k T^n w(x) = 2^{k/2} w\left (2^k(x-2^{-k}n)\right ) .
\]

\begin{center}
{\bf Same support as $\{s^k_n(x)\}$}
\end{center}

\end{frame}

%slide 15 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Comments
}    
\end{center}

\begin{center}
{\bf
Orthonormal basis for $L^2(\mathbb{R})$:
}    
\end{center}  

\[{\color{red}
\{ s^k_n(x)\}_{n=-\infty}^{\infty}\}
\cup 
\{ w^l_n(x)\}_{n=-\infty,l\geq k}^{\infty}}
\]

\begin{center}
{\bf 
All basis functions constructed from a single function ${\color{red}s(x)}$.
}
\end{center} 

\begin{center}
{\bf 
The condition
}
\end{center} 
\[
\int x^m \sum_{l=0}^{2K-1} g_l s(x-l) = 0 \qquad m=0,1,\cdots K  
\]
\begin{center}
{\bf 
that determines the  $h_l$ is equivalent to the  requirement
}
\end{center} 
\[
{\color{red}\int x^m w^k_n (x)dx=0} \qquad \forall k,n \qquad\mbox{and}\qquad m=0,1,\cdots ,K-1
\]


\end{frame}

%slide 16 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Local {\color{red}pointwise} polynomial condition 
}    
\end{center}


\begin{center}
{\bf  
Completeness implies
}
\end{center}
\[
f(x) = \sum_n c_n s^k_n(x) + \sum_{n,l\geq k} d_{nl} w^l_n (x)
\]
\begin{center}
{\bf
For $f(x) = x^m$, $m<K$ {\color{red}all} the $d_{nl}=0$ which means
}
\end{center}  
\[
x^m = \sum_n c_n s^k_n(x)  \qquad m < K 
\]
\begin{center}
{\bf
where for any $x$ only a {\color{red}finite} number of the $s^k_n(x)$ are non-zero.
}
\end{center}

\begin{center}
{\bf
This requires that both sides of the equation agree at every $x$
}
\end{center}  

\end{frame}  

%slide 17 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{itemize}
{\bf
\item[$\bullet$] This explains why JPEG works.  All of coefficients $d_n^k$
associated with structures that are smooth on scale $2^{-k}$ vanish,
or are very small, 
resulting in an efficient representation of the data.

\vfill

\item[$\bullet$] The transformation relating ${\cal S}_{k+n}$ and 
${\cal S}_{k} \oplus {\cal W}_{k}\oplus \cdots \oplus {\cal W}_{k+n} $
is a real orthogonal transformations, called the {\color{red}wavelet transform},
that can be performed very efficiently.    

\vfill


\item[$\bullet$] The wavelet transformation can be used to recover
an approximation to the original function.

\vfill

\item[$\bullet$] Basis allows natural resolution and volume truncations.
}
\end{itemize}  


\end{frame}
%slide 18 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
\begin{frame}

\begin{center}
{\bf
Weird stuff!
}
\end{center}  

\[
1 = \underbrace{\int s(x) dx}_{\mbox{setting scale}} =
\underbrace{\int s^2(x) dx}_{\mbox{normalization}} 
\]

\begin{itemize}
{\bf 
\item[$\bullet$] Locally finite sums of fractal functions can be
differentiable.  This means that a FINITE number of these functions
with complex fractal boundaries fit together like a jigsaw puzzle!

\vfill

\item[$\bullet$] Under any magnification the functions do not look
like straight lines, but they are differentiable! 
}
\end{itemize}  


\end{frame}  
%slide 19 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{DaubFun_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 20 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
  \begin{center}
    {\bf As $K$ increases:}
  \end{center}
  \vfill
  \begin{itemize}
    {\bf 
\item[] Support increases,
\vfill
\item[] Smoothness increases,
\vfill
\item[] Number of non-zero functions at a point increases
      }
\end{itemize}
\end{frame}

%slide 21 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{transdial_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 22 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{DaubExp_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 21 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Kernel, $K(x,y)$  of a scattering integral equation after wavelet transform\\
$
F(x) = D(x) + \int K(x,y) F(y) dy
$
}
\end{center}  

\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{sparse_matrix_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 23 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


\begin{frame}

\begin{center}
{\bf     
Advantages of Daubechies wavelets
}
\end{center}

{\bf
\begin{itemize}  
\item[{\color{red}$\bullet$}] Basis for $L^2(\mathbb{R})$.
\item[{\color{red}$\bullet$}] Basis functions have compact support.
\item[{\color{red}$\bullet$}] Infinite number of basis functions with support in
 any open set.
\item[{\color{red}$\bullet$}] Basis functions have limited (controllable) smoothness.
\item[{\color{red}$\bullet$}] $2^{k/2} s^k_n(x)$ are locally finite partitions of unity.
\item[{\color{red}$\bullet$}] Efficient at representing problems with
multiple scales.
\item[{\color{red}$\bullet$}] In many applications basis functions do not
have to be computed.
\end{itemize}
}
\end{frame}  

%slide 24 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf     
To do numerical analysis with fractal functions
you need to be able to:
}
\end{center}

{\bf
\begin{itemize}  
\item[{\color{red}$\bullet$}] Evaluate basis functions.
\item[{\color{red}$\bullet$}] Evaluate expansion coefficients.
\item[{\color{red}$\bullet$}] Integrals of  polynomials $\times$ basis functions.
\item[{\color{red}$\bullet$}] Implement boundary conditions.
\item[{\color{red}$\bullet$}] Integrate products of basis functions $\times$ polynomials.
\item[{\color{red}$\bullet$}] Evaluate derivatives of basis functions.
\item[{\color{red}$\bullet$}] Evaluate products of basis functions, derivatives of basis functions and polynomials.
\item[{\color{red}$\bullet$}] Evaluate singular integrals.  
\item[{\color{red}$\bullet$}] Evaluate integrals with moving singularities.
\end{itemize}
}
\end{frame}  

%slide 25 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Standard methods cannot be used to satisfy these requirements.   
}
\end{center}
\vfill
\begin{center}
{\bf
The renormalization group equation is a new computational tool!      
}
\end{center}
\vfill
\begin{center}
{\bf
The renormalization group equation and the scale fixing condition provide
a means to solve
the problems on the previous slide!
}
\end{center}

\end{frame}  

%slide 26 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Computation of fractal basis functions?
}    
\end{center}  

\begin{center}
{\bf
Approximation by iteration.\\
Pick any function $f_1(x)$ satisfying $\int f_1(x) dx =1$.
}    
\end{center}  
\[
f_n(x) = \sum_l h_l D T^l f_{n-1}(x) 
\]
\[
s(x) = \lim_{n\to \infty} f_n(x)
\]

\begin{center}
{\bf
Exact calculation at dyadic rationals, \\
{\color{red}use the renormalization group equation} at dyadic rational points:
}    
\end{center}  
\[
s(n) = \sqrt{2} \sum_l h_l s(2n-l) \qquad \sum_n s(n) =1 \qquad n=1,2,\cdots  2K-2
\]
\[
s(r/2) = \sqrt{2} \sum_l h_l s(r-l) \qquad r= {m \over 2^k}
\]

\end{frame}

%slide 27 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Calculations of integrals {\color{red}use the renormalization group equation} and
the scale fixing condition.  
}    
\end{center}  

\begin{center}
{\bf
All moments of $s^k_n(x)$ and $w^l_n(x)$ can be computed exactly.
}    
\end{center}  

\[
<x^m>_{s^k_n} := \int x^m s_n^k (x) dx = ({1 \over 2})^{{3nm\over 2}} \int (x+n)^m s(x) dx
\]
\vfill
\[
<x^m>_{w^k_n} := \int x^m w_n^k (x) dx =({1 \over 2})^{{3nm\over 2}} \int (x+n)^m w(x) dx
\]


\end{frame}

%slide 28 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Calculating moments:
}    
\end{center}  
\[
<x^m>_{s} = \int s (x) x^m dx 
\qquad
<x^m>_{w} = \int w (x) x^m dx .
\]

\[
<x^0>_{s} = (x^0, s) = \int dx s (x) = 1
\]
\begin{center}
{\bf   
{\color{red}Using the renormalization group equation:}
}
\end{center}
\[
<x^m>_{s} = (x^m ,s )  = (D^{-1}x^m ,D^{-1}s ) 
\]
\[
={1 \over \sqrt{2}}{1 \over 2^m} \sum_l h_l (x^m , T^l s)   
\]
\[
={1 \over \sqrt{2}}{1 \over 2^m} \sum_l h_l ((x+l)^m , s) 
\]
\[
={1 \over \sqrt{2}}{1 \over 2^m} \sum_l h_l \sum_{k=0}^m
{ m! \over k! (m-k)!} l^{m-k} <x^k>_{s} .
\]

\end{frame}

%slide 29 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Using $\sum_{l} h_l = \sqrt{2}$, and moving the $k=m$ term to the
left side of the above equation gives the recursion relation:
}
\end{center}
{
\[
<x^m>_{s} =  {1 \over 2^m-1}{ 1 \over \sqrt{2}}  \sum_{k=0}^{{\color{red}m-1}}
{ m! \over k! (m-k)!} \left (\sum_{l=1}^{2K-1} h_l l^{m-k}\right ) 
<x^k>_{s}. 
\]
}
\begin{center}
{\bf
Note $k<m$
}    
\end{center}  


\[
<x>_s := \int x s (x) dx = {1 \over \sqrt{2}} \sum_{l=0}^{2K-1} lh_l.
\]
\begin{center}
{\bf
With this definition:
}
\end{center}
\[
\int (a+bx) s (x) dx  = a+b <x>_s .
\]

\end{frame}

%slide 30 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Boundary conditions:
}
\end{center}

\[
<x^m>_{s_n:0} = \int_{{\color{red}0}}^\infty x^m s_n(x) dx    
\]

\begin{center}
{\bf
The {\color{red}renormalization group equation}  relates these
endpoint partial moments to ordinary moments.
}
\end{center}

\begin{center}
{\bf
The computation reduces to linear algebra.
}
\end{center}

\end{frame}

%slide 31 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Integrating functions with small support
}    
\end{center}  

\begin{center}
{\bf
One-point quadrature
}    
\end{center}  

\begin{center}
{\bf
Property of moments
}    
\end{center}  

\[
\int x^2 s(x) dx = (\int x s(x) dx)^2
\]

