Unformatted text preview: nt]π 2
π/
2n
n
−π sin(nπ/2) cos(nπ/2)
1
+
=
− 2 cos(nπ ).
2
2n
n
n
= Thus when n ≥ 1,
an = 2
[2 cos(nπ/2) − 1 − 1(−1)n ],
πn2 and the Fourier sine series of f (t) is
f (t) = π
2
2
2
− cos 2t −
cos 6t −
cos 10t − . . . .
4
π
9π
25π (3.17) Note that we have expanded exactly the same function f (t) on the interval [0, π ]
as either a superposition of sines in (3.16) or as a superposition of cosines in
(3.17).
Exercises:
3.3.1.a. Find the Fourier sine series of the following function deﬁned on the
interval [0, π ]:
4t,
for 0 ≤ t < π/2,
f (t) =
4π − 4t, for π/2 ≤ t ≤ π .
b. Find the Fourier cosine series of the same function.
3.3.2.a. Find the Fourier sine series of the following function deﬁned on the
interval [0, π ]:
f (t) = t(π − t).
b. Find the Fourier cosine series of the same function.
3.3.3. Find the Fourier sine series of the following function deﬁned on the
interval [0, 10]:
t,
for 0 ≤ t < 5,
f (t) =
10 − t, for 5 ≤ t ≤...
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This document was uploaded on 01/12/2014.
 Winter '14
 Equations

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