calculus-4c-1.pdf

Leif Mejlbro
Examples of Fourier series
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Leif Mejlbro
Calculus 4c-1

Examples of Fourier series – Calculus 4c-1
© 2008 Leif Mejlbro & Ventus Publishing ApS
ISBN 978-87-7681-380-2

4
Contents
Contents
Introduction
1. Sum function of Fourier series
2. Fourier series and uniform convergence
3. Parseval’s equation
4. Fourier series in the theory of beams
5
6
62
101
115
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5
Introduction
Introduction
Here we present a collection of examples of applications of the theory of Fourier series. The reader is
also referred to Calculus 4b as well as to Calculus 3c-2.
It should no longer be necessary rigourously to use the ADIC-model, described in Calculus 1c and
Calculus 2c, because we now assume that the reader can do this himself.
Even if I have tried to be careful about this text, it is impossible to avoid errors, in particular in the
ﬁrst edition. It is my hope that the reader will show some understanding of my situation.
Leif Mejlbro
20th May 2008

6
1 Sum function of Fourier series
A general remark. In some textbooks the formulation of the main theorem also includes the
unnecessary assumption that the graph of the function does not have vertical half tangents. It should
be replaced by the claim that f ∈ L2
over the given interval of period. However, since most people
only know the old version, I have checked in all examples that the graph of the function does not have
half tangents. Just in case . . . ♦
Example 1.1 Prove that cos nπ = (−1)n
, n ∈ N0. Find and prove an analogous expression for
cos n
π
2
and for sin n
π
2
.
(Hint: check the expressions for n = 2p, p ∈ N0, and for n = 2p − 1, p ∈ N).
0
–3/2*Pi
-Pi
Pi/2
(cos(t),sin(t))
–1
–0.5
0.5
1
–1 –0.5 0.5 1
One may interpret (cos t, sin t) as a point on the unit circle.
The unit circle has the length 2π, so by winding an axis round the unit circle we see that nπ always
lies in (−1, 0) [rectangular coordinates] for n odd, and in (1, 0) for n even.
It follows immediately from the geometric interpretation that
cos nπ = (−1)n
.
We get in the same way that at
cos n
π
2
=

0 for n ulige,
(−1)n/2
for n lige,
and
sin n
π
2
=

(−1)(n−1)/2
for n ulige,
0 for n lige.
Sum function of Fourier series

7
Example 1.2 Find the Fourier series for the function f ∈ K2π, which is given in the interval ]−π, π]
by
f(t) =

0 for − π t ≤ 0,
1 for 0 t ≤ π,
and find the sum of the series for t = 0.
1
–4 –2 2 4
x
Obviously, f(t) is piecewise C1
without vertical half tangents, so f ∈ K∗
2π. Then the adjusted function
f∗
(t) is defined by
f∗
(t) =

f(t) for t = pπ, p ∈ Z,
1/2 for t = pπ, p ∈ Z.
The Fourier series is pointwise convergent everywhere with the sum function f∗
(t). In particular, the
sum of the Fourier series at t = 0 is
f∗
(0) =
1
2
, (the last question).
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8
The Fourier coefficients are then
a0 =
1
π
π
−π
f(t) dt =
1
π
π
0
dt = 1,
an =
1
π
π
−π
f(t) cos nt dt =
1
π
π
0
cos nt dt =
1
nπ
[sin nt]π
0 = 0, n ≥ 1,
bn =
1
π
π
−π
f(t) sin nt dt =
1
π
π
0
sin nt dt = −
1
nπ
[cos nt]π
0 =
1−(−1)n
nπ
,
hence
b2n = 0 og b2n+1 =
2
π
·
1
2n + 1
.
The Fourier series is (with = instead of ∼)
f∗
(t) =
1
2
a0 +
∞

n=1
{an cos nt + bn sin nt} =
1
2
+
2
π
∞

n=0
1
2n + 1
sin(2n + 1)t.
Example 1.3 Find the Fourier series for the function f ∈ K2π, given in the interval ]− π, π] by
f(t) =
⎧
⎨
⎩
0 for − π t ≤ 0,
sin t for 0 t ≤ π,
and find the sum of the series for t = pπ, p ∈ Z.
1
–4 –2 2 4
x
The function f is piecewise C1
without any vertical half tangents, hence f ∈ K∗
2π. Since f is contin-
uous, we even have f∗
(t) = f(t), so the symbol ∼ can be replaced by the equality sign =,
f(t) =
1
2
a0 +
∞

n=1
{an cos nt + bn sin nt}.
It follows immediately (i.e. the last question) that the sum of the Fourier series at t = pπ, p ∈ Z, is
given by f(pπ) = 0, (cf. the graph).
The Fourier coefficients are
a0 =
1
π
π
−π
f(t) dt =
1
π
π
0
sin t dt =
1
π
[− cos t]π
0 =
2
π
,
a1 =
1
π
π
0
sin t · cos t dt =
1
2π

sin2
t
π
0
= 0,

9
an =
1
π
π
0
sin t · cos nt dt =
1
2π
π
0
{sin(n + 1)t − sin(n − 1)t}dt
=
1
2π
1
n − 1
cos(n − 1)t −
1
n + 1
cos(n + 1)t
π
0
=
1
2π

1
n − 1
(−1)n−1
− 1 −
1
n + 1
(−1)n+1
− 1

= −
1
π
·
1 + (−1)n
n2 − 1
for n 1.
Now,
1 + (−1)n
=

2 for n even,
0 for n odd,
hence a2n+1 = 0 for n ≥ 1, and
a2n = −
2
π
·
1
4n2 − 1
, n ∈ N, (replace n by 2n).
Analogously,
b1 =
1
π
π
0
sin2
t dt =
1
π
·
1
2
π
0
{cos2
t + sin2
t}dt =
1
2
,
and for n 1 we get
bn =
1
π
π
0
sin t · sin nt dt =
1
2π
π
0
{cos(n − 1)t − cos(n + 1)t}dt = 0.
Summing up we get the Fourier series (with =, cf. above)
f(t) =
1
2
a0 +
∞

n=1
{an cos nt + bn sin nt} =
1
π
+
1
2
sin t −
2
π
∞

n=1
1
4n2 − 1
cos 2nt.
Repetition of the last question. We get for t = pπ, p ∈ Z,
f(pπ) = 0 =
1
π
−
2
π
∞

n=1
1
4n2 − 1
,
hence by a rearrangement
∞

n=1
1
4n2 − 1
=
1
2
.
We can also prove this result by a decomposition and then consider the sectional sequence,
sN =
N

n=1
1
4n2 − 1
=
N

n=1
1
(2n − 1)(2n + 1)
=
1
2
N

n=1

1
2n − 1
−
1
2n + 1

=
1
2

1 −
1
2N + 1

→
1
2

10
for N → ∞, hence
∞

n=1
1
4n2 − 1
= lim
N→∞
sN =
1
2
.
Example 1.4 Let the periodic function f : R → R, of period 2π, be given in the interval ]− π, π] by
f(t) =
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
0, for t ∈ ] −π, −π/2[ ,
sin t, for t ∈ [−π/2, π/2] ,
0 for t ∈ ]π/2, π] .
Find the Fourier series of the function and its sum function.
–1
–0.5
0.5
1
–3 –2 –1 1 2 3
x
without vertical half tangents, hence f ∈ K∗
2π. According to the main
theorem, the Fourier theorem is then pointwise convergent everywhere, and its sum function is
f∗
(t) =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
−1/2 for t = −
π
2
+ 2pπ, p ∈ Z,
1/2 for t =
π
2
+ 2pπ, p ∈ Z,
f(t) ellers.
Since f(t) is discontinuous, the Fourier series cannot be uniformly convergent.
Clearly, f(−t) = −f(t), so the function is odd, and thus an = 0 for every n ∈ N0, and
bn =
2
π
π
0
f(t) sin nt dt =
2
π
π/2
0
sin t · sin nt dt =
1
π
π/2
0
{cos((n − 1)t) − cos((n + 1)t)}dt.
In the exceptional case n = 1 we get instead
b1 =
1
π
π/2
0
(1 − cos 2t)dt =
1
π
t −
1
2
sin 2t
π/2
0
=
1
2
,
and for n ∈ N {1} we get
bn =
1
π
1
n − 1
sin((n − 1)t) −
1
n + 1
sin((n + 1)t)
π/2
0
=
1
π

1
n − 1
sin

n − 1
2
π

−
1
n + 1
sin

n + 1
2
π

.

11
It follows immediately that if n 1 is odd, n = 2p + 1, p ≥ 1, then b2p+1 = 0 (note that b1 =
1
2
has
been calculated separately) and that (for n = 2p even)
b2p =
1
π

1
2p − 1
sin

pπ −
π
2

−
1
2p + 1
sin

pπ +
π
2

=
1
π

1
2p − 1

− cos(pπ) · sin
π
2

−
1
2p + 1

cos pπ · sin
π
2

=
1
π
(−1)p+1

1
2p − 1
+
1
2p + 1

=
1
π
(−1)p+1
·
4p
4p2 − 1
.
By changing variable p → n, it follows that f has the Fourier series
f ∼
1
2
sin t +
∞

n=1
1
π
(−1)n−1
·
4n
4n2 − 1
sin 2nt = f∗
(t),
where we already have proved that the series is pointwise convergent with the adjusted function f∗
(t)
as its sum function.
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12
Example 1.5 Find the Fourier series for the periodic function f ∈ K2π, given in the interval ]π, π]
by
f(t) = | sin t|.
Then find the sum of the series
∞

n=1
(−1)n+1
4n2 − 1
.
0
1
–4 –2 2 4
x
It follows from the figure that f is piecewise differentiable without vertical half tangents, hence f ∈
K∗
2π. Since f is also continuous, we have f∗
(t) = f(t) everywhere. Then it follows by the main
theorem that the Fourier series is pointwise convergent everywhere so we can replace ∼ by =,
f(t) =
1
2
a0 +
∞

n=1
{an cos nt + bn sin nt}.
Calculation of the Fourier coefficients. Since f(−t) = f(t) is even, we have bn = 0 for every
n ∈ N, and
an =
2
π
π
0
sin t · cos nt dt =
1
π
π
0
{sin(n + 1)t − sin(n − 1)t}dt.
Now, n − 1 = 0 for n = 1, so we have to consider this exceptional case separately:
a1 =
1
π
π
0
sin 2t dt =
1
2π
[− cos 2t]π
0 = 0.
We get for n = 1,
an =
1
π
π
0
{sin(n + 1)t − sin(n − 1)t}dt
=
1
π
−
1
n + 1
cos(n + 1)t +
1
n − 1
cos(n − 1)t
π
0
=
1
π

1 + (−1)n
n + 1
−
1 + (−1)n
n − 1

= −
2
π
·
1 + (−1)n
n2 − 1
.
Now,
1 + (−1)n
=

2 for n even,
0 for n odd,
so we have to split into the cases of n even and n odd,
a2n+1 = 0 for n ≥ 1 (and for n = 0 by a special calculation),

13
and
a2n = −
4
π
·
1
4n2 − 1
for n ≥ 0, especially a0 = +
4
π
.
Then the Fourier series can be written with = instead of ∼,
(1) f(t) = | sin t| =
2
π
−
4
π
∞

n=1
1
4n2 − 1
cos 2nt.
Remark 1.1 By using a majoring series of the form c
∞
n=1
1
n2
, it follows that the Fourier series is
uniformly convergent.
We shall ﬁnd the sum of
∞
n=1(−1)n+1
/(4n2
− 1). When this is compared with the Fourier series, we
see that they look alike. We only have to choose t, such that cos 2nt gives alternatingly ±1.
By choosing t =
π
2
, it follows by the pointwise result (1) that
f
π
2

= 1 =
2
π
−
4
π
∞

n=1
1
4n2 − 1
cos nπ =
2
π
−
4
π
∞

n=1
(−1)n
4n2 − 1
=
2
π
+
4
π
∞

n=1
(−1)n+1
4n2 − 1
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14
thus
4
π
∞

n=1
(−1)n+1
4n2 − 1
= 1 −
2
π
=
π − 2
π
,
and hence
∞

n=1
(−1)n+1
4n2 − 1
=
π − 2
4
.
Example 1.6 Let the periodic function f : R → R of period 2π, be given by
f(t) =
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
0, for t ∈ ]−π, −π/4[ ,
1, for t ∈ [−π/4, π/4] ,
0 for t ∈ ]π/4, π] .
1) Prove that f has the Fourier series
1
4
+
2
π
∞

n=1
1
n
sin
nπ
4

cos nt.
2) Find the sum of the Fourier series for t =
π
4
, and then ﬁnd the sum of the series
∞

n=1
1
n
sin
nπ
2

1
–4 –2 2 4
x
Clearly, f is piecewise C1
(with f
= 0, where the derivative is deﬁned), hence f ∈ K∗
2π. According
to the main theorem, the Fourier series is then pointwise convergent everywhere with the adjusted
function as its sum function,
f∗
(t) =
⎧
⎪
⎨
⎪
⎩
1
2
for t = ±
π
4
+ 2pπ, p ∈ Z,
f(t) otherwise.
Since f(t) is not continuous, the Fourier series cannot be uniformly convergent.

15
1) Since f is even, we have bn = 0 for every n ∈ N, and
an =
2
π
π
0
f(t) cos nt dt =
2
π
π/4
0
1 · cos nt dt =
2
πn
sin
nπ
4

for n ∈ N. For n = 0 we get instead
a0 =
2
π
π/4
0
1 dt =
2
π
·
π
4
=
1
2
,
so
f ∼
1
2
a0 +
∞

n=1
an cos nt =
1
4
+
2
π
∞

n=1
1
n
sin
nπ
4

cos nt.
2) When t =
π
4
we get from the beginning of the example,
f∗
π
4

=
1
2
=
1
4
+
2
π
∞

n=1
1
n
sin
nπ
4

· cos
nπ
4

=
1
4
+
1
π
∞

n=1
1
n
sin
nπ
2

.
Then by a rearrangement,
∞

n=1
1
n
sin
nπ
2

=
π
4
.
Alternatively,
∞

n=1
1
n
sin
nπ
2

=
∞

p=1
1
2p − 1
sin

pπ −
π
2

=
∞

p=1
(−1)p−1
2p − 1
= Arctan 1 =
π
4
.
Example 1.7 Let f : ]0, 2[ → R be the function given by f(t) = t in this interval.
1) Find a cosine series with the sum f(t) for every t ∈ ]0, 2[.
2) Find a sine series with the sum for every t ∈ ]0, 2[.
The trick is to extend f as an even, or an odd function, respectively.
1) The even extension is
F(t) = |t| for t ∈ [−2, 2], continued periodically.
It is obviously piecewise C1
and without vertical half tangents, hence F ∈ K∗
4 . The periodic
continuation is continuous everywhere, hence it follows by the main theorem (NB, a cosine
series) with equality that
F(t) =
1
2
a0 +
∞

n=1
an cos

nπt
2

,

16
0
0.5
1
1.5
2
–3 –2 –1 1 2 3
x
where
an =
4
4
2
0
t · cos

n ·
2π
4
t

dt =
2
0
t cos

nπt
2

dt, n ∈ N0.
Since we must not divide by 0, we get n = 0 as an exceptional case,
a0 =
2
0
t dt =
t2
2
2
0
= 2.
For n 0 we get by partial integration,
an =
2
0
t cos
nπ
2
t

dt =
2
nπ
t sin
nπ
2
t
2
0
−
2
nπ
2
0
sin
nπ
2
t

dt
=
4
π2n2

cos
nπ
2
t
2
0
=
4
π2n2
{(−1)n
− 1}.
For even indices = 0 we get a2n = 0.
For odd indices we get
a2n+1 =
2
π2(2n + 1)2
{(−1)2n+1
− 1} = −
8
π2
·
1
(2n + 1)2
, n ∈ N0.
The cosine series is then
F(t) = 1 −
8
π2
∞

n=0
1
(2n + 1)2
cos

nπ +
π
2

t,
and in particular
f(t) = t = 1 −
8
π2
∞

n=0
1
(2n + 1)2
cos

n +
1
2

πt, t ∈ [0, 2].
2) The odd extension becomes
G(t) = t for t ∈ ] − 2, 2[.
We adjust by the periodic extension by G(2p) = 0, p ∈ Z. Clearly, G ∈ K∗
4 , and since G is odd
and adjusted, it follows from the main theorem with equality that
G(t) =
∞

n=1
bn sin

nπt
2

,

17
–2
–1
1
2
–3 –2 –1 1 2 3
x
where
bn =
2
0
t sin

nπt
2

dt =
2
nπ
−t cos

nπt
2
2
0
+
2
nπ
2
0
cos

nπt
2

dt
=
2
nπ
{−2 cos(nπ) + 0} + 0 = (−1)n+1
·
4
nπ
.
The sine series becomes (again with = instead of ∼ )
G(t) =
∞

n=1
(−1)n+1
·
4
nπ
sin

nπt
2

.
what‘s missing in this equation?
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18
Thus, in the interval ]0, 2[ we have
G(t) = f(t) = t =
4
π
∞

n=1
(−1)n−1
n
sin

nπt
2

, t ∈ ]0, 2[.
It is no contradiction that f(t) = t, t ∈ ]0, 2[, can be given two diﬀerent expressions of the same sum.
Note that the cosine series is uniformly convergent, while the sine series is not uniformly convergent.
3mm
In the applications in the engineering sciences the sine series are usually the most natural ones.
Example 1.8 A periodic function f : R → R of period 2π is given in the interval ]π, π] by
f(t) = t sin2
t, t ∈ ]− π, π].
1) Find the Fourier series of the function. Explain why the series is pointwise convergent and ﬁnd
its sum function.
2) Prove that the Fourier series for f is uniformly convergent on R.
–1
0
1
–4 –2 2 4
x
1) Clearly, f is piecewise C1
without vertical half tangents (it is in fact of class C1
; but to prove
this will require a fairly long investigation), so f ∈ K∗
2π. Then by the main theorem the Fourier
series is pointwise convergent with the sum function f∗
(t) = f(t), because f(t) is continuous.
Now, f(t) is odd, so an = 0 for every n ∈ N0, and
bn =
2
π
π
0
t sin2
t sin nt dt =
1
π
π
0
t(1−cos 2t) sin nt dt
=
1
2π
π
0
t{2 sin nt − sin(n + 2)t − sin(n − 2)t}dt.
Then we get for n = 2 (thus n − 2 = 0)
bn =
1
2π
t

−
2
n
cos nt+
1
n+2
cos(n+2)t+
1
n−2
cos(n−2)t
π
0
−
1
2π
π
0

−
2
n
cos nt+
1
n+2
cos(n+2)t+
1
n−2
cos(n−2)t

dt
=
1
2π
· π

−
2
n
(−1)n
+
(−1)n
n+2
+
(−1)n
n−2

=
(−1)n
2

2n
n2 −4
−
2
n

= (−1)n

n
n2 − 4
−
1
n

= (−1)n
·
4
n(n2 − 4)
.