%\[
%\sum_m  m s(x-m) = x- <x>  
%\]
%\begin{center}
%{\bf
%which gives
%}    
%\end{center}  
%\[
%0 = \sum_m m  \int x s (x) s (x-m) dx =
%\sum_m m  \int x s (x) s (x+m) dx =
%\sum_m  \int  s (x) x ( x-<x>_s ) dx
%\]
\[
<x^2>_s = <x>_s^2 
\]
\vfill
\[
{\color{red}\int p(x)s(x) dx}  =  \int (a +bx +cx^2) s(x) dx = 
\]
\vfill
\[
a +b\langle x \rangle_s+ c \langle x \rangle_s^2
= {\color{red} p(\langle x \rangle_s)}
\]
\[
\langle x \rangle_s ={1 \over \sqrt{2}} \sum_{l=0}^{2K-1} lh_l.
\]
\end{frame}

%slide 32 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


\begin{frame}

\begin{center}
{\bf
General methods - integrals of products of basis functions:
}    
\end{center}  

\begin{center}
{\bf
Step 1: Increasing resolution $2^{-k}$:
}    
\end{center}  

\[
s^k_n(x) = \sum_{l=0}^{2K-1} h_l s^{k+1}_{2n+l} (x)
\]

\begin{center}
{\bf
Step 2: Replacing wavelets by scaling functions:
}    
\end{center}  

\[
w^k_n(x) = \sum_{l=0}^{2K-1} g_l s^{k+1}_{2n+l} (x)
\]

\begin{center}
{\bf
Step 3: Changing scale: $2^{-k} \to 0$
}    
\end{center}  

\[
\int s^k_{n_1}(x)\cdots s^k_{n_m}(x) dx =
2^{{km\over 2}-k} \int s^0_{n_1}(x)\cdots s^0_{n_m}(x) dx 
\]

\end{frame}

%slide 33 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Calculating the $k=0$ integrals 
}
\end{center}  
  
\begin{center}
{\bf
Step 4:{\color{red}Use the renormalization group equation} 
}
\end{center}  

\[
\Gamma_{n_1, \cdots , n_k} :=
\int s^0_{n_1}(x)\cdots s^0_{n_m}(x) dx \qquad \mbox{use }n_1=0 
\]
\begin{center}
{\bf 
Homogeneous equation (RG equation):
}
\end{center}
\[
\Gamma_{n_1, \cdots n_m} =
\sum 2^{m/2-1} h_{l_1}\cdots h_{l_m}
\Gamma_{2n_1+l_1, \cdots , 2n_m+l_m} 
\]
\begin{center}
{\bf Step 5:
Inhomogeneous equation: use $\sum s_n(x) =1$:
}
\end{center}
\[
\sum_{n_1} \Gamma_{n_1, \cdots n_m} =
\Gamma_{n_2, \cdots n_m}
\]
\begin{center}
  {\bf
Step 6:    
Solve the finite linear system
}
\end{center}

\end{frame}  

%slide 34 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf 
Integrating products of scaling functions and polynomials
}
\end{center}
\begin{center}
{\bf
Homogeneous equations {\color{red}(RG equation)}   
}
\end{center}
\[
I^m_{n_1 \cdots n_k} :=
\int x^m s_{n_1}(x) \cdots s_{n_k}(x) dx =
\]
\[
2^{-{2m+k\over 2}}\sum h_{l_1} \cdots h_{l_k} 
\int x^m s_{2n_1+l_1}(x) \cdots s_{2n_k+l_k}(x) dx =
\]
\[
2^{-m-k/2}\sum h_{l_1} \cdots h_{l_k} 
I^m_{2n_1+l_1, \cdots 2n_k+l_k}
\]
\begin{center}
{\bf
Inhomogeneous equations    
}
\end{center}
\[
\sum_{n_1} I^m_{n_1 \cdots n_k} = I^m_{n_2 \cdots n_k} 
\]

\begin{center}
{\bf
Solve using linear algebra and recursion on $k$.    
}
\end{center}

\end{frame}  

%slide 35 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf 
A necessary condition for the solution of the {\color{red} RG equation} to have 
$k$ derivatives can be obtained by differentiating the {\color{red} RG equation}
$k$ times, which gives
}
\end{center}
\[
{d^ks(x) \over dx^k} (x) = \sqrt{2} 2^k \sum_l h_l {d^ks(x) \over dx^k} (2x-l)
\]
\begin{center}
{\bf 
Letting $x=m$ and $n=2m-l$ gives the eigenvalue equation 
}
\end{center}
\[
{d^ks \over dx^k} (m) = \sqrt{2} 2^k \sum_n h_{2m-n} {d^ks \over dx^k} (n)
\]

\[
\sum_n H_{mn}  {d^ks \over dx^k} (n) = 2^{-k-{1 \over 2}} {d^ks \over dx^k} (m) 
\]

\begin{center}
{\bf 
The matrix is $(2K-2) \times (2K-2)$ which limits the
number of eigenvalues (again - calculus replaced by
linear algebra).
}
\end{center}

\end{frame}

%slide 36 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Derivatives of basis functions
}
\end{center}
\[
{ds(x) \over dx} = 2 \sum h_l D T^l {ds(x) \over dx}
\]
\begin{center}
{\bf
{\color{red}
Replaces renormalization group equation}
}
\end{center}

\begin{center}
{\bf
{
Differentiate}
}
\end{center}
\[
x=<x>_{s_n} + \sum_n n s_n(x)
\]
\begin{center}
{\bf
to get a normalization condition 
}
\end{center}
\[
1 = \sum_n n {ds_n(x) \over dx}
\]
\end{frame}

%slide 37 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Example: Integral of product of derivatives $(K=3)$
}
\end{center}

\begin{center}
{\bf
$0$ unless the support of $s_m$ and $s_n$ overlap
}
\end{center}

\[
D_{mn} = D_{m-n,0} = \int {d s_m(x) \over dx}{d s_n(x) \over dx} dx
\]
\begin{center}
{\bf
non-zero $D_{m,0}$ have exact rational values
}
\end{center}
\[
D_{40}=D_{-40}=- 3/560 
\]
%\label{a:15}
%\eeq
%\beq
\[
D_{30}=D_{-30}= - 4/35 
\]
%\label{a:16}
%\eeq
%\beq
\[
D_{20}= D_{-20} =  92/105 
\]
%\label{a:17}
%\eeq
%\beq
\[
D_{10}= D_{-10} = -356/105 
\]
%\label{a:18}
%\eeq
%\beq
\[
D_{00} = 295/56 .
\]
\end{frame}

%slide 38 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Singular integrals
}
\end{center}

\[
S_{n}^+ := \int {s_n(x) dx\over x+i0^+}
\]

\begin{center}
{\bf
{\color{red}Renormalization group equation}
}
\end{center}

\[
S^+_{n} := \sqrt{2} \sum_{l} h_l S_{2n+l}^+
\]

\begin{center}
{\bf
Treatment of singularity (partition of unity)
}
\end{center}


\[
-i \pi = \sum_n \int_{-a}^a {dx s_n(x) \over x+i0^+} =\sum_n S^+_{n:a}
\]

\end{frame}

%slide 39 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
{\color{red}RG equation couples integrals over $x=0$ to integrals with support far from
$x=0$}.  The integrals far from the singularity can be approximated in terms
of moments.  Rigorous error bounds can be computed.
The evaluation of the integrals reduce to algebra.
}
\end{center}

\[
S^+_{n:a} = \int_{-a}^a {s_n(x)dx \over x+i0^+} =
{1 \over n} \int_{-a-n}^{a-n} {s(x)dx \over 1+ x/n}=
\]
\[
{1\over n} \sum_{k=0}^\infty ({-1 \over n})^k \int_{-a-n}^{a-n} x^k s(x)dx
\]

\begin{center}
{\bf
{\color{red}RG equation couples integrals over $x=0$ to integrals with support far from
$x=0$}.  The integrals far from the singularity can be approximated in terms
of moments.  Rigorous error bounds can be computed.
The evaluation of the integrals reduce to algebra.
}
\end{center}



\end{frame}
%slide 40 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx  
\begin{frame}


\[
S^+_{n} = \int {s_n(x)dx \over x +i0^+} \qquad n=-1,-2,-3,-4
\]
\vfill
\begin{center}
\begin{table} % [hbt]
{\bf Table 2: Singular integrals $(K=3)$} \\[1.0ex]
\begin{tabular}{|l|l|}
\hline
$S_{-1+}$& -0.1717835441734- i4.041140804162\\
$S_{-2+}$& -1.7516314066967+ i1.212142562305\\
$S_{-3+}$& -0.3025942645356- i0.299291822651\\
$S_{-4+}$& -0.3076858066180- i0.013302589081\\
\hline
\end{tabular}
% \label{tabspec}
\end{table}
\end{center}
\end{frame}

%slide 41 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf Integrals with natural logs}
\end{center}

\[
L(n) := \int_0^\infty s_n (x) \ln (x) dx  
\]
\begin{center}
{\bf {\color{red}The renormalization group equations gives}}
\end{center}
\[
L(n) =
{1 \over \sqrt{2} }\left ( \sum_l h_l L(2n+l) - \ln (2) \right ).
\]
\begin{center}
{\bf $L(n)$ for large n can be expressed in terms of moments}
\end{center}

\[
L(n) = \int s_n (x) \ln (x) dx = \int s (y) \ln (n(1+y/n)) dy
\]
\[
=\ln (n)- \sum_{m=1}^\infty {(-1)^m \over m}{ <x^m >_{s} \over   
n^m }.
\] 
\end{frame}

%slide 42 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

  
\begin{center}
\begin{table} % [hbt]
{\bf Table 2: Log integrals } \\[1.0ex]
\begin{tabular}{|l|l|l|}
\hline
$K=2$  &    & \\
\hline		
$n=-2$ & $\int s_n(x)\ln \vert x\vert dx$ & $0.456927033732831$ \\
$n=-1$ & $\int s_n(x)\ln \vert x\vert dx$ & $-1.64215549088219$ \\
\hline
$K=3$  &    & \\
\hline		
$n=-4$ & $\int s_n(x)\ln \vert x\vert dx $ & $ 1.15737952417967$ \\
$n=-3$ & $\int s_n(x)\ln \vert x\vert dx $ & $ 0.750468355278047$ \\
$n=-2$ & $\int s_n(x)\ln \vert x\vert dx $ & $ 0.315624303943019$ \\
$n=-1$ & $\int s_n(x)\ln \vert x\vert dx $ & $ -1.83646456399118$ \\
\hline
\end{tabular}
% \label{tabspec}
\end{table}
\end{center}
\end{frame}

