12 pág.

# Séries de Fourier de Senos e de Cossenos de Índices Ímpares

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sen
(
2kpit
2L
)
= \u2212h(t)
Assim segue da aplicac¸a\u2dco do item (a) que b2k = 0.
Para h(t) = f (t) sen (2k+1)pit2L temos que
h(2L\u2212 t) = f (2L\u2212 t) sen
(2k+ 1)pi(2L\u2212 t)
2L
= f (t) sen
(
(2k+ 1)pi \u2212
(2k+ 1)pit
2L
)
= f (t) sen
(
pi \u2212
(2k+ 1)pit
2L
)
= f (t) sen
(
(2k+ 1)pit
2L
)
= h(t)
Assim segue da aplicac¸a\u2dco do item (b) que
b2k+1 =
2
L
\u222b L
0
f (t) sen
(2k+ 1)pit
2L
dt para k = 0, 1, 2, . . .
(d) Para h(t) = f (t) cos 2kpit2L temos que
h(2L\u2212 t) = f (2L\u2212 t) cos
2kpi(2L\u2212 t)
2L
= \u2212 f (t) cos
(
2kpi \u2212
2kpit
2L
)
= \u2212 f (t) cos
(
\u2212
2kpit
2L
)
= \u2212 f (t) cos
(
2kpit
2L
)
= \u2212h(t)
Assim segue da aplicac¸a\u2dco do item (a) que a2k = 0.
Para h(t) = f (t) cos (2k+1)pit2L temos que
h(2L\u2212 t) = f (2L\u2212 t) cos
(2k+ 1)pi(2L\u2212 t)
2L
= \u2212 f (t) cos
(
(2k+ 1)pi \u2212
(2k+ 1)pit
2L
)
= \u2212 f (t) cos
(
pi \u2212
(2k+ 1)pit
2L
)
= f (t) cos
(
((2k+ 1)pit
2L
)
= h(t)
Assim segue da aplicac¸a\u2dco do item (b) que
a2k+1 =
2
L
\u222b L
0
f (t) cos
(2k+ 1)pit
2L
dt para k = 0, 1, 2, . . .
12
2. Lembrando que a integrac¸a\u2dco deve ser feita no intervalo [0, 2L]:
a2k+1 =
L
2
a2k+1( f
(0)
0, 14
)\u2212 a2k+1( f
(1)
0, 14
)
=
L
2
· 4 ·
1
(2k+ 1)pi
sen s
\u2223\u2223\u2223
(2k+1)pi
4
0
\u2212 4 ·
2L
(2k+ 1)2pi2
(s sen s+ cos s)
\u2223\u2223\u2223
(2k+1)pi
4
0
=
8L
(2k+ 1)2pi2
(
1\u2212 cos
(2k+ 1)pi
4
)
f (t) =
8L
pi
2
\u221e
\u2211
k=0
1\u2212 cos
(2k+1)pi
4
(2k+ 1)2
cos
(2k+ 1)pit
2L
b2k+1 =
L
2
b2k+1( f
(0)
0, 14
)\u2212 b2k+1( f
(1)
0, 14
)
=
L
2
· 4 ·
\u22121
(2k+ 1)pi
cos s
\u2223\u2223\u2223
(2k+1)pi
4
0
\u2212 4 ·
2L
(2k+ 1)2pi2
(\u2212s cos s+ sen s)
\u2223\u2223\u2223
(2k+1)pi
4
0
=
2L
(2k+ 1)2pi2
(
(2k+ 1)pi \u2212 4 sen
(2k+ 1)pi
4
)
f (t) =
2L
pi
2
\u221e
\u2211
k=0
(2k+ 1)pi \u2212 4 sen (2k+1)pi4
(2k+ 1)2
sen
(2k+ 1)pit
2L