19
We get for the exceptional case n = 2 that
b2 =
1
2π
π
0
t(2 sin 2t − sin 4t) dt
=
1
2π
t

−cos 2t+
1
4
cos 4t
π
0
+
1
2π
π
0

cos 2t−
1
4
cos 4t

dt
=
1
2π
· π

−1 +
1
4

+ 0 = −
3
8
.
Hence the Fourier series for f is (with pointwise convergence, thus equality sign)
f(t) =
4
3
sin t −
3
8
sin 2t +
∞

n=3
(−1)n
·
4
n(n2 − 4)
sin nt.
2) Since the Fourier series has the convergent majoring series
4
3
+
3
8
+
∞

n=3
4
n(n2 − 4)
=
41
24
+
∞

n=3
(−1)n
·
4
(n + 1)(n2 + 2n − 3)
≤
∞

n=1
4
n3
,
the Fourier series is uniformly convergent on R.
Example 1.9 We deﬁne an odd function f ∈ K2π by
f(t) = t(π − t), t ∈ [0, π].
8
π
∞

p=1
sin(2p − 1)t
(2p − 1)3
, t ∈ R.
2) Explain why the sum function of the Fourier series is f(t) for every t ∈ R, and ﬁnd the sum of the
series
∞

p=1
(−1)p−1
(2p − 1)3
The graph of the function is an arc of a parabola over [0, π] with its vertex at

π
2
,
π2
4

. The odd
continuation is continuous and piecewise C1
2π. Then by the
main theorem the Fourier series is pointwise convergent with the sum function f∗
(t) = f(t).
1) Now, f is odd, so an = 0. Furthermore, by partial integration,
bn =
2
π
π
0
t(π − t) sin nt dt = −
2
πn
[t(π − t) cos nt]π
0 +
2
πn
π
0
(π − 2t) cos nt dt
= 0 +
2
πn2
[(π − 2t) sin nt]π
0 +
4
πn2
π
0
sin nt dt = 0 −
4
πn3
[cos nt]π
0 =
4
π
·
1
n3
{1 − (−1)n
}.

20
–2
–1
0
1
2
–3 –2 –1 1 2 3
x
It follows that b2p = 0, and that
b2p−1 =
8
π
·
1
(2p − 1)3
,
hence the Fourier series becomes
f(t) =
8
π
∞

p=1
sin(2p − 1)t
(2p − 1)3
where we can use = according to the above.
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21
–3
–2
–1
1
2
3
–6 –4 –2 2 4 6 8
x
2) The ﬁrst question was proved in the beginning of the example.
If we choose t =
π
2
, then
f
π
2

=
π2
4
=
8
π
∞

p=1
1
(2p−1)3
sin

pπ−
π
2

=
8
π
∞

p=1
(−1)p−1
(2p − 1)3
.
∞

p=1
(−1)p−1
(2p − 1)3
=
π3
32
.
Example 1.10 Let the function f ∈ K2π be given on the interval ]− π, π] by
f(t) = t cos t.
1) Explain why the Fourier series is pointwise convergent in R, and sketch the graph of its sum
function in the interval ]− π, 3π].
−
1
2
sin t +
∞

n=2
(−1)n
·
2n
n2 − 1
sin nt, t ∈ R.
1) Since f is piecewise C1
without vertical half tangents, we have f ∈ K∗
2π. Then by the main
theorem, the Fourier series is pointwise convergent everywhere and its sum function is
f∗
(t) =

0 for t = π + 2pπ, p ∈ Z,
f(t) otherwise.
2) Since f(t) os (almost) odd, we have an = 0, and
bn =
2
π
π
0
t · cos t · sin nt dt =
1
π
π
0
t {sin(n+1)t+sin(n−1)t} dt.
For n = 1 we get
b1 =
1
π
π
0
t sin 2t dt = −
1
2π
[t cos 2t]π
0 +
1
2π
π
0
cos 2t dt = −
1
2
.

22
For n 1 we get by partial integration
bn =
1
π
t

−
cos(n+1)t
n + 1
−
cos(n−1)t
n − 1
π
0
+
1
π
π
0

cos(n+1)t
n + 1
+
cos(n−1)t
n − 1

dt
=
1
π
· π

−
cos(n + 1)π
n + 1
−
cos(n − 1)π
n − 1

+ 0 = (−1)n

1
n + 1
+
1
n − 1

= (−1)n
·
2n
n2 − 1
.
Hence the Fourier series is with pointwise equality
f∗
(t) = −
1
2
sin t +
∞

n=2
(−1)n
·
2n
n2 − 1
sin nt.
Example 1.11 A 2π-periodic function is given in the interval ]− π, π] by
f(t) = 2π − 3t.
1) Explain why the Fourier series is pointwise convergent for every t ∈ R, and sketch the graph of its
sum function s(t).
2) Find the Fourier series for f.
1
2
3
4
–6 –4 –2 2 4 6
x
without vertical half tangents, we get f ∈ K∗
theorem the Fourier series is pointwise convergent with the sum function
s(t) =

2π for t = π + 2pπ, p ∈ Z,
f(t) otherwise.
The graph of the function f(t) is sketched on the ﬁgure.
2) Now, f(t) = 2π − 3t =
1
2
a0 − 3t is split into its even and its odd part, so it is seen by inspection
that a0 = 4π, and that the remainder part of the series is a sine series, so an = 0 for n ≥ 1, and
bn =
2
π
π
0
(−3t) sin nt dt =
6
πn
[t cos nt]π
0 −
6
πn
π
0
cos nt dt = (−1)n
·
6
n
,
hence (with equality sign instead of ∼ )
s(t) = 2π +
∞

n=1
(−1)n
·
6
n
sin nt, t ∈ R.

23
Example 1.12 Let f : [0, π] → R denote the function given by
f(t) = t2
− 2t.
1) Find the cosine series the sum of which for every t ∈ [0, π] is equal to f(t).
2) Find a sine series the sum of which for every t ∈ [0, π[ is equal to f(t).
Since f is piecewise C1
2π. The even extension is
continuous, hence the cosine series is by the main theorem equal to f(t) in [0, π].
–3
–2
–1
0
1
2
3
–4 –2 2 4
x
The odd extension is continuous in the half open interval [0, π[, hence the main theorem only shows
that the sum function is f(t) in the half open interval [0, π[.
1) Cosine series. From
a0 =
2
π
π
0
f(t) dt =
2
π
π
0
(t2
−2t)dt =
2
π
t3
3
−t2
π
0
=
2π2
3
− 2π,
and for n ∈ N,
an =
2
π
π
0
(t2
− 2t) cos nt dt =
2
πn

(t2
− 2t) sin nt
π
0
−
4
πn
π
0
(t − 1) sin nt dt
= 0 +
4
πn2
[(t − 1) cos nt]π
0 −
4
πn2
π
0
cos nt dt =
4
πn2
{(π − 1) · (−1)n
+ 1} + 0,
we get by the initial comments with equality sign
f(t) = t2
− 2t =
π2
3
− π +
∞

n=1
4
πn2
{1 + (−1)n
(π − 1)} cos nt
for t ∈ [0, π]

24
2) Sine series. Since
bn =
2
π
π
0
(t2
− 2t) sin nt dt = −
2
πn
(t2
−2t) cos nt
π
0
+
4
πn
π
0
(t−1) cos nt dt
= −
2
πn
π(π−2) · (−1)n
+
4
πn2
[(t−1) sin nt]π
0 −
4
πn2
π
0
sin nt dt
=
2
n
(π − 2) · (−1)n−1
+ 0 +
4
πn3
[cos nt]π
0 =
2(π − 2)
n
· (−1)n−1
+
4
πn3
{(−1)n
− 1} ,
we get by the initial comments with equality sign,
f(t)=t2
−2t=
∞

n=1

2(π−2)
n
(−1)n−1
−
4
πn3
[1−(−1)n
]

sin nt
for t ∈ [0, π[.

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25
Example 1.13 Find the Fourier series of the periodic function of period 2π, given in the interval
]− π, π] by
f(t) =
⎧
⎨
⎩
t sin t, for t ∈ [0, π],
−t sin t, for t ∈ ]− π, 0[,
and find for every t ∈ R the sum of the series.
Then find for every t ∈ [0, π] the sum of the series
∞

n=1
n2
(2n + 1)2(2n − 1)2
cos 2nt.
Finally, find the sum of the series
∞

n=1
n2
(2n + 1)2(2n − 1)2
.
Since f is continuous and piecewise C1
without vertical half tangents, we see that f ∈ K∗
2π. Then by
the main theorem the Fourier series is pointwise convergent with the sum f∗
(t) = f(t).
–1
0
1
–4 –2 2 4
x
Since f(t) is odd, the Fourier series is a sine series, hence an = 0, and
bn =
2
π
π
0
t · sin t · sin nt dt =
1
π
π
0
t{cos(n−1)t−cos(n+1)t}dt.
We get for n = 1,
b1 =
1
π
π
0
t{1 − cos 2t}dt =
1
π
t2
2
π
0
−
1
2π
[t sin 2t]π
0 +
1
2π
π
0
sin 2t dt =
π
2
.
For n 1 we get instead,
bn =
1
π
t

sin(n−1)t
n − 1
−
sin(n+1)t
n + 1
π
0
−
1
π
π
0

sin(n−1)t
n − 1
−
sin(n+1)t
n + 1

dt
= 0 +
1
π
cos(n−1)t
(n − 1)2
−
cos(n+1)t
(n + 1)2
π
0
=
1
π

1
(n − 1)2
−
1
(n + 1)2

·

(−1)n−1
−1

.
It follows that b2n+1 = 0 for n ≥ 1, and that
b2n = −
2
π

1
(2n−1)2
−
1
(2n+1)2

=−
1
π
·
16n
(2n−1)2(2n+1)2
for n ∈ N.

26
Hence, the Fourier series is (with an equality sign according to the initial comments)
f(t) =
π
2
sin t −
16
π
∞

n=1
n
(2n−1)2(2n+1)2
sin 2nt.
When we compare with the next question we see that a) we miss a factor n, and b) we have sin 2nt
occurring instead of cos 2nt. However, the formally diﬀerentiated series
π
2
cos t −
32
π
∞

n=1
n2
(2n − 1)2(2n + 1)2
cos 2nt
has the right structure. Since it has the convergent majoring series
π
2
+
32
π
∞

n=1
n2
(2n − 1)2(2n + 1)2
,
(the diﬀerence between the degree of the denominator and the degree of the numerator is 2, and

n−2
is convergent), it is absolutely and uniformly convergent, and its derivative is given by
f
(t) =
⎧
⎨
⎩
sin t + t cos t, for t ∈ ]0, π[,
− sin t − t cos t, for t ∈ ]− π, 0[,
where
lim
t→0+
f
(t) = lim
t→0−
f
(t) = 0 and lim
t→π−
f
(t) = lim
t→−π+
f
(t) = −π
The continuation of f
(t) is continuous, hence we conclude that
f
(t) =
π
2
cos t −
32
π
∞

n=1
n2
(2n − 1)2(2n + 1)2
cos 2nt,
and thus by a rearrangement,
∞

n=1
n2
(2n − 1)2(2n + 1)2
cos 2nt =
π2
64
cos t−
π
32
f
(t) =
π2
64
cos t−
π
32
sin t−
π
32
t cos t for t ∈ [0, π].
Finally, insert t = 0, and we get
∞

n=1
n2
(2n − 1)2(2n + 1)2
=
π2
64
.
Alternatively, the latter sum can be calculated by a decomposition and the application of the sum
of a known series. In fact, it follows from
n2
(2n − 1)2(2n + 1)2
=
1
16
·
{(2n − 1) + (2n + 1)}2
(2n − 1)2(2n + 1)2
=
1
16

(2n+1)2
+(2n−1)2
+ 2(2n+1)(2n−1)
(2n − 1)2(2n + 1)2

=
1
16

1
(2n−1)2
+
1
(2n+1)2
2
(2n−1)(2n+1)

=
1
16

1
(2n−1)2
+
1
(2n+1)2
+
1
2n−1
−
1
2n+1


27
that
∞

n=1
n2
(2n − 1)2(2n + 1)2
=
1
16
∞

n=1
1
(2n−1)2
+
1
16
∞

n=1
1
(2n+1)2
+
1
16
lim
N→∞
N

n=1

1
2n − 1
−
1
2n + 1

=
1
16

2
∞

n=1
1
(2n−1)2
− 1

+
1
16
lim
N→∞

1 −
1
2N + 1

=
1
8
∞

n=1
1
(2n − 1)2
.
Since
π2
6
=
∞

n=1
1
n2
=
∞

n=1
1
(2n − 1)2

1 +
1
22
+
1
24
+
1
26
+ · · ·

=
∞

n=1
1
(2n − 1)2
·
∞

k=0

1
4
k
=
1
1 −
1
4
∞

n=1
1
(2n − 1)2
=
4
3
∞

n=1
1
(2n − 1)2
,
we get
∞

n=1
n2
(2n − 1)2(2n + 1)2
=
1
8
∞

n=1
1
(2n − 1)2
=
1
8
·
3
4
·
π2
6
=
π2
64
.
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28
Example 1.14 The odd and periodic function f of period 2π, is given in the interval ]0, π[ by
f(t) = cos 2t, t ∈ ]0, π[.
2) Indicate the sum of the series for t =
7π
6
.
3) Find the sum of the series
∞

n=0
(−1)n+1
·
2n + 1
(2n − 1)(2n + 3)
.
–1
0
1
–4 –2 2 4
x
Since f(t) is piecewise C1
2π, so the Fourier series
converges according to the main theorem pointwise towards the adjusted function f∗
(t). Since f is
odd, it is very important to have a ﬁgure here. The function f∗
(t) is given in [−π, π] by
f∗
(t) =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
0 for t = −π,
− cos 2t for t ∈ ]− π, 0[,
0 for t = 0,
cos 2t for t ∈ ]0, π[,
0 for t = π,
continued periodically.
1) Now, f is odd, so an = 0, and
bn =
2
π
π
0
cos 2t · sin nt dt =
1
π
π
0
{sin(n+2)t+sin(n−2)t}dt.
Since sin(n − 2)t = 0 for n = 2, this is the exceptional case. We get for n = 2,
b2 =
1
π
π
0
sin 4t dt =
1
π
−
cos 4t
4
π
0
= 0.
Then for n = 2,
bn =
1
π
−
cos(n+2)t
n + 2
−
cos(n−2)t
n − 2
π
0
= −
1
π

1
n+2
+
1
n−2

{(−1)n
−1}.

29
It follows that b2n = 0 for n 1 (and also for n = 1, by the earlier investigation of the exceptional
case), and that
b2n+1 = −
1
π

1
2n+3
+
1
2n−1

· (−2) =
4
π
·
2n + 1
(2n−1)(2n+3)
.
Summing up we get the Fourier series (with an equality sign instead of the difficult one, ∼ )
(2) f∗
(t) =
4
π
∞

n=0
2n + 1
(2n − 1)(2n + 3)
sin(2n + 1)t.
2) This question is very underhand, cf. the figure. It follows from the periodicity that the sum of the
series for t =
7π
6
π, is given by
f

7π
6

= f

7π
6
− 2π

= f

−
5π
6

= − cos

−
5π
3

= − cos
π
3
= −
1
2
.
3) The coefficient of the series is the same as in the Fourier series, so we shall only choose t in such
a way that sin(2n + 1)t becomes equal to ±1.
We get for t =
π
2
,
sin(2n + 1)
π
2
= sin nπ · cos
π
2
+ cos nπ · sin
π
2
= (−1)n
,
hence by insertion into (2),
f∗
π
2

= cos π = −1 =
4
π
∞

n=0
2n + 1
(2n − 1)(2n + 3)
(−1)n
,
and finally by a rearrangement,
∞

n=0
(−1)n+1
·
2n + 1
(2n − 1)(2n + 3)
=
π
4
.
Remark 1.2 The last question can also be calculated by means of a decomposition and a considera-
tion of the sectional sequence (an Arctan series). The sketch of this alternative proof is the following,
∞

n=0
(−1)n+1
·
2n + 1
(2n − 1)(2n + 3)
= · · · =
∞

n=0
(−1)n
2n + 1
= Arctan 1 =
π
4
.
The details, i.e. the dots, are left to the reader.

30
Example 1.15 Find the Fourier series the function f ∈ K2π, which is given in the interval [−π, π]
by
f(t) = t · sin t.
Find by means of this Fourier series the sum function of the trigonometric series
∞

n=2
(−1)n
sin nt
(n − 1)n(n + 1)
for t ∈ [−π, π].
Since f is continuous and piecewise C1
2π. Then by
the main theorem, the Fourier series is pointwise convergent everywhere and its sum function is
f∗
(t) = f(t).
1
–4 –2 2 4
x
Since f is even, the Fourier series is a cosine series, thus bn = 0, and
an =
2
π
π
0
t sin t cos nt dt =
1
π
π
0
t{sin(n+1)t−sin(n−1)t}dt.
The exceptional case is n = 1, in which sin(n − 1)t = 0 identically. For n = 1 we calculate instead,
a1 =
1
π
π
0
t sin 2t dt =
1
2π
[−t cos 2t]π
0 +
1
2π
π
0
cos 2t dt =
−π
2π
= −
1
2
.
For n = 1 we get
an =
1
π
t

−
cos(n + 1)t
n + 1
+
cos(n − 1)t
n − 1
π
0
+
1
π
π
0

cos(n + 1)t
n + 1
−
cos(n − 1)t
n − 1

dt
=
1
π
· π

−
1
n + 1
+
1
n − 1

· (−1)n+1
= (−1)n+1
·
2
(n − 1)(n + 1)
.
According to the initial remarks we get with pointwise equality sign,
f(t) = t sin t = 1 −
1
2
cos t+2
∞

n=2
(−1)n−1
(n−1)(n+1)
cos nt, for t ∈ [−π, π].
The Fourier series has the convergent majoring series
1 +
1
2
+ 2
∞

n=2
1
n2 − 1
,

31
hence it is uniformly convergent. We may therefore integrate it term by term,
t
0
f(τ) dτ =t −
1
2
sin t+2
∞

n=2
(−1)n−1
(n−1)n(n+1)
sin nt,
for t ∈ [−π, π].
Hence by a rearrangement for t ∈ [−π, π],
∞

n=2
(−1)n
sin nt
(n − 1)n(n + 1)
=
1
2
t −
1
4
sin t −
1
2
t
0
f(τ) dτ =
1
2
t −
1
4
sin t −
1
2
t
0
τ sin τ dτ
=
1
2
t −
1
4
sin t −
1
2
[−τ cos τ + sin τ]t
0 =
1
2
t −
1
4
sin t +
1
2
t cos t −
1
2
sin t
=
1
2
t +
1
2
t cos t −
3
4
sin t =
1
2
t(1 + cos t) −
3
4
sin t.
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32
Example 1.16 Prove that for every n ∈ N,
π
0
t2
cos nt dt = (−1)n
·
2π
n2
.
Find the Fourier series for the function f ∈ K2π, given in the interval [−π, π] by
f(t) = t2
sin t.
Then write the derivative f
(t) by means of a trigonometric series and ﬁnd the sum of the series
∞

n=1
(−1)n−1
·
(2n)2
(2n − 1)2(2n + 1)2
.
–4
–2
2
4
–4 –2 2 4
x
We get by partial integration,
π
0
t2
cos nt dt =
1
n
t2
sin nt
π
0
−
2
n
π
0
t sin nt dt = 0+
2t
n2
cos nt
π
0
−
2
n2
π
0
cos nt dt = (−1)n
·
2π
n2
.
The function f is continuous and piecewise C1
2π. By
the main theorem the Fourier series is pointwise convergent everywhere and its sum function is
f∗
(t) = f(t).
Since f is odd, its Fourier series is a sine series, thus an = 0, and
bn =
2
π
π
0
t2
sin t sin nt dt =
1
π
π
0
t2
{cos(n−1)t−cos(n+1)t}dt.
For n = 1 we get by the result above,
b1 =
1
π
π
0
t2
(1 − cos 2t)dt =
1
π
·
π3
3
−
1
π
· (−1)2
·
2π
4
=
π2
3
−
1
2
.

