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MATH 675 – HOMEWORK
SOLUTIONS
Exercise 1. Given Euler’s formula eix^ = cos(x) + i sin(x) derive the trig identities
2 sin(θ) sin(φ) = cos(θ − φ) − cos(θ + φ),
2 sin(θ) cos(φ) = sin(θ − φ) + sin(θ + φ).
Solution:
2 sin(θ) sin(φ) = 2
( (^) eiθ (^) − e−iθ 2 i
) ( (^) eiφ (^) − e−iφ 2 i
)
(ei(θ+φ)^ − ei(θ−φ)^ − e−i(θ−φ)^ + e−i(θ+φ))
=
( (^) ei(θ−φ) (^) + e−i(θ−φ) 2
) −
( (^) ei(θ+φ) (^) + e−i(θ+φ) 2
)
= cos(θ − φ) − cos(θ + φ)
A similar calculation gives the other identity.
Exercise 2. Exercise 5, page 26 in Stein and Shakarchi.
Solution: Since for every integer n, cos(nx) and sin(nx) have period 2π, so does einx^ = cos(nx) + i sin(nx). Now, if n = 0 then einx^ = 1 for all x, so that 1 2 π
∫ (^) π −π
einx^ dx =
2 π
∫ (^) π −π
dx = 1.
If n 6 = 0 then 1 2 π
∫ (^) π −π
einx^ dx = einx 2 πin
∣∣ ∣∣ ∣
π
−π
since einx^ has period 2π. This proves the first assertion in the problem. Following the first part of the hint, observe that (using the fact that cos(x) is even and sin(x) is odd)
einx^ e−imx^ + einx^ eimx^ = (cos(nx) + i sin(nx))(cos(mx) − i sin(mx)) +(cos(nx) + i sin(nx))(cos(mx) + i sin(mx)) = (cos(nx) cos(mx) + sin(nx) sin(mx)
- cos(nx) cos(mx) − sin(nx) sin(mx)) +i (sin(nx) cos(mx) − cos(nx) sin(mx)
- sin(nx) cos(mx) + cos(nx) sin(mx)) = 2 cos(nx) cos(mx) + 2i sin(nx) cos(mx)
Therefore 1 2 π
∫ (^) π −π
(einx^ e−imx+einx^ eimx) dx =
π
∫ (^) π −π
cos(nx) cos(mx) dx+i
π
∫ (^) π −π
sin(nx) cos(mx) dx.
By the first part of the problem and the fact that n + m is always a nonzero integer 1 2 π
∫ (^) π −π (einx^ e−imx^ + einx^ eimx) dx =
2 π
∫ (^) π −π (ei(n−m)x^ + ei(n+m)x) dx
=
2 π
∫ (^) π −π ei(n−m)x^ dx
=
{ 0 if n 6 = m 1 if n = m.
Equating real and imaginary parts, 1 π
∫ (^) π −π
cos(nx) cos(mx) dx =
{ 0 if n 6 = m 1 if n = m
and 1 π
∫ (^) π −π
sin(nx) cos(mx) dx = 0
for all n, m ≥ 1. This gives the second and fourth assertions in the problem. Following the second part of the hint in a similar fashion gives the third assertion in the problem.
Exercise 3. (a) Show that if F and G are twice differentiable functions then the function
u(x, t) = F (x + t) + G(x − t) (1)
is a solution to the wave equation utt = uxx. This solution is known as the travelling wave solution to the wave equation. (b) Show that the standing wave solution derived in class, namely
u(x, t) =
∑^ ∞ m=
(am cos(mt) + bm sin(mt)) sin(mx)
f (x) = u(x, 0) =
∑^ ∞ m=
am sin(mx) g(x) = ut(x, 0) =
∑^ ∞ m=
mbm sin(mx)
satisfies (1) with
F (x) + G(x) = f (x) and F ′(x) − G′(x) = g(x).
(Hint: For these calculations you may manipulate all infinite series formally without worrying about convergence.)
Clearly the sine terms in the series calculated in part (a) all vanish so we are left with a cosine series.
Part (c) is virtually the same calculation and note that in the series of part (a) the cosine terms all vanish so we get a sine series.
For part (d) note that, using the periodicity of f (θ),
f̂ (n) = 1 2 π
∫ (^) π −π f (θ) e−inθ^ dθ
=
2 π
∫ (^) π −π
f (θ + π) e−in(θ+π)^ dθ
=
2 π
∫ (^) π −π
f (θ) e−inπ^ e−inθ^ dθ
= (−1)n^
2 π
∫ (^) π −π
f (θ) e−inθ^ dθ = (−1)n^ f̂ (n)
If n is odd then f̂ (n) = − f̂ (n) so that f̂ (n) = 0.
For part (e) note that f (θ) is real-valued if and only if f (θ) = f (θ) for all θ. So if f is real-valued then
f̂ (n) = 1 2 π
∫ (^) π −π
f (θ) e−inθ^ dθ
=
2 π
∫ (^) π −π f (θ) e−inθ^ dθ
=
2 π
∫ (^) π −π f (θ) einθ^ dθ
=
2 π
∫ (^) π −π
f (θ) einθ^ dθ = f̂ (−n).
For the converse let f (θ) = u(θ) + i v(θ) with u and v real-valued functions. We showed in class (and in any case it is a straightforward calculation using the first part of this proof) that
̂ u(n) =
̂ f (n) + f̂ (−n) 2
and ̂v(n) =
̂ f (n) − f̂ (−n) 2 i
Therefore if f̂ (n) = f̂ (−n) then v̂(n) = 0 for all n. By the Uniqueness Theorem, this implies that v(θ) = 0 for almost every θ and hence that f (θ) is real-valued for almost every θ.
