commit 91480ea035367ec1289fad6a166cefae242a1b16
parent 73568521e9e777443b7149310ae4e7207dfd559e
Author: miksa234 <milutin@popovic.xyz>
Date: Wed, 2 Mar 2022 19:07:52 +0100
new setup
Diffstat:
39 files changed, 1899 insertions(+), 1883 deletions(-)
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diff --git a/appl_ana/prb1.tex b/appl_ana/prb1.tex
@@ -0,0 +1,331 @@
+\include{preamble.tex}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 1}
+
+\subsection{Fall from high}
+We consider a free fall ($\dot{x}(t=0)=0$) of an object with mass $20\
+\text{kg}$ from a height $x(0) = h = 20\; \text{km}$, such that the
+gravitational force depends on the height $x(t)$ in the following way
+\begin{align}\label{eq: free fall}
+ \ddot{x}(t) = -g\frac{R^2}{(x(t) + R)^2},
+\end{align}
+where $R$ is the radius of the earth $R \approx 6000\; \text{km}$ and $g
+\approx 9.81\ \frac{m}{s^2}$ is the gravitational acceleration on the surface
+of the earth. For this problem there are two possible non-dimensionalisations,
+but first let us rewrite the variables in terms of non-dimensional variables
+and some dimensional constants, a priori let
+\begin{align}
+ t &= t_c \tau \;\;\; \text{and}\\
+ x &= x_c \xi.
+\end{align}
+With the above ansatz we get the following second derivative in
+time
+\begin{align}
+ \frac{d^2}{dt^2} &= \frac{1}{t_c^2}\frac{d^2}{d\tau^2} \\
+ \Rightarrow \frac{d^2x}{dt^2} &= \frac{x_c}{t_c^2}
+ \frac{d^2\xi}{d\tau^2},
+\end{align}
+and thus the initial conditions can be rewritten as
+\begin{align}
+ \xi(0) = \frac{h}{x_c},\\
+ \dot{\xi} = 0.
+\end{align}
+Now we can rewrite the equation of the free fall in \ref{eq: free fall} in
+terms of $\xi(\tau)$ as
+\begin{align}
+ \frac{x_c}{gt_c^2} \ddot{\xi} = -\frac{1}{(\frac{x_c}{R}\xi +1)^2}.
+\end{align}
+Thereby we have three dimensional constants constants $\Pi_1, \Pi_2, \Pi_3$,
+as follows
+\begin{align}
+ \Pi_1 = \frac{x_c}{R}, \;\;\;\; \Pi_2 = \frac{h}{x_c}, \;\;\;\;
+ \Pi_3 = \frac{x_c}{gt_c^2}.
+\end{align}
+
+The first scaling is done by reducing $\Pi_1$ and $\Pi_3$ to 1, by setting
+\begin{align}
+ x_c = R, \;\;\;\; t_c = \sqrt{\frac{R}{g}},
+\end{align}
+reformulating the initial problem in equation \ref{eq: free fall} to
+\begin{align}
+ &\ddot{\xi} = -\frac{1}{(\xi + 1)^2},\;\;\;\;
+ \text{with} \nonumber\\
+ &\xi(0) = \frac{h}{R}, \;\;\;\; \dot{\xi}(0) = 0.
+\end{align}
+Reducing the problem, meaning if $\frac{h}{R} \rightarrow 0$ makes the first
+initial condition $\xi(0) \rightarrow 0$. We can conclude that this scaling
+is bad since it changes the initial condition in the reduced problem.
+
+The second scaling option reduces $\Pi_2$ and $\Pi_3$ to 1, by setting
+\begin{align}
+ x_c = h, \;\;\;\; t_c = \sqrt{\frac{h}{g}},
+\end{align}
+reformulating the initial problem in equation \ref{eq: free fall} to
+\begin{align}
+ &\ddot{\xi} = -\frac{1}{(\frac{h}{R}\xi + 1)^2},\;\;\;\;
+ \text{with} \nonumber\\
+ &\xi(0) = 1, \;\;\;\; \dot{\xi}(0) = 0.
+\end{align}
+By letting $R \rightarrow \infty$ we get the following reduced problem
+\begin{align}\label{eq: free fall reduced}
+ \ddot{\xi} = -1.
+\end{align}
+Integrating and solving for $\xi(\tau = T\sqrt{\frac{g}{h}}) = 0$ for when the
+object hits the ground we get a familiar solution
+\begin{align}
+ T = \sqrt{\frac{2h}{g}}
+\end{align}
+Note that in the reduced problem the time until the object hits the ground is
+\textbf{(much) shorter} since the acceleration is at its maximum $\ddot{x}(t) =
+g$ for all $t$. Yet in the original problem the acceleration (gravitation
+force) \textbf{increases} as the object comes \textbf{closer} to earth . For
+instance, if we let an object fall down from the height $h = R$ then its
+gravitational force (acceleration) at that height would be $\ddot{x}(0) =
+g/2$ and upon landing on earth the gravitational force $\ddot{x}(T) = g$,
+while in the reduced solution its gravitational force would be $\ddot{x}(t) =
+g$ for all $t$.
+
+Additionally we can calculate the velocity at impact we need to integrate the
+reduced problem \ref{eq: free fall reduced} once and put in the initial
+condition
+\begin{align}
+ \dot{\xi}(\tau = \frac{T}{t_c}) &= -\tau = -\sqrt{2} \\
+ \text{and} \;\;\; \dot{x} &= \frac{x_c}{t_c}\dot{\xi} =
+ \sqrt{gh}\; \dot{\xi}\\
+ \Rightarrow \dot{x}(T) &= -\sqrt{2gh},
+\end{align}
+The result is exactly the same as we would get from energy conservation
+\begin{align}
+ \frac{m}{2}\dot{x}^2 = mgh \quad \Rightarrow \quad \dot{x} = \sqrt{2gh}.
+\end{align}
+The vertical throw allows for an additional scaling because the
+initial conditions are different, $x(0) = 0$ and $\dot{x}(0) = v$. Thus
+the solution too.
+
+To summarize, the assumptions that used for modeling and simplifying the
+equation are
+\begin{itemize}
+ \item no relativistic influence,
+ \item closed system, no outside influence (gravitation of the sun, air
+ resistance),
+ \item spherical symmetry of the earth (thereby center of mass can be
+ set in the middle of earth).
+\end{itemize}
+By looking at our assumptions a question arises:\textbf{Is it a good
+approximation to replace the attractive force of the earth by the attraction
+of the whole mass concentrated at the center?}.
+
+To answer this question more or less simply we look at the Poisson's equation
+for gravity,
+\begin{align}
+ \ddot{\vec{x}}(\vec{r}) = -\nabla \phi(\vec{r}) \\
+ \Delta \phi = 4\pi G\varrho(\vec{r}).
+\end{align}
+for a gravitational potential $\phi$ and the mass density of earth
+$\varrho$. We assume that \textbf{the earth can be approximated by a sphere}
+and then we integrate both sides along the sphere (and use the Gaussian law
+for integration)
+\begin{align}
+ \int_{S} \nabla \ddot{\vec{x}}\ dS =
+ \int_{\partial S}\ddot{\vec{x}}\ d\vec{s} = -4\pi
+ G \int_S\varrho(\vec{r})\ ds = -4\pi GM.
+\end{align}
+Obviously $\ddot{\vec{x}}$ and $d\vec{s}$ point in the same direction. We
+choose (rotate) the coordinate system such hat $\ddot{\vec{x}} =
+\ddot{x}\ \mathbf{\hat{n}}$ and $d\vec{s} = \mathbf{\hat{n}}\ ds$, thereby
+we get
+\begin{align}
+ &\ddot{x}\int_{\partial S} ds = 4\pi r^2 \ddot{x},\\
+ \Rightarrow &\ddot{x} = -\frac{GM}{r^2}.
+\end{align}
+The further derivation to get the exact equation of motion as in \ref{eq:
+free fall}, we have to keep in mind that $r = x + R$, because by our
+assumptions we are not in the sphere only outside or on the border $R$.
+Lastly by reformulating the constants $gR^2 = GM$ gets us to our equation of
+motion
+\begin{align}
+ \ddot{x}(t) = -g\frac{R^2}{(x(t) + R)^2}.
+\end{align}
+
+\subsection{Scaling The Van der Pol equation}
+The Van der Pol equation is a perturbation of the oscillation equation
+\begin{align}\label{eq: vanderpol}
+ LC\frac{d^2I}{dt^2} + (-g_1C +3g_3CI^2)\frac{dI}{dt} = -I
+\end{align}
+with initial conditions
+\begin{align}\label{eq: van initial}
+ I(0) = I_0,\;\;\;\; \dot{I}(0) = 0.
+\end{align}
+where $I(t)$ is the current at a time $t$, $C$ is the capacity, $L$ is the
+inductivity and $g_1, g_3$ are some parameters. The units of all the
+parameters are
+\begin{align}
+ [LC] &= s^2\\
+ [g_1C] &= s\\
+ [g_3C] &= sA^{-2}
+\end{align}
+The oscillation equation is
+\begin{align}
+ CL\ddot{I} + I = 0.
+\end{align}
+Solvable by the exponential ansatz of $I = Ae^{\lambda t}$, where $\lambda=
+\pm i \sqrt{\frac{1}{LC}}$, thereby
+\begin{align}
+ I(t) = A_1 e^{i\sqrt{\frac{1}{LC}}t} + A_2 e^{-i\sqrt{\frac{1}{LC}}t}.
+\end{align}
+With the initial conditions in equation \ref{eq: van initial} we get $A_1 =
+A_2$ and thus the solution to the oscillation equation is
+\begin{align}
+ I(t) = I_0\cos(\frac{t}{\sqrt{LC}})
+\end{align}
+Now that we know the reduced problem and the solution to it, we may work with
+the Van-Der-Pol equation \ref{eq: vanderpol}, by determining all possible
+non-dimensionalisations. Let us begin by setting
+\begin{align}
+ I(t) = I_c\psi,\\
+ t = t_c \tau,
+\end{align}
+where $I_c$ and $t_c$ have the dimension of $I(t)$ and $t$ accordingly and
+$\psi(\tau)$ and $\tau$ are dimensionless
+The \textbf{first} and second derivative in time is
+\begin{align}
+ \frac{d}{dt} &= \frac{1}{t_c}\frac{d}{d\tau}\\
+ \frac{d^2}{dt^2} &= \frac{1}{t_c^2}\frac{d^2}{d\tau^2}.
+\end{align}
+We can rewrite the Van-Der-Pol equation in terms of $\psi$ and $\tau$
+\begin{align}
+ &\frac{LC}{t_c^2}\ddot{\psi} - \left(\frac{3g_3I_c^2}{g_1}\psi^2 -
+ 1\right)\frac{g_1C}{t_c}\dot{\psi}= -\psi\\
+ &\psi(0) = \frac{I_0}{I_c} \;\;\;\; \dot{\psi}(0) = 0
+\end{align}
+There are a total of four constants that we can eliminate
+\begin{align}
+ \Pi_1 &= \frac{I_0}{I_c}, \qquad
+ \Pi_2 = \frac{LC}{t_c^2},\nonumber\\
+ \Pi_3 &= \frac{3g_3I_c^2}{g_1}, \qquad
+ \Pi_4 = \frac{g_1C}{t_c}.
+\end{align}
+The \textbf{first} scaling is
+\begin{align}
+ I_c = \sqrt{\frac{g_1}{3g_3}},\;\;\; t_c=\sqrt{LC}.
+\end{align}
+Thereby we get the following problem
+\begin{align}
+ \ddot{\psi} + (\psi^2 - 1)\eps \dot{\psi} = -\psi, \qquad \psi(0) =
+ \sqrt{\frac{3g_3}{g_1}}I_0,
+\end{align}
+where $\eps = g_1\sqrt{\frac{C}{L}}$.
+
+The \textbf{second} scaling is
+\begin{align}
+ I_c = \sqrt{\frac{g_1}{3g_3}},\;\;\; t_c=g_1C.
+\end{align}
+Thereby we get the following
+\begin{align}
+ \eps \psi'' +(\psi^2 +1)\psi' = -\psi, \qquad \psi(0) =
+ \sqrt{\frac{3g_3}{g_1}}I_0,
+\end{align}
+where $\eps = \frac{L}{g_1^2C}$. We could also consider scaling $I_c = I_0$
+with $t_c = \sqrt{LC}$ or $t_c = g_1C$ but they wouldn't develop significant
+model hierarchies like the above two scaling.
+\subsection{Scale the Schrödinger Equation}
+The well known Schrödinger equation that describes quantum physics of the one
+particle system is
+\begin{align}
+ &i\hbar \partial_t\psi = -\frac{\hbar^2}{2m}\Delta \psi + V\psi \nonumber\\
+ &\psi(t=0) =\psi_0
+\end{align}
+where $\hbar$ is the reduced Plank constant, $\psi=\psi(x, t)$ the wave function,
+$m$ the mass and $V = V(x)$ the potential in which the wave function is. The
+dimensions are
+\begin{align}
+ [\hbar] = js, \;\;\;\; [V] = j, \;\;\;\; [\psi]= m^{-d/2}
+\end{align}
+where $d$ is the spacial dimension. The standard scaling ansatz is
+\begin{align}
+ &\psi = \psi_c \phi \\
+ &t = t_c \tau \;\;\;\; x = x_c \xi,
+\end{align}
+by that we get the following derivatives in time and in space
+\begin{align}
+ \partial_{x_i} &=\frac{1}{x_{(i)c}} \partial_{\psi_i} \\
+ \partial^2_{x_i} &=\frac{1}{x_{(i)c}^2} \partial_{\psi_i}^2\\
+ \partial_{t} &=\frac{1}{t_c} \partial_{\psi_i} \\
+\end{align}
+for $i = 1, 2, 3$, or depending on the dimension we are dealing with.
+
+Let us consider $x\in \mathbb{R}^3$ and $V = 0$ to scale the equation. First
+we now have
+\begin{align}
+ i \partial_t\psi = -\frac{\hbar}{2m}\Delta \psi
+\end{align}
+with the initial condition $\phi(0) = \frac{\psi_0}{\psi_c}$. With our scaling the equation turns out to be
+\begin{align}
+ \frac{i\hbar t_c}{2m}\frac{1}{||\vec{x}_c||^2}\Delta_{\vec{\xi}}\ \phi =
+ \partial_\tau\ \phi.
+\end{align}
+The constants we get are
+\begin{align}
+ \Pi_1 = \frac{t_c\hbar}{2m}\frac{1}{||\vec{x}_c||^2}, \;\;\;\; \Pi_2 =
+ \frac{\psi_0}{\psi_c}.
+\end{align}
+
+The simple choice of
+\begin{align}
+ \frac{1}{||\vec{x}_c||^2} = 1, \;\;\;\; \psi_c = \psi_0, \;\;\;\; t_c =
+ \frac{2m}{\hbar}||\vec{x}_c||^2,
+\end{align}
+simplifies the Schrodinger equation without the potential to
+\begin{align}
+ i\Delta_{\vec{\xi}}\ \phi = \partial_\tau \phi,
+\end{align}
+with the initial condition $\phi(\tau=0) = 1$.
+.
+
+Now consider $V = 0$, $x\in[0, L]$ and $t \in [0, T]$, the Schrodinger
+equation is the same only with one spacial dimension as above, we can set
+\begin{align}
+ \psi_c = \psi_0, \;\;\;\; x_c =L, \;\;\;\; t_c = \frac{2mL^2}{\hbar}.
+\end{align}
+Thus we get
+\begin{align}
+ i\partial_{\xi}^2 \phi = \partial_\tau \phi,
+\end{align}
+with the initial condition $\phi(\tau=0) = 1$, where $\xi \in [0, 1]$ and
+$\tau \in [0, \frac{\hbar T}{2mL^2}]$.
+.
+
+In the last example let us consider the quantum harmonic oscillator
+represented by the potential $V(x) = m\omega^2 x^2$ for $x\in \mathbb{R}$,
+where $\omega$ is the frequency. The equation is the following
+\begin{align}
+ i \hbar \partial_t \psi = -\frac{\hbar^2}{2m}\partial^2_x \psi
+ +m\omega^2x^2 \psi.
+\end{align}
+By inserting the standard scaling ansatz we get
+\begin{align}
+ i\partial_\tau \phi = -\frac{t_c\hbar}{2mx_c^2}\partial_\xi^2 \phi
+ +\frac{t_cm\omega^2x_c^2}{\hbar} \xi^2 \phi,
+\end{align}
+The dimensional constants are
+\begin{align}
+ \Pi_1 = \frac{t_0\hbar}{mx_c^2},\;\;\;\;\Pi_2 =
+ \frac{m\omega^2x_c^2t_c}{\hbar},\;\;\;\; \Pi_3 = \frac{\psi_0}{\psi_c}.
+\end{align}
+The choice of scaling is
+\begin{align}
+ \psi_c = \psi_0, \;\;\;\; t_c = \frac{1}{\omega}, \;\;\;\; x_c =
+ \sqrt{\frac{\hbar}{m\omega}}.
+\end{align}
+Thereby getting the following problem
+\begin{align}
+ i\partial_\tau \phi = -\frac{1}{2} \partial_\xi^2 \phi +\xi^2 \phi
+\end{align}
+with $\phi(\tau = 0) = 1$.
+
+%\printbibliography
+\end{document}
diff --git a/appl_ana/prb2.tex b/appl_ana/prb2.tex
@@ -0,0 +1,249 @@
+\include{preamble.tex}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 2}
+\subsection{Problem 4}
+We consider a quadratic equation with two ways to perturb it by $\eps$:
+\begin{align}
+ x^2 + 2\eps x -1 = 0, \label{eq: (1)}\\
+ \nonumber \\
+ \eps x^2 + 2x - 1 = 0.\label{eq: (2)}
+\end{align}
+Equation \ref{eq: (2)} is singular, because the reduced problem ($\eps
+\rightarrow 0$) has only one solution at $x = \frac{1}{2}$. While the reduced
+problem in \ref{eq: (1)} has two solutions for $x = \pm 1$, which is the case
+for this non reduced equation. Let us thereby calculate the asymptotic
+expansion of the regular case up to $O(\eps^2)$, we take the ansatz for the
+asymptotic expansion
+\begin{align}\label{eq: p4 ansatz}
+ x_\eps = x_0 + \eps x_1 + \eps^2 x_2 + O(\eps^3).
+\end{align}
+By substituting $x_\eps$ into \ref{eq: (1)} and factoring out the orders of
+$\eps$ we get
+\begin{align}
+ \eps^0 (x_0^2 - 1) + \eps^1(2x_0 + 2x_0x_1) + \eps^2(x_1^2+2x_2x_0
+ +2x_1) + O(\eps^3) = 0
+\end{align}
+By solving the equations in order of $\eps$, for the coefficients
+$x_0$, $x_1$ and $x_2$ we get
+\begin{align}
+ x_0 = \pm 1, \;\;\;\; x_1 = -1, \;\;\;\; x_2 = \pm \frac{1}{2}.
+\end{align}
+By substituting into the equation \ref{eq: p4 ansatz} we get
+\begin{align}
+ x_\eps = \pm 1 - \eps \pm \frac{1}{2} \eps + O(\eps^3).
+\end{align}
+For $\eps = 0.001$ we get
+\begin{align}
+ &x_\eps = -1.0010005 + O(\eps^3), &x_\eps = 0.9990005 + O(\eps^3),\\
+ &x_\eps = -1.001 + O(\eps^2), &x_\eps = 0.999 + O(\eps^2).
+\end{align}
+\subsection{Problem 5}
+Consider the following equations
+\begin{align}
+ \label{eq: p5 1}\eps y' + y = x \;\;\;\;\;\; y(0) = 1\\
+ \label{eq: p5 2}\eps y' + y = x \;\;\;\;\;\; y(0) = 0\\
+ \nonumber\\
+ \label{eq: p5 3}\eps y' + y = x \;\;\;\;\;\; y(0) = \eps\\
+ \label{eq: p5 4}\eps^2 y' + y = x \;\;\;\;\;\; y(0) = \eps\\
+ \nonumber\\
+ \label{eq: p5 5}y' + \eps y = x \;\;\;\;\;\; y(0) = 1\\
+ \label{eq: p5 6}y' + y = \eps x \;\;\;\;\;\; y(0) = 1
+\end{align}
+We will go through the equations and elaborate on if the perturbation is
+regular or singular, if regular we will compute the asymptotic expansion up
+to second order.
+Let us begin with equation \ref{eq: p5 1}. By the first look, the reduced
+problem does not agree with the boundary condition
+\begin{align}
+ y_0 = x \;\;\;\;\; y_0(0) = 1,
+\end{align}
+is a contradiction in $y_0(0) = 0 \neq 1$, thereby equation \ref{eq: p5 1} is
+\textbf{singularly perturbed}.
+
+The reduced problem of equation \ref{eq: p5 2} on the other hand agrees with
+the boundary condition, since
+\begin{align}
+ y_0 = x \;\;\;\;\; y_0(0) = 0.
+\end{align}
+But by doing the ansatz for the asymptotic expansion
+\begin{align}
+ y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3),
+\end{align}
+plugging in into \ref{eq: p5 2} and separating coefficients in terms of
+$\eps$, we get
+\begin{align}
+ \eps^0 (y_0 -x) + \eps^1(y_0' + y_1) + \eps^2(y_1' + y_2) + O(\eps^3) = 0
+\end{align}
+The solutions to these equations are
+\begin{align}
+ y_0 = x, \;\;\;\; y_1 = 1, \;\;\;\; y_2 = 0,
+\end{align}
+which is a contradiction to the boundary condition of $y_1(0) = 1 \neq 0$.
+Thereby we can conclude that equation \ref{eq: p5 2} is \textbf{singularly
+perturbed}.
+
+Next up is equation \ref{eq: p5 3}, where by the asymptotic expansion the
+first order coefficient of $\eps$, $y_2$, has the boundary condition $y_2(0)
+= 0$. But by applying the ansatz of the asymptotic expansion and plugging
+into the equation we get
+\begin{align}
+ \eps^2(y_0 - x) + \eps^1 (y_0' + y_1) + \eps^2(y_1' + y_2) + O(\eps^3) =
+ 0.
+\end{align}
+Solving these equations we get
+\begin{align}
+ y_0 = 0, \;\;\;\; y_1 = 1 \;\;\;\; y_2 = 0 ,
+\end{align}
+which is a contradiction $y_1(0) = 1 \neq 0 $, thus the equation \ref{eq: p5
+3} is \textbf{singularly perturbed}.
+
+The next equation \ref{eq: p5 4} is also singularly perturbed, we
+can see this by plugging the asymptotic expansion into the equation
+\begin{align}
+ \eps^0 ( y_0 - x) + \eps^1(y_1) + \eps^2(y_0' + y_2) = O(\eps^3),
+\end{align}
+solving for the coefficients we get
+\begin{align}
+ y_0 = x, \;\;\;\; y_1 = 0, \;\;\;\;\; y_2 = -1,
+\end{align}
+which is contradiction by the boundary condition $y_2(0) = -1 \neq 0$,
+thereby \ref{eq: p5 4} is \textbf{singularly perturbed}.
+
+Equation \ref{eq: p5 5} on the first sight does not indicate for any
+contradictions, we may plug the ansatz of the asymptotic expansion into the
+equation and see what happens
+\begin{align}
+ \eps^0(y_0 -x) + \eps^1(y_1' + y_0) + \eps^2(y_2' +y_1) + O(\eps^2) = 0,
+\end{align}
+with the initial conditions $y_0(0) = 1$, $y_1(0) = y_2(0) = 0$.