33
For n 1 we also get by the result above,
bn =
1
π
· 2π

(−1)n−1
(n − 1)2
−
(−1)n+1
(n + 1)2

= (−1)n−1
· 2 ·
(n+1)2
−(n−1)2
(n − 1)2(n + 1)2
= (−1)n−1
·
8n
(n−1)2(n+1)2
.
According to the initial comments we have equality sign for t ∈ [−π, π],
f(t)=t2
sin t=

π2
3
−
1
2

sin t+
∞

n=2
(−1)n−1
· 8n
(n−1)2(n+1)2
sin nt.
By a formal termwise diﬀerentiation of the Fourier series we get

π2
3
−
1
2

cos t + 8
∞

n=2
(−1)n−1
·
n2
(n − 1)2(n + 1)2
cos nt.
This has the convergent majoring series
π2
3
−
1
2
+ 8
∞

n=2
n2
(n − 1)2(n + 1)2
,
hence it is uniformly convergent and its sum function is
f
(t) = t2
cos t + 2t sin t =

π2
3
−
1
2

cos t+8
∞

n=2
(−1)n−1
· n2
(n − 1)2(n + 1)2
cos nt.
When we insert t =
π
2
, we get
f
π
2

= 0 + π = π = 0 + 8
∞

n=2
(−1)n−1
· n2
(n − 1)2(n + 1)2
cos

n
π
2

= 8
∞

n=1
(−1)2n−1
· (2n)2
(2n − 1)2(2n + 1)2
cos(nπ) + 0
= 8
∞

n=1
(−1)n−1
·
(2n)2
(2n − 1)2(2n + 1)2
,
hence by a rearrangement,
∞

n=1
(−1)n−1 (2n)2
(2n − 1)2(2n + 1)2
=
π
8
.
Alternatively, we get by a decomposition,
(2n)2
(2n−1)2(2n+1)2
=
1
4

1
(2n−1)2
+
1
(2n+1)2
+
1
2n−1
−
1
2n+1

,

34
thus
∞

n=1
(−1)n−1 (2n)2
(2n − 1)2
=
1
4
∞

n=1
(−1)n−1
(2n − 1)2
+
∞

n=1
(−1)n−1
(2n + 1)2
+ lim
N→∞
N

n=1
(−1)n−1

1
2n − 1
−
1
2n + 1

=
1
4
∞

n=1
(−1)n−1
(2n − 1)2
−
∞

n=2
(−1)n−1
(2n − 1)2

+
1
4
lim
N→∞
N

n=1
(−1)n−1
2n − 1
+
N+1

n=2
(−1)n−1
2n − 1

=
1
4
+
1
4
lim
N→∞

2
N

n=1
(−1)n−1
2n − 1
− 1 +
(−1)N
2N + 1

=
1
2
∞

n=1
(−1)n−1
2n − 1
=
1
2
Arctan 1 =
π
8
.
Example 1.17 The odd and periodic function f of period 2π is given in the interval [0, π] by
f(t) =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
sin t, for t ∈

0,
π
2

,
− sin t, for t ∈
π
2
, π

1) Find the Fourier series of the function. Explain why the series is pointwise convergent, and ﬁnd
its sum for every t ∈ [0, π].
∞

n=0
(−1)n
(2n + 1)
(4n + 1)(4n + 3)
.
Since f is piecewise C1
2π. By the main theorem the
Fourier series is pointwise convergent and its sum is
f∗
(t) =

0 for t =
π
2
+ pπ, p ∈ Z,
f(t) otherwise.
–1
0
1
–4 –2 2 4
x

35
1) Since f is odd, we have an = 0, and for n 1 we get
bn =
2
π
π
0
f(t) sin nt dt =
2
π
π/2
0
sin t sin nt dt −
2
π
π
π/2
sin t sin nt dt
=
1
π
π/2
0
{cos(n − 1)t − cos(n + 1)t}dt −
1
π
π
π/2
{cos(n − 1)t − cos(n + 1)t}dt
=
1
π

sin(n − 1)t
n − 1
−
sin(n + 1)t
n + 1
π/2
0
+
sin(n − 1)t
n − 1
−
sin(n + 1)t
n + 1
π/2
π

=
2
π
⎧
⎨
⎩
sin(n − 1)
π
2
n − 1
−
sin(n + 1)
π
2
n + 1
⎫
⎬
⎭
.
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36
Hence bn+1 = 0 for n ≥ 1, and
b2n =
2
π
⎧
⎨
⎩
sin

nπ −
π
2

2n − 1
−
sin

nπ +
π
2

2n + 1
⎫
⎬
⎭
=
2
π
(−1)n−1

1
2n − 1
+
1
2n + 1

= (−1)n−1
·
2
π
·
4n
(2n − 1)(2n + 1)
, n ∈ N.
For n = 1 (the exceptional case) we get
b1 =
2
π
π/2
0
sin2
t dt−
π
π/2
sin2
t dt

=
2
π
π/2
0
sin2
t dt−
π/2
0
sin2
t dt

= 0.
Summing up we get the Fourier series
f ∼
∞

n=1
(−1)n−1
·
8
π
·
n
(2n − 1)(2n + 1)
sin 2nt.
The sum is in [0, π] given by
8
π
∞

n=1
(−1)n−1
n
(2n − 1)(2n + 1)
=
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
sin t for t ∈

0,
π
2

,
0 for t =
π
2
,
− sin t for t ∈
π
2
, π

.
2) When we put t =
π
4
into the Fourier series, we get
sin
π
4
=
√
2
2
=
8
π
∞

n=1
(−1)n−1
·
n
(2n − 1)(2n + 1)
sin

n
π
2

=
8
π
∞

p=0
(−1)2p+1−1
·
2p + 1
(4p + 1)(4p + 3)
· (−1)p
,
hence
∞

n=0
(−1)n
·
2n + 1
(4n + 1)(4n + 3)
=
π
√
2
8
.

37
Example 1.18 1) Given the infinite series
a)
∞

n=1
(−1)n+1
n
n2 + 1
, b)
∞

n=1
n
n2 + 1
, c)
∞

n=1
(−1)n+1
(2n − 1)
(2n − 1)2 + 1
.
Explain why the series of a) and c) are convergent, while the series of b) is divergent.
2) Prove for the series of a) that the difference between its sum s and its n-th member of its sectional
sequence sn is numerically smaller than 10−1
, when n ≥ 9.
3) Let a function f ∈ K2π be given by
f(t) = sinh t for − π t ≤ π.
Prove that the Fourier series for f is
2 sinh π
π
∞

n=1
(−1)n+1
n
n2 + 1
sin nt.
4) Find by means of the result of (3) the sum of the series c) in (1).
1) Since
n
n2 + 1
∼
1
n
, and
∞
n=1
1
n
is divergent, it follows from the criterion of equivalence that
b) is divergent. It also follows that neither a) nor c) can be absolutely convergent. Since an → 0
for n → ∞, we must apply Leibniz’s criterion. Clearly, both series are alternating. If we put
ϕ(x) =
x
x2 + 1
, er ϕ
(x) =
x2
+1−2x2
(x2 +1)2
=
1 − x2
(1+x2)2
0
for x 1, then ϕ(x) → 0 decreasingly for x → ∞, x 1. Then it follows from Leibniz’s criterion
that both a) and c) are (conditionally) convergent.
2) Since a) is alternating, the error is at most equal to the first neglected term, hence
|s − sn| ≤ |s − s9| ≤ |a10| =
10
102 + 1
=
10
101

1
10
for n ≥ 9.
theorem, the Fourier series is pointwise convergent and its sum function is
f∗
(t) =
⎧
⎨
⎩
0 for t = (2p + 1)π, p ∈ Z
f(t) ellers.
Since f is odd, we have an = 0, and
bn =
2
π
π
0
sinh t · sin nt dt = −
2
πn
[sinh t · cos nt]π
0 +
2
πn
π
0
cosh t · cos nt dt
=
2
πn
sinh π · (−1)n+1
+
2
πn2
[cosh t · sin nt]π
0 −
2
πn2
π
0
sinh t · sin nt dt
= (−1)n+1
·
2
n
·
sinh π
π
−
1
n2
bn,

38
–10
–5
0
5
10
–4 –2 2 4
x
bn =

1 +
1
n2
−1
· (−1)n+1
·
2
n
·
sinh π
π
=
2 sinh π
π
·
(−1)n+1
n
n2 + 1
.
The Fourier series is (with equality sign, cf. the above)
f∗
(t) = sinh t =
2 sinh π
π
∞

n=1
(−1)n+1
n
n2 + 1
sin nt for t ∈ ]− π, π[.
4) When we put t =
π
2
into the Fourier series, we get
sinh
π
2
= 2
sinh π
π
∞

n=1
(−1)n+1
n
n2 + 1
sin

n
π
2

= 2
sinh π
π
∞

p=1
(−1)2p
(2p − 1)
(2p − 1)2 + 1
sin

pπ −
π
2

= 4
sinh
π
2
cosh
π
2
π
∞

n=1
(−1)n+1
(2n − 1)
(2n − 1)2 + 1
,
∞

n=1
(−1)n+1
(2n − 1)
(2n − 1)2 + 1
=
π
4 cosh
π
2
.

39
Example 1.19 The even and periodic function f of period 2π is given in the interval [0, π] by
f(x) =
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
k − k2
x, x ∈ 0,
1
k
,
0, x ∈
1
k
, π ,
where k ∈
1
π
, ∞ .
1) Find the Fourier series of the function. Explain why the series is uniformly convergent, and find
its sum for x =
1
k
.
2) Explain why the series
∞

n=1
cos n
n2
and
∞

n=1
cos2
n
n2
are convergent, and prove by means of (1) that
∞

n=1
cos2
n
n2
=
1
4
+
∞

n=1
cos n
n2
.
3) In the Fourier series for f we denote the coefficient of cos nx by an(k), n ∈ N. Prove that
limk→∞ an(k) exists for every n ∈ N and that it does not depend on n.
1) It follows by a consideration of the figure that f ∈ K∗
2π and that f is continuous. Then by the
main theorem, f is the sum function for its Fourier series.
0
0.5
1
1.5
2
–3 –2 –1 1 2 3
x
Since f is even, we get bn = 0, and for n ∈ N we find
an =
2
π
1/k
0
(k − k2
x) cos nx dx =
2
πn

(k − k2
x) sin nx
1/k
0
+
2k2
πn
1/k
0
sin nx dx
=
2k2
πn2

1 − cos
n
k

.
Since
a0 =
2
π
1/k
0
(k − k2
x)dx =
2
π
kx −
1
2
k2
x2
1/k
0
=
1
π
,

40
the Fourier series becomes (with equality sign, cf. the above)
f(x) =
1
2π
+
2k2
π
∞

n=1
1
n2

1 − cos
n
k

cos nx.
Since
1
2π
+
2k2
π
∞
n=1
1
n2
is a convergent majoring series, the Fourier series is uniformly convergent.
When x =
1
k
, the sum is equal to
(3) f

1
k

= 0 =
1
2π
+
2k2
π
∞

n=1
1
n2

1 − cos
n
k

cos
n
k
.
2) Since
2k2
π
∞
n=1
1
n2
is a convergent majoring series, the series of (3) can be split. Then by a
rearrangement,
∞

n=1
1
n2
cos2
n
k

=
1
4k2
+
∞

n=1
1
n2
cos
n
k

for every k
1
π
.
If we especially choose k = 1
1
π
, we get
∞

n=1
1
n2
cos2
n =
1
4
+
∞

n=1
1
n2
cos n.
3) Clearly,
a0(k) =
1
π
→
1
π
for k → ∞.
For n 0 it follows by a Taylor expansion,
an(k) =
2k2
πn2

1 − cos
n
k

=
2k2
πn2

1 −

1 −
1
2
n2
k2
+
n2
k2
ε
n
k

=
1
π
+ ε
n
k

→
1
π
for k → ∞.

41
Example 1.20 Given the function f ∈ K2π, where
f(t) = cos
t
2
, −π t ≤ π.
1) Sketch the graph of f.
2
π
+
1
π
∞

n=1
(−1)n+1
n2 −
1
4
cos nt,
and explain why the Fourier series converges pointwise towards f on R.
∞

n=1
(−1)n+1
n2 −
1
4
.
1) Clearly, f is piecewise C1
2π, and we can apply the main
theorem. Now, f(t) is continuous, hence the adjusted function is f(t) itself, and we have with an
equality sign,
f(t) =
1
2
a0 +
∞

n=1
an cos nt,
where we have used that f(t) is even, so bn = 0. We have thus proved (1) and the latter half of
(2).
0.4
0.8
y
–6 –4 –2 2 4 6
x

42
2) Calculation of the Fourier coeﬃcients. It follows from the above that bn = 0. Furthermore,
a0 =
2
π
π
0
cos
t
2
dt =
4
π
sin
t
2
π
0
=
4
π
,
and
an =
2
π
π
0
cos
t
2
cos nt dt =
1
π
π
0

cos

n+
1
2

t+cos

n−
1
2

t

dt
=
1
π
1
n+ 1
2
sin

n+
1
2

t+
1
n− 1
2
sin

n−
1
2

t
π
0
=
1
π

(−1)n
n+ 1
2
−
(−1)n
n− 1
2

=
(−1)n
π
·
(n − 1
2 ) − (n + 1
2 )
n2 − 1
4
=
(−1)n+1
π
·
1
n2 −
1
4
.
Hence, the Fourier series is (with equality, cf. (1))
f(t) =
1
2
a0 +
∞

n=1
an cos nt =
2
π
+
1
π
∞

n=1
(−1)n+1
n2 −
1
4
cos nt.
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43
Alternative proof of the convergence. Since the Fourier series has the convergent majoring
series
2
π
+
1
π
∞

n=1
1
n2 −
1
4
,
it is uniformly convergent, hence also pointwise convergent.
3) The sum function is f(t), hence for t = 0,
f(0) = cos 0 = 1 =
2
π
+
1
π
∞

n=1
(−1)n+1
n2 −
1
4
,
and we get by a rearrangement,
∞

n=1
(−1)n+1
n2 −
1
4
= π − 2.
Example 1.21 The even and periodic function f of period 2π ia given in the interval [0, π] by
⎧
⎨
⎩
(t − (π/2))2
, t ∈ [0, π/2],
0, t ∈ ]π/2, π].
1) Sketch the graph of f in the interval [−π, π] and explain why f is everywhere pointwise equal its
Fourier series.
2) Prove that
f(t) =
π2
24
+ 2
∞

n=1

1
n2
−
2
πn3
sin n
π
2

cos nt, t ∈ R.
3) Find by using the result of (2) the sum of the series
∞

p=1
(−1)p−1
p2
.

Hint: Insert t =
π
2

.
without vertical half tangents, we see that f ∈ K∗
2π. Since f is continuous,
we have f∗
= f. Since f is even, it follows that bn = 0, hence we have with equality sign by the
main theorem that
f(t) =
1
2
a0 +
∞

n=1
an cos nt,
where
an =
2
π
π
0
f(t) cos nt dt =
2
π
π/2
0

t −
π
2
2
cos nt dt, n ∈ N0.

44
0
0.5
1
1.5
2
2.5
y
–3 –2 –1 1 2 3
x
2) We have bn = 0 (an even function), and
a0 =
2
π
π/2
0

t −
π
2
2
dt =
2
π
1
3

t −
π
2
3 π/2
0
=
π2
12
.
For n ∈ N we get by partial integration
an =
2
π
π/2
0

t −
π
2
2
cos nt dt =
2
π

t −
π
2
2
·
sin nt
n
π/2
0
−
4
πn
π/2
0

t −
π
2

sin nt dt
= 0 +
4
πn

t −
π
2

·
cos nt
n
π/2
0
−
4
πn2
π/2
0
cos nt dt
= −
4
πn

−
π
2

·
1
n
−
4
πn2
sin nt
n
π/2
0
=
2
n2
−
4
πn3
sin n
π
2
.
Hence the Fourier series is
(4) f(t) =
1
2
a0 +
∞

n=1
an cos nt =
π2
24
+ 2
∞

n=1

1
n2
−
2
πn3
sin n
π
2

cos nt, t ∈ R.

45
3) When we insert t =
π
2
into (4) we get
f
π
2

= 0 =
π2
24
+ 2
∞

n=1

1
n2
−
2
πn3
sin n
π
n

cos n
π
2
=
π2
24
+ 2
∞

n=1

1
n2
cos n
π
2
−
2 sin nπ
2 cos nπ
2
πn3

=
π2
24
+ 2
∞

n=1

1
n2
cos n
π
2
−
1
πn3
sin nπ

=
π2
24
+ 2
∞

n=1
1
n2
cos n
π
2
=
π2
24
+ 2
∞

p=0
1
(2p+1)2
cos
π
2
+pπ

+2
∞

p=1
1
(2p)2
cos pπ
=
π2
24
+
1
2
∞

p=1
(−1)p
p2
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46
because cos
π
2
+ pπ

= 0 for p ∈ Z.
∞

p=1
(−1)p−1
p2
=
π2
12
.
Example 1.22 An even function f ∈ K4 is given in the interval [0, 2] by
f(t) =
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
1 for 0 ≤ t ≤ /2,
1/2 for /2 t ≤ 3/2,
0 for 3/2 t ≤ 2.
1) Sketch the graph of f in the interval −3 ≤ t ≤ 3, and find the angular frequency ω.
When we answer the next question, the formula at the end of this example may be helpful.
2) a) Give reasons for why the Fourier series for f is of the form
f ∼
1
2
a0 +
∞

n=1
an cos

nπt
2

,
and find the value of a0.
b) Prove that an = 0 for n = 2, 4, 6, · · · .
c) Prove that for n odd an may be written as
an =
2
nπ
sin

n
π
4

.
3) It follows from the above that
f ∼
1
2
+
√
2
π

cos
πt
2
+
1
3
cos
3πt
2
−
1
5
cos
5πt
2
−
1
7
cos
7πt
2
+ · · ·

.
Apply the theory of Fourier series to find the sum of the following two series,
(1) 1 +
1
3
−
1
5
−
1
7
+
1
9
+
1
11
−
1
13
− · · · ,
(2) 1 −
1
3
+
1
5
−
1
7
+
1
9
−
1
11
+
1
13
− · · · .
The formula to be used in (2):
sin u + sin v = 2 sin

u + v
2

cos

u − v
2

.

