diff --git a/extra/cap01@notas_de_aula.tex b/extra/cap01@notas_de_aula.tex index b4031b9..dab7ffc 100644 --- a/extra/cap01@notas_de_aula.tex +++ b/extra/cap01@notas_de_aula.tex @@ -51,7 +51,7 @@ \chapter{S\'{e}ries de Fourier} Temos que $\left\{ \cos\left( n x \right), \sin\left( n x \right) \right\}$ s\~{a}o fun\c{c}\~{o}es peri\'{o}dicas com $T = 2\pi$ e portanto $f(x)$ \'{e} uma fun\c{c}\~{a}o peri\'{o}dica, i.e., $f(x + 2\pi) = f(x)$. \end{obs} -\begin{exem} +\begin{exem} \label{exem:fourier:x^2} Para $f(x) = x^2$ temos que \begin{align*} a_0 &= \frac{1}{\pi} \int_{-\pi}^\pi x^2 \id{x} = \frac{1}{\pi} \left. \frac{x^3}{3} \right|_{-\pi}^\pi = \frac{2 \pi^2}{3}, \\ @@ -60,13 +60,14 @@ \chapter{S\'{e}ries de Fourier} &= \frac{-1}{n \pi} \left[ \left. \frac{-2 x \cos\left( n x \right)}{n} \right|_{-\pi}^\pi + \frac{1}{n} \int_{-\pi}^\pi 2 \cos\left( n x \right) \id{x} \right] \\ &= \frac{-1}{n \pi} \left[ \frac{-2 \pi \cos\left( n \pi) \right)}{n} + \frac{2 (-\pi) \cos(n) (-\pi)}{n} + \frac{2}{n} \underbrace{\left. \frac{\sin\left( n x \right)}{n} \right|_{-\pi}^\pi}_{= 0} \right] \\ &= \frac{-1}{n \pi} \left[ \frac{- 4 \pi}{n} \cos\left( n \pi \right) \right] \\ - &= \frac{4}{n^2} (-1)^n && n > 0 \\ + &= \frac{4}{n^2} (-1)^n, \\ b_n &= \frac{1}{n} \int_{-\pi}^\pi x^2 \sin\left( n x \right) \id{x} \\ &= \frac{1}{\pi} \left[ \left. \frac{- x^2 \cos\left( n x \right)}{n} \right|_{-\pi}^\pi + \frac{1}{n} \int_{-\pi}^\pi 2 x \cos\left( n x \right) \id{x} \right] \\ &= \frac{1}{n} \left[ \frac{-\pi^2 \cos\left( n \pi \right)}{n} + \frac{\pi \cos\left( n (-\pi) \right)}{n} + \frac{1}{n} \int_{-\pi}^\pi 2 x \cos\left( n x \right) \id{x} \right] \\ &= \frac{1}{n} \left[ \frac{1}{n} \left[ \underbrace{\left. \frac{2 x \sin\left( n x \right)}{n} \right|_{-\pi}^\pi}_{= 0} - \frac{1}{n} \underbrace{\int_{-\pi}^\pi 2 \sin\left( n x \right) \id{x}}_{= 0} \right] \right] \\ - &= 0 + &= 0, \end{align*} + onde $n > 0$. Logo, \begin{align*} @@ -213,7 +214,8 @@ \chapter{S\'{e}ries de Fourier} \begin{align*} \begin{split} \int_{-\pi}^\pi f(x) \cos\left( m x \right) \id{x} &= \frac{A_0}{2} \underbrace{\int_{-\pi}^\pi \cos\left( m x \right) \id{x}}_{= 0} \\ - &\quad {}+ \sum_{n = 1}^\infty \left( A_n \underbrace{\int_{-\pi}^\pi \cos\left( n x \right) \cos\left( m x \right) \id{x}}_{= 0 (n \neq m)} + B_n \underbrace{\int_{-\pi}^\pi \sin\left( n x \right) \cos\left( m x \right) \id{x}}_{= 0} \right) + &\quad {}+ \sum_{n = 1}^\infty \left( A_n \underbrace{\int_{-\pi}^\pi \cos\left( n x \right) \cos\left( m x \right) \id{x}}_{= 0 (n \neq m)} \right. \\ + &\quad \left. {}+ B_n \underbrace{\int_{-\pi}^\pi \sin\left( n x \right) \cos\left( m x \right) \id{x}}_{= 0} \right) \end{split} \end{align*} onde usamos a proposi\c{c}\~{a}o anterior, de modo que @@ -300,8 +302,8 @@ \section{Forma complexa} \begin{align*} \int_{-L}^L \exp\left( -i m \pi x / L \right) \exp\left( i n \pi x / L \right) \id{x} &= \int_{-L}^L \exp\left( i (n - m) \pi x / L \right) \id{x} \\ &= \begin{cases} - \int_{-L}^L \id{x} = 2 L, & n = m, \\ - \left. \exp\left( i (n - m) \pi x / L \right) \left( i (n - m) \pi / L \right)^{-1} \right|_{-L}^L = 0 + \int_{-L}^L \id{x} = 2 L, n = m, \\ + \left. \exp\left( \frac{i (n - m) \pi x}{L} \right) \left( \frac{i (n - m) \pi}{L} \right)^{-1} \right|_{-L}^L = 0, \end{cases} \\ &= 2 L \delta_{mn}. \end{align*} @@ -339,7 +341,10 @@ \section{Forma complexa} \end{align*} Tamb\'{e}m \'{e} poss\'{i}vel \begin{align*} - f(x) &= \frac{1}{2 L} + \sum_{n = 1}^\infty \frac{\sin\left( n \pi a / L \right)}{2 n \pi a} \exp\left( i n \pi a / L \right) + \sum_{n = 1}^\infty \frac{\sin\left( (-n) \pi a / L \right)}{2 (-n) \pi a} \exp\left( i (-n) \pi a / L \right) \\ + \begin{split} + f(x) &= \frac{1}{2 L} + \sum_{n = 1}^\infty \frac{\sin\left( n \pi a / L \right)}{2 n \pi a} \exp\left( i n \pi a / L \right) \\ + &\quad {}+ \sum_{n = 1}^\infty \frac{\sin\left( (-n) \pi a / L \right)}{2 (-n) \pi a} \exp\left( i (-n) \pi a / L \right) + \end{split} \\ &= \frac{1}{2L} + \sum_{n = 1}^\infty \frac{\sin\left( n \pi a / L \right)}{2 n \pi a} \left( \exp\left( i n \pi a / L \right) + \exp\left( -i n \pi a / L \right) \right) \\ &= \frac{1}{2L} + \sum_{n = 1}^\infty \frac{\sin\left( n \pi a / L \right)}{n \pi a} \cos\left( n \pi a / L \right). \end{align*} @@ -377,7 +382,7 @@ \section{Propriedades de Paridade: S\'{e}ries em Seno e Cosseno} &= \begin{cases} 2 \int_0^L f_\pm(x) \id{x}, & \text{para } f_+, \\ 0, & \text{para } f_-. - \end{cases} + \end{cases} \qedhere \end{align*} \end{proof} @@ -463,7 +468,10 @@ \section{Propriedades de Paridade: S\'{e}ries em Seno e Cosseno} \begin{align*} b_n &= \frac{2}{L} \int_0^L \sin\left( n \pi x / L \right) \left( \frac{x}{2} + \frac{1}{2} \right) \id{x} \\ &= \frac{1}{L^2} \int_0^L x \sin\left( n \pi x / L \right) \id{x} + \frac{1}{L} \int_0^L \sin\left( n \pi x / L \right) \id{x} \\ - &= \frac{1}{L^2} \left[ \left. \frac{- x \cos\left( n \pi x / L \right)}{n \pi / L} \right|_0^L + \frac{L}{n \pi} \int_0^L \cos\left( n \pi x / L \right) \id{x} \right] + \frac{1}{L} \left. \frac{- \cos\left( n \pi x / L \right)}{n \pi / L} \right|_0^L \\ + \begin{split} + &= \frac{1}{L^2} \left[ \left. \frac{- x \cos\left( n \pi x / L \right)}{n \pi / L} \right|_0^L + \frac{L}{n \pi} \int_0^L \cos\left( n \pi x / L \right) \id{x} \right] \\ + &\quad {}+ \frac{1}{L} \left. \frac{- \cos\left( n \pi x / L \right)}{n \pi / L} \right|_0^L + \end{split} \\ &= \frac{-1}{n \pi} \cos\left( n \pi \right) - \frac{1}{n \pi} \cos\left( n \pi \right) + \frac{1}{n \pi} \\ &= \frac{1 - 2 (-1)^n}{n \pi}. \end{align*} @@ -480,7 +488,10 @@ \section{Propriedades de Paridade: S\'{e}ries em Seno e Cosseno} &= \frac{3}{2}. \\ a_n &= \frac{2}{L} \int_0^L \left( \frac{x}{2L} + \frac{1}{2} \right) \cos\left( n \pi x / L \right) \\ &= \frac{1}{L^2} \int_0^L x \cos\left( n \pi x / L \right) \id{x} + \frac{1}{L} \int_0^L \cos\left( n \pi x / L \right) \id{x} \\ - &= \frac{1}{L^2} \left[ \left. \frac{x \sin\left( n \pi x / L \right)}{n \pi / L} \right|_0^L - \frac{L}{n \pi} \int_0^L \sin\left( n \pi x / L \right) \id{x} \right] + \frac{1}{L} \frac{1}{n \pi / L} \left. \sin\left( n \pi x / L \right) \right|_0^L \\ + \begin{split} + &= \frac{1}{L^2} \left[ \left. \frac{x \sin\left( n \pi x / L \right)}{n \pi / L} \right|_0^L - \frac{L}{n \pi} \int_0^L \sin\left( n \pi x / L \right) \id{x} \right] \\ + &\quad {}+ \frac{1}{L} \frac{1}{n \pi / L} \left. \sin\left( n \pi x / L \right) \right|_0^L + \end{split} \\ &= \frac{1}{L n \pi} \left. \frac{\cos\left( n \pi x / L \right)}{n \pi / L} \right|_0^L \\ &= \frac{1}{n^2 \pi^2} \left[ (-1)^n - 1 \right]. \end{align*} @@ -497,9 +508,14 @@ \section{Teorema de Fourier} \begin{align*} f(x) &= \frac{a_0}{2} + \sum_{n = 1}^\infty \left( a_n \cos\left( n x \right) + b_n \sin\left( n x \right) \right) \\ \begin{split} - &= \frac{1}{2} \left[ \frac{1}{\pi} \int_{-\pi}^\pi f(\xi) \id{\xi} \right] \\ &\quad {}+ \sum_{n = 1}^\infty \left[ \left( \frac{1}{\pi} \int_{-\pi}^\pi f(\xi) \cos\left( n \xi \right) \id{\xi} \right) \cos\left( n x \right) + \left( \frac{1}{\pi} \int_{-\pi}^\pi f(\xi) \sin\left( n \xi \right) \id{\xi} \right) \sin\left( n x \right) \right] + &= \frac{1}{2} \left[ \frac{1}{\pi} \int_{-\pi}^\pi f(\xi) \id{\xi} \right] \\ + &\quad {}+ \sum_{n = 1}^\infty \left[ \left( \frac{1}{\pi} \int_{-\pi}^\pi f(\xi) \cos\left( n \xi \right) \id{\xi} \right) \cos\left( n x \right) \right. \\ + &\quad \left. {}+ \left( \frac{1}{\pi} \int_{-\pi}^\pi f(\xi) \sin\left( n \xi \right) \id{\xi} \right) \sin\left( n x \right) \right] \end{split} \\ - &= \frac{1}{2\pi} \int_{-\pi}^\pi f(\xi) \id{\xi} + \frac{1}{\pi} \sum_{n = 1}^\infty \int_{-\pi}^\pi f(\xi) \left[ \cos\left( n \xi \right) \cos\left( n x \right) + \sin\left( n \xi \right) \sin\left( n x \right) \right] \id{\xi}. + \begin{split} + &= \frac{1}{2\pi} \int_{-\pi}^\pi f(\xi) \id{\xi} \\ + &\quad {}+ \frac{1}{\pi} \sum_{n = 1}^\infty \int_{-\pi}^\pi f(\xi) \left[ \cos\left( n \xi \right) \cos\left( n x \right) + \sin\left( n \xi \right) \sin\left( n x \right) \right] \id{\xi}. + \end{split} \end{align*} Portanto, \begin{align} @@ -507,7 +523,7 @@ \section{Teorema de Fourier} \label{eq:serie_fourier} \end{align} -\begin{teo}[Fourier] +\begin{teo}[Fourier] \label{teo:fourier} Seja $f(x)$ uma fun\c{c}\~{a}o cont\'{i}nua por partes e com derivadas laterais no intervalo $(-\pi, \pi)$ e peri\'{o}dica com per\'{i}odo $2\pi$. Ent\~{a}o sua s\'{e}rie de Fourier, \eqref{eq:serie_fourier}, converge para o valor \begin{align*} \frac{1}{2} \left[ f(x + 0) + f(x - 0) \right] @@ -607,7 +623,70 @@ \subsection{Derivadas Laterais} Portanto, $f'(x) = 2 x \sin\left( 1 / x \right) - \cos\left( 1 / x \right)$ que implica n\~{a}o existir $f'(0+)$ e nem $f'(0-)$ de maneira que n\~{a}o existe $f'(0)$. Neste caso, $f_+'(0) \neq f'(0+)$ e $f_-'(0) \neq f'(0-)$. \end{exem} -% TODO Incluir arquivo M2S12-3.pdf +\subsection{Lemas auxiliares} +\begin{lem} \label{lem:cont} + Se $F$ \'{e} cont\'{i}nua por partes em $[a,b]$, ent\~{a}o + \begin{enumerate} + \item \label{enum:cont:sin} $\lim_{k \to \infty} \int_a^b F(x) \sin\left( k x \right) \id{x} = 0$, + \item \label{enum:cont:cos} $\lim_{k \to \infty} \int_a^b F(x) \cos\left( k x \right) \id{x} = 0$. + \end{enumerate} +\end{lem} +\begin{proof} + Vamos dividir $(a,b)$ em intervalos onde $F$ \'{e} cont\'{i}nua. Vamos denotar um desses intervalos por $[p,q]$. Ent\~{a}o vale o item~\ref{enum:cont:sin} se + \begin{align*} + I = \lim_{k \to \infty} \int_p^q F(x) \sin\left( k x \right) \id{x} = 0. + \end{align*} + % TODO Terminar a demonstra\c{c}\~{a}o do Lema. +\end{proof} + +\begin{lem} \label{lem:sin(x)/x} + $\int_0^\infty \sin\left( x \right) / x \id{x} = \pi / 2$. +\end{lem} +\begin{proof} + Temos que + \begin{align*} + \int_0^\infty \frac{\sin\left( x \right)}{x} \id{x} = F(0), + \end{align*} + onde $F(x) = \int_0^\infty \exp\left( -t x \right) \sin\left( x \right) / x \id{x}$ para $t \geq 0$. + + Seja $S(x) = \sin\left( x \right) / x$. + % TODO Terminar a demonstra\c{c}\~{a}o do Lema. +\end{proof} + +\begin{lem} \label{lem:F_+'(x)} + Se $F$ \'{e} cont\'{i}nua por partes em $[0,b]$ e tem derivada \`{a} direita $F_+'(0)$, ent\~{a}o + \begin{align*} + \lim_{k \to \infty} \int_0^b F(x) \frac{\sin\left( k x \right)}{x} \id{x} &= \frac{\pi}{2} F(0+). + \end{align*} +\end{lem} +\begin{proof} + Temos que + \begin{align*} + I(k) &= \int_0^b F(x) \frac{\sin\left( k x \right)}{x} \id{x} \\ + &= \lim_{k \to \infty} \int_0^{k b} \frac{\sin\left( t \right)}{t} \id{t} \\ + &= \int_0^\infty \frac{\sin\left( t \right)}{t} \id{t} \\ + \intertext{e pelo Lema~\ref{lem:sin(x)/x}} + I(k) &= \pi / 2. + \end{align*} + % TODO Terminar a demonstra\c{c}\~{a}o do Lema. +\end{proof} + +\begin{lem} + Seja $F$ uma fun\c{c}\~{a}o cont\'{i}nua por partes em $(a,b)$ e que tem derivadas laterias \`{a} esquerda e \`{a} direita em um ponto $x_0$ tal que $a < x_0 < b$. Ent\~{a}o + \begin{align*} + \lim_{k \to \infty} \int_a^ F(x) \frac{\sin\left( k \left( x - x_0 \right) \right)}{x - x_0} \id{x} &= \pi \frac{F\left( x_0 + 0 \right) + F\left( x_0 - 0 \right)}{2}. + \end{align*} +\end{lem} +\begin{proof} + Temos que + \begin{align*} + I(k) &= \int_a^b F(x) \frac{\sin\left( k \left( x - x_0 \right) \right)}{x - x_0} \id{x} + \end{align*} + % TODO Terminar a demonstra\c{c}\~{a}o do Lema. +\end{proof} + +\subsection{Demonstra\c{c}\~{a}o do Teorema de Fourier} +% TODO Terminar de incluir o arquivo M2S12-3.pdf. Interrompido na p\'{a}gina 8. \section{Converg\^{e}ncia na M\'{e}dia} Da express\~{a}o para o produto escalar em $\mathcal{L}^2(a, b)$ definimos uma norma atrav\'{e}s de @@ -634,14 +713,18 @@ \section{Converg\^{e}ncia na M\'{e}dia} &= \int_{-\pi}^\pi \left[ f(x) - \frac{A_0}{2} - \sum_{n = 1}^N \left( A_n \cos\left( n x \right) + B_n \sin\left( n x \right) \right) \right]^2 \id{x} \\ \begin{split} &= \int_{-\pi}^\pi \left[ f(x) \right]^2 \id{x} + \int_{-\pi}^\pi \left( \frac{A_0}{2}^2 \right) \id{x} + \int_{-\pi}^\pi \sum_{n = 1}^N \sum_{m = 1}^N A_n A_m \cos\left( n x \right) \cos\left( m x \right) \id{x} \\ - &\quad {}+ \int_{-\pi}^\pi \sum_{n = 1}^N \sum_{m = 1}^N B_n B_m \sin\left( n x \right) \sin\left( m x \right) \id{x} - 2 \int_{-\pi}^\pi f(x) \frac{A_0}{2} \id{x} - 2 \int_{-\pi}^\pi f(x) \sum_{n = 1}^N \cos\left( n x \right) \id{x} \\ - &\quad {}- 2 \int_{-\pi}^\pi f(x) \sum_{n = 1}^N B_n \sin\left( n x \right) \id{x} + 2 \int_{-\pi}^\pi \frac{A_0}{2} \cos\left( n x \right) \id{x} + 2 \int_{-\pi}^\pi \frac{A_0}{2} \sum){n = 1}^N B_n \sin\left( n x \right) \id{x} \\ + &\quad {}+ \int_{-\pi}^\pi \sum_{n = 1}^N \sum_{m = 1}^N B_n B_m \sin\left( n x \right) \sin\left( m x \right) \id{x} - 2 \int_{-\pi}^\pi f(x) \frac{A_0}{2} \id{x} \\ + &\quad {}- 2 \int_{-\pi}^\pi f(x) \sum_{n = 1}^N \cos\left( n x \right) \id{x} \\ + &\quad {}- 2 \int_{-\pi}^\pi f(x) \sum_{n = 1}^N B_n \sin\left( n x \right) \id{x} + 2 \int_{-\pi}^\pi \frac{A_0}{2} \cos\left( n x \right) \id{x} \\ + &\quad {}+ 2 \int_{-\pi}^\pi \frac{A_0}{2} \sum){n = 1}^N B_n \sin\left( n x \right) \id{x} \\ &\quad {}+ 2 \int_{-\pi}^\pi \sum_{n = 1}^N \sum_{m = 1}^N A_n B_n \cos\left( n x \right) \sin\left( m x \right) \id{x} \end{split} \\ \begin{split} &= \int_{-\pi}^\pi \left[ f(x) \right]^2 \id{x} + \frac{A_0^2}{4} 2 \pi + \sum_{n = 1}^N \sum_{m = 1}^N A_n A_m \underbrace{\int_{-\pi}^\pi \cos\left( n x \right) \cos\left( m x \right) \id{x}}_{\pi \delta_{nm}} \\ - &\quad {}+ \sum_{n = 1}^N \sum_{m = 1}^N B_n B_m \underbrace{\int_{-\pi}^\pi \sin\left( n x \right) \sin\left( m x \right) \id{x}}_{\pi \delta_{nm}} - A_0 \int_{-\pi}^\pi f(x) \id{x} - 2 \sum_{n = 1}^N A_n \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} \\ - &\quad {}- 2 \sum_{n = 1}^N B_n \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} + A_0 \sum_{n = 1}^N A_n \underbrace{\int_{-\pi}^\pi \cos\left( n x \right) \id{x}}_{=0} + A_0 \sum_{n = 1}^N B_n \underbrace{\int_{-\pi}^\pi \sin\left( n x \right) \id{x}}_{=0} \\ + &\quad {}+ \sum_{n = 1}^N \sum_{m = 1}^N B_n B_m \underbrace{\int_{-\pi}^\pi \sin\left( n x \right) \sin\left( m x \right) \id{x}}_{\pi \delta_{nm}} - A_0 \int_{-\pi}^\pi f(x) \id{x} \\ + &\quad {}- 2 \sum_{n = 1}^N A_n \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} \\ + &\quad {}- 2 \sum_{n = 1}^N B_n \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} + A_0 \sum_{n = 1}^N A_n \underbrace{\int_{-\pi}^\pi \cos\left( n x \right) \id{x}}_{=0} \\ + &\quad {}+ A_0 \sum_{n = 1}^N B_n \underbrace{\int_{-\pi}^\pi \sin\left( n x \right) \id{x}}_{=0} \\ &\quad {}+ 2 \sum_{n = 1}^N \sum_{m = 1}^N A_n B_m \underbrace{\int_{-\pi}^\pi \cos\left( n x \right) \sin\left( m x \right) \id{x}}_{=0} \end{split} \end{align*} @@ -649,7 +732,8 @@ \section{Converg\^{e}ncia na