+\begin{align}
+ y_0 = \frac{x^2}{2} + 1, \;\;\;\;
+ y_1 = -\frac{x^3}{6} + x, \;\;\;\;
+ y_2 = \frac{x^4}{24} + \frac{x^2}{2}.
+\end{align}
+Finally we get
+\begin{align}
+ y_\eps(x) = (\frac{x^2}{2}+1) + \eps(-\frac{x^3}{6} -x)
+ +\eps^2(\frac{x^4}{24} + \frac{x^2}{2}) + O(\eps^3).
+\end{align}
+Thereby we can conclude that \ref{eq: p5 5} is \textbf{regularly perturbed}.
+
+The last equation \ref{eq: p5 6} is also regular, let us do the asymptotic
+expansion of the equation and order the equation in orders of $\eps$.
+\begin{align}
+ \eps^0(y_0' + y_0) + \eps^1(y_1' + y_1 -x) + \eps^2(y_2' + y_2)+
+ O(\eps^3) = 0 .
+\end{align}
+by solving these differential equations with the boundary conditions $y_0(0)
+= 1$, $y_1(0) = y_2(0) = 0$ we get
+\begin{align}
+ y_0 = e^{-x} \;\;\;\; y_1 = (x-1) + e^{-x} \;\;\;\; y_2 = 0.
+\end{align}
+The equation we get
+\begin{align}
+ y_\eps(x) = e^{-x} + \eps(x-1 +e^{-x}) + O(\eps^3).
+\end{align}
+Thereby we can conclude that the last equation \ref{eq: p5 6} is
+\textbf{regularly perturbed}.
+\subsection{Problem 6}
+In this section we will calculate the asymptotic expansion of a regularly
+perturbed equation in two ways, by doing the regular expansion ansatz and by
+substituting and expanding in terms of $\eps$. The ordinary differential
+equation we are dealing with is
+\begin{align}
+ y' = -y + \eps y^2 \;\;\;\;\; y(0) = 1,
+\end{align}
+where $t > 0$ and $0 < \eps \ll 1$. The standard expansion ansatz is
+\begin{align}
+y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3).
+\end{align}
+The ODE then expands to
+\begin{align}
+ \eps^0(y_0' + y_0) + \eps(y_1' + y_1 - y_0^2) + \eps^2(y_2' + y_2 -
+ 2y_0y_1) + O(\eps^3) = 0.
+\end{align}
+Equations in order of $\eps$ and $\eps^2$ are non-homogeneous ODE's. The
+solution to these three coefficients with the boundary conditions $y_0(0) =
+1$, $y_1(0) =
+y_2(0) = 0$ we get
+\begin{align}
+ y_0 = e^{-x}, \;\;\;\; y_1 = -e^{-2x} + e^{-x}, \;\;\;\; y_2 = e^{-3x} -
+ 2e^{-2x} + e^{-x}.
+\end{align}
+The expansion of $y$ is then
+\begin{align}
+ y_\eps (x) = e^{-x} + \eps(-e^{-2x} + e^{-x}) + \eps^2(e^{-3x} - 2e^{-2x}
++ e^{-x}) + O(\eps^3). \end{align}
+
+The second ansatz, considers the substitution $z = \frac{1}{y}$, by
+calculating the first derivative and substituting the original problem we
+get
+\begin{align}
+ z' &= \frac{-y'}{y^2} = \frac{y-\eps y^2}{y^2} = \frac{1}{y} - \eps = z -
+ \eps. \\
+ z(0) &= \frac{1}{y(0)} = 1.
+\end{align}
+The solution is
+\begin{align}
+ z(x) = \eps + (1-\eps) e^x.
+\end{align}
+By substituting this into $y = \frac{1}{z}$ and expanding we get
+\begin{align}
+ y(x) &= \frac{1}{\eps+(1-\eps)e^x} = e^{-x} \frac{1}{1 - (1- e^{-x})\eps}
+ \\
+ &= e^{-x} \sum_{n\geq 0} \eps^n(1-e^{-x})^n.
+\end{align}
+which is the geometric series.
+\subsection{Problem 7}
+The last problem consists of a perturbation of a partial differential
+equation (heat equation).
+\begin{align}
+ &\partial_t u(x, t) + \partial_x^2 u(x,t) - \eps u(x, t)^2 = 0
+ &x\in (0, 1),\; t>0,\\
+ &u(x, 0) = \tilde{u}_0(x) &x\in(0, 1), \\
+ &u(0, t) = u(1, t) = 0 & t>0.
+\end{align}
+The problem is regular because the reduced solution is the regular heat
+equation in the one special dimension on $x\in (0, 1)$, we know this is
+solvable. By doing the expansion ansatz we can derive the first equations
+for the first three terms, the ansatz is always the same
+\begin{align}
+ u_\eps = u_0 + \eps u_1 + \eps^2 u_2 + O(\eps^3).
+\end{align}
+Plugging this into the perturbed problem problem and factoring out the terms
+in the order of $\eps$ we get
+\begin{align}
+ &\eps^0 (\partial_t u_0 + \partial_x^2 u_0) + \\
+ &\eps^1 (\partial_t u_1 + \partial_x^2 u_1 - u_0^2) +\\
+ &\eps^2 (\partial_t u_2 + \partial_x^2 u_2 - 2u_1u_0) + O(\eps^3) = 0.
+\end{align}
+
+We can solve the reduced problem with the initial condition $\tilde{u}_0 =
+\sin(\pi x)$ by separation of variables. Setting $u(x, t) = \psi(x) \phi(t)$
+and substituting into the equation we get two ordinary differential equation
+\begin{align}
+ \underbrace{\frac{\psi_{xx}}{\psi}}_{=k}
+ +\underbrace{\frac{\phi_t}{\phi}}_{=-k} = 0,
+\end{align}
+for some $k$. Solving these two by the exponential ansatz.
+\begin{align}
+ \psi(x) &= A_1 e^{\sqrt{k} x} +A_2 e^{-\sqrt{k} x},\\
+ \phi(t) &= A_3 e^{-kt}.
+\end{align}
+With the initial condition we get the conditions that
+\begin{align}
+ A_1A_3 &= -A_2 A_3,\\
+ k &= \pi^2
+\end{align}
+we choose $A_1 = A_3 = 1$ , $A_2 = -1$. We get the following solution to the
+PDE
+\begin{align}
+ u(x, t) = \psi(x)\phi(t) = \sin(\pi x) e^{-\pi^2 t}.
+\end{align}
+
+%\printbibliography
+\end{document}
diff --git a/appl_ana/prb3.tex b/appl_ana/prb3.tex
@@ -0,0 +1,198 @@
+\include{preamble.tex}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 3}
+\subsection{Problem 8}
+Let us look at functions $f: \mathcal{D} \mapsto \mathbb{R}$ that show
+boundary layer behavior at the following manifolds.
+
+The \textbf{first} for $\mathcal{D} = \mathbb{R}^2$ and $S = \{0\}$ we have a
+function e.g.
+\begin{align}
+ f_{\eps}(x, y) = e^{-\frac{x}{\eps}} + y,
+\end{align}
+with the reduced equation
+\begin{align}
+ \lim_{\eps \rightarrow 0} f_{\eps}(x, y) =
+ \begin{cases}
+ y \;\;\;\;\;\;\;\;\;\; x > 0\\
+ 1+y \;\;\;\; x = 0\\
+ \end{cases}
+\end{align}
+
+The \textbf{second} example is $\mathcal{D} = \mathbb{R}^n$ and $S = \{|x| = 1\}$.
+\begin{align}
+ f_\eps(x_1,\dots,x_n) = \tanh\left(\frac{|x| - 1}{\eps} \right),
+\end{align}
+with the reduced equation
+\begin{align}
+ \lim_{\eps \rightarrow 0} f_{\eps}(x_1,\dots, x_n) =
+ \begin{cases}
+ -1 \;\;\;\; |x| < 0\\
+ 1 \;\;\;\;\;\;\; |x| > 0\\
+ \end{cases}
+\end{align}
+
+The \textbf{third} example is $\mathcal{D} = \mathbb{R}^3$ and $S = \{x_1 =
+1\}$
+\begin{align}
+ f_\eps(x_1, x_2, x_3) = \tanh\left(\frac{x_1 - 1}{\eps}\right)+x_2x_3
+\end{align}
+with the reduced equation
+\begin{align}
+ \lim_{\eps \rightarrow 0} f_{\eps}(x_1,x_2,x_3) =
+ \begin{cases}
+ -1 + x_2x_3 \;\;\;\; x_1 < 0\\
+ 1 + x_2x_3 \;\;\;\;\;\;\; x_1 > 0\\
+ \end{cases}
+\end{align}
+\subsection{Problem 9}
+Consider a linear BVP
+\begin{align}
+ Lu := -\eps u'' + b(x)u' + c(x)u = f(x),\\
+ u(0) = u(1) = 0,
+\end{align}
+for $0 < \eps \ll \eps_0$ and $b, c, f \in C([0,1])$ with the conditions
+\begin{align}
+ c(x) \geq 0, \qquad b(x) \geq \beta > 0 \qquad x\in[0, 1]
+\end{align}
+We are to show that for all $x\in[0, 1)$ the reduced solution $u_0$ of the
+above BVP satisfies
+\begin{align}
+ \lim_{\eps \rightarrow 0} u_\eps(x) = u_0(x)))),
+\end{align}
+where the reduced solution $u_0$ is the solution to the following
+differential equation
+\begin{align}
+ b(x)u' + c(x)u = f(x), \quad u(0) = 0.
+\end{align}
+The hint was given: Set
+\begin{align}
+ w_1(x) = e^{\beta x} \quad w_2(x) = e^{-\beta\frac{1-x}{\eps}},
+\end{align}
+such that $Lw_1 \geq \gamma > 0$ for some suitable $\gamma$ and $Lw_2 \geq
+0$. Then for
+\begin{align}
+ v = \pm (u_\eps - u_0), \qquad w = A\eps w_1 + B\eps w_2,
+\end{align}
+for some suitable $A, B$. The following comparison principal is applicable:
+IF
+\begin{align}
+ &Lv(x) \leq Lw(x) \quad \forall x \in (0, 1) \label{eq:cond1}\\
+ &v(0) \leq w(0) \label{eq:cond2}\\
+ &v(1) \leq w(1) \label{eq:cond3}\\
+\end{align}
+then
+\begin{align}
+ \Longrightarrow v(x) \leq w(x) \quad \forall x\in(0, 1)
+\end{align}
+which holds for $u, v \in C^2((0, 1)) \cap C([0, 1])$. Thus a boundary layer
+is possible only at $x=1$. On the other hand, for $b(x) \leq \beta < 0$ it
+follows that the boundary layer is possible only at $x=0$.
+
+We shall go through the chronological order of the conditions\ref{eq:cond1},
+\ref{eq:cond2}, \ref{eq:cond3} and check them. So for \ref{eq:cond1}
+we have that
+\begin{align}
+ Lw(x) &= A\eps Lw_1(x) + B Lw_2(x) \\
+ &\geq A\eps Lw_1(x) = A\eps e^{\beta x} \left(-\eps \beta^2 -
+ b(x)\beta+c(x)\right)\\
+ &\geq A\eps \beta e^{\beta x} \left(1-\eps\right)\\
+ &\geq \eps A\beta^2 e^{\beta}(1-\eps) = \gamma > 0
+\end{align}
+And obviously
+\begin{align}
+ Lv(x) \leq 0 ,
+\end{align}
+by that we have that
+\begin{align}
+ Lv(x) \leq \gamma \leq Lw(x).
+\end{align}
+For the condition \ref{eq:cond2} we have
+\begin{align}
+ w(0) &= A\eps w_1(0) Bw_2(0) = Be^{-\frac{\beta}{\eps}},\\
+ v(0) &= \pm\left(u_\eps(0) - u_0(0)\right) = 0.
+\end{align}
+By the simple choice $B \geq 0$ we satisfy the condition
+\begin{align}
+ v(0) \leq w(0).
+\end{align}
+Now for the last condition \ref{eq:cond3} we have
+\begin{align}
+ w(1) &= A\eps e^\beta + B \geq A\eps e^\beta,\\
+ v(1) &= \mp u_0(1) = 0.
+\end{align}
+And choose $A = \frac{\pm u(1)}{\eps} e^{-\beta}$, which satisfies the last
+condition
+\begin{align}
+ v(1) \leq w(1).
+\end{align}
+Thereby we have
+\begin{align}
+ &v(x) &\leq w(x)\\
+ &\Rightarrow \lim_{\eps \rightarrow 0} v(x) &\leq \lim_{\eps \rightarrow
+ 0} w(x) = 0\\
+ &\Rightarrow \lim_{\eps \rightarrow 0} v(x) = 0
+\end{align}
+uniformly on $(0, 1)$.
+\subsection{Problem 10}
+Consider the following BVP
+\begin{align}
+ -\eps u'' + (1 + x)u' + u = 2, \qquad u(0) = u(1) - 0,
+\end{align}
+for $0 < \eps \ll 1$. \textbf{Where can this problem have a boundary layer?}
+To answer this question we need to look at the reduced problem
+\begin{align}
+ -(1+x)u' + u = 2.
+\end{align}
+The solution to the equation is
+\begin{align}
+ \bar{u}(x) = 2 + A(x+1).
+\end{align}
+According to the boundary conditions it is unclear what the value of the
+constant is, according to $\bar{u}(0)=0$ we get $A = -2$ or according to
+$\bar{u}(1)=0$ we get $A = -1$. Ultimately this means that there exists a
+boundary layer near $x=1$ or $x=0$. We choose $x=0$ and according to this the
+local variable $\xi = x\eps^{-\alpha}$ ($x = \xi \eps^{-\alpha}$). The
+derivatives of $u$ are calculated using the chain rule
+\begin{align}
+ \frac{du}{dx}&= \frac{du}{d\xi}\frac{d\xi}{dx} = \eps^{-\alpha} \dot{u}\\
+ \frac{d^2u}{dx^2}&= \eps^{-\alpha} \frac{d^2u}{d\xi^2}\frac{d\xi}{dx} =
+ \eps^{-2\alpha} \ddot{u}.
+\end{align}
+The BVP transforms as follows
+\begin{align}
+ -\eps^{1-\alpha}\ddot{u} - \dot{u} + \eps(u - \xi\dot{u} - 2) =
+ \begin{cases}
+ -\ddot{u} - \dot{u} = 0 \;\;\;\;\; \alpha=1\\
+ -\dot{u} = 0 \;\;\;\;\;\;\;\; 0<\alpha<1
+ \end{cases}
+\end{align}
+Choosing $\alpha = 1$ for a reasonable solution
+\begin{align}
+ \hat{u}(\xi) = Be^{-\xi},
+\end{align}
+which converges in the local limit (!). Thereby we have a asymptotic
+representation up to the degree of $\eps$
+
+\begin{align}
+ u_\eps(x) &= \bar{u}(x) + \hat{u}(\psi) + O(\eps)\\
+ &= 2 + A(1+x) + B e^{-\frac{x}{\eps}} + O(\eps)
+\end{align}
+And by the boundary conditions
+\begin{align}
+ u_\eps(0) = 2+A+B=0, \qquad u_\eps(1) = 2+2A+B=0,
+\end{align}
+we get that the constants are
+\begin{align}
+ A = -4, \qquad B = 2.
+\end{align}
+The asymptotic representation is thereby
+\begin{align}
+ u_\eps(x) = 2 - 4(1+x) + 2 e^{-\frac{x}{\eps}} + O(\eps)
+\end{align}
+%\printbibliography
+\end{document}
diff --git a/appl_ana/prb4.tex b/appl_ana/prb4.tex
@@ -0,0 +1,141 @@
+\include{preamble.tex}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 4}
+
+\subsection{Fourier Series}
+The Fourier series of a $p$ periodic function $f$, integrable on
+$[-\frac{p}{2}, \frac{p}{2}]$ is
+\begin{align}
+ f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(\frac{2\pi n x}{p})
+ b_n sin(\frac{2\pi n x}{p})\right).
+\end{align}
+The coefficients $a_n$ and $b_n$ are called the Fourier coefficients of $f$
+and are given by
+\begin{align}
+ a_n &= \frac{2}{p} \int_{-\frac{p}{2}}^{\frac{p}{2}} f(x) \sin(\frac{2\pi
+ n x}{p}) dx, \;\;\;\;\; n\geq 0 \\
+ b_n &= \frac{2}{p} \int_{-\frac{p}{2}}^{\frac{p}{2}} f(x) \cos(\frac{2\pi
+ n x}{p}) dx, \;\;\;\;\; n\geq 1
+\end{align}
+Let us compute the Fourier series of $f(t) = t$ for $t \in [-\frac{1}{2},
+\frac{1}{2}]$. The Fourier coefficients are
+\begin{align}
+ a_n &= 2\int_{-\frac{1}{2}}^{\frac{1}{2}} t \cos(2\pi n t)\ dt = 0
+ \;\;\;\;\; \text{(odd: g(-t) = -g(t))},\\
+ \nonumber\\
+ b_n &= 2\int_{-\frac{1}{2}}^{\frac{1}{2}} t \sin(2\pi n t)\ dt = \\
+ &= 2 \left(-\frac{1}{2\pi n} \cos(2\pi n
+ t)\bigg|_{-\frac{1}{2}}^{\frac{1}{2}}
+ +\int_{\frac{1}{2}}^{\frac{1}{2}} \frac{1}{2 \pi n}\cos(2\pi n t)\ dt
+ \right) =\\
+ &= -\frac{1}{\pi n}\left( -\cos(\pi n) + \frac{1}{\pi n }\sin(\pi
+ n)\right) =
+ \frac{\sin(\pi n) - \pi n \cos(\pi n)}{(\pi n)^2}.
+\end{align}
+Thereby the Fourier series of $f(t) = t$ is
+\begin{align}
+ f(t) = \sum_{n=1}^\infty \left(\frac{\sin(\pi n) - \pi n \cos(\pi n)}{(\pi
+ n)^2}\right) \sin(2\pi n t) = t
+\end{align}
+\subsection{Truncation Error}
+The truncation error of the trigonometric polynomial $(Sf_N)$ of degree $N$ is
+\begin{align}
+ \sum_{|k| > N} |\hat{f}(k)|^2 = \lVert f - S_N\rVert_2^2 =
+ \int_{-\frac{1}{2}}^{\frac{1}{2}} |E_N(t)|^2 dt.
+\end{align}
+Computations for $N = 3$ and $N = 9$ were done in python with a integration error of
+around $10^{-15}$, resulting in the overall truncation errors of
+\begin{align}
+ \sum_{|k| > 3} |\hat{f}(k)|^2 = 0.0053,\\
+ \sum_{|k| > 9} |\hat{f}(k)|^2 = 0.0143.
+\end{align}
+To achieve $\lVert E_N\rVert^2_2 < 0.1 \lVert f \rVert^2_2$, the number of
+coefficients needed are about $61$. This was done using a while loop and
+evaluating $\lVert E_N\rVert^2_2$ for $N$ until the above condition is met.
+
+\subsection{Orthonormal Bases}
+Here we will go through the most important properties of orthonormal bases.
+So let $\{b_n\}_{n\in \mathbb{N}}$ be an ONB of a vector space $\mathcal{H}$,
+then for every $x\in \mathcal{H}$ we may write
+\begin{align}
+ x = \sum_{b_n} \langle b_n, x\rangle b_n,
+\end{align}
+and
+\begin{align}
+ \lVert x \rVert^2 = \sum_{b_n} |\langle b_n, x\rangle|^2.
+\end{align}
+For any $x, y \in \mathcal{H}$ we can write the scalar product as
+\begin{align}
+ \langle x, y\rangle = \sum_{b_n} \langle b_n, x\rangle \langle b_n,
+ y\rangle,
+\end{align}
+Furthermore there exists a linear projection $\Phi\ : \mathcal{H}
+\rightarrow l^2(\{b_n\}_n)$ such that
+\begin{align}
+ \langle \Phi(x), \Phi(y)\rangle = \langle x, y \rangle\;\;\; \forall x, y
+ \in \mathcal{H}.
+\end{align}
+
+An example of an orthonormal basis, which spans $L^2([-\frac{p}{2},
+\frac{p}{2}])$ is $\mathcal{T}_p = \{e_n := \frac{e^{2\pi i
+\frac{n}{p}x}}{\sqrt{p}}\}_{n\in\mathbb{Z}}$. The $e_n$'s are orthonormal in
+$L^2$ which can be easily seen by using the scalar product of $L^2$, so for
+$n, m \in \mathbb{Z}$
+\begin{align}
+ \langle e_n, e_m\rangle_{L^2([-\frac{p}{2}, \frac{p}{2})} &=
+ \frac{1}{p}\int_{[-\frac{p}{2}, \frac{p}{2}]}e_n \cdot e_m^* \ dx=\\
+ &=\frac{1}{p}\int_{[-\frac{p}{2}, \frac{p}{2}]} e^{2\pi i \frac{(n-m)}{p} x} \ dx=\\
+ &=\frac{\sin(\pi (n-m))}{\pi(n-m)} =
+ \begin{cases}
+ 0 \;\;\;\; n\neq m\\
+ 1 \;\;\;\; n=m
+ \end{cases}
+\end{align}
+\subsection{Dirichlet Kernel}
+The function
+\begin{align}
+ D_t(x) := \sum_{\lVert k \rVert_\infty \leq t} e_k(x), \;\;\;\;\; x\in
+ \mathbb{R}^d
+\end{align}
+is called the Dirichlet Kernel. For $0 < t \in \mathbb{N}$ we have
+\begin{align}
+ (S_tf)(x) = \int_{I^d} f(y) D_t(x-y) dy,
+\end{align}
+where $S_t$ represents the orthogonal projection onto the trigonometric
+polynomials $\Pi_t$ of degree $t$, by
+\begin{align}
+ &S_t:\ L^1(\mathbb{T}^d) \rightarrow \Pi_t \\
+ &f \mapsto \sum_{\lVert k \rVert \leq t} \langle f,
+ e_k\rangle_{L^2(\mathbb{T}^d)} e_k \;\;\;\;\; k \in \mathbb{Z}^d
+\end{align}
+And furthermore the Dirichlet Kernel satisfies
+\begin{align}
+ D_t(x) = \prod_{i=1}^d \frac{e_{t+1}(x_i) - e_{-t}(x_i)}{e_1(x_i) - 1}
+\end{align}
+To show the convolution property, we start off by applying the orthogonal
+projection into the trigonometric polynomials $S_t$ onto a function $f \in
+L(\mathbb{T}^d)$
+\begin{align}
+ (S_tf) &= \sum_{\lvert k\rVert_\infty \leq t} \int_{I^d} f(y) e^{-2\pi i
+ \langle k, y\rangle}\ dy\ e^{2\pi i\langle k, x\rangle} =\\
+ &= \int_{I^d}f(y) \sum_{\lvert k\rVert_\infty \leq t} e^{2\pi i \langle
+ k, (x- y)\rangle}\ dy =\\
+ &= (f * D_t) (x) = \int_{I^d} f(y) D_t(x - y)\ dy.
