%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Short Sectioned Assignment
% LaTeX Template
% Version 1.0 (5/5/12)
%
% This template has been downloaded from:
% http://www.LaTeXTemplates.com
%
% Original author:
% Frits Wenneker (http://www.howtotex.com)
%
% License:
% CC BY-NC-SA 3.0 (http://creativecommons.org/licenses/by-nc-sa/3.0/)
%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%----------------------------------------------------------------------------------------
% PACKAGES AND OTHER DOCUMENT CONFIGURATIONS
%----------------------------------------------------------------------------------------
\documentclass[paper=a4, fontsize=11pt]{scrartcl} % A4 paper and 11pt font size
\usepackage[T1]{fontenc} % Use 8-bit encoding that has 256 glyphs
\usepackage{fourier} % Use the Adobe Utopia font for the document - comment this line to return to the LaTeX default
\usepackage[english]{babel} % English language/hyphenation
\usepackage{amsmath,amsfonts,amsthm,amssymb} % Math packages
\usepackage{lipsum} % Used for inserting dummy 'Lorem ipsum' text into the template
\usepackage{sectsty} % Allows customizing section commands
\allsectionsfont{\centering \normalfont\scshape} % Make all sections centered, the default font and small caps
\usepackage{fancyhdr} % Custom headers and footers
\pagestyle{fancyplain} % Makes all pages in the document conform to the custom headers and footers
\fancyhead{} % No page header - if you want one, create it in the same way as the footers below
\fancyfoot[L]{} % Empty left footer
\fancyfoot[C]{} % Empty center footer
\fancyfoot[R]{\thepage} % Page numbering for right footer
\renewcommand{\headrulewidth}{0pt} % Remove header underlines
\renewcommand{\footrulewidth}{0pt} % Remove footer underlines
\setlength{\headheight}{13.6pt} % Customize the height of the header
\numberwithin{equation}{section} % Number equations within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)
\numberwithin{figure}{section} % Number figures within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)
\numberwithin{table}{section} % Number tables within sections (i.e. 1.1, 1.2, 2.1, 2.2 instead of 1, 2, 3, 4)
\setlength\parindent{0pt} % Removes all indentation from paragraphs - comment this line for an assignment with lots of text
%----------------------------------------------------------------------------------------
% TITLE SECTION
%----------------------------------------------------------------------------------------
\newcommand{\horrule}[1]{\rule{\linewidth}{#1}} % Create horizontal rule command with 1 argument of height
\title{
\normalfont \normalsize
\textsc{university of texas at austin} \\ [25pt] % Your university, school and/or department name(s)
\horrule{0.5pt} \\[0.4cm] % Thin top horizontal rule
\huge Methods of Applied Mathematics Problem Set 1 \\ % The assignment title
\horrule{2pt} \\[0.5cm] % Thick bottom horizontal rule
}
\author{Qi Lei} % Your name
\date{\normalsize\today} % Today's date or a custom date
\begin{document}
\maketitle % Print the title
%----------------------------------------------------------------------------------------
% PROBLEM 1
%----------------------------------------------------------------------------------------
\section{Exercise 6.1}
Compute he Fourier transform of $e^{-|x|}$ for $x\in \mathbb{R}$.\\
Solution: \begin{eqnarray*}
\hat{f}(\xi)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-|x|}e^{-ix\xi}dx\\
&=&\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}e^{-x-ix\xi}dx+\int_{-\infty}^0e^{x-ix\xi}dx\\
&=&\frac{1}{\sqrt{2\pi}}\int_{0}^{\infty}(e^{-x-ix\xi}-e^{-x+ix\xi})dx\\
&=&\frac{1}{\sqrt{2\pi}}[\frac{1}{-(1+i\xi)}(-1)-\frac{1}{-1+i\xi}(-1)]\\
&=&\frac{1}{\sqrt{2\pi}}[\frac{1-i\xi}{1+\xi^2}+\frac{-(1+i\xi)}{1+\xi^2}]\\
&=&\frac{1}{\sqrt{2\pi}}\frac{-2i\xi}{1+\xi^2}\\
&=&-\sqrt{\frac{2}{\pi}}\frac{i\xi}{1+\xi^2}
\end{eqnarray*}
\\
\section{Exercise 6.2}
Compute the Fourier transform of $e^{-a|x|^2},~a>0$, directly, where $x\in \mathbb{R}$.\\
Solution:
\begin{eqnarray*}
\hat{f}(\xi)&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a|x|^2}e^{-ix\xi}dx\\
&=&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}e^{-a(x+\frac{i\xi}{2a})^2+\frac{-\xi^2}{4a}}dx~~~~~~~~x'\doteq x+\frac{i\xi}{2a}\\
&=&\frac{1}{\sqrt{2\pi}}e^{-\frac{\xi^2}{4a}}\int_{-\infty}^{\infty}e^{-ax^2}dx\\
&=&\frac{e^{-\frac{\xi^2}{4a}}}{2a}
\end{eqnarray*}
\\
\section{Exercise 6.4}
$f\in L_1(\mathbb{R}^d),$ and $f(x)=g(|x|)$ for some $g$, show that $\tilde{f}(\xi)=h(|\xi|)$ for some $h$.
Solution:
\begin{eqnarray}
\nonumber
\hat{f}(\xi)&=&\frac{1}{(2\pi)^{d/2}}\int_{\mathbb{R}^d}f(x)e^{-ix\xi}dx~~~~~~~~~~~~~(\text{polar coordinate transformation})\\
\nonumber
&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}\int_{\partial S^d}f(r\alpha)e^{-ir\alpha\cdot\xi} r^{d-1}d\alpha dr\\
\label{1.1}
&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}g(r)r^{d-1}\int_{\partial S^d}e^{-ir\alpha\cdot\xi}d\alpha dr
\end{eqnarray}
For any $\xi$, $\exists$ rotation $Q$, such that $\xi=|\xi|Qe_1$, so that $\alpha\cdot \xi=\alpha^T\xi=\alpha^T(Qe_1)|\xi|=|\xi|(Q^T\alpha)^Te_1$\\
\begin{eqnarray*}
\therefore~~(\ref{1.1})&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}g(r)r^{d-1}\int_{\partial S^d}e^{-ir(Q^T\alpha)\cdot e_1|\xi|}d\alpha dr\\
&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}g(r)r^{d-1}\int_{\partial S^d}e^{-ir(Q^T\alpha)\cdot e_1|\xi|}d(Q^T\alpha) dr~~~~\beta\doteq Q^T\alpha\\
&=&\frac{1}{(2\pi)^{d/2}}\int_0^{\infty}g(r)r^{d-1}\int_{\partial S^d}e^{-ir\beta_1|\xi|}d\beta dr\\
&=&h(|\xi|)
\end{eqnarray*}
only depends on $|\xi|$\\
\section{Exercise 6.10}
Let the field be complex and define $T: L^2(\mathbb{R}^d)\rightarrow L^2(\mathbb{R}^d)$ by
\begin{equation*}
Tf(x)=\int e^{-|x-y|^2/2}f(y)dy
\end{equation*}
Show that $T$ is positive, injective operator, but not surjective.\\
Solution:
\begin{eqnarray*}
\text{Positive}&\Longleftrightarrow& \langle Tf(x),f(x)\rangle\geq 0, \forall f\in L^2(\mathbb{R}^d)\\
&\Longleftrightarrow& \int \int e^{-|x-y|^2/2}f(y)dy f(x)dx\geq 0,\forall f\in L^2(\mathbb{R}^d)\\
\end{eqnarray*}
%----------------------------------------------------------------------------------------
\end{document}