%slide 43 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
\begin{frame}
\begin{center}
{\bf
Autocorrelation function
}
\end{center}
\[
A(x)  := \int s(x-y) s(y) dy
\]
\begin{center}
{\bf
{\color{red}Renormalization group equation for $A(x)$}
}
\end{center}
\[
A(x) = \sum_{m,n}h_mh_n A(2x-m-n) 
\]
\[
a_l={1 \over \sqrt{2}}\sum_n h_{l-n} h_n
\]
\[
A(x) = \sum_{l=0}^{4K-2} a_l D T^l A(x)
\]
\begin{center}
{\bf
Scale fixing for A(x)
}
\end{center}
\[
\int A(x)=1
\]
\end{frame}

%slide 44 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Autocorrelation function A(x) (K=3)
}
\end{center}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{aucto_3.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 45 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf Integrals with moving singularities}
\end{center}


\[
J_{k-m-n}:= \int {s_m(x) s_n(y) dx dy \over k-x-y+i0^+}
=
\int {A(x) dx \over k-m-n-x +i0^+}  
\]
\begin{center}
{\bf
{\color{red}Renormalization group equation for $A(x)$}
}
\end{center}
\[
J_n = \sqrt{2} \sum_l a_l J_{2n-l}
\]

\[
1 =\sum_n A(x+n)
\]
\[
i\pi = \sum_n \int_{-a}^a {A(x+n) dx \over -x+i0^+} = \sum_n J_{n:a}
\]
\end{frame} 

%slide 46 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Applications (my interests)
}
\end{center}

\begin{itemize}
{\bf 
\item[{\color{red}$\bullet$}] Solving singular integral equations for
scattering and imaging.

\vfill  
  
\item[{\color{red}$\bullet$}] Renormalizing equations of quantum field theory.

\vfill
\begin{itemize}
\item[]   Detemine how parameters of theory behave as a function of
resolution and volume for fixed measured quantities.
Looking for fixed point.
\end{itemize}

\vfill
  
\item[{\color{red}$\bullet$}] Path integral representations of quantum field theory.

\vfill
\begin{itemize}
\item[]  Replaces time evolution of system with an infinite number of degrees of freedom by an infinite dimensional integral.  Important for quantum computing. 
\end{itemize}
  
}
\end{itemize}
\end{frame}  

%slide 47 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Conclusions
}
\end{center}
\begin{itemize}
{\bf 
\item[{\color{red}$\bullet$}] Daubechies wavelets are a useful basis for
  problems involving multiple scales.
  
\item[{\color{red}$\bullet$}] Standard numerical methods do not work very well
  when applied to fractal functions.

\item[{\color{red}$\bullet$}] In most cases the standard numerical methods
  can be replaced by new methods based on the renormalization group
  equation.

\item[{\color{red}$\bullet$}] The calculations of derivatives and integrals
are replaced by linear algebra.

\item[{\color{red}$\bullet$}] Basis provides control over resolution
and volume while remaining efficient.  

  
}  
\end{itemize}  

\end{frame}   

%slide 48 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf
Local references:
}
\end{center}

\vfill
\begin{center}
{\bf
{\color{red}Palle J{\o}rgensen}
: Analysis and Probability\\
Wavelets, Signals, Fractals
}
\end{center}


\vfill
{\bf
O. Bratelli and  {\color{red}P. J{\o}rgensen}
: Wavelets through a looking glass
}

\vfill

\begin{center}
{\bf
Zhihua Zhang \& {\color{red}Palle J{\o}rgensen}\\
Hermite-Wavelet Transforms of Multivariate Functions on [0,1]\\
Acta Applicandae Mathematicae 170,773(2020)
}
\end{center}


\end{frame}   

\end{document}

\begin{frame}
\begin{center}
{\bf
Kernel, $K(x,y)$  of a scattering integral equation after wavelet transform\\
$
F(x) = D(x) + \int K(x,y) F(y) dy
$
}
\end{center}  

\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{sparse_matrix_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}



%slide 49 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Example: $D_{\mathbf{m}\mathbf{n}}$
}
\end{center}

\begin{center}
{\bf
$0$ unless the support of $s_m$ and $s_n$ overlap
}
\end{center}

\[
D_{mn} = D_{m-n,0} = \int {d s_m(x) \over dx}{d s_n(x) \over dx} dx
\]
\begin{center}
{\bf
non-zero solutions have exact rational values
}
\end{center}
\[
D_{40}=D_{-40}=- 3/560 
\]
%\label{a:15}
%\eeq
%\beq
\[
D_{30}=D_{-30}= - 4/35 
\]
%\label{a:16}
%\eeq
%\beq
\[
D_{20}= D_{-20} =  92/105 
\]
%\label{a:17}
%\eeq
%\beq
\[
D_{10}= D_{-10} = -356/105 
\]
%\label{a:18}
%\eeq
%\beq
\[
D_{00} = 295/56 .
\]
\end{frame}

\begin{frame}
%{\bf   
\[
U(\lambda ) = e^{K(\lambda)}, \qquad K(\lambda) = [ G(\lambda), H(\lambda)]
\]
\vfill
%\label{feq.7}
%\eeq
{\bf where $G(\lambda)$ is the part of $H(\lambda)$ with the {\color{red}operators
that couple different scales} turned off.  With this choice 
$G(\lambda)=G^{\dagger}(\lambda)$ so $K(\lambda)$ is anti-Hermetian.

  \vfill
  
It follows that} \vfill
\[
{dH(\lambda) \over d\lambda}= [K(\lambda),H(\lambda)] =
%[[G(\lambda),H(\lambda)],H(\lambda)] =
[H(\lambda),[H(\lambda),G(\lambda)]] .   
\]
\vfill
{\bf $H(\lambda)$ is block diagonal when 
%\beq
$
[H(\lambda),G(\lambda)] = 0 
$.}
%}
\end{frame}

%slide 36 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{full0_0.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 37 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{full_02.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 38 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{full_2.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 39 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{full20.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 340 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}{Approximation Space}
{
\color{blue}

\begin{itemize}
\item For {\color{magenta}large} $m$ and {\color{magenta}smooth} $f(x)$
\[
f(x) \approx  \sum_n f_n  \phi_{-mn} (x) 
\]
\[
f_n = \int \phi_{-mn} (x) f(x) dx \approx 2^{-m/2} f({n \over 2^m})
\]
\item Expansion coefficients proportional to function values on 
support of $\phi_{-mn}$.
\end{itemize}  
}
\end{frame}


\begin{frame}{$K=3$ Wavelet and Scaling Function}     

\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{2.5in}{!}{
\includegraphics{d3_split_level.pdf}}
}
\end{figure}
\end{center}
\end{frame}


\begin{frame}{Unit Translates}     

\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{2.5in}{!}{
\includegraphics{translates.pdf}}
}
\end{figure}
\end{center}
\end{frame}


\begin{frame}{Constant Function}     

\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{2.5in}{!}{
\includegraphics{representing_constant.pdf}}
}
\end{figure}
\end{center}
\end{frame}

\begin{frame}{Linear Function}     

\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{2.5in}{!}{
\includegraphics{representing_linear.pdf}}
}
\end{figure}
\end{center}
\end{frame}


\begin{frame}{Local Polynomials}
%\begin{itemize}
{\color{blue}
\[
L^2(R) = \cdots \oplus {\cal W}_{m-2} \oplus {\cal W}_{m-1} 
\cdots \oplus {\cal W}_{m} \oplus {\cal V}_{m}
\]
\[
\int \psi_{mn} x^l =0 \qquad \forall m,n; \quad l=0,1,\cdots , K-1
\]
\[
\Downarrow
\]
\[
\boxed{
\therefore \qquad x^l = \sum_n c_n \phi_{mn} (x)  
\qquad \mbox{\scriptsize {\color{magenta}pointwise}}
}
\]
}
%\end{itemize}
\end{frame}

\begin{frame}{Computation of $\phi (x)$ and $\psi (x)$}

\begin{itemize}     
\item $\phi (x)$  at integer points can be obtained by 
solving:
\[
\phi (n) = \sum_{l=0}^{2K-1}  \sqrt{2} h_l \phi (2n-l)   
\qquad
\sum_{n} \phi (n) =1 
\]
\item The support of $\phi (x) \in [0,2K-1]$.
\[
\Downarrow
\]
\[
\phi (n) = 0 \qquad n \leq 0 \quad \mbox{or} \quad n \geq 2K-1 
\]
\end{itemize}     
\end{frame}

\begin{frame}{Computation of $\phi (x)$ and $\psi (x)$}
\begin{itemize}     
\item The scaling equation generates $\psi$ and $\phi$ 
recursively at all dyadic rationals:
\[
\phi ({n\over 2^k} ) = \sum_{l=0}^{2K-1}  \sqrt{2} h_l 
\phi ({n\over 2^{k-1}}-l)   
\]
\[
\psi ({n\over 2^k} ) = \sum_{l=0}^{2K-1}  \sqrt{2} g_l 
\phi ({n\over 2^{k-1}}-l)   
\]
\end{itemize}     
\end{frame}
\begin{frame}{$K=3$ Wavelet and Scaling Function}     

\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{2.5in}{!}{
\includegraphics{d3_split_level.pdf}}
}
\end{figure}
\end{center}
\end{frame}

\begin{frame}{Approximation Spaces}
\begin{itemize}
\item There are two approximation spaces related by a fast 
orthogonal transformation $(o(N))$.
\[ 
{\cal V}_m  \Leftrightarrow  {\cal W}_{m+1} \oplus {\cal W}_{m+2}
\oplus \cdots \oplus 
{\cal W}_{m+k} \oplus {\cal V}_{m+k} 
\]
\item[] with orthonormal bases 
\end{itemize}  
\[
\lbrace \phi_{mn} (x) \rbrace_{n=-\infty}^{\infty}
\]
\[
\Updownarrow
\]
\[
\lbrace \phi_{m+k\, n} (x) \rbrace_{n=-\infty}^{\infty}
\cup 
\lbrace \psi_{ln} (x) \rbrace_{n=-\infty,l=m+1}^{\infty,m+k}
\]
\end{frame}



\begin{frame}{Interesting Properties}

\begin{itemize}{} 
\item Wavelets are fractals.

