Section 7.1
Periodic Functions
151
Solutions to Exercises 7.1
1. (a) cos x has period 2π.
(b) cos πx has period T = 2π
(c) cos 23 x has period
π = 2.
2π
T = 2/3 = 3π. (d) cos x has period 2π, cos 2x has period π, 2π, 3π,.̇. A common period of cos x
and cos 2x is 2π. So cos x + cos 2x has period 2π.
5. The function is periodic with period 1. To describe it, we use any interval of length one period.
Let us use the interval [0, 1). On that interval, the function is given by f(x) = x. Thus, the complete
description of the function is
(
x
if 0 ≤ x < 1,
f(x) =
f(x + 1) otherwise.
9. (a) Suppose that f and g are T -periodic. Then f(x + T ) · g(x + T ) = f(x) · g(x), and so f · g is
T periodic. Similarly,
f(x)
f(x + T )
=
,
g(x + T )
g(x)
and so f/g is T periodic.
(b) Suppose that f is T -periodic and let h(x) = f(x/a). Then
x
x + aT
h(x + aT ) = f
=f
+T
a
a
x
= f
(because f is T -periodic)
a
= h(x).
Thus h has period aT . Replacing a by 1/a, we find that the function f(ax) has period T /a.
(c) Suppose that f is T -periodic. Then g(f(x + T )) = g(f(x)), and so g(f(x)) is also T -periodic.
13. We have
Z
π/2
f(x) dx =
−π/2
Z
π/2
1 dx = π/2.
0
R2
17. By Exercise 16, F is 2 periodic, because 0 f(t) dt = 0 (this is clear from the graph of f). So
it is enough to describe F on any interval of length 2. For 0 < x < 2, we have
Z x
t2 x
x2
F (x) =
(1 − t) dt = t − = x − .
2 0
2
0
For all other x, F (x + 2) = F (x). (b) The graph of F over the interval [0, 2] consists of the arch of
a parabola looking down, with zeros at 0 and 2. Since F is 2-periodic, the graph is repeated over
and over.
Fourier Series
Solutions to Exercises 7.2
1. The graph of the Fourier series is identical to the graph of the function, except at the points of
discontinuity where the Fourier series is equal to the average of the function at these points, which
is 12 .
5. We compute the Fourier coefficients using he Euler formulas. Let us first note that since
f(x) = |x| is an even function on the interval −π < x < π, the product f(x) sin nx is an odd
function. So
Z odd function
1 π z }| {
bn =
|x| sin nx dx = 0,
π −π
because the integral of an odd function over a symmetric interval is 0. For the other coefficients, we
have
Z π
Z π
1
1
a0 =
f(x) dx =
|x| dx
2π −π
2π −π
Z 0
Z π
1
1
= =
(−x) dx +
x dx
2π −π
2π 0
Z π
1
1 2π
π
=
x dx =
x = .
π 0
2π 0
2
In computing an (n ≥ 1), we will need the formula
Z
cos(a x) x sin(a x)
x cos ax dx =
+
+C
a2
a
(a 6= 0),
which can be derived using integration by parts. We have, for n ≥ 1,
Z
Z
1 π
1 π
an =
f(x) cos nx dx =
|x| cos nx dx
π −π
π −π
Z π
2
=
x cos nx dx
π 0
2 x
1
π
=
sin nx + 2 cos nx π n
n
0
n
2
2 (−1)
1
(−1)n − 1
=
− 2 =
2
2
π
n
n
πn
0
if n is even
=
4
− πn
if n is odd.
2
Thus, the Fourier series is
∞
π
1
4X
cos(2k + 1)x .
−
2
π
(2k + 1)2
k=0
Section 7.2
s n_, x_ : Pi 2
4 Pi Sum 1
partialsums Table s
x
2 Pi Floor
f x_
g x_
Abs f x
Plot g x , x, 3 Pi,
Plot Evaluate g x ,
2k
1 ^ 2 Cos
n, x , n, 1, 7
x Pi
2 Pi
3 Pi
partialsums
2k
Fourier Series
153
1 x , k, 0, n
;
, x,
2 Pi, 2 Pi
The function g(x) = | x |
and its periodic extension
Partial sums of
the Fourier series. Since we are
summing over the odd integers,
when n = 7, we are actually summing
the 15th partial sum.