+\end{align}
+To show the reformulation of the Dirichlet kernel, we need to simply
+calculate it directly
+\begin{align}
+ \sum_{\lVert k \rVert_\infty \leq t} e^{2\pi i \langle k , x\rangle} &=
+ \prod_{j=1}^d \sum_{k_j = -t}^t e^{2\pi i k_j x_j} =\\
+ &= \prod_{j=1}^d e^{-2\pi i t x_j} \sum_{k_j = 0}^{2t} e^{2\pi i k_j
+ x_j}=;\;\;\;\; \text{(trigonometric series)}\\
+ &= \prod_{j=1}^d e^{-2\pi i t x_j} \frac{e^{2\pi i (2t + 1)x_j} -
+ 1}{e^{2\pi i x_j} - 1} =\\
+ &= \prod_{j = 1} \frac{e_{t+1}(x_j) - e_{-t}(x_j)}{e_1(x_j) - 1}.
+\end{align}
+%\printbibliography
+\end{document}
diff --git a/appl_ana/prb5.tex b/appl_ana/prb5.tex
@@ -0,0 +1,92 @@
+\include{preamble.tex}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 5}
+\subsection{Fourier Transform}
+In this section we prove the linearity of the Fourier Transform $\mathcal{F}$ on
+$L^1(\mathbb{R}^d)$. For $f, g \in L^1(\mathbb{R}^d)$ and $\lambda, \mu \in
+\mathbb{R}$ the linearity condition for $\mathcal{F}$ is the following
+\begin{align}
+ \mathcal{F}(\lambda f + \mu g) = \lambda \mathcal{F}(f) + \mu
+ \mathcal{F}(g).
+\end{align}
+We start by using the Fourier transform definition for $x, \xi \in \mathbb{R}^d$
+\begin{align}
+ \mathcal{F}(\lambda f + \mu g)(\xi) &= \int_{\mathbb{R}^d} (\lambda f(x)+
+ \mu g(x)) e^{-2\pi i \langle x, \xi\rangle}\ dx =\\
+ &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle} + \mu
+ g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\
+ &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+
+ \int_{\mathbb{R}^d} \mu
+ g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\
+ &= \lambda \int_{\mathbb{R}^d} f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+
+ \mu \int_{\mathbb{R}^d}
+ g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\
+ &= \lambda \mathcal{F}(f)(\xi) + \mu \mathcal{F}(g)(\xi)
+\end{align}
+\subsection{Identities of the Fourier transform}
+The following are three identities of the Fourier transform
+
+\begin{table}[h!]
+\centering
+\begin{tabular}{| l | c | c |}
+\hline
+ & $g(x)$ & $\hat{g}(\xi)$ \\ \hline \hline
+(1) & $f(x-x_0)$ & $e^{-2\pi ix_0 \xi} \hat{f}(\xi)$ \\ \hline
+(2) & $e^{2\pi i \xi_0 x} f(x)$ & $f(\xi - \xi_0)$ \\ \hline
+(3) & $f(ax)$ & $\frac{1}{a} \hat{f}(\frac{\xi}{a})$\\ \hline
+\end{tabular}
+ \caption{Identities of the Fourier transform for $a > 0,
+ \xi_0, x \in \mathbb{R}$}
+\end{table}
+We start with (1)
+\begin{align}
+ \widehat{f(x-x_0)}
+ &= \int_\mathbb{R} f(x-x_0) e^{-2\pi i x \xi}\ dx=
+ \;\;\;\;\;\; (y = x-x_0)\\
+ &= \int_\mathbb{R} f(y) e^{-2\pi i (y+x_0) \xi}\
+ dy=\\
+ &= e^{-2\pi i x_0 \xi} \int_\mathbb{R}f(y)e^{-2\pi i y
+ \xi}\ dy=\\
+ &= e^{-2\pi i x_0 \xi} \hat{f}(\xi).
+\end{align}
+For (2) we have
+\begin{align}
+ \widehat{e^{2\pi i x \xi_0} f(x)}
+ &= \int_\mathbb{R} e^{2\pi i x \xi_0} f(x) e^{-2\pi i x \xi}\ dx =\\
+ &= \int_\mathbb{R} f(x) e^{-2\pi i x (\xi -\xi_0)}\ dx=\\
+ &= \hat{f}(\xi - \xi_0).
+\end{align}
+For (3) we have
+\begin{align}
+ \widehat{f(ax)}
+ &= \int_\mathbb{R} f(ax) e^{-2\pi i \xi x}\ dx = \qquad \text{sub:
+ $(y=ax)$}\\
+ &= \int_\mathbb{R} \frac{1}{a}f(y) e^{-2\pi i \frac{\xi}{a} y}\ dy=\\
+ &= \frac{1}{a} \hat{f}\left(\frac{\xi}{a}\right).
+\end{align}
+\subsection{The Box-Function}
+Consider the following Box-Function
+\begin{align}
+ \Pi(x) :=
+ \begin{cases}
+ 1\;\;\;\;\;\; -\frac{3}{2} < x < \frac{1}{2}\\
+ 0\;\;\;\;\; \text{else}
+ \end{cases}
+\end{align}
+The Fourier transform of this function is
+\begin{align}
+ \widehat{\Pi(x)}
+ &= \int_\mathbb{R} \Pi(x) e^{-2\pi i x\xi}\ dx=\\
+ &= \int_{-\frac{3}{2}}^{\frac{1}{2}} e^{-2\pi i x \xi}\ dx
+ =\frac{-1}{2\pi i \xi} e^{-2\pi i x\xi}
+ \bigg|_{-\frac{3}{2}}^{\frac{1}{2}}=\\
+ &= \frac{1}{2\pi i \xi} \left(e^{3\pi i \xi} - e^{-\pi i \xi}\right)=\\
+ &= \frac{e^{\pi i \xi}\sin(2\pi\xi)}{\pi \xi}.
+\end{align}
+
+%\printbibliography
+\end{document}
diff --git a/appl_ana/prb6.tex b/appl_ana/prb6.tex
@@ -0,0 +1,139 @@
+\include{preamble.tex}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 6}
+\subsection{Fourier Transform of the convolution}
+Consider the function $f(x)$, which has a Fourier Transform $\hat{f}(\xi)$,
+now let us compute the Fourier transform of
+\begin{align}
+ h(x) = f(3x-1) \sin(x) .
+\end{align}
+We know that the Fourier transform of the convolution is (we use somewhat of
+the inverse convolution theorem).
+\begin{align}
+ \widehat{(f(3x-1)*g(x))} = \widehat{f(3x-1)} \cdot \hat{g}(\xi).
+\end{align}
+The Fourier transform of $f(3x-1)$ is simply done by substituting a new
+variable
+\begin{align}
+ \widehat{f(3x-1)} = \frac{1}{3}e^{2\pi i\frac{\xi}{3}}\
+ f\left(\frac{\xi}{3}\right).
+\end{align}
+The Fourier transform of $\sin(x)$ can be calculated when looking at the
+Fourier transform of the Dirac-delta function
+\begin{align}
+ \widehat{\delta(ax-b)}
+ &=\int_\mathbb{R} \delta(ax-b) e^{-2\pi i x \xi}\ dx
+ \;\;\;\;\;\;\; (y = ax-b)\\
+ &=\int_\mathbb{R} \delta(y) e^{-2\pi i (y+b)\frac{\xi}{a}}\frac{dy}{a}\\
+ &=\frac{1}{a} e^{-2\pi i \xi \frac{b}{a}}.
+\end{align}
+We may plug in $\sin(x)$ in the definition of the Fourier transformation and
+observe where we can use the Dirac-delta to to the inverse Fourier transform
+\begin{align}
+ \widehat{\sin(x)}
+ &=\int_\mathbb{R} \sin(x)e^{-2\pi i x\xi}\ dx=\\
+ &=\frac{1}{2i}\int_\mathbb{R} (e^{ix} - e^{-ix})e^{-2\pi i \xi x}\ dx\\
+ &=\frac{1}{2i}\left(
+ \int_\mathbb{R} e^{ix} e^{-2\pi i \xi x}\ dx+
+ \int_\mathbb{R} e^{-ix} e^{-2\pi i \xi x}\ dx
+ \right).
+\end{align}
+Here we may use the above formula for the Fourier transform of the Dirac
+delta. We choose $a=1$, $b= \pm \frac{1}{2\pi}$ and do some $y=-x$
+substitutions and thereby get the following result
+\begin{align}
+ \widehat{\sin(x)} = \frac{1}{2i} \left(
+ \delta(\xi - \frac{1}{2\pi})
+ -\delta(\xi + \frac{1}{2\pi})
+ \right)
+\end{align}
+The whole result is thereby
+\begin{align}
+ \widehat{f(3x-1)} * \widehat{sin(x)}
+ =& \frac{1}{6i} \bigg(
+ e^{2\pi
+ i(\frac{\xi}{3}-\frac{1}{6\pi})}\hat{f}\big(\frac{\xi}{3}-\frac{1}{6\pi}\big)-
+ e^{2\pi
+ i(\frac{\xi}{3}+\frac{1}{6\pi})}\hat{f}\big(\frac{\xi}{3}+\frac{1}{6\pi}\big)
+ \bigg)
+\end{align}
+\subsection{More Fourier Transforms}
+Consider the function
+\begin{align}
+ f(x) = e^{-|x|}
+\end{align}
+The Fourier transform of this function is
+\begin{align}
+ \hat{f}(\xi)
+ &=\int_\mathbb{R} e^{-|x| e^{-2\pi i x \xi}}\ dx\\
+ &= \int_{-\infty}^0 e^x e^{-2\pi i x \xi}\ dx
+ + \int_0^\infty e^{-x} e^{-2\pi i x \xi}\ dx=\\
+ &= \frac{1}{1-2\pi i \xi} e^{(1-2\pi i \xi) x}\bigg|_{-\infty}^0+
+ \frac{-1}{1+2\pi i \xi} e^{-(1+2\pi i \xi) x}\bigg|_{-\infty}^0 = \\
+ &= \frac{1}{1-2\pi i \xi} + \frac{1}{1 + 2\pi i \xi} =\\
+ &= \frac{2}{1+(2\pi \xi)^2}.
+\end{align}
+Let us use this result to solve the following integral
+\begin{align}
+ \int_\mathbb{R} \frac{\cos(a\xi)}{(2\pi \xi)^2 + 1}\ d\xi =
+ \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx,\\
+\end{align}
+where we used the fact that $\text{Re}(e^{ia\xi}) = \cos(a\xi)$ and
+$\hat{f}(\xi) = \frac{2}{1+(2\pi \xi)^2}$, thereby
+\begin{align}
+ \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx
+ &= \frac{1}{2}\text{Re}\left(
+ \int_\mathbb{R}\hat{f}(\xi)e^{ia\xi}\ d\xi
+ \right)=\\
+ &= \frac{1}{2}\text{Re}\left(
+ \int_\mathbb{R} \hat{f}(\xi) e^{2\pi i \frac{a}{2\pi}\xi}\ d\xi
+ \right)=\\
+ &= \frac{1}{2}\text{Re}\left(f(\frac{a}{2\pi})\right)=\\
+ &= \frac{1}{2} e^{-\frac{|a|}{2\pi}}.
+\end{align}
+\subsection{Finite discrete Fourier transform}
+Consider $s\in \mathbb{C}^N$ with entries
+\begin{align}
+ s[n] = \sin\left(2\pi\xi_0\frac{n}{N}\right),
+\end{align}
+for same $0 < \xi_0 < N$. The finite discrete Fourier transform of $s$ is
+\begin{align}
+ \hat{s}[k] &= \frac{1}{N} \sum_{n=0}^{N-1} \sin\left(2\pi\xi_0\frac{n}{N}\right)
+ e^{-2\pi i \frac{k}{N} n} =\\
+ &=\frac{1}{2iN}\left(
+ \sum_{n=0}^{N-1}e^{2\pi i \frac{n}{N}(\xi_0 -k)} - e^{-2\pi i
+ \frac{n}{N}(\xi_0 +k)}
+ \right).
+\end{align}
+If we consider $\xi_0 \in \mathbb{Z}$, we have
+\begin{align}
+ \hat{s}[k] =
+ \begin{cases}
+ \frac{1}{2i}\;\;\;\;\;\; \xi_0 = k\\
+ -\frac{1}{2i}\;\;\;\;\;\; \xi_0 = -k\\
+ 0 \;\;\;\;\;\; \text{else}
+ \end{cases}
+\end{align}
+\subsection{Discrete Matrix Notation}
+The convolution of two vectors $f, g \in \mathbb{C}^N$, can be expressed by a
+circulate matrix applied to f
+\begin{align}
+ (f * g) [n] = \sum_{k=0}^{N-1} f[k] g[n-k].
+\end{align}
+Consider $g=s$, then the matrix takes the following values
+\begin{align}
+ s[n-k] = s_{nk} = \sin\left(2\pi \xi_0 \frac{n-k}{N}\right).
+\end{align}
+The convolution with an impulse input $f=\delta_{0k}$, a vector that is $1$
+for $k=0$ and else 0 reads
+\begin{align}
+ \sum_k s_{nk}f_k &= \sum_k s_{nk} \delta_{0k} =\\
+ &= \sin\left(2\pi \xi_0 \frac{n}{N}\right).
+\end{align}
+
+%\printbibliography
+\end{document}
diff --git a/appl_ana/prb7.tex b/appl_ana/prb7.tex
@@ -0,0 +1,138 @@
+\include{preamble.tex}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 7}
+\subsection{Dirac Comb}
+The Dirac train or Dirac comb on defined in the following way
+\begin{align}
+ \Sha_m[n] =
+ \begin{cases}
+ 1\;\;\;\;\;\; n = 0, \pm m, \pm 2m,\dots\\
+ 0\;\;\;\;\;\; \text{else}
+ \end{cases}
+\end{align}
+The dirac comb can be represented in a series of discrete dirac delta's
+\begin{align}
+ \Sha_m[n] = \sum_{l=-N}^N \delta[n - lm],
+\end{align}
+where $\delta[s] = 1$ if $s = 0$ else $0$, for $s \in \mathbb{Z}$.
+The discrete Fourier transform of the Dirac comb in $\mathbb{C}^N$ is
+\begin{align}
+ \widehat{\Sha_m[n]}
+ &=\frac{1}{N}\sum_{n=0}^{N-1} \Sha_m[n] e^{-2\pi i \frac{k}{N}n}=\\
+ &=\frac{1}{N}\sum_{n=0}^{N-1}
+ \left(
+ \sum_{l=-N}^N \delta(n-lm)
+ \right)
+ e^{-2\pi i \frac{k}{N}n},
+\end{align}
+where the summation happens exactly $\frac{N}{m}$ times, then
+\begin{align}
+ &\frac{1}{m}\sum_{l=-N}^N e^{-2\pi i \frac{k}{N}lm}=\\
+ &= \frac{1}{m} \sum_{l=-N}^N \delta[k - l\cdot \frac{N}{m}]\qquad
+ \text{(Poisson's summation formula)} \\
+ &= \frac{1}{m}\Sha_{\frac{N}{m}}[k]
+\end{align}
+\subsection{Schwartz Space}
+The Schwartz space $\mathcal{S}(\mathbb{R}^d)$, for $d \in \mathbb{N}$ is
+defined as
+\begin{align}
+ &\mathcal{S} :=
+\bigg\{
+ f\in\mathcal{C}^\infty(\mathbb{R}^d):
+ \forall\alpha,\beta\in\mathbb{N}^d\;\; \lVert f \rVert_{\alpha,\beta}
+ < \infty
+\bigg\},\\
+&\lVert f \rVert_{\alpha, \beta} :=
+\sup_{x\in\mathbb{R}^d}\left|x^\alpha (D^\beta f) (x) \right|.
+\end{align}
+Our aim is to show that if $f\in\mathcal{S}(\mathbb{R})$ then $\hat{f} \in
+\mathcal{S}(\mathbb{R})$. The condition is obviously
+\begin{align}
+ &\lVert \hat{f} \rVert_{\alpha, \beta} =
+ \sup_{\xi\in\mathbb{R}}\left|\xi^\alpha (D^\beta \hat{f}) (\xi)
+ \right|<\infty,
+\end{align}
+for all $\alpha, \beta \in \mathbb{N}$.
+We can start with what we know about the Fourier transform
+\begin{align}
+ \xi^\alpha \hat{f}(\xi) &= \mathcal{F}\left(\frac{1}{(2\pi
+ i)^\alpha}(D^{\alpha}f)(x)\right)\\
+ D^{\beta}\hat{f}(\xi) &= \mathcal{F}\left(
+ (-2\pi i x)^\beta f(x)
+ \right).
+\end{align}
+Combining the two relations above we get
+\begin{align}
+ \xi^\alpha (D^\beta \hat{f})(\xi) =
+ \mathcal{F}\left(\frac{(-2\pi i x)^\beta}{(2\pi
+ i)^\alpha}x^\beta(D^{\alpha}f)(x)\right)=: \mathcal{F}(g(x))\\
+\end{align}
+If we call this function $g$, then $g\in\mathcal{S}(\mathbb{R})$ and
+$g\in L^1(\mathbb{R})$. Applying the Riemann-Lebesgue Lemma we get
+\begin{align}
+ \hat{g}(\xi) = \int_\mathbb{R} g(x) e^{-2\pi i x \xi}\ dx \longrightarrow 0
+ \;\;\; \text{as $|\xi| \rightarrow \infty$ }
+\end{align}
+Thereby $\hat{g} \in \mathcal{S}(\mathbb{R})$ and thus $\hat{f} \in
+\mathcal{S}(\mathbb{R})$.
+\subsection{Tempered Distributions}
+Tempered distributions are the elements of
+\begin{align}
+ \mathcal{S}'(\mathbb{R}^d) :=
+ \bigg\{
+ L: \mathcal{S}(\mathbb{R}^d) \rightarrow \mathbb{C} | \text{$L$ is
+ linear and continuous}
+ \bigg\}.
+\end{align}
+Consider $\xi$ as a tempered distribution, buy acting on $\varphi \in
+\mathcal{S}(\mathbb{R})$ we have
+\begin{align}
+ \xi(\phi) = \int_\xi \xi \varphi(\xi)\ d\xi.
+\end{align}
+The Fourier transform of $\xi$ is
+\begin{align}
+ \hat{\xi}(\varphi)
+ &=\xi(\hat{\varphi})
+ = \int_\mathbb{R} \xi \hat{\varphi}(\xi)\ d\xi\\
+ &= \int_{\mathbb{R}^2}\xi \varphi(x) e^{2\pi i\xi x}\ dxd\xi\\
+ &= \int_{\mathbb{R}^2}\varphi(x) \xi e^{2\pi i \xi x}\ dxd\xi\\
+ &=\int_{\mathbb{R}^2}\varphi(x)\frac{i}{2\pi} \frac{\partial}{\partial x}
+ e^{2\pi i \xi x}\ dxd\xi =\\
+ &=\frac{i}{2\pi}\int_{\mathbb{R}^2}\varphi(x)\delta'(x)\ dx=\\
+ &=\frac{i}{2\pi} \delta'(\varphi).
+\end{align}
+\subsection{Fourier transform of the Dirac Comb}
+The general case of the Dirac Comb as a distribution is
+\begin{align}
+ \Sha_T = \sum_{n \in \mathbb{Z}} \delta_{nT}.
+\end{align}
+The Fourier transform of the $\Sha_T$ distribution for $\varphi \in
+\mathcal{S}(\mathbb{R})$ is
+\begin{align}
+ \widehat{\Sha_T}(\varphi)
+ &= \sum_{n\in\mathbb{Z}} \hat{\delta}_{nT}(\varphi)\\
+ &= \sum_{n\in\mathbb{Z}} \delta_{n\omega_0}(\varphi)\\
+ &=\Sha_{\omega_0}(\varphi).
+\end{align}
+The Fourier transform, transforms the period of the combs.
+\subsection{Shannon Sampling}
+The Fourier transform of $1_{[-\frac{a}{2}, \frac{a}{2}]}(x)$ is
+\begin{align}
+ \mathcal{F}\left(1_{[-\frac{a}{2}, \frac{a}{2}]}\right)(\xi)
+ &= \int_\mathbb{R} 1_{[-\frac{a}{2}, \frac{a}{2}]} e^{-2\pi i x \xi}\
+ dx\\
+ &= \int_{-\frac{a}{2}}^{\frac{a}{2}} e^{-2\pi i x\xi}\ dx\\
+ &= \frac{-1}{2\pi i \xi} e^{-2\pi i x
+ \xi}\bigg|_{-\frac{a}{2}}^{\frac{a}{2}}\\
+ &= \frac{1}{\pi \xi} \frac{1}{2i}\left(
+ e^{pi i a \xi} - e^{-\pi i a \xi}
+ \right)\\
+ &= \frac{\sin(\pi \xi a)}{\pi \xi}
+\end{align}
+
+%\printbibliography
+\end{document}
diff --git a/appl_ana/prb8.tex b/appl_ana/prb8.tex
@@ -0,0 +1,158 @@
+\include{preamble.tex}
+
+\begin{document}
+\maketitle
+\tableofcontents
+
+\section{Sheet 8}
+\subsection{Finite Discrete Fourier Transform (FDFT)}
+Consider the vector $\begin{pmatrix}a & b & c & d\end{pmatrix}^T \in
+\mathbb{C}^4$ with a FDFT $\begin{pmatrix}A & B & C & D\end{pmatrix}^T$. We
+can show that the vector
+\begin{align}
+ \begin{pmatrix}a & 0 & b & 0 & c & 0 & d & 0\end{pmatrix}^T,
+\end{align}
+has the FDFT of
+\begin{align}
+ \frac{1}{2}\begin{pmatrix}A & 0 & B & 0 & C & 0 & D & 0\end{pmatrix}^T.
+\end{align}
+For the $N=4$, $n\in\{0,\dots,3\}$ the coefficients $a, b, c, d$ are denoted in
+$f[n]$. The FDFT is
+\begin{align}
+ \hat{f}[k] &= \frac{1}{4} * \sum_{n=0}^3 f[n] e^{-2\pi i \frac{n}{4}k} \\
+ &=\frac{1}{4}\left(
+ a + be^{-\pi i \frac{k}{2}}
+ + ce^{-\pi i k}+ de^{-\frac{3\pi i k}{2}}
+ \right) = \\
+ (&=\begin{pmatrix}A & B & C & D\end{pmatrix}^T)
+\end{align}
+for $k \in \{0,\dots, 3\}$ accordingly. For the $N=8$, $\mathbb{C}^8$ case
+ we have $f_2[n]$ for $n \in \{0,\dots 7\}$,
+ \begin{align}
+ \hat{f}_2[k] &= \frac{1}{8} * \sum_{n=0}^7 f_2[n] e^{-2\pi i \frac{n}{8}k} \\
+ &=\frac{1}{2}\frac{1}{4}\left(
+ a + be^{-\pi i \frac{k}{2}}
+ + ce^{-\pi i k}+ de^{-\frac{3\pi i k}{2}}
+ \right) = \\
+ (&=\frac{1}{2}\begin{pmatrix}A & B & C & D & A & B & C & D\end{pmatrix}^T)
+ \end{align}
+for $k \in \{0,\dots, 7\}$ accordingly. We may generalize now for
+$\mathbb{C}^{4N}$, and the sequence for $a, b, c, d, 0$ represented by the
+function $g[n]$ for $n \in \{0,\dots, 4N-1\}$,
+\begin{align}
+ g[n] =\begin{cases}
+ f[n] \qquad n\in \{0, N, 2N, 3N\}\\
+ 0 \qquad \text{else}
+ \end{cases}.
