Encontre a Série de Fourier correspondente a cada função dada.
f(x)=\left\{\begin{array}{rc}0, & -2<x<-1 \\\\ -2, & -1 \leq x<0 \\\\ 1, & 0 \leq x<1 \\\\ 0, & 1 \leq x<2\end{array}, \quad f(x+4)=f(x)\right.

Passo 1

Resumindo o enunciado, temos a função periódica f\) de período \(4\) definida em um período por \[f(x)= \begin{cases} 0,&-2<x<-1,\\\\ -2,&-1\le x<0,\\\\ 1,&0\le x<1,\\\\ 0,&1\le x<2, \end{cases}
e estendida por f(x+4)=f(x)\]. O conteúdo aqui é Série de Fourier de uma função de período \(2L=4, isto é, L=2. Vamos usar as fórmulas dos coeficientes
a_0=\frac{1}{2L}\int_{-L}^{L}f(x)\,dx,\quad a_n=\frac{1}{L}\int_{-L}^{L}f(x)\cos\Bigl(\frac{n\pi x}{L}\Bigr)\,dx,\quad b_n=\frac{1}{L}\int_{-L}^{L}f(x)\sin\Bigl(\frac{n\pi x}{L}\Bigr)\,dx.

Passo 2

Cálculo de a_0:
a_0=\frac{1}{4}\int_{-2}^{2}f(x)\,dx =\frac{1}{4}\Bigl(\int_{-1}^{0}(-2)\,dx+\int_{0}^{1}1\,dx\Bigr) =\frac{1}{4}\bigl((-2)\cdot1+1\cdot1\bigr) =-\frac{1}{4}.

Passo 3

Cálculo de a_n:
a_n=\frac{1}{2}\int_{-2}^{2}f(x)\cos\Bigl(\frac{n\pi x}{2}\Bigr)\,dx =\frac{1}{2}\Bigl(-2\int_{-1}^{0}\cos\Bigl(\frac{n\pi x}{2}\Bigr)\,dx +\int_{0}^{1}\cos\Bigl(\frac{n\pi x}{2}\Bigr)\,dx\Bigr).
Chamando \alpha=\tfrac{n\pi}{2},
\int_{-1}^{0}\cos(\alpha x)\,dx =\frac{\sin(\alpha x)}{\alpha}\Big|_{-1}^{0} =\frac{\sin(0)-\sin(-\alpha)}{\alpha} =\frac{\sin(\alpha)}{\alpha},
\int_{0}^{1}\cos(\alpha x)\,dx =\frac{\sin(\alpha x)}{\alpha}\Big|_{0}^{1} =\frac{\sin(\alpha)}{\alpha}.
Logo
a_n =\frac{1}{2}\Bigl(-2\frac{\sin(\alpha)}{\alpha} +\frac{\sin(\alpha)}{\alpha}\Bigr) =-\frac{1}{2}\,\frac{\sin(\alpha)}{\alpha} =-\frac{1}{n\pi}\sin\Bigl(\frac{n\pi}{2}\Bigr).

Passo 4

Cálculo de b_n:
b_n=\frac{1}{2}\int_{-2}^{2}f(x)\sin\Bigl(\frac{n\pi x}{2}\Bigr)\,dx =\frac{1}{2}\Bigl(-2\int_{-1}^{0}\sin(\alpha x)\,dx +\int_{0}^{1}\sin(\alpha x)\,dx\Bigr).
Usando
\int\sin(\alpha x)\,dx=-\frac{\cos(\alpha x)}{\alpha},
temos
\int_{-1}^{0}\sin(\alpha x)\,dx =-\frac{\cos(\alpha x)}{\alpha}\Big|_{-1}^{0} =-\frac{1-\cos(-\alpha)}{\alpha} =-\frac{1-\cos(\alpha)}{\alpha},
\int_{0}^{1}\sin(\alpha x)\,dx =-\frac{\cos(\alpha x)}{\alpha}\Big|_{0}^{1} =-\frac{\cos(\alpha)-1}{\alpha} =\frac{1-\cos(\alpha)}{\alpha}.
Assim
b_n =\frac{1}{2}\Bigl( -2\bigl(-\tfrac{1-\cos(\alpha)}{\alpha}\bigr) +\frac{1-\cos(\alpha)}{\alpha} \Bigr) =\frac{1}{2}\,\frac{6\,(1-\cos(\alpha))}{\alpha} =\frac{3}{n\pi}\Bigl(1-\cos\Bigl(\frac{n\pi}{2}\Bigr)\Bigr).

Passo 5

Juntando tudo, a Série de Fourier de f é
f(x) =a_0+\sum_{n=1}^{\infty}\Bigl[a_n\cos\Bigl(\frac{n\pi x}{2}\Bigr) +b_n\sin\Bigl(\frac{n\pi x}{2}\Bigr)\Bigr]
com
a_0=-\frac14,\quad a_n=-\frac{1}{n\pi}\sin\Bigl(\frac{n\pi}{2}\Bigr), \quad b_n=\frac{3}{n\pi}\Bigl(1-\cos\Bigl(\frac{n\pi}{2}\Bigr)\Bigr).

Resposta

\boxed{ f(x)=-\frac14 +\sum_{n=1}^{\infty}\Bigl[ -\frac{1}{n\pi}\sin\Bigl(\frac{n\pi}{2}\Bigr)\cos\Bigl(\frac{n\pi x}{2}\Bigr) +\frac{3}{n\pi}\Bigl(1-\cos\Bigl(\frac{n\pi}{2}\Bigr)\Bigr)\sin\Bigl(\frac{n\pi x}{2}\Bigr) \Bigr] }.

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