47
0
0.2
0.4
0.6
0.8
1
y
–3 –2 –1 1 2 3
x
1) The angular frequency is ω =
2π
T
=
2π
4
=
π
2
.
Since f is piecewise constant, f is piecewise C1
without vertical half tangents, thus f ∈ K∗
4.
According to the main theorem, the Fourier series is pointwise convergent everywhere with
the adjusted function f∗
(t) as its sum function. Here f∗
(t) = f(t), with the exception of the
discontinuities of f, in which the value is the mean value.
2) a) Since f is even and ω =
π
2
, the Fourier series has the structure
f ∼
1
2
a0 +
∞

n=0
an cos
nπt
2
= f∗
(t),
where
a0 =
4
T
T/2
0
f(t) dt =
1

2
0
f(t) dt =
1

1 ·

2
+
1
2
·

= 1.
b) If we put n = 2p, p ∈ N, then
a2p =
1

2
0
f(t) cos
2pπt
2
dt =
1

2
0
f(t) cos
pπt

dt
=
1

1 ·
/2
0
cos
pπt

dt +
1
2
3/2
/2
cos
pπt

dt

=
1

pπ
sin
pπt

/2
t=0
+
1
2
·

pπ
sin
pπt

3/2
/2

=
1
2pπ

sin
pπ
2
+ sin p ·
3π
2

=
1
2pπ
· 2 sin(pπ) · cos p ·
π
2
= 0.

48
c) If instead n = 2p + 1, p ∈ N0, then
a2p+1 =
1

/2
0
f(t) cos
(2p+1)πt
2
dt +
1
2
3/2
/2
f(t) cos
(2p+1)πt
2
dt

=
1
(2p+1)π

sin

(2p+1)π ·
1
4

+ sin

(2p+1)π ·
3
4

=
1
nπ

sin
nπ
4

+ sin

n

π −
π
4

,
where we have put 2p + 1 = n. Since n is odd, we get
sin

nπ − n
π
4

= cos nπ · sin

−n
π
4

= + sin

n
π
4

.
Then by insertion,
an =
2
nπ
sin

n
π
4

for n odd.
3) Since sin

n
π
4

=
1
√
2
, and sin

n
π
4

for n odd has changing “double”-sign (two pluses follows
by two minuses and vice versa), we get all things considered that
f∗
(t) =
1
2
+
√
2
π

cos
πt
2
+
1
3
cos
3πt
2
−
1
5
cos
5πt
2
−
1
7
cos
7πt
2
+ + − − · · ·

.
When t = 0 we get in particular,
f∗
(0) = 1 =
1
2
+
√
2
π

1 +
1
3
−
1
5
−
1
7
+
1
9
+
1
11
− − · · ·

,
1 +
1
3
−
1
5
−
1
7
+ + − − · · · =
π
√
2

1 −
1
2

=
π
2
√
2
.
We get for t =

2
the adjusted value f∗

2

=
3
4
, thus
3
4
= f∗

2

=
1
2
+
√
2
π

cos
π
4
+
1
3
cos
3π
4
−
1
5
cos
5π
4
−
1
7
cos
7π
4
+ · · ·

=
1
2
+
1
π

1 −
1
3
+
1
5
−
1
7
+
1
9
− · · ·

,
and by a rearrangement,
1 −
1
3
+
1
5
−
1
7
+
1
9
−
1
11
+ · · · =
π
4
.
Remark 1.3 The result is in agreement with that the series on the left hand side is the series for
Arctan 1 =
π
4
.

49
Example 1.23 The periodic function f of period 2π os given in the interval ]− π, π] by
f(t) =
1
π4
(t2
− π2
)2
, t ∈ ] − π, π].
1) Sketch the graph of f in the interval [−π, π].
2) Prove that the Fourier series for f is given by
8
15
+
48
π4
∞

n=1
(−1)n−1
n4
cos nt, t ∈ R.
Hint: It may be used that
π
0
(t2
− π2
)2
cos nt dt = 24π ·
(−1)n−1
n4
, n ∈ N.
∞

n=1
1
n4
by using the result of (2).
1) The function f(t) is continuous and piecewise C1
without vertical half tangents. It follows from
f(−π) = f(π) = 0 and f
(t) =
4t
π4
(t2
− π2
) where f
(−π+) = f
(π−) = 0 that we even have that
f(t) is everywhere C1
, so f ∈ K∗
2π. It follows from the main theorem that the Fourier series for
f(t) is everywhere pointwise convergent and its sum function is f(t).
0
0.2
0.4
0.6
0.8
1
y
–3 –2 –1 1 2 3
x

50
2) Since f(t) is an even function, the Fourier series is a cosine series. We get for n = 0,
a0 =
2
π
π
0
1
π4
(t2
− π2
)2
dt =
2
π5
π
0
(t4
− 2π2
t2
+ π4
) dt
=
2
π5
1
5
t5
−
2
3
π2
t3
+ π4
t
π
0
= 2

1
5
−
2
3
+ 1

=
16
15
,
hence
1
2
a0 =
8
15
.
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Furthermore, for n ∈ N,
an =
2
π
π
0
1
π4
(t2
− π2
)2
cos nt dt =
2
π5
π
0
(t2
− π2
)2
cos nt dt
=
2
π5
(t2
−π2
)2
·
1
n
sin nt
π
0
−
2
π5
·
4
n
π
0
t(t2
−π2
) sin nt dt
= 0 +
8
π5
·
1
n2
[t(t2
−π2
) cos nt]π
0 −
8
π5
·
1
n2
π
0
(3t2
−π2
) cos nt dt
= 0 −
8
π5
·
1
n3
[(3t2
−π2
) sin nt]π
0 +
48
π5
π
0
t sin nt dt
= −
48
π5
·
1
n4
[t cos nt]π
0 +
48
π5
·
1
n4
π
0
cos nt dt
=
48
π4
·
1
n4
· (−1)n−1
+ 0 =
48
π4
·
1
n4
· (−1)n−1
.
We have proved that the Fourier series is pointwise convergent with an equality sign, cf. (1),
(5) f(t) =
8
15
+
48
π4
∞

n=1
(−1)n−1
n4
cos nt, t ∈ R.

52
3) In particular, if we choose t = π in (5), then
0 = f(π) =
8
15
+
48
π4
∞

n=1
(−1)n−1
n4
cos nπ =
8
15
−
48
π4
∞

n=1
1
n4
.
Finally, by a rearrangement,
∞

n=1
1
n4
=
π4
48
·
8
15
=
π4
90
.
Example 1.24 1) Sketch the graph of the function f(t) = sin
t
2
, t ∈ R, in the interval [−2π, 2π].
2) Prove that
f(t) =
2
π
−
4
π
∞

n=1
cos nt
4n2 − 1
, t ∈ R.
Hint: One may use without proof that

sin
t
2
cos nt dt =
4
4n2 −1

n sin
t
2
sin nt +
1
2
cos
t
2
cos nt

,
for t ∈ R and n ∈ N0.
3) Find, by using the result of (2), the sum of the series
(a)
∞

n=1
1
4n2 − 1
og
∞

n=1
(−1)n−1
4n2 − 1
.
1) The function f(t) is continuous and piecewise C∞
without vertical half tangents. It is also even
and periodic with the interval of period [−π, π[. Then by the main theorem the Fourier series
for f(t) is pointwise convergent everywhere and f(t) is its sum function. Since f(t) is even, the
Fourier series is a cosine series.
2) It follows from the above that bn = 0 and (cf. the hint)
an =
2
π
π
0
sin
t
2
cos nt dt =
2
π
π
0
sin
t
2
· cos nt dt
=
2
π
·
4
4n2 − 1
n sin
t
2
· sin nt +
1
2
cos
t
2
cos nt
π
0
=
4
π
·
−1
4n2 − 1
.
In particular,
1
2
a0 =
1
π
π
0
sin
t
2
dt =
1
π
− cos
t
2
π
0
=
2
π
,
so we get the Fourier expansion with pointwise equality sign, cf. (1),
f(t) =
2
π
−
4
π
∞

n=1
cos nt
4n2 − 1
, t ∈ R.

53
0
0.4
0.8
y
–6 –4 –2 2 4 6
x
3) a) If we insert t = 0 into the Fourier series, we get
f(0) = 0 =
2
π
−
4
π
∞

n=1
1
4n2 − 1
=
4
π

1
2
−
∞

n=1
1
4n2 − 1

,
∞

n=1
1
4n2 − 1
=
1
2
.
Alternatively, it follows by a decomposition that
1
4n2 − 1
=
1
(2n − 1)(2n + 1)
=
1
2
·
1
2n − 1
−
1
2
·
1
2n + 1
.
The corresponding segmental sequence is then
sN =
N

n=1
1
4n2 − 1
=
1
2
∞

n=1
1
2n − 1
−
1
2
∞

n=1
1
2n + 1
=
1
2
N

n=1
1
2n − 1
−
1
2
N+1

n=2
1
2n − 1
=
1
2
−
1
2
·
1
2N + 1
→
1
2
for N → ∞,
and the series is convergent with the sum
∞

n=1
= lim
N→∞
sN =
1
2
.

54
b) When we insert t = π into the Fourier series, we get
f(π) = 1 =
2
π
−
4
π
∞

n=1
(−1)n
4n2 − 1
=
4
π

1
2
+
∞

n=1
(−1)n−1
4n2 − 1

.
Hence by a rearrangement,
∞

n=1
(−1)n−1
4n2 − 1
=
π
4
−
1
2
.
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Alternatively, we get (cf. the decomposition above) the segmental sequence,
sN =
N

n=1
(−1)n−1
4n2 − 1
=
1
2
N

n=1
(−1)n−1
2n − 1
−
1
2
N

n=1
(−1)n−1
2n + 1
=
1
2
N−1

n=0
(−1)n
2n + 1
+
1
2
N

n=1
(−1)n
2n + 1
=
N−1

n=0
(−1)n
2n + 1
· 12n+1
−
1
2
+
1
2
·
(−1)N
2N + 1
→ Arctan 1 −
1
2
=
π
4
−
1
2
for N → ∞.
The series is therefore convergent with the sum
∞

n=1
(−1)n−1
4n2 − 1
= lim
N→∞
N

n=1
(−1)n−1
4n2 − 1
=
π
4
−
1
2
.
Example 1.25 Let the function f : R → R be given by
f(x) =
1
5 − 3 cos x
, x ∈ R.
Prove that f(x) has the Fourier series
1
4
+
1
2
∞

n=1
1
3n
cos nx, x ∈ R.
Let the function g : R → R be given by
g(x) =
sin x
5 − 3 cos x
, x ∈ R.
Prove that g(x) has the Fourier series
2
3
∞

n=1
1
3n
sin nx, x ∈ R.
1) Explain why the Fourier series for f can be differentiated termwise, and find the sum of the
differentiated series for x = π
2 .
2) Find by means of the power series for ln(1 − x) the sum of the series
∞

n=1
1
n · 3n
.
3) Prove that the Fourier series for g can be integrated termwise in R.
4) Finally, find the sum of the series
∞

n=1
1
n · 3n
cos nx, x ∈ R.

56
Since
eix
3
=
1
3
1 for every x ∈ R, we get by the complex quotient series (the proof for the legality
of this procedure is identical with the proof in the real case),
∞

n=1
1
3n
einx
=
∞

n=1

eix
3
n
=
eix
3
·
1
1 −
eix
3
=
eix
3 − eix
·
3 − e−ix
3 − e−ix
=
3eix
− 1
9 − 6 cos x + 1
=
1
2
·
3 cos x − 1 + 3i sin x
5 − 3 cos x
.
Hence
–0.2
–0.1
0
0.1
0.2
y
–8 –6 –4 –2 2 4 6 8
x
1
4
+
1
2
∞

n=1
1
3n
cos nx =
1
4
+
1
2
Re
∞

n=1
1
3n
einx

=
1
4
+
1
2
·
1
2
·
3 cos x − 1
5 − 3 cos x
=
1
4
·
(5 − 3 cos x) + (3 cos x − 1)
5 − 3 cos x
=
1
5 − 3 cos x
= f(x),
and
2
3
∞

n=1
1
3n
sin nx =
2
3
Im
∞

n=1
1
3n
einx

=
2
3
·
1
2
·
3 sin x
5 − 3 cos x
=
sin x
5 − 3 cos x
= g(x).

57
1) From
1
4
+
1
2
∞

n=1
1
3n
cos nx ≤
1
4
+
1
2
∞

n=1

1
3
n
=
1
4
+
1
4
=
1
2
,
follows that the Fourier series has a convergent majoring series, so it is uniformly convergent with
the continuous sum function
1
4
+
1
2
∞

n=1
1
3n
cos nx =
1
5 − 3 cos x
= f(x).
The termwise differentiated series,
−
1
2
∞

n=1
n
3n
sin nx,
is also uniformly convergent, because

n/3n
∞ is a convergent majoring series. Then it follows
from the theorem of differentiation of series that the differentiated series is convergent with the
sum function
−
1
2
∞

n=1
n
3n
sin nx = f
(x) =
d
dx

1
5 − 3 cos x

= −
3 sin x
(5 − 3 cos x)2
.
Hence for x =
π
2
,
−
3
25
= −
1
2
∞

n=1
n
3n
sin
nπ
2
= −
1
2
∞

m=0
4m + 1
34m+1
+
1
2
∞

m=0
4m + 3
34m+3
= −
1
2
∞

n=0
(−1)n
·
2n + 1
32n+1
=
1
6
∞

n=0
(−1)n
·
2n + 1
9n
.
2) It follows from
ln

1
1 − x

=
∞

n=1
xn
n
, |x| 1,
that we for x =
1
3
have
∞

n=1
1
n · 3n
= ln
3
2
.
3) Since also 2
3

3−n
sin nx is uniformly convergent (same argument as in (1), i.e. the obvious ma-
joring series is convergent), it follows that the Fourier series for g can be integrated termwise in
R.

58
4) We get by termwise integration that
x
0
g(t) dt =
x
0
sin t
5 − 3 cos t
dt =
1
3
[ln(5 − 3 cos t)]x
0 =
1
3
ln(5 − 3 cos x) −
1
3
ln 2
=
2
3
∞

n=1
1
3n
x
0
sin nx dx = −
2
3
∞

n=1
1
n · 3n
(cos nx − 1),
∞

n=1
1
n · 3n
cos nx =
∞

n=1
1
n · 3n
+
1
2
ln 2 −
1
2
ln(5 − 3 cos x) = ln
3
2
+
1
2
ln 2 −
1
2
ln(5 − 3 cos x)
= ln 3 −
1
2
ln 2 −
1
2
ln(5 − 3 cos x).
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Example 1.26 Let f ∈ K2π be given by
f(t) =
⎧
⎨
⎩
t, for 0 t ≤ π,
0, for π t ≤ 2π.
1) Sketch the graph of f in the interval [−2π, 2π].
π
4
+
∞

n=1

(−1)n
−1
πn2
cos nt +
(−1)n−1
n
sin nt

, t ∈ R.
Hint: One may without proof apply that for every n ∈ N,

t cos nt dt =
1
n2
(nt sin nt + cos nt),

t sin nt dt =
1
n2
(−nt cos nt + sin nt).
3) Find the sum of the Fourier series for t = π.
1) The adjusted function is
f∗
(t) =
⎧
⎨
⎩
t, for 0 t π,
π/2, for t = π,
0, for π t ≤ 2π,
continued periodically.
0
0.5
1
1.5
2
2.5
3
y
–6 –4 –2 2 4 6
x
Since f(t) is piecewise C1
,
f
(t) =

1 for 0 t π,
0 for π t 2π,

60
without vertical half tangents, it follows from the main theorem that the Fourier series is point-
wise convergent with the adjusted function f∗
(t) as its sum function. In particular, f ∼ can be
replaced by f∗
(t) = .
2) The Fourier series is pointwise
f∗
(t) =
1
2
a0 +
∞

n=1
{an cos nt + bn sin nt},
where
an =
1
π
2π
0
f(t) cos nt dt =
1
π
π
0
t cos nt dt =
1
πn2
[nt sin nt + cos nt]π
0
=
1
πn2
{0 + cos nπ − 1} =
(−1)n
− 1
πn2
for n ∈ N,
and
bn =
1
π
2π
0
f(t) sin nt dt =
1
π
π
0
t sin nt dt =
1
πn2
[−nt cos nt + sin nt]π
0
=
1
πn2
{−nπ cos nπ + 0 + 0 − 0} =
1
n
· (−1)n−1
for n ∈ N.
Finally, we consider the exceptional value n = 0, where
a0 =
1
π
2π
0
f(t) dt =
1
π
π
0
t dt =
1
π
t2
2
∞
0
=
π
2
.
Hence by insertions,
f∗
(t) =
π
4
+
∞

n=1

(−1)n
− 1
πn2
cos nt +
(−1)n−1
n
sin nt

, t ∈ R.
3) The argument is given in (1), so the sum is
f∗
(π) =
π
2
.
Alternatively,
π
4
+
∞

n=1

(−1)n
− 1
πn2
cos nπ +
(−1)n−1
n
sin nπ

=
π
4
+
1
π
∞

p=0
−2
(2p + 1)2
cos(2p + 1)π
=
π
4
+
2
π
∞

p=0
1
(2p + 1)2
.
Notice that every n ∈ N is uniquely written in the form n = (2p + 1) · 2q
, thus
π2
6
=
∞

n=1
1
n2
=
∞

p=0
1
(2p + 1)2
∞

q=0
1
(22)q
=
1
1 −
1
4
·
∞

p=0
1
(2p + 1)2
=
4
3
∞

p=0
1
(2p + 1)2
,

61
and hence
∞

p=0
1
(2p + 1)2
=
3
4
·
π2
6
=
π2
8
.
We get by insertion the sum
π
4
+
2
π
∞

p=0
1
(2p + 1)2
=
π
4
+
2
π
·
π2
8
=
π
4
+
π
4
=
π
2
.
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62
2 Fourier series and uniform convergence
Example 2.1 The function f ∈ K2π is given by
f(t) = π2
− t2
, −π t ≤ π.
2) Find the sum function of the Fourier series and prove that the Fourier series is uniformly conver-
gent in R.
0
2
4
6
8
10
–6 –4 –2 2 4 6
x
No matter the formulation of the problem, it is always a good idea to start by sketching the graph of
the function over at periodic interval and slightly into the two neighbouring intervals.
Then check the assumptions of the main theorem: Clearly, f ∈ C1
(]− π, π[) without vertical half
tangents, hence f ∈ K∗
2π.
The Fourier series is pointwise convergent everywhere, so ∼ can be replaced by = when we use the
adjusted function
f∗
(t) =
f(t+) + f(t−)
2
as our sum function.
It follows from the graph that f(t) is continuous everywhere, hence f∗
(t) = f(t), and we have obtained
without any calculation that we have pointwise everywhere
f(t) =
1
2
a0 +
∞

n=1
{an cos nx + bn sin nx}.
After this simple introduction with lots of useful information we start on the task itself.
1) The function f is even, (f(−t) = f(t)), so bn = 0, and
an =
2
π
π
0
(π2
−t2
) cos nt dt.
Fourier series and uniform convergence

63
We must not divide by 0, so let n = 0. Then we get by a couple of partial integrations,
an =
2
π
π
0
(π2
−t2
) cos nt dt =
2
πn