+\end{align}
+Now we can compute the FDFT for $k \in \{0,\dots, 4N-1\}$
+\begin{align}
+ \hat{g}[k] &= \frac{1}{4N}\sum_{n=0}^{4N-1} g[n]e^{-2\pi i
+ \frac{n}{4N}k}\\
+ &=\frac{4}{N}\sum_{n=0}{3}f[n]e^{-2\pi i \frac{n}{4}k}\\
+ &=\frac{1}{N}\left(\frac{1}{4}\sum_{n=0}^3 f[n] e^{-2\pi i
+ \frac{n}{4}k} \right) \\
+ &= \frac{1}{N} \underbrace{\begin{pmatrix}A & B & C & D & \dots &
+ \dots & A & B & C & D\end{pmatrix}^T)}_{\text{$4N$ entries, $N$
+ sequences}}.
+\end{align}
+\subsection{More FDFT}
+Consider the discrete complex exponential with frequency of $1Hz$ in
+$\mathbb{C}^8$, for $n \in \{0, \dots , 7\}$,
+\begin{align}
+ \exp[n] = e^{2\pi i n/8}.
+\end{align}
+The FDFT for $k \in \{0, \dots, 7\}$ is
+\begin{align}
+ \hat{\exp}[k] &= \frac{1}{8}\sum_{n=0}^7 e^{2\pi i \frac{n}{8}}e^{-2\pi i
+ n \frac{k}{8}} \\
+ &= \frac{1}{8} \sum_{n=0}^7e^{-2\pi i (k-1)\frac{n}{8}}\\
+ &=
+ \begin{cases}
+ 1\quad k=1\\
+ 0 \qquad k\neq 1
+ \end{cases}.
+\end{align}
+\begin{figure}[H]
+ \centering
+ \includegraphics[width=0.49\textwidth]{./figures/fdft.png}
+ \includegraphics[width=0.49\textwidth]{./figures/normal.png}
+ \caption{Test in Julia}
+\end{figure}
+\subsection{Sampling Sinusoids}
+Consider the following continuous signal
+\begin{align}
+ f(t) = sin(20\pi t) + sin(40\pi t)
+\end{align}
+with frequencies $\omega = 2\pi \nu$, $\nu_1 = 10\ \text{Hz}$ and $\nu_2 = 20\
+\text{Hz}$. Sketching its Fourier transform would be something like this
+\begin{figure}[H]
+ \centering
+\begin{tikzpicture}[
+ axisline/.style={very thick, -stealth},
+ xscale = 1.5,
+ yscale = 1.5
+ ]
+ \draw[axisline] (-3,0)--(3,0) node[right]{$\nu$};
+ \draw[axisline] (0,-1.5)--(0,1.5) node[above]{$\hat{f}$};
+ \draw[->] (-1,0) -- (-1, -1) node[below] {$-\delta(\nu - 10)$};
+ \draw[->] (-2,0) -- (-2, 1) node[above] {$\delta(\nu - 20)$};
+ \draw[->] (1,0) -- (1, 1) node[above] {$\delta(\nu - 10)$};
+ \draw[->] (2,0) -- (2, 1) node[above] {$\delta(\nu - 20)$};
+\end{tikzpicture}
+\end{figure}
+The Nyquist frequency for sampling would be
+\begin{align}
+ \nu_{\text{Nyquist}} = 2\nu_\text{max} = 2\nu_2 = 40\ \text{Hz},
+\end{align}
+If we choose $50\ \text{Hz}$ for sampling we would get aliasing with the
+following frequencies
+\begin{align}
+ n \cdot 50\ \text{Hz} - 20\ \text{Hz} = 30\ \text{Hz},80\ \text{Hz}, 130\
+ \text{Hz}, \dots
+\end{align}
+\subsection{Short-Time Fourier Transform (STFT)}
+The Definition of the STFT is
+\begin{align}
+ \text{STFT}\{f\} &= S_\varphi f(\tau, \omega) = \int_\mathbb{R} f(t)
+ \overline{\text{M}_\omega \text{T}_\tau \varphi}dt \\
+ &=\int_\mathbb{R} f(t)
+ \bar{\varphi}(t - \tau)e^{-2\pi i \omega t}\ dt \\
+\end{align}
+Then we have the following identity
+\begin{align}
+ S_\varphi(\text{T}_u\text{M}_\eta f)(x,\omega)
+ &= \int_\mathbb{R}
+ \left(\text{T}_u \text{M}_\eta f(t)\right) \bar{\varphi}(t-x) e^{-2\pi i
+ \omega t}\ dt\\
+ &= \int_\mathbb{R} e^{2\pi i \eta(t-u)}f(t-u) e^{-2\pi i \omega
+ t}\bar{\varphi}(t-x)\ dt \qquad \text{(sub: $s = t-u$)}\\
+ &= \int_\mathbb{R} f(s)\bar{\varphi}(s-(x-u))e^{2\pi i \eta s}e^{-2\pi i
+ \omega s} e^{-2\pi i \omega u}\ ds \\
+ &=e^{-2\pi i \omega u}\int_\mathbb{R} f(s) \bar{\varphi}(s-(x-u))e^{-2\pi i
+ (\omega - \eta)s}\ ds\\
+ &=e^{-2\pi i \omega u}\int_\mathbb{R}
+ f(s)\overline{ \text{M}_{(\omega-\eta)} \text{T}_{(x-u)}\varphi(s)}\ ds\\
+ &=e^{-2\pi i \omega u} S_\varphi f\left(x-u,\ \omega -\eta\right).
+\end{align}
+The second identity we can show
+\begin{align}
+ S_\varphi f(x, \omega)
+ &= \langle f, \overline{\text{M}_\omega \text{T}_x \varphi}\rangle \\
+ &= \langle\mathcal{F} f, \mathcal{F} \overline{\text{M}_\omega \text{T}_x \varphi}\rangle \\
+ &= \int_\xi \hat{f}(\xi)\int_t \overline{\text{M}_\omega \text{T}_x
+ \varphi}(t) e^{-2\pi i \xi t}\ dt\ d\xi \\
+ &= \int_\xi \hat{f}(\xi)\int_t \hat{\bar{\varphi}}(t-x) e^{2\pi i \omega
+ t} e^{-2\pi i \xi t}\ dt\ d\xi \\
+ &= \int_\xi \hat{f}(\xi)\int_t \hat{\bar{\varphi}}(t-x)e^{-2\pi i (\xi
+ -\omega)t}\ dt\ d\xi \qquad \text{sub $u=t-x$}\\
+ &= \int_\xi \hat{f}(\xi)\int_t \hat{\bar{\varphi}}(u)e^{-2\pi i (\xi
+ -\omega)u}e^{-2\pi i (\xi -\omega)x} \ dt\ d\xi\\
+ &= \int_\xi \hat{f}(\xi)e^{-2\pi i (\xi -\omega)x}\int_t
+ \hat{\varphi}(u)e^{-2\pi i (\xi -\omega)u} \ dt\ d\xi\\ &= e^{2\pi i
+ \omega x}\int_\xi \hat{f}(\xi) \hat{\bar{\varphi}}(\xi - \omega) e^{-2\pi
+ i \xi x}d\xi\\
+ &= e^{2\pi i \omega x} S_{\hat{\varphi}} \hat{f}(\omega, -x).
+\end{align}
+% printbibliography
+\end{document}
diff --git a/appl_ana/preamble.tex b/appl_ana/preamble.tex
@@ -0,0 +1,59 @@
+\documentclass[a4paper]{article}
+
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{mlmodern}
+
+%\usepackage{ngerman} % Sprachanpassung Deutsch
+
+\usepackage{graphicx}
+\usepackage{geometry}
+\geometry{a4paper, top=15mm}
+
+\usepackage{subcaption}
+\usepackage[shortlabels]{enumitem}
+\usepackage{amssymb}
+\usepackage{amsthm}
+\usepackage{mathtools}
+\usepackage{braket}
+\usepackage{bbm}
+\usepackage{graphicx}
+\usepackage{float}
+\usepackage{yhmath}
+\usepackage{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+\usetikzlibrary{calc,decorations.markings}
+
+%\usepackage[backend=biber, sorting=none]{biblatex}
+%\addbibresource{uni.bib}
+
+\usepackage[framemethod=TikZ]{mdframed}
+
+\tikzstyle{titlered} =
+ [draw=black, thick, fill=white,%
+ text=black, rectangle,
+ right, minimum height=.7cm]
+
+
+\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
+\usepackage[parfill]{parskip}
+\usepackage{lipsum}
+
+
+\usepackage{tcolorbox}
+\tcbuselibrary{skins,breakable}
+
+\pagestyle{myheadings}
+
+\newcommand{\eps}{\varepsilon}
+\usepackage[OT2,T1]{fontenc}
+\DeclareSymbolFont{cyrletters}{OT2}{wncyr}{m}{n}
+\DeclareMathSymbol{\Sha}{\mathalpha}{cyrletters}{"58}
+
+\markright{Popović\hfill Applied Analysis\hfill}
+
+
+\title{University of Vienna\\ Faculty of Mathematics\\
+\vspace{1cm}Applied Analysis Problems
+}
+\author{Milutin Popovic}
diff --git a/appl_ana/sesh1/main.pdf b/appl_ana/sesh1/main.pdf
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diff --git a/appl_ana/sesh1/main.tex b/appl_ana/sesh1/main.tex
@@ -1,387 +0,0 @@
-\documentclass[a4paper]{article}
-
-
-\usepackage[T1]{fontenc}
-\usepackage[utf8]{inputenc}
-\usepackage{mlmodern}
-
-%\usepackage{ngerman} % Sprachanpassung Deutsch
-
-\usepackage{graphicx}
-\usepackage{geometry}
-\geometry{a4paper, top=15mm}
-
-\usepackage{subcaption}
-\usepackage[shortlabels]{enumitem}
-\usepackage{amssymb}
-\usepackage{amsthm}
-\usepackage{mathtools}
-\usepackage{braket}
-\usepackage{bbm}
-\usepackage{graphicx}
-\usepackage{float}
-\usepackage{yhmath}
-\usepackage{tikz}
-\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
-\usetikzlibrary{calc,decorations.markings}
-
-%\usepackage[backend=biber, sorting=none]{biblatex}
-%\addbibresource{uni.bib}
-
-\usepackage[framemethod=TikZ]{mdframed}
-
-\tikzstyle{titlered} =
- [draw=black, thick, fill=white,%
- text=black, rectangle,
- right, minimum height=.7cm]
-
-
-\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
-\usepackage[parfill]{parskip}
-\usepackage{lipsum}
-
-
-\usepackage{tcolorbox}
-\tcbuselibrary{skins,breakable}
-
-\pagestyle{myheadings}
-
-\newcommand{\eps}{\varepsilon}
-
-\markright{Popović\hfill Applied Analysis\hfill}
-
-
-\title{University of Vienna\\ Faculty of Mathematics\\
-\vspace{1cm}Applied Analysis Problems
-}
-\author{Milutin Popovic}
-
-\begin{document}
-\maketitle
-\tableofcontents
-
-\section{Sheet 1}
-
-\subsection{Fall from high}
-We consider a free fall ($\dot{x}(t=0)=0$) of an object with mass $20\
-\text{kg}$ from a height $x(0) = h = 20\; \text{km}$, such that the
-gravitational force depends on the height $x(t)$ in the following way
-\begin{align}\label{eq: free fall}
- \ddot{x}(t) = -g\frac{R^2}{(x(t) + R)^2},
-\end{align}
-where $R$ is the radius of the earth $R \approx 6000\; \text{km}$ and $g
-\approx 9.81\ \frac{m}{s^2}$ is the gravitational acceleration on the surface
-of the earth. For this problem there are two possible non-dimensionalisations,
-but first let us rewrite the variables in terms of non-dimensional variables
-and some dimensional constants, a priori let
-\begin{align}
- t &= t_c \tau \;\;\; \text{and}\\
- x &= x_c \xi.
-\end{align}
-With the above ansatz we get the following second derivative in
-time
-\begin{align}
- \frac{d^2}{dt^2} &= \frac{1}{t_c^2}\frac{d^2}{d\tau^2} \\
- \Rightarrow \frac{d^2x}{dt^2} &= \frac{x_c}{t_c^2}
- \frac{d^2\xi}{d\tau^2},
-\end{align}
-and thus the initial conditions can be rewritten as
-\begin{align}
- \xi(0) = \frac{h}{x_c},\\
- \dot{\xi} = 0.
-\end{align}
-Now we can rewrite the equation of the free fall in \ref{eq: free fall} in
-terms of $\xi(\tau)$ as
-\begin{align}
- \frac{x_c}{gt_c^2} \ddot{\xi} = -\frac{1}{(\frac{x_c}{R}\xi +1)^2}.
-\end{align}
-Thereby we have three dimensional constants constants $\Pi_1, \Pi_2, \Pi_3$,
-as follows
-\begin{align}
- \Pi_1 = \frac{x_c}{R}, \;\;\;\; \Pi_2 = \frac{h}{x_c}, \;\;\;\;
- \Pi_3 = \frac{x_c}{gt_c^2}.
-\end{align}
-
-The first scaling is done by reducing $\Pi_1$ and $\Pi_3$ to 1, by setting
-\begin{align}
- x_c = R, \;\;\;\; t_c = \sqrt{\frac{R}{g}},
-\end{align}
-reformulating the initial problem in equation \ref{eq: free fall} to
-\begin{align}
- &\ddot{\xi} = -\frac{1}{(\xi + 1)^2},\;\;\;\;
- \text{with} \nonumber\\
- &\xi(0) = \frac{h}{R}, \;\;\;\; \dot{\xi}(0) = 0.
-\end{align}
-Reducing the problem, meaning if $\frac{h}{R} \rightarrow 0$ makes the first
-initial condition $\xi(0) \rightarrow 0$. We can conclude that this scaling
-is bad since it changes the initial condition in the reduced problem.
-
-The second scaling option reduces $\Pi_2$ and $\Pi_3$ to 1, by setting
-\begin{align}
- x_c = h, \;\;\;\; t_c = \sqrt{\frac{h}{g}},
-\end{align}
-reformulating the initial problem in equation \ref{eq: free fall} to
-\begin{align}
- &\ddot{\xi} = -\frac{1}{(\frac{h}{R}\xi + 1)^2},\;\;\;\;
- \text{with} \nonumber\\
- &\xi(0) = 1, \;\;\;\; \dot{\xi}(0) = 0.
-\end{align}
-By letting $R \rightarrow \infty$ we get the following reduced problem
-\begin{align}\label{eq: free fall reduced}
- \ddot{\xi} = -1.
-\end{align}
-Integrating and solving for $\xi(\tau = T\sqrt{\frac{g}{h}}) = 0$ for when the
-object hits the ground we get a familiar solution
-\begin{align}
- T = \sqrt{\frac{2h}{g}}
-\end{align}
-Note that in the reduced problem the time until the object hits the ground is
-\textbf{(much) shorter} since the acceleration is at its maximum $\ddot{x}(t) =
-g$ for all $t$. Yet in the original problem the acceleration (gravitation
-force) \textbf{increases} as the object comes \textbf{closer} to earth . For
-instance, if we let an object fall down from the height $h = R$ then its
-gravitational force (acceleration) at that height would be $\ddot{x}(0) =
-g/2$ and upon landing on earth the gravitational force $\ddot{x}(T) = g$,
-while in the reduced solution its gravitational force would be $\ddot{x}(t) =
-g$ for all $t$.
-
-Additionally we can calculate the velocity at impact we need to integrate the
-reduced problem \ref{eq: free fall reduced} once and put in the initial
-condition
-\begin{align}
- \dot{\xi}(\tau = \frac{T}{t_c}) &= -\tau = -\sqrt{2} \\
- \text{and} \;\;\; \dot{x} &= \frac{x_c}{t_c}\dot{\xi} =
- \sqrt{gh}\; \dot{\xi}\\
- \Rightarrow \dot{x}(T) &= -\sqrt{2gh},
-\end{align}
-The result is exactly the same as we would get from energy conservation
-\begin{align}
- \frac{m}{2}\dot{x}^2 = mgh \quad \Rightarrow \quad \dot{x} = \sqrt{2gh}.
-\end{align}
-The vertical throw allows for an additional scaling because the
-initial conditions are different, $x(0) = 0$ and $\dot{x}(0) = v$. Thus
-the solution too.
-
-To summarize, the assumptions that used for modeling and simplifying the
-equation are
-\begin{itemize}
- \item no relativistic influence,
- \item closed system, no outside influence (gravitation of the sun, air
- resistance),
- \item spherical symmetry of the earth (thereby center of mass can be
- set in the middle of earth).
-\end{itemize}
-By looking at our assumptions a question arises:\textbf{Is it a good
-approximation to replace the attractive force of the earth by the attraction
-of the whole mass concentrated at the center?}.
-
-To answer this question more or less simply we look at the Poisson's equation
-for gravity,
-\begin{align}
- \ddot{\vec{x}}(\vec{r}) = -\nabla \phi(\vec{r}) \\
- \Delta \phi = 4\pi G\varrho(\vec{r}).
-\end{align}
-for a gravitational potential $\phi$ and the mass density of earth
-$\varrho$. We assume that \textbf{the earth can be approximated by a sphere}
-and then we integrate both sides along the sphere (and use the Gaussian law
-for integration)
-\begin{align}
- \int_{S} \nabla \ddot{\vec{x}}\ dS =
- \int_{\partial S}\ddot{\vec{x}}\ d\vec{s} = -4\pi
- G \int_S\varrho(\vec{r})\ ds = -4\pi GM.
-\end{align}
-Obviously $\ddot{\vec{x}}$ and $d\vec{s}$ point in the same direction. We
-choose (rotate) the coordinate system such hat $\ddot{\vec{x}} =
-\ddot{x}\ \mathbf{\hat{n}}$ and $d\vec{s} = \mathbf{\hat{n}}\ ds$, thereby
-we get
-\begin{align}
- &\ddot{x}\int_{\partial S} ds = 4\pi r^2 \ddot{x},\\
- \Rightarrow &\ddot{x} = -\frac{GM}{r^2}.
-\end{align}
-The further derivation to get the exact equation of motion as in \ref{eq:
-free fall}, we have to keep in mind that $r = x + R$, because by our
-assumptions we are not in the sphere only outside or on the border $R$.
-Lastly by reformulating the constants $gR^2 = GM$ gets us to our equation of
-motion
-\begin{align}
- \ddot{x}(t) = -g\frac{R^2}{(x(t) + R)^2}.
-\end{align}
-
-\subsection{Scaling The Van der Pol equation}
-The Van der Pol equation is a perturbation of the oscillation equation
-\begin{align}\label{eq: vanderpol}
- LC\frac{d^2I}{dt^2} + (-g_1C +3g_3CI^2)\frac{dI}{dt} = -I
-\end{align}
-with initial conditions
-\begin{align}\label{eq: van initial}
- I(0) = I_0,\;\;\;\; \dot{I}(0) = 0.
-\end{align}
-where $I(t)$ is the current at a time $t$, $C$ is the capacity, $L$ is the
-inductivity and $g_1, g_3$ are some parameters. The units of all the
-parameters are
-\begin{align}
- [LC] &= s^2\\
- [g_1C] &= s\\
- [g_3C] &= sA^{-2}
-\end{align}
-The oscillation equation is
-\begin{align}
- CL\ddot{I} + I = 0.
-\end{align}
-Solvable by the exponential ansatz of $I = Ae^{\lambda t}$, where $\lambda=
-\pm i \sqrt{\frac{1}{LC}}$, thereby
-\begin{align}
- I(t) = A_1 e^{i\sqrt{\frac{1}{LC}}t} + A_2 e^{-i\sqrt{\frac{1}{LC}}t}.
-\end{align}
-With the initial conditions in equation \ref{eq: van initial} we get $A_1 =
-A_2$ and thus the solution to the oscillation equation is
-\begin{align}
- I(t) = I_0\cos(\frac{t}{\sqrt{LC}})
-\end{align}
-Now that we know the reduced problem and the solution to it, we may work with
-the Van-Der-Pol equation \ref{eq: vanderpol}, by determining all possible
-non-dimensionalisations. Let us begin by setting
-\begin{align}
- I(t) = I_c\psi,\\
- t = t_c \tau,
-\end{align}
-where $I_c$ and $t_c$ have the dimension of $I(t)$ and $t$ accordingly and
-$\psi(\tau)$ and $\tau$ are dimensionless
-The \textbf{first} and second derivative in time is
-\begin{align}
- \frac{d}{dt} &= \frac{1}{t_c}\frac{d}{d\tau}\\
- \frac{d^2}{dt^2} &= \frac{1}{t_c^2}\frac{d^2}{d\tau^2}.
-\end{align}
-We can rewrite the Van-Der-Pol equation in terms of $\psi$ and $\tau$
-\begin{align}
- &\frac{LC}{t_c^2}\ddot{\psi} - \left(\frac{3g_3I_c^2}{g_1}\psi^2 -
- 1\right)\frac{g_1C}{t_c}\dot{\psi}= -\psi\\
- &\psi(0) = \frac{I_0}{I_c} \;\;\;\; \dot{\psi}(0) = 0
-\end{align}
-There are a total of four constants that we can eliminate
-\begin{align}
- \Pi_1 &= \frac{I_0}{I_c}, \qquad
- \Pi_2 = \frac{LC}{t_c^2},\nonumber\\
- \Pi_3 &= \frac{3g_3I_c^2}{g_1}, \qquad
- \Pi_4 = \frac{g_1C}{t_c}.
-\end{align}
-The \textbf{first} scaling is
-\begin{align}
- I_c = \sqrt{\frac{g_1}{3g_3}},\;\;\; t_c=\sqrt{LC}.
-\end{align}
-Thereby we get the following problem
-\begin{align}
- \ddot{\psi} + (\psi^2 - 1)\eps \dot{\psi} = -\psi, \qquad \psi(0) =
- \sqrt{\frac{3g_3}{g_1}}I_0,
-\end{align}
-where $\eps = g_1\sqrt{\frac{C}{L}}$.
-
-The \textbf{second} scaling is
-\begin{align}
- I_c = \sqrt{\frac{g_1}{3g_3}},\;\;\; t_c=g_1C.
-\end{align}
-Thereby we get the following
-\begin{align}
- \eps \psi'' +(\psi^2 +1)\psi' = -\psi, \qquad \psi(0) =
- \sqrt{\frac{3g_3}{g_1}}I_0,
-\end{align}
-where $\eps = \frac{L}{g_1^2C}$. We could also consider scaling $I_c = I_0$
-with $t_c = \sqrt{LC}$ or $t_c = g_1C$ but they wouldn't develop significant
-model hierarchies like the above two scaling.
-\subsection{Scale the Schrödinger Equation}
-The well known Schrödinger equation that describes quantum physics of the one
-particle system is
-\begin{align}
- &i\hbar \partial_t\psi = -\frac{\hbar^2}{2m}\Delta \psi + V\psi \nonumber\\
- &\psi(t=0) =\psi_0
-\end{align}
-where $\hbar$ is the reduced Plank constant, $\psi=\psi(x, t)$ the wave function,
-$m$ the mass and $V = V(x)$ the potential in which the wave function is. The
-dimensions are
-\begin{align}
- [\hbar] = js, \;\;\;\; [V] = j, \;\;\;\; [\psi]= m^{-d/2}
-\end{align}
-where $d$ is the spacial dimension. The standard scaling ansatz is
-\begin{align}
- &\psi = \psi_c \phi \\
- &t = t_c \tau \;\;\;\; x = x_c \xi,
-\end{align}
-by that we get the following derivatives in time and in space
-\begin{align}
- \partial_{x_i} &=\frac{1}{x_{(i)c}} \partial_{\psi_i} \\
- \partial^2_{x_i} &=\frac{1}{x_{(i)c}^2} \partial_{\psi_i}^2\\
- \partial_{t} &=\frac{1}{t_c} \partial_{\psi_i} \\
-\end{align}
-for $i = 1, 2, 3$, or depending on the dimension we are dealing with.
