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lista3.tex
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% Filename: lista3.tex
%
% This code is part of 'Solutions for MS650, Métodos de Matemática
% Aplicada II, and F620, Métodos Matemáticos da F\'{i}sica II'
%
% Description: This file corresponds to the solutions of homework sheet 3.
%
% Created: 14.07.12 11:13:03 AM Last Change: 14.07.12 11:13:03 AM
%
% Authors:
% - Raniere Silva (2012): initial version
%
% Copyright (c) 2012 Raniere Silva <[email protected]>
%
% This work is licensed under the Creative Commons Attribution-ShareAlike 3.0
% Unported License. To view a copy of this license, visit
% http://creativecommons.org/licenses/by-sa/3.0/ or send a letter to Creative
% Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA.
%
% This work is distributed in the hope that it will be useful, but WITHOUT ANY
% WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR
% A PARTICULAR PURPOSE.
%
\documentclass[a4paper,12pt, leqno, answers]{exam}
% Customização da classe exam
\newcommand{\mycheader}{Lista 3 - Transformada de Fourier}
\header{MS560, F560}{\mycheader}{\thepage/\numpages}
\headrule
\footer{Dispon\'{i}vel em \\\input{repository.tex}}{}{Reportar erros para
\\\input{maintainer.tex}}
\footrule
\pagestyle{headandfoot}
\renewcommand{\solutiontitle}{\noindent\textbf{Solução:}\enspace}
\SolutionEmphasis{\slshape}
\unframedsolutions
\pointname{}
\input{paper_size.tex}
\input{packages.tex}
\DeclareMathOperator{\sgn}{sgn}
\begin{document}
%cover
\thispagestyle{empty}
\input{cover.tex}
\newpage
\setcounter{page}{1}
\begin{questions}
\question Mostre que a transformada de Fourier das funções $f(x)$
abaixo são dadas pelas correspondentes funções $F(k)$ onde
$\sgn(x) = 2 H(x) - 1$ é a função sinal (ou seja, $\sgn(x) = 1$
se $x > 0$ e $\sgn(x) = -1$ se $x < 0$) e $H(x)$ é a função
escada (ou seja, $H(x) = 1$ se $x > 0$ e $H(x) = 0$ se $x < 0$).
\begin{center}
\begin{tabular}{|c|c|}
\hline
$f(x)$ & $F(x)$ \\ \hline
$(1 - x^2) H(1 - |x|)$ & $\left( -1 / \sqrt{2 \pi} \right) \left(
4 / k^3 \right) \left( k \cos(k) - \sin(k) \right)$ \\ \hline
$\sin\left( \alpha x \right)$ & $i \sqrt{\pi / 2} \left( \delta(k -
\alpha) - \delta(k + \alpha) \right)$ \\ \hline
$1 / x$ & $i \sqrt{\pi/2} \sgn(k) = i \sqrt{2 \pi} \left( H(k) - 1/2
\right)$ \\ \hline
$H(x)$ & $ \left( 1 / \sqrt{2 \pi} \right) \left( \pi \delta(k) + 1
/ k \right)$ \\ \hline
\end{tabular}
\end{center}
\begin{solution}
% TODO Escrever solução.
\end{solution}
\question Usando $\delta(x) = H'(x)$ e a expressão acima para a
transformada de Fourier de $H(x)$, mostre que $\mathcal{F}[\delta(x)] = 1 /
\sqrt{2 \pi}$.
\begin{solution}
% TODO Escrever solução.
\end{solution}
\question Mostre que
\begin{align*}
\int_0^\infty \left( \frac{x \cos(x) - \sin(x)}{x^3} \right) \cos(x/2)
\id{x} &= \frac{3 \pi}{16}.
\end{align*}
\begin{solution}
% TODO Escrever solução.
\end{solution}
\question Use a f\'{o}rmula integral de Fourier para mostrar que
\begin{parts}
\part $\int_0^\infty \left[ \left( \cos(xy) \right) / \left( 1 + y^2
\right) \right] \id{x} = \left( 2 / \pi \right) \exp(-|x|),$
\begin{solution}
% TODO Escrever solução.
