1.
Parseval's identity of the Fourier sine series1
is expressed as
⌠ ⌡
π
−π
{f(x)}2dx = π
∞ ∑ m=1
bm2,
where f(x) is an odd function and bm is the Fourier sine coefficient.
(1) Prove the above identity.
(2) Obtain the Fourier series of f(x) = x (−π < x < π).
(3) Applying Parseval's identity to the above, obtain
1 +
1
22
+
1
32
+
1
42
+ …
2.
Solve the following integral equation for u(x).
⌠ ⌡
∞
−∞
u(x − y) u(y) dy = e− x2.
Hint:
F(e−ax2 ) =
⎛ √
π
a
e−ω2/(4 a).
Solution
1.
(a)
⌠ ⌡
π
−π
{f(x)}2dx =
∞ ∑ m=1
∞ ∑ n=1
bmbn
⌠ ⌡
π
−π
sin mx sin nxdx = π
∞ ∑ m=1
bm2.
Note that
⌠ ⌡
π
−π
sin mx sin nxdx =
⎧ ⎪ ⎨
⎪ ⎩
π
m = n
0
m ≠ n
(b)
bm =
1
π
⌠ ⌡
π
−π
x sin mxdx = −
2 (−1)m
m
,
so
f(x) = (−2)
∞ ∑ m=1
(−1)m
m
sin mx.
(c)
⌠ ⌡
π
−π
x2dx = 4 π
∞ ∑ m=1
1
m2
,
hence
∞ ∑ m=1
1
m2
=
π2
6
.
2.
The given equation is be written as
u(x) * u(x) = e−x2,
where * is the Fourier convolution.
Fourier transform both sides of the above to get
