程序代写代做代考 Homework 1 Solutions

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.

1

Chapter 2, Exercise 4b

Since f is odd by definition, it follows that

f̂(n) =
i

∫ 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.

4

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.

5