9. Just some hints:
(1) f is even, so all the bn ’s are zero.
(2)
a0 =
1
π
Z
π
x2 dx =
0
π2
.
3
(3) Establish the identity
Z
−2 + a2 x2 sin(a x)
2 x cos(a x)
x cos(ax) dx =
+
+C
a2
a3
2
(a 6= 0),
using integration by parts.
13. You can compute directly as we did in Example 1, or you can use the result of Example 1 as
follows. Rename the function in Example 1 g(x). By comparing graphs, note that f(x) = −2g(x+π).
Now using the Fourier series of g(x) from Example, we get
f(x) = −2
∞
∞
X
X
sin n(π + x)
(−1)n+1
=2
sin nx.
n
n
n=1
n=1
17. (a) Setting x = π in the Fourier series expansion in Exercise 9 and using the fact that the
Fourier series converges for all x to f(x), we obtain
π2 = f(π) =
∞
∞
X
X
(−1)n
1
π2
π2
cos
nπ
=
,
+4
+
4
2
2
3
n
3
n
n=1
n=1
Fourier Series
where we have used cos nπ = (−1)n. Simplifying, we find
∞
X
1
π2
.
=
2
6
n
n=1
(b) Setting x = π2 in the Fourier series expansion in Exercise 13 and using the fact that the Fourier
series converges to f(x) = x for −π < x < π, we obtain
π
2
=
=
=
f(π/2) = 2
∞
X
(−1)n+1
nπ
sin
n
2
n=1
(−1)1+1
π (−1)2+1
2π (−1)3+1
3π
(−1)4+1
4π
sin +
sin
+
sin
++
sin
+···
1
2
2
2
3
2
4
4
1
1
2 1 + 0 + (−1) + +0 + · · · = 2 1 − + · · · .
3
3
2
Dividing by 2 both sides, we get the desired identity.
21. We simply repeat the solution of Example 6, making some obvious simplifications. The idea is
to realize that the function
1
f(θ) =
3 + cos θ
is the restriction to the unit circle of a function that is analytic on the unit circle. This function
has a Laurent series expansion, that, when restricted to the unit circle, gives the Fourier series of f.
−1
Write cos θ = z+z2 , where z = eiθ . Then
1
2z
1
= 2
=
= g(z).
z+z −1
3 + cos θ
z
+
6z + 1
3+ 2
We have
z 2 + 6z + 1 = (z − z1 )(z − z2 )
√
wherez1 = −3 −
√
8 and z2 = −3 +
8.
Since |z2| < 1 < |z1 |, the function g(z) is analytic on the annulus |z2 | < |z| < |z1 |, that contains the
unit circle. We next find the Laurent series expansion of g in that annulus. The computation is not
straightforward, but it is done in complete detail in Example 6. I will simply recall the result of this
example (but you should go over the details leading to the cited result). We have, in the annulus
|z2| < |z| < |z1|, (take a = 3 and z1 and z2 as previously)
"
#
∞
X
2z
1
1
z2n z n + n
= √
1+
z 2 + 6z + 1
z
9−1
n=1
"
#
∞ X
√ n n
1
1
= √ 1+
z + n
−3 + 8
z
8
n=1
Now set z = eiθ (which is a complex number in the domain of the Laurent series), then
1
3 + cos θ
=
=
=
2z
+ 6z + 1
#
"
∞ X
√ n inθ
1
−inθ
√ 1+
−3 + 8
e +e
8
n=1
#
"
∞ X
√ n
1
√ 1+2
−3 + 8 cos nθ ,
8
n=1
z2
Section 7.2
Fourier Series
155
which is the desired Fourier series.
