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 \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}
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