MATH 675 – HOMEWORK #2
SOLUTIONS
Exercise 1. Exercise 4, page 59 in Stein and Shakarchi.
SOLUTION: For part (b) note that since f(θ) is odd,
b
f(n) = 1
2πZπ
−πf(θ)e−inθ dθ =−i
πZπ
0f(θ) sin(nθ)dθ =−i
πZπ
0θ(π−θ) sin(nθ)dθ.
Continuing and using a standard integral table and the fact that sin(nπ) = 0 for
all n,Zπ
0θ(π−θ) sin(nθ)dθ =πZπ
0θsin(nθ)dθ −Zπ
0(π−θ) sin(nθ)dθ
=πµsin(nθ)
n2−θcos(nθ)
n¯¯¯¯¯
π
0¶−µ2θsin(nθ)
n2+2 cos(nθ)
n3−θ2cos(nθ)
n¯¯¯¯¯
π
0¶
=−π2cos(nθ)
n−µ2 cos(nπ)
n3−π2cos(nπ)
n−2
n3¶=2(1 −(−1)n)
n3.
Consequently b
f(k) = −4i
π k3if kis odd and zero otherwise so that in particular
b
f(−k) = −b
f(k) for every k.
By part (a) of Exercise 2, we have that
f(θ)∼X
n≥1
2ib
f(n) sin(nθ) = X
k odd, k≥1
2i−4i
π k3sin(kθ) = 8
πX
k odd, k≥1
sin(kθ)
k3.
Exercise 2. Exercise 5, page 59 in Stein and Shakarchi.
SOLUTION: Here f(θ) is even so that as above
b
f(n) = 1
πZπ
0f(θ)cos(nθ)dθ =1
πZδ
0(1 −θ/δ)cos(nθ)dθ.
Now for n6= 0, and integrating by parts (with the help of an integral table)
Zδ
0(1 −θ/δ) cos(nθ)dθ =µsin(nθ)
n−cos(nθ)
δ n2+θsin(nθ)
δ n ¶¯¯¯¯¯
δ
0
=sin(nδ)
n−cos(nδ)
δ n2−sin(nδ)
n+1
δ n2
=1−cos(n δ)
δ n2
Consequently b
f(n) = 1−cos(nδ )
δ π n2for n6= 0 and (obviously) b
f(0) = δ
2π.
By part (a) of Exercise 2, we have that
f(θ)∼b
f(0) + X
n≥1
2b
f(n) cos(nθ) = δ
2π+ 2
∞
X
n=1
1−cos(n δ)
δ π n2cos(nθ).
