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      1 \include{preamble.tex}
      2 
      3 \begin{document}
      4 \maketitle
      5 \tableofcontents
      6 
      7 \section{Sheet 4}
      8 
      9 \subsection{Fourier Series}
     10 The Fourier series of a $p$ periodic function $f$, integrable on
     11 $[-\frac{p}{2}, \frac{p}{2}]$ is
     12 \begin{align}
     13     f(x) = \frac{a_0}{2} + \sum_{n=1}^\infty \left(a_n \cos(\frac{2\pi n x}{p})
     14     b_n sin(\frac{2\pi n x}{p})\right).
     15 \end{align}
     16 The coefficients $a_n$ and $b_n$ are called the Fourier coefficients of $f$
     17 and are given by
     18 \begin{align}
     19     a_n &= \frac{2}{p} \int_{-\frac{p}{2}}^{\frac{p}{2}} f(x) \sin(\frac{2\pi
     20     n x}{p}) dx, \;\;\;\;\; n\geq 0 \\
     21     b_n &= \frac{2}{p} \int_{-\frac{p}{2}}^{\frac{p}{2}} f(x) \cos(\frac{2\pi
     22     n x}{p}) dx, \;\;\;\;\; n\geq 1
     23 \end{align}
     24 Let us compute the Fourier series of $f(t) = t$ for $t \in [-\frac{1}{2},
     25 \frac{1}{2}]$. The Fourier coefficients are
     26 \begin{align}
     27     a_n &= 2\int_{-\frac{1}{2}}^{\frac{1}{2}} t \cos(2\pi n t)\ dt = 0
     28     \;\;\;\;\; \text{(odd: g(-t) = -g(t))},\\
     29     \nonumber\\
     30     b_n &= 2\int_{-\frac{1}{2}}^{\frac{1}{2}} t \sin(2\pi n t)\ dt = \\
     31         &= 2 \left(-\frac{1}{2\pi n} \cos(2\pi n
     32         t)\bigg|_{-\frac{1}{2}}^{\frac{1}{2}}
     33         +\int_{\frac{1}{2}}^{\frac{1}{2}} \frac{1}{2 \pi n}\cos(2\pi n t)\ dt
     34             \right) =\\
     35         &= -\frac{1}{\pi n}\left( -\cos(\pi n) + \frac{1}{\pi n }\sin(\pi
     36             n)\right) =
     37         \frac{\sin(\pi n) - \pi n \cos(\pi n)}{(\pi n)^2}.
     38 \end{align}
     39 Thereby the Fourier series of $f(t) = t$ is
     40 \begin{align}
     41     f(t) = \sum_{n=1}^\infty \left(\frac{\sin(\pi n) - \pi n \cos(\pi n)}{(\pi
     42     n)^2}\right) \sin(2\pi n t) = t
     43 \end{align}
     44 \subsection{Truncation Error}
     45 The truncation error of the trigonometric polynomial $(Sf_N)$ of degree $N$ is
     46 \begin{align}
     47     \sum_{|k| > N} |\hat{f}(k)|^2 = \lVert f - S_N\rVert_2^2 =
     48     \int_{-\frac{1}{2}}^{\frac{1}{2}} |E_N(t)|^2 dt.
     49 \end{align}
     50 Computations for $N = 3$ and $N = 9$ were done in python with a integration error of
     51 around $10^{-15}$, resulting in the overall truncation errors of
     52 \begin{align}
     53     \sum_{|k| > 3} |\hat{f}(k)|^2 = 0.0053,\\
     54     \sum_{|k| > 9} |\hat{f}(k)|^2 = 0.0143.
     55 \end{align}
     56 To achieve $\lVert E_N\rVert^2_2 < 0.1 \lVert f \rVert^2_2$, the number of
     57 coefficients needed are about $61$. This was done using a while loop and
     58 evaluating $\lVert E_N\rVert^2_2$ for $N$ until the above condition is met.
     59 
     60 \subsection{Orthonormal Bases}
     61 Here we will go through the most important properties of orthonormal bases.
     62 So let $\{b_n\}_{n\in \mathbb{N}}$ be an ONB of a vector space $\mathcal{H}$,
     63 then for every $x\in \mathcal{H}$ we may write
     64 \begin{align}
     65     x = \sum_{b_n} \langle b_n, x\rangle b_n,
     66 \end{align}
     67 and
     68 \begin{align}
     69     \lVert x \rVert^2 = \sum_{b_n} |\langle b_n, x\rangle|^2.
