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Leif Mejlbro

Examples of Fourier series

Calculus 4c-1

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Examples of Fourier series – Calculus 4c-1

© 2008 Leif Mejlbro & Ventus Publishing ApS

ISBN 978-87-7681-380-2

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Examples of Fourier series

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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Examples of Fourier series

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

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Examples of Fourier series

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

sinn

π

2

=

{

(−1)(n−1)/2

for n ulige,

0

for n lige.

Sum function of Fourier series

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Examples 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).

Sum function of Fourier series

what‘s missing in this equation?

www.maersk.com/mitas

Examples of Fourier series

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

[sinnt]π0 = 0, n ≥ 1,

bn =

1

π

∫ π

−π

f(t)sin nt dt =

1

π

∫ π

0

sinnt dt = − 1

nπ

[cos nt]π0 =

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} = 12 +

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 tdt =

1

π

[− cos t]π0 =

2

π

,

a1 =

1

π

∫ π

0

sin t · cos tdt = 1

2π

[

sin2 t

]π

0

= 0,

Sum function of Fourier series

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Examples of Fourier series

9

Sum function of Fourier series

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 tdt =

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

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Examples of Fourier series

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

The function f is piecewise C1 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

)}

.

Sum function of Fourier series

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Examples of Fourier series

11

It follows immediately that if n > 1 is odd, n = 2p + 1, p ≥ 1, then b2p+1 = 0 (note that b1 = 12 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.

Sum function of Fourier series

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Examples of Fourier series

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 2tdt =

1

2π

[− cos 2t]π0 = 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),

Sum function of Fourier series

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Examples of Fourier series

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 cosnπ =

2

π

− 4

π

∞∑

n=1

(−1)n

4n2 − 1 =

2

π

+

4

π

∞∑

n=1

(−1)n+1

4n2 − 1 ,

Sum function of Fourier series

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Examples of Fourier series

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.

Sum function of Fourier series

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Examples of Fourier series

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

)

,

Sum function of Fourier series

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Examples of Fourier series

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

tdt =

[

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

)

,

Sum function of Fourier series

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Examples of Fourier series

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

)

.

Sum function of Fourier series

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Examples of Fourier series

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{2sin 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) .

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Examples of Fourier series

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].

1) Prove that f has the Fourier series

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 without vertical half tangents, so f ∈ K∗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]π0 +

2

πn

∫ π

0

(π − 2t)cos nt dt

= 0 +

2

πn2

[(π − 2t)sin nt]π0 +

4

πn2

∫ π

0

sin nt dt = 0 − 4

πn3

[cos nt]π0 =

4

π

· 1

n3

{1 − (−1)n}.

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Examples of Fourier series

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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Examples of Fourier series

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 .

Then by a rearrangement,

∞∑

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].

2) Prove that f has the Fourier series

−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 · sinnt 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 2tdt = − 1

2π

[t cos 2t]π0 +

1

2π

∫ π

0

cos 2tdt = −1

2

.

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Examples of Fourier series

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 sinnt.

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

1) Since f is piecewise C1 without vertical half tangents, we get f ∈ K∗2π. Then by the main

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]π0 −

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.

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Examples of Fourier series

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 without vertical half tangents, we have f ∈ K∗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]π0 −

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 = π

2

3

− π +

∞∑

n=1

4

πn2

{1 + (−1)n(π − 1)} cos nt

for t ∈ [0,π]

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Examples of Fourier series

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) cosnt dt

= − 2

πn

π(π−2) · (−1)n + 4

πn2

[(t−1) sinnt]π0 −

4

πn2

∫ π

0

sinnt dt

=

2

n

(π − 2) · (−1)n−1 + 0 + 4

πn3

[cosnt]π0 =

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]

}

sinnt

for t ∈ [0,π[.

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2009

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Examples of Fourier series

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]π0 +

1

2π

∫ π

0

sin 2tdt =

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.

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Examples of Fourier series

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

}

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Examples of Fourier series

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

· π

2

6

=

π2

64

.

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Examples of Fourier series

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[.

1) Find the Fourier series for f .

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 without vertical half tangents, we have f ∈ K∗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 4tdt =

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}.

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Examples of Fourier series

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

= sinnπ · 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.

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Examples of Fourier series

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 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) = 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 2tdt =

1

2π

[−t cos 2t]π0 +

1

2π

∫ π

0

cos 2tdt =

−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 ,

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Examples of Fourier series

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) sinnt,

for t ∈ [−π, π].