(π2
−t2
) sin nt
π
0
+
4
πn
π
0
t sin nt dt
= 0 +
4
πn2
[−t cos nt]π
0 +
4
πn2
π
0
cos nt dt = −
4
πn2
π cos nπ + 0 =
4(−1)n+1
n2
.
a0 =
2
π
π
0
(π2
−t2
)dt =
2
π
π2
x −
x3
3
π
0
=
4π3
3π
=
4π2
3
.
The Fourier series is then, where we already have argued for the equality sign,
(6) f(t) =
1
2
a0 +
∞

n=1
an cos nt =
2π2
3
+
∞

n=1
(−1)n−1
·
4
n2
cos nt.
2) The estimate (−1)n−1 4
n2
cos nt ≤
4
n2
shows that
2π2
3
+ 4
∞

n=1
1
n2
=
2π2
3
+ 4 ·
π2
6
=
4π2
3
is a convergent majoring series. Hence the Fourier series is uniformly convergent.
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64
Remark 2.1 We note that if we put t = 0 into (6), then
f(0) = π2
=
2π2
3
+ 4
∞

n=1
(−1)n−1
n2
,
and hence by a rearrangement,
∞

n=1
(−1)n−1
n2
=
π2
12
.
Example 2.2 A function f ∈ K2π is given in the interval ]0, 2π] by
f(t) = t2
.
Notice the given interval!
1) Sketch the graph of f in the interval ] 2π, 2π].
2) Sketch the graph of the sum function of the Fourier series in the interval ]− 2π, 2π], and check if
the Fourier series is uniformly convergent in R.
3) Explain why we have for every function F ∈ K2π,
π
−π
F(t) dt =
2π
0
F(t) dt.
The ﬁnd the Fourier series for f.
10
20
30
40
–6 –4 –2 2 4 6
x

65
1) The graph is sketch on the ﬁgure. It is not easy to sketch the adjusted function f∗
(1) in MAPLE,
so we shall only give the deﬁnition,
f∗
(t) =
⎧
⎨
⎩
f(t) for t − 2nπ ∈ ]0, 2π[, n ∈ Z,
2π2
for t = 2nπ, n ∈ Z.
theorem the Fourier series is pointwise convergent everywhere, and its sum function is the adjusted
function f∗
(t).
Each term of the Fourier series is continuous, while the sum function f∗
(t) is not continuous.
Hence, it follows that the Fourier series cannot be uniformly convergent in R.
3) When F ∈ K2π, then F is periodic of period 2π, hence
π
−π
F(t) dt =
π
0
F(t) dt +
0
−π
F(t + 2π) dt =
π
0
F(t) dt +
2π
π
F(t) dt =
2π
0
F(t) dt.
In particular,
an =
1
π
2π
0
t2
cos nt dt, og bn =
1
π
2π
0
t2
sin nt dt.
Thus
a0 =
1
π
2π
0
t2
dt =
8π3
3π
=
8π2
3
,
and
an =
1
π
2π
0
t2
cos nt dt=
1
πn

t2
sin t
2π
0
−
2
πn
2π
0
t sin nt dt
= 0+
2
πn2
[t cos nt]2π
0 −
2
πn2
2π
0
cos nt dt=
2
πn2
· 2π =
4
n2
for n ≥ 1, and
bn =
1
π
2π
0
t2
sin nt dt =
1
πn

−t2
cos nt
2π
0
+
2
πn
2π
0
t cos nt dt
= −
4π2
πn
+
2
πn2
[t sin nt]2π
0 −
2
πn2
2π
0
sin nt dt = −
4π
n
−
2
πn3
[cos nt]2π
0 = −
4π
n
.
The Fourier series is (NB. Remember the term
1
2
a0)
f ∼
4π2
3
+
∞

n=1

4
n2
cos nt −
4π
n
sin nt

(convergence in “energy”) and
f∗
(t) =
4π2
3
+
∞

n=1

4
n2
cos nt −
4π
n
sin nt

(pointwise convergence).

66
Example 2.3 Let the function f ∈ K2π be given by
f(t) = et
sin t for − π t ≤ π.
sinh π
π

1
2
+
∞

n=1

(−1)n
(4−2n2
)
n4 + 4
cos nt+
(−1)n−1
4n
n4 + 4
sin nt

.
We may use the following formulæ without proof:
π
−π
et
cos mt dt =
2(−1)m
sinh π
1 + m2
, m ∈ N0,
π
π
et
sin mt dt =
2m(−1)m+1
sinh π
1 + m2
, m ∈ N0.
2) Prove that the Fourier series in 1. is uniformly convergent.
3) Find by means of the result of 1. the sum of the series
∞

n=1
n2
− 2
n4 + 4
.
2
4
6
–6 –4 –2 2 4 6
x
The function f(t) is continuous and piecewise C1
2π. Then
by the main theorem the Fourier series is pointwise convergent everywhere, and its sum function is
f∗
(t) = f(t).
1) By using complex calculations, where
sin t =
1
2i
eit
− e−it
,

67
and b0 = 0, we get that
an + ibn =
1
π
π
−π
et
sin t · eint
dt =
1
2iπ
π
−π

e(1+i(n+1))t
− e(1+i(n−1))t

dt
=
1
2πi
1
1+ i(n+1)
{(−1)n+1
(eπ
−e−π
)} −
1
1+ i(n−1)
{(−1)n−1
(eπ
−e−π
)}
=
sinh π
π
· i(−1)n

1
1 + i(n + 1)
−
1
1 + i(n − 1)

=
sinh π
π
· (−1)n
i ·
1 + i(n − 1) − {1 + i(n + 1)}
1 − (n2 − 1) + i 2n
=
sinh π
π
(−1)n
· i ·
−2i
−(n2 − 2) + 2in
=
sinh π
π
· (−1)n
· 2 ·
−(n2
− 2) − 2in
(n2 − 2)2 + 4n2
=
sinh π
π
· (−1)n
·
4 − 2n2
− 4in
n4 + 4
.

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68
When we split into the real and the imaginary part we ﬁnd
an =
sinh π
π
· (−1)n
·
4 − 2n2
n4 + 4
, n ≥ 0,
and
bn =
sinh π
π
· (−1)n
·
4n
n4 + 4
, n ≥ 1.
Alternatively we get by real computations,
an =
1
π
π
−π
et
sin t cos nt dt =
1
2π
π
−π
et
{sin(n + 1)t − sin(n − 1)t} dt,
=
1
2π

2(n + 1) · (−1)n+2
sinh π
1 + (n + 1)2
−
2(n − 1)(−1)n
sinh π
1 + (n − 1)2

=
sinh π
π
(−1)n (n+1)(n2
−2n+2)−(n−1)(n2
+2n+2)
(n2 +2−2n)(n2 +2+2n)
=
sinh π
π
· (−1)n
·
4 − 2n2
n4 + 4
,
and
bn =
1
π
π
−π
et
sin t sin nt dt =
1
2π
π
−π
et
{cos(n − 1)t − cos(n + 1)t} dt
=
1
2π

2(−1)n−1
sinh π
1 + (n − 1)2
−
2(−1)n−1
sinh π
1 + (n + 1)2

=
sinh π
π
· (−1)n−1
·
n2
+2n+2−(n2
−2n+2)
(n2 −2n+2)(n2 +2n+2)
=
sinh π
π
· (−1)n−1
·
4n
n4 + 4
.
In both cases we see that a0 =
sinh π
π
, hence we get the Fourier series (with equality by the remarks
above)
f(t) =
sinh π
π

1
2
+
∞

n=1

(−1)n
(4 − 2n2
)
n4 + 4
cos nt +
(−1)n−1
4n
n4 + 4
sin nt

.
2) The Fourier series has the convergent majoring series
sinh π
π

1
2
+
∞

n=1
2n2
+ 4
n4 + 4
+
∞

n=1
4n
n4 + 4

(the diﬀerence of the degrees of the denominator and the numerator is ≥ 2), hence the Fourier
series is uniformly convergent.
3) By choosing t = π we get the pointwise result,
f(π) = 0 =
sinh π
π

1
2
− 2
∞

n=1
n2
− 2
n4 + 4

,

69
∞

n=1
n2
− 2
n4 + 4
=
1
4
.
Alternatively it follows by a decomposition,
n2
− 2
n4
=
n2
− 2
(n2 − 2n + 2)(n2 + 2n + 2)
=
1
2
n − 1
1 + (n − 1)2
−
1
2
n + 1
1 + (n + 1)2
,
so the sequential sequence is a telescoping sequence,
sN =
N

n=1
n2
− 2
n4 +4
=
1
2
N

n=1
n − 1
1 + (n − 1)2
−
N

n=1
n + 1
1 + (n + 1)2
=
1
2
N−1

n=0
(n=1)
n
1 + n2
−
1
2
N+1

n=2
n
1 + n2
=
1
2
·
1
1 + 12
−
1
2
·
N
1 + N2
−
1
2
·
N + 1
1 + (N + 1)2
→
1
4
for N → ∞,
and it follows by the definition that
∞

n=1
n2
− 2
n4 + 4
= lim
N→∞
sN =
1
4
.
Example 2.4 Find the Fourier series for the function f ∈ K2π, which is given in the interval ]−π, π]
by
f(t) =
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
0, for − π t −π/2,
cos t, for − π/2 ≤ t ≤ π/2,
0, for π/2 t ≤ π.
Prove that the series is absolutely and uniformly convergent in the interval R and find for t ∈ [−π, π]
the sum of the termwise integrated series from 0 to t. Then find the sum of the series
∞

n=0
(−1)n
(4n + 1)(4n + 2)(4n + 3)
2π. The Fourier
series is by the main theorem pointwise convergent with the sum function f∗
(t) = f(t).

70
0
1
–3 –2 –1 1 2 3
x
Since f(t) is even, all bn = 0. For n = 1 we get
an =
2
π
π/2
0
cos t · cos nt dt =
1
π
π/2
0
{cos(n+1)t+cos(n−1)t}dt
=
1
π

1
n+1
sin

(n+1)
π
2

+
1
n−1
sin

(n−1)
π
2

=
1
π

1
n+1
cos
nπ
2

−
1
n−1
cos
nπ
2

= −
2
π
1
n2 −1
cos
nπ
2

.
It follows that a2n+1 = 0 for n ∈ N and that
a2n = −
2
π
(−1)n
4n2 −1
, for n ∈ N0, in particular a0 =
2
π
for n = 0.
a1 =
2
π
π/2
0
cos2
t dt =
1
π
π/2
0
{cos 2t + 1} dt =
1
2
.
The Fourier series becomes with an equality sign according to the above,
f(t) =
1
π
+
1
2
cos t +
∞

n=1
(−1)n−1
4n2 − 1
cos 2nt.
1
π
+
1
2
+
2
π
∞

n=1
1
4n2 − 1
,
so it is absolutely and uniformly convergent. Therefore, we can integrate it termwise, and we get
t
0
f(τ) dτ =
t
π
+
1
2
sin t +
2
π
∞

n=1
(−1)n−1
(2n − 1)2n(2n + 1)
sin 2nt,
which also is equal to
t
0
f(τ) dτ =
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
−1, for − π t −
π
2
,
sin t, for −
π
2
≤ t ≤
π
2
,
1, for
π
2
t π.

71
By choosing t =
π
4
we get
π/4
0
f(τ) dτ =sin
π
4
=
√
2
2
=
1
4
+
√
2
4
+
2
π
∞

n=1
(−1)n−1
(2n−1)2n(2n+1)
sin
nπ
2

=
1
4
+
√
2
4
+
2
π
∞

n=0
(−1)n
(2n + 1)(2n + 2)(2n + 3)
sin

n
π
2
+
π
2

=
1
4
+
√
2
4
+
2
π
∞

p=0
(−1)2p
4p + 1)(4p + 2)(4p + 3)
sin

pπ +
π
2

=
1
4
+
√
2
4
+
2
π
∞

n=0
(−1)n
(4n + 1)(4n + 2)(4n + 3)
,
where we have a) changed index, n → n + 1, and b) noticed that we only get contributions for n = 2p
even.
Finally, we get by a rearrangement,
∞

n=0
(−1)n
(4n + 1)(4n + 2)(4n + 3)
=
π
2
!√
2
2
−
1
4
−
√
2
4

=
π(
√
2 − 1)
8
.
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Example 2.5 Find the Fourier series for the function f ∈ K2π, which is given in the interval ]−π, π[
by
f(t) = t(π2
− t2
).
Prove that the Fourier series is uniformly convergent in the interval R, and ﬁnd the sum of the series
∞

n=1
(−1)n+1
(2n − 1)3
.
The function f ∈ C∞
(]− π, π[) is without vertical half tangents, so f ∈ K∗
2π. Furthermore, f is odd,
so the Fourier series is a sine series, thus an = 0. The periodic continuation is continuous, so the
adjusted function f∗
(t) = f(t) is by the main theorem the pointwise sum function for the Fourier
series, and we can replace ∼ by an equality sign,
f(t) =
∞

n=1
bn sin nt, t ∈ R.
–10
–5
0
5
10
–4 –2 2 4
x

73
We obtain by some partial integrations,
bn =
2
π
π
0
(π2
− t3
) sin nt dt = −
2
πn

t(π2
− t2
) cos nt
π
0
+
2
πn
π
0
(π2
− 3t2
) cos nt dt
=
2
πn2

π2
− 3t2
sin nt
π
0
+
12
πn2
π
0
t sin nt dt =
12
πn3
[−t cos nt]π
0 +
12
πn3
π
0
cos nt dt
=
12π
πn3
· (−1)n+1
=
12
n3
· (−1)n+1
.
The Fourier series is then
f(t) = 12
∞

n=1
(−1)n+1
n3
sin nt.
12
∞

n=1
1
n3
,
so it is uniformly convergent in R.
If we put t =
π
2
, we get
f
π
2

=
π
2

π2
−
π2
4

=
3π2
8
= 12
∞

n=1
(−1)n+1
n3
sin n
π
n
= 12
∞

p=1
(−1)2p
(2p − 1)3
sin

pπ −
π
2

= 12
∞

n=1
(−1)n+1
(2n − 1)3
.
∞

n=1
(−1)n+1
(2n − 1)3
=
π3
32
.

74
Example 2.6 Let f ∈ K2π be given in the interval [−π, π] by
f(t) =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
sin 2t, for |t| ≤
π
2
,
0, for
π
2
|t| ≤ π.
1
2
sin 2t +
4
π
∞

n=0
(−1)n+1
(2n − 1)(2n + 3)
sin(2n + 1)t,
and prove that it is uniformly convergent in the interval R.
∞

n=0
(−1)n
(2n − 1)(2n + 1)(2n + 3)
.
2π. The Fourier
series is then by the main theorem pointwise convergent with sum f∗
(t) = f(t).
–1
–0.5
0
0.5
1
–3 –2 –1 1 2 3
x
1) Since f is odd, we have an = 0, and
bn =
2
π
π/2
0
sin 2t sin nt dt=
1
π
π/2
0
{cos(n−2)t−cos(n+2)t}dt.
For n = 2 we get in particular,
b2 =
1
π
π/2
0
(1 − cos 4t)dt =
1
π
·
π
2
− 0 =
1
2
.
For n ∈ N {2} we get
bn =
1
π
sin(n − 2)t
n − 2
−
sin(n + 2)t
n + 2
π/2
0
=
1
π
⎧
⎨
⎩
sin
n
2
− 1

π
n − 2
−
sin
n
2
+ 1

π
n + 2
⎫
⎬
⎭
.
In particular, b2n = 0 for n ≥ 2, and
b2n+1 =
1
π
⎧
⎪
⎪
⎨
⎪
⎪
⎩
sin

n −
1
2

π
2n − 1
−
sin

n + 2 −
1
2

π
2n + 3
⎫
⎪
⎪
⎬
⎪
⎪
⎭
=
(−1)n+1
π

1
2n − 1
−
1
2n + 3

=
4
π
· (−1)n+1
·
1
(2n − 1)(2n + 3)
for n ≥ 0.

75
The Fourier series is (with equality, cf. the above)
f(t) =
1
2
sin 2t +
4
π
∞

n=0
(−1)n+1
(2n − 1)(2n + 3)
sin(2n + 1)t.
4
π
∞

n=0
1
(2n − 1)(2n + 3)
,
so it is uniformly convergent.
2) By a comparison we see that we are missing a factor 2n+1 in the denominator. We can obtain this
by a termwise integration of the Fourier series, which is legal now due to the uniform convergence),
t
0
f(τ) dτ =
1
4
−
1
4
cos 2t +
4
π
∞

n=0
(−1)n
cos(2n + 1)t
(2n − 1)(2n + 1)(2n + 3)
−
4
π
∞

n=0
(−1)n
2n − 1)(2n + 1)(2n + 3)
.
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76
By choosing t =
π
2
, the ﬁrst series is 0, hence
π/2
0
f(τ) dτ =
π/2
0
sin 2τ dτ = −
1
2
cos 2τ
π/2
0
= 1
=
1
4
+
1
4
+ 0 −
4
π
∞

n=0
(−1)n
(2n − 1)(2n + 1)(2n + 3)
,
and by a rearrangement,
∞

n=0
(−1)n
(2n + 1)(2n + 1)(2n + 3)
=
π
4

1
2
− 1

= −
π
8
.
Alternatively we may apply the following method (only sketched here):
a) We get by a decomposition,
1
(2n−1)(2n+1)(2n+3)
=
1
8

1
2n−1
−
1
2n+1

−

1
2n + 3
−
1
2n+3

.
b) The segmental sequence becomes
sN =
N

n=0
(−1)n
(2n − 1)(2n + 1)(2n + 3)
= −
1
4
+
1
2
N

n=1
(−1)n
2n − 1
+
1
4
+
1
4
·
(−1)N+1
2N + 1
+
1
8
· (−1)N+1

1
2N + 1
−
1
2N + 3

.
c) Finally, by taking the limit,
∞

n=0
(−1)n
(2n − 1)(2n + 1)(2n + 3)
=
1
2
∞

n=1
(−1)n
2n − 1
−
1
2
Arctan 1 = −
π
8
.

77
Example 2.7 Given the periodic function f : R → R of period 2π, which is given in the interval
[−π, π] by
f(t) =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
| sin 2t|, 0 ≤ |t| ≤
π
2
,
0,
π
2
|t| ≤ π.
1) Prove that the Fourier series for f can be written
1
2
a0 +a1 cos t+
∞

n=1
{a4n−1 cos(4n−1)t+a4n cos 4nt+a4n+1 cos(4n+1)t},
and ﬁnd a0 and a1 and a4n−1, a4n and a4n+1, n ∈ N.
2) Prove that the Fourier series is uniformly convergent in R.
3) Find for t ∈ [−π, π] the sum of the series which is obtained by termwise integration from 0 to t of
the Fourier series.
2π. Then the
Fourier series is by the main theorem convergent with the sum function f∗
(t) = f(t).
0
1
–3 –2 –1 1 2 3
x
1) Now, f is even, so bn = 0, and
an =
2
π
π/2
0
sin 2t · cos nt dt=
1
π
π/2
0
{sin(n+2)t−sin(n−2)t}dt.
We get for n = 2,
an =
1
π
−
1
n+2
cos(n+2)t +
1
n−1
cos(n−2)t
π/2
0
=
1
π

−
1
n+2

cos
nπ
2
+π

−1

+
1
n−2

cos
nπ
2
−π

−1

=
1
π

1
n+2

1+cos
nπ
2

−
1
n−2

1+cos
nπ
2

=
1
π

1 + cos
nπ
2

·
−4
n2 − 4
.
For n = 2,
a2 =
1
π
π/2
0
sin 4t dt = −
1
4π
cos 4t
π/2
0
= 0.