\end{solution}
\part $\int_0^\infty \left[ \left( y \sin(xy) \right) / \left( 1 + y^2
\right) \right] \id{y} = \left( 2/\pi \right) \sgn(x) \exp(-|x|)$.
\begin{solution}
% TODO Escrever solução.
\end{solution}
\end{parts}
\question Use a identidade de Parseval para mostrar que
\begin{parts}
\part $\int_0^\infty \left[ 1 / \left( 1 + x^2 \right)^2 \right] \id{x}
= \pi / 4$,
\begin{solution}
% TODO Escrever solução.
\end{solution}
\part $\int_0^\infty \left[ \left( x \cos(x) - \sin(x) \right)^2 / x^6
\right] \id{x} = \pi / 15$.
\begin{solution}
% TODO Escrever solução.
\end{solution}
\end{parts}
\question Verifique a validade da convolução para as funções
\begin{align*}
f(x) = g(x) &= \begin{cases}
1, & |x| < 1, \\
0, & |x| > 1.
\end{cases}
\end{align*}
\begin{solution}
% TODO Escrever solução.
\end{solution}
\question Ilustre o Princ\'{i}pio de Incerteza de Heisenberg para a
função
\begin{align*}
f(x) &= \frac{a}{x^2 + a^2},
\end{align*}
com $a > 0$, mostrando que nesse caso $\Delta x = a$ e $\Delta k = 1 /
\sqrt{2} a$.
\begin{solution}
% TODO Escrever solução.
\end{solution}
\question Seja $\psi(x, t)$ uma função dada na forma
\begin{align*}
\psi(x, t) = \int_{-\infty}^\infty \phi(k) \exp\left( i (k x - k^2 t / 2
\right) \id{k}.
\end{align*}
Sabendo que
\begin{align*}
\psi(x, 0) &= \exp(-x^2 / 2),
\end{align*}
mostre que
\begin{align*}
\psi(x, t) &= \frac{1}{(1 + i t)^{1/2}} \exp\left( -x^2 / \left[ 2 (1 +
i t) \right] \right).
\end{align*}
\question[T3 de 2011] Seja $f(x)$ dada por
\begin{align*}
f(x) &= \begin{cases}
1 - \left( 1/2 \right) |x|, & |x| \leq 2, \\
0, & |x| > 2.
\end{cases}
\end{align*}
\begin{parts}
\part Mostre que a transformada de Fourier $F(k)$ de $f(x)$ é dada
por
\begin{align*}
F(k) &= \sqrt{\frac{2}{\pi}} \left( \frac{\sin(k)}{k} \right)^2.
\end{align*}
\begin{solution}
Temos que
\begin{align*}
F(k) &= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty f(x)
\exp\left( i k x \right) \id{x} \\
&= \frac{1}{\sqrt{2 \pi}} \int_{-2}^2 \left( 1 - |x| / 2 \right)
\exp\left( i k x \right) \id{x} \\
\begin{split}
&= \frac{1}{\sqrt{2 \pi}} \left[ \int_{-2}^0 \left( 1 + x / 2
\right) \exp\left( i k x \right) \id{x} \right. \\
&\quad \left. {}+ \int_0^2 \left( 1 - x
/ 2 \right) \exp\left( i k x \right) \id{x} \right] \\
\end{split} \\
\begin{split}
&= \frac{1}{\sqrt{2 \pi}} \left[ \left. \left(
\frac{\exp(i k x)}{i k} + \frac{x}{2} \frac{\exp(i k x}{i k}
- \frac{\exp(i k x)}{2 (i k)^2} \right) \right|_{-2}^0
\right. \\
&\quad \left. {}+ \left. \left( \frac{\exp(i k x}{i k} -
\frac{x}{2} \frac{\exp(i k x)}{i k} + \frac{\exp(i k x}{2 (i
k)^2} \right) \right|_0^2 \right]
\end{split} \\
&= \frac{1}{2 \sqrt{2 \pi}} \left[ \left( \frac{\exp(i k)}{i k}
\right)^2 - \frac{2}{(i k)^2} + \left( \frac{\exp(-i k)}{i k}
\right)^2 \right] \\
&= \frac{1}{2 \sqrt{2 \pi}} \left( \frac{\exp(i k) + \exp(-i
k)}{i k} \right)^2 \\
&= \sqrt{\frac{2}{\pi}} \left( \frac{\sin(k)}{k} \right)^2.