25. The Fourier series in Example 6 is valid for all θ. Replacing θ by θ − π in the Fourier series
of Example 6, we obtain the desired Fourier series, since cos(θ − π) = − cos θ and cos[n(θ − π)] =
(−1)n cos nθ.
Fourier Series
Solutions to Exercises 7.3
1. (a) and (b) Since f is odd, all the an’s are zero and
Z
2 p
nπ
dx
bn =
sin
p 0
p
−2
−2 nπ π
=
cos
(−1)n − 1
=
nπ
p 0
nπ
0
if n is even,
=
4
if n is odd.
nπ
4X
1
(2k + 1)π
sin
x. At the points of discontinuity, the Fourier
π
(2k + 1)
p
k=0
series converges to the average value of the function. In this case, the average value is 0 (as can be
seen from the graph.
∞
Thus the Fourier series is
5. (a) and (b) The function is even. It is also continuous for all x. All the bn s are 0. Also, by
computing the area between the graph of f and the x-axis, from x = 0 to x = p, we see that a0 = 0.
Now, using integration by parts, we obtain
an
p
Z
v0
p
u
{
z }| { z }|
nπ
(x − p/2) cos
x dx
p

2c
nπ
4c
−
(x − p/2) cos
x dx = − 2
p
p
p 0
0

=0
z
}|
{
Z p

4c  p
p
nπ
nπ p
= − 2
−
sin
(x
−
p/2)
sin
x
x dx


p
nπ
p x=0 nπ 0
p
=
2
p
Z
4c p2
nπ p
4c
= − 2 2 2 cos
= 2 2 (1 − cos nπ)
x
p n π
p x=0
n π
(
0
if n is even,
=
8c
if n is odd.
n2 π 2
Thus the Fourier series is
h
i
∞ cos (2k + 1) π x
X
p
8c
f(x) = 2
.
π
(2k + 1)2
k=0
9. The function is even; so all the bn ’s are 0,
Z
p 1 − e−cp
1 p −cx
1
a0 =
e
dx = − e−cx =
;
p 0
cp
cp
0
and with the help of the integral formula from Exercise 15, Section 2.2, for n ≥ 1,
Z
2 p −cx
nπx
an =
e
cos
dx
p 0
p
2
nπx
nπx p
1
−cx
2 −cx
=
nπpe
sin
ce
cos
−
p
p n2 π2 + p2c2
p
p
0
2pc
1 − (−1)ne−cp .
=
n2π2 + p2c2
Thus the Fourier series is
∞
X
1
1
nπ
(1 − e−cp (−1)n ) cos( x) .
(1 − e−cp ) + 2cp
2 p2 + (nπ)2
pc
c
p
n=1
Section 7.3
Fourier Series of Functions with Arbitrary Periods
157
13. Take p = 1 in Exercise 1, call the function in Exercise 1 f(x) and the function in this exercise
g(x). By comparing graphs, we see that
g(x) =
1
(1 + f(x)) .
2
Thus the Fourier series of g is
∞
1
4X
1+
sin(2k + 1)πx
π
(2k + 1)
1
2
!
k=0
f x_
Which x 0, 0, 0 x 1, 1, x 1, 0
s n_, x_
1 2 2 Pi Sum 1
2 k 1 Sin 2 k
, x, 1, 1
Plot Evaluate f x , s 20, x
Which x
0, 0, 0
∞
1
1 2X
+
sin(2k + 1)πx.
2 π
(2k + 1)
=
x
1, 1, x
k=0
1
Pi x ,
k, 0, n
;
1, 0
The 41st partial sum of the Fourier series
and the function on the interval (-1, 1).
17. (a) Take x = 0 in the Fourier series of Exercise 4 and get
0=
∞
p2
4p2 X (−1)n−1
− 2
3
π n=1
n2
⇒
∞
X
(−1)n−1
π2
.
=
12 n=1
n2
(b) Take x = p in the Fourier series of Exercise 4 and get
p2 =
∞
p2 4p2 X (−1)n−1(−1)n
− 2
3
π n=1
n2
⇒
∞
X
1
π2
.
