AE2 Mathematics
Solutions to Example Sheet 2: Fourier Series
f (x) = | sin x| on (−π, π) with L = π: f (x) is
an even function so bn = 0. On [0, π] we have
| sin x| = sin x.
Z
Z
2 π
2 π
| sin x| dx =
sin x dx
a0 =
π 0
π 0
= 4/π
| sin x|
1)
where cos nπ = (−1)n .
x
−π
0
π
Z
2
an =
π
π
sin x cos nx dx
0
We also know that 2 sin x cos nx = sin[(n + 1)x] − sin[(n − 1)x] so for n ≥ 2 (note: a1 = 0)
Z
Z
1 π
1 π
sin[(n + 1)x] dx −
sin[(n − 1)x] dx
an =
π 0
π 0
1 cos[(n + 1)x] π 1 cos[(n − 1)x] π
= −
+
π
n+1
π
n−1
0
0 n+1
2
(−1)
−1
1
1
1
n+1
=
−
= − (−1)
−1
π
n+1 n−1
π(n2 − 1)
Therefore
an =
(
0
− π(4m42 −1)
and so
| sin x| =
n = 2m + 1 (odd)
n = 2m (even)
∞
2
4 X cos 2mx
−
π π
(4m2 − 1)
m=1
f (x) =
f (x)
2)
x(1 − x)
0
L = 1 & the function is neither odd nor even.
Z 1
Z 1
f (x) dx =
x(1 − x) dx = 1/6
a0 =
−1
0
x
−1
0
an =
1
Z
1
0
x(1 − x) cos nπx dx
bn =
By evaluating these integrals, one finds
an =
0≤x≤1
−1≤x≤0
0
− 2m12 π2
n = 2m + 1
n = 2m
(odd)
(even)
bn =
1
Z
(
1
0
x(1 − x) sin nπx dx
4
(2m+1)3 π 3
0
n = 2m + 1
n = 2m
(odd)
(even)
thus giving the answer. The odd extension
R 1of f (x) = x(1 − x), originally defined on 0 ≤ x ≤ 1,
on the range −1 ≤ x ≤ 1, has bn = 2 0 x(1 − x) sin(nπx) dx. Thus the answer is an odd
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sine-series with coefficient twice that above, namely (2m+1)
3 π3 .
Rπ
3) x sin x is an even function over (−π, π) so bn = 0 and an = π2 0 x sin x cos nx dx . Using the
fact that 2 sin x cos nx = sin[(n + 1)x] − sin[(n − 1)x], we have (except for n = 1)
Z
2(−1)n+1
1 π
x sin[(n + 1)x] − sin[(n − 1)x] dx =
by parts
an =
n2 − 1
π 0
Thus a0 = 2 and a1 is
∴
a1 =
2
π
Z
π
x sin x cos x dx =
0
1
π
Z
π
∞
X
(−1)n+1
x sin x = 1 − cos x + 2
1
2
4)
x sin 2x dx = − 12
0
n=2
n2 − 1
(by parts)
cos nx
f (x)
f (x) = x
−π/2
−π
π/2
x
π
f (x) = x − π
The figure shows that f (x) is even about x = 0; thus L = π and bn = 0 and
)
(Z
Z π
π/2
2
x dx +
(x − π) dx = 0
a0 =
π
0
π/2
an =
=
2
π
2
π
(Z
Z
π/2
x cos nx dx +
0
π
0
x cos nx dx − 2
Z
Z
π
π/2
(x − π) cos nx dx
)
π
cos nx dx
π/2
By parts
Z
π
x cos nx dx =
0
1
2
(−1)n − 1
π
[nx
sin
nx
+
cos
nx]
=
=−
0
2
2
n
n
(2m + 1)2
when n = 2m + 1 and zero when n is even. Moreover,
Z π
sin 12 nπ
1
(−1)m
cos nx dx = [sin nx]ππ/2 = −
=−
2m + 1
n
n
π/2
thus giving the answer as advertised.
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