\item Basis functions are generated from a single ``Mother'' function
by translations and dyadic scale changes.

\item ``Mother'' function constructed from solution (``Father'' function
$=$ scaling function) of a linear renormalization-group equation.

\end{itemize}
\end{frame}


\begin{frame}{Useful Numerical Properties}
{
\footnotesize
\begin{itemize}{} 

\item Basis functions have compact support. 

\item Basis functions are orthonormal.

\item Basis functions never have to be computed. 

\item The wavelet transform automatically eliminates unimportant 
basis functions.

\item Basis functions can locally pointwise represent polynomials.

\item Wavelets lead to efficient treatment of scattering singularities.

\item There is an efficient one-point quadrature rule. 
\end{itemize}
}
\end{frame}

\begin{frame}{Why are Wavelets Interesting ?}
%\small

\vfill

\begin{itemize}{} 
\item Efficient representation of information.

\item Used in the FBI's fingerprint archive.

\item Used in the JPEG2000 image compression algorithm.

\item Fast reconstruction of information.

\item Natural basis for functions with smooth structures on multiple scales.

\end{itemize}
\end{frame}


\begin{frame}

{Structure of equation:}
{
\color{blue}
\footnotesize
\[ 
f(x) = g(x) + \int {K(x,y) \over y}f(y) dy
\]
\[
f= \sum f_n \phi_n (x) \qquad x_n = \langle x^1 \rangle_{\phi_n}  
\]
\[
f_m = g(x_m) + \sum_n \left ({K(x_m ,y_n )-K(x_m ,0 )\over y_n}
+ K(x_m ,0 ) I_n \right ) f_n
\]
\[
f(x) = g(x) + \sum_n \left ({K(x ,y_n )-K(x ,0 )\over y_n}
+ K(x ,0 ) I_n \right ) f_n
\]   
}
\end{frame}


\begin{frame}

{Transformed Kernel}     

\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{2.5in}{!}{
\includegraphics{sparse_matrix.pdf}}
}
\end{figure}
\end{center}
\end{frame}



\end{document}
\begin{frame}


\begin{frame}
%{\bf   
\[
U(\lambda ) = e^{K(\lambda)}, \qquad K(\lambda) = [ G(\lambda), H(\lambda)]
\]
\vfill
%\label{feq.7}
%\eeq
{\bf where $G(\lambda)$ is the part of $H(\lambda)$ with the {\color{red}operators
that couple different scales} turned off.  With this choice 
$G(\lambda)=G^{\dagger}(\lambda)$ so $K(\lambda)$ is anti-Hermetian.

  \vfill
  
It follows that} \vfill
\[
{dH(\lambda) \over d\lambda}= [K(\lambda),H(\lambda)] =
%[[G(\lambda),H(\lambda)],H(\lambda)] =
[H(\lambda),[H(\lambda),G(\lambda)]] .   
\]
\vfill
{\bf $H(\lambda)$ is block diagonal when 
%\beq
$
[H(\lambda),G(\lambda)] = 0 
$.}
%}
\end{frame}

xxxxx

  
\[
I_m(n) = \int {D^m T^n \phi (x) \over x} dx = 
\int  T^n \phi (x)  D^{-m}{ 1\over x} dx =
2^{-{m \over 2}} \int \phi (x) T^{-n} { 1\over x} dx 
\]
\[
=2^{-{m \over 2}} \int \phi (x) { 1\over x+n } dx =
2^{-{m \over 2}} {1 \over n} \sum_{l=0}^{\infty} \left ( -{1\over n} \right )^l
\langle x^l \rangle_{\phi} . 
\]

For singularities of the form $1/ (x \pm i \epsilon)$  
equation (\ref{eq:pva}) is replaced by 
\[
\int_{-m}^{m} {dx \over x\pm i 0^+} =  \mp i \pi .
\]
Using this with the wavelet expansion 
\[
\sum_n \phi(x) = 1
\]

\end{frame}
\begin{frame}

provides the needed additional equation,
\[
\mp 2 \pi i  = \sum_n I_{m}^k(n) ,
\]
which replaces (\ref{eq:sing}). 
The result of solving these linear equations is accurate approximations to 
the integrals 
\[
\int {\phi_n (x) \over x \pm i0^+} .
\]


\end{frame}
\begin{frame}

Similar methods can be used to treat other types of integrable 
singularities.  For example, for logarithmic singularities define 
\[
I_0(n) := \int \phi_n (x) \ln (x) dx  
\]
The scaling equation gives the linear relations
\[
I_0(n) = (T^n \phi , \ln ) =  (DT^n \phi , D\ln )
\]
\[
={1 \over \sqrt{2}} [(T^{2n} D \phi ,\ln) - \ln (2) (T^{2n} D \phi ,1)] 
\]

\end{frame}
\begin{frame}


\[
=
{1 \over \sqrt{2} }\left ( \sum_l h_l I_0(2n+l) - \ln (2) \right ).  
\]
In this case, because the singularity is integrable and the value of 
the integral is unambiguous, we do not need an additional equation
to specify the treatment of the singularity; however the function is
multiply valued, so the computation of the input integrals far from the
singularity should reflect the choice of Riemann sheet. 
 
The linear equations above relate the integrals $I_0(n)$ that overlap the 
singularity to integrals far away from the singularity, which may or 
may not have support containing the endpoints $\pm a$.  These terms 
serve as input to the linear system and can be computed with the 
moments and partial moments using expansion methods.  For the case
that the support of $\phi_n (x)$ does not contain $\pm a$ the following 
expansion can be used when the support of $\phi_n (x)$ contains 
positive values of $x$:

\end{frame}
\begin{frame}

\[
I_0(n) = \int \phi_n (x) \ln (x) dx = \int \phi (y) \ln (n(1+y/n)) dy
\]
\[
=\ln (n)- \sum_{m=1}^\infty {(-1)^m \over m}{ <x^m >_{\phi} \over n^m }.
\] 
This expansion converges rapidly for large $n$. When $\vert n \vert $ is
large with $n<0$ we use 
\[
\ln (n) = ln \vert n \vert + i (2m+1) \pi
\]

\end{frame}
\begin{frame}

A necessary condition for the solution of the scaling equation to have 
$k$ derivatives can be obtained by differentiating the scaling equation 
$k$ times, which gives
\[
\phi^{(k)} (x) = \sqrt{2} 2^k \sum_l h_l \phi^{k} (2x-l)
\]
Letting $x=m$ and $n=2m-l$ gives the equation 
\[
\phi^{(k)} (m) = \sqrt{2} 2^k \sum_l h_{2m-n} \phi^{k} (n)
\]

\end{frame}
\begin{frame}

where the non-zero values of $\phi^k (n)$ satisfy $0 \leq n \leq 2K-1$.
For this equation of have a solution the matrix $H_{mn}:= h_{2m-n}$
must have eigenvalue $2^{-(k+1/2)}$.  This is a necessary condition 
for the basis to have $k$ derivatives.   When the $k$-th derivative 
exists, it can be computed up to normalization by iterating the Fourier 
transform of equation (\ref{eq:der}).  The method used to compute 
wavelets at dyadic points can also be used with the above equation to
compute the $k$-th derivatives of scaling functions and wavelets 
at dyadic points.

The scaling equation can be used to exactly compute all of the 
expansion coefficients.  In order 
to exhibit the key relations it is useful to use operators:

\end{frame}
\begin{frame}

\[
Df(x) = {1 \over \sqrt{2}} f ({x \over 2})
\] 
\[
Tf(x) = f (x -1)
\] 
\[
\Delta f(x) = {df \over dx}(x).
\] 


\end{frame}
\begin{frame}

Direct computation shows the following intertwining relations
\[
\Delta D = { 1 \over 2} D \Delta 
\]
\[
D T = T^2  D  
\]
\[
\Delta T = T \Delta 
\]
\[
\Delta^{\dagger} = - \Delta \qquad T^{\dagger} = T^{-1}
\qquad D^{\dagger} = D^{-1} .
\]

\end{frame}
\begin{frame}

We also have the scaling equations:
\[
D\phi = \sum_l h_l T^l \phi
\]
\[
D\psi = \sum_l g_l T^l \phi .
\]

Using the operator relations above give
\[
D \Delta^r \phi = 2^r \sum_l h_l T^l \Delta^r \phi
\]
\[
D \Delta^r \psi = 2^r \sum_l g_l T^l \Delta^r \phi .
\]

\end{frame}
\begin{frame}

\begin{center}
\begin{table} % [hbt]
{\bf Table 2: Singular Integrals } \\[1.0ex]
\begin{tabular}{|l|l|l|}
\hline
$K=2$  &    & \\
\hline		
$n=-2$ & $\int \phi(x-n)\ln \vert x\vert dx$ & $0.456927033732831$ \\
$n=-1$ & $\int \phi(x-n)\ln \vert x\vert dx$ & $-1.64215549088219$ \\
\hline
$K=3$  &    & \\
\hline		
$n=-4$ & $\int \phi(x-n)\ln \vert x\vert dx $ & $ 1.15737952417967$ \\
$n=-3$ & $\int \phi(x-n)\ln \vert x\vert dx $ & $ 0.750468355278047$ \\
$n=-2$ & $\int \phi(x-n)\ln \vert x\vert dx $ & $ 0.315624303943019$ \\
$n=-1$ & $\int \phi(x-n)\ln \vert x\vert dx $ & $ -1.83646456399118$ \\
\hline
\end{tabular}
% \label{tabspec}
\end{table}
\end{center}