=
6
n2
n=1
Summing over the even and odd integers separately, we get
∞
∞
∞
X
X
X
π2
1
1
1
=
+
.
=
2
2
2
6
n
(2k
+
1)
(2k)
n=1
k=0
But
P∞
1
k=1 (2k)2
=
1
4
P∞
1
k=1 k 2
=
1 π2
.
4 6
So
∞
π2 X
1
π2
+
=
6
(2k + 1)2
24
k=0
k=1
⇒
∞
X
k=0
1
π2
π2
π2
=
−
=
.
(2k + 1)2
6
24
8
Fourier Series
21. From the graph, we have
f(x) =
So
f(−x) =
−1 − x
1+x
1−x
−1 + x
hence
fe (x) =
if − 1 < x < 0,
if 0 < x < 1.
f(x) + f(−x)
=
2
and
f(x) − f(−x)
fo (x) =
=
2
if − 1 < x < 0,
if 0 < x < 1;
−x if − 1 < x < 0,
x
if 0 < x < 1,
−1 if − 1 < x < 0,
1
if 0 < x < 1.
Note that, fe (x) = |x| for −1 < x < 1. The Fourier series of f is the sum of the Fourier series of fe
and fo . From Example 1 with p = 1,
∞
1
1
4 X
fe (x) = − 2
cos[(2k + 1)πx].
2 π
(2k + 1)2
k=0
From Exercise 1 with p = 1,
fo (x) =
∞
4X 1
sin[(2k + 1)πx].
π
2k + 1
k=0
Hence
∞ 1
cos[(2k + 1)πx] sin[(2k + 1)πx]
4X
f(x) = +
−
+
.
2 π
π(2k + 1)2
2k + 1
k=0
25. Since f is 2p-periodic and continuous, we have f(−p) = f(−p + 2p) = f(p). Now
Z p
p
1
1
1
0
a0 =
f 0 (x) dx =
f(x) =
(f(p) − f(−p)) = 0.
−p
2p −p
2p
2p
Integrating by parts, we get
a0n
=
1
p
Z
p
f 0 (x) cos
−p
nπx
dx
p
bn
=
=
=0
z Z
}|
{
z
}|
{
1
nπx p
nπ 1 p
nπx
f(x) cos
f(x) sin
dx
+
p
p −p p p −p
p
nπ
bn.
p
Similarly,
b0n
=
=
=
1
p
Z
p
f 0 (x) sin
−p
=0
nπx
dx
p
an
z Z
}|
{
}|
{
1
nπx p
nπ 1 p
nπx
f(x) sin
f(x) cos
dx
−
p
p −p p p −p
p
nπ
− an .
p
z
Section 7.3
Fourier Series of Functions with Arbitrary Periods
159
29. The function in Exercise 8 is piecewise smooth and continuous, with a piecewise smooth
derivative. We have
 c
 − d if 0 < x < d,
0
if d < |x| < p,
f 0 (x) =
 c
if − d < x < 0.
d
The Fourier series of f 0 is obtained by differentiating term by term the Fourier series of f (by
Exercise 26). Now the function in this exercise is obtained by multiplying f 0 (x) by − dc . So the
desired Fourier series is
dnπ dnπ
∞
∞
d 0
nπ
d 2cp X 1 − cos p
nπ
2 X 1 − cos p
nπ
− f (x) = −
−
sin
x
=
sin
x.
2
2
c
c dπ n=1
n
p
p
π n=1
n
p
33. The function F (x) is continuous and piecewise smooth with F 0(x) = f(x) at all the points
where f is continuous (see Exercise 25, Section 2.1). So, by Exercise 26, if we differentiate the
Fourier series of F , we get the Fourier series of f. Write
F (x) = A0 +
and
∞ X
nπ
nπ
An cos
x + Bn sin
x
p
p
n=1
∞ X
nπ
nπ
f(x) =
an cos
x + bn sin
x .
p
p
n=1
Note that the a0 term of the Fourier series of f is 0 because by assumption
Differentiate the series for F and equate it to the series for f and get
R 2p
0
f(x) dx = 0.