78
Then
a4n+2 = −
4
(4n+2)2 −4
·
1
π
(1 + cos π) = 0 for n ∈ N,
and it follows that the Fourier series has the right structure.
Then by a calculation,
a0 =
1
π
(1+1) ·
−4
02 −4
=
2
π
, a1 =
1
π
· (1+0) ·
−4
12 −4
=
4
3π
,
a4n−1 =
1
π

1+cos

−
π
2

·
−4
(4n−1)2 −4
= −
4
π
·
1
(4n−1)2 −4
=−
4
π
·
1
(4n−3)(4n+1)
,
a4n =
1
π
(1+1) ·
−4
16n2 −4
=−
2
π
·
1
4n2 −1
=−
2
π
·
1
(2n−1)(2n+1)
,
a4n+1 = −
1
π

1+cos
π
2

·
−4
(4n+1)2 −4
= −
4
π
·
1
(4n+1)2 −4
=−
4
π
·
1
(4n−1)(4n+3)
.
The Fourier series is (with equality sign, cf. the above)
f(t) =
1
π
+
4
3π
cos t −
2
π
∞

n=1

2 cos(4n−1)t
(4n−3)(4n−1)
+
cos 4nt
(2n−1)(2n+1)
+
2 cos(4n+1)t
(4n−1)(4n+3)

.
2) Clearly, the Fourier series has a majoring series which is equivalent to the convergent series
c
∞
n=1
1
n2
. This implies that the Fourier series is absolutely and uniformly convergent.
3) The Fourier series being uniformly convergent, it can be termwise integrated for x ∈ [−π, π],
x
0
f(t) dt =
x
π
+
4
3π
sin t −
1
2π
∞

n=1

8 sin(4n − 1)t
(4n−3)(4n−1)(4n+1)
+
sin 4nt
(2n−1)n(2n+1)
+
8 sin(4n + 1)t
(4n−1)(4n+1)(4n+3)

,
where
x
0
f(t) dt =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
1 for x ∈
π
2
, π

,
1
2
(1 − cos 2x) for x ∈

0,
π
2

,
−
1
2
(1 − cos 2x) for x ∈

−
π
2
, 0

,
−1 for x ∈

−π, −
π
2

.

79
Example 2.8 Given the trigonometric series
∞

n=1
cos(nx)
n2(n2 + 1)
, x ∈ R.
1) Prove that it is pointwise convergent for every x ∈ R. The sum function of the series is denoted
by g(x), x ∈ R.
2) Prove that the trigonometric series
∞

n=1
cos(nx)
n2 + 1
is uniformly convergent in R.
3) Find an expression of g
(x) as a trigonometric series.
It is given that the function f, f ∈ K2π, given by
f(x) =
x2
4
−
πx
2
, 0 ≤ x ≤ 2π,
has the Fourier series
−
π2
6
+
∞

n=1
cos(nx)
n2
.
4) Prove that g is that solution of the diﬀerential equation
d2
y
dx2
− y = −f(x) −
π2
6
, 0 ≤ x ≤ 2π,
for which g
(0) = g
(π) = 0, and ﬁnd an expression as an elementary function of g for
0 ≤ x ≤ 2π.
5) Find the exact value of
∞

n=1
1
n2 + 1
.
1) Since
∞
n=1
cos(nx)
n2(n2 + 1)
has the convergent majoring series
∞

n=1
1
n2(n2 + 1)
,
the Fourier series is uniformly convergent and thus also pointwise convergent everywhere.

80
2) The series
∞
n=1
cos(nx)
n2 + 1
∞

n=1
1
n2 + 1
,
so it is also uniformly convergent.
3) Finally, the termwise diﬀerentiated series,
∞
n=1
− sin(nx)
n(n2 + 1)
∞

n=1
1
n(n2 + 1)
,
so it is also uniformly convergent. By another termwise diﬀerentiation we get from (2),
g
(x) = −
∞

n=1
cos(nx)
n2 + 1
.
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81
Intermezzo. We note that f is of class C∞
in ]0, 2π[ and without vertical half tangents and that
f(0) = f(2π) = 0.
Since the Fourier series for f by (2) is uniformly convergent, we have
f(x) = −
π2
6
+
∞

n=1
cos(nx)
n2
,
both pointwise and uniformly. The graph of f is shown on the figure.
–2.5
–2
–1.5
–1
–0.5
0
y
1 2 3 4 5 6
x
4) The trigonometric series of g, g
and g
are all uniformly convergent. When they are inserted into
the differential equation, we get
d2
y
dx2
− y = −
∞

n=1
cos nx
n2 + 1
−
∞

n=1
cos nx
n2(n2 + 1)
= −
∞

n=1
n2
+ 1
n2(n2 + 1)
cos nx
= = −
∞

n=1
1
n2
cos nx = −f(x) −
π2
6
,
and we have shown that they fulfil the differential equation.
Now,
g
(x) = −
∞

n=1
sin nx
n(n2 + 1)
,
so g
(0) = g
(π) = 0. It follows that g is a solution of the boundary value problem
(7)
d2
y
dx2
− y = −f(x) −
π2
6
= −
1
4
x2
+
π
2
x −
π2
6
,
with the boundary conditions y
(0) = y
(π) = 0 [notice, over half of the interval].

82
The corresponding homogeneous equation without the boundary conditions has the complete so-
lution,
y = c1 cosh x + c2 sinh x.
Then we guess a particular solution of the form
y = ax2
+ bx + c.
When this is put into the left hand side of the equation we get
d2
y
dx2
− y = 2a−ax2
−bx−c = −ax2
−bx+(2a−c).
This equal to
−f(x) −
π2
6
= −
1
4
x2
+
π
2
x −
π2
6
,
if a =
1
4
, b = −
π
2
, c − 2a = c −
1
2
=
π2
6
, thus c =
1
2
+
π2
6
.
The complete solution of (7) is
y =
1
4
x2
−
π
2
x +
1
2
+
π2
6
+ c1 cosh x + c2 sinh x.
Since
y
=
1
2
x −
π
2
+ c1 sinh x + c2 cosh x,
it follows from the boundary conditions that
y
(0) = −
π
2
+ c2 = 0, i.e. c2 =
π
2
,
and
y
(π) =
π
2
−
π
2
+ c1 sinh π + c2 cosh π = c1 sinh π +
π
2
cosh π = 0,
hence c1 = −
π
2
coth π and c2 =
π
2
.
We see that the solution of the boundary value problem is unique. Since g(x) is also a solution,
we have obtained two expressions for g(x), which must be equal,
(8) g(x) =
1
4
x2
−
π
2
x +
1
2
+
π2
6
−
π
2
coth π · cosh x +
π
2
sinh x =
∞

n=1
cos nx
n2(n2 + 1)
.
5) Put x = 0 into (8). Then we get by a decomposition
g(0) =
1
2
+
π2
6
−
π
2
coth π =
∞

n=1
1
n2(n2 + 1)
=
∞

n=1
1
n2
−
∞

n=1
1
n2 + 1
=
π2
6
−
∞

n=1
1
n2 + 1
,
so by a rearrangement,
∞

n=1
1
n2 + 1
=
π
2
coth π −
1
2
.

83
f(t) = |t|3
for − π t ≤ π.
2) Prove that
f(t) =
π3
4
+ 6π
∞

n=1

(−1)n
n2
+
2(1 − (−1)n
)
π2n4

cos nt, t ∈ R.
Hint: We may use without proof that
π
0
t3
cos nt dt = 3π2 (−1)n
n2
+ 6
1 − (−1)n
n4
for n ∈ N.
3) Prove that the Fourier series is uniformly convergent.
4) Apply the result of (2) to prove that
∞

p=1
1
(2p − 1)4
=
π4
96
.
Hint: Put t = π, and exploit that
∞
n=1
1
n2
=
π2
6
.
1) It follows from the graph that the function is continuous. It is clearly piecewise C1
without vertical
half tangents, so according to the main theorem the Fourier series for f is pointwise convergent
with sum function f(t), and we can even write = instead of ∼.
0
5
10
15
20
25
30
y
–6 –4 –2 2 4 6
x

84
2) Since f(t) is even, all bn = 0, and
an =
2
π
π
0
t3
cos nt dt, n ∈ N0.
For n = 0 (the exceptional case) we get
a0 =
2
π
π
0
t3
dt =
π4
2π
=
π3
2
.
When n 0, we either use the hint, or partial integration. For completeness, the latter is shown
below:
an =
2
π
π
0
t3
cos nt dt =
2
πn

t3
sin nt
π
0
− 3
π
0
t2
sin nt dt

=
6
πn2

t2
cos nt
π
0
− 2
π
0
t cos nt dt

=
6
πn2
π2
· (−1)n
−
12
πn3

[t sin nt]π
0 −
π
0
sin nt dt

=
6π2
n2
(−1)n
−
12
πn4
[cos nt]π
0 = 6π

(−1)n
n2
+
2(1 − (−1)n
)
π2n4

.
Then by insertion (remember 1
2 a0) and application of the equality sign in stead of ∼ we therefore
get
f(t) =
π3
4
+ 6π
∞

n=1

(−1)n
n2
+
2(1 − (−1)n
)
π2n4

cos nt, t ∈ R.

85
3) The Fourier series has clearly the majoring series
π3
4
+ 6π
∞

n=1

1
n2
+
4
π2n4

π3
4
+ 12π
∞

n=1
1
n2
.
This is convergent, so it follows that the Fourier series is uniformly convergent.
4) If we put t = π, then
f(π) = π3
=
π3
4
+ 6π
∞

n=1

(−1)n
n2
+
2(1 − (−1)n
)
π2n4

(−1)n
=
π3
4
+ 6π
∞

n=1
1
n2
+
12π
π2
∞

n=1
1 − (−1)n
n4
· (−1)n
=
π3
4
+ 6π ·
π2
6
+
12
π
∞

p=1
2
(2p − 1)4
(−1)2p−1
=
π3
4
+ π3
−
24
π
∞

p=1
1
(2p − 1)4
,
∞

p=1
1
(2p − 1)4
=
π
24

π3
4
+ π3
− π3

=
π4
96
.
Example 2.10 The periodic function f : R → R of period 2π is deﬁned by
f(t) =
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
1, t ∈ ]− π, −π/2],
0, t ∈ ]− π/2, π/2[,
1, t ∈ [π/2, π].
Sketch the graph of f. Prove that f has the Fourier series
f ∼
1
2
+
2
π
∞

n=1
(−1)n
2n − 1
cos(2n − 1)t.
Find the sum of the Fourier series and check if the Fourier series is uniformly convergent.
The function is even and piecewise C1
2π, and the Fourier
series is a cosine series, bn = 0. According to the main theorem we then have pointwise,
f∗
(t) =
1
2
a0 +
∞

n=1
an cos nt,

86
here the adjusted function f∗
(t) is given by
f∗
(t) =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
1, for t ∈ ]− π, −π/2[,
1/2, for t = −π/2,
0, for t ∈]− π/2, π/2[,
1/2, for t = π/2,
1, for t ∈ ]π/2, π],
continued periodically, cf. ﬁgure.
0.4
0.8
y
–6 –4 –2 2 4 6
x
The Fourier coeﬃcients are
an =
2
π
π
0
f(t) cos nt dt =
2
π
π
π/2
cos nt dt.
We get for n = 0,
a0 =
2
π
π
π/2
1 dt = 1, thus
1
2
a0 =
1
2
.
Then for n ∈ N,
an =
2
π
π
π/2
cos nt dt =
2
πn
[sin nt]π
π/2 = −
2
πn
sin n
π
2
.
If we split into the cases of n even or odd, we get
a2p = −
2
π · 2p
· sin pπ = 0,

87
a2p−1 = −
2
π(2p − 1)
· sin(2p − 1)
π
2
=
2
π
·
(−1)p
2p − 1
.
We get by insertion the given Fourier series (with equality sign for the adjusted function)
f∗
(t) =
1
2
+
2
π
∞

p=1
(−1)p
2p − 1
cos(2p − 1)t.
Since all terms cos(2p − 1)t are continuous, and f∗
(t) [or f(t) itself] is not, the convergence cannot be
uniform.
f(t) = e|t|
1) Sketch the graph of f and explain why f ∈ K∗
2π.
1
π
(eπ
− 1) −
2
π
∞

n=1
1 − eπ
(−1)n
n2 + 1
cos nt.
3) Prove that the Fourier series is uniformly convergent.
1) Since f is piecewise C∞
2π. Now, f is continuous,
cf. the ﬁgure, so it follows by the main theorem that f(t) is pointwise equal its Fourier series.
Since f is an even function, the Fourier series is a cosine series, thus
(9) f(t) =
1
2
a0 +
∞

n=1
an cos nt,
cf. the ﬁgure.
2) Then we get by successive partial integrations,
an =
2
π
π
0
et
cos nt dt =
2
π

et
cos nt
π
0
+
2n
π
π
0
et
sin nt dt
=
2
π
{(−1)n
eπ
− 1} +
2n
π

et
sin nt
π
0
−
2n2
π
π
0
et
cos nt dt
=
2
π
{(−1)n
eπ
− 1} + 0 − n2
an,
an =
2
π
·
(−1)n
eπ
− 1
n2 + 1
, n ∈ N0, specielt a0 =
2
π
{eπ
− 1}.

88
5
10
15
20
y
–4 –2 0 2 4
x
Alternatively it follows directly by complex calculations with cos nt = Re eint
that
an =
2
π
π
0
et
cos nt dt =
2
π
Re
π
0
e(1+in)t
dt =
2
π
Re
1
1 + in
e(1+in)t
π
0
=
2
π
·
1
1 + n2
Re[(1 − in){eπ
(−1)n
− 1}] =
2
π
·
(−1)n
eπ
− 1
n2 + 1
.
Then by insertion into (9) we get (pointwise equality by (1)) that
f(t) =
eπ
− 1
π
−
2
π
∞

n=1
1 − eπ
(−1)n
n2 + 1
cos nt.
3) The Fourier series has the convergent majoring series
eπ
− 1
π
+
2
π
(eπ
+ 1)
∞

n=1
1
n2 + 1

eπ
− 1
π
+
2
π
(eπ
+ 1)
∞

n=1
1
n2
∞,
so the Fourier series is uniformly convergent.

89
f(t) = (t − π)2
2) Prove that the Fourier series is convergent for every t ∈ R, and sketch the graph of the sum
function in the interval [−π, π].
3) Explain why the Fourier series is not uniformly convergent.
4π2
3
+ 4
∞

n=1

(−1)n
n2
cos nt +
(−1)n
π
n
sin nt

, t ∈ R.
(1) and (2) Since f is piecewise C1
2π. Then by the
main theorem the Fourier series is pointwise convergent with the adjusted function f∗
(t) as its
sum function, where
f∗
(t) =
⎧
⎨
⎩
f(t) for t = (2p + 1)π, p ∈ Z,
2π2
for t = (2p + 1)π, p ∈ Z.
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90
0
10
20
30
40
y
–4 2 4
x
(3) Since f∗
(t) is not continuous, the Fourier series cannot be uniformly convergent. In fact, if it was
uniformly convergent, then the sum function should also be continuous, which it is not.
(4) We only miss the derivation of the Fourier series itself. For n 0 we get by partial integration,
an =
1
π
π
−π
(t − π)2
cos nt dt =
1
π
1
n
sin nt · (t − π)2
π
−π
−
2
πn
π
−π
(t − π) sin nt dt
= 0 +
2
πn
1
n
cos nt · (t − π)
π
−π
−
2
πn2
π
−π
cos nt dt
=
2
πn2
{0 − (−1)n
· (−2π)} − 0 =
4
n2
(−1)n
,
and for n = 0
a0 =
1
π
π
−π
(t − π)2
dt =
1
π
(t − π)3
3
π
−π
=
1
π
·
8π3
3
=
8π2
3
,
and for n ∈ N,
bn =
1
π
π
−π
(t − π)2
sin nt dt =
1
π
−
1
π
cos nt · (t − π)2
π
−π
+
2
πn
π
−π
(t − π) cos nt dt
=
1
πn
· (−1)n
· 4π2
+
2
πn
[sin nt · (t − π)]π
−π −
2
πn
π
−π
sin nt dt =
4π
n
· (−1)n
.

91
Summing up we get with pointwise equality,
f∗
(t) =
4π2
3
+ 4
∞

n=1

(−1)n
n2
cos nt +
(−1)n
π
n
sin nt

, t ∈ R.
f(t) =
⎧
⎨
⎩
sin t, t ∈ ]− π, 0],
cos t, t ∈ ]0, π].
It is given that f has the Fourier series
f ∼ −
1
π
+
cos t + sin t
2
+
2
π
∞

n=1
cos 2nt + 2n sin 2nt
4n2 − 1
.
1) Sketch the graph of for f.
2) Prove that the coefficients of cos nt, n ∈ N0, in the Fourier series for f are as given above.
3) Find the sum function of the Fourier series and check if the Fourier series is uniformly convergent.
Let
f+
(t) =
f(t) + f(−t)
2
, f−
(t) =
f(t) − f(−t)
2
be the even and the odd part of f, respectively.
4) Find the Fourier series for f+
, and check if it is uniformly convergent.
5) Find the Fourier series for f−
, and check if it is uniformly convergent.
1) We note that f is piecewise differentiable without vertical half tangents. Then by the main
theorem the Fourier series is pointwise convergent with the adjusted function ˜
f as its sum function.
2) The coefficients an are defined by
an =
1
π
π
−π
f(t) cos nt dt =
1
π
0
−π
sin t cos nt dt +
1
π
π
0
cos t cos nt dt
=
1
2π
0
−π
{sin(n + 1)t − sin(n − 1)t}dt +
1
2π
π
0
{cos(n + 1)t + cos(n − 1)t}dt.
In order to divide unawarely by 0, we immediately calculate separately the case n = 1:
a1 =
1
2π
0
−π
sin 2t dt +
1
2π
π
0
{cos 2t + 1}dt = 0 +
1
2π
{0 + π} =
1
2
.