\end{frame}

\begin{frame}
Table 1.
\begin{center}
\begin{table} % [hbt]
{\bf Table 1: Scaling Coefficients } \\[1.0ex]
\begin{tabular}{|l|l|l|l|}
\hline
$h_l$ & K=1 & K=2 & K=3   \\
\hline					      		      
$h_0$ &$1 / \sqrt{2}$  & $(1+\sqrt{3})/4\sqrt{2}$ &$(1+\sqrt{10}+\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
$h_1$ &$1 / \sqrt{2}$  & $(3+\sqrt{3})/4\sqrt{2}$ & $(5+\sqrt{10}+3\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
$h_2$ &$0$  & $(3-\sqrt{3})/4\sqrt{2}$ & $(10-2\sqrt{10}+2\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
$h_3$ &$0$  & $(1-\sqrt{3})/4\sqrt{2}$ & $ (10-2\sqrt{10}-2\sqrt{5+2\sqrt{10}})/16\sqrt{2} $ \\
$h_4$ &$0$  & $0$ & $(5+\sqrt{10}-3\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
$h_5$ &$0$  & $0$ & $(1+\sqrt{10}-\sqrt{5+2\sqrt{10}})/16\sqrt{2}$ \\
\hline
\end{tabular}
% \label{tabspec}
\end{table}
\end{center}
\end{frame}
%slide 6 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf Scaling functions, $s^k_n(x)$}
\end{center}


\begin{center}
{\bf Rescale and translate fixed point, $s(x)$}
\end{center}
\[
s^k_n (x) := D^k T^n s(x) = 2^{k/2} s\left (2^k(x-2^{-k}n)\right ) .
\]

\begin{center}
{\bf ${\cal S}_k :=$ resolution $2^{-k}$ subspace, \qquad $\{s^k_n(x)\}$ basis
}
\end{center}

\[
{\cal S}_k := \{ f(x) \vert f(x)= \sum_{n=-\infty}^\infty c_n s^k_n(x),
\quad \sum_{n=-\infty}^\infty \vert c_n \vert^2 < \infty \}.
\]

\[
{\cal S}_k := D^k {\cal S}_0\qquad
\]

\[
{\cal S}_{k} \subset {\cal S}_{k+n} \qquad n\geq 0 
\]

\[
{\cal S}_{k+1} = {\cal S}_{k} \oplus {\cal W}_{k} \qquad {\cal W}_k \not= \{\emptyset \}.  
\]

\end{frame}

%slide 8 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf Multi-resolution decomposition of Hilbert space 
}
\end{center}

\[
{\cal S}_{k+1} = {\cal S}_{k} \oplus {\cal W}_{k} 
\]
 

\[
L^2 (\mathbb{R}) = {\cal S}_{k}\oplus {\cal W}_{k} \oplus {\cal W}_{k+1}
\oplus {\cal W}_{k+2} \oplus {\cal W}_{k+3} \oplus  \cdots  =
\]

\[
\cdots \oplus {\cal W}_{k-2}\oplus {\cal W}_{k-1} \oplus {\cal W}_{k}
\oplus {\cal W}_{k+1} \oplus {\cal W}_{k+2} \oplus  \cdots
\]

\begin{center}
{\bf Wavelets (basis for ${\cal W}_k$)}
\end{center}

\[
w(x) :=\sum_{l=0}^{2K-1} g_l T^l s(x)
\qquad g_l = (-)^l h_{2K-1-l}
\]

\[
w^k_n (x) := D^k T^n w(x) = 2^{k/2} w\left (2^k(x-2^{-k}n)\right ) .
\]

\end{frame}

%slide 9 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf Multi-resolution basis}
\end{center}
\[
\boxed{
\{\xi_m(x)\}= \{ s^k_n(x)\}_{n=-\infty}^\infty \cup
\{ w^l_n(x)\}_{n=-\infty}^\infty{}_{l=k}^\infty 
}
\]
\begin{itemize} 
{\bf 
\item[$\bullet$] Complete, orthonormal
\vfill
\item[$\bullet$] Limited smoothness (increases with $K$) 
\vfill
\item[$\bullet$] Compact support $(s^k_n(x),w^k_n(x))\subset {\color{red}
2^{-k}[n,(n+2K-1)]}$  
\vfill
\item[$\bullet$] Partition of unity $1 = 2^{k/2} \sum_m s^k_m(x)$ 
\vfill
\item[$\bullet$] $x^n =\sum_m c_m s^k_m(x)$ (pointwise, $n<K$)
\vfill
\item[$\bullet$] $s_n^k(x)$ resolution $2^{-k}$, $w_n^l(x)$ resolution $2^{-{l+1}}$ not in $2^{-l}$
\vfill
\item[$\bullet$] $\int dx x^m w_n^l(x) =0, \qquad  m<K$   
\vfill
\item[$\bullet$] Basis functions are fractal  
}
\end{itemize}

\end{frame}

%slide 10 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf Multi-resolution decomposition of canonical fields}
\end{center}
\[
\xi_n(x) \in \{ s_n^k(x), w^l_n(x) \}
\qquad
\xi_{\mathbf{n}}(\mathbf{x}):= \xi_{n_1}(x)\xi_{n_2}(y) \xi_{n_3}(z)
\]
\vfill
\[
\Phi (\mathbf{x},t) =
\sum_{\mathbf{n}} \Phi^k (\mathbf{n},t) \xi_{\mathbf{n}}(\mathbf{x})   
\qquad
\Phi^k (\mathbf{n},t) = \int d\mathbf{x}\xi_{\mathbf{n}}(\mathbf{x}) 
\Phi (\mathbf{x},t)
\]
\vfill
\[
\Pi (\mathbf{x},t) =
\sum_{\mathbf{n}} \Pi^k (\mathbf{n},t)\xi_{\mathbf{n}}(\mathbf{x})  
\qquad
\Pi^k (\mathbf{n},t) = \int d\mathbf{x}\xi_{\mathbf{n}}(\mathbf{x}) 
\Pi (\mathbf{x},t)
\]
\vfill
\[
[\Phi (\mathbf{n},t), \Pi (\mathbf{m},t)]=
i \delta_{\mathbf{n},\mathbf{m}}
\]
\vfill
\[
[\Phi (\mathbf{n},t), \Phi (\mathbf{m},t)]=
[\Pi (\mathbf{n},t), \Pi (\mathbf{m},t)]=0 , 
\]

\end{frame}

%slide 11 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{itemize} 
{\bf 
\item[$\bullet$] Expansion is exact.
\vfill
\item[$\bullet$] Operator valued distributions replaced by infinite
  sums of well-defined discrete field operators.
\vfill
\item[$\bullet$] Products of discrete fields are well defined.
\vfill
\item[$\bullet$] Discrete fields satisfy canonical commutation relations.
\vfill
\item[$\bullet$] Decomposes the field into local observables by resolution.
\vfill
\item[$\bullet$] Natural resolution (limit $l$) and volume (limit $n$) truncations.
}
\end{itemize}
\end{frame}

%slide 12 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


\begin{frame}
\begin{center}
{\bf
Resolution and volume truncated fields
}
\end{center}
\[
\Phi_T (\mathbf{x},t) =
\sum_{\mathbf{n}\in {\cal I}} \Phi^k (\mathbf{n},t) \xi_{\mathbf{n}}(\mathbf{x})   
\]
\vfill
\[
\Pi_T (\mathbf{x},t) =
\sum_{\mathbf{n}\in {\cal I}} \Pi^k (\mathbf{n},t)\xi_{\mathbf{n}}(\mathbf{x})  
\]
\vfill
\begin{itemize} 
{\bf 
\item[$\bullet$] Restrict index set to finite subset, ${\cal I}$ (finite volume, finite resolution).
  \vfill
\item[$\bullet$] Truncated fields are still differentiable functions of $\mathbf{x}$
}
\end{itemize}
\[
{\cal I} = \{k \leq l \leq l_{max}, -n_{l,max} \leq n_l \leq n_{l,max} \}
\]

\end{frame}


%slide 13 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx


\begin{frame}
\begin{center}
{\bf Hamiltonians (exact multi-scale representation)}
\end{center}

\begin{center}
{\bf Mass and kinetic terms}
\end{center}

\[
\mu^2 \int d\mathbf{x} \Phi^2 (\mathbf{x},t) =
\mu^2 \sum \Phi (\mathbf{m},t)^2 
\qquad
\int d\mathbf{x} \Pi^2 (\mathbf{x},t) =
\sum \Pi (\mathbf{m},t)^2 
\]

\begin{center}
{\bf Derivative terms}
\end{center}
\[
\int d\mathbf{x} \pmb{\nabla} \Phi (\mathbf{x},t) \cdot \pmb{\nabla} \Phi (\mathbf{x},t) =
\sum \Phi (\mathbf{m},t)D_{\mathbf{m}\mathbf{n}}\Phi (\mathbf{n},t) 
\]
\[
D_{\mathbf{m}\mathbf{n}}:=
\int d\mathbf{x} \pmb{\nabla} \xi_{\mathbf{m}} (\mathbf{x}) \cdot \pmb{\nabla} 
\xi_{\mathbf{n}} (\mathbf{x}) 
\]

\begin{center}
{\bf Local interactions}
\end{center}
\[
\int d\mathbf{x} \Phi^N (\mathbf{x},t) =
\sum \Gamma_{\mathbf{n}_1 \cdots \mathbf{n}_N}
\Phi (\mathbf{n}_1,t) \cdots \Phi (\mathbf{n}_N,t) 
\]
\[
\Gamma_{\mathbf{n}_1 \cdots \mathbf{n}_N} =
\int d\mathbf{x}
\xi_{\mathbf{n}_1} (\mathbf{x}) \cdots \xi_{\mathbf{n}_N} (\mathbf{x}) 
\]

\begin{center}
{\bf Terms almost local due to support conditions}
\end{center}

\end{frame}

%slide 14 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
\begin{frame}
\begin{center}
{\bf
All coefficients can be computed analytically using the
renormalization group equation, scale fixing condition,
and weight coefficients, $h_l$.  For example the non-zero 
matrix elements
}    
\end{center}
\[
D^0_{mn} := \int {d s^0_m(x) \over dx}{d s^0_n(x) \over dx} dx
\]
\begin{center}
{\bf
have the following exact rational values
}
\end{center}
\[
D^0_{s;40}=D^0_{s;-40}=- 3/560 
\]
%\label{a:15}
%\eeq
%\beq
\[
D^0_{s;30}=D^0_{s;-30}= - 4/35 
\]
%\label{a:16}
%\eeq
%\beq
\[
D^0_{s;20}= D^0_{s;-20} =  92/105 
\]
%\label{a:17}
%\eeq
%\beq
\[
D^0_{s;10}= D^0_{s;-10} = -356/105 
\]
%\label{a:18}
%\eeq
%\beq
\[
D^0_{s;00} = 295/56 .
\]
\end{frame}

\begin{frame}

\begin{center}
{\bf Truncated Hamiltonian \\
replace fields by truncated fields}
\end{center}

\begin{center}
{\bf Dynamics}
\end{center}
\[
\dot{\Phi}_{\mathbf{n}}(t) = i [H_T, {\Phi}_{\mathbf{n}}(t)]  
\qquad
\dot{\Pi}_{\mathbf{n}}(t) = i [H_T, {\Pi}_{\mathbf{n}}(t)]   
\]
%\eeq

\begin{center}
{\bf Initial conditions}
\end{center}
\[
[\Phi (\mathbf{n},0), \Pi (\mathbf{m},0)]=
i \delta_{\mathbf{n},\mathbf{m}}
\]
\[
[\Phi (\mathbf{n},0), \Phi (\mathbf{m},0)]=
[\Pi (\mathbf{n},0), \Pi (\mathbf{m},0)]=0 , 
\]

\begin{center}
{\bf Hilbert space - the truncated problem well-defined in the free field Fock space}
\end{center}