\end{align*}
\end{solution}
\part Use a identidade de Parseval para calcular a integral
\begin{align*}
\int_{-\infty}^\infty \left( \frac{\sin(k)}{k} \right)^4 \id{k}.
\end{align*}
\begin{solution}
Temos, pela identidade de Parseval, que
\begin{align*}
\int_{-\infty}^\infty |f(x)|^2 \id{x} &= \int_{-\infty}^\infty
|F(k)|^2 \id{k}.
\end{align*}
Logo,
\begin{align*}
\int_{-\infty}^\infty \left( \frac{\sin(k)}{k} \right)^4 \id{k}
&= \frac{\pi}{2} \int_{-\infty}^\infty \left( F(k) \right)^2
\id{k} \\
&= \frac{\pi}{2} \int_{-\infty}^\infty \left( f(x) \right)^2
\id{x} \\
&= \frac{\pi}{2} \left[ \int_{-2}^0 \left( 1 +
\frac{x}{2} \right)^2 \id{x} + \int_0^2 \left( 1 -
\frac{x}{2} \right)^2 \id{x} \right] \\
\begin{split}
&= \frac{\pi}{2} \left[ \left. \left( x +
\frac{x^2}{2} + \frac{x^3}{12} \right) \right|_{-2}^0
\right. \\
&\quad \left. {}+ \left. \left( x - \frac{x^2}{12} +
\frac{x^3}{12} \right) \right|_0^2 \right]
\end{split} \\
&= \frac{\pi}{2} \frac{16}{12} \\
&= 2 \pi / 3.
\end{align*}
\end{solution}
\end{parts}
\question[P1 de 2011]
\begin{parts}
\part Mostre que a transformada de Fourier de $f(x) = \exp(-a x^2)$
é dada por
\begin{align*}
F(k) &= \frac{1}{\sqrt{2 a}} \exp\left( -k^2 / (4a) \right).
\end{align*}
\begin{solution}
Temos que
\begin{align*}
F(k) &= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \exp\left(
-a x^2 \right) \exp\left( i k x \right) \id{x} \\
&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^\infty \exp\left[ -
\left( \sqrt{a}x - i k / (2 \sqrt{a}) \right)^2 \exp\left( - k^2
/ (4 a) \right) \right] \id{x} \\
&= \frac{\exp\left( -k^2 / (4a) \right)}{\sqrt{2 \pi}}
\int_{-\infty}^\infty \exp\left[ -\left( \sqrt{a} x - i k /
(2\sqrt{a}) \right)^2 \right] \id{x} \\
&= \frac{\exp\left( -k^2 / 4a \right)}{\sqrt{2 \pi}}
\frac{1}{\sqrt{a}} \int_{-\infty}^\infty \exp\left( -u^2
\right) \id{u} \\
&= \frac{\exp(-k^2 / (4a)}{\sqrt{2 a}}.
\end{align*}
\end{solution}
\part Discuta o comportamente de $f(x)$ e $F(k)$ para diferentes valores
de $a$ e no limite em que $a \to 0$.
\begin{solution}
Temos que $\lim_{a \to 0} f(x) = \exp(0) = 1$ e
\begin{align*}
F[1] &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \exp(i k x)
\id{x} \\
&= \sqrt{2 \pi} \frac{1}{2 \pi} \int_{-\infty}^\infty \exp(i k
x) \id{x} \\
&= \sqrt{2 \pi} \delta(k).
\end{align*}
Logo, $\lim_{a \to 0}\left( \exp(-k^2 / (4a)) / \sqrt{2a} \right) =
\sqrt{2 \pi} \delta(k)$.
\end{solution}
\end{parts}
\end{questions}
% \bibliographystyle{plain}
% \bibliography{bibliography}
\end{document}