92
–1
–0.5
0.5
1
y
–3 –2 –1 1 2 3
x
Then we get for n ≥ 0, n = 1,
an =
1
2π
−
cos(n + 1)t
n + 1
+
cos(n − 1)t
n − 1
0
−π
+
1
2π
sin(n + 1)t
n + 1
+
sin(n − 1)t
n − 1
π
0
=
1
2π

−
1
n + 1
[1 − (−1)n+1
] +
1
n − 1
[1 − (−1)n−1
]

=
1
2π
·
2
n2 − 1
{1 + (−1)n
} =
1
π
·
1
n2 − 1
{1 + (−1)n
}.
It follows immediately, that if n = 2p + 1, p ∈ N, is odd and 1, then
a2p+1 = 0.
When n is replaced by 2n, we get
a2n =
1
π
·
2
4n2 − 1
.
In particular we ﬁnd for n = 0 that
1
2
a0 = −
1
π
.
Summing up we get
1
2
a0 = −
1
π
, a1 =
1
2
, a2n =
1
π
·
1
4n2 − 1
, a2n+1 = 0, for n ∈ N,
in agreement with the given Fourier series.
3) According to (1) we have pointwise convergence with the sum
˜
f(t) =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎩
sin t, t ∈ ]− π, 0[,
1/2, t = 0,
cos t, t ∈ ]0, π[,
−1/2, t = π,

93
which is continued periodically.
Since f (and ˜
f) is not continuous, the Fourier series for f cannot be uniformly convergent.
4) The Fourier series for f+
is the even part of the Fourier series for f, thus
f+
∼ −
1
π
+
1
2
cos t +
2
π
∞

n=1
1
4n2 − 1
cos 2nt.
This is clearly uniformly convergent, because it has the convergent majoring series
1
π
+
1
2
+
2
π
∞

n=1
1
4n2 − 1
≤ 1 +
2
π
∞

n=1
1
n2
= 1 +
π
3
.
Remark 2.2 It follows that
˜
f+
(t) =
⎧
⎪
⎪
⎨
⎪
⎪
⎩
{sin t + cos t}/2 for t ∈ ]− π, 0[,
1/2 for t = 0,
{− sin t + cos t}/2 for t ∈ ]0, π[,
−1/2 for t = π,
hence the continuation of f+
is continuous and piecewise C1
without vertical half tangents.
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94
5) The Fourier series for f−
is the odd part of the Fourier series for f, thus
f−
∼
1
2
+
4
π
∞

n=1
n
4n2 − 1
sin 2nt.
If this series was uniformly convergent, then f−
should be continuous, and hence also f = f+
+f−
continuous. ¡vspace3mm
However, f is not continuous, so the Fourier series for f−
is not uniformly convergent.
f(t) =
1
4π2
t(4π − t), t ∈ [0, 2π[.
1) Sketch the graph of f in the interval [−2π, 2π[.
2) Explain why the Fourier series is pointwise convergent for every t ∈ R, and sketch the graph of the
sum function in the interval [−2π, 2π[.
3) Show that the Fourier series is not uniformly convergent.
2
3
−
1
π2
∞

n=1

1
n2
cos nt +
π
n
sin nt

, t ∈ R.
Hint: One may use without proof that

t(4π−t) cos nt dt =
2
n2
(2π−t) cos nt +

2
n3
+
t(4π−t)
n

sin nt,
and

t(4π−t) sin nt dt =
2
n2
(2π−t) sin nt −

2
n3
+
t(4π−t)
n

cos nt,
for n ∈ N.
(1) and (2) It follows from the rearrangement,
f(t) =
1
4π2
{4π2
− (t − 2π)2
}
that the graph of f(t) in [0, 2π[ is a part of an arc of a parabola with its vertex at (2π, 1). The
normalized function f∗
(t) is equal to
1
2
for t = 2pπ, p ∈ Z, and = f(t) at any other point. Since
f(t) is of class C∞
in ]0, 2π[ and without vertical half tangents, the Fourier series is by the main
theorem pointwise convergent with f∗
(t) as its sum function.
(3) Since f∗
(t) is not continuous, it follows that the Fourier series cannot be uniformly convergent.

95
0
0.4
0.8
y
–6 –4 –2 2 4 6
x
(4) We are now only missing the Fourier coeﬃcients. We calculate there here without using the hints
above:
a0 =
1
π
2π
0
1
4π2
(4πt−t2
)dt =
1
4π3
2πt2
−
1
3
t3
2π
0
= 2 −
1
3
· 2 =
4
3
.
We get for n ∈ N,
an =

1
π
2π
0
1
4π2
(4πt−t2
) cos nt dt
=
1
4π3
1
n
t(4π−t) sin nt
2π
0
−
1
4π3n
2π
0
(4π−2t) sin nt dt
=
2
4π3n2
[(2π−t) cos nt]2π
0 +
1
2π2n2
2π
0
cos nt dt = −
2π
2π3n2
= −
1
π2
·
1
n2
,
and
bn =
1
π
2π
0
1
4π2
(4πt−t2
) sin nt dt
=
1
4π3
−
1
n
t(4π−t) cos nt
2π
0
+
1
2π3n
2π
0
(2π−t) cos nt dt
=
1
4π3n
(−2π · 2π) +
1
2π3n2
[(2π−t) sin nt]2π
0 +
1
2π3n2
2π
0
sin nt dt = −
π
π2n
.
Hence we get the Fourier series with its sum function f∗
(t),
f∗
(t) =
1
2
a0 +
∞

n=1
{an cos nt + bn sin nt}
=
2
3
−
1
π2
∞

n=1

1
n2
cos nt +
π
n
sin nt

, t ∈ R.

96
Example 2.15 We define for every fixed r, 0 r 1, the function fr : R → R, by
fr(t) = ln(1 + r2
− 2r cos t), t ∈ R.
1) Explain why fr ∈ K∗
2π.
Prove that fr has the Fourier series
(10) −2
∞

n=1
r2
n
cos(nt).
2) Prove that the Fourier series (10) is uniformly convergent for t ∈ R, and find its sum function.
3) Calculate the value of each of the integrals
2π
0
fr(t) dt,
2π
0
fr(t) cos(5t) dt,
π
−π
fr(t) sin(5t) dt.
4) Find the sum of each of the series
∞

n=1
1
2n · n
og
∞

n=1
(−1)n
3n · n
.
5) Prove that the series which is obtained by termwise differentiation (with respect to t) of (10), is
uniformly convergent for t ∈ R, and find the sum of the differentiated series in −π ≤ t ≤ π.
1) Clearly, fr(t) is defined and C∞
in t, when 0 r 1, because we have
1 + r2
− 2r cos t ≥ 1 + r2
− 2r = (1 − r)2
0.
Since it is also periodic of period 2π, it follows that fr ∈ K∗
2π. Then by the main theorem the
Fourier series for each fr(t), 0 r 1, is pointwise convergent with fr(t) as its sum function.
Then we prove that the Fourier series becomes
fr(t) = ln(1 + r2
− 2r cos t) = −2
∞

n=1
rn
n
cos(nt), 0 r 1,
where we have earlier noted that the equality sign is valid.
First note that the quotient series expansion
1
1 − z
=
∞

n=0
zn
, for |z| 1,
also holds for complex z ∈ C, if only |z| 1.
Then put z = reit
, thus |z| = r ∈ ]0, 1[, and we get by Moivre’s formula that
zn
= rn
eint
= rn
{cos nt + i sin nt}.

97
Then we get for 0 r 1 by insertion into the quotient series that
1
1 − z
=
1
1 − reit
=
∞

n=0
zn
=
∞

n=0
rn
{cos nt + i sin nt}.
Here we take two times the imaginary part,
2
∞

n=1
rn
sin nt = 2 Im

1
1 − reit

= 2 Im

1
1 − reit
·
1 − re−it
1 − re−it

=
2r sin t
1 + r2 − r(eit + e−it)
=
2r sin t
1 + r2 − 2r cos t
.
The series has the convergent majoring series
2
∞

n=1
rn
=
2r
1 − r
∞,
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98
so it is uniformly convergent. We may therefore perform termwise integration,
−2
∞

n=1
rn
n
cos nt = ln(1 + r2
− 2r cos t) + c,
where the constant c is fixed by putting t = 0 and then apply the logarithmic series,
ln(1 + r2
− 2r) + c = ln{(1 − r)2
} + c = 2 ln(1 − r) + c
= −2
∞

n=1
rn
n
= 2
∞

n=1
(−1)n+1
n
(−r)n
= 2 ln(1 − r).
We get c = 0, and we have proved that we have uniformly that
ln(1 + r2
− 2r cos t) = −2
∞

n=1
rn
n
cos nt.
This intermezzo contains latently (2) and (5); but we shall not use this fact here.
2) If 0 r 1 is kept fixed, then we have the trivial estimate
−2
∞

n=1
rn
n
cos nt ≤ 2
∞

n=1
rn
n
= 2 ln
1
1 − r
∞.
The Fourier series has a convergent majoring series, so it is uniformly convergent.
3) This question may be answered in many different ways.
a) First variant. It follows from the definition of a Fourier series that
fr(t) ∼ −2
∞

n=1
rn
n
cos nt =
1
2
a0 +
∞

n=1
{an cos nt + bn sin nt},
where
an =
1
π
2π
0
fr(t) cos nt dt,
bn =
1
π
2π
0
fr(t) sin nt dt =
1
π
π
−π
fr(t) sin nt dt.
Then by identification,
2π
0
fr(t) dt = πa0 = 0,
2π
0
fr(t) cos(5t)dt = πa5 = −
2πr5
5
,
π
−π
fr(t) sin(5t)dt = πb5 = 0.

99
b) Second variant. Since the series expansion
fr(t) = −2
∞

n=1
rn
n
cos(nt), 0 r 1,
is uniformly convergent, it follows by interchanging the summation and integration that
2π
0
fr(t) dt = −2
∞

n=1
rn
n
2π
0
cos nt dt = 0,
2π
0
fr(t) cos(5t) dt = −2
∞

n=1
rn
n
2π
0
cos nt · cos(5t)dt
= −2 ·
r5
5
2π
0
cos2
(5t)dt = −
2πr5
5
,
π
−π
fr(t) sin(5t)dt = −2
∞

n=1
rn
n
π
−π
cos nt · sin 5t dt = 0.
Remark 2.3 A direct integration of e.g.
2π
0
fr(t) cos(5t)dt =
2π
0
ln(1+r2
−2r cos t) · cos(5t)dt
does not look promising and my pocket calculator does not either like this integral.
4) Here, we also have two variants.
a) First variant. Since
∞

n=1
rn
n
cos nt = −
1
2
fr(t) = −
1
2
ln(1 + r2
− 2r cos t),
we get by choosing r =
1
2
and t = 0 that
∞

n=1
1
n2n
= −
1
2
f1/2(0) = −
1
2
ln

1+
1
4
−2 ·
1
2

= −
1
2
ln

1
4

= ln 2.
If we instead choose r =
1
3
and t = π, we get
∞

n=1
(−1)n
n3n
= −
1
2
f1/3(π) = −
1
2
ln

1+
1
9
+
2
3

= −
1
2
ln
16
9
= ln
3
4
.
b) Second variant. If we instead use the series expansion
∞

n=1
(−1)n+1
rn
n
= ln(1 + r), r ∈ ]− 1, 1[,

100
we obtain for r = −
1
2
that
−
∞

n=1
1
n2n
= ln

1−
1
2

= − ln 2, dvs.
∞

n=1
1
n2n
= ln 2.
Then for r =
1
3
,
−
∞

n=1
(−1)n
n3n
= ln

1+
1
3

= ln
4
3
, dvs.
∞

n=1
(−1)n
n3n
= ln
3
4
.
5) When we perform termwise diﬀerentiation of the Fourier series. we get
2
∞

n=1
rn
sin nt.
If 0 r 1, then 2
∞
n=1 rn
=
2r
1 − r
∞ is a convergent majoring series. Consequently, the
diﬀerentiated series is uniformly convergent with the sum function f
r(t), thus
2
∞

n=1
rn
sin nt =
d
dt
ln(1+r2
−2r cos t) =
2t sin t
1+r2 −2r cos t
.
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101
3 Parseval’s equation
Example 3.1 A function f ∈ K2π is given in the interval ]− π, π] by
f(t) =
⎧
⎪
⎨
⎪
⎩
2π
3
− |t|, for |t| ≤
2π
3
,
0 otherwise.
1) Sketch the graph of f in the interval ]π, π].
Prove that f has the Fourier series
2π
9
+
∞

n=1
2
πn2

1 − cos

n
2π
3

cos nt, t ∈ R.
2) Given that
π
π
f(t)2
dt =
16
81
π3
,
ﬁnd the sum of the series
1
14
+
1
24
+
1
44
+
1
54
+
1
74
+
1
84
+
1
104
+
1
114
+ · · · .
0
0.5
1
1.5
2
–3 –2 –1 1 2 3
x
1) Since f is continuous and piecewise C1
2π. The
Fourier series is then by the main theorem pointwise convergent and its sum function is f∗
(t) =
f(t), because f(t) is continuous. Now, f(t) is even, so bn = 0, and
a0 =
2
π
2π/3
0

2π
3
−t

dt=−
1
π
#
2π
3
−t
2
$2π/3
0
=
1
π

2π
3
2
=
4π
9
.
We get for n 1,
an =
2
π
2π/3
0

2π
3
− t

cos nt dt =
2
π
1
n

2π
3
− t

sin nt
2π/3
0
+
2
πn
2π/3
0
sin nt dt
=
2
πn2
[− cos nt]
2π/3
0 =
2
πn2

1 − cos

2π
3
n

.
The Fourier series is then with an equality sign, cf. the above,
f(t) =
2π
9
+
∞

n=1
2
πn2

1 − cos

n
2π
3

cos nt.
Parseval’s equation

102
2) By Parseval’s equation we get
(11)
1
π
π
−π
f(t)2
dt =
1
2
a2
0 +
∞

n=1
a2
n =
1
2

4π
9
2
+
4
π2
∞

n=1

1 − cos

2πn
3
2
·
1
n4
.
Here
π
−π
f(t)2
dt = 2
2π/3
0

2π
3
− t
2
dt = −
2
3
#
2π
3
− t
3
$2π/3
0
=
2
3

2π
3
3
=
16π3
81
,
and

1−cos

2πn
3
2
=

2 sin3
πn
n
2
=4 sin4

n
π
3

=
⎧
⎪
⎨
⎪
⎩
4
!
±
√
3
2
4
=
9
4
n = 3p,
0 n = 3p.
Then by insertion into (11),
16π2
81
=
8π2
81
+
4
π2
·
9
4
∞

n=1
1
n4
−
∞

n=1
1
(3n)4

,
1
14
+
1
24
+
1
44
+
1
54
+
1
74
+
1
84
+ · · ·
=
∞

n=1
1
n4
−
∞

n1
1
(3n)4
=

1 −
1
81
∞

n=1
1
n4
=
80
81
∞

n=1
1
n4
=
π2
9
·
8π2
81
=
8π4
729
.
Note that it follows from the above that
∞

n=1
1
n4
=
81
80
·
8π4
729
=
π4
90
.

103
Example 3.2 A periodic function f : R → R of period 2π is given in the interval ]− π, π] by
f(t) =

π|t| − t2
2
, t ∈ ]− π, π].
π4
30
−
∞

n=1
3
n4
cos 2nt.
∞
n=1
1
n8
.
3) Prove that the series which is obtained by termwise diﬀerentiation of the Fourier series above is
uniformly convergent in R. Find the sum of the termwise diﬀerentiated series for t ∈ ]− π, π].
2π. The
Fourier series is by the main theorem pointwise convergent and its sum function is f∗
(t) = f(t).
0
1
2
3
4
5
6
–3 –2 –1 1 2 3
x
1) Since π|t| − t2
is even, f(t) is also even, and the series is a cosine series, hence bn = 0, and
a0 =
2
π
π
0
f(t) dt =
2
π
π
0
πt − t2 2
dt
=
2
π
π
0
t4
− 2πt3
+ π2
t2
dt =
2
π

π5
5
−
1
2
π5
+
1
3
π5

=
2π4
30
(6 − 15 + 10) =
π4
15
.
If n ∈ N, we get instead
an =
2
π
π
0
(t4
− 2πt3
+ π2
t2
) cos nt dt
=
2
πn

t2
−πt
2
sin nt
π
0
−
2
πn
π
0
(4t3
−6πt2
+2π2
t) sin nt dt

104
so
an = 0+
4
πn2

2t3
−3πt2
+π2
t2 π
0
−
4
πn2
π
0
(6t2
−6πt+π2
) cos nt, dt
= 0−
4
πn3

(6t2
−6πt+π2
) sin nt
π
0
+
24
n3π
π
0
(2t−π) sin nt dt
= 0 −
24
πn4
[(2t − π) cos nt]π
0 +
48
πn4
π
0
cos nt dt
= −
24
πn4
{π(−1)n
+ π} = −
24
n4
{1 + (−1)n
} .
Thus we get a2n+1 = 0 for n ≥ 0, and
a2n = −
24
(2n)4
· 2 = −
3
n4
for n ∈ N.
Summing up the Fourier series is with equality sign (by the beginning of the example),
f(t) =

π|t| − t2
2
=
π4
30
−
∞

n=1
3
n4
cos 2nt, t ∈ [−π, π].
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105
2) We get by Parseval’s equation that
1
2
a2
0 +
∞

n=1
(a2
n + b2
n) =
1
π
π
−π
f(t)2
dt.
In the present case,
1
2
a2
0 +
∞

n=1
(a2
n + b2
n) =
1
2

π4
15
2
+ 9
∞

n=1
1
n8
=
π8
450
+ 9
∞

n=1
1
n8
,
and
1
π
π
−π
f(t)2
dt =
2
π
π
0
t4
(π − t)4
dt =
2
π
t5
5
(π − t)4
π
0
+
2
π
·
4
5
π
0
t5
(π − t)3
dt
= 0 +
2
π
·
4
5
1
6
t6
(π−t)3
π
0
+
2
π
·
4
5
·
3
6
π
0
t6
(π−t)2
dt
= 0 +
2
π
·
4
5
·
3
6
1
7
t7
(π−t)2
π
0
+
2
π
·
4
5
·
3
6
·
2
7
π
0
t7
(π−t)dt
= 0 +
2
π
·
4
5
·
3
6
·
2
7
1
8
t8
(π−t)
π
0
+
2
π
·
4
5
·
3
6
·
2
7
·
1
8
π
0
t8
dt
=
2
π
·
4
5
·
3
6
·
2
7
·
1
8
·
π9
9
=
π8
5 · 7 · 9
=
π8
315
.
By insertion into Parseval’s equation we get
π8
450
+ 9
∞

n=1
1
n8
=
π8
315
,
∞

n=1
1
n8
=
π8
9

1
315
−
1
450

=
135
315 · 450
·
π8
9
=
9 · 15
3 · 105 · 3 · 150
·
π8
9
=
π8
9450
.
3) The termwise diﬀerentiated series
6
∞

n=1
1
n3
sin 2nt
∞
n=1
6
n3
, hence the series is uniformly convergent.
Its sum is for t ∈ ]0, π[ given by
f
(t) =
d
dt
πt − t2 2
=
d
dt
t4
− 2πt3
+ π2
t2
= 4t3
− 6πt2
+ 2π2
t.
Analogously, we get for t ∈ ]− π, 0[,
f
(t) = 4t3
+ 6πt2
+ 2πt.