-
-Let us consider $x\in \mathbb{R}^3$ and $V = 0$ to scale the equation. First
-we now have
-\begin{align}
- i \partial_t\psi = -\frac{\hbar}{2m}\Delta \psi
-\end{align}
-with the initial condition $\phi(0) = \frac{\psi_0}{\psi_c}$. With our scaling the equation turns out to be
-\begin{align}
- \frac{i\hbar t_c}{2m}\frac{1}{||\vec{x}_c||^2}\Delta_{\vec{\xi}}\ \phi =
- \partial_\tau\ \phi.
-\end{align}
-The constants we get are
-\begin{align}
- \Pi_1 = \frac{t_c\hbar}{2m}\frac{1}{||\vec{x}_c||^2}, \;\;\;\; \Pi_2 =
- \frac{\psi_0}{\psi_c}.
-\end{align}
-
-The simple choice of
-\begin{align}
- \frac{1}{||\vec{x}_c||^2} = 1, \;\;\;\; \psi_c = \psi_0, \;\;\;\; t_c =
- \frac{2m}{\hbar}||\vec{x}_c||^2,
-\end{align}
-simplifies the Schrodinger equation without the potential to
-\begin{align}
- i\Delta_{\vec{\xi}}\ \phi = \partial_\tau \phi,
-\end{align}
-with the initial condition $\phi(\tau=0) = 1$.
-.
-
-Now consider $V = 0$, $x\in[0, L]$ and $t \in [0, T]$, the Schrodinger
-equation is the same only with one spacial dimension as above, we can set
-\begin{align}
- \psi_c = \psi_0, \;\;\;\; x_c =L, \;\;\;\; t_c = \frac{2mL^2}{\hbar}.
-\end{align}
-Thus we get
-\begin{align}
- i\partial_{\xi}^2 \phi = \partial_\tau \phi,
-\end{align}
-with the initial condition $\phi(\tau=0) = 1$, where $\xi \in [0, 1]$ and
-$\tau \in [0, \frac{\hbar T}{2mL^2}]$.
-.
-
-In the last example let us consider the quantum harmonic oscillator
-represented by the potential $V(x) = m\omega^2 x^2$ for $x\in \mathbb{R}$,
-where $\omega$ is the frequency. The equation is the following
-\begin{align}
- i \hbar \partial_t \psi = -\frac{\hbar^2}{2m}\partial^2_x \psi
- +m\omega^2x^2 \psi.
-\end{align}
-By inserting the standard scaling ansatz we get
-\begin{align}
- i\partial_\tau \phi = -\frac{t_c\hbar}{2mx_c^2}\partial_\xi^2 \phi
- +\frac{t_cm\omega^2x_c^2}{\hbar} \xi^2 \phi,
-\end{align}
-The dimensional constants are
-\begin{align}
- \Pi_1 = \frac{t_0\hbar}{mx_c^2},\;\;\;\;\Pi_2 =
- \frac{m\omega^2x_c^2t_c}{\hbar},\;\;\;\; \Pi_3 = \frac{\psi_0}{\psi_c}.
-\end{align}
-The choice of scaling is
-\begin{align}
- \psi_c = \psi_0, \;\;\;\; t_c = \frac{1}{\omega}, \;\;\;\; x_c =
- \sqrt{\frac{\hbar}{m\omega}}.
-\end{align}
-Thereby getting the following problem
-\begin{align}
- i\partial_\tau \phi = -\frac{1}{2} \partial_\xi^2 \phi +\xi^2 \phi
-\end{align}
-with $\phi(\tau = 0) = 1$.
-
-%\printbibliography
-\end{document}
diff --git a/appl_ana/sesh2/main.pdf b/appl_ana/sesh2/main.pdf
Binary files differ.
diff --git a/appl_ana/sesh2/main.tex b/appl_ana/sesh2/main.tex
@@ -1,302 +0,0 @@
-\documentclass[a4paper]{article}
-
-
-\usepackage[T1]{fontenc}
-\usepackage[utf8]{inputenc}
-\usepackage{mlmodern}
-
-%\usepackage{ngerman} % Sprachanpassung Deutsch
-
-\usepackage{graphicx}
-\usepackage{geometry}
-\geometry{a4paper, top=15mm}
-
-\usepackage{subcaption}
-\usepackage[shortlabels]{enumitem}
-\usepackage{amssymb}
-\usepackage{amsthm}
-\usepackage{mathtools}
-\usepackage{braket}
-\usepackage{bbm}
-\usepackage{graphicx}
-\usepackage{float}
-\usepackage{yhmath}
-\usepackage{tikz}
-\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
-\usetikzlibrary{calc,decorations.markings}
-
-%\usepackage[backend=biber, sorting=none]{biblatex}
-%\addbibresource{uni.bib}
-
-\usepackage[framemethod=TikZ]{mdframed}
-
-\tikzstyle{titlered} =
- [draw=black, thick, fill=white,%
- text=black, rectangle,
- right, minimum height=.7cm]
-
-
-\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
-\usepackage[parfill]{parskip}
-\usepackage{lipsum}
-
-
-\usepackage{tcolorbox}
-\tcbuselibrary{skins,breakable}
-\newcommand{\eps}{\varepsilon}
-\pagestyle{myheadings}
-
-\markright{Popović\hfill Applied Analysis\hfill}
-
-
-
-\title{Applied Analysis Problems}
-\author{Milutin Popović}
-
-\begin{document}
-\maketitle
-\tableofcontents
-
-\section{Sheet 2}
-\subsection{Problem 4}
-We consider a quadratic equation with two ways to perturb it by $\eps$:
-\begin{align}
- x^2 + 2\eps x -1 = 0, \label{eq: (1)}\\
- \nonumber \\
- \eps x^2 + 2x - 1 = 0.\label{eq: (2)}
-\end{align}
-Equation \ref{eq: (2)} is singular, because the reduced problem ($\eps
-\rightarrow 0$) has only one solution at $x = \frac{1}{2}$. While the reduced
-problem in \ref{eq: (1)} has two solutions for $x = \pm 1$, which is the case
-for this non reduced equation. Let us thereby calculate the asymptotic
-expansion of the regular case up to $O(\eps^2)$, we take the ansatz for the
-asymptotic expansion
-\begin{align}\label{eq: p4 ansatz}
- x_\eps = x_0 + \eps x_1 + \eps^2 x_2 + O(\eps^3).
-\end{align}
-By substituting $x_\eps$ into \ref{eq: (1)} and factoring out the orders of
-$\eps$ we get
-\begin{align}
- \eps^0 (x_0^2 - 1) + \eps^1(2x_0 + 2x_0x_1) + \eps^2(x_1^2+2x_2x_0
- +2x_1) + O(\eps^3) = 0
-\end{align}
-By solving the equations in order of $\eps$, for the coefficients
-$x_0$, $x_1$ and $x_2$ we get
-\begin{align}
- x_0 = \pm 1, \;\;\;\; x_1 = -1, \;\;\;\; x_2 = \pm \frac{1}{2}.
-\end{align}
-By substituting into the equation \ref{eq: p4 ansatz} we get
-\begin{align}
- x_\eps = \pm 1 - \eps \pm \frac{1}{2} \eps + O(\eps^3).
-\end{align}
-For $\eps = 0.001$ we get
-\begin{align}
- &x_\eps = -1.0010005 + O(\eps^3), &x_\eps = 0.9990005 + O(\eps^3),\\
- &x_\eps = -1.001 + O(\eps^2), &x_\eps = 0.999 + O(\eps^2).
-\end{align}
-\subsection{Problem 5}
-Consider the following equations
-\begin{align}
- \label{eq: p5 1}\eps y' + y = x \;\;\;\;\;\; y(0) = 1\\
- \label{eq: p5 2}\eps y' + y = x \;\;\;\;\;\; y(0) = 0\\
- \nonumber\\
- \label{eq: p5 3}\eps y' + y = x \;\;\;\;\;\; y(0) = \eps\\
- \label{eq: p5 4}\eps^2 y' + y = x \;\;\;\;\;\; y(0) = \eps\\
- \nonumber\\
- \label{eq: p5 5}y' + \eps y = x \;\;\;\;\;\; y(0) = 1\\
- \label{eq: p5 6}y' + y = \eps x \;\;\;\;\;\; y(0) = 1
-\end{align}
-We will go through the equations and elaborate on if the perturbation is
-regular or singular, if regular we will compute the asymptotic expansion up
-to second order.
-Let us begin with equation \ref{eq: p5 1}. By the first look, the reduced
-problem does not agree with the boundary condition
-\begin{align}
- y_0 = x \;\;\;\;\; y_0(0) = 1,
-\end{align}
-is a contradiction in $y_0(0) = 0 \neq 1$, thereby equation \ref{eq: p5 1} is
-\textbf{singularly perturbed}.
-
-The reduced problem of equation \ref{eq: p5 2} on the other hand agrees with
-the boundary condition, since
-\begin{align}
- y_0 = x \;\;\;\;\; y_0(0) = 0.
-\end{align}
-But by doing the ansatz for the asymptotic expansion
-\begin{align}
- y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3),
-\end{align}
-plugging in into \ref{eq: p5 2} and separating coefficients in terms of
-$\eps$, we get
-\begin{align}
- \eps^0 (y_0 -x) + \eps^1(y_0' + y_1) + \eps^2(y_1' + y_2) + O(\eps^3) = 0
-\end{align}
-The solutions to these equations are
-\begin{align}
- y_0 = x, \;\;\;\; y_1 = 1, \;\;\;\; y_2 = 0,
-\end{align}
-which is a contradiction to the boundary condition of $y_1(0) = 1 \neq 0$.
-Thereby we can conclude that equation \ref{eq: p5 2} is \textbf{singularly
-perturbed}.
-
-Next up is equation \ref{eq: p5 3}, where by the asymptotic expansion the
-first order coefficient of $\eps$, $y_2$, has the boundary condition $y_2(0)
-= 0$. But by applying the ansatz of the asymptotic expansion and plugging
-into the equation we get
-\begin{align}
- \eps^2(y_0 - x) + \eps^1 (y_0' + y_1) + \eps^2(y_1' + y_2) + O(\eps^3) =
- 0.
-\end{align}
-Solving these equations we get
-\begin{align}
- y_0 = 0, \;\;\;\; y_1 = 1 \;\;\;\; y_2 = 0 ,
-\end{align}
-which is a contradiction $y_1(0) = 1 \neq 0 $, thus the equation \ref{eq: p5
-3} is \textbf{singularly perturbed}.
-
-The next equation \ref{eq: p5 4} is also singularly perturbed, we
-can see this by plugging the asymptotic expansion into the equation
-\begin{align}
- \eps^0 ( y_0 - x) + \eps^1(y_1) + \eps^2(y_0' + y_2) = O(\eps^3),
-\end{align}
-solving for the coefficients we get
-\begin{align}
- y_0 = x, \;\;\;\; y_1 = 0, \;\;\;\;\; y_2 = -1,
-\end{align}
-which is contradiction by the boundary condition $y_2(0) = -1 \neq 0$,
-thereby \ref{eq: p5 4} is \textbf{singularly perturbed}.
-
-Equation \ref{eq: p5 5} on the first sight does not indicate for any
-contradictions, we may plug the ansatz of the asymptotic expansion into the
-equation and see what happens
-\begin{align}
- \eps^0(y_0 -x) + \eps^1(y_1' + y_0) + \eps^2(y_2' +y_1) + O(\eps^2) = 0,
-\end{align}
-with the initial conditions $y_0(0) = 1$, $y_1(0) = y_2(0) = 0$.
-\begin{align}
- y_0 = \frac{x^2}{2} + 1, \;\;\;\;
- y_1 = -\frac{x^3}{6} + x, \;\;\;\;
- y_2 = \frac{x^4}{24} + \frac{x^2}{2}.
-\end{align}
-Finally we get
-\begin{align}
- y_\eps(x) = (\frac{x^2}{2}+1) + \eps(-\frac{x^3}{6} -x)
- +\eps^2(\frac{x^4}{24} + \frac{x^2}{2}) + O(\eps^3).
-\end{align}
-Thereby we can conclude that \ref{eq: p5 5} is \textbf{regularly perturbed}.
-
-The last equation \ref{eq: p5 6} is also regular, let us do the asymptotic
-expansion of the equation and order the equation in orders of $\eps$.
-\begin{align}
- \eps^0(y_0' + y_0) + \eps^1(y_1' + y_1 -x) + \eps^2(y_2' + y_2)+
- O(\eps^3) = 0 .
-\end{align}
-by solving these differential equations with the boundary conditions $y_0(0)
-= 1$, $y_1(0) = y_2(0) = 0$ we get
-\begin{align}
- y_0 = e^{-x} \;\;\;\; y_1 = (x-1) + e^{-x} \;\;\;\; y_2 = 0.
-\end{align}
-The equation we get
-\begin{align}
- y_\eps(x) = e^{-x} + \eps(x-1 +e^{-x}) + O(\eps^3).
-\end{align}
-Thereby we can conclude that the last equation \ref{eq: p5 6} is
-\textbf{regularly perturbed}.
-\subsection{Problem 6}
-In this section we will calculate the asymptotic expansion of a regularly
-perturbed equation in two ways, by doing the regular expansion ansatz and by
-substituting and expanding in terms of $\eps$. The ordinary differential
-equation we are dealing with is
-\begin{align}
- y' = -y + \eps y^2 \;\;\;\;\; y(0) = 1,
-\end{align}
-where $t > 0$ and $0 < \eps \ll 1$. The standard expansion ansatz is
-\begin{align}
-y_\eps(x) = y_0 + \eps y_1 + \eps^2 y_2 + O(\eps^3).
-\end{align}
-The ODE then expands to
-\begin{align}
- \eps^0(y_0' + y_0) + \eps(y_1' + y_1 - y_0^2) + \eps^2(y_2' + y_2 -
- 2y_0y_1) + O(\eps^3) = 0.
-\end{align}
-Equations in order of $\eps$ and $\eps^2$ are non-homogeneous ODE's. The
-solution to these three coefficients with the boundary conditions $y_0(0) =
-1$, $y_1(0) =
-y_2(0) = 0$ we get
-\begin{align}
- y_0 = e^{-x}, \;\;\;\; y_1 = -e^{-2x} + e^{-x}, \;\;\;\; y_2 = e^{-3x} -
- 2e^{-2x} + e^{-x}.
-\end{align}
-The expansion of $y$ is then
-\begin{align}
- y_\eps (x) = e^{-x} + \eps(-e^{-2x} + e^{-x}) + \eps^2(e^{-3x} - 2e^{-2x}
-+ e^{-x}) + O(\eps^3). \end{align}
-
-The second ansatz, considers the substitution $z = \frac{1}{y}$, by
-calculating the first derivative and substituting the original problem we
-get
-\begin{align}
- z' &= \frac{-y'}{y^2} = \frac{y-\eps y^2}{y^2} = \frac{1}{y} - \eps = z -
- \eps. \\
- z(0) &= \frac{1}{y(0)} = 1.
-\end{align}
-The solution is
-\begin{align}
- z(x) = \eps + (1-\eps) e^x.
-\end{align}
-By substituting this into $y = \frac{1}{z}$ and expanding we get
-\begin{align}
- y(x) &= \frac{1}{\eps+(1-\eps)e^x} = e^{-x} \frac{1}{1 - (1- e^{-x})\eps}
- \\
- &= e^{-x} \sum_{n\geq 0} \eps^n(1-e^{-x})^n.
-\end{align}
-which is the geometric series.
-\subsection{Problem 7}
-The last problem consists of a perturbation of a partial differential
-equation (heat equation).
-\begin{align}
- &\partial_t u(x, t) + \partial_x^2 u(x,t) - \eps u(x, t)^2 = 0
- &x\in (0, 1),\; t>0,\\
- &u(x, 0) = \tilde{u}_0(x) &x\in(0, 1), \\
- &u(0, t) = u(1, t) = 0 & t>0.
-\end{align}
-The problem is regular because the reduced solution is the regular heat
-equation in the one special dimension on $x\in (0, 1)$, we know this is
-solvable. By doing the expansion ansatz we can derive the first equations
-for the first three terms, the ansatz is always the same
-\begin{align}
- u_\eps = u_0 + \eps u_1 + \eps^2 u_2 + O(\eps^3).
-\end{align}
-Plugging this into the perturbed problem problem and factoring out the terms
-in the order of $\eps$ we get
-\begin{align}
- &\eps^0 (\partial_t u_0 + \partial_x^2 u_0) + \\
- &\eps^1 (\partial_t u_1 + \partial_x^2 u_1 - u_0^2) +\\
- &\eps^2 (\partial_t u_2 + \partial_x^2 u_2 - 2u_1u_0) + O(\eps^3) = 0.
-\end{align}
-
-We can solve the reduced problem with the initial condition $\tilde{u}_0 =
-\sin(\pi x)$ by separation of variables. Setting $u(x, t) = \psi(x) \phi(t)$
-and substituting into the equation we get two ordinary differential equation
-\begin{align}
- \underbrace{\frac{\psi_{xx}}{\psi}}_{=k}
- +\underbrace{\frac{\phi_t}{\phi}}_{=-k} = 0,
-\end{align}
-for some $k$. Solving these two by the exponential ansatz.
-\begin{align}
- \psi(x) &= A_1 e^{\sqrt{k} x} +A_2 e^{-\sqrt{k} x},\\
- \phi(t) &= A_3 e^{-kt}.
-\end{align}
-With the initial condition we get the conditions that
-\begin{align}
- A_1A_3 &= -A_2 A_3,\\
- k &= \pi^2
-\end{align}
-we choose $A_1 = A_3 = 1$ , $A_2 = -1$. We get the following solution to the
-PDE
-\begin{align}
- u(x, t) = \psi(x)\phi(t) = \sin(\pi x) e^{-\pi^2 t}.
-\end{align}
-
-%\printbibliography
-\end{document}
diff --git a/appl_ana/sesh3/main.pdf b/appl_ana/sesh3/main.pdf
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diff --git a/appl_ana/sesh3/main.tex b/appl_ana/sesh3/main.tex
@@ -1,250 +0,0 @@
-\documentclass[a4paper]{article}
-
-
-\usepackage[T1]{fontenc}
-\usepackage[utf8]{inputenc}
-\usepackage{mlmodern}
-
-%\usepackage{ngerman} % Sprachanpassung Deutsch
-
-\usepackage{graphicx}
-\usepackage{geometry}
-\geometry{a4paper, top=15mm}
-
-\usepackage{subcaption}
-\usepackage[shortlabels]{enumitem}
-\usepackage{amssymb}
-\usepackage{amsthm}
-\usepackage{mathtools}
-\usepackage{braket}
-\usepackage{bbm}
-\usepackage{graphicx}
-\usepackage{float}
-\usepackage{yhmath}
-\usepackage{tikz}
-\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
-\usetikzlibrary{calc,decorations.markings}
-
-%\usepackage[backend=biber, sorting=none]{biblatex}
-%\addbibresource{uni.bib}
-
-\usepackage[framemethod=TikZ]{mdframed}
-
-\tikzstyle{titlered} =
- [draw=black, thick, fill=white,%
- text=black, rectangle,
- right, minimum height=.7cm]
-
-
-\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
-\usepackage[parfill]{parskip}
-\usepackage{lipsum}
-
-
-\usepackage{tcolorbox}
-\tcbuselibrary{skins,breakable}
-\newcommand{\eps}{\varepsilon}
-\pagestyle{myheadings}
-
-\markright{Popović\hfill Applied Analysis\hfill}
-
-
-
-\title{Applied Analysis Problems}
-\author{Milutin Popović}
-
-\begin{document}
-\maketitle
-\tableofcontents
-
-\section{Sheet 3}
-\subsection{Problem 8}
-Let us look at functions $f: \mathcal{D} \mapsto \mathbb{R}$ that show
-boundary layer behavior at the following manifolds.
-
-The \textbf{first} for $\mathcal{D} = \mathbb{R}^2$ and $S = \{0\}$ we have a
-function e.g.
-\begin{align}
- f_{\eps}(x, y) = e^{-\frac{x}{\eps}} + y,
-\end{align}
-with the reduced equation
-\begin{align}
- \lim_{\eps \rightarrow 0} f_{\eps}(x, y) =
- \begin{cases}
- y \;\;\;\;\;\;\;\;\;\; x > 0\\
- 1+y \;\;\;\; x = 0\\
- \end{cases}
-\end{align}
-
-The \textbf{second} example is $\mathcal{D} = \mathbb{R}^n$ and $S = \{|x| = 1\}$.
-\begin{align}
- f_\eps(x_1,\dots,x_n) = \tanh\left(\frac{|x| - 1}{\eps} \right),
-\end{align}
-with the reduced equation
-\begin{align}
- \lim_{\eps \rightarrow 0} f_{\eps}(x_1,\dots, x_n) =
- \begin{cases}
- -1 \;\;\;\; |x| < 0\\
- 1 \;\;\;\;\;\;\; |x| > 0\\
- \end{cases}
-\end{align}
-
-The \textbf{third} example is $\mathcal{D} = \mathbb{R}^3$ and $S = \{x_1 =
-1\}$
-\begin{align}
- f_\eps(x_1, x_2, x_3) = \tanh\left(\frac{x_1 - 1}{\eps}\right)+x_2x_3
-\end{align}
-with the reduced equation
-\begin{align}
- \lim_{\eps \rightarrow 0} f_{\eps}(x_1,x_2,x_3) =
- \begin{cases}
- -1 + x_2x_3 \;\;\;\; x_1 < 0\\
- 1 + x_2x_3 \;\;\;\;\;\;\; x_1 > 0\\
- \end{cases}
-\end{align}
-\subsection{Problem 9}
-Consider a linear BVP
-\begin{align}
- Lu := -\eps u'' + b(x)u' + c(x)u = f(x),\\
- u(0) = u(1) = 0,
-\end{align}
-for $0 < \eps \ll \eps_0$ and $b, c, f \in C([0,1])$ with the conditions
-\begin{align}
- c(x) \geq 0, \qquad b(x) \geq \beta > 0 \qquad x\in[0, 1]
-\end{align}
-We are to show that for all $x\in[0, 1)$ the reduced solution $u_0$ of the
-above BVP satisfies
-\begin{align}
- \lim_{\eps \rightarrow 0} u_\eps(x) = u_0(x)))),
-\end{align}
-where the reduced solution $u_0$ is the solution to the following
-differential equation
-\begin{align}
- b(x)u' + c(x)u = f(x), \quad u(0) = 0.
-\end{align}
-The hint was given: Set
-\begin{align}
- w_1(x) = e^{\beta x} \quad w_2(x) = e^{-\beta\frac{1-x}{\eps}},
-\end{align}
-such that $Lw_1 \geq \gamma > 0$ for some suitable $\gamma$ and $Lw_2 \geq
-0$. Then for
-\begin{align}
- v = \pm (u_\eps - u_0), \qquad w = A\eps w_1 + B\eps w_2,
-\end{align}
-for some suitable $A, B$. The following comparison principal is applicable:
-IF
-\begin{align}
- &Lv(x) \leq Lw(x) \quad \forall x \in (0, 1) \label{eq:cond1}\\
- &v(0) \leq w(0) \label{eq:cond2}\\
- &v(1) \leq w(1) \label{eq:cond3}\\
-\end{align}
-then
-\begin{align}
- \Longrightarrow v(x) \leq w(x) \quad \forall x\in(0, 1)
-\end{align}
-which holds for $u, v \in C^2((0, 1)) \cap C([0, 1])$. Thus a boundary layer
-is possible only at $x=1$. On the other hand, for $b(x) \leq \beta < 0$ it
-follows that the boundary layer is possible only at $x=0$.