X
∞ ∞ X
nπ
nπ
nπ
nπ
nπ
nπ
−An
an cos
sin
x+
Bn cos
x =
x + bn sin
x .
p
p
p
p
p
p
n=1
n=1
Equate the nth Fourier coefficients and get
nπ
= bn
p
nπ
Bn
= an
p
−An
p
bn ;
nπ
p
⇒ Bn =
an.
nπ
⇒ An = −
This derives the nth Fourier
R x coefficients of F for n ≥ 1. To get A0 , note that F (0) = 0 because of
the definition of F (x) = 0 f(t) dt. So
0 = F (0) = A0 +
∞
X
n=1
An = A0 +
∞
X
n=1
−
p
bn ;
nπ
P∞
p
and so A0 = n=1 nπ
bn. We thus obtained the Fourier series of F in terms of the Fourier coefficients
of f; more precisely,
∞
∞ p X bn X
p
nπ
nπ
p
F (x) =
− bn cos
+
x+
an sin
x .
π n=1 n
nπ
p
nπ
p
n=1
The point of this result is to tell you that, in order to derive the Fourier series of F , you can integrate
the Fourier series of f term by term. Furthermore, the only assumption on f is that it is piecewise
Fourier Series
smooth and integrates to 0 over one period (to guarantee the periodicity of F .) Indeed, if you start
with the Fourier series of f,
∞ X
nπ
nπ
an cos
t + bn sin
t ,
f(t) =
p
p
n=1
and integrate term by term, you get
F (x) =
=
=
Z
x
Z
∞ X
an
f(t) dt =
x
nπ
cos
t dt + bn
p
Z
x
nπ
sin
t dt
p
0
0
0
n=1
∞ p p X
nπ x
nπ x
an
sin
t dt + bn −
cos
t
nπ
p 0
nπ
p 0
n=1
∞
∞ p X bn X
p
nπ
nπ
p
− bn cos
+
x+
an sin
x ,
π n=1 n
nπ
p
nπ
p
n=1
as derived earlier. See the following exercise for an illustration.
Section 7.4
Half-Range Expansions: The Cosine and Sine Series
161
Solutions to Exercises 7.4
1. The even extension is the function that is identically 1. So the cosine Fourier series is just the
constant 1. The odd extension yields the function in Exercise 1, Section 2.3, with p = 1. So the sine
series is
∞
4 X sin((2k + 1)πx)
.
π
2k + 1
k=0
This is also obtained by evaluating the integral in (4), which gives
bn = 2
Z
1
sin(nπx) dx = −
0
9. We have
bn = 2
Z
1
2
2
cos nπx =
(1 − (−1)n ).
nπ
nπ
0
1
x(1 − x) sin(nπx) dx.
0
To evaluate this integral, we will use integration by parts to derive the following two formulas: for
a 6= 0,
Z
x cos(a x) sin(a x)
x sin(ax) dx = −
+ C,
+
a
a2
and
Z
x2 sin(ax) dx =
2 cos(a x) x2 cos(a x) 2 x sin(a x)
−
+ C.
+
a3
a
a2
So
Z
x(1 − x) sin(ax) dx
=
−2 cos(a x) x cos(a x) x2 cos(a x) sin(a x) 2 x sin(a x)
−
−
+ C.
+
+
a3
a
a
a2
a2
Applying the formula with a = nπ, we get
Z
1
x(1 − x) sin(nπx) dx
0
=
=
=
−2 cos(nπ x) x cos(nπ x) x2 cos(nπ x) sin(nπ x) 2 x sin(nπ x) 1
−
−
+
+
(nπ)3
nπ
nπ
(nπ)2
(nπ)2
0
n
n
n
n
−2 ((−1) − 1) (−1)
(−1)
−2 ((−1) − 1)
−
+
=
(nπ)3
nπ
nπ
(nπ)3
4
if n is odd,
(nπ)3
0
if n is even.
Thus
bn =
8
(nπ)3
0
if n is odd,
if n is even,
Hence the sine series in
∞
8 X sin(2k + 1)πx
.
