MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
1.
f (x) = sin x
0≤x≤π
Extend to an even periodic function with period 2π (note that it also has period
π, but we will treat it as having period 2π).
(a) Draw the graph of this function for −π ≤ x ≤ π.
(b) Calculate the amplitudes a0 , a1 , b1 , a2 , b2 of its real Fourier series.
(c) Calculate the energy of f (x).
(d) Using the fact that
π 2 ∼ 10,
how many terms of the Fourier series do you need to capture 97% of the
energy?
Hint: you may want to rewrite your sin and cos functions in terms of eix type
functions, in order to calculate the integrals. Or, if you prefer, try integration by
parts.
Solution:
(a) See figure 1 on the following page.
(b) The bm are all 0, because the function is even.
2
a0 =
π
a1 = 0
4
a2 = −
3π
(c)
Z 2π
kf k2 =
f (x)2 dx = π
0
(d) The energy in these terms is
88 1
T 2
T a20 +
a1 + a22 =
2
9 π
so that the percentage of energy is
88 1
9.777 . . .
∼
= 97.77 . . .
9 π2
10
more than 97% of the energy.
Date: December 7, 2000.
1
2
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
1
0.8
0.6
0.4
0.2
–6
–4
0
–2
2
x
4
6
Figure 1. The even extension of sin x
2.
If g(x) is a function with period T , show that
Z
g(x) dx
has period T precisely when the average value of g(x) is zero. Hint: write out a
complex Fourier series for g(x) and integrate it.
Solution:
Suppose that
g(x) =
X
ck e2πikx/T
k
Then
Z
g(x) dx =
X
Z
ck
e2πikx/T dx
k
(
=c+
X
ck
k
= c + c0 x +
e2πikx/T
2πik/T
x
X
k6=0
if k 6= 0
if k = 0
ck
e2πikx/T
2πik/T
where c is a constant of integration. To kill the nonperiodic c0 x term, we need
c0 = 0. All the other terms are T periodic. But c0 is the average value of g(x),
so we need g(x) to have 0 average value.
3.
What is the smallest period of
1
h(x) = e 2 cos x
Solution:
T = 2π
MATH 3150: PDE FOR ENGINEERS
4.
MIDTERM TEST #1
3
Write f (x) = (1 − x)2 as the sum of an even function with an odd function.
Solution:
You could just write:
even
odd
z
}|
{ z
}|
{
2
2
2
2
(1
−
x)
+
(1
+
x)
(1
−
x)
−
(1
+
x)
(1 − x)2 =
+
2
2
or you could expand it out:
even
odd
z}|{ z }| {
(1 − x) = −2x + 1 + x2
2
5. Among the functions f (x) in figure 2 on the next page, which one of them has
am amplitudes decaying fastest as function of m? Which has the slowest decay in
m?
Solution:
(a) am ∼ 1/m in between.
(b) am ∼ 1/m2 decaying fastest.
(c) am ∼ 1/log(m + 1) decaying slowest.
4
MATH 3150: PDE FOR ENGINEERS
MIDTERM TEST #1
0.8
2
0.6
1.5
0.4
1
0.2
0.5
–3
–3
–2
–1
0
1
x
2
–2
–1
1
x
2
3
–0.2
(a)
(b)
6
4
2
–3
–2
–1
0
1
x
2
3
(c)
Figure 2. Three even periodic functions
3
MATH 3150: PDE FOR ENGINEERS
6.
MIDTERM TEST #1
5
You may assume that
Z 1
2
(1 − x2 ) cos(αx) dx = 3 (sin α − α cos α)
α
0
for α any non-zero constant. Let f (x) = 1 − x2 for 0 ≤ x ≤ 1 and let f (x) be
even and periodic with period 2.
(a) Calculate the real amplitudes for f (x).
(b) Calculate the energy of f (x).
(c) Find the energy of f (x) stored in the amplitudes am and bm in terms of
m.
Solution:
(a) The function is even, so bm = 0.
Z
1 L
a0 =
f (x) dx
L 0
Z 1
=
(1 − x2 ) dx
0
am
= 2/3
Z
πmx 2 L
=
f (x) cos
dx
L 0
L
Z 1
=2
(1 − x2 ) cos (πmx) dx
0
= (−1)m+1
4
π 2 m2
(b)
kf k2 =
Z
1
f (x)2 dx
−1
1
Z
=
(1 − x2 )2 dx
−1
=
16
15
(c)
2
2
8
=
3
9
16
= 4 4
π m
T a20 = 2 ·
T 2
a
2 m