Hence by a rearrangement for t ∈ [−π, π],

∞∑

n=2

(−1)n sinnt

(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 ]t0 =

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.

Sum function of Fourier series

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Examples of Fourier series

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 sinnt

]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 without vertical half tangents, hence f ∈ K∗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

· π

3

3

− 1

π

· (−1)2 · 2π

4

=

π2

3

− 1

2

.

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Examples of Fourier series

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

}

,

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Examples of Fourier series

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].

2) Find the sum of the series

∞∑

n=0

(−1)n(2n + 1)

(4n + 1)(4n + 3)

.

Since f is piecewise C1 without vertical half tangents, we have f ∈ K∗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

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Examples of Fourier series

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 sinnt dt − 2

π

∫ π

π/2

sin t sinnt 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

⎫⎬

⎭ .

Sum function of Fourier series

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Examples of Fourier series

36

Hence bn+1 = 0 for n ≥ 1, and

b2n =

2

π

⎧⎨

⎩

sin

(

nπ − π

2

)

2n − 1

−

sin

(

nπ +

π

2

)

2n + 1

⎫⎬

⎭ = 2π (−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 tdt−

∫ π

π/2

sin2 tdt

}

=

2

π

{∫ π/2

0

sin2 tdt−

∫ π/2

0

sin2 tdt

}

= 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

.

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Examples of Fourier series

37

Example 1.18 1) Given the infinite series

a)

∞∑

n=1

(−1)n+1n

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

2sinhπ

π

∞∑

n=1

(−1)n+1n

n2 + 1

sinnt.

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.

3) 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 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]π0 +

2

πn

∫ π

0

cosh t · cos nt dt

=

2

πn

sinhπ · (−1)n+1 + 2

πn2

[cosh t · sinnt]π0 −

2

πn2

∫ π

0

sinh t · sin nt dt

= (−1)n+1 · 2

n

· sinhπ

π

− 1

n2

bn,

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Examples of Fourier series

38

–10

–5

0

5

10

–4

–2

2

4

x

hence by a rearrangement,

bn =

(

1 +

1

n2

)−1

· (−1)n+1 · 2

n

· sinhπ

π

=

2sinhπ

π

· (−1)

n+1n

n2 + 1

.

The Fourier series is (with equality sign, cf. the above)

f∗(t) = sinh t =

2sinhπ

π

∞∑

n=1

(−1)n+1n

n2 + 1

sinnt for t ∈ ]− π, π[.

4) When we put t =

π

2

into the Fourier series, we get

sinh

π

2

= 2

sinhπ

π

∞∑

n=1

(−1)n+1n

n2 + 1

sin

(

n

π

2

)

= 2

sinhπ

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

,

hence by a rearrangement

∞∑

n=1

(−1)n+1(2n − 1)

(2n − 1)2 + 1 =

π

4cosh

π

2

.

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Examples of Fourier series

39

Example 1.19 The even and periodic function f of period 2π is given in the interval [0,π] by

f(x) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

k − k2x, 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 − k2x)cos nxdx = 2

πn

[

(k − k2x)sinnx]1/k

0

+

2k2

πn

∫ 1/k

0

sinnxdx

=

2k2

πn2

{

1 − cos

(n

k

)}

.

Since

a0 =

2

π

∫ 1/k

0

(k − k2x)dx = 2

π

[

kx − 1

2

k2x2

]1/k

0

=

1

π

,

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Examples of Fourier series

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 → ∞.

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Examples of Fourier series

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) Prove that f has the Fourier series

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.

3) Find the sum of the series

∞∑

n=1

(−1)n+1

n2 − 1

4

.

1) Clearly, f is piecewise C1 without vertical half tangents, so f ∈ K∗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

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Examples of Fourier series

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

cosnt dt =

1

π

∫ π

0

{

cos

(

n+

1

2

)

t+cos

(

n− 1

2

)

t

}

dt

=

1

π

[

1

n+ 12

sin

(

n+

1

2

)

t+

1

n− 12

sin

(

n− 1

2

)

t

]π

0

=

1

π

{

(−1)n

n+ 12

− (−1)

n

n− 12

}

=

(−1)n

π

· (n −

1

2 ) − (n + 12 )

n2 − 14

=

(−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 cosnt =

2

π

+

1

π

∞∑

n=1

(−1)n+1

n2 − 1

4

cosnt.