π3
(2k + 1)3
k=0
Fourier Series
b k_
8 Pi ^ 3 1
2 k 1 ^ 3;
ss n_, x_ : Sum b k Sin 2 k 1 Pi x , k, 0, n ;
partialsineseries Table ss n, x , n, 1, 5 ;
x 1 x
f x_
Plot Evaluate partialsineseries, f x
, x, 0, 1
0.25
0.2
0.15
0.1
0.05
0.2
0.4
0.6
0.8
1
Perfect!
13. We have
1
sin 2πx.
2
This yields the desired 2-periodic sine series expansion.
sin πx cos πx =
17. (b) Sine series expansion:
Z
Z
2 ah
nπx
2 p h
nπx
bn =
x sin
dx +
(x − p) sin
dx
p 0 a
p
p a a−p
p
Z a
2h h
p
nπx i
p
nπx a
=
cos
−x
cos
dx
+
ap
nπ
p 0 nπ 0
p
Z p
2h h
p
(−p)
n π x p
nπx i
+
(x − p)
cos(
) +
cos
dx
a
(a − p)p
nπ
p
p
a nπ
2h h −ap
nπa i
nπa
p2
=
sin
cos
+
pa nπ
p
(nπ)2
p
h
2h
nπa i
p
nπa
p2
+
sin
(a − p) cos
−
(a − p)p nπ
p
(nπ)2
p
2hp
nπa 1
1 =
sin
−
(nπ)2
p a a−p
2hp2
nπa
=
sin
.
2
(nπ) (p − a)a
p
Hence, we obtain the given Fourier series.
Section 7.5
Complex Form of Fourier Series
Solutions to Exercises 7.5
1. From Example 1, for a 6= 0, ±i, ±2i, ±3i, . . .,
eax =
∞
sinh πa X (−1)n inx
e
π
a − in
n=−∞
(−π < x < π);
consequently,
e−ax =
∞
sinh πa X (−1)n inx
e
π
a + in
n=−∞
(−π < x < π),
and so, for −π < x < π,
cosh ax
=
=
=
eax + e−ax
2
∞
sinh πa X
1
1
n
(−1)
+
einx
2π n=−∞
a + in a − in
∞
a sinh πa X (−1)n inx
e .
π
n2 + a2
n=−∞
2. From Example 1, for a 6= 0, ±i, ±2i, ±3i, . . .,
eax =
∞
sinh πa X (−1)n inx
e
π
a − in
n=−∞
(−π < x < π);
consequently,
e−ax =
∞
sinh πa X (−1)n inx
e
π
a + in
n=−∞
(−π < x < π),
and so, for −π < x < π,
sinh ax =
=
=
eax − e−ax
2
∞
sinh πa X
2π
(−1)n
n=−∞
∞
X
1
1
−
a − in a + in
einx
n
i sinh πa
(−1)n 2
einx.
π
n
+
a2
n=−∞
5. Use identities (1); then
e2ix + e−2ix
e3ix + e−3ix
+2
2
2
−2ix
2ix
e
e
= e−3ix +
+
+ e3ix .
2
2
cos 2x + 2 cos 3x =
9. If m = n then
cm = cn =
1
2π
Z
Z π
z}|{
1
einx = f(x)e−inx dx =
dx = 1.
2π −π
−π
π
163
Fourier Series
If m 6= n, then
cm
1
=
2π
Z
π
inx −imx
e
e
dx =
−π
=
=
=
1
2π
Z
π
ei(n−m)x dx
−π
π
−i
ei(n−m)x −π
2(n − m)π
−i
ei(n−m)π − e−i(n−m)π
2(n − m)π
−i
(cos[(n − m)π] − cos[−(n − m)π]) = 0.
2(n − m)π
Thus all the Fourier coefficients are 0 except cn = n. Hence einx is its own Fourier series.
(b) Because of the linearity of the Fourier coefficients and by part (a), the function is its own Fourier
series.
