Determine a série de Fourier de f\) no intervalo indicado. \(11 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 define a função periódica f\) de período \(T=4\) no intervalo \(-2<x<2\) por \[f(x)= \begin{cases} 0,&-2<x<-1,\\\\ -2,&-1\le x<0,\\\\ 1,&0\le x<1,\\\\ 0,&1\le x<2, \end{cases}
e pede a sua série de Fourier de período 4\]. Como \(L=2, usamos as fórmulas
a_0=\frac{1}{4}\int_{-2}^{2}f(x)\,dx,\quad a_n=\frac{1}{2}\int_{-2}^{2}f(x)\cos\!\Bigl(\frac{n\pi x}{2}\Bigr)\,dx,\quad b_n=\frac{1}{2}\int_{-2}^{2}f(x)\sin\!\Bigl(\frac{n\pi x}{2}\Bigr)\,dx.

Passo 2

Calcular a_0. Como f só é não-zero em -1\le x<0 e 0\le x<1,
\int_{-2}^{2}f(x)\,dx =\int_{-1}^{0}(-2)\,dx+\int_{0}^{1}1\,dx =(-2)\cdot1+1\cdot1=-2+1=-1.
Logo
a_0=\frac{1}{4}(-1)=-\frac{1}{4}.

Passo 3

Calcular a_n.
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].
Fazendo as integrais:
\int_{-1}^{0}\cos\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx =\frac{2}{n\pi}\bigl[\sin(\tfrac{n\pi x}{2})\bigr]_{-1}^{0} =\frac{2}{n\pi}\sin\!\bigl(\tfrac{n\pi}{2}\bigr),
\int_{0}^{1}\cos\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx =\frac{2}{n\pi}\sin\!\bigl(\tfrac{n\pi}{2}\bigr).
Então
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{\sin(\tfrac{n\pi}{2})}{n\pi}.

Passo 4

Calcular b_n.
b_n=\frac{1}{2}\Bigl[\int_{-1}^{0}(-2)\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx +\int_{0}^{1}\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx\Bigr].
Compute cada parte:
\int_{-1}^{0}\sin\!\bigl(\tfrac{n\pi x}{2}\bigr)\,dx =-\frac{2}{n\pi}\bigl[\cos(\tfrac{n\pi x}{2})\bigr]_{-1}^{0} =\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 x}{2})\bigr]_{0}^{1} =\frac{2}{n\pi}\bigl(1-\cos\!\tfrac{n\pi}{2}\bigr).
Somando com os fatores:
b_n=\frac{1}{2}\Bigl[-2\cdot\frac{2}{n\pi}(1-\cos\!\tfrac{n\pi}{2}) +\frac{2}{n\pi}(1-\cos\!\tfrac{n\pi}{2})\Bigr] =\frac{3}{n\pi}\bigl(1-\cos\!\tfrac{n\pi}{2}\bigr).

Resposta

A série de Fourier de f de período 4 fica
f(x)=-\tfrac14 +\sum_{n=1}^{\infty}\Bigl[ -\frac{\sin(\tfrac{n\pi}{2})}{n\pi}\cos\!\bigl(\tfrac{n\pi x}{2}\bigr) +\frac{3\bigl(1-\cos(\tfrac{n\pi}{2})\bigr)}{n\pi}\, \sin\!\bigl(\tfrac{n\pi x}{2}\bigr) \Bigr].

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