M\'{e}dia} \begin{align*} \begin{split} \delta_N &= \int_{-\pi}^\pi \left[ f(x) \right]^2 \id{x} + \frac{\pi}{2} A_0^2 + \sum_{n = 1}^N \pi A_n^2 + \sum_{n = 1}^N \pi B_n^2 \\ - &\quad {}- A_0 \int_{-\pi}^\pi f(x) \id{x} - 2 \sum_{n = 1}^N A_n \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} - 2 \sum_{n = 1}^N B_n \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} + &\quad {}- A_0 \int_{-\pi}^\pi f(x) \id{x} - 2 \sum_{n = 1}^N A_n \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} \\ + &\quad {}- 2 \sum_{n = 1}^N B_n \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} \end{split} \end{align*} Exigindo agora que @@ -738,11 +822,20 @@ \section{Converg\^{e}ncia na M\'{e}dia} Temos que \begin{align*} &= \int_{-\pi}^\pi f(x) \left[ \frac{A_0}{2} + \sum_{n = 1}^N \left( A_n \cos\left( n x \right) + B_n \sin\left( n x \right) \right) \right] \id{x} \\ - &= \frac{A_0}{2} \int_{-\pi}^\pi f(x) \id{x} + \sum_{n = 1}^N A_n \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} + \sum_{n = 1}^N B_n \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} \\ + \begin{split} + &= \frac{A_0}{2} \int_{-\pi}^\pi f(x) \id{x} + \sum_{n = 1}^N A_n \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} \\ + &\quad {}+ \sum_{n = 1}^N B_n \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} + \end{split} \\ &= \frac{\pi}{2} A_0 a_0 + \pi \sum_{n = 1}^N \left( a_n A_n + b_n B_n \right) \\ \intertext{e} - &= \int_{-\pi}^\pi \left[ \frac{a_0}{2} + \sum_{n = 1}^N \left( a_n \cos\left( n x \right) + b_n \sin\left( n x \right) \right) \right] \left[ \frac{A_0}{2} + \sum_{m = 1}^N \left( A_m \cos\left( m x \right) + B_m \sin\left( m x \right) \right) \right] \id{x} \\ - &= \frac{\pi}{2} A_0 a_0 + \pi \sum_{n =1}^N\left( a_n A_n + b_n B_n \right) \qedhere + &= \int_{-\pi}^\pi \bigstar \id{x} \\ + \intertext{onde} + \begin{split} + \bigstar &= \left[ \frac{a_0}{2} + \sum_{n = 1}^N \left( a_n \cos\left( n x \right) + b_n \sin\left( n x \right) \right) \right] \\ + &\quad \left[ \frac{A_0}{2} + \sum_{m = 1}^N \left( A_m \cos\left( m x \right) + B_m \sin\left( m x \right) \right) \right] + \end{split} \\ + \intertext{e portanto} + &= \frac{\pi}{2} A_0 a_0 + \pi \sum_{n =1}^N\left( a_n A_n + b_n B_n \right) \qedhere \end{align*} \end{proof} @@ -826,7 +919,9 @@ \subsection{Coverg\^{e}ncia na M\'{e}dia \textit{versus} Converg\^{e}ncia Pontua \end{align*} Ent\~{a}o, \begin{align*} - \| f_n \|^2 = \int_0^1 f_n^2(x) \id{x} = \int_0^{1/n^3} n^2 \id{x} = n^2 n^{-3} = n^{-1}, \\ + \| f_n \|^2 &= \int_0^1 f_n^2(x) \id{x} \\ + &= \int_0^{1/n^3} n^2 \id{x} = n^2 n^{-3} \\ + &= n^{-1}, \\ \| f_n \| &= n^{-1/2}, \\ \lim_{n \to \infty} \| f_n \| &= 0 \end{align*} @@ -934,29 +1029,383 @@ \subsection{Coverg\^{e}ncia na M\'{e}dia \textit{versus} Converg\^{e}ncia Pontua Resumindo os tipos de converg\^{e}ncia e suas condi\c{c}\~{o}es suficientes, temos: \begin{center} \begin{tikzpicture} - \node[draw=black, text width=6cm] (U) at (0,0) {Converg\^{e}ncia Uniforme + \node[draw=black, text width=5.5cm] (U) at (0,0) {Converg\^{e}ncia Uniforme \begin{enumerate} \item $f$ \'{e} cont\'{i}nua; \item $f(\pi) = f(-\pi)$; \item $f'$ \'{e} cont\'{i}nua por partes. \end{enumerate}}; - \node[draw=black, text width=6cm] (P) at (-5,-6) {Converg\^{e}ncia Pontual + \node[draw=black, text width=5.5cm] (P) at (-4.5,-6) {Converg\^{e}ncia Pontual \begin{enumerate} \item $f$ cont\'{i}nua por partes e peri\'{o}dica; \item Existem as derivadas laterais; \item \textbf{OBS:} Converge para $1/2 \left[ f(x + 0) + f(x - 0) \right]$. \end{enumerate}}; - \node[draw=black, text width=6cm] (M) at (5,-6) {Converg\^{e}ncia na M\'{e}dia + \node[draw=black, text width=5.5cm] (M) at (4.5,-6) {Converg\^{e}ncia na M\'{e}dia \begin{enumerate} \item $f \in \mathcal{L}^2(-\pi, \pi)$; \item $f$ \'{e} peri\'{o}dica. \end{enumerate}}; - \draw[->, line width=.2cm] (U.south) -- (P.north); - \draw[->, line width=.2cm] (U.south) -- (M.north); + \draw[->, line width=.2cm] (U.south west) -- (P.north); + \draw[->, line width=.2cm] (U.south east) -- (M.north); \draw[<->, line width=.2cm] (P.east) -- (M.west) node[midway, below, text width=2.5cm] {Exceto em um conjunto de medida nula}; \end{tikzpicture} \end{center} -% TODO Incluir arquivo M2S12-5.pdf +\section{M\'{e}todo de Fej\'{e}r} +\begin{defi} + Dadas as somas parciais $S_k(x) = \sum_{i = 0}^k \mu_i(x)$ para $k = 0, 1, \ldots, N - 1$, a soma de Ces\`{a}ro (ou m\'{e}dia C-1) \'{e} definida como a meia aritm\'{e}tica dessas somas parciais: + \begin{align*} + \sigma_N(x) &= \frac{1}{N} \sum_{k = 0}^{N - 1} S_k(x). + \end{align*} + Dizemos que uma s\'{e}rie \'{e} C-1 som\'{a}vel se existir o $\lim_{N \to \infty} \sigma_N(x)$. +\end{defi} +\begin{obs} + Podemos definir a m\'{e}dia C-2, etc, como a m\'{e}dia aritm\'{e}tica das somas de Ces\`{a}ro, ou seja, + \begin{align*} + \sigma_N^{(2)}(x) &= \frac{1}{N} \sum_{k = 0}^{N - 1} \sigma_N(x), + \end{align*} + e dizer que a s\'{e}rie \'{e} C-2 som\'{a}vel se existir $\lim_{N \to \infty} \sigma_N^{(2)}(x)$. +\end{obs} +\begin{exem} + Considere $S_k(x) = \sum_{i = 0}^{k - 1} (-1)^i$, ent\~{a}o + \begin{align*} + S_0 &= 1, \\ + S_1 &= 1 - 1 = 0, \\ + S_2 &= 1 - 1 + 1 = 1, + \vdots + \end{align*} + e + \begin{align*} + S_0 &= 1, \\ + S_0 + S_1 &= 1, \\ + S_0 + S_1 + S_2 &= 2, \\ + \vdots + \end{align*} + Logo, + \begin{align*} + \sigma_0 &= \frac{1}{1} 1 = 1, & \sigma_1 &= \frac{1}{2} 1 = \frac{1}{2}, \\ + \sigma_2 &= \frac{1}{3} 2 = \frac{2}{3}, & \sigma_3 &= \frac{1}{4} 2 = \frac{1}{2}, \\ + \sigma_4 &= \frac{1}{5} 3 = \frac{3}{5}, & \sigma_5 &= \frac{1}{6} 3 = \frac{1}{2}, \\ + \vdots && \vdots \\ + \sigma_{2n} &= \frac{n + 1}{2n + 1}, & \sigma_{2n + 1} &= \frac{1}{2}. + \end{align*} + e como + \begin{align*} + & \lim_{n \to \infty} \sigma_{2n + 1} = 1/2, \\ + & \lim_{n \to \infty} \sigma_{2n} = 1/2, + \end{align*} + temos que a s\'{e}rie $\sum_{i = 0}^\infty (-1)^i$ \'{e} C-1 som\'{a}vel e o resultado \'{e} $1/2$. +\end{exem} +\begin{obs} + Usar a soma de Ces\`{a}ro da s\'{e}rie de Fourier. + \begin{align*} + \sigma_1 &= \frac{a_0}{2}, \\ + \sigma_2 &= \frac{1}{2} \left[ \frac{a_0}{2} + \left[ \frac{a_0}{2} + \left( a_1 \cos\left( x \right) + b_1 \sin\left( x \right) \right) \right] \right], \\ + \begin{split} + \sigma_3 &= \frac{1}{3} \left[ \frac{a_0}{2} + \left[ \frac{a_0}{2} + \sum_{k = 1}^1 \left( a_k \cos\left( k x \right) + b_k \sin\left( k x \right) \right) \right] \right. \\ + &\quad \left. {}+ \left[ \frac{a_0}{2} + \sum_{k = 1}^2 \left( a_k \cos\left( k x \right) + b_k \sin\left( k x \right) \right) \right] \right], + \end{split} \\ + \vdots \\ + \begin{split} + \sigma_N &= \frac{1}{N} \left[ \frac{a_0}{2} + \left[ \frac{a_0}{2} + \sum_{k = 1}^1 \left( a_k \cos\left( k x \right) + b_k \sin\left( k x \right) \right) \right] \right. \\ + &\quad {}+ \left[ \frac{a_0}{2} + \sum_{k = 1}^2 \left( a_k \cos\left( k x \right) + b_k \sin\left( k x \right) \right) \right] \\ + &\quad \vdots \\ + &\quad \left. {}+ \left[ \frac{a_0}{2} + \sum_{k = 1}^{N - 1} \left( a_k \cos\left( k x \right) + b_k \sin\left( k x \right) \right) \right]\right] \\ + &= \left( \frac{N}{N} \right) \left( \frac{a_0}{2} \right) + \left( \frac{N - 1}{N} \right) \left( a_1 \cos\left( x \right) + b_1 \sin\left( x \right) \right) \\ + &\quad {}+ \left( \frac{N - 2}{N} \left( a_2 \cos\left( 2 x \right) + b_2 \sin\left( 2 x \right) \right) \right) \\ + &\quad {}+ \cdots \\ + &\quad {}+ \left( \frac{N - \left( N - 1 \right)}{N} \right) \left( a_{N - 1} \cos\left( \left( N - 1 \right) x \right) + b_{N - 1} \sin\left( \left( N - 1 \right) x \right) \right) \\ + \end{split} + \end{align*} + Portanto, + \begin{align*} + \sigma_N(x) &= \frac{a_0}{2} + \sum_{k = 1}^{N - 1} \left( \alpha_k^N \cos\left( k x \right) + \beta_k^N \sin\left( k x \right) \right), \\ + \alpha_k^N &= \left( 1 - \frac{k}{N} \right) a_k, \\ + \beta_k^N &= \left( 1 - \frac{k}{N} \right) b_k. + \end{align*} + + Por outro lado, sabemos que + \begin{align*} + S_N(x) &= \int_{-\pi}^\pi f(\xi) D_N(\xi - x) \id{\xi}, + \end{align*} + onde $D_N$ denota o n\'{u}cleo de Dirichlet. Logo + \begin{align*} + \sigma_N(x) &= \frac{1}{N} \sum_{k = 0}^{N - 1} \int_{-\pi}^\pi f(\xi) D_k(\xi - x) \id{\xi} + \end{align*} + ou ainda + \begin{align*} + \sigma_N(x) &= \int_{-\pi}^\pi f(\xi) F_N(\xi - x) \id{\xi}, + \end{align*} + onde + \begin{align*} + F_N(\mu) &= \frac{1}{N} \sum_{k = 0}^{N - 1} D_k(\mu). + \end{align*} + + Lembrando que $D_k(\mu) = \left[ \sin\left( k + 1/2 \right) \mu \right] / \left[ 2 \pi \sin\left( \mu/2 \right) \right]$, temos + \begin{align*} + \sin^2\left( \mu/2 \right) F_N(\mu) &= \frac{1}{N} \sin^2\left( \mu/2 \right) \sum_{k = 0}^{N - 1} \frac{\sin\left( k + 1/2 \right) \mu}{2 \pi \sin\left( \mu/2 \right)} \\ + &= \frac{1}{2 \pi N} \sum_{k = 0}^{N - 1} \sin\left( \mu/2 \right) \sin\left( k + 1/2 \right) \mu \\ + &= \frac{1}{2 \pi N} \sum_{k = 0}^{N - 1} \frac{1}{2} \left[ \cos\left( k + 1/2 - 1/2 \right) \mu - \cos\left( k + 1/2 + 1/2 \right) \mu \right] \\ + &= \frac{1}{4 \pi N} \sum_{k = 0}^{N - 1} \left( \cos\left( k \mu \right) - \cos\left( \left( k + 1 \right) \mu \right) \right) \\ + &= \frac{1}{4 \pi N} \left( 1 - \cos\left( N \mu \right) \right) \\ + &= \frac{1}{2 \pi N} \left( \frac{1 - \cos\left( N \mu \right)}{2} \right) \\ + &= \frac{1}{2 \pi N} \sin^2\left( N \mu / 2 \right). + \end{align*} +\end{obs} +\begin{defi} + O n\'{u}cleo de Fej\'{e}r $F_N(\mu)$ \'{e} definido como + \begin{align*} + F_N(\mu) &= \frac{1}{N} \sum_{k = 0}^{N - 1} D_k(\mu) = \frac{\sin^2\left( N \mu / 2 \right)}{2 \pi N \sin^2\left( \mu/2 \right)}. + \end{align*} +\end{defi} + +Antes de prosseguir, \'{e} importante notermos que: +\begin{lem} + Temos que + \begin{align*} + & \int_{-\pi}^\pi D_k(\mu) \id{\mu} = 1, \\ + & \int_{-\pi}^\pi F_k\left( \mu \right) \id{\mu} = 1. + \end{align*} +\end{lem} +\begin{proof} + Sabendo que + \begin{align*} + D_k(\mu) &= \frac{\sin\left( k + 1/2 \right) \mu}{2 \pi \sin\left( \mu/2 \right)} \\ + &= \left( 1 + 2 \sum_{j = 1}^{k - 1} \cos\left( j \mu \right) \right) \frac{1}{2 \pi} + \end{align*} + temos que + \begin{align*} + \int_{-\pi}^\pi D_k\left( \mu \right) \id{\mu} &= \frac{1}{2 \pi} \left[ \pi - \left( -\pi \right) + 2 \sum_{j = 1}^{k - 1} \left. \frac{\sin\left( j \mu \right)}{j} \right|_{-\pi}^\pi \right] = 1 + \end{align*} + e + \begin{align*} + \int_{-\pi}^\pi F_k(\mu) \id{\mu} &= \frac{1}{k} \sum_{j = 0}^{k - 1} \int_{-\pi}^\pi D_j(\mu) \id{\mu} = \frac{1}{k} k = 1. \qedhere + \end{align*} +\end{proof} +\begin{teo}[Fej\'{e}r] \label{teo:fejer} + Seja $f(x)$ uma fun\c{c}\~{a}o cont\'{i}nua e peri\'{o}dica (de per\'{i}odo $2\pi$) e seja $\sigma_N(x)$ a soma de Ces\`{a}ro da s\'{e}rie de Fourier de $f(x)$, ou seja, + \begin{align*} + \sigma_N(x) &= \frac{a_0}{2} + \sum_{k = 1}^{N - 1} \left( \alpha_k^N \cos\left( k x \right) + \beta_k^N \sin\left( k x \right) \right), + \end{align*} + onde $\sigma_k^N = \left( 1 - k/N \right) a_k$ e $\beta_k^N = \left( 1 - k/N \right) b_k$. Ent\~{a}o a sequ\^{e}ncia de fun\c{c}\~{o}es $\sigma_n(x)$ converge para $f(x)$. +\end{teo} +\begin{proof} + (mais f\'{a}cil que Fourier pois $F_N(\mu) \geq 0$) + \begin{align*} + \sigma_N(x) &= \int_{-\pi}^\pi f(\xi) F_N(\xi - x) \id{\xi} \\ + &= \int_{a - \pi}^{a + \pi} f(\xi) F_{N}(\xi - x) \id{\xi} \\ + &= \int_{x - \pi}^{x + \pi} f(\xi) F_N(\xi - x) \id{\xi} \\ + &= \int_{-\pi}^\pi f(x + \mu) F_N(\mu) \id{\mu} \\ + \sigma_N(x) - f(x) &= \int_{-\pi}^\pi f(x + \mu) F_N(\mu) \id{\mu} - f(x) \int_{-\pi}^\pi F_N(\mu) \id{\mu} \\ + &= \int_{-\pi}^\pi \left[ f(x + \mu) - f(x) \right] F_N(\mu) \id{\mu} \\ + &= \frac{1}{\pi N} \int_{-\pi}^\pi \left[ f(x + \mu) - f(x) \right] \frac{\sin^2\left( N \mu/2 \right)}{2 \sin^2\left( \mu/2 \right)} \id{\mu} + \end{align*} + Logo, como $f$ \'{e} cont\'{i}nua ent\~{a}o $\forall \epsilon > 0, \exists \delta > 0$ tal que $| f(x + \mu) - f(x)| < \epsilon$ para $| x + \mu - x| = | \mu | < \delta$. + \begin{align*} + | \sigma_N(x) - f(x)| &= \left| \int_{-\pi}^\pi \left( \ldots \right) \right| \\ + &= \left| \int_{-\pi}^{-\delta} \left( \ldots \right) + \int_{-\sigma}^\sigma \left( \ldots \right) + \int_\sigma^\pi \left( \ldots \right) \right| \\ + &\leq \underbrace{\left| \int_{-\pi}^{-\delta} \left( \ldots \right) \right|}_{|I_1|} + \underbrace{\left| \int_{-\sigma}^\sigma \left( \ldots \right) \right|}_{|I_2|} + \underbrace{\left| \int_\sigma^\pi \left( \ldots \right) \right|}_{|I_3|} + \end{align*} + Mas + \begin{align*} + |I_2| &\leq \int_{-\delta}^\delta \underbrace{| f(x + \mu) - f(x) |}_{< \epsilon} F_N(\mu) \id{\mu} \\ + &< \epsilon \int_{-\delta}^\delta \underbrace{F_N(\mu)}_{\geq 0} \id{\mu} \\ + &< \epsilon \underbrace{\int_{-\delta}^\delta F_N(\mu) \id{\mu}}_{=1} \\ + &< \epsilon, \\ + |I_1| &\leq \frac{1}{\pi N} \int_{-\pi}^{-\delta} | f(x + \mu) - f(x) | \frac{\sin^2\left( N \mu/2 \right)}{2 \sin^2\left( \mu/2 \right)} \id{\mu} \\ + &\leq \frac{2 M}{\pi N} \int_{-\pi}^{-\delta} \frac{\sin^2\left( N \mu/2 \right)}{2 \sin^2\left( \mu/2 \right)} \id{\mu} \\ + \intertext{onde $M = \max |f(x)|$} + &\leq \frac{2 M}{\pi N} \int_{-\pi}^{-\delta} \frac{1}{2 \sin^2\left( \mu/2 \right)} \id{\mu} \\ + &\leq \frac{2 M}{\pi N} \int_{-\pi}^{-\delta} \frac{1}{2 \sin^2\left( \delta/2 \right)} \id{\mu}\\ + &= \frac{M}{\pi N \sin^2\left( \delta/2 \right)} \underbrace{\int_{-\pi}^{-\delta} \id{\mu}}_{\pi - \delta < \pi} \\ + &\leq \frac{M \pi}{\pi N \sin^2\left( \delta/2 \right)} \\ + &\leq \frac{M}{N \sin^2\left( \delta/2 \right)}, \\ + \intertext{e analogamente} + |I_3| &\leq \frac{M}{N \sin^2\left( \delta/2 \right)}. + \end{align*} + Portanto, + \begin{align*} + & |\sigma_N(x) - f(x)| \leq \epsilon + \frac{2 M}{N \sin^2\left( \delta/2 \right)}, \\ + & \lim_{N \to \infty} | \sigma_N(x) - f(x) | \leq \epsilon + \end{align*} + de onde conlu\'{i}mos que + \begin{align*} + \lim_{N \to \infty} \sigma_N(x) &= f(x) \qedhere + \end{align*} +\end{proof} + +\section{Diferencia\c{c}\~{a}o e Integra\c{c}\~{a}o} +\begin{teo} + Seja $f(x)$ uma fun\c{c}\~{a}o cont\'{i}nua em $[-\pi,\pi]$ tal que $f(-\pi) = f(\pi)$ e seja $f'(x)$ cont\'{i}nua por partes e com derivadas laterais nesse intervalo. Ent\~{a}o a s\'{e}rie de Fourier de $f(x)$ \'{e} diferencia\'{a}vel e + \begin{align*} + f'(x) &= \sum_{n = 1}^\infty n \left( -a_n \sin\left( n x \right) + b_n \cos\left( n x \right) \right) + \end{align*} +\end{teo} +\begin{proof} + Como $f'(x)$ satisfaz as condi\c{c}\~{o}es do Teorema de Fourier (p\'{a}gina~\pageref{teo:fourier}), ela possui a s\'{e}rie de Fourier + \begin{align*} + f'(x) &= \frac{\alpha_0}{2} + \sum_{n = 1}^\infty \left( \alpha_n \cos\left( n x \right) + \beta_n \sin\left( n x \right) \right). + \end{align*} + Mas + \begin{align*} + \alpha_0 &= \frac{1}{\pi} \int_{-\pi}^\pi f'(x) \id{x} \\ + &= \frac{1}{\pi} \left[ f(\pi) - f(-\pi) \right] \\ + &= 0, \\ + \alpha_n &= \frac{1}{\pi} \left[ \underbrace{\left. f(x) \cos\left( n x \right) \right|_{-\pi}^\pi}_{=0} + n \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} \right] \\ + &= n b_n \\ + \beta_n &= \frac{1}{\pi} \left[ \underbrace{\left. f(x) \sin\left( n x \right) \right|_{-\pi}^\pi}_{=0} - n \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} \right] \\ + &= -n a_n + \end{align*} + de modo que + \begin{align*} + f'(x) &= \sum_{n = 1}^\infty \left( n b_n \cos\left( n x \right) - n a_n \sin\left( n x \right) \right). \qedhere + \end{align*} +\end{proof} +\begin{obs} + Note que o teorema fornece apenas uma condi\c{c}\~{a}o suficiente. Al\'{e}m disso, podemos ter s\'{e}rie de Fourier para $f(x)$ e n\~{a}o para $f'(x)$ (nesse caso o termo $n$ no numerador diminui a taxa de converg\^{e}ncia da s\'{e}rie). +\end{obs} +\begin{exem} + A fun\c{c}\~{a}o $f(x) = x$ possui a s\'{e}rie de Fourier + \begin{align*} + x = 2 \sum_{n = 1}^\infty (-1)^{n + 1} \frac{\sin\left( n x \right)}{n}. + \end{align*} + Por\'{e}m $f(-\pi) = -\pi \neq \pi f(\pi)$, de modo que as condi\c{c}\~{o}es do teorema n\~{a}o s\~{a}o satisfeitas e nada podemos afirmar acerca da s\'{e}rie de Fourier de $f'(x)$ usando esse teorema. Note que nesse caso que a derivada da s\'{e}rie de Fourier n\~{a}o converge! + \begin{align*} + 1 \neq 2 \sum_{n = 1}^\infty (-1)^{n + 1} \cos\left( n x \right). + \end{align*} +\end{exem} +\begin{obs} + Dentro das condi\c{c}\~{o}es do teorema anterior, uma vez que $\lim_{n \to \infty} \alpha_n = 0$ e $\lim_{n \to \infty} \beta_n = 0$ (crit\'{e}rio do termo geral), temos + \begin{align*} + \lim_{n \to \infty} n a_n &= 0, & \lim_{n \to \infty} n b_n &= 0. + \end{align*} + Note que a s\'{e}rie do exemplo anterior n\~{a}o satisfaz essas condi\c{c}\~{o}es. +\end{obs} +\begin{teo} + Seja $f(x)$ uma fun\c{c}\~{a}o cont\'{i}nua por partes em $(-\pi,\pi)$. Ent\~{a}o, independentemente da s\'{e}rie de Fourier de $f(x)$ convergir ou n\~{a}o, vale a seguinte igualdade: + \begin{align*} + \int_{-\pi}^x f(\xi) \id{\xi} &= \frac{1}{2} a_0 \left( x + \pi \right) + \sum_{n = 1}^\infty \frac{1}{n} \left[ a_n \sin\left( n x \right) - b_n \left( \cos\left( n x \right) - \cos\left( n \pi \right) \right) \right], + \end{align*} + quando $x \in [-\pi,\pi]$. +\end{teo} +\begin{proof} + Seja + \begin{align*} + F(x) &= \int_{-\pi}^x f(\xi) \id{\xi} - \frac{a_0}{2} x. + \end{align*} + Se $f(x)$ \'{e} cont\'{i}nua por partes ent\~{a}o $F(x)$ \'{e} cont\'{i}nua. Al\'{e}m disso $F'(x) = f(x) - a_0/2$, de modo que $F'(x)$ \'{e} cont\'{i}nua por partes. Podemos ainda notar que $F(-\pi) = -a_0 / 2 (-\pi) = \pi a_0 / 2$ e + \begin{align*} + F(\pi) &= \int_{-\pi}^\pi f(\xi) \id{\xi} - \frac{a_0}{2} \pi \\ + &= a_0 \pi - \frac{a_0}{2} \pi \\ + &= \frac{a_0 \pi}{2} + \end{align*} + ou seja, $F(\pi) = F(-\pi)$. Temos com isso que $F(x)$ satisfaz as condi\c{c}\~{o}es do teorema de Fourier e possui portanto a s\'{e}rie de Fourier + \begin{align*} + F(x) &= \frac{A_0}{2} + \sum_{n = 1}^\infty \left( A_n \cos\left( n x \right) + B_n \sin\left( n x \right) \right), \\ + A_n &= \frac{1}{\pi} \int_{-\pi}^\pi F(x) \cos\left( n x \right) \id{x}, \\ + B_n &= \frac{1}{\pi} \int_{-\pi}^\pi F(x) \sin\left( n x \right) \id{x}. + \end{align*} + Para $n \neq 0$, + \begin{align*} + A_n &= \frac{1}{n} \int_{-\pi}^\pi \left[ \int_{-\pi}^x f(\xi) \id(\xi) - \frac{a_0}{2} x \right] \cos\left( n x \right) \id{x} \\ + &= \frac{1}{\pi} \underbrace{\left. F(x) \frac{\sin\left( n x \right)}{n} \right|_{-\pi}^\pi}_{=0} - \frac{1}{\pi} \int_{-\pi}^\pi \frac{\sin\left( n x \right)}{n} \left[ f(x) - \frac{a_0}{2} \right] \id{x} \\ + &= - \frac{1}{n \pi} \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} + \frac{a_0}{2 n \pi} \underbrace{\int_{-\pi}^\pi \sin\left( n x \right) \id{x}}_{=0} \\ + &= -\frac{1}{n} b_n, \\ + B_n &= \frac{1}{n} \int_{-\pi}^\pi \left[ \int_{-\pi}^x f(\xi) \id{\xi} - \frac{a_0 x}{2} \right] \sin\left( n x \right) \id{x} \\ + &= \underbrace{-\frac{1}{\pi} \left. F(x) \frac{\cos\left( n x \right)}{n} \right|_{-\pi}^\pi}_{=0} + \frac{1}{n \pi} \int_{-\pi}^\pi \cos\left( n x \right) \left[ f(x) - \frac{a_0}{2} \right] \id{x} \\ + &= \frac{1}{n \pi} \int_{-\pi}^\pi \cos\left( n x \right) f(x) \id{x} - \frac{a_0}{2 n \pi} \underbrace{\int_{-\pi}^\pi \cos\left( n x \right) \id{x}}_{=0} \\ + &= \frac{1}{n} a_n + \end{align*} + e portanto + \begin{align*} + F(x) &= \frac{A_0}{2} + \sum_{n = 1}^\infty \left( \frac{-b_n}{n} \cos\left( n x \right) + \frac{a_n}{n} \sin\left( n x \right) \right). + \end{align*} + Tomando $x = \pi$, + \begin{align*} + F(\pi) &= \frac{a_0 \pi}{2} = \frac{A_0}{2} \sum_{n = 1}^\infty \left( \frac{-b_n}{n} \cos\left( n \pi \right) + \frac{a_n}{n} \underbrace{\sin\left( n \pi \right)}_{=0} \right) + \end{align*} + e portanto + \begin{align*} + \frac{A_0}{2} &= \frac{a_0 \pi}{2} + \sum_{n = 1}^\infty \frac{b_n}{n} \cos\left( n \pi \right) + \end{align*} + de modo que + \begin{align*} + \int_{-\pi}^x f(\xi) \id{\xi} - \frac{a_0 x}{2} &= \frac{a_0 \pi}{2} + \sum_{n = 1}^\infty \frac{1}{n} \left[ a_n \sin\left( n x \right) - b_n \left( \cos\left( n x \right) - \cos\left( n \pi \right) \right) \right]. \qedhere + \end{align*} +\end{proof} +\begin{obs} + Note que no caso da integra\c{c}\~{a}o o termo $n$ no denominador aumenta a taxa de converg\^{e}ncia. +\end{obs} +\begin{exem} + A s\'{e}rie de $f(x) = x$ \'{e} + \begin{align*} + x = 2 \sum_{n = 1}^\infty (-1)^{n + 1} \frac{\sin\left( n x \right)}{n}. + \end{align*} + Tomando $\int_0^x f(\xi) \id{\xi}$ temos + \begin{align*} + x^2 &= 4 \sum_{n = 1}^\infty \frac{(-1)^n}{n^2} \cos\left( n x \right) - 4 \sum_{n = 1}^\infty \frac{(-1)^n}{n^2} + \end{align*} + que \'{e} justamente a s\'{e}rie obtida na p\'{a}gina~\pageref{exem:fourier:x^2} uma vez que $\sum_{n = 1}^\infty (-1)^n / (n^2) = -\pi^2 / 12$. +\end{exem} + +\section{Fen\^{o}meno de Gibbs} +Para ilustrar o fen\^{o}meno de Gibbs, vamos considerar uma onda quadrada peri\'{o}dica representada por +\begin{align*} + f(x) &= \begin{cases} + h/2, & 0 < x < \pi, \\ + -h/2, & -\pi < x < 0. + \end{cases} + 1;3P +\end{align*} +e ilustrada na Figura +\begin{figure}[!htb] + \centering + \includegraphics[width=.5\textwidth]{figuras/gibbs_phenomenon.png} + \begin{flushright} + Imagem em dom\'{i}nio p\'{u}blico de Johannes Rössel e dispon\'{i}vel em \url{http://en.wikipedia.org/wiki/File:Gibbs_phenomenon_10.svg}. + \end{flushright} + \caption{Ilustra\c{c}\~{a}o do Fen\^{o}meno de Gibbs para onda quadrada.