\end{frame}

%slide 15 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}{\bf 
Advantage of wavelet basis
}
\end{center}
  
\begin{center}{\bf 
Scaling properties of coefficients 
(dim $= 1+1$) 
}
\end{center}
\vfill
\[
{\cal S}_{k+m} = {\cal S}_{k} \oplus {\cal W}_{k} \oplus \cdots \oplus 
{\cal W}_{k+m}
\]

\[
\sum D^k_{mn} \Phi_m^k \Phi_n^k =
2^{2k} \sum D^0_{mn} \Phi_m^k \Phi_n^k 
\]
\vfill
\begin{center}
{\bf 
and}
\end{center}
\vfill
\[
\sum \Gamma^k_{n+1 \cdots n_N} \Phi_{n_1}^k \cdots \Phi_{n_N}^k=
2^{3k({n\over 2}-1)} \sum \Gamma^0_{n+1 \cdots n_N} \Phi_{n_1}^k \cdots \Phi_{n_N}^k
\]

\end{frame}

\begin{frame}
\[
\Downarrow
\]
  
\begin{center}{\bf 
Operator renormalization group equation relates
$H$ truncated at different scales (rescale $+$ canonical transformation)    
}
\end{center}

\vfill

\[
H^k(\Phi^k, \Pi^k, m^{2k} ,\gamma_N^k) =
2^k H^0(\Phi^0, \Pi^0, 2^{-2k}m^{0} , 2^{k(N-4)}\gamma_N^0) 
\]

\begin{center}{\bf 
Canonical transformation
}
\end{center}

\[
\Phi^k = \eta \Phi^0 \qquad \Pi^k = \eta^{-1} \Pi^0 \qquad \eta= 2^{-k/2}  
\]

\begin{center}{\bf 
can be realized as a unitary transformation
}
\end{center}


\end{frame}

%slide 16 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf The problem of constructing a coarse-scale effective
theory that includes the effects of eliminated fine-scale degrees of freedom
is studied first using a free field theory in 1+1 dimension
}
\end{center}

\[
H= 
{1 \over 2}\int ({\Pi}({x},0){\Pi}({x},0)
+\pmb{\nabla}\Phi ({x},0) \cdot \pmb{\nabla}\Phi ({x},0)
+ \mu^2 \Phi({x},0)\Phi({x},0))d{x} ,
\]
\begin{center}
{\bf Spatial derivatives give non-trivial scale coupling
}
\end{center}

\[
H_{Ts}^{k+n} = H^k_{Ts} + H_{Tw} +H_{Tsw}
\]

\begin{itemize}
{\bf
\item[$\bullet$] Resolution $1/2^{n+k}$ Hamiltonian $=$ resolution
$1/2^k$ Hamiltonian $+$ fine scale corrections $+$ scale coupling terms

\item[$\bullet$] Goal - block diagonalize truncated Hamiltonian by resolution
}    
\end{itemize}  
%\[
%H^k_{Ts} :=  
%{1 \over 2} (\sum_{{n}} {\Pi}^k(s,{n},0){\Pi}^k(s,n,0)
%+\sum_{{m}{n}} \Phi^k (s,{m} ,0) D^k_{s;{m}{n}} \Phi^k (s,{n},0)
%\]
%\[+ \mu^2 \sum_{{n}} {\Phi}^k(s,{n},0){\Phi}^k(s,{n},0)),
%\]

\end{frame}  

%slide 17 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\[
H^k_T :=  
{1 \over 2} (\sum_{{n}} {\Pi}^k(s,{n},0){\Pi}^k(s,n,0)
+\sum_{{m}{n}} \Phi^k (s,{m} ,0) D^k_{s;{m}{n}} \Phi^k (s,{n},0)
\]
\[+ \mu^2 \sum_{{n}} {\Phi}^k(s,{n},0){\Phi}^k(s,{n},0)),
\]

\[
H_w :=  
{1 \over 2} (\sum_{{n},l} {\Pi}^{l}(w,{n},0)
{\Pi}^l(w,{n},0)
+ \sum_{{m},l,{n},j}\Phi^l (w,{m} ,0) 
D^{lj}_{w;{m}{n}}
\Phi^j (w,{n},0)
\]
%\beq
\[
+ \mu^2 \sum_{l,{n}} {\Phi}^l(w,{n},0)
{\Phi}^l(w,{n},0)),
\]

\[
H_{sw} :=  
{1 \over 2} 
\sum_{{m},l,{n}}\Phi^l (w,{m} ,0) 
{\color{red}D^{lk}_{sw;{m}{n}}}
\Phi^k (s,{n},0) .
\]
\end{frame}

%slide 18 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
  {\bf Use similarity renormalization group evolution to decouple scales
  (eliminate ${\color{red}D^{lk}_{sw;{m}{n}}}$)
}
\end{center}


\begin{center}
{\bf Generate unitarily equivalent Hamiltonians
}
\end{center}

\[
H(\lambda) = U(\lambda)H(0)U^{\dagger}(\lambda) 
\]

\[
{dU(\lambda) \over d\lambda} = {dU(\lambda) \over d\lambda}
U^{\dagger}(\lambda) U(\lambda) =
K(\lambda) U(\lambda)     
\]

\[
K(\lambda) = {dU(\lambda) \over d\lambda}U^{\dagger}(\lambda) = -
K^{\dagger}(\lambda) 
\]


\begin{center}
{\bf Use a generator $K(\lambda)$ of the form
}
\end{center}

\[
K(\lambda) =
[G(\lambda),H(\lambda)]
\]

\[
{dH(\lambda) \over d\lambda}= [K(\lambda),H(\lambda)] = [[G(\lambda),H(\lambda)],H(\lambda)] =
[H(\lambda),[H(\lambda),G(\lambda)]]
\]


\end{frame}

%slide 19 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{center}
{\bf 
Initial conditions
}
\end{center}
\[
H(0) = H_T^{k+n}
\qquad 
G(0) = H_{Tsw} 
\]
\[
G(\lambda) = \mbox{\bf part of } H(\lambda) \quad \mbox{\bf that couples scales}
\]
\[
H(\lambda) = H_{block}(\lambda) + G(\lambda)
\]

\begin{center}
{\bf 
Equations separate (free field case)
}
\end{center}

\[
{dH_{block}(\lambda)\over d\lambda} =
[G(\lambda),[G(\lambda),H_{block}(\lambda)]]
\]
\[
{dG(\lambda)\over d\lambda} =
-[H_{block}(\lambda),[H_{block}(\lambda),G(\lambda)]]
\]

\end{frame}

%slide 20 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Expand first equation in basis of eigenstates of $H_C:=G(\lambda)$,
second in eigenstates of $H_B:=H_{block}(\lambda)$
}
\end{center}


\[
{dH_{Bmn} (\lambda) \over d\lambda} =
(e_{cm}(\lambda)-e_{cn}(\lambda))^2 H_{Bmn}(\lambda)
\]
{\bf and}
\[
{dH_{Cmn} (\lambda) \over d\lambda} =
-(e_{bm}(\lambda)-e_{bn}(\lambda))^2 H_{Cmn}(\lambda) .
\]

\begin{center}
{\bf
Integrating
}
\end{center}
\[
H_{Bmn}(\lambda) = e^{\int_0^\lambda (e_{cm}(\lambda')-e_{cn}(\lambda'))^2 d\lambda'}
H_{Bmn}(0)
\]
\[
H_{Cmn}(\lambda) = e^{-\int_0^\lambda (e_{bm}(\lambda')-e_{bn}(\lambda'))^2 d\lambda'}
H_{Cmn}(0).
\]
\end{frame}   

%slide 21 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}

\begin{itemize}
{\bf 
\item [${\color{red}\bullet}$] Formal solution shows coupling terms exponentially suppressed.
\vfill
\item [${\color{red}\bullet}$] Evolves to decoupled effective Hamiltonians
involving degrees of freedom on different scales.
\vfill
\item [${\color{red}\bullet}$] Can stall when eigenvalues get close or cross.
\vfill
\item [${\color{red}\bullet}$] The choice of generator is specific to the
free field case.
\vfill
\item [${\color{red}\bullet}$] Evolution easily computed - Hamiltonian
has discrete canonical fields with constant coefficients.
\vfill
\item  [${\color{red}\bullet}$] Free field case allows a detailed analysis
  of the evolution of scales.
}
\end{itemize}
\end{frame}

%slide 22 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Test case: 16 scaling functions, 16 wavelets, BC - vanish at boundary    
}
\end{center}    
\vfill
\begin{center}
{\bf
$H(\lambda)$ has 16 quadratic types obtained by taking
products of $\Phi_s (n,0)$, $\Pi_s (m,0)$, $\Phi_w (n,0)$,
and $\Pi_w (m,0)$.
}    
\end{center}
\vfill
\begin{center}
{\bf
The $\lambda$ evolution should drive the coefficients of the
$\Phi_s(n,0) \Phi_w(m,0), \cdots$ to $0$. 
}    
\end{center}

\begin{center}
{\bf
  Plots show the Hilbert-Schmidt norms of the coefficient matrices of each
  type of operator as a function of $\lambda$.
}    
\end{center}

\end{frame}


%slide 35 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Observations
}
\end{center}  
\vfill
\begin{itemize}
{\bf     
\item[$\bullet$] Coefficients exhibit expected behavior.
\vfill
\item[$\bullet$] The decay of the coefficients of the scale coupling
terms is initially fast, but slows significantly.
\vfill
\item[$\bullet$] Truncated free fields are equivalent to coupled oscillators;
how does the SRG evolution distribute the normal modes among the blocks?   
}
\end{itemize}  
\end{frame}  

