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

Passo 1

O enunciado quer a série de Fourier da função periódica de período T=4\) definida em \([-2,2)\) 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}
Aqui (L=2], e usamos a forma padrão
f(x)=\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 2

Calcular a_{0}. Como f(x)=0 em (-2,-1)\cup(1,2), só contribuem os intervalos [-1,0) e [0,1):
\int_{-2}^{2}f(x)\,dx =\int_{-1}^{0}(-2)\,dx+\int_{0}^{1}1\,dx =(-2)\cdot1+1\cdot1=-1.
Logo
a_{0}=\frac{1}{2}(-1)=-\frac12,
e o termo constante na série é \tfrac{a_{0}}{2}=-\tfrac14.

Passo 3

Calcular a_{n} para n\ge1:
a_{n} =\frac{1}{2}\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].
Usando \int\cos(ax)\,dx=\frac{1}{a}\sin(ax) com a=\tfrac{n\pi}{2}, obtém-se
\int_{-1}^{0}\cos\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx =\frac{2}{n\pi}\bigl[\sin(0)-\sin(-\tfrac{n\pi}{2})\bigr] =\frac{2}{n\pi}\sin\!\bigl(\tfrac{n\pi}{2}\bigr)
e
\int_{0}^{1}\cos\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx =\frac{2}{n\pi}\sin\!\bigl(\tfrac{n\pi}{2}\bigr).
Assim
a_{n} =\frac{1}{2}\Bigl[-2\cdot\frac{2}{n\pi}\sin\!\bigl(\tfrac{n\pi}{2}\bigr) +\frac{2}{n\pi}\sin\!\bigl(\tfrac{n\pi}{2}\bigr)\Bigr] =-\frac{1}{n\pi}\sin\!\bigl(\tfrac{n\pi}{2}\bigr).

Passo 4

Calcular b_{n} para n\ge1:
b_{n} =\frac{1}{2}\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].
Com \int\sin(ax)\,dx=-\tfrac{1}{a}\cos(ax) e a=\tfrac{n\pi}{2}, venho
\int_{-1}^{0}\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx =-\frac{2}{n\pi}\bigl[\cos(0)-\cos(-\tfrac{n\pi}{2})\bigr] =\frac{2}{n\pi}\bigl[1-\cos\!\tfrac{n\pi}{2}\bigr],
\int_{0}^{1}\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx =-\frac{2}{n\pi}\bigl[\cos(\tfrac{n\pi}{2})-\cos(0)\bigr] =\frac{2}{n\pi}\bigl[1-\cos\!\tfrac{n\pi}{2}\bigr].
Logo
b_{n} =\frac12\Bigl[-2\cdot\frac{2}{n\pi}\bigl(1-\cos\!\tfrac{n\pi}{2}\bigr) +\frac{2}{n\pi}\bigl(1-\cos\!\tfrac{n\pi}{2}\bigr)\Bigr] =\frac{3}{n\pi}\Bigl(1-\cos\!\tfrac{n\pi}{2}\Bigr).

Passo 5

Substituindo em
f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\Bigl(a_{n}\cos\!\tfrac{n\pi x}{2} +b_{n}\sin\!\tfrac{n\pi x}{2}\Bigr)
temos
f(x)=-\tfrac14 +\sum_{n=1}^{\infty}\Biggl[-\frac{\sin\!\tfrac{n\pi}{2}}{n\pi} \cos\!\Bigl(\tfrac{n\pi x}{2}\Bigr) +\frac{3}{n\pi}\Bigl(1-\cos\!\tfrac{n\pi}{2}\Bigr) \sin\!\Bigl(\tfrac{n\pi x}{2}\Bigr)\Biggr].

Resposta

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

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