} + \label{fig:fenom_gibbs} +\end{figure} +% TODO Terminar de digitar a se\c{c}\~{a}o sobre o Fen\^{o}meno de Gibbs do arquivo M2S12-5. + +\section{Teorema da Aproxima\c{c}\~{a}o de Weierstrass} +\begin{teo} + Seja $f(x)$ cont\'{i}nua em $a \leq x \leq b$. Ent\~{a}o para todo $\epsilon > 0$ existe um polin\^{o}mio $P(x)$ tal que $|P(x) - f(x)| \leq \epsilon, \forall x \in [a,b]$. +\end{teo} +\begin{proof} + Dado $f(x)$, $x \in [a,b]$, podemos definir $g(t)$ para $t \in [-\pi/2, \pi/2]$ atrav\'{e}s de + \begin{align*} + g(t) &= f\left[ \left( \frac{b - a}{\pi} t + \left( \frac{b - a}{2} \right) \right) \right]. + \end{align*} + Podemos ainda definir uma extens\~{a}o de $g(t)$ para $x \in [-\pi, \pi]$, que denotaremos por $G(t)$, e de modo que $G(-\pi) = G(\pi)$. Podemos al\'{e}m disso considerar para os demais pontos a extens\~{a}o peri\'{o}dica de $G(t)$. Nessas condi\c{c}\~{o}es o Teorema de Fej\'{e}r (p\'{a}gina~\pageref{teo:fejer}) nos garante que a sequ\^{e}ncia $\sigma_N(t)$, + \begin{align*} + \sigma_N(t) &= \frac{a_0}{2} + \sum_{k = 1}^{N - 1} \left( \alpha_k^N \cos\left( k t \right) + \beta_k^N \sin\left( k t \right) \right), + \end{align*} + converge para $G(t)$, ou seja, $\forall \epsilon > 0$, existe $N_0 > 0$ tal que + \begin{align*} + | \sigma_N(t) - G(t) | < \epsilon/2 + \end{align*} + para $N > N_0$. Por outro lado, pelo Teorema de Taylor, existem polin\^{o}mios $R_n^k(t)$ e $S_n^k(t)$ de grau $n$ tais que + \begin{align*} + | \cos\left( k t \right) - R_n^k(t) | &< \epsilon', & | \sin\left( k t \right) - S_n^k(t) | &< \epsilon'', + \end{align*} + para $n > N_1$. Portanto, existe um polin\^{o}mio $Q_N(t)$, que \'{e} uma combina\c{c}\~{a}o dos polin\^{o}mios $R_n^k(t)$ e $S_n^k(t)$, tal que + \begin{align*} + | \sigma_N(t) - Q_N(t) | < \epsilon/2. + \end{align*} + Com isso, + \begin{align*} + | G(t) - Q_N(t) | &= | G(t) - \sigma_N(t) + \sigma_N(t) - Q_N(t) | \\ + &\leq \underbrace{| G(t) - \sigma_N(t) |}_{< \epsilon/2} + \underbrace{| \sigma_N(t) - Q_N(t) |}_{< \epsilon/2} + \end{align*} + e portanto $| G(t) - Q_N(t) | < \epsilon$ para $t \in [-\pi, \pi]$ e $| g(t) - Q_N(t) | < \epsilon$ para $t \in [-\pi/2, \pi/2]$ ou, definindo $P_N(x) = Q_N\left[ \left( \pi / \left( b - a \right) \right) x - \left( \pi / 2 \right) \left( \left( b + a \right) / \left( b - a \right) \right) \right]$, + \begin{align*} + | f(x) - P_N(x) | < \epsilon + \end{align*} + para $x \in [a, b]$. +\end{proof} + % TODO Incluir arquivo M2S12-6.pdf % TODO Incluir arquivo M2S12-7.pdf diff --git a/extra/figuras/gibbs_phenomenon.png b/extra/figuras/gibbs_phenomenon.png new file mode 100644 index 0000000..b484ab1 Binary files /dev/null and b/extra/figuras/gibbs_phenomenon.png differ diff --git a/extra/makefile b/extra/makefile new file mode 100644 index 0000000..984e525 --- /dev/null +++ b/extra/makefile @@ -0,0 +1,21 @@ +all: notas_de_aula + +ereader: + -sed -i '20s/^/% /' paper_size.tex + -sed -i '23,24s/^% //' paper_size.tex + -make + -sed -i '20s/^% //' paper_size.tex + -sed -i '23,24s/^/% /' paper_size.tex + +tablet: + -sed -i '20s/^/% /' paper_size.tex + -sed -i '27,28s/^% //' paper_size.tex + -make + -sed -i '20s/^% //' paper_size.tex + -sed -i '27,28s/^/% /' paper_size.tex + +notas_de_aula: notas_de_aula.tex + -pdflatex -interaction nonstopmode -shell-escape notas_de_aula + -bibtex lista1 + -pdflatex -interaction nonstopmode -shell-escape notas_de_aula + -pdflatex -interaction nonstopmode -shell-escape notas_de_aula diff --git a/extra/notas_de_aula.tex b/extra/notas_de_aula.tex index b6d7857..444c04f 100644 --- a/extra/notas_de_aula.tex +++ b/extra/notas_de_aula.tex @@ -16,9 +16,7 @@ % 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[12pt,a4paper,leqno, openright,twoside]{book} -\usepackage[top=3cm,left=2cm,right=2cm,bottom=3cm]{geometry} -\usepackage[brazil]{babel} -\usepackage{indentfirst} +\input{paper_size.tex} \input{package.tex} \begin{document} \title{Notas de aula n\~{a}o oficiais de MS650 e F620} diff --git a/extra/package.tex b/extra/package.tex index 2ba1a7d..bec7e25 100644 --- a/extra/package.tex +++ b/extra/package.tex @@ -1,5 +1,5 @@ % Filename: package.tex -% This code is part of 'LaTeX with Vim'. +% This code is part of 'Notas de aula n\~{a}o oficiais de MS650 e F620' % % Description: This file correspond to the packages to be used. % @@ -11,15 +11,14 @@ % % Copyright (c) 2012, Raniere Silva. All rights reserved. % -% 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 licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/3.0/. % % 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. % \usepackage[utf8]{inputenc} \usepackage[T1]{fontenc} -% \usepackage[top=3cm,left=2cm,right=2cm,bottom=3cm]{geometry} % Set in the file. -% \usepackage[brazil]{babel} % Set in the file. -% \usepackage{indentfirst} % Set in the file. +\usepackage[brazil]{babel} +\usepackage{indentfirst} % Text \usepackage{enumerate} @@ -40,6 +39,7 @@ \usepackage{amsthm} \usepackage{amsfonts} \usepackage{amssymb} +\allowdisplaybreaks[4] \newtheorem{defi}{Defini\c{c}\~{a}o} \newtheorem{prop}{Proposi\c{c}\~{a}o} diff --git a/extra/paper_size.tex b/extra/paper_size.tex new file mode 100644 index 0000000..e604809 --- /dev/null +++ b/extra/paper_size.tex @@ -0,0 +1,28 @@ +% Filename: paper_size.tex +% +% This code is part of 'Notas de aula n\~{a}o oficiais de MS650 e F620' +% +% Description: This file corresponds to the paper size output. +% +% Created: 14.07.12 11:13:03 AM +% Last Change: 22.08.12 08:24:31 AM +% +% Authors: +% - Raniere Silva (2012): initial version +% +% Copyright (c) 2012 Raniere Silva +% +% This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivs 3.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/3.0/. +% +% 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. +% +% Para impress\~{a}o +\usepackage[top=3cm, bottom=3cm, left=2cm, right=2cm]{geometry} + +% Para ereaders (Kindle, Nook, Kobo, ...) +% \usepackage[papersize={160mm,200mm},margin=5mm]{geometry} +% \sloppy + +% Para tablets (iPad, GalaxyTab, ...) +% \usepackage[papersize={140mm,190mm},margin=5mm]{geometry} +% \sloppy diff --git a/lista1.tex b/lista1.tex index 0ec2234..c0457cc 100644 --- a/lista1.tex +++ b/lista1.tex @@ -39,6 +39,13 @@ \input{cover.tex} \newpage \setcounter{page}{1} + +Algumas express\~{o}es eventualmente \'{u}teis: +\begin{align} + & f(x) = \frac{a_0}{2} + \sum_{r = 1}^\infty \left[ a_r \cos\left( \frac{2 \pi r x}{L} \right) + b_r \sin\left( \frac{2 \pi r x}{L} \right) \right] \label{eq:serie_fourier} \\ + & a_r = \frac{2}{L} \int_{x_0}^{x_0 + L} f(x) \cos\left( \frac{2 \pi r x}{L} \right) \id{x} \label{eq:serie_fourier_a} \\ + & b_r = \frac{2}{L} \int_{x_0}^{x_0 + L} f(x) \sin\left( \frac{2 \pi r x}{L} \right) \id{x} \label{eq:serie_fourier_b} +\end{align} \begin{questions} \question Escreva a s\'{e}rie de Fourier no intervalor $(-\pi, \pi)$ das seguintes fun\c{c}\~{o}es e esboce os gr\'{a}fico das fun\c{c}\~{o}es representadas por essas s\'{e}ries para todo $x$: \begin{parts} @@ -47,11 +54,74 @@ x, & 0 < x < \pi. \end{cases}$ \begin{solution} - % TODO Escrever solu\c{c}\~{a}o. + Temos que + \begin{align*} + a_0 &= \frac{1}{\pi} \int_{-\pi}^\pi f(x) \id{x} \\ + &= \frac{1}{\pi} \int_{-\pi}^0 \left( -\pi \right) \id{x} + \frac{1}{\pi} \int_0^\pi x \id{x} \\ + &= \frac{1}{\pi} \left( -\pi \right) \left. x \right|_{-\pi}^0 + \frac{1}{\pi 2} \left. x^2 \right|_0^\pi \\ + &= -\left( 0 - \left( -\pi \right) \right) + \frac{1}{2 \pi} \left( \pi^2 - 0 \right) \\ + &= -\pi + \frac{\pi}{2} \\ + &= \frac{-\pi}{2} \\ + a_n &= \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} \\ + &= \frac{1}{\pi} \left[ \int_{-\pi}^0 \left( -\pi \right) \cos\left( n x \right) \id{x} + \int_0^\pi x \cos\left( n x \right) \id{x} \right] \\ + &= - \int_{-\pi}^0 \cos\left( n x \right) \id{x} + \frac{1}{\pi} \int_0^\pi x \cos\left( n x \right) \id{x} \\ + &= - \left. \frac{\sin\left( n x \right)}{n} \right|_{-\pi}^0 + \frac{1}{\pi} \left[ \left. \frac{x \sin\left( n x \right)}{n} \right|_0^\pi - \int_0^\pi \frac{\sin\left( n x \right)}{n} \id{x} \right] \\ + &= - \left( 0 - \frac{\sin\left( -n \pi \right)}{n} \right) + \frac{1}{\pi} \left[ \pi \frac{\sin\left( n \pi \right)}{n} - 0 + \frac{1}{n} \int_0^\pi \left( - \sin\left( n x \right) \right) \id{x} \right] \\ + &= \frac{1}{n \pi}\left. \frac{\cos\left( n x \right)}{n} \right|_0^\pi \\ + &= \frac{1}{n^2 \pi} \left( \cos\left( n \pi \right) - \cos(0) \right) \\ + &= \frac{1}{n^2 \pi} \left( (-1)^n - 1 \right) \\ + &= \frac{(-1)^n - 1}{n^2 \pi} \\ + b_n &= \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin\left( n x \right) \id{x} \\ + &= \frac{1}{\pi} \left[ \int_{-\pi}^0 \left( -\pi \right) \sin\left( n x \right) \id{x} + \int_0^\pi x \sin\left( n x \right) \id{x} \right] \\ + &= \int_{-\pi}^0 \left( -\sin\left( n x \right) \right) \id{x} + \frac{1}{\pi} \int_0^\pi x \sin\left( n x \right) \id{x} \\ + &= \left. \frac{\cos\left( n x \right)}{n} \right|_{-\pi}^0 + \frac{1}{\pi} \left[ \left. x \frac{\left( -\cos\left( n x \right) \right)}{n} \right|_0^\pi + \int_0^\pi \frac{\cos\left( n x \right)}{n} \id{x} \right] \\ + &= \frac{1}{n} \left( 1 - \cos\left( - n \pi \right) \right) - \frac{1}{n \pi} \left( \pi \cos\left( n \pi \right) - 0 \right) + \frac{1}{n} \left. \frac{\sin\left( n x \right)}{n} \right|_0^\pi \\ + &= \frac{1 - \left( -1 \right)^n}{n} - \frac{1}{n} (-1)^n + \frac{1}{n^2} \left( \sin\left( n \pi \right) - 0 \right) \\ + &= \frac{1 - (-1)^n - (-1)^n}{n} \\ + &= \frac{1 - 2 (-1)^n}{n}. + \end{align*} + Portanto, + \begin{align*} + f(x) &= \frac{a_0}{2} + \sum_{n = 1}^\infty \left( a_n \cos\left( n x \right) + b_n \sin\left( n x \right) \right) \\ + &= \frac{-\pi}{4} + \sum_{n = 1}^\infty \left( \frac{(-1)^n - 1}{n^2 \pi} \cos\left( n x \right) + \frac{1 - 2 (-1)^n}{n} \sin\left( n x \right) \right). + \end{align*} + % TODO Incluir gr\'{a}fico. \end{solution} - \part $f(x) = | \sin x |$. + \part $f(x) = | \sin x |, -\pi < x < \pi$. \begin{solution} + + Temos que $f(x) = | \sin x |, -\pi < x < \pi$ equivale a + \begin{align*} + f(x) &= \begin{cases} + \sin x, 0 < x < \pi, \\ + -\sin x, -\pi < x < \pi, + \end{cases} + \end{align*} + e, portanto, + \begin{align*} + a_0 &= \frac{1}{\pi} \int_{-\pi}^\pi f(x) \id{x} \\ + &= \frac{1}{\pi} \left[ \int_{-\pi}^0 \left( - \sin(x) \right) \id{x} + \int_0^\pi \sin(x) \id{x} \right] \\ + &= \frac{1}{\pi} \left[ \left. \cos(x) \right|_{-\pi}^0 + \left. \left( -\cos(x) \right) \right|_0^\pi \right] \\ + &= \frac{1}{\pi} \left[ 1 - (-1) - \left( -1 - 1 \right) \right] \\ + &= \frac{4}{\pi}, \\ + a_n &= \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos\left( n x \right) \id{x} \\ + &= \frac{1}{\pi} \left[ -\int_{-\pi}^0 \sin\left( x \right) \cos\left( n x \right) \id{x} + \int_0^\pi \sin\left( x \right) \cos\left( n x \right) \id{x} \right] \\ + \begin{split} + &= - \frac{1}{\pi} \int_{-\pi}^0 \frac{1}{2} \left( \sin\left( (1 + n) x \right) + \sin\left( (1 - n) x \right) \right) \id{x} \\ + &\quad {}+ \frac{1}{\pi} \int_0^\pi \frac{1}{2} \left( \sin\left( (1 + n) x \right) + \sin\left( (1 - n) x \right) \right) \id{x} \\ + \end{split} \\ + \begin{split} + &= - \frac{1}{2 \pi} \int_{-\pi}^0 \sin\left( (1 + n) x \right) \id{x} - \frac{1}{2 \pi} \int_{-\pi}^0 \sin\left( (1 - n) x \right) \id{x} \\ + &\quad {}+ \frac{1}{2 \pi} \int_0^\pi \sin\left( (1 + n) x \right) \id{x} + \frac{1}{2 \pi} \int_0^\pi \sin\left( (1 - n) x \right) \id{x} + \end{split} \\ + \begin{split} + &= \frac{1}{2 \pi} \left. \frac{\cos\left( (1 + n) x \right)}{1 + n} \right|_{-\pi}^0 + \frac{1}{2 \pi} \left. \frac{\cos\left( (1 - n) x \right)}{1 - n} \right|_{-\pi}^0 \\ + &\quad {}- \frac{1}{2 \pi} \left. \frac{\cos\left( (1 + n) x \right)}{1 + n} \right|_0^\pi - \frac{1}{2 \pi} \left. \frac{\cos\left( (1 - n) x \right)}{1 - n} \right|_0^\pi + \end{split} \\ + % TODO Incluir passagens intermedi\'{a}rias. + &= \frac{1 - (-1)^{n + 1}}{\pi} \left[ \frac{1}{1 + n} + \frac{1}{1 - n} \right]. + \end{align*} % TODO Escrever solu\c{c}\~{a}o. \end{solution} @@ -63,7 +133,7 @@ % TODO Escrever solu\c{c}\~{a}o. \end{solution} - \part $f(x) = \cosh x$. + \part $f(x) = \cosh x, -\pi < x < \pi$. \begin{solution} % TODO Escrever solu\c{c}\~{a}o. \end{solution} @@ -148,6 +218,156 @@ % TODO Escrver solu\c{c}\~{a}o. \end{solution} + \question[P1 de 2006] Considdere a fun\c{c}\~{a}o $f(x) = x^3 + 1$. + % TODO Escrever integra\c{c}\~{a}o. + \begin{parts} + \part Encontre sua s\'{e}rie de Fourier no intervalo $(-\pi,\pi)$; + \begin{solution} + Temos que + \begin{align*} + a_0 &= \frac{1}{\pi} \int_{-\pi}^\pi \left( x^3 + 1 \right) \id{x} \\ + &= \frac{1}{4} \left. \left( \frac{x^4}{4} + x \right) \right|_{-\pi}^\pi \\ + &= 2 \pi / \pi = 2, \\ + a_n &= \frac{1}{\pi} \int_{-\pi}^\pi \left( x^3 + 1 \right) \cos\left( n x \right) \id{x} \\ + \begin{split} + &= \frac{1}{\pi} \left. \frac{\sin\left( n x \right)}{n} \right|_{-\pi}^\pi + \frac{1}{\pi} \frac{3 \pi^2 \cos\left( n \pi \right)}{n^2} - \frac{1}{\pi} \frac{3 (-\pi)^2 \cos\left( -n \pi \right)}{n^2} \\ + &\quad {}- \frac{1}{\pi} \frac{6}{n^4} \cos\left( n \pi \right) + \frac{6}{n^4} \cos\left( -n \pi \right) + \end{split} \\ + &= 0, \\ + b_n &= \frac{1}{\pi} \int_{-\pi}^\pi \left( x^3 + 1 \right) \sin\left( n x \right) \id{x} \\ + \begin{split} + &= \frac{1}{\pi} \left. \frac{- \cos\left( n x \right)}{n} \right|_{-\pi}^\pi - \frac{1}{\pi} \frac{\pi^3 \cos\left( n \pi \right)}{n} - \frac{1}{\pi} \frac{- (-\pi)^3 \cos\left( - n \pi \right)}{n} \\ + &\quad {}+ \frac{1}{\pi} \frac{6 \pi \cos\left( n \pi \right)}{n^3} - \frac{1}{\pi} \frac{6 (-\pi) \cos\left( n (-\pi) \right)}{n^3} + \end{split} \\ + &= \frac{2 (-1)^n [6 - \pi^2 n^2]}{n^3}. + \end{align*} + Portanto, + \begin{align*} + F[f](x) &= 1 + \sum_{n = 1}^\infty \frac{2 (-1)^n (6 - \pi^2 n^2)}{n^3} \sin\left( n x \right). + \end{align*} + \end{solution} + + \part Encontre sua s\'{e}rie de Fourier em senos no intervalo $(0,\pi)$; + \begin{solution} + Temos que + \begin{align*} + b_n &= \frac{2}{\pi} \int_0^\pi \left( x^3 + 1 \right) \sin\left( n x \right) \id{x} \\ + &= \frac{2}{\pi} \left[ \left. \frac{-\cos\left( n x \right)}{n} \right|_0^\pi - \frac{\pi^3 \cos\left( n \pi \right)}{n} + \frac{6 \pi \cos\left( n \pi \right)}{n^3} \right] \\ + &= \frac{2 (-1)^n}{\pi n} \left[ -1 + (-1)^n - \pi^3 + \frac{6 \pi}{n^2} \right]. + \end{align*} + Portanto, + \begin{align*} + F_s[f](x) &= \sum_{n = 1}^\infty \frac{2 (-1)^n}{\pi n} \left[ (-1)^n - 1 - \frac{\pi}{n^2} \left( 6 + \pi^2 n^2 \right) \right] \sin\left( n x \right). + \end{align*} + \end{solution} + + \part Enconre sua s\'{e}rie de Fourier em cossenos no intervalo $(-\pi,0)$; + \begin{solution} + Temos que + \begin{align*} + a_0 &= \frac{2}{\pi} \int_{-\pi}^0 \left( x^3 + 1 \right) \id{x} \\ + &= \frac{2}{\pi} \left. \left( \frac{x^4}{4} + x \right) \right|_{-\pi}^0 \\ + &= 2 \left( 1 - \frac{\pi^3}{4} \right), \\ + a_n &= \frac{2}{\pi} \int_{-\pi}^0 \left( x^3 + 1 \right) \cos\left( n x \right) \id{x} \\ + &= \frac{2}{\pi} \left[ \left. \frac{\sin\left( n x \right)}{n} \right|_{-\pi}^0 - \frac{3 \pi^2 \cos\left( n \pi \right)}{n^2} - \frac{6}{n^4} + \frac{6 (-1)^n}{n^4} \right] \\ + &= \frac{12}{\pi n^4} \left[ (-1)^n \left( 1 - \frac{\pi^2 n^2}{2} \right) - 1 \right]. + \end{align*} + Portanto, + \begin{align*} + F_c[f](x) &= \left( 1 - \frac{\pi^3}{4} \right) + \sum_{n = 1}^\infty \frac{12}{\pi n^4} \left[ (-1)^n \left( 1 - \frac{\pi^2 n^2}{2} \right) - 1 \right] \cos\left( n x \right). + \end{align*} + \end{solution} + + \part Fa\c{c}a um esbo\c{c}o do gr\'{a}fico das fun\c{c}\~{o}es representadas pelas s\'{e}ries