13. (a) This is straightforward. Start with the Fourier series in Exercise 1: For a 6= 0, ±i, ±2i, ±3i, . . .,
and −π < x < π, we have
∞
a sinh πa X (−1)n inx
cosh ax =
e .
π
n2 + a2
n=−∞
On the left side, we have
Z π
1
cosh2(ax) dx =
2π −π
=
Z
1 π cosh(2ax) + 1
dx
π 0
2
π
1 1 1
1
x+
sinh(2ax) =
π+
sinh(2aπ) .
0
2π
2a
2π
2a
On the right side of Parseval’s identity, we have
∞
(a sinh πa)2 X
1
.
2
2
π
(n + a2 )2
n=−∞
Hence
∞
(a sinh πa)2 X
1 1
1
.
π+
sinh(2aπ) =
2
2
2π
2a
π
(n + a2)2
n=−∞
Simplifying, we get
∞
X
n=−∞
(n2
1
1
π
π+
=
sinh(2aπ) .
2
2
2
+a )
2(a sinh πa)
2a
(b) This part is similar to part (a). Start with the Fourier series of Exercise 2: For a 6= 0, ±i, ±2i, ±3i, . . .,
and −π < x < π, we have
sinh ax =
On the left side, we have
Z π
1
sinh2(ax) dx =
2π −π
=
∞
i sinh πa X
n
(−1)n 2
einx.
2
π
n
+
a
n=−∞
Z
1 π cosh(2ax) − 1
dx
π 0
2
π
1 1 1
1
−x+
sinh(2ax) =
−π+
sinh(2aπ) .
0
2π
2a
2π
2a
Section 7.5
Complex Form of Fourier Series
165
On the right side of Parseval’s identity, we have
∞
sinh2 πa X
n2
.
π2
(n2 + a2)2
n=−∞
Hence
∞
sinh2 (πa) X
1 n2
1
.
−π+
sinh(2aπ) =
2π
2a
π2
(n2 + a2 )2
n=−∞
Simplifying, we get
∞
X
n=−∞
(n2
n2
π
1
=
sinh(2aπ) .
−π+
2
2
2
+a )
2a
2 sinh (πa)
17. In the Taylor series
∞
X
zm
m!
m=1
ez =
(all z),
take z = eiθ . Then, for all θ,
iθ
ee
=
∞
X
eimθ
,
m!
m=1
iθ
which is the complex form of the Fourier series of ee = ecos θ+i sin θ .
21. To find the complex Fourier coefficients of f, we use (6) and the fact that the (usual) Fourier
n+1
coefficients of f are an = 0 for n and bn = 2(−1)
(see Exercise 13, Section 7.2). Thus
n
i
−i(−1)n+1
cn = − bn =
2
n
if n > 0 and
cn =
i
i(−1)n+1
bn =
2
n
if n < 0. Thus the Fourier coefficients of f ∗ f are
−
1
n2
for all n 6= 0.
These coefficients are 4 times the coefficients of the convolution that we have in Example 4. Thus
the convolution in this exercise is 4 times the function in Example 6. Thus, for x in (0, 2π),
f ∗ f(x) =
1 π2
− (x − π)2 .
2 3
25. (a) At x = π, the Fourier series converges to the average of the function,
(f(π+) + f(π−))/2 =
eaπ + e−aπ
= cosh aπ
2
Fourier Series
(this is clear on Figure 1). Thus
=(−1)n
cosh aπ
=
∞
z}|{
sinh aπ X (−1)n
einπ
(a
+
in)
2 + n2
π
a
n=−∞
=
−1
∞
sinh aπ X (a + in) 1 X (−1)n
+
(a + in)einx
+
2 + n2
2 + n2
π
a
a
a
n=−∞
n=1
=
∞
∞
1 sinh aπ X (a − in) X (a + in)
+
+
a
π
a2 + n2 n=1 a2 + n2
n=1
=
∞
1 sinh aπ X 2a
.
+
a
π
a2 + n2
n=1
(b) Multiply both sides by
aπ
sinh aπ
and the desired identity follows.