Determinar as series de Fourier de Cosenos para a seguinte função. Encontrar também a função para a qual a serie converge.
f(t)=\left\{\begin{array}{ccc} 0 & \text { se } & 0<t<\pi \\\\ 1 & \text { se } & \pi<t<2 \pi \\\\ 2 & \text { se } & 2 \pi<t<3 \pi \end{array}\right.

Passo 1

Temos a função definida em 0<t<3\pi\) por \[f(t)= \begin{cases} 0, & 0<t<\pi,\\\\ 1, & \pi<t<2\pi,\\\\ 2, & 2\pi<t<3\pi. \end{cases}
Queremos a série de Fourier de cossenos em [0,3\pi]\]. Lembrando que, para \(L=3\pi, a expansão é
f(t)\sim \frac{a_{0}}{2} + \sum_{n=1}^{\infty}a_{n}\cos\!\bigl(\tfrac{n\pi t}{L}\bigr) =\frac{a_{0}}{2} + \sum_{n=1}^{\infty}a_{n}\cos\!\bigl(\tfrac{n t}{3}\bigr),
com
a_{0}=\frac{2}{L}\int_{0}^{L}f(t)\,\mathrm{d}t, \quad a_{n}=\frac{2}{L}\int_{0}^{L}f(t)\cos\!\bigl(\tfrac{n t}{3}\bigr)\,\mathrm{d}t.

Passo 2

Calcular a_{0}.
\int_{0}^{3\pi}f(t)\,\mathrm{d}t =\int_{0}^{\pi}0\,\mathrm{d}t +\int_{\pi}^{2\pi}1\,\mathrm{d}t +\int_{2\pi}^{3\pi}2\,\mathrm{d}t =0 + \pi + 2\pi = 3\pi.
Logo
a_{0} =\frac{2}{3\pi}\,3\pi = 2,
e o termo constante da série é \tfrac{a_{0}}{2}=1.

Passo 3

Calcular a_{n}.
a_{n} =\frac{2}{3\pi}\Bigl[ \underbrace{\int_{0}^{\pi}0\cdot\cos(\tfrac{n t}{3})\,\mathrm{d}t}_{0} +\underbrace{\int_{\pi}^{2\pi}1\cdot\cos(\tfrac{n t}{3})\,\mathrm{d}t}_{I_{1}} +\underbrace{\int_{2\pi}^{3\pi}2\cdot\cos(\tfrac{n t}{3})\,\mathrm{d}t}_{2I_{2}} \Bigr].
Calculemos
I_{1} =\int_{\pi}^{2\pi}\cos(\tfrac{n t}{3})\,\mathrm{d}t =\frac{3}{n}\Bigl[\sin(\tfrac{n t}{3})\Bigr]_{\pi}^{2\pi} =\frac{3}{n}\bigl(\sin(\tfrac{2n\pi}{3})-\sin(\tfrac{n\pi}{3})\bigr),
I_{2} =\int_{2\pi}^{3\pi}\cos(\tfrac{n t}{3})\,\mathrm{d}t =\frac{3}{n}\Bigl[\sin(\tfrac{n t}{3})\Bigr]_{2\pi}^{3\pi} =\frac{3}{n}\bigl(\sin(n\pi)-\sin(\tfrac{2n\pi}{3})\bigr) =-\frac{3}{n}\sin(\tfrac{2n\pi}{3}).
Substituindo,
a_{n} =\frac{2}{3\pi}\cdot\frac{3}{n}\Bigl[\sin(\tfrac{2n\pi}{3})-\sin(\tfrac{n\pi}{3}) -2\,\sin(\tfrac{2n\pi}{3})\Bigr] =-\frac{2}{\pi n}\Bigl[\sin(\tfrac{n\pi}{3})+\sin(\tfrac{2n\pi}{3})\Bigr].

Passo 4

Podemos usar a soma de senos:
\sin(\tfrac{n\pi}{3})+\sin(\tfrac{2n\pi}{3}) =2\sin(\tfrac{n\pi}{2})\cos(\tfrac{n\pi}{6}),
logo
a_{n} =-\frac{2}{\pi n}\,2\sin(\tfrac{n\pi}{2})\cos(\tfrac{n\pi}{6}) =-\frac{4}{\pi n}\,\sin\!\bigl(\tfrac{n\pi}{2}\bigr)\cos\!\bigl(\tfrac{n\pi}{6}\bigr).

Passo 5

A série de Fourier de cossenos fica
f(t)\sim 1 +\sum_{n=1}^{\infty}\Bigl[ -\frac{4}{\pi n}\,\sin\!\bigl(\tfrac{n\pi}{2}\bigr)\cos\!\bigl(\tfrac{n\pi}{6}\bigr) \Bigr]\cos\!\bigl(\tfrac{n t}{3}\bigr).

Passo 6

Essa série converge para a extensão par e 6\pi\)-periódica de \(f\). Isto é, definindo \[g(t)= \begin{cases} f(t), & 0<t<3\pi,\\\ f(-t),& -3\pi<t<0, \end{cases} \quad g(t+6\pi)=g(t),
a série converge para (g(t)] em todos os pontos, e nos saltos converge ao valor médio.

Resposta

f(t)\sim 1 +\sum_{n=1}^{\infty}\Bigl[ -\frac{4}{\pi n}\,\sin\!\bigl(\tfrac{n\pi}{2}\bigr)\cos\!\bigl(\tfrac{n\pi}{6}\bigr) \Bigr]\cos\!\bigl(\tfrac{n t}{3}\bigr),
que converge para a extensão par e 6\pi-periódica de f.

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