obtidas nos itens anteriores para todo $x \in \mathbb{R}$. + \begin{solution} + % TODO Escrever solu\c{c}\~{a}o. + \end{solution} + \end{parts} + + \question[P1 de 2006] Seja $f(x)$ uma fun\c{c}\~{a}o satisfazendo a propriedade + \begin{align*} + f(x + L) &= -f(x), && L > 0, + \end{align*} + para todo $x \in \mathbb{R}$. Mostre que todos os coeficientes pares da sua s\'{e}rie de Fourier no intervalor $(a, a + 2 L)$ s\~{a}o nulos, ou seja, + \begin{align*} + & a_0 = a_2 = a_4 = \ldots = 0, \\ + & b_2 = b_4 = b_6 = \ldots = 0. + \end{align*} + \begin{solution} + % TODO Escrever solu\c{c}\~{a}o. + Do formul\'{a}rio temos que $\alpha = a$ e $\beta = a + 2 l$, portanto + \begin{align*} + a_0 &= \frac{1}{L} \int_a^{a + 2 L} f(x) \id{x} \\ + &= \frac{1}{L} \int_a^{a + L} f(x) \id{x} + \frac{1}{L} \int_{a + L}^{a + 2 L} f(x) \id{x} \\ + &= \frac{1}{L} \int_a^{a + L} f(x) \id{x} + \frac{1}{L} \int_a^{a + L} f(\xi + L) \id{\xi} \\ + &= 0, \\ + a_n &= \frac{1}{L} \int_a^{a + 2 L} f(x) \cos\left( \frac{n \pi x}{L} \right) \id{x} \\ + &= \frac{1}{L} \int_a^{a + L} f(x) \cos\left( \frac{n \pi x}{L} \right) \id{x} + \frac{1}{L} \int_{a + L}^{a + 2 L} f(x) \cos\left( \frac{n \pi x}{L} \right) \id{x} \\ + &= \frac{1}{L} \int_a^{a + L} f(x) \cos\left( \frac{n \pi x}{L} \right) \id{x} + \frac{1}{L} \int_a^{a + L} f(\xi + L) \cos\left( \frac{n \pi \xi}{L} + n \pi \right) \id{\xi} \\ + &= \frac{1}{L} \int_a^{a + L} f(x) \cos\left( \frac{n \pi x}{L} \right) \id{x} - (-1)^n \frac{1}{L} \int_a^{a + L} f(\xi) \cos\left( \frac{n \pi \xi}{L} \right) \id{\xi} \\ + &= \begin{cases} + 0, & n \text{ \'{e} par}, \\ + 2(\ldots), & n \text{ \'{e} \'{i}mpar}, + \end{cases} \\ + b_n &= \frac{1}{L} \int_a^{a + 2 L} f(x) \sin\left( \frac{n \pi x}{L} \right) \id{x} \\ + &= \frac{1}{L} \int_a^{a + L} f(x) \sin\left( \frac{n \pi x}{L} \right) \id{x} + \frac{1}{L} \int_{a + L}^{a + 2 L} f(x) \sin\left( \frac{n \pi x}{L} \right) \id{x} \\ + &= \frac{1}{L} \int_a^{a + L} f(x) \sin\left( \frac{n \pi x}{L} \right) \id{x} + \frac{1}{L} \int_a^{a + L} f(\xi + L) \sin\left( \frac{n \pi \xi}{L} + n \pi \right) \id{\xi} \\ + &= \frac{1}{L} \int_a^{a + L} f(x) \sin\left( \frac{n \pi x}{L} \right) \id{x} - (-1)^n \frac{1}{L} \int_a^{a + L} f(\xi) \sin\left( \frac{n \pi \xi}{L} \right) \id{\xi} \\ + &= \begin{cases} + 0, & n \text{ \'{e} par}, \\ + 2 (\ldots), & n \text{ \'{e} \'{i}mpar}. + \end{cases} + \end{align*} + \end{solution} + + \question[T2 de 2008] Calcule $\int_{-\infty}^\infty \sin\left( x \right) / x \id{x}$. + \begin{solution} + Considere o caminho no plano complexo representado na figura abaixo. + \begin{center} + \begin{tikzpicture} + \draw[->] (-4.2,0) -- (4.2,0); + \draw[->] (0,-.2) -- (0,4.2); + \draw[line width=2, ->] (4,0) arc (0:60:4) node[above]{$C_R$}; + \draw[line width=2, ->] (60:4) arc (60:120:4); + \draw[line width=2, ->] (120:4) arc (120:180:4); + \draw[line width=2, ->] (-4,0) node[below]{$-R$} -- (-1,0) node[below]{$-r$}; + \draw[line width=2, ->] (-1,0) arc (180:120:1); + \draw[line width=2, ->] (120:1) arc (120:60:1) node[above]{$C_r$}; + \draw[line width=2, ->] (60:1) arc (60:0:1); + \draw[line width=2, ->] (1,0) node[below]{$r$} -- (4,0) node[below]{$R$}; + \node[below left] at (30:4) {$\Gamma$}; + \end{tikzpicture} + \end{center} + + Do Teorema de Cauchy, temos que + \begin{align*} + \int_\Gamma \frac{\exp\left( i z \right)}{x} \id{z} &= \int_{-R}^{-r} \frac{\exp\left( i z \right)}{z} \id{z} + \int_{C_r} \frac{\exp\left( i z \right)}{z} \id{z} + \int_r^R \frac{\exp\left( i z \right)}{z} \id{z} + \int_{C_R} \frac{\exp\left( i z \right)}{z} = 0. + \end{align*} + Por\'{e}m, note que + \begin{align*} + \int_{-R}^{-r} \frac{\exp\left( i z \right)}{z} \id{z} + \int)_r^R \frac{\exp\left( i z \right)}{z} \id{z} &= i \int_{-R}^{-r} \frac{\sin\left( z \right)}{z} \id{z} + i \int_r^R \frac{\sin\left( z \right)}{z} \id{z}. + \end{align*} + Como $\sin\left( x \right) / x$ \'{e} cont\'{i}nuo em $x = 0$, tem-se + \begin{align*} + i \int_{-\infty}^\infty \frac{\sin\left( x \right)}{x} \id{x} = - \lim_{r \to 0} \int_{C_r} \frac{\exp\left( i z \right)}{z} \id{z} - \lim_{R \to \infty} \int_{C_R} \frac{\exp\left( i z \right)}{z} \id{z}. + \end{align*} + Fazendo-se uma integra\c{c}\~{a}o por partes + \begin{align*} + \int_{C_R} \frac{\exp\left( i z \right)}{z} \id{z} &= \left. \frac{\exp\left( i z \right)}{i z} \right|_R^{-R} - \int_{C_R} \frac{\exp\left( i z \right)}{z^2} \id{z} \\ + &= - \frac{\cos\left( R \right)}{i R} + i \int_0^\pi \frac{\exp\left( i R \exp\left( i \theta \right) \right)}{R \exp\left( 2 i \theta \right)} \id{\theta} \\ + &= 0, + \end{align*} + para $R \to \infty$. Para o trecho $C_r$ + \begin{align*} + \int_{C_r} \frac{\exp\left( i z \right)}{z} \id{z} &= - i \int_0^\pi \exp\left( i r \exp\left( i \theta \right) \right) \id{\theta} \\ + &= - i \pi, + \end{align*} + para $r = 0$, de onde finalmente tem-se + \begin{align*} + \int_{-\infty}^\infty \frac{\sin\left( x \right)}{x} \id{x} = \pi. + \end{align*} + \end{solution} + \question[T1 de 2011] Encontre a s\'{e}rie em cossenos de Fourier sobre o intervalo $(0, \pi)$ para a fun\c{c}\~{a}o $f(x) = \sin\left( \kappa x \right)$, onde $\kappa$ \'{e} um n\'{u}mero inteiro positivo. \begin{solution} Temos para $f(x) = \sin\left( \kappa x \right)$, $\kappa = 1, 2, 3 \ldots$ que diff --git a/makefile b/makefile index d704c52..225c393 100644 --- a/makefile +++ b/makefile @@ -1,12 +1,19 @@ all: lista1 lista2 lista3 lista4 lista5 lista6 lista7 lista8 -ebook: +ereader: -sed -i '20s/^/% /' paper_size.tex -sed -i '23,24s/^% //' paper_size.tex -make -sed -i '20s/^% //' paper_size.tex -sed -i '23,24s/^/% /' paper_size.tex +tablet: + -sed -i '20s/^/% /' paper_size.tex + -sed -i '27,28s/^% //' paper_size.tex + -make + -sed -i '20s/^% //' paper_size.tex + -sed -i '27,28s/^/% /' paper_size.tex + lista1: lista1.tex -pdflatex -interaction nonstopmode -shell-escape lista1 -bibtex lista1 diff --git a/packages.tex b/packages.tex index b884c27..3154471 100644 --- a/packages.tex +++ b/packages.tex @@ -24,6 +24,7 @@ \usepackage{amssymb} \usepackage{hyperref} \usepackage{graphicx} +\usepackage{tikz} % Customiza\c{c}\~{a}o do pacote amsmath \allowdisplaybreaks[4] diff --git a/paper_size.tex b/paper_size.tex index e31c793..682e9d4 100644 --- a/paper_size.tex +++ b/paper_size.tex @@ -1,11 +1,11 @@ % Filename: paper_size.tex % -% This code is part of 'Solutions for MS550, M\'{e}todos de Matem\'{a}tica Aplicada I, and F520, M\'{e}todos Matem\'{a}ticos da F\'{i}sica I' +% This code is part of 'Solutions for MS650, M\'{e}todos de Matem\'{a}tica Aplicada II, and F620, M\'{e}todos Matem\'{a}ticos da F\'{i}sica II' % % Description: This file corresponds to the paper size output. % -% Created: 07.03.12 04:00:00 PM -% Last Change: 30.05.12 04:40:25 PM +% Created: 14.07.12 11:13:03 AM +% Last Change: 22.08.12 08:24:31 AM % % Authors: % - Raniere Silva (2012): initial version @@ -19,6 +19,10 @@ % Para impress\~{a}o \usepackage[top=3cm, bottom=3cm, left=2cm, right=2cm]{geometry} -% Para ereaders (Kindle, Nook, Kobo, ...) and tablets (iPad, GalaxyTab, ...) -% \usepackage[papersize={180mm,240mm},margin=2mm]{geometry} +% Para ereaders (Kindle, Nook, Kobo, ...) +% \usepackage[papersize={160mm,200mm},margin=2mm]{geometry} +% \sloppy + +% Para tablets (iPad, GalaxyTab, ...) +% \usepackage[papersize={140mm,190mm},margin=2mm]{geometry} % \sloppy