     70 \end{align}
     71 For any $x, y \in \mathcal{H}$ we can write the scalar product as
     72 \begin{align}
     73     \langle x, y\rangle = \sum_{b_n} \langle b_n, x\rangle \langle b_n,
     74     y\rangle,
     75 \end{align}
     76 Furthermore there exists a linear projection $\Phi\ : \mathcal{H}
     77 \rightarrow l^2(\{b_n\}_n)$ such that
     78 \begin{align}
     79     \langle \Phi(x), \Phi(y)\rangle = \langle x, y \rangle\;\;\; \forall x, y
     80     \in \mathcal{H}.
     81 \end{align}
     82 
     83 An example of an orthonormal basis, which spans $L^2([-\frac{p}{2},
     84 \frac{p}{2}])$ is $\mathcal{T}_p = \{e_n := \frac{e^{2\pi i
     85 \frac{n}{p}x}}{\sqrt{p}}\}_{n\in\mathbb{Z}}$. The $e_n$'s are orthonormal in
     86 $L^2$ which can be easily seen by using the scalar product of $L^2$, so for
     87 $n, m \in \mathbb{Z}$
     88 \begin{align}
     89     \langle e_n, e_m\rangle_{L^2([-\frac{p}{2}, \frac{p}{2})} &=
     90     \frac{1}{p}\int_{[-\frac{p}{2}, \frac{p}{2}]}e_n \cdot e_m^* \ dx=\\
     91     &=\frac{1}{p}\int_{[-\frac{p}{2}, \frac{p}{2}]} e^{2\pi i \frac{(n-m)}{p} x} \ dx=\\
     92     &=\frac{\sin(\pi (n-m))}{\pi(n-m)} =
     93     \begin{cases}
     94         0  \;\;\;\; n\neq m\\
     95         1 \;\;\;\;  n=m
     96     \end{cases}
     97 \end{align}
     98 \subsection{Dirichlet Kernel}
     99 The function
    100 \begin{align}
    101     D_t(x) := \sum_{\lVert k \rVert_\infty \leq t} e_k(x), \;\;\;\;\; x\in
    102     \mathbb{R}^d
    103 \end{align}
    104 is called the Dirichlet Kernel. For $0 < t \in \mathbb{N}$ we have
    105 \begin{align}
    106     (S_tf)(x) = \int_{I^d} f(y) D_t(x-y) dy,
    107 \end{align}
    108 where $S_t$ represents the orthogonal projection onto the trigonometric
    109 polynomials $\Pi_t$ of degree $t$, by
    110 \begin{align}
    111     &S_t:\ L^1(\mathbb{T}^d) \rightarrow \Pi_t \\
    112     &f \mapsto \sum_{\lVert k \rVert \leq t} \langle f,
    113     e_k\rangle_{L^2(\mathbb{T}^d)} e_k \;\;\;\;\; k \in \mathbb{Z}^d
    114 \end{align}
    115 And furthermore the Dirichlet Kernel satisfies
    116 \begin{align}
    117     D_t(x) = \prod_{i=1}^d \frac{e_{t+1}(x_i) - e_{-t}(x_i)}{e_1(x_i) - 1}
    118 \end{align}
    119 To show the convolution property, we start off by applying the orthogonal
    120 projection into the trigonometric polynomials $S_t$ onto a function $f \in
    121 L(\mathbb{T}^d)$
    122 \begin{align}
    123     (S_tf) &= \sum_{\lvert k\rVert_\infty \leq t} \int_{I^d} f(y) e^{-2\pi i
    124     \langle k, y\rangle}\ dy\ e^{2\pi i\langle k, x\rangle} =\\
    125     &= \int_{I^d}f(y) \sum_{\lvert k\rVert_\infty \leq t} e^{2\pi i \langle
    126     k, (x- y)\rangle}\ dy =\\
    127     &= (f * D_t) (x) = \int_{I^d} f(y) D_t(x - y)\ dy.
    128 \end{align}
    129 To show the reformulation of the Dirichlet kernel, we need to simply
    130 calculate it directly
    131 \begin{align}
    132     \sum_{\lVert k \rVert_\infty \leq t} e^{2\pi i \langle k , x\rangle} &=
    133     \prod_{j=1}^d \sum_{k_j = -t}^t e^{2\pi i k_j x_j} =\\
    134     &= \prod_{j=1}^d e^{-2\pi i t x_j} \sum_{k_j = 0}^{2t} e^{2\pi i k_j
    135         x_j}=;\;\;\;\; \text{(trigonometric series)}\\
    136     &= \prod_{j=1}^d e^{-2\pi i t x_j} \frac{e^{2\pi i (2t + 1)x_j} -
    137     1}{e^{2\pi i x_j} - 1} =\\
    138     &= \prod_{j = 1} \frac{e_{t+1}(x_j) - e_{-t}(x_j)}{e_1(x_j) - 1}.
    139 \end{align}
    140 %\printbibliography
    141 \end{document}