-
-We shall go through the chronological order of the conditions\ref{eq:cond1},
-\ref{eq:cond2}, \ref{eq:cond3} and check them. So for \ref{eq:cond1}
-we have that
-\begin{align}
- Lw(x) &= A\eps Lw_1(x) + B Lw_2(x) \\
- &\geq A\eps Lw_1(x) = A\eps e^{\beta x} \left(-\eps \beta^2 -
- b(x)\beta+c(x)\right)\\
- &\geq A\eps \beta e^{\beta x} \left(1-\eps\right)\\
- &\geq \eps A\beta^2 e^{\beta}(1-\eps) = \gamma > 0
-\end{align}
-And obviously
-\begin{align}
- Lv(x) \leq 0 ,
-\end{align}
-by that we have that
-\begin{align}
- Lv(x) \leq \gamma \leq Lw(x).
-\end{align}
-For the condition \ref{eq:cond2} we have
-\begin{align}
- w(0) &= A\eps w_1(0) Bw_2(0) = Be^{-\frac{\beta}{\eps}},\\
- v(0) &= \pm\left(u_\eps(0) - u_0(0)\right) = 0.
-\end{align}
-By the simple choice $B \geq 0$ we satisfy the condition
-\begin{align}
- v(0) \leq w(0).
-\end{align}
-Now for the last condition \ref{eq:cond3} we have
-\begin{align}
- w(1) &= A\eps e^\beta + B \geq A\eps e^\beta,\\
- v(1) &= \mp u_0(1) = 0.
-\end{align}
-And choose $A = \frac{\pm u(1)}{\eps} e^{-\beta}$, which satisfies the last
-condition
-\begin{align}
- v(1) \leq w(1).
-\end{align}
-Thereby we have
-\begin{align}
- &v(x) &\leq w(x)\\
- &\Rightarrow \lim_{\eps \rightarrow 0} v(x) &\leq \lim_{\eps \rightarrow
- 0} w(x) = 0\\
- &\Rightarrow \lim_{\eps \rightarrow 0} v(x) = 0
-\end{align}
-uniformly on $(0, 1)$.
-\subsection{Problem 10}
-Consider the following BVP
-\begin{align}
- -\eps u'' + (1 + x)u' + u = 2, \qquad u(0) = u(1) - 0,
-\end{align}
-for $0 < \eps \ll 1$. \textbf{Where can this problem have a boundary layer?}
-To answer this question we need to look at the reduced problem
-\begin{align}
- -(1+x)u' + u = 2.
-\end{align}
-The solution to the equation is
-\begin{align}
- \bar{u}(x) = 2 + A(x+1).
-\end{align}
-According to the boundary conditions it is unclear what the value of the
-constant is, according to $\bar{u}(0)=0$ we get $A = -2$ or according to
-$\bar{u}(1)=0$ we get $A = -1$. Ultimately this means that there exists a
-boundary layer near $x=1$ or $x=0$. We choose $x=0$ and according to this the
-local variable $\xi = x\eps^{-\alpha}$ ($x = \xi \eps^{-\alpha}$). The
-derivatives of $u$ are calculated using the chain rule
-\begin{align}
- \frac{du}{dx}&= \frac{du}{d\xi}\frac{d\xi}{dx} = \eps^{-\alpha} \dot{u}\\
- \frac{d^2u}{dx^2}&= \eps^{-\alpha} \frac{d^2u}{d\xi^2}\frac{d\xi}{dx} =
- \eps^{-2\alpha} \ddot{u}.
-\end{align}
-The BVP transforms as follows
-\begin{align}
- -\eps^{1-\alpha}\ddot{u} - \dot{u} + \eps(u - \xi\dot{u} - 2) =
- \begin{cases}
- -\ddot{u} - \dot{u} = 0 \;\;\;\;\; \alpha=1\\
- -\dot{u} = 0 \;\;\;\;\;\;\;\; 0<\alpha<1
- \end{cases}
-\end{align}
-Choosing $\alpha = 1$ for a reasonable solution
-\begin{align}
- \hat{u}(\xi) = Be^{-\xi},
-\end{align}
-which converges in the local limit (!). Thereby we have a asymptotic
-representation up to the degree of $\eps$
-\begin{align}
- u_\eps(x) &= \bar{u}(x) + \hat{u}(\psi) + O(\eps)\\
- &= 2 + A(1+x) + B e^{-\frac{x}{\eps}} + O(\eps)
-\end{align}
-And by the boundary conditions
-\begin{align}
- u_\eps(0) = 2+A+B=0, \qquad u_\eps(1) = 2+2A+B=0,
-\end{align}
-we get that the constants are
-\begin{align}
- A = -4, \qquad B = 2.
-\end{align}
-The asymptotic representation is thereby
-\begin{align}
- u_\eps(x) = 2 - 4(1+x) + 2 e^{-\frac{x}{\eps}} + O(\eps)
-\end{align}
-%\printbibliography
-\end{document}
diff --git a/appl_ana/sesh4/main.pdf b/appl_ana/sesh4/main.pdf
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diff --git a/appl_ana/sesh4/main.tex b/appl_ana/sesh4/main.tex
@@ -1,195 +0,0 @@
-\documentclass[a4paper]{article}
-
-
-\usepackage[T1]{fontenc}
-\usepackage[utf8]{inputenc}
-\usepackage{mlmodern}
-
-%\usepackage{ngerman} % Sprachanpassung Deutsch
-
-\usepackage{graphicx}
-\usepackage{geometry}
-\geometry{a4paper, top=15mm}
-
-\usepackage{subcaption}
-\usepackage[shortlabels]{enumitem}
-\usepackage{amssymb}
-\usepackage{amsthm}
-\usepackage{mathtools}
-\usepackage{braket}
-\usepackage{bbm}
-\usepackage{graphicx}
-\usepackage{float}
-\usepackage{yhmath}
-\usepackage{tikz}
-\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
-\usetikzlibrary{calc,decorations.markings}
-
-%\usepackage[backend=biber, sorting=none]{biblatex}
-%\addbibresource{uni.bib}
-
-\usepackage[framemethod=TikZ]{mdframed}
-
-\tikzstyle{titlered} =
- [draw=black, thick, fill=white,%
- text=black, rectangle,
- right, minimum height=.7cm]
-
-
-\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
-\usepackage[parfill]{parskip}
-\usepackage{lipsum}
-
-
-\usepackage{tcolorbox}
-\tcbuselibrary{skins,breakable}
-
-\pagestyle{myheadings}
-
-\markright{Popović\hfill Applied Analysis\hfill}
-
-
-\title{University of Vienna\\ Faculty of Mathematics\\
-\vspace{1cm}Applied Analysis Problems
-}
-\author{Milutin Popovic}
-
-\begin{document}
-\maketitle
-\tableofcontents
-
-\section{Sheet 4}
-
-\subsection{Fourier Series}
-The Fourier series of a $p$ periodic function $f$, integrable on
-$[-\frac{p}{2}, \frac{p}{2}]$ is
-\begin{align}
- f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(\frac{2\pi n x}{p})
- b_n sin(\frac{2\pi n x}{p})\right).
-\end{align}
-The coefficients $a_n$ and $b_n$ are called the Fourier coefficients of $f$
-and are given by
-\begin{align}
- a_n &= \frac{2}{p} \int_{-\frac{p}{2}}^{\frac{p}{2}} f(x) \sin(\frac{2\pi
- n x}{p}) dx, \;\;\;\;\; n\geq 0 \\
- b_n &= \frac{2}{p} \int_{-\frac{p}{2}}^{\frac{p}{2}} f(x) \cos(\frac{2\pi
- n x}{p}) dx, \;\;\;\;\; n\geq 1
-\end{align}
-Let us compute the Fourier series of $f(t) = t$ for $t \in [-\frac{1}{2},
-\frac{1}{2}]$. The Fourier coefficients are
-\begin{align}
- a_n &= 2\int_{-\frac{1}{2}}^{\frac{1}{2}} t \cos(2\pi n t)\ dt = 0
- \;\;\;\;\; \text{(odd: g(-t) = -g(t))},\\
- \nonumber\\
- b_n &= 2\int_{-\frac{1}{2}}^{\frac{1}{2}} t \sin(2\pi n t)\ dt = \\
- &= 2 \left(-\frac{1}{2\pi n} \cos(2\pi n
- t)\bigg|_{-\frac{1}{2}}^{\frac{1}{2}}
- +\int_{\frac{1}{2}}^{\frac{1}{2}} \frac{1}{2 \pi n}\cos(2\pi n t)\ dt
- \right) =\\
- &= -\frac{1}{\pi n}\left( -\cos(\pi n) + \frac{1}{\pi n }\sin(\pi
- n)\right) =
- \frac{\sin(\pi n) - \pi n \cos(\pi n)}{(\pi n)^2}.
-\end{align}
-Thereby the Fourier series of $f(t) = t$ is
-\begin{align}
- f(t) = \sum_{n=1}^\infty \left(\frac{\sin(\pi n) - \pi n \cos(\pi n)}{(\pi
- n)^2}\right) \sin(2\pi n t) = t
-\end{align}
-\subsection{Truncation Error}
-The truncation error of the trigonometric polynomial $(Sf_N)$ of degree $N$ is
-\begin{align}
- \sum_{|k| > N} |\hat{f}(k)|^2 = \lVert f - S_N\rVert_2^2 =
- \int_{-\frac{1}{2}}^{\frac{1}{2}} |E_N(t)|^2 dt.
-\end{align}
-Computations for $N = 3$ and $N = 9$ were done in python with a integration error of
-around $10^{-15}$, resulting in the overall truncation errors of
-\begin{align}
- \sum_{|k| > 3} |\hat{f}(k)|^2 = 0.0053,\\
- \sum_{|k| > 9} |\hat{f}(k)|^2 = 0.0143.
-\end{align}
-To achieve $\lVert E_N\rVert^2_2 < 0.1 \lVert f \rVert^2_2$, the number of
-coefficients needed are about $61$. This was done using a while loop and
-evaluating $\lVert E_N\rVert^2_2$ for $N$ until the above condition is met.
-
-\subsection{Orthonormal Bases}
-Here we will go through the most important properties of orthonormal bases.
-So let $\{b_n\}_{n\in \mathbb{N}}$ be an ONB of a vector space $\mathcal{H}$,
-then for every $x\in \mathcal{H}$ we may write
-\begin{align}
- x = \sum_{b_n} \langle b_n, x\rangle b_n,
-\end{align}
-and
-\begin{align}
- \lVert x \rVert^2 = \sum_{b_n} |\langle b_n, x\rangle|^2.
-\end{align}
-For any $x, y \in \mathcal{H}$ we can write the scalar product as
-\begin{align}
- \langle x, y\rangle = \sum_{b_n} \langle b_n, x\rangle \langle b_n,
- y\rangle,
-\end{align}
-Furthermore there exists a linear projection $\Phi\ : \mathcal{H}
-\rightarrow l^2(\{b_n\}_n)$ such that
-\begin{align}
- \langle \Phi(x), \Phi(y)\rangle = \langle x, y \rangle\;\;\; \forall x, y
- \in \mathcal{H}.
-\end{align}
-
-An example of an orthonormal basis, which spans $L^2([-\frac{p}{2},
-\frac{p}{2}])$ is $\mathcal{T}_p = \{e_n := \frac{e^{2\pi i
-\frac{n}{p}x}}{\sqrt{p}}\}_{n\in\mathbb{Z}}$. The $e_n$'s are orthonormal in
-$L^2$ which can be easily seen by using the scalar product of $L^2$, so for
-$n, m \in \mathbb{Z}$
-\begin{align}
- \langle e_n, e_m\rangle_{L^2([-\frac{p}{2}, \frac{p}{2})} &=
- \frac{1}{p}\int_{[-\frac{p}{2}, \frac{p}{2}]}e_n \cdot e_m^* \ dx=\\
- &=\frac{1}{p}\int_{[-\frac{p}{2}, \frac{p}{2}]} e^{2\pi i \frac{(n-m)}{p} x} \ dx=\\
- &=\frac{\sin(\pi (n-m))}{\pi(n-m)} =
- \begin{cases}
- 0 \;\;\;\; n\neq m\\
- 1 \;\;\;\; n=m
- \end{cases}
-\end{align}
-\subsection{Dirichlet Kernel}
-The function
-\begin{align}
- D_t(x) := \sum_{\lVert k \rVert_\infty \leq t} e_k(x), \;\;\;\;\; x\in
- \mathbb{R}^d
-\end{align}
-is called the Dirichlet Kernel. For $0 < t \in \mathbb{N}$ we have
-\begin{align}
- (S_tf)(x) = \int_{I^d} f(y) D_t(x-y) dy,
-\end{align}
-where $S_t$ represents the orthogonal projection onto the trigonometric
-polynomials $\Pi_t$ of degree $t$, by
-\begin{align}
- &S_t:\ L^1(\mathbb{T}^d) \rightarrow \Pi_t \\
- &f \mapsto \sum_{\lVert k \rVert \leq t} \langle f,
- e_k\rangle_{L^2(\mathbb{T}^d)} e_k \;\;\;\;\; k \in \mathbb{Z}^d
-\end{align}
-And furthermore the Dirichlet Kernel satisfies
-\begin{align}
- D_t(x) = \prod_{i=1}^d \frac{e_{t+1}(x_i) - e_{-t}(x_i)}{e_1(x_i) - 1}
-\end{align}
-To show the convolution property, we start off by applying the orthogonal
-projection into the trigonometric polynomials $S_t$ onto a function $f \in
-L(\mathbb{T}^d)$
-\begin{align}
- (S_tf) &= \sum_{\lvert k\rVert_\infty \leq t} \int_{I^d} f(y) e^{-2\pi i
- \langle k, y\rangle}\ dy\ e^{2\pi i\langle k, x\rangle} =\\
- &= \int_{I^d}f(y) \sum_{\lvert k\rVert_\infty \leq t} e^{2\pi i \langle
- k, (x- y)\rangle}\ dy =\\
- &= (f * D_t) (x) = \int_{I^d} f(y) D_t(x - y)\ dy.
-\end{align}
-To show the reformulation of the Dirichlet kernel, we need to simply
-calculate it directly
-\begin{align}
- \sum_{\lVert k \rVert_\infty \leq t} e^{2\pi i \langle k , x\rangle} &=
- \prod_{j=1}^d \sum_{k_j = -t}^t e^{2\pi i k_j x_j} =\\
- &= \prod_{j=1}^d e^{-2\pi i t x_j} \sum_{k_j = 0}^{2t} e^{2\pi i k_j
- x_j}=;\;\;\;\; \text{(trigonometric series)}\\
- &= \prod_{j=1}^d e^{-2\pi i t x_j} \frac{e^{2\pi i (2t + 1)x_j} -
- 1}{e^{2\pi i x_j} - 1} =\\
- &= \prod_{j = 1} \frac{e_{t+1}(x_j) - e_{-t}(x_j)}{e_1(x_j) - 1}.
-\end{align}
-%\printbibliography
-\end{document}
diff --git a/appl_ana/sesh5/main.pdf b/appl_ana/sesh5/main.pdf
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-\documentclass[a4paper]{article}
-
-
-\usepackage[T1]{fontenc}
-\usepackage[utf8]{inputenc}
-\usepackage{mlmodern}
-
-%\usepackage{ngerman} % Sprachanpassung Deutsch
-
-\usepackage{graphicx}
-\usepackage{geometry}
-\geometry{a4paper, top=15mm}
-
-\usepackage{subcaption}
-\usepackage[shortlabels]{enumitem}
-\usepackage{amssymb}
-\usepackage{amsthm}
-\usepackage{mathtools}
-\usepackage{braket}
-\usepackage{bbm}
-\usepackage{graphicx}
-\usepackage{float}
-\usepackage{yhmath}
-\usepackage{tikz}
-\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
-\usetikzlibrary{calc,decorations.markings}
-
-%\usepackage[backend=biber, sorting=none]{biblatex}
-%\addbibresource{uni.bib}
-
-\usepackage[framemethod=TikZ]{mdframed}
-
-\tikzstyle{titlered} =
- [draw=black, thick, fill=white,%
- text=black, rectangle,
- right, minimum height=.7cm]
-
-
-\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
-\usepackage[parfill]{parskip}
-\usepackage{lipsum}
-
-
-\usepackage{tcolorbox}
-\tcbuselibrary{skins,breakable}
-
-\pagestyle{myheadings}
-
-\markright{Popović\hfill Applied Analysis\hfill}
-
-
-\title{University of Vienna\\ Faculty of Mathematics\\
-\vspace{1cm}Applied Analysis Problems
-}
-\author{Milutin Popovic}
-
-\begin{document}
-\maketitle
-\tableofcontents
-
-\section{Sheet 5}
-\subsection{Fourier Transform}
-In this section we prove the linearity of the Fourier Transform $\mathcal{F}$ on
-$L^1(\mathbb{R}^d)$. For $f, g \in L^1(\mathbb{R}^d)$ and $\lambda, \mu \in
-\mathbb{R}$ the linearity condition for $\mathcal{F}$ is the following
-\begin{align}
- \mathcal{F}(\lambda f + \mu g) = \lambda \mathcal{F}(f) + \mu
- \mathcal{F}(g).
-\end{align}
-We start by using the Fourier transform definition for $x, \xi \in \mathbb{R}^d$
-\begin{align}
- \mathcal{F}(\lambda f + \mu g)(\xi) &= \int_{\mathbb{R}^d} (\lambda f(x)+
- \mu g(x)) e^{-2\pi i \langle x, \xi\rangle}\ dx =\\
- &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle} + \mu
- g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\
- &= \int_{\mathbb{R}^d} \lambda f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+
- \int_{\mathbb{R}^d} \mu
- g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\
- &= \lambda \int_{\mathbb{R}^d} f(x) e^{-2\pi i\langle x,\xi\rangle}\ dx+
- \mu \int_{\mathbb{R}^d}
- g(x) e^{-2\pi i\langle x,\xi\rangle}\ dx =\\
- &= \lambda \mathcal{F}(f)(\xi) + \mu \mathcal{F}(g)(\xi)
-\end{align}
-\subsection{Identities of the Fourier transform}
-The following are three identities of the Fourier transform
-
-\begin{table}[h!]
-\centering
-\begin{tabular}{| l | c | c |}
-\hline
- & $g(x)$ & $\hat{g}(\xi)$ \\ \hline \hline
-(1) & $f(x-x_0)$ & $e^{-2\pi ix_0 \xi} \hat{f}(\xi)$ \\ \hline
-(2) & $e^{2\pi i \xi_0 x} f(x)$ & $f(\xi - \xi_0)$ \\ \hline
-(3) & $f(ax)$ & $\frac{1}{a} \hat{f}(\frac{\xi}{a})$\\ \hline
-\end{tabular}
- \caption{Identities of the Fourier transform for $a > 0,
- \xi_0, x \in \mathbb{R}$}
-\end{table}
-We start with (1)
-\begin{align}
- \widehat{f(x-x_0)}
- &= \int_\mathbb{R} f(x-x_0) e^{-2\pi i x \xi}\ dx=
- \;\;\;\;\;\; (y = x-x_0)\\
- &= \int_\mathbb{R} f(y) e^{-2\pi i (y+x_0) \xi}\
- dy=\\
- &= e^{-2\pi i x_0 \xi} \int_\mathbb{R}f(y)e^{-2\pi i y
- \xi}\ dy=\\
- &= e^{-2\pi i x_0 \xi} \hat{f}(\xi).
-\end{align}
-For (2) we have
-\begin{align}
- \widehat{e^{2\pi i x \xi_0} f(x)}
- &= \int_\mathbb{R} e^{2\pi i x \xi_0} f(x) e^{-2\pi i x \xi}\ dx =\\
- &= \int_\mathbb{R} f(x) e^{-2\pi i x (\xi -\xi_0)}\ dx=\\
- &= \hat{f}(\xi - \xi_0).
-\end{align}
-For (3) we have
-\begin{align}
- \widehat{f(ax)}
- &= \int_\mathbb{R} f(ax) e^{-2\pi i \xi x}\ dx = \qquad \text{sub:
- $(y=ax)$}\\
- &= \int_\mathbb{R} \frac{1}{a}f(y) e^{-2\pi i \frac{\xi}{a} y}\ dy=\\
- &= \frac{1}{a} \hat{f}\left(\frac{\xi}{a}\right).
-\end{align}
-\subsection{The Box-Function}
-Consider the following Box-Function
-\begin{align}
- \Pi(x) :=
- \begin{cases}
- 1\;\;\;\;\;\; -\frac{3}{2} < x < \frac{1}{2}\\
- 0\;\;\;\;\; \text{else}
- \end{cases}
-\end{align}
-The Fourier transform of this function is
-\begin{align}
- \widehat{\Pi(x)}
- &= \int_\mathbb{R} \Pi(x) e^{-2\pi i x\xi}\ dx=\\
- &= \int_{-\frac{3}{2}}^{\frac{1}{2}} e^{-2\pi i x \xi}\ dx
- =\frac{-1}{2\pi i \xi} e^{-2\pi i x\xi}
- \bigg|_{-\frac{3}{2}}^{\frac{1}{2}}=\\
- &= \frac{1}{2\pi i \xi} \left(e^{3\pi i \xi} - e^{-\pi i \xi}\right)=\\
- &= \frac{e^{\pi i \xi}\sin(2\pi\xi)}{\pi \xi}.
-\end{align}
-
-%\printbibliography
-\end{document}
diff --git a/appl_ana/sesh6/main.pdf b/appl_ana/sesh6/main.pdf
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-\documentclass[a4paper]{article}
-
-
-\usepackage[T1]{fontenc}
-\usepackage[utf8]{inputenc}
-\usepackage{mlmodern}
-
-%\usepackage{ngerman} % Sprachanpassung Deutsch
-
-\usepackage{graphicx}
-\usepackage{geometry}
-\geometry{a4paper, top=15mm}
-
-\usepackage{subcaption}
-\usepackage[shortlabels]{enumitem}
-\usepackage{amssymb}
-\usepackage{amsthm}
-\usepackage{mathtools}
-\usepackage{braket}
-\usepackage{bbm}
-\usepackage{graphicx}
-\usepackage{float}
-\usepackage{yhmath}
-\usepackage{tikz}
-\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
-\usetikzlibrary{calc,decorations.markings}
-
-%\usepackage[backend=biber, sorting=none]{biblatex}
-%\addbibresource{uni.bib}
-
-\usepackage[framemethod=TikZ]{mdframed}
-
-\tikzstyle{titlered} =
- [draw=black, thick, fill=white,%
- text=black, rectangle,
- right, minimum height=.7cm]
-
-
-\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
-\usepackage[parfill]{parskip}
-\usepackage{lipsum}
-
-
-\usepackage{tcolorbox}
-\tcbuselibrary{skins,breakable}
-
-\pagestyle{myheadings}
-
-\markright{Popović\hfill Applied Analysis\hfill}
-
-
-\title{University of Vienna\\ Faculty of Mathematics\\
-\vspace{1cm}Applied Analysis Problems
-}
-\author{Milutin Popovic}
-
-\begin{document}
-\maketitle
-\tableofcontents
-
-\section{Sheet 6}
-\subsection{Fourier Transform of the convolution}
-Consider the function $f(x)$, which has a Fourier Transform $\hat{f}(\xi)$,
-now let us compute the Fourier transform of
-\begin{align}
- h(x) = f(3x-1) \sin(x) .
