18.085 Computational Science and Engineering
April 23, 2010
Answers to Problem Set 9
TAs: Dorian Croitoru, Nan Li
sw(x) = π4 sin x + sin33x + ... ,
´π
2π = −π |sw(x)|2 dx = π · π162 (12 + ( 13 )2 + ...),
4.1.2. The Fourier series of the square wave function is
hence Parseval's formula implies
2
1 + ( 31 )2 + ( 15 )2 + ... = π8 .
4.1.3. f is even, hence bk = 0. We
´π
k > 0, ak = π1 −π f (x) cos(kx)dx =
hence
for
4.1.4.
and for
1
T
´ T /2
−T /2
k
f (x)e
´ π/2
a0 =
1
2π
´π
f (x)dx = 12 , and
−π
2
2
cos kxdx = πk
sin( πk
).
π 0
´2
1 T /2
By the orthogonality of sines and cosines we have a0 =
T −T /2 f (x)dx,
´
´
2 T /2
2πkx
2 T /2
> 0, ak = T −T /2 f (x) cos T dx, bk = T −T /2 f (x) sin 2πkx
T dx, ck =
−2πikx/T
compute
dx.
4.1.5. The decay rate is cubic, since the function is quadratic.
4.1.7. Since f is 0 on [−π, 0], we just think of f as sin x on [0, π]. Hence
´π
´π
´π
1
1
sin xdx = π1 , and for k > 0, ak = π1 0 sin x cos kxdx = 2π
sin(k +
a0 = 2π
0´
0
π
1
1
1
1)xdx− 2π
sin(k−1)xdx.
Hence ak = 0 for odd k , and ak =
−
π(k+1)
π(k−1) =
0
2
.
π(1−k2 ) for even k
´
´
1 π
1 π
1
2
Next, b1 =
π 0 (sin x) dx = 2 and for k > 1, bk = π 0 sin x sin kxdx = 0.
Therefore, the Fourier series expansion of f is
1
2
1
1
1
1
cos 2x +
cos 4x +
cos 6x + ... .
f (x) = + sin x −
π 2
π 1·3
3·5
5·7
P
1 1
1 1
1 2
T
4.1.8. a) For u = (1, , , ...) and v = (1, , , ...) we have u u =
k≥0 ( 2k ) =
2
4
3
9
P
P
4
1 2
9
1
1
6
T
T
k≥0 ( 3k ) = 8 and u v =
k≥0 2k · 3k = 5 . Hence, the Cauchy3, v v =
4 9
2
Schwartz inequality asserts that (6/5) ≤
3 · 8.
b) This follows immediately from a) and Parseval's identity.
4.1.9.
The solution u(r, θ) has the form u = a0 + ra1 cos θ + rb1 sin θ +
´π
´π
1
r2 a2 cos 2θ+r2 b2 sin 2θ+.... We compute a0 = 2π
θdθ = 0, ak = π1 −π θ cos kθdθ =
−π
´π
kπ
2 sin 2θ
0 and bk = π1 −π θ sin kθdθ = −2 cos
. Hence, u(r, θ) = 2(r sin θ − r
+
k
2
3 sin 3θ
r 3 − ...). The Taylor series for 2 log(1 + z) on the unit circle z = eiθ is
2 log(1 + eiθ ) = 2(eiθ − 21 e2iθ + 13 e3iθ − ...). Taking its imaginary part, we obtain
exactly the above series of u(r, θ) for r = 1.
1