%slide 36 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
To get a more detailed understanding of the effects of the
$\lambda$ flow note that the truncated Hamiltonian has the form
}
\end{center}  
\vfill
\[
H_T = {1 \over 2} 
[
(\Pi^s ,\Pi^w) 
\left (
\begin{array}{cc}
I_s & 0\\
0 & I_w 
\end{array}
\right ) 
\left (
\begin{array}{c}
\Pi^s \\
\Pi^w  
\end{array}
\right ) +
\]
\[
(\Phi^s ,\Phi^w) 
\left (
\begin{array}{cc}
\mu^2I+D_{s} & {\color{red}D_{sw}}\\
{\color{red}D_{ws}}  & \mu^2I + D_{w} 
\end{array}
\right ) 
\left (
\begin{array}{c}
\Phi^s \\
\Phi^w  
\end{array}
\right ) 
]
\]
\end{frame}

%slide 37 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\[
M:= 
\left (
\begin{array}{cc}
\mu^2I+D_{s} & {\color{red}D_{sw}}\\
{\color{red}D_{ws}}  & \mu^2I + D_{w} 
\end{array}
\right ) 
\]
\[
O^t M O = 
%\left (
%\begin{array}{cc}
%Im^2 + D^{ss} & D^{sw}0\\
%D^{ws} & Im^2 + D^{ww}  
%\end{array}
%\right ) 
%O
\left (
\begin{array}{cc}
m^s & 0\\
0 & m^w  
\end{array}
\right )
%\label{an.1}
\]
\begin{center}
{\bf   
where $m^s$ and $m^w$ are diagonal matrices consisting 
of eigenvalues of the
matrix $M$.
}
\end{center}
\begin{center}
{\bf   
Transformed discrete fields are related by the canonical
transformation
}
\end{center}
\[
\left (
\begin{array}{c}
\Phi^{\prime s} \\
\Phi^{\prime w}  
\end{array}
\right )  := 
O^t
\left (
\begin{array}{c}
\Phi^s \\
\Phi^w  
\end{array}
\right ) 
\qquad \mbox{and} \qquad
\left (
\begin{array}{c}
\Pi^{\prime s} \\
\Pi^{\prime w}  
\end{array}
\right )  := 
O^t \left (
\begin{array}{c}
\Pi^s \\
\Pi^w  
\end{array}
\right ) .
\]
\end{frame}
\begin{frame}
\[
H'= UHU^{\dagger} = {1 \over 2} 
[
(\Pi^{\prime s} ,\Pi^{\prime w})
\left (
\begin{array}{cc}
I & 0\\
0 & I 
\end{array}
\right ) 
\left (
\begin{array}{c}
\Pi^{\prime s} \\
\Pi^{\prime w}  
\end{array}
\right ) +
\]
\[
(\Phi^{\prime s} ,\Phi^{\prime w}) 
\left (
\begin{array}{cc}
m^{s} & 0\\
0  & m^{w} 
\end{array}
\right ) 
\left (
\begin{array}{c}
\Phi^{\prime s} \\
\Phi^{\prime w}  
\end{array}
\right ) 
]
\]

\begin{itemize}
{\bf
\item[$\bullet$] Transformed Hamiltonian is the Hamiltonian for uncoupled
harmonic oscillators
with frequencies $\sqrt{m_i}$

\item[$\bullet$] $\Pi \Pi$ coefficients in flow-evolved Hamiltonian are
approximately $(1/2)(I_s \oplus I_w)$ so $\sqrt{}$ of eigenvalues of
$M$ correspond to normal mode frequencies. 

\item[$\bullet$] A general $O$ will separate the eigenvalues of $M$ into two
distinct groups - there is no general relation between normal mode
frequencies and scale.
}
\end{itemize}
\end{frame}


%slide 39 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
{\footnotesize
\color{black}    
\begin{table}[t]
\caption{\color{black}Normal mode frequencies}
\begin{tabular}{|l|l|l|l|}  
\hline		
$\lambda=20, \mu=1$& truncated & exact 1:16 & exact 17:32 \\
\hline
1.037e+00 &  1.037e+00 &  1.041e+00 &  1.665e+01 \\   
1.145e+00 &  1.146e+00 &  1.153e+00 &  1.925e+01 \\
1.326e+00 &  1.333e+00 &  1.340e+00 &  2.208e+01 \\
1.583e+00 &  1.609e+00 &  1.604e+00 &  2.512e+01 \\
1.919e+00 &  1.995e+00 &  1.947e+00 &  2.834e+01 \\
2.341e+00 &  2.525e+00 &  2.373e+00 &  3.167e+01 \\
2.861e+00 &  3.236e+00 &  2.890e+00 &  3.507e+01 \\
3.493e+00 &  4.161e+00 &  3.508e+00 &  3.846e+01 \\
4.263e+00 &  5.317e+00 &  4.243e+00 &  4.178e+01 \\
5.201e+00 &  6.689e+00 &  5.112e+00 &  4.495e+01 \\
6.346e+00 &  8.232e+00 &  6.134e+00 &  4.789e+01 \\
7.722e+00 &  9.859e+00 &  7.332e+00 &  5.053e+01 \\
9.309e+00 &  1.145e+01 &  8.729e+00 &  5.279e+01 \\
1.102e+01 &  1.289e+01 &  1.034e+01 &  5.462e+01 \\
1.274e+01 &  1.403e+01 &  1.219e+01 &  5.597e+01 \\
1.435e+01 &  1.476e+01 &  1.429e+01 &  5.679e+01 \\
\hline
\end{tabular}
\label{omega}
\end{table}
}
\end{frame}


%slide 38 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf   
The table shows that the SRG flow equation puts the lowest normal modes in coarse scale
Hamiltonian and the highest normal mode frequencies in the fine scale Hamiltonian
}
\end{center}
\end{frame}




%slide 40 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Conclusions
}  
\end{center}
  
\begin{itemize}
{\bf 

\item[1.] Flow equation methods with a suitable
generator can be used to construct an effective field theory
with coarse scale degrees of freedom.

\item[2.] The generator used separates both energy and distance
scales.

\item[3.] Increasing the truncated volume generated new low frequency
modes, while increasing the resolution increased the separation
between modes.  The mass sets a lower bound on the normal mode frequencies.

\item[4.] The flow equation also exhibited convergence for mass 0.

}
\end{itemize}  
\end{frame}

\begin{frame}
\begin{center}
{\bf
Summary
}  
\end{center}
  
\begin{itemize}
{\bf 

\item[5.] For this problem, the flow equation was successfully applied
directly to the Hamiltonian, without projecting on a subspace.

\item[6.] We found the that flow equation could be integrated using the
Euler method, but perturbation theory failed to converge.

\item[7.] The final evolved Hamiltonian was approximately local.   

\item[8.] The spectral properties suggest the in order to approach
the continuous spectrum of the exact theory, the volume and resolution
truncations need to be removed together.  

\item[9.] In our test the coefficients of the coupling terms initially
fell off quickly, but the rate of fall off slowed down significantly
as the flow parameter increased.  The method reduced the coupling
coefficients by a factors of about 100 for a modest value of the
flow parameter.

\item[10.] The convergence of the flow equation will slow down as
the separation between normal modes decreases.
}
\end{itemize}
\end{frame}

%slide 42 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
{\bf
Outlook
}  
\end{center}

\begin{itemize}
{\bf 
\item[1] It has been shown that the truncated correlation functions
converge to the exact free field Wightman functions as the resolution
is increased (Singh-Brennan Arxiv 1606:050686).

\item[2] Basis functions have only a finite number of derivatives - but
the Fourier transform of the test functions are entire.

\item[3] Solution of Heisenberg equations of motion + ground state
can be used to construct space-real-time correlation functions
as functions of continuous variables

\item[4] In interacting theories integrating the flow equation generates
infinite numbers of operators.

\item[5] Fields differentiable functions of $x$ - mathematics discrete.
}
\end{itemize}
\end{frame}


\begin{frame}
\begin{center}
{\bf Basis functions (Daubechies wavelets)}
\end{center}
\vfill  
\begin{center}
{\bf Generated by the fixed point, ${\color{red}s(x)}$, of a\\
renormalization group equation }
\end{center}
\[
{\color{red}s(x)} = \underbrace{D
\underbrace{(\sum_{l=0}^{2K-1} h_l T^l {\color{red}s(x)})}_{\mbox{block average}}
}_{\mbox{rescale}} .
\]
\vfill
\begin{center}
{\bf Unitary translations \qquad Unitary scale transformations}
\end{center}
\vfill
\[
Ts(x) = s(x-1) \qquad Ds(x) = \sqrt{2} s(2x) .
\]
\vfill
\begin{center}
{\bf Scale fixing}
\end{center}
\[
\int dx s(x) =1
\]
\end{frame}

%slide 5 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
{\footnotesize
\begin{center}
{\bf{\color{black}
Weights $h_l$ (constants) determine properties of basis
}
}
\end{center}    
\begin{table}[t]
{\color{black}  
\caption{\bf \color{black} Filter weights for Daubechies K=3 Wavelets}
\begin{tabular}{|l|l|}
\hline				      		      
$h_0$ & $(1+\sqrt{10}+\sqrt{5+2\sqrt{10}}\,)/16\sqrt{2}$ \\
$h_1$ & $(5+\sqrt{10}+3\sqrt{5+2\sqrt{10}}\,)/16\sqrt{2}$ \\
$h_2$ & $(10-2\sqrt{10}+2\sqrt{5+2\sqrt{10}}\,)/16\sqrt{2}$ \\
$h_3$ & $ (10-2\sqrt{10}-2\sqrt{5+2\sqrt{10}}\,)/16\sqrt{2} $ \\
$h_4$ & $(5+\sqrt{10}-3\sqrt{5+2\sqrt{10}}\,)/16\sqrt{2}$ \\
$h_5$ & $(1+\sqrt{10}-\sqrt{5+2\sqrt{10}}\,)/16\sqrt{2}$ \\
\hline
\end{tabular}
\label{coef}
}
\end{table}
\vfill 
\begin{itemize}
{\color{black} \bf
\item[] Determined uniquely (up to reflection) by requiring 
orthonormal basis functions that can pointwise
represent polynomials of degree 2. 
}
\end{itemize}
}
\end{frame}


\begin{frame}
The {\bf Haar scaling function} is defined by
\[
\phi (x) := \left\{
\begin{array}{ll}
0 & x \leq 0 \\
1 & 0<x\leq 1 \\
0 & x>1 
\end{array} 
\right. \,.
\]