-\end{align}
-We know that the Fourier transform of the convolution is (we use somewhat of
-the inverse convolution theorem).
-\begin{align}
- \widehat{(f(3x-1)*g(x))} = \widehat{f(3x-1)} \cdot \hat{g}(\xi).
-\end{align}
-The Fourier transform of $f(3x-1)$ is simply done by substituting a new
-variable
-\begin{align}
- \widehat{f(3x-1)} = \frac{1}{3}e^{2\pi i\frac{\xi}{3}}\
- f\left(\frac{\xi}{3}\right).
-\end{align}
-The Fourier transform of $\sin(x)$ can be calculated when looking at the
-Fourier transform of the Dirac-delta function
-\begin{align}
- \widehat{\delta(ax-b)}
- &=\int_\mathbb{R} \delta(ax-b) e^{-2\pi i x \xi}\ dx
- \;\;\;\;\;\;\; (y = ax-b)\\
- &=\int_\mathbb{R} \delta(y) e^{-2\pi i (y+b)\frac{\xi}{a}}\frac{dy}{a}\\
- &=\frac{1}{a} e^{-2\pi i \xi \frac{b}{a}}.
-\end{align}
-We may plug in $\sin(x)$ in the definition of the Fourier transformation and
-observe where we can use the Dirac-delta to to the inverse Fourier transform
-\begin{align}
- \widehat{\sin(x)}
- &=\int_\mathbb{R} \sin(x)e^{-2\pi i x\xi}\ dx=\\
- &=\frac{1}{2i}\int_\mathbb{R} (e^{ix} - e^{-ix})e^{-2\pi i \xi x}\ dx\\
- &=\frac{1}{2i}\left(
- \int_\mathbb{R} e^{ix} e^{-2\pi i \xi x}\ dx+
- \int_\mathbb{R} e^{-ix} e^{-2\pi i \xi x}\ dx
- \right).
-\end{align}
-Here we may use the above formula for the Fourier transform of the Dirac
-delta. We choose $a=1$, $b= \pm \frac{1}{2\pi}$ and do some $y=-x$
-substitutions and thereby get the following result
-\begin{align}
- \widehat{\sin(x)} = \frac{1}{2i} \left(
- \delta(\xi - \frac{1}{2\pi})
- -\delta(\xi + \frac{1}{2\pi})
- \right)
-\end{align}
-The whole result is thereby
-\begin{align}
- \widehat{f(3x-1)} * \widehat{sin(x)}
- =& \frac{1}{6i} \bigg(
- e^{2\pi
- i(\frac{\xi}{3}-\frac{1}{6\pi})}\hat{f}\big(\frac{\xi}{3}-\frac{1}{6\pi}\big)-
- e^{2\pi
- i(\frac{\xi}{3}+\frac{1}{6\pi})}\hat{f}\big(\frac{\xi}{3}+\frac{1}{6\pi}\big)
- \bigg)
-\end{align}
-\subsection{More Fourier Transforms}
-Consider the function
-\begin{align}
- f(x) = e^{-|x|}
-\end{align}
-The Fourier transform of this function is
-\begin{align}
- \hat{f}(\xi)
- &=\int_\mathbb{R} e^{-|x| e^{-2\pi i x \xi}}\ dx\\
- &= \int_{-\infty}^0 e^x e^{-2\pi i x \xi}\ dx
- + \int_0^\infty e^{-x} e^{-2\pi i x \xi}\ dx=\\
- &= \frac{1}{1-2\pi i \xi} e^{(1-2\pi i \xi) x}\bigg|_{-\infty}^0+
- \frac{-1}{1+2\pi i \xi} e^{-(1+2\pi i \xi) x}\bigg|_{-\infty}^0 = \\
- &= \frac{1}{1-2\pi i \xi} + \frac{1}{1 + 2\pi i \xi} =\\
- &= \frac{2}{1+(2\pi \xi)^2}.
-\end{align}
-Let us use this result to solve the following integral
-\begin{align}
- \int_\mathbb{R} \frac{\cos(a\xi)}{(2\pi \xi)^2 + 1}\ d\xi =
- \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx,\\
-\end{align}
-where we used the fact that $\text{Re}(e^{ia\xi}) = \cos(a\xi)$ and
-$\hat{f}(\xi) = \frac{2}{1+(2\pi \xi)^2}$, thereby
-\begin{align}
- \frac{1}{2}\int_\mathbb{R} \hat{f}(\xi) \text{Re}(e^{ia\xi})\ dx
- &= \frac{1}{2}\text{Re}\left(
- \int_\mathbb{R}\hat{f}(\xi)e^{ia\xi}\ d\xi
- \right)=\\
- &= \frac{1}{2}\text{Re}\left(
- \int_\mathbb{R} \hat{f}(\xi) e^{2\pi i \frac{a}{2\pi}\xi}\ d\xi
- \right)=\\
- &= \frac{1}{2}\text{Re}\left(f(\frac{a}{2\pi})\right)=\\
- &= \frac{1}{2} e^{-\frac{|a|}{2\pi}}.
-\end{align}
-\subsection{Finite discrete Fourier transform}
-Consider $s\in \mathbb{C}^N$ with entries
-\begin{align}
- s[n] = \sin\left(2\pi\xi_0\frac{n}{N}\right),
-\end{align}
-for same $0 < \xi_0 < N$. The finite discrete Fourier transform of $s$ is
-\begin{align}
- \hat{s}[k] &= \frac{1}{N} \sum_{n=0}^{N-1} \sin\left(2\pi\xi_0\frac{n}{N}\right)
- e^{-2\pi i \frac{k}{N} n} =\\
- &=\frac{1}{2iN}\left(
- \sum_{n=0}^{N-1}e^{2\pi i \frac{n}{N}(\xi_0 -k)} - e^{-2\pi i
- \frac{n}{N}(\xi_0 +k)}
- \right).
-\end{align}
-If we consider $\xi_0 \in \mathbb{Z}$, we have
-\begin{align}
- \hat{s}[k] =
- \begin{cases}
- \frac{1}{2i}\;\;\;\;\;\; \xi_0 = k\\
- -\frac{1}{2i}\;\;\;\;\;\; \xi_0 = -k\\
- 0 \;\;\;\;\;\; \text{else}
- \end{cases}
-\end{align}
-\subsection{Discrete Matrix Notation}
-The convolution of two vectors $f, g \in \mathbb{C}^N$, can be expressed by a
-circulate matrix applied to f
-\begin{align}
- (f * g) [n] = \sum_{k=0}^{N-1} f[k] g[n-k].
-\end{align}
-Consider $g=s$, then the matrix takes the following values
-\begin{align}
- s[n-k] = s_{nk} = \sin\left(2\pi \xi_0 \frac{n-k}{N}\right).
-\end{align}
-The convolution with an impulse input $f=\delta_{0k}$, a vector that is $1$
-for $k=0$ and else 0 reads
-\begin{align}
- \sum_k s_{nk}f_k &= \sum_k s_{nk} \delta_{0k} =\\
- &= \sin\left(2\pi \xi_0 \frac{n}{N}\right).
-\end{align}
-
-%\printbibliography
-\end{document}
diff --git a/appl_ana/sesh7/main.pdf b/appl_ana/sesh7/main.pdf
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diff --git a/appl_ana/sesh7/main.tex b/appl_ana/sesh7/main.tex
@@ -1,195 +0,0 @@
-\documentclass[a4paper]{article}
-
-
-\usepackage[T1]{fontenc}
-\usepackage[utf8]{inputenc}
-\usepackage{mlmodern}
-
-%\usepackage{ngerman} % Sprachanpassung Deutsch
-
-\usepackage{graphicx}
-\usepackage{geometry}
-\geometry{a4paper, top=15mm}
-
-\usepackage{subcaption}
-\usepackage[shortlabels]{enumitem}
-\usepackage{amssymb}
-\usepackage{amsthm}
-\usepackage{mathtools}
-\usepackage{braket}
-\usepackage{bbm}
-\usepackage{graphicx}
-\usepackage{float}
-\usepackage{yhmath}
-\usepackage{tikz}
-\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
-\usetikzlibrary{calc,decorations.markings}
-
-%\usepackage[backend=biber, sorting=none]{biblatex}
-%\addbibresource{uni.bib}
-
-\usepackage[framemethod=TikZ]{mdframed}
-
-\tikzstyle{titlered} =
- [draw=black, thick, fill=white,%
- text=black, rectangle,
- right, minimum height=.7cm]
-
-
-\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
-\usepackage[parfill]{parskip}
-\usepackage{lipsum}
-
-\usepackage[OT2,T1]{fontenc}
-\DeclareSymbolFont{cyrletters}{OT2}{wncyr}{m}{n}
-\DeclareMathSymbol{\Sha}{\mathalpha}{cyrletters}{"58}
-
-\usepackage{tcolorbox}
-\tcbuselibrary{skins,breakable}
-
-\pagestyle{myheadings}
-
-\markright{Popović\hfill Applied Analysis\hfill}
-
-
-\title{University of Vienna\\ Faculty of Mathematics\\
-\vspace{1cm}Applied Analysis Problems
-}
-\author{Milutin Popovic}
-
-\begin{document}
-\maketitle
-\tableofcontents
-
-\section{Sheet 7}
-\subsection{Dirac Comb}
-The Dirac train or Dirac comb on defined in the following way
-\begin{align}
- \Sha_m[n] =
- \begin{cases}
- 1\;\;\;\;\;\; n = 0, \pm m, \pm 2m,\dots\\
- 0\;\;\;\;\;\; \text{else}
- \end{cases}
-\end{align}
-The dirac comb can be represented in a series of discrete dirac delta's
-\begin{align}
- \Sha_m[n] = \sum_{l=-N}^N \delta[n - lm],
-\end{align}
-where $\delta[s] = 1$ if $s = 0$ else $0$, for $s \in \mathbb{Z}$.
-The discrete Fourier transform of the Dirac comb in $\mathbb{C}^N$ is
-\begin{align}
- \widehat{\Sha_m[n]}
- &=\frac{1}{N}\sum_{n=0}^{N-1} \Sha_m[n] e^{-2\pi i \frac{k}{N}n}=\\
- &=\frac{1}{N}\sum_{n=0}^{N-1}
- \left(
- \sum_{l=-N}^N \delta(n-lm)
- \right)
- e^{-2\pi i \frac{k}{N}n},
-\end{align}
-where the summation happens exactly $\frac{N}{m}$ times, then
-\begin{align}
- &\frac{1}{m}\sum_{l=-N}^N e^{-2\pi i \frac{k}{N}lm}=\\
- &= \frac{1}{m} \sum_{l=-N}^N \delta[k - l\cdot \frac{N}{m}]\qquad
- \text{(Poisson's summation formula)} \\
- &= \frac{1}{m}\Sha_{\frac{N}{m}}[k]
-\end{align}
-\subsection{Schwartz Space}
-The Schwartz space $\mathcal{S}(\mathbb{R}^d)$, for $d \in \mathbb{N}$ is
-defined as
-\begin{align}
- &\mathcal{S} :=
-\bigg\{
- f\in\mathcal{C}^\infty(\mathbb{R}^d):
- \forall\alpha,\beta\in\mathbb{N}^d\;\; \lVert f \rVert_{\alpha,\beta}
- < \infty
-\bigg\},\\
-&\lVert f \rVert_{\alpha, \beta} :=
-\sup_{x\in\mathbb{R}^d}\left|x^\alpha (D^\beta f) (x) \right|.
-\end{align}
-Our aim is to show that if $f\in\mathcal{S}(\mathbb{R})$ then $\hat{f} \in
-\mathcal{S}(\mathbb{R})$. The condition is obviously
-\begin{align}
- &\lVert \hat{f} \rVert_{\alpha, \beta} =
- \sup_{\xi\in\mathbb{R}}\left|\xi^\alpha (D^\beta \hat{f}) (\xi)
- \right|<\infty,
-\end{align}
-for all $\alpha, \beta \in \mathbb{N}$.
-We can start with what we know about the Fourier transform
-\begin{align}
- \xi^\alpha \hat{f}(\xi) &= \mathcal{F}\left(\frac{1}{(2\pi
- i)^\alpha}(D^{\alpha}f)(x)\right)\\
- D^{\beta}\hat{f}(\xi) &= \mathcal{F}\left(
- (-2\pi i x)^\beta f(x)
- \right).
-\end{align}
-Combining the two relations above we get
-\begin{align}
- \xi^\alpha (D^\beta \hat{f})(\xi) =
- \mathcal{F}\left(\frac{(-2\pi i x)^\beta}{(2\pi
- i)^\alpha}x^\beta(D^{\alpha}f)(x)\right)=: \mathcal{F}(g(x))\\
-\end{align}
-If we call this function $g$, then $g\in\mathcal{S}(\mathbb{R})$ and
-$g\in L^1(\mathbb{R})$. Applying the Riemann-Lebesgue Lemma we get
-\begin{align}
- \hat{g}(\xi) = \int_\mathbb{R} g(x) e^{-2\pi i x \xi}\ dx \longrightarrow 0
- \;\;\; \text{as $|\xi| \rightarrow \infty$ }
-\end{align}
-Thereby $\hat{g} \in \mathcal{S}(\mathbb{R})$ and thus $\hat{f} \in
-\mathcal{S}(\mathbb{R})$.
-\subsection{Tempered Distributions}
-Tempered distributions are the elements of
-\begin{align}
- \mathcal{S}'(\mathbb{R}^d) :=
- \bigg\{
- L: \mathcal{S}(\mathbb{R}^d) \rightarrow \mathbb{C} | \text{$L$ is
- linear and continuous}
- \bigg\}.
-\end{align}
-Consider $\xi$ as a tempered distribution, buy acting on $\varphi \in
-\mathcal{S}(\mathbb{R})$ we have
-\begin{align}
- \xi(\phi) = \int_\xi \xi \varphi(\xi)\ d\xi.
-\end{align}
-The Fourier transform of $\xi$ is
-\begin{align}
- \hat{\xi}(\varphi)
- &=\xi(\hat{\varphi})
- = \int_\mathbb{R} \xi \hat{\varphi}(\xi)\ d\xi\\
- &= \int_{\mathbb{R}^2}\xi \varphi(x) e^{2\pi i\xi x}\ dxd\xi\\
- &= \int_{\mathbb{R}^2}\varphi(x) \xi e^{2\pi i \xi x}\ dxd\xi\\
- &=\int_{\mathbb{R}^2}\varphi(x)\frac{i}{2\pi} \frac{\partial}{\partial x}
- e^{2\pi i \xi x}\ dxd\xi =\\
- &=\frac{i}{2\pi}\int_{\mathbb{R}^2}\varphi(x)\delta'(x)\ dx=\\
- &=\frac{i}{2\pi} \delta'(\varphi).
-\end{align}
-\subsection{Fourier transform of the Dirac Comb}
-The general case of the Dirac Comb as a distribution is
-\begin{align}
- \Sha_T = \sum_{n \in \mathbb{Z}} \delta_{nT}.
-\end{align}
-The Fourier transform of the $\Sha_T$ distribution for $\varphi \in
-\mathcal{S}(\mathbb{R})$ is
-\begin{align}
- \widehat{\Sha_T}(\varphi)
- &= \sum_{n\in\mathbb{Z}} \hat{\delta}_{nT}(\varphi)\\
- &= \sum_{n\in\mathbb{Z}} \delta_{n\omega_0}(\varphi)\\
- &=\Sha_{\omega_0}(\varphi).
-\end{align}
-The Fourier transform, transforms the period of the combs.
-\subsection{Shannon Sampling}
-The Fourier transform of $1_{[-\frac{a}{2}, \frac{a}{2}]}(x)$ is
-\begin{align}
- \mathcal{F}\left(1_{[-\frac{a}{2}, \frac{a}{2}]}\right)(\xi)
- &= \int_\mathbb{R} 1_{[-\frac{a}{2}, \frac{a}{2}]} e^{-2\pi i x \xi}\
- dx\\
- &= \int_{-\frac{a}{2}}^{\frac{a}{2}} e^{-2\pi i x\xi}\ dx\\
- &= \frac{-1}{2\pi i \xi} e^{-2\pi i x
- \xi}\bigg|_{-\frac{a}{2}}^{\frac{a}{2}}\\
- &= \frac{1}{\pi \xi} \frac{1}{2i}\left(
- e^{pi i a \xi} - e^{-\pi i a \xi}
- \right)\\
- &= \frac{\sin(\pi \xi a)}{\pi \xi}
-\end{align}
-
-%\printbibliography
-\end{document}
diff --git a/appl_ana/sesh8/main.pdf b/appl_ana/sesh8/main.pdf
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diff --git a/appl_ana/sesh8/main.tex b/appl_ana/sesh8/main.tex
@@ -1,215 +0,0 @@
-\documentclass[a4paper]{article}
-
-
-\usepackage[T1]{fontenc}
-\usepackage[utf8]{inputenc}
-\usepackage{mlmodern}
-
-%\usepackage{ngerman} % Sprachanpassung Deutsch
-
-\usepackage{graphicx}
-\usepackage{geometry}
-\geometry{a4paper, top=15mm}
-
-\usepackage{subcaption}
-\usepackage[shortlabels]{enumitem}
-\usepackage{amssymb}
-\usepackage{amsthm}
-\usepackage{mathtools}
-\usepackage{braket}
-\usepackage{bbm}
-\usepackage{graphicx}
-\usepackage{float}
-\usepackage{yhmath}
-\usepackage{tikz}
-\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
-\usetikzlibrary{calc,decorations.markings}
-
-%\usepackage[backend=biber, sorting=none]{biblatex}
-%\addbibresource{uni.bib}
-
-\usepackage[framemethod=TikZ]{mdframed}
-
-\tikzstyle{titlered} =
- [draw=black, thick, fill=white,%
- text=black, rectangle,
- right, minimum height=.7cm]
-
-
-\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
-\usepackage[parfill]{parskip}
-\usepackage{lipsum}
-
-\usepackage[OT2,T1]{fontenc}
-\DeclareSymbolFont{cyrletters}{OT2}{wncyr}{m}{n}
-\DeclareMathSymbol{\Sha}{\mathalpha}{cyrletters}{"58}
-
-\usepackage{tcolorbox}
-\tcbuselibrary{skins,breakable}
-
-\pagestyle{myheadings}
-
-\markright{Popović\hfill Applied Analysis\hfill}
-
-
-\title{University of Vienna\\ Faculty of Mathematics\\
-\vspace{1cm}Applied Analysis Problems
-}
-\author{Milutin Popovic}
-
-\begin{document}
-\maketitle
-\tableofcontents
-
-\section{Sheet 8}
-\subsection{Finite Discrete Fourier Transform (FDFT)}
-Consider the vector $\begin{pmatrix}a & b & c & d\end{pmatrix}^T \in
-\mathbb{C}^4$ with a FDFT $\begin{pmatrix}A & B & C & D\end{pmatrix}^T$. We
-can show that the vector
-\begin{align}
- \begin{pmatrix}a & 0 & b & 0 & c & 0 & d & 0\end{pmatrix}^T,
-\end{align}
-has the FDFT of
-\begin{align}
- \frac{1}{2}\begin{pmatrix}A & 0 & B & 0 & C & 0 & D & 0\end{pmatrix}^T.
-\end{align}
-For the $N=4$, $n\in\{0,\dots,3\}$ the coefficients $a, b, c, d$ are denoted in
-$f[n]$. The FDFT is
-\begin{align}
- \hat{f}[k] &= \frac{1}{4} * \sum_{n=0}^3 f[n] e^{-2\pi i \frac{n}{4}k} \\
- &=\frac{1}{4}\left(
- a + be^{-\pi i \frac{k}{2}}
- + ce^{-\pi i k}+ de^{-\frac{3\pi i k}{2}}
- \right) = \\
- (&=\begin{pmatrix}A & B & C & D\end{pmatrix}^T)
-\end{align}
-for $k \in \{0,\dots, 3\}$ accordingly. For the $N=8$, $\mathbb{C}^8$ case
- we have $f_2[n]$ for $n \in \{0,\dots 7\}$,
- \begin{align}
- \hat{f}_2[k] &= \frac{1}{8} * \sum_{n=0}^7 f_2[n] e^{-2\pi i \frac{n}{8}k} \\
- &=\frac{1}{2}\frac{1}{4}\left(
- a + be^{-\pi i \frac{k}{2}}
- + ce^{-\pi i k}+ de^{-\frac{3\pi i k}{2}}
- \right) = \\
- (&=\frac{1}{2}\begin{pmatrix}A & B & C & D & A & B & C & D\end{pmatrix}^T)
- \end{align}
-for $k \in \{0,\dots, 7\}$ accordingly. We may generalize now for
-$\mathbb{C}^{4N}$, and the sequence for $a, b, c, d, 0$ represented by the
-function $g[n]$ for $n \in \{0,\dots, 4N-1\}$,
-\begin{align}
- g[n] =\begin{cases}
- f[n] \qquad n\in \{0, N, 2N, 3N\}\\
- 0 \qquad \text{else}
- \end{cases}.
-\end{align}
-Now we can compute the FDFT for $k \in \{0,\dots, 4N-1\}$
-\begin{align}
- \hat{g}[k] &= \frac{1}{4N}\sum_{n=0}^{4N-1} g[n]e^{-2\pi i
- \frac{n}{4N}k}\\
- &=\frac{4}{N}\sum_{n=0}{3}f[n]e^{-2\pi i \frac{n}{4}k}\\
- &=\frac{1}{N}\left(\frac{1}{4}\sum_{n=0}^3 f[n] e^{-2\pi i
- \frac{n}{4}k} \right) \\
- &= \frac{1}{N} \underbrace{\begin{pmatrix}A & B & C & D & \dots &
- \dots & A & B & C & D\end{pmatrix}^T)}_{\text{$4N$ entries, $N$
- sequences}}.
-\end{align}
-\subsection{More FDFT}
-Consider the discrete complex exponential with frequency of $1Hz$ in
-$\mathbb{C}^8$, for $n \in \{0, \dots , 7\}$,
-\begin{align}
- \exp[n] = e^{2\pi i n/8}.
-\end{align}
-The FDFT for $k \in \{0, \dots, 7\}$ is
-\begin{align}
- \hat{\exp}[k] &= \frac{1}{8}\sum_{n=0}^7 e^{2\pi i \frac{n}{8}}e^{-2\pi i
- n \frac{k}{8}} \\
- &= \frac{1}{8} \sum_{n=0}^7e^{-2\pi i (k-1)\frac{n}{8}}\\
- &=
- \begin{cases}
- 1\quad k=1\\
- 0 \qquad k\neq 1
- \end{cases}.
