Math 2XX3
Hints for Practice Problems # 10 and 11
Practice Problems # 10
1. For powers and product of sin and cos, you get Fourier Series by trig identity: sin x cos x =
1
2
sin(2x), which is a Fourier Series with ak = 0 ∀k = 0, 1, 2, . . . , bk = 0 for all k 6= 1, and
b2 = 12 . So
1 1
1 1
1
3 1
1
sin4 x = ( − cos(2x))2 = − cos(2x) + cos2 (2x) = − cos(2x) + cos(4x),
2 2
4 2
4
8 2
8
with a0 = 34 , a2 = − 12 , a4 = 18 , and all other ak , bk = 0.
2. When we make sin(x/2) periodic, it might create a jump discontinuity at mπ for odd m.
So we check: (note correction!) f (π−) = limx→π− sin(x/2) = 1 and
f (π+) = lim f (x) = lim f (x) = lim sin(x/2) = −1.
x→π+
x→−π+
x→−π+
So the periodic extension f (x) is discontinuous, and S(π) = 0 = 12 (f (π−) + f (π+)).
3. For f (x) = cosh x, −π < x < π, the 2π-periodic extension is continuous (f is even, and so
even continuous functions extend to continuous periodic functions on a symmetric interval.)
On the other hand, sinh x is odd, and sinh(−π) = − sinh(π) 6= 0, so its periodic extension
will have a jump discontinuity. The Fourier Series of sinh x has a jump, and converges to
S(±π) = 0, the midpoint of the jump values.
4. This is an odd function, the odd 2π-periodic extension of f (x) = cos x, x ∈ (0, π) to
R. So it will already be discontinuous at x = 0, (f (0+) = 1, f (0−) = −1), and similary at
even multiples of π, and also discontinuous at odd multiples of π, where it jumps between
−1 and +1 (but the other direction.) The series only has sin(kx) terms (ak = 0), and takes
the value S(mπ) = 0, m ∈ Z.
5.
For any orthogonal family, ck = hf, φk i/kφk k2 , so the only thing to do is to calculate
kφk k2 = π/2, k = 1, 2, 3, . . . , kφ0 k2 = π. Since cos(kx) is an even function for each k, so is any
series in that family of functions, so S(x) is even. The even periodic extension of f (x) = x,
x ∈ (0, π) is the same as |x| when (−π, π), and then we extend to all R by periodicity. The
resulting function is continuous, so S(x) = f (x) (by the Pointwise Convergence Theorem).
Practice Problems # 11
1
2
1. bk = 0 for all k, since f (x) is even, and ak = 2[(−1)k − 1]/πk 2 , k = 1, 2, 3, . . . , which
vanishes when k is even and a2k−1 = −4/π(2k − 1)2 , k = 1, 2, 3, . . . describes the odd
coefficients. We also calculate a0 = π. So
∞
π X
4
f (x) ∼ S(x) = −
cos((2k − 1)x).
2 k=1 π(2k − 1)2
Since
4
4 1
|a2k−1 cos((2k − 1)x)| ≤ |a2k−1 | =
≤
,
2
π(2k − 1)
π k2
P∞ 1
for all x and all k, and k=1 k2 converges, by the M-Test the convergence is uniform (and
absolutely convergent, by comparison test.)
[Note that the last inequality is because (2k − 1) ≥ (2k − k) = k for k ≥ 1. You could
P
1
also use the “Limit Comparison Test” to show that ∞
k=1 (2k−1)2 converges without using
inequalities.]
Since f (x) is contiuous as a 2π-periodic function on R, by the Pointwise Convergence Theorem we have
∞
4
π X
,
f (0) = |0| = 0 = S(0) = −
2 k=1 π(2k − 1)2
and so you can solve for the series and get
f ( π2 ) =
π
2
π2
.
8
The alternating series is obtained by using
= S( π2 ). The last series is obtained using Parseval’s Identity.
3. f (x) is odd, so ak = 0 for all k. bk =
√
16 2(−1)k k
,
π(16k2 −1)
k −1, 2, 3, . . . . As a 2π-periodic function,
√
f (x) is discontinuous at odd multiples of π, where it jumps between ±
the midpoint of the jump interval. However,
S( π2 )
=
f ( π2 )
2
,
2
so S(±π) = 0,
by the Pointwise Convergence
Theorem and you get the first series by evaluating the terms at x = π2 . The second series is
obtained from Parseval’s Identity.
√
16 2 k
The series converges conditionally (it does not converge absolutely,) since |bk | = π(16k
2 −1) ≥
√
P
2
1
, and ∞
k=1 k diverges. And it does not converge uniformly, since the limit S(x) is disπk
continuous (and uniform convergence of continuous functions always yields a continuous
function in the limit.)
4. For (a)and (b),follow what I did in class notes. For (c), you just need to verify that
u(x) = cos 2(k−1)πx
solves u00 (x) = λu(x) with λk = −( 2(k−1)π
)2 , and u0 (0) = 0 = u(L).
2L
2L