Determine a série de Fourier de f no intervalo indicado:
f(x)=\left\{\begin{aligned} 0, & -2<x<-1 \\\\ -2, & -1 \leq x<0 \\\\ 1, & 0 \leq x<1 \\\\ 0, & 1 \leq x<2\end{aligned} \quad\right.

Passo 1

O enunciado pede a série de Fourier da função periódica de período T=4\) definida em \(-2<x<2\) por \[f(x)= \begin{cases} 0, & -2<x<-1,\\\\ -2, & -1\le x<0,\\\\ 1, & 0\le x<1,\\\\ 0, & 1\le x<2. \end{cases}
Para resolver, vamos usar as fórmulas padrão de série de Fourier num intervalo simétrico [-L,L]\] com \(L=2.

Passo 2

Recordando as fórmulas, a expansão é
f(x)\sim \frac{a_0}{2} \;+\;\sum_{n=1}^{\infty}\Bigl(a_n\cos\!\bigl(\tfrac{n\pi x}{2}\bigr)+b_n\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\Bigr),
com
a_0=\frac{1}{2}\int_{-2}^{2}f(x)\,dx,\quad a_n=\frac{1}{2}\int_{-2}^{2}f(x)\cos\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx,\quad b_n=\frac{1}{2}\int_{-2}^{2}f(x)\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx.

Passo 3

Calcular a_0:
a_0=\frac12\Bigl( \int_{-2}^{-1}0\,dx +\int_{-1}^{0}(-2)\,dx +\int_{0}^{1}1\,dx +\int_{1}^{2}0\,dx \Bigr) =\frac12\bigl(-2+1\bigr) =-\frac12,
logo
\frac{a_0}{2}=-\frac{1}{4}.

Passo 4

Calcular a_n separando apenas os trechos onde f\neq0:
a_n =\frac12\Bigl(\int_{-1}^{0}(-2)\cos\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx +\int_{0}^{1}1\cdot\cos\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx\Bigr).
Fazendo as integrais dá
a_n =-\frac{1}{n\pi}\,\sin\!\bigl(\tfrac{n\pi}{2}\bigr).

Passo 5

Calcular b_n de modo análogo:
b_n =\frac12\Bigl(\int_{-1}^{0}(-2)\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx +\int_{0}^{1}1\cdot\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx\Bigr) =\frac{3}{n\pi}\Bigl(1-\cos\!\bigl(\tfrac{n\pi}{2}\bigr)\Bigr).

Passo 6

Reunindo tudo, a série de Fourier fica
f(x)\sim -\tfrac14 +\sum_{n=1}^{\infty}\Bigl( -\frac{1}{n\pi}\,\sin\!\bigl(\tfrac{n\pi}{2}\bigr)\cos\!\bigl(\tfrac{n\pi x}{2}\bigr) +\frac{3}{n\pi}\bigl(1-\cos\!\bigl(\tfrac{n\pi}{2}\bigr)\bigr)\sin\!\bigl(\tfrac{n\pi x}{2}\bigr) \Bigr).

Resposta

f(x)\sim -\tfrac14 +\sum_{n=1}^{\infty}\Bigl( -\frac{1}{n\pi}\,\sin\!\bigl(\tfrac{n\pi}{2}\bigr)\cos\!\bigl(\tfrac{n\pi x}{2}\bigr) +\frac{3}{n\pi}\bigl(1-\cos\!\bigl(\tfrac{n\pi}{2}\bigr)\bigr)\sin\!\bigl(\tfrac{n\pi x}{2}\bigr) \Bigr).

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