It satisfies the normalization conditions: 
\[
(\phi,\phi) := \int^{\infty}_{-\infty} \phi^* (x) \phi (x) dx =
\int_0^1 \phi (x) dx =1.
\]

  
\end{frame}  

\begin{frame}
\[
(T^m \phi , T^n \phi ) = ( \phi , T^{n-m}\phi)=
\int^{\infty}_{-\infty} \phi^* (x) \phi (x-n+m) dx =
\]
\[
\int^{1}_{0} \phi (x-n+m) dx =
\delta_{nm}    
\]
This means the functions 
\[
\phi_n (x) := (T^n \phi)(x) = \phi (x-n) 
\]
are orthonormal.
\end{frame}

\begin{frame}
\[
DT = T^2 D . 
\]

\[
\psi(x) = D^{-1} \sum_l g_l T^l \phi (x)
\]
where we will see later that
\[
\qquad g_l = (-)^kh_{k-l}  
\]

\end{frame}
\begin{frame}

\[
\tilde{\phi}(k)= {1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i k x} \phi (x) dx = \sum_l \sqrt{2} h_l 
{1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i k x} \phi (2x-l) dx.
\]

Changing variables $x \to 2x-l$ on the right-hand side gives
\[
{1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i k x} \phi (x) dx = \sum_l {1 \over \sqrt{2}} h_l 
{1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i (k/2) (x+l) } \phi (x) dx
\]
or 
\[
\tilde{\phi} (k) = \tilde{\phi}\left ({k \over 2}\right )  
\tilde{h}\left ( {k \over 2}\right )  
\]
where  
\[
\tilde{h}({k})  =
\sum_l {h_l \over \sqrt{2}} e^{-i k l }.
\]

\end{frame}
\begin{frame}

This form of the scaling equation can be iterated $n$ times to get:
\[
\tilde{\phi} (k) = \tilde{\phi}\left ({k \over 2^n}\right ) \prod_{m=1}^n 
\tilde{h}\left ({k \over 2^m}\right ) 
\]

\[
\tilde{\phi}(k)= {1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i k x} \phi (x) dx = \sum_l \sqrt{2} h_l 
{1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i k x} \phi (2x-l) dx.
\]

Changing variables $x \to 2x-l$ on the right-hand side gives
\[
{1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i k x} \phi (x) dx = \sum_l {1 \over \sqrt{2}} h_l 
{1 \over \sqrt{2 \pi} }
\int_{-\infty}^{\infty}  e^{-i (k/2) (x+l) } \phi (x) dx
\]
or 
\[
\tilde{\phi} (k) = \tilde{\phi}\left ({k \over 2}\right )  
\tilde{h}\left ( {k \over 2}\right )  
\]
where  
\[
\tilde{h}({k})  =
\sum_l {h_l \over \sqrt{2}} e^{-i k l }.
\]
\end{frame}
\begin{frame}

This form of the scaling equation can be iterated $n$ times to get:
\[
\tilde{\phi} (k) = \tilde{\phi}\left ({k \over 2^n}\right ) \prod_{m=1}^n 
\tilde{h}\left ({k \over 2^m}\right ) 
\label{eq:iah}
\]

\[
\tilde{\phi} (k) = \lim_{n\to \infty} \tilde{\phi}_0({k \over 2^n}) 
\prod_{l=1}^n  \tilde{h}({k \over 2^l}) 
\]
\[
=\tilde{\phi}(0) 
\prod_{l=1}^{\infty}  \tilde{h}({k \over 2^l}) .
\]

\[
1 = \prod_{l=0}^\infty  \tilde{h}(0). 
\]  
Now using (\ref{eq:iag}) gives 
\[
\tilde{h} (0) = 1 =  \sum_l {h_l \over \sqrt{2}}
\]
or
\[
\sum_l h_l =  \sqrt{2}.
\]
\end{frame}
\begin{frame}

\[
\sum_l h_{l-2m} h_{l} = \delta_{m0} .
\]

\[
\phi(x) = {\tilde{\phi}(0)  \over \sqrt{2 \pi}} \int_{-\infty}^{\infty} 
e^{i k x} \prod_{m=1}^{\infty}  \tilde{h}({k \over 2^m}) dk 
\]
\[
={1 \over 2 \pi} \int_{-\infty}^{\infty} 
e^{i k x} \prod_{m=1}^{\infty} \left (\sum_{n_m=0}^{N-1} {h_{n_m} \over \sqrt{2}} 
e^{-i k {n_m}/2^m }\right ) dk
\]
\[
=\lim_{m\to \infty}  
\sum_{n_1=0}^{N-1} \cdots \sum_{n_m=0}^{N-1} \left ( \prod_{k=1}^m 
{h_{n_k} \over \sqrt{2}}\right )
\delta (x -\sum_{k=1}^m  n_k/2^m ).
\]
\end{frame}
\begin{frame}

This defines the scaling function as a distribution.  
This is not a useful representation for computation, however it indicates
that if a scaling function has $N$ non-zero coefficients $h_l$ then the 
scaling function has support on 
\[
[0,(N-1)({1 \over 2} + {1 \over 4} + {1 \over 8} \cdots )] = [0,N-1]
\]
where $N$ is the number of non-zero scaling coefficients.

This shows that the coefficients $h_l$, satisfying  
\[
\sum_l h_l =  \sqrt{2}.
\]
\[
\sum_l h_{l-2m} h_{l} = \delta_{m0} 
\]
with $g_k$ defined by 
\[
g_k := (-1)^k h_{l-k} \qquad l \qquad \mbox{odd}
\]
give a multi-resolution analysis, scaling function, and a mother function. 
\end{frame}
\begin{frame}

The Daubechies order-$K$ wavelets are defined by 
the  
conditions
\[
\int x^n \psi (x) dx = 0, \qquad n=0,1, \cdots ,K-1.
\]

\[
0 = (x^n, \psi) = (Dx^n, D\psi)
\]

\[
\sum m^2  g_m=0, \to \sum_m m^2(-1)^m h_{l-m} =0,
\]

\end{frame}
\begin{frame}

\[
\int \psi (x) x^l dx =0 \qquad l=0 \cdots K-1,
\]
has important consequences. 
Eq. (\ref{eq:wavmom}) implies  
\[
\int \psi_{0m} (x) x^l dx = \int \psi (x-m) x^l dx = \int \psi (y) (y+m)^l
dy  
\]
\[
=\sum_{k=0}^l {l! \over k! (l-k)!} m^{l-k} \int \psi (y) y^k dy=0 
\qquad l=0 \cdots K-1 ,
\]
which means that first $K-1$ moments of the unit translates of the 
mother wavelet function vanish.
\end{frame}
\begin{frame}


Similarly, changing scale gives
\[
\int \psi_{10} (x) x^l =
\int D \psi (x) x^l dx = {1 \over \sqrt{2}} \int \psi (x/2) x^l dx 
\]
\[
=2^{l+1/2} \int \psi(y) y^l dy =0 
\qquad l=0 \cdots K-1 .
\]
It straight forward to proceed inductively to show for all $m$ and $n$ that  
\[
\int \psi_{nm} (x) x^l =0 \qquad l=0 \cdots K-1 .
\]


\[
p(x)b(x) = \sum_{mn} c_{mn} \psi_{mn} (x) =  
\sum_{n} d_{n} \phi_{kn} (x) + \sum_{n}\sum_{m \leq k} c_{mn} \psi_{mn} (x) 
\]
\end{frame}

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{DaubFun_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}



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\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{d2_split_level_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}


\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{d3_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{representing_constant_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{representing_linear_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{translates_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}


\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{DaubFun_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{3.0in}{!}{
\includegraphics{{sparse_gimp.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}



\end{document}

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\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{hs5x.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

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\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{hs6x.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

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\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{hs7x.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

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\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{hs8x.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

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\begin{frame}
\begin{figure}
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[angle=000,scale=.25]{mass_0_s_p2.pdf}
\caption{Full matrix, $\lambda$=0.2, mass=0}
\label{fig:8}
\end{minipage}
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[angle=000,scale=.25]{mass_0_s_2.pdf}
\caption{Full matrix, $\lambda$=2.0, mass=0}
\label{fig:9}
\end{minipage}
\end{figure}
\end{frame}

%slide 34 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{figure}
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[angle=000,scale=.25]{mass_4_s_p2.pdf}
\caption{Full matrix, $\lambda$=0.2, mass=4}
\label{fig:8}
\end{minipage}
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[angle=000,scale=.25]{mass_4_s_2.pdf}
\caption{Full matrix, $\lambda$=2.0, mass=4}
\label{fig:9}
\end{minipage}
\end{figure}
\end{frame}

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\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{hs1x.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 24 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{hs2x.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

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\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{hs3x.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}

%slide 26 xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx

\begin{frame}
\begin{center}
\begin{figure} 
\rotatebox{00}{\resizebox{4.0in}{!}{
\includegraphics{{hs4x.pdf}}}
%\includegraphics{fig18.pdf}}
}
\end{figure}
\end{center}
\end{frame}


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\begin{frame}
\begin{figure}
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[angle=000,scale=.25]{mass_1_s_0.pdf}
\caption{Full matrix, $\lambda$=0, mass=1}
\label{fig:8}
\end{minipage}
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[angle=000,scale=.25]{mass_1_s_p2.pdf}
\caption{Full matrix, $\lambda$=0.2, mass=1}
\label{fig:9}
\end{minipage}
\end{figure}
\end{frame}

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\begin{frame}
\begin{figure}
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[angle=000,scale=.25]{mass_1_s_2.pdf}
\caption{Full matrix, $\lambda$=2.0, mass=1}
\label{fig:8}
\end{minipage}
\begin{minipage}[t]{.45\linewidth}
\centering
\includegraphics[angle=000,scale=.25]{mass_1_s_20.pdf}
\caption{Full matrix, $\lambda$=20.0, mass=1}
\label{fig:9}
\end{minipage}
\end{figure}
\end{frame}






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