Math 118 :: Winter 2009 :: Midterm solutions
1.
• Split e−ikx = cos(kx) − i sin(kx). Since the function is odd, the cosine terms give a zero
contribution. We have
Z
Z π
cos(kx) π
1 π
cos(kx)dx − x
sin(kx)xdx =
.
k −π
k
−π
−π
The integral vanishes. Notice that cos(kx) = (−1)k . What is left is
−2π
(−1)k
,
k
Hence the Fourier series expansion is x =
fˆk = −i
Z
k 6= 0.
1
2π
P∞
ikx fˆ ,
k
k=−∞ e
π
sin(kx)x dx = 2iπ
−π
(−1)k
,
k
with fˆ0 = 0 and
k 6= 0.
• By comparing areas, we see that
N
X
k
−1
N +1
Z
≥
x−1 dx.
1
1
The latter integral diverges, because log(N + 1) → ∞ as N → ∞.
• The FS coeffcients decay like a constant times k −1 in modulus, so they are not absolutely
summable.
• Consider g(x) = πx for x ∈ [−1, 1]. The trick for Chebyshev expansions is to view g as
h(θ) = g(cos θ) = π cos θ,
θ ∈ [0, 2π],
and to perform a FS of h with cosine terms only (by evenness in θ.) Now use the formulas
Z
ĥk =
2π
cos (kθ)h(θ) dθ,
0
h(θ) =
∞
1
1X
cos (kθ)ĥk ,
ĥ0 +
2π
π
k ∈ Z+ .
k=1
For h(θ) = π cos θ the expansion has only one term – h itself – corresponding to k = 1.
Another way to see this is to compute the inner product of cos θ times cos(kθ), and to
notice that it is nonzero only for k = 1.
We identify ĥ1 = π 2 , and all the other ĥk are zero. Back to g(x) = πx, we have the
Chebyshev expansion
1
πx = ĥ1 T1 (x) = πT1 (x),
π
with T1 (x) = x.
• The Chebyshev expansion has only one term, vs. a FS that decays like 1/k, so the
Chebyshev expansion converges a lot faster. Because of the x = cos θ trick, the fast
decay of the Cheb. expansion is inherited from the fast decay of the FS of a smooth
periodic function. On the other hand x is not periodic in any interval, so the truncated
FS suffers from the Gibbs effect.
2. We saw in class that
Z
sin k
=
k
∞
e−ikx
−∞
1
χ
(x) dx.
2 [−1,1]
So by Plancherel,
Z
1
2π
∞
−∞
Hence
Z
∞
sin k
k
−∞
3.
ĝk = h
2
1
1
dx.
4
−1
sin k
k
N
X
Z
dk =
2
dk = π.
e−ikjh (−1)j fj .
j=1
Thanks to the relation π/h = N/2, and −1 = eiπ , we can combine two factors as
e−ikjh (−1)j = e−ij(k−N/2)h .
So we have ĝk = fˆk−N/2 . Or ĝk = fˆk+N/2 is also correct.
4.
•
Z
ix
|e − 1| = |(−i)
•
So
|
f (x) − f (y)
|≤
x−y
Z
|e−iky | |
iy
Z
e dy| ≤
0
f (x) − f (y)
=
x−y
x
Z
x
1dy = x.
0
e−ikx − e−iky ˆ
f (k) dk.
x−y
e−ik(x−y) − 1 ˆ
| |f (k)| dk ≤
x−y
Z
|k| |fˆ(k)| dk.