-\end{align}
-\begin{figure}[H]
- \centering
- \includegraphics[width=0.49\textwidth]{./fdft.png}
- \includegraphics[width=0.49\textwidth]{./normal.png}
- \caption{Test in Julia}
-\end{figure}
-\subsection{Sampling Sinusoids}
-Consider the following continuous signal
-\begin{align}
- f(t) = sin(20\pi t) + sin(40\pi t)
-\end{align}
-with frequencies $\omega = 2\pi \nu$, $\nu_1 = 10\ \text{Hz}$ and $\nu_2 = 20\
-\text{Hz}$. Sketching its Fourier transform would be something like this
-\begin{figure}[H]
- \centering
-\begin{tikzpicture}[
- axisline/.style={very thick, -stealth},
- xscale = 1.5,
- yscale = 1.5
- ]
- \draw[axisline] (-3,0)--(3,0) node[right]{$\nu$};
- \draw[axisline] (0,-1.5)--(0,1.5) node[above]{$\hat{f}$};
- \draw[->] (-1,0) -- (-1, -1) node[below] {$-\delta(\nu - 10)$};
- \draw[->] (-2,0) -- (-2, 1) node[above] {$\delta(\nu - 20)$};
- \draw[->] (1,0) -- (1, 1) node[above] {$\delta(\nu - 10)$};
- \draw[->] (2,0) -- (2, 1) node[above] {$\delta(\nu - 20)$};
-\end{tikzpicture}
-\end{figure}
-The Nyquist frequency for sampling would be
-\begin{align}
- \nu_{\text{Nyquist}} = 2\nu_\text{max} = 2\nu_2 = 40\ \text{Hz},
-\end{align}
-If we choose $50\ \text{Hz}$ for sampling we would get aliasing with the
-following frequencies
-\begin{align}
- n \cdot 50\ \text{Hz} - 20\ \text{Hz} = 30\ \text{Hz},80\ \text{Hz}, 130\
- \text{Hz}, \dots
-\end{align}
-\subsection{Short-Time Fourier Transform (STFT)}
-The Definition of the STFT is
-\begin{align}
- \text{STFT}\{f\} &= S_\varphi f(\tau, \omega) = \int_\mathbb{R} f(t)
- \overline{\text{M}_\omega \text{T}_\tau \varphi}dt \\
- &=\int_\mathbb{R} f(t)
- \bar{\varphi}(t - \tau)e^{-2\pi i \omega t}\ dt \\
-\end{align}
-Then we have the following identity
-\begin{align}
- S_\varphi(\text{T}_u\text{M}_\eta f)(x,\omega)
- &= \int_\mathbb{R}
- \left(\text{T}_u \text{M}_\eta f(t)\right) \bar{\varphi}(t-x) e^{-2\pi i
- \omega t}\ dt\\
- &= \int_\mathbb{R} e^{2\pi i \eta(t-u)}f(t-u) e^{-2\pi i \omega
- t}\bar{\varphi}(t-x)\ dt \qquad \text{(sub: $s = t-u$)}\\
- &= \int_\mathbb{R} f(s)\bar{\varphi}(s-(x-u))e^{2\pi i \eta s}e^{-2\pi i
- \omega s} e^{-2\pi i \omega u}\ ds \\
- &=e^{-2\pi i \omega u}\int_\mathbb{R} f(s) \bar{\varphi}(s-(x-u))e^{-2\pi i
- (\omega - \eta)s}\ ds\\
- &=e^{-2\pi i \omega u}\int_\mathbb{R}
- f(s)\overline{ \text{M}_{(\omega-\eta)} \text{T}_{(x-u)}\varphi(s)}\ ds\\
- &=e^{-2\pi i \omega u} S_\varphi f\left(x-u,\ \omega -\eta\right).
-\end{align}
-The second identity we can show
-\begin{align}
- S_\varphi f(x, \omega)
- &= \langle f, \overline{\text{M}_\omega \text{T}_x \varphi}\rangle \\
- &= \langle\mathcal{F} f, \mathcal{F} \overline{\text{M}_\omega \text{T}_x \varphi}\rangle \\
- &= \int_\xi \hat{f}(\xi)\int_t \overline{\text{M}_\omega \text{T}_x
- \varphi}(t) e^{-2\pi i \xi t}\ dt\ d\xi \\
- &= \int_\xi \hat{f}(\xi)\int_t \hat{\bar{\varphi}}(t-x) e^{2\pi i \omega
- t} e^{-2\pi i \xi t}\ dt\ d\xi \\
- &= \int_\xi \hat{f}(\xi)\int_t \hat{\bar{\varphi}}(t-x)e^{-2\pi i (\xi
- -\omega)t}\ dt\ d\xi \qquad \text{sub $u=t-x$}\\
- &= \int_\xi \hat{f}(\xi)\int_t \hat{\bar{\varphi}}(u)e^{-2\pi i (\xi
- -\omega)u}e^{-2\pi i (\xi -\omega)x} \ dt\ d\xi\\
- &= \int_\xi \hat{f}(\xi)e^{-2\pi i (\xi -\omega)x}\int_t
- \hat{\varphi}(u)e^{-2\pi i (\xi -\omega)u} \ dt\ d\xi\\ &= e^{2\pi i
- \omega x}\int_\xi \hat{f}(\xi) \hat{\bar{\varphi}}(\xi - \omega) e^{-2\pi
- i \xi x}d\xi\\
- &= e^{2\pi i \omega x} S_{\hat{\varphi}} \hat{f}(\omega, -x).
-\end{align}
-% printbibliography
-\end{document}
diff --git a/num_ana/build/prb1.pdf b/num_ana/build/prb1.pdf
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diff --git a/num_ana/prb1.tex b/num_ana/prb1.tex
@@ -0,0 +1,335 @@
+\include{preamble.tex}
+
+\usepackage{scratch}
+
+\begin{document}
+\maketitle
+\tableofcontents
+\section{Sheet 1}
+\subsection{Problem 1}
+Consider the following two matrices $A, L_1 \in \mathbb{R}^{4 \times 4}$
+defined in as
+\begin{align}
+ A :=
+ \begin{pmatrix}
+ 2 & 1 & 1 & 0 \\
+ 4 & 3 & 3 & 1 \\
+ 8 & 7 & 9 & 5 \\
+ 6 & 7 & 9 & 8
+ \end{pmatrix},
+ \qquad
+ L_1 :=
+ \begin{pmatrix}
+ 1 & 0 & 0 & 0\\
+ x & 1 & 0 & 0\\
+ y & 0 & 1 & 0\\
+ z & 0 & 0 & 1
+ \end{pmatrix}
+,\end{align}
+for $x, y, z \mathbb{R}$.
+
+To show that $A$ is invertible, we need to show it has maximal rank, that
+is $\text{rank}(A) = 4$. We can do this by doing Gaussian elimination
+steps until $A$ is of the form of a upper triangular matrix
+\begin{gather}
+ \begin{bmatrix}
+ 2 & 1 & 1 & 0 \\
+ 4 & 3 & 3 & 1 \\
+ 8 & 7 & 9 & 5 \\
+ 6 & 7 & 9 & 8
+ \end{bmatrix}
+ \begin{matrix}
+ \\
+ -2 \cdot I\\
+ -4 \cdot I\\
+ -3 \cdot I
+ \end{matrix} \quad
+ \longrightarrow
+ \begin{bmatrix}
+ 2 & 1 & 1 & 0 \\
+ 0 & 1 & 1 & 1 \\
+ 0 & 3 & 5 & 5 \\
+ 0 & 4 & 6 & 8
+ \end{bmatrix}
+ \begin{matrix}
+ \\
+ \\
+ -3 \cdot II\\
+ -4 \cdot II
+ \end{matrix} \quad
+ \longrightarrow
+ \begin{bmatrix}
+ 2 & 1 & 1 & 0 \\
+ 0 & 1 & 1 & 1 \\
+ 0 & 0 & 2 & 2 \\
+ 0 & 0 & 2 & 4
+ \end{bmatrix}
+ \begin{matrix}
+ \\
+ \\
+ \\
+ -1 \cdot III
+ \end{matrix} \quad
+ \label{eq: gelim1}
+ \\
+ \longrightarrow
+ \begin{bmatrix}
+ 2 & 1 & 1 & 0 \\
+ 0 & 1 & 1 & 1 \\
+ 0 & 0 & 2 & 2 \\
+ 0 & 0 & 0 & 2
+ \end{bmatrix} \quad
+ \underset{\text{det}}{\longrightarrow} \quad 8
+ \label{eq: gelim2}
+.\end{gather}
+
+Next we will determine $x, y$ and $z$, s.t. $(L_1A)_{\cdot, 1} =
+\begin{pmatrix} 2 & 0 & 0 & 0 \end{pmatrix} $ by solving the linear
+system
+\begin{align}
+ L_1 A =
+ \begin{pmatrix}
+ 2 & 1 & 1 & 0 \\
+ 2x+4 & x+3 & x+3 & 1\\
+ 2y+8 & y+7 & y+9 & 5\\
+ 2z+6 & z+7 & z+9 & 8\\
+ \end{pmatrix}
+,\end{align}
+we get $x = -2$, $y = -4$ and $z=-3$ and thereby
+\begin{align}
+ L_1 A =
+ \begin{pmatrix}
+ 1 & 0 & 0 & 0\\
+ -2 & 1 & 0 & 0\\
+ -4 & 0 & 1 & 0\\
+ -3 & 0 & 0 & 1
+ \end{pmatrix}
+ \begin{pmatrix}
+ 2 & 1 & 1 & 0 \\
+ 4 & 3 & 3 & 1 \\
+ 8 & 7 & 9 & 5 \\
+ 6 & 7 & 9 & 8
+ \end{pmatrix}=
+ \begin{pmatrix}
+ 2 & 1 & 1 & 0 \\
+ 0 & 1 & 1 & 1\\
+ 0 & 3 & 5 & 5\\
+ 0 & 3 & 5 & 8\\
+ \end{pmatrix}
+.\end{align}
+In an analogous structure we may define $L_2, L_3 \in \mathbb{R}^{4\times4}$,
+s.t.
+\begin{align}
+ L_3L_2L_1A=U,
+\end{align}
+where $U$ is an upper triangular matrix. We may notice that this is an LU
+decompositions of a matrix and can be determined by the inversion of a
+single step of Gaussian elimination. By that the three steps needed to
+achieve the upper triangular by Gaussian elimination are introduced
+in \ref{eq: gelim1} and \ref{eq: gelim2}, that is also why $-2, -4, -3$ aligns up with $L_1$.
+To summarize, by looking at \ref{eq: gelim1} and \ref{eq: gelim2} the matrices $L_2, L_3$ are
+the following
+\begin{align}
+ L_2 =
+ \begin{pmatrix}
+ 1 & 0 & 0 & 0\\
+ 0 & 1 & 0 & 0\\
+ 0 & -3 & 1 & 0\\
+ 0 & -4 & 0 & 1
+ \end{pmatrix}, \qquad
+ L_3 =
+ \begin{pmatrix}
+ 1 & 0 & 0 & 0\\
+ 0 & 1 & 0 & 0\\
+ 0 & 0 & 1 & 0\\
+ 0 & 0 & -1 & 1
+ \end{pmatrix}.
+\end{align}
+And by no calculation we know that $U$ needs to be the upper triangular
+found in \ref{eq: gelim2}, i.e.
+\begin{align}
+ L_3L_2L_1A = U =
+ \begin{pmatrix}
+ 2 & 1 & 1 & 0 \\
+ 0 & 1 & 1 & 1 \\
+ 0 & 0 & 2 & 2 \\
+ 0 & 0 & 0 & 2
+ \end{pmatrix}.
+\end{align}
+We have indeed preformed an LU decomposition of $A$, which is indeed
+useful for solving a linear system of the form
+\begin{align}
+ A x &= b \qquad \text{and} \quad L_3L_2L_1A = U,\\
+ (L_3L_2L_1A)x = Ux &= L_3L_2L_1b = y\\
+ \Rightarrow Ux &= y,
+\end{align}
+where the system is recursively solvable as $U$ is the upper triangular
+and no additional transformation steps are required only ''plug and
+play``.
+\subsection{Problem 2}
+Next we consider $A_\varepsilon \in \mathbb{R}^{2 \times 2}$ defined as
+\begin{align}
+ A_\varepsilon :=
+ \begin{pmatrix}
+ \varepsilon & 1\\
+ 1 & 1
+ \end{pmatrix},
+\end{align}
+for $\varepsilon > 0$. The inverse of $A_\varepsilon$ is
+\begin{align}
+ A_\varepsilon^{-1} = \frac{1}{\text{det}(A_\varepsilon)}
+ \text{adj}(A_\varepsilon) =
+ \frac{1}{\varepsilon - 1} \begin{pmatrix} 1 & -1 \\ -1 & \varepsilon \end{pmatrix}
+\end{align}
+Now let $\|x\|_\infty = \max\{|x_1|, |x_2|\}$ be the maximum norm of $x \in
+\mathbb{R}^{2}$, and $\|A_\varepsilon\|_\infty$ the induced matrix norm of
+$A_\varepsilon$. We can show that
+\begin{align}
+ \lim_{\varepsilon \to 0} K(A_\varepsilon) = 4,
+\end{align}
+where $K(A_\varepsilon) = \|A_\varepsilon\|_\infty
+\|A_\varepsilon^{-1}\|_\infty$ is the condition number of $A_\varepsilon$.
+\begin{align}
+ \|A_\varepsilon\|_\infty &= \|\begin{pmatrix} \varepsilon + 1 & 1 + 1
+ \end{pmatrix} \|_\infty = 2\\
+ \|A_\varepsilon^{-1}\|_\infty &=
+ \|\begin{pmatrix} \mid -\frac{2}{\varepsilon-1} \mid & 1 \end{pmatrix} \|_\infty
+ = \frac{2}{1-\varepsilon},
+\end{align}
+and thereby
+\begin{align}
+ \lim_{\varepsilon \to 0} K(A_\varepsilon)=\lim_{\varepsilon \to 0} 2\cdot
+ \frac{2}{1-\varepsilon} = 4
+\end{align}
+If we preformed an LU decomposition of $A_\varepsilon$ like in the first
+problem to get an upper diagonal the decompostion would be
+\begin{align}
+ LA_\varepsilon &=
+ \begin{pmatrix} 1 & 0 \\ -\frac{1}{\varepsilon} & 1 \end{pmatrix}
+ \begin{pmatrix} \varepsilon & 1 \\ 1 & 1 \end{pmatrix} \\
+ &=
+ \begin{pmatrix} \varepsilon & 1 \\ 0 & 1 - \frac{1}{\varepsilon}
+ \end{pmatrix} = U_\varepsilon,
+\end{align}
+with the inverse
+\begin{align}
+ U_\varepsilon^{-1}= \frac{1}{\varepsilon -1}
+ \begin{pmatrix}1-\frac{1}{\varepsilon} & -1 \\ 0 & \varepsilon
+ \end{pmatrix} .
+\end{align}
+The condition number of the resulting upper triangular matrix
+$U_\varepsilon$, $K(U_\varepsilon)$ as $\varepsilon \rightarrow 0$ is
+\begin{align}
+ \|U_\varepsilon\|_\infty &= \|\begin{pmatrix} \varepsilon + 1 &
+ \mid 1-\frac{1}{\varepsilon} \mid \end{pmatrix} \|_\infty = \frac{1}{\varepsilon} -
+ 1\\
+ \|U_\varepsilon^{-1}\|_\infty &=
+ \|\begin{pmatrix} \mid \frac{1- \frac{1}{\varepsilon}}{\varepsilon
+ -1} \mid & \mid\frac{\varepsilon}{\varepsilon-1} \mid \end{pmatrix}
+ \|_\infty = \frac{1}{\varepsilon(\varepsilon - 1)}\\
+ \Longrightarrow
+ \lim_{\varepsilon \to 0}K(U_\varepsilon) &= \lim_{\varepsilon \to 0}
+ \frac{1-\varepsilon}{\varepsilon}\frac{1}{\varepsilon(1 -
+ \varepsilon)} = \infty.
+\end{align}
+But if we on the other hand considered a pivoting step in which we exchange
+the rows of $A_\varepsilon$
+\begin{align}
+ PA_\varepsilon = A_\varepsilon' =
+ \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix}
+ \begin{pmatrix} \varepsilon & 1 \\ 1 & 1 \end{pmatrix}
+ =
+ \begin{pmatrix}1 & 1 \\ \varepsilon & 1 \end{pmatrix}
+\end{align}
+Then the P-LU decomposition is
+\begin{align}
+ L'A_\varepsilon' =
+ \begin{pmatrix} 1 & 0 \\ -\varepsilon & 1 \end{pmatrix}
+ \begin{pmatrix}1 & 1 \\ \varepsilon & 1 \end{pmatrix}
+ =
+ \begin{pmatrix}1 & 1 \\ 0 & 1 - \varepsilon \end{pmatrix} =
+ U_\varepsilon',
+\end{align}
+with the inverse
+\begin{align}
+ (U_\varepsilon')^{-1} = \frac{1}{1-\varepsilon}
+ \begin{pmatrix} 1-\varepsilon & - 1\\ 0 & 1 \end{pmatrix}.
+\end{align}
+Then the condition number as $\varepsilon \rightarrow 0$
+\begin{align}
+ \|U_\varepsilon'\|_\infty
+ &= \|\begin{pmatrix} 2 & 1-\varepsilon \end{pmatrix} \|_\infty = 2\\
+ \|\left(U_\varepsilon'\right)^{-1} \|_\infty
+ &= \|\begin{pmatrix}
+ \frac{1-\varepsilon + 1}{1 - \varepsilon} & \frac{1}{1-\varepsilon}
+ \end{pmatrix} \| = \frac{2-\varepsilon}{1-\varepsilon}\\
+ \Longrightarrow
+ \lim_{\varepsilon \to 0}K(U_\varepsilon') &= \lim_{\varepsilon \to 0}
+ 2\cdot \frac{2-\varepsilon}{1-\varepsilon} = 2\cdot2 = 4
+\end{align}
+\subsection{Problem 3}
+Let $v \in \mathbb{R}^n$, $n \in \mathbb{N}$ and $v \neq 0$. We define the Housholder
+matrix
+\begin{align}
+ H = \text{Id} - \frac{2}{\langle v, v \rangle}v v^T.
+\end{align}
+Indeed $H$ is an orthogonal matrix, it satisfies $H H^T = H^T H = \text{Id}$.
+\begin{align}
+ H H^T
+ &=
+ \left( \text{Id} - \frac{2}{\langle v, v \rangle}vv^T\right)
+ \left( \text{Id} - \frac{2}{\langle v, v \rangle}vv^T\right)^T\\
+ &=
+ \left( \text{Id} - \frac{2}{\langle v, v \rangle}vv^T\right)
+ \left( \text{Id} - \frac{2}{\langle v, v \rangle}(vv^T)^T\right)\\
+ &=
+ \left( \text{Id} - \frac{2}{\langle v, v \rangle}vv^T\right)
+ \left( \text{Id} - \frac{2}{\langle v, v \rangle}vv^T\right)\\
+ &= \text{Id} - \frac{4}{\langle v, v \rangle} vv^T + \frac{4}{\langle v,
+ v \rangle^2} (v v^T)(v v^T)\\
+ &= \text{Id} - \frac{4}{\langle v, v \rangle} vv^T + \frac{4}{\langle v,
+ v \rangle} (v v^T) = \text{Id}
+ \\
+ \nonumber\\
+ H^T H &=
+ \left( \text{Id} - \frac{2}{\langle v, v \rangle}vv^T\right)
+ \left( \text{Id} - \frac{2}{\langle v, v \rangle}vv^T\right)\\
+ &= \text{Id}
+\end{align}
+Let us look at the projection of some $x \in \mathbb{R}^n$ onto $v$
+\begin{align}
+ x_v = x - \frac{\langle v, x \rangle}{\langle v, v \rangle} v,
+\end{align}
+to get the projective inversion onto $v$ we would have to subtract the vector
+$\frac{\langle v, x \rangle}{\langle v, v \rangle} v$ twice, graphically it would look like this
+\begin{figure}[H]
+ \centering
+ \begin{tikzpicture}[
+ xscale = 1.5,
+ yscale = 1.5
+ ]
+ \draw[->] (0, 0) -- (1, 1) node[right] {$x$};
+ \draw[->] (0, 0) -- (2, 0) node[right] {$v$};
+ \draw[dotted, very thick] (1, 1) -- (1, 0) node[below] {$x_v$};
+ \draw[->] (0, 0) -- (1, 0) node[below] {$x_v$}; \draw[->] (0, 0) -- (-1, 0) node[below] {$-x_v$}; \end{tikzpicture}
+\end{figure}
+The Household matrix acting on a vector $x$, $Hx$ is exactly the above case
+since vector multiplication is associative we have
+\begin{align}
+ Hx &= x - \frac{2}{\langle v, v \rangle} vv^T x\\
+ &= x - 2\frac{\langle v, x\rangle}{\langle v, v \rangle} v
+\end{align}
+The condition number of an orthogonal matrix $A$ in the $\|\cdot\|_2$ induced
+norm is
+\begin{align}
+ K(A) = \|A\|_2 \|A^{-1}\|_2 = 1,
+\end{align}
+because the orthogonal matrix preserves distance, i.e. $\|Ax\|_2 = \|x\|_2$
+for all $x$. Also $A^{-1} =A^T$ is orthogonal as well
+\begin{align}
+ \|A\|_2 = \sup \frac{\|Ax\|_2}{\|x\|_2} = \sup \frac{\|x\|_2}{\|x\|_2} =
+ 1
+\end{align}
+%\printbibliography
+\end{document}
+
diff --git a/num_ana/preamble.tex b/num_ana/preamble.tex
@@ -0,0 +1,59 @@
+\documentclass[a4paper]{article}
+
+\usepackage[T1]{fontenc}
+\usepackage[utf8]{inputenc}
+\usepackage{mlmodern}
+
+%\usepackage{ngerman} % Sprachanpassung Deutsch
+
+\usepackage{graphicx}
+\usepackage{geometry}
+\geometry{a4paper, top=15mm}
+
+\usepackage{subcaption}
+\usepackage[shortlabels]{enumitem}
+\usepackage{amssymb}
+\usepackage{amsthm}
+\usepackage{mathtools}
+\usepackage{braket}
+\usepackage{bbm}
+\usepackage{graphicx}
+\usepackage{float}
+\usepackage{yhmath}
+\usepackage{tikz}
+\usetikzlibrary{patterns,decorations.pathmorphing,positioning}
+\usetikzlibrary{calc,decorations.markings}
+
+%\usepackage[backend=biber, sorting=none]{biblatex}
+%\addbibresource{uni.bib}
+
+\usepackage[framemethod=TikZ]{mdframed}
+
+\tikzstyle{titlered} =
+ [draw=black, thick, fill=white,%
+ text=black, rectangle,
+ right, minimum height=.7cm]
+
+
+\usepackage[colorlinks=true,naturalnames=true,plainpages=false,pdfpagelabels=true]{hyperref}
+\usepackage[parfill]{parskip}
+\usepackage{lipsum}
+
+
+\usepackage{tcolorbox}
+\tcbuselibrary{skins,breakable}
+
+\pagestyle{myheadings}
+
+\newcommand{\eps}{\varepsilon}
+\usepackage[OT2,T1]{fontenc}
+\DeclareSymbolFont{cyrletters}{OT2}{wncyr}{m}{n}
+\DeclareMathSymbol{\Sha}{\mathalpha}{cyrletters}{"58}
+
+\markright{Popović\hfill Applied Analysis\hfill}
+
+
+\title{University of Vienna\\ Faculty of Mathematics\\
+\vspace{1cm}Applied Analysis Problems
+}
+\author{Milutin Popovic}
diff --git a/num_ana/sheets/sheet_1.pdf b/num_ana/sheets/sheet_1.pdf
Binary files differ.
