Homework 1 Solutions
Chapter 2, Exercise 1
Part 1: By substituting u = x + 2π and then using periodicity of f , we see
that ∫ b+2π
a+2π
f(x)dx =
∫ b
a
f(u− 2π)du =
∫ b
a
f(u)du,
which proves the first equality. For the second equality, we can use a simi-
lar argument, or we can simply apply the first equality with (a, b) replaced by
(a− 2π, b− 2π).
Part 2: Note that∫ π+a
−π+a
f(x)dx =
∫ −π
−π+a
f(x)dx+
∫ π
−π
f(x)dx+
∫ π+a
π
f(x)dx.
By Part 1 of the problem, the first and third terms cancel out, thus proving
that ∫ π
−π
f(x)dx =
∫ π+a
−π+a
f(x)dx =
∫ π
−π
f(u+ a)du,
where we made a substitution u = x− a in the final equality.
Remark: In the case that f is continuous, there is an alternative solution to
Part 2, namely we consider the function F (r) =
∫ π+r
−π+r f(x)dx, then note that
F ′(r) = f(π + r)− f(−π + r) = 0, hence F must be constant.
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Chapter 2, Exercise 4b
Since f is odd by definition, it follows that
f̂(n) =
i
2π
∫ π
−π
f(θ) sin(nθ)dθ =
i
π
∫ π
0
θ(π − θ) sin(nθ)dθ,
where we used the fact that f(θ) sin(nθ) is an even function in the last equality.
When n is even, the substitution x = π− θ reveals that the above integral is its
own negative, hence 0. Thus we now suppose that n is odd. Using integration-
by-parts:∫ π
0
θ sin(nθ)dθ = −
θ
n
cos(nθ)
∣∣∣∣π
0
+
1
n
∫ π
0
cos(nθ)dθ =
π
n
,∫ π
0
θ2 sin(nθ)dθ = −
θ2
n
cos(nθ)
∣∣∣∣π
0
+
2
n
∫ π
0
θ cos(nθ)dθ
=
π2
n
+
2
n
[
θ
n
sin(nθ)
∣∣π
0
−
1
n
∫ π
0
sin(nθ)dθ
]
=
π2
n
+
4
n3
.
Subtracting π times the first integral minus the second one, then multiplying
by i/π, we obtain that when n is odd,
f̂(n) = −
4i
πn3
.
Using the fact that f is odd, we know that f̂(n) = −f̂(−n) so that
f(θ) =
∑
k
f̂(k)eikθ =
∑
k>0,odd
2if̂(k) sin(kθ) =
8
π
∑
k>0,odd
sin(kθ)
k3
.
2
Chapter 2, Exercise 6
Part b: Let f(θ) = |θ|, and let g be the function from Exercise 4. Now note
that g is a C1 function whose (periodic) derivative is g′(θ) = π − 2|θ|. Now
recall from class that for a 2π-periodic C1 function, the fourier coefficients of
the derivative are related to the fourier coefficients of the original function by
ĝ′(n) = inĝ(n).
Since the addition of constant terms only affect the zeroth fourier coefficient, and
since g′ = π−2f , it follows that for n 6= 0 one has f̂(n) = − 1
2
ĝ′(n) = − 1
2
inĝ(n),
which is −2/(πn2) for odd n and zero for even n 6= 0. When n = 0, it is easily
computed that f̂(0) = π/2.
Part c:
f(θ) =
π
2
−
4
π
∑
k>0,odd
cos(kθ)
k2
.
Part d: Plugging in θ = 0, we see
0 =
π
2
−
4
π
∑
k>0,odd
1
k2
=⇒
∑
k>0,odd
1
k2
=
π2
8
.
Now let S =
∑
k>0 k
−2. We find that
S =
∑
k>0,odd
1
k2
+
∑
k>0,even
1
k2
=
π2
8
+
1
4
S =⇒ S =
π2
6
.
3
Chapter 2, Exercise 8
The computation of the Fourier coefficients can be done using repeated integration-
by-parts as in exercise 4.
So we still need to verify that the given series of functions satisfies the con-
ditions of the Dirichlet test at each point x. Since 1
n
is a monotone decreasing
sequence, we just need to show that for every x 6= 0, there exists a constant
C(x) such that
sup
N∈N
∣∣∣∣ ∑
|n|≤N
e−inx
∣∣∣∣ ≤ C(x).
To this end, we note by the geometric series identity that
∑
|n|≤N
einx =
ei(N+
1
2
)x − e−i(N+
1
2
)x
eix/2 − e−ix/2
.
By the triangle inequality, the numerator is always bounded-in-magnitude by
2, meanwhile the denominator is simply 2i sin(x/2), and hence we find that if
x 6= 0 then
sup
N∈N
∣∣∣∣ ∑
|n|≤N
e−inx
∣∣∣∣ ≤ 1| sin(x/2)| ,
thus proving the claim.
When x = 0, the positive partial sums cancel out the negative partial sums,
hence the limit clearly exists and is zero.
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Chapter 2, Exercise 10
Recall from class that if f is of class C1, then we have that f̂ ′(n) = inf̂(n).
This was proved using integration-by-parts.
Now if f is of class Ck, then induction on the above fact reveals that
f̂ (k)(n) = iknkf̂(n).
Let g = f (k). By assumption g is continuous (and thus bounded) on [−π, π],
therefore |ĝ(k)| ≤
∫
|g(x)|dx ≤ C, for some constant C. Hence we find that
f̂(n) =
ĝ(n)
iknk
is bounded in magnitude by C/nk.
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