106
Then by a continuous continuation, f
(π) = f
(−π) = 0 and f
(0) = 0, so
6
∞

n=1
1
n3
sin 2nt =
⎧
⎨
⎩
4t3
+ 6πt2
+ 2π2
t for t ∈ [−π, 0],
4t3
− 6πt2
+ 2π2
t for t ∈ [0, π].
Example 3.3 Find the Fourier series for the function f ∈ K8, which is given in the interval ]− 4, 4]
by
f(t) =
⎧
⎨
⎩
−t for − 4 t ≤ 0,
t for 0 t ≤ 4.
Then apply Parseval’s equation in order to ﬁnd the sum of the series
∞

n=1
1
(2n − 1)4
.
1
2
3
4
–6 –4 –2 2 4 6
x
As in Example 1.7 we see that f ∈ K∗
8 . The function f is continuous and even, so it follows from the
main theorem that the symbol ∼ can be replaced by an equality sign. Furthermore, the series is a
cosine series,
f(t) =
1
2
a0 +
∞

n=1
an cos
nπt
4
,
where we have for n ∈ N,
an =
4
8
4
0
t cos

2πn
8
t

dt =
1
2
4
0
t cos

nπt
4

dt =
1
2
4
nπ
t · sin

nπt
4
4
0
−
2
nπ
4
0
sin

nπt
t

dt
= 0 +
8
n2π2
cos

nπt
4
4
0
=
8
n2π2
{(−1)n
− 1} .
For n = 0 we get instead
a0 =
1
2
4
0
t dt =
1
2
t2
2
4
0
= 4.
Since
(−1)n
− 1 =
⎧
⎨
⎩
0 for n even,
−2 for n odd,

107
we get a2n = 0 for n ∈ N (however, a0 = 4 for n = 0), and
a2n−1 = −
16
π2
·
1
(2n − 1)2
, for n ∈ N.
Summing up we get the Fourier series with equality sign instead of the symbol ∼)
f(t) = 2 −
16
π2
∞

n=1
1
(2n − 1)2
cos(2n − 1)
π
4
t.
Whenever Parseval’s equation is applied, it is always a good strategy ﬁrst to identify all coeﬃcients.
In particular, we must be very careful with a0, because it due to the factor
1
2
plays a special role:
a0 = 4, a2n−1 = −
16
π2
·
1
(2n − 1)2
, a2n = 0, bn = 0 for n ∈ N.
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108
Then by Parseval’s equation,
1
2
a2
0 +
∞

n=1
{a2
n +b2
n}=
2
8
4
−4
t2
dt=
1
2
4
0
t2
dt=
1
2
t3
3
4
0
=
32
3
,
hence
32
3
=
1
2
a2
0 +
∞

n=1
a2
2n−1 =
16
2
+
162
π4
∞

n=1
1
(2n − 1)4
,
and we get by a rearrangement,
∞

n=1
1
(2n − 1)4
=
π4
162

32
3
−
16
2

=
π4
16

2
3
−
1
2

=
π4
16
·
1
6
=
π4
96
.
Example 3.4 A periodic function f : R → R of period 2π is deﬁned by
f(t) =
⎧
⎨
⎩
cos 2qt, for t ∈ [0, π],
0, for t ∈ ]− π, 0[,
where q ∈ N is a constant.
1) Find the Fourier series of the function.
2) Use the Fourier series to ﬁnd the sum of the series
∞

p=1
(2p − 1)2
[(2p − 1)2 − 4q2]
2
er
1
16
π2
.
2π. The Fourier series is then
by the main theorem convergent, and its sum function is
f∗
(t) =
⎧
⎪
⎨
⎪
⎩
1
2
for t = pπ, p ∈ Z,
f(t) ellers.
–1
0
1
–4 –2 2 4
x

109
1) Now,
an =
1
π
π
0
cos 2qt cos nt dt=
1
2π
π
0
{cos(2q+n)t+cos(2q−n)t}dt,
so an = 0 for n = 2q, and
a2q =
1
2π
π
0
{cos 4qt + 1}dt =
1
2
.
Furthermore,
bn =
1
π
2π
0
cos 2qt sin nt dt=
1
2π
π
0
{sin(n+2q)t+sin(n−2q)t}dt.
If n = 2q, then
bn =
1
2π
−
cos(2q + n)t
2q + n
−
cos(n − 2q)t
n − 2q
π
0
=
1
2π

−
(−1)2q+n
2q + n
−
(−1)n−2q
n − 2q
+
1
2q + n
+
1
n − 2q

=
1
2π

1
n + 2q
+
1
n − 2q

{1 − (−1)n
} .
If 2n = 2q, we immediately get a2n = 0.
If 2n = 2q, then
b2q =
1
2π
π
0
sin 4qt dt =
1
2π
−
1
4q
cos 4qt
π
0
= 0,
hence b2n = 0 for every n ∈ N.
Finally,
b2n−1 =
1
π

1
(2n−1)+2q
+
1
(2n−1)−2q

=
2
π
·
2n − 1
(2n−1)2 −4q2
,
and the Fourier series becomes with an equality sign, cf. the beginning,
f∗
(t) =
1
2
cos 2qt +
2
π
∞

n=1
2n − 1
(2n−1)2 −4q2
sin(2n−1)t.
2) Now, 2q = 0, so applying Parseval’s equation,
1
4
+
4
π2
∞

n=1
(2n − 1)2
{(2n−1)2 −4q2}
2 =
2
2π
π
0
cos2
2qt dt =
1
2π
· π =
1
2
,
∞

n=1
(2n − 1)2
{(2n − 1)2 − 4q2}
2 =
π2
4

1
2
−
1
4

=
π2
16
.
Alternatively, the latter sum can also be found in the following traditional way. It is, however,
rather diﬃcult, so I shall only sketch the solution in the following.

110
a) First by a decomposition,
(2n − 1)2
{(2n−1)2 −4q2}
2 =

2n − 1
(2n−1)2 −4q2
2
=

1
2

1
2n−2q−1
+
1
2n+2q−1
2
=
1
4

1
(2b−2q−1)2
+
1
(2n+2q−1)2
+
1
2q

1
2n−2q− 1
−
1
2n+2q− 1

.
b) Then we obtain after a very long calculation that
s
n =
N

n=1

1
2n − 2q − 1
−
1
2n + 2q − 1

= −
N+q

p=N−q+1
1
2p − 1
→ 0 for N → ∞,
because
N+q

p=N−q+1
1
2p − 1
≤ 2q ·
1
2N − 2q + 1
→ 0 for N → ∞.
(We have 2q terms which are all smaller than or equal to the ﬁrst term).
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111
c) Then it follows from (a) and (b) that
∞

n=1
(2n − 1)2
{(2n−1)2 −4q2}
2 =
1
4
∞

n=1
1
(2n−2q−1)2
+
1
4
∞

n=1
1
(2n+2q+1)2
=
1
4
∞

p=−q+1
1
(2p−1)2
+
1
4
∞

p=q+1
1
(2p−1)2
=· · · =
1
2
∞

p=1
1
(2p−1)2
=
1
2
∞

p=1
1
(2p−1)2

1+
1
22
+
1
24
+
1
26
+· · ·

·
# ∞

p=0

1
4
p
$−1
=
1
2
∞

p=1
1
p2
·
⎧
⎪
⎨
⎪
⎩
1
1 −
1
4
⎫
⎪
⎬
⎪
⎭
−1
=
1
2
·
3
4
·
π2
6
=
π2
16
.
Remark 3.1 For q = 0 it follows immediately that
∞

n=1
(2n − 1)2
{(2n − 1)2 − 4q2}
2 =
∞

n=1
1
(2n − 1)2
= 2 ·
π2
16
=
π2
8
,
and thus not
π2
16
, which one might expect.
f(t) = t2
, t ∈ ]− π, π].
It can be proved that f has the Fourier series
(12) f ∼
π2
3
+
∞

n=1
4
n2
(−1)n
cos nt.
1) Prove that this Fourier series (12) is uniformly convergent, and ﬁnd its sum function.
2) Prove by applying Parseval’s equation that
∞
n=1
1
n4
=
π4
90
.
3) Prove that
∞
n=1
(−1)n
n2
= −
π2
12
.
4) Find an integer N, such that −
π2
12
−
N
n=1
(−1)n
n2
≤ 10−4
.
Introduction. The function is continuous and piecewise C1
without vertical half tangents, hence
f ∈ K∗
2π. The Fourier series is by the main theorem convergent with f(t) itself as its sum function.
Cf. the ﬁgure.
1) It follows from
π2
3
+
∞

n=1
4
n2
|(−1)n
cos nt| ≤
π2
3
+
∞

n=1
4
n2
=
π2
3
+ 4 ·
π2
6
= π2
∞,
that the Fourier series has a convergent majoring series, hence it is uniformly convergent with the
sum function f(t).

112
2
4
6
8
10
–10 –5 5 10
x
2) First we set up Parseval’s equation:
1
2
a2
0 +
∞

n=1
(a2
n + b2
n) =
1
π
π
−π
[f(t)]2
dt.
It follows from the Fourier series that
1
2
a0 =
π2
3
, dvs. a0 =
2
3
π2
, og an =
4(−1)n
n2
, n ∈ N.
Then by an insertion,
1
2

2
3
π
2
+
∞

n=1

4
n2
2
=
1
π
π
−π
t4
dt =
2
5
π4
,
and thus by a rearrangement,
∞

n=1
1
n4
=
1
16

2
5
π4
−
4
18
π4

=
π4
8

1
5
−
1
9

=
π4
8
·
4
45
=
π4
90
.
3) If we put t = 0 into (12), we get since f(t) is the sum function that
f(0) = 0 =
π2
3
+ 4
∞

n=1
(−1)n
n2
,
(13)
∞

n=1
(−1)n
n2
= −
π2
12
.
4) The series (13) is alternating. Since
(−1)n
n2
=
1
n2
tends decreasingly towards 0, it follows from
Leibniz’s criterion that we have the following error estimate,
−
π2
12
−
N

n=1
(−1)n
n2
≤
(−1)N+1
(N + 1)2
=
1
(N + 1)2
.

113
Then we have
1
(N + 1)2
≤ 10−4
≤
1
1002
for N + 1 ≥ 100, so we get by the error estimate that we can choose N ≥ 99.
f(t) = sin
t
2
, t ∈ ]− π, π].
1) Sketch the graph of f in the interval ]− 3π, 3π].
(Hint: It is given without proof that
π
0
sin
t
2
sin nt dt = (−1)n−1
·
4n
4n2 − 1
, n ∈ N

.
3) Find the sum of the Fourier series at the point t =
7π
3
.
4) Explain why the Fourier series is not uniformly convergent.
5) Apply Parseval’s equation in order to prove that
∞

n=1

n
4n2 − 1
2
=
π2
64
.
1) The function is odd and piecewise C∞
without vertical half tangents, and with discontinuities at
t = (2p + 1)π, p ∈ Z. It therefore follows from the main theorem that the Fourier series is
convergent with the sum function
f∗
(t) =
⎧
⎨
⎩
f(t) for t = (2p + 1)π, p ∈ Z,
0 for t = (2p + 1)π, p ∈ Z.
2) The function f is odd, so an = 0, and
bn =
2
π
π
0
sin

t
2

sin(nt)dt =
2
π
·
4n cos(nπ)
1 − 4n2
, n ∈ N,
and the Fourier series is given with its sum function by
f∗
(t) =
8
π
∞

n=1
n(−1)n
1 − 4n2
sin(nt).

114
–1
–0.5
0.5
1
y
–8 –6 –4 –2 2 4 6 8
x
3) It follows from the periodicity that
f∗

7π
3

= f∗

7π
3
− 2π

= f∗
π
3

= sin
π
6
=
1
2
=
8
π
∞

n=1
n(−1)n
1 − 4n2
sin
nπ
3
.
4) The sum function f∗
(t) is not continuous, hence the convergence cannot be uniform. [In fact, if
the convergence was uniform, then the sum function should be continuous, which it is not].
5) Then we get by Parseval’s equation,
1
π
π
−π
sin2

t
2

dt = 1 =
∞

n=1
b2
n =
64
π2
∞

n=1

n
1 − 4n2
2
,
∞

n=1

n
4n2 − 1
2
=
π2
64
.

115
4 Fourier series in the theory of beams
Example 4.1 A periodic function f of period 2 is given in the interval ]− , [ by
f(t) =
⎧
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎩
0 for − t −

2
,
−q0 for −

2
t 0,
q0 for 0 t

2
,
0 for

2
t ≤ ,
where q0 is a positive constant.
We define in points of discontinuity f(t) =
1
2
{f(t+) + f(t−)}.
1) Sketch the graph of f(t) in ]− , [, and prove that the Fourier series for f is given by
2q0
π
∞

n=1
1
2n−1
sin(2n−1)
π

t+
1−(−1)n
2n
sin 2n
π

t .
2) A simply supported beam of length and of bending stiffness EI is loaded (constant load q0 on the
first half of the beam).
a) The linearized boundary value problem for the bending u(x) of the beam by the load q(x) is
d4
u
dx4
=
q(x)
EI
, u(0) = u() = u
(0) = u
() = 0.
b) Find the bending u(x) in the form of a Fourier series of the type
∞

n=1

b2n−1 sin(2n−1)
π

x + b2n sin 2n
π

x

,
where the boundary value problem in (a) is solved by means of the method of Fourier series,
and where the result of (1) is applied.
c) Prove that the bending u

2

can be written as the series
2qo4
EIπ5
∞

n=1
(−1)n+1
(2n − 1)5
,
and explain why the series is convergent. Find an approximative value of u

2

with an error
which is smaller than
2q04
EIπ5
·
1
75
.
1) Clearly, f ∈ K∗
2π, because f is piecewise constant. The function f is already adjusted. and since
f is odd, we get an = 0, and the Fourier series is a sine series, which by the main theorem has
the sum function f(t).
Fourier series in the theory of beams

116
–1
–0.5
0
0.5
1
–2 –1 1 2
x
By a calculation,
bn =
2

0
f(t) sin
nπt

dt =
2q0

/2
0
sin
nπ

dt
=
2q0

·

nπ
− cos

nπt

/2
0
=
2q0
nπ

1 − cos
nπ
2

,
so
b2n−1 =
2q0
π
·
1
2n − 1
og b2n =
2q0
π
·
1 − (−1)n
2n
.

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117
The Fourier series is with an equality sign, cf. the beginning, given as required by
f(t)=
2q0
π
∞

n=1

1
2n−1
sin(2n−1)
π

t +
1− (−1)n
2n
sin 2n
π

t

.
2) Now, q(x) = f(x), so it follows from (1) that
q(x)
EI
=
2q0
πEI
∞

n=1
1
2n−1
sin(2n−1)
π

t +
1−(−1)n
2n
sin 2n
π

t

.
This series is not uniformly convergent (because q(x) is not continuous), so strictly speaking we
are not allowed to perform termwise integration. Nevertheless, if we perform termwise integration
four times, we get
u(t) =
2q04
π5EI
∞

n=1
1
(2n−1)5
sin(2n−1)
π

t +
1−(−1)n
(2n)5
sin 2n
π

t

+ c0 + c1t + c2t2
+ c3t3
,
where we later shall come back to this inconsistency with the usual theory.
Then by the boundary conditions,
u(0) = 0 = c0 and u() = 0 = (c1 + c2 + c32
),
u
(0) = 0 = 2c2 and u
() = 0 = 2c2 + 6c3,
hence c0 = c1 = c2 = c3 = 0. Thus
u(t)=
2q04
π5EI
∞

n=1
1
(2n−1)5
sin(2n−1)
π

t +
1−(−1)n
(2n)5
sin 2n
π

t

.
If t =

2
, then sin

2n
π

·

2

= 0 and sin

(2n − 1)
π

·

2

= (−1)n+1
, hence
u

2

=
2q04
EIπ5
∞

n=1
(−1)n+1
(2n − 1)5
.
This series obviously has the convergent majoring series
2q04
EIπ5
∞

n=1
1
n5
.
The series is alternating and
1
(2n − 1)5
is decreasing, so we get the estimate
∞

n=1
(−1)n+1
(2n − 1)5
−
3

n=1
(−1)n+1
(2n − 1)5
≤
1
(2 · 4 − 1)5
=
1
75
,
whence,
u

2

≈
2q04
EIπ5

1
15
−
1
35
+
1
55

=
1512986
759375π5
·
q04
EI
≈ 0, 006511 ·
q04
EI
.

118
We can now repair the “hole” in the argument above by directly solve the equation
d4
u
dx4
=
q(x)
EI
, u(0) = u() = u
(0) = u
() = 0
with four (diﬃcult) integrations. First we get from
d4
u
dx4
=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
q0
EI
for x ∈ 0,

2
,
0 for x ∈

2
, ,
that
d3
u
dx3
=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
q0
EI
x + c1 for x ∈ 0,

2
,
q0
2EI
+ c1 for x ∈

2
, ,
hence
d2
u
dx2
=
⎧
⎪
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎪
⎩
q0
2EI
x2
+ c1 + c2, for x ∈ 0,

2
,
q02
8EI
+
c1
2
+c2 +

q0
2EI
+c1

x−

2

, for x ∈

2
, .
Now, u
(0) = 0 so c2 = 0, and since u
() = 0 we get
0 =
q02
8EI
+
c1
2
+
q02
4EI
+
c1
2
=
3q2
8EI
+ c1,
thus
c1 = −
3q0
8EI
.
By insertion and reduction,
d2
u
dx2
=
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
q0
2EI
x2
−
3q0
8EI
x for 0 ≤ x ≤

2
,
−
q02
16EI
+
q0
8EI

x −

2

for

2
x ≤ ,
so
du
dx
=
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
q0
6EI
x3
−
3q0
16EI
x2
+ c3, for 0 ≤ x ≤

2
,
q03
48EI
−
3q03
64EI
+ c3 −
q02
16EI

x −

2

+
q0
16EI

x −

2
2
, for

2
x ≤ .

119
Since u(0) = 0, we therefore get for 0 ≤ x ≤

2
that
u(x) =
q0x4
24EI
−
q0x3
16EI
+ c3x for 0 ≤ x ≤

2
.
If then x ∈

2
, we have
u(x) =
q04
24 · 16EI
−
q04
16 · 8EI
+ c3

2
+

c3 −
5q03
192EI

x −

2

−
q02
32EI

x −

2
2
+
q0
48EI

x −

2
3
,
where
u() = 0 =
q04
16EI

1
24
−
1
8
−
1
2
·
5
12
−
1
4
·
1
2
+
1
8 · 3

+ c3
=
q04
16EI
·
1−3−5−3+1
24
+ c3 = −
3q04
128EI
+ c3,
so
c3 =
3
128
·
q04
EI
.
Then by insertion of c3 and some further calculations we ﬁnally obtain that
u(x) =
⎧
⎪
⎪
⎪
⎨
⎪
⎪
⎪
⎩
q0
24EI
x4
−
q0
16EI
x3
+
3
128
q03
EI
x, for 0 ≤ x ≤

2
,
7
384
·
q03
EI
( − x) −
1
48
·
q0
EI
( − x)3
, for

2
≤ x ≤ ,
where it is more convenient for x ∈

2
, to use − x as the variable.
If x =

2
, then
u

2

=
5
768
·
q04
EI
=
2q04
EIπ5
∞

n=1
(−1)n+1
(2n − 1)5
,
hence
∞

n=1
(−1)n−1
(2n − 1)5
=
5
768
·
π5
2
=
5π5
1536
.

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