Sum function of Fourier series

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3400 Hillerød

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Examples of Fourier series

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

)

.

1) Since f is piecewise C1 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.

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Examples of Fourier series

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

)

sinnt 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

sinn

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

sinn

π

2

)

cos nt,

t ∈ R.

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Examples of Fourier series

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

sinn

π

n

)

cosn

π

2

=

π2

24

+ 2

∞∑

n=1

(

1

n2

cosn

π

2

− 2 sinn

π

2 cosn

π

2

πn3

)

=

π2

24

+ 2

∞∑

n=1

(

1

n2

cosn

π

2

− 1

πn3

sinnπ

)

=

π2

24

+ 2

∞∑

n=1

1

n2

cosn

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

,

Sum function of Fourier series

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Examples of Fourier series

46

because cos

(π

2

+ pπ

)

= 0 for p ∈ Z.

Then by a rearrangement,

∞∑

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):

sinu + sin v = 2 sin

(

u + v

2

)

cos

(

u − v

2

)

.

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Examples of Fourier series

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

· 2sin(pπ) · cos p · π

2

= 0.

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Examples of Fourier series

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

− −···

}

,

hence by a rearrangement,

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

.

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Examples of Fourier series

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.

3) Find the sum of the series

∞∑

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

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Examples of Fourier series

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 − 2π2t2 + π4) dt

=

2

π5

[

1

5

t5 − 2

3

π2t3 + π4t

]π

0

= 2

{

1

5

− 2

3

+ 1

}

=

16

15

,

hence

1

2

a0 =

8

15

.

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Examples of Fourier series

51

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

sinnt

]π

0

− 2

π5

· 4

n

∫ π

0

t(t2−π2)sin nt dt

= 0 +

8

π5

· 1

n2

[t(t2−π2)cos nt]π0 −

8

π5

· 1

n2

∫ π

0

(3t2−π2)cos nt dt

= 0 − 8

π5

· 1

n3

[(3t2−π2)sin nt]π0 +

48

π5

∫ π

0

t sinnt dt

= −48

π5

· 1

n4

[t cos nt]π0 +

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.

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Examples of Fourier series

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 t2

∣∣∣∣, 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

sinnt +

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 t2

∣∣∣∣ cos nt dt = 2π

∫ 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.

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Examples of Fourier series

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

}

,

hence by a rearrangement,

∞∑

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

.

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Examples of Fourier series

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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Examples of Fourier series

55

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 − 3cos 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 − 3cos 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 = π2 .

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.

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Examples of Fourier series

56

Since

∣∣∣∣eix3

∣∣∣∣ = 13 < 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 − e

ix

3

=

eix

3 − eix ·

3 − e−ix

3 − e−ix

=

3eix − 1

9 − 6cos x + 1 =

1

2

· 3cos x − 1 + 3i sin x

5 − 3cos 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

· 3cos x − 1

5 − 3cos x

=

1

4

· (5 − 3cos x) + (3 cos x − 1)

5 − 3cos x

=

1

5 − 3cos x = f(x),

and

2

3

∞∑

n=1

1

3n

sin nx =

2

3

Im

{ ∞∑

n=1

1

3n

einx

}

=

2

3

· 1

2

·

3sin x

5 − 3cos x =

sinx

5 − 3cos x = g(x).

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Examples of Fourier series

57

1) From

∣∣∣∣∣14 + 12

∞∑

n=1

1

3n

cos nx

∣∣∣∣∣ ≤ 14 + 12

∞∑

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 − 3cos x = f(x).

The termwise differentiated series,

−1

2

∞∑

n=1

n

3n

sinnx,

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

sinnx = f ′(x) =

d

dx

(

1

5 − 3cos x

)

= −

3sin x

(5 − 3cos 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 23

∑

3−n sinnx 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.

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Examples of Fourier series

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)]x0 =

1

3

ln(5 − 3 cos x) − 1

3

ln 2

=

2

3

∞∑

n=1

1

3n

∫ x

0

sinnx dx = −2

3

∞∑

n=1

1

n · 3n (cosnx − 1),

hence by a rearrangement,

∞∑

n=1

1

n · 3n cosnx =

∞∑

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).

Sum function of Fourier series

what‘s missing in this equation?

Examples of Fourier series

59

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].

2) Prove that the Fourier series for f is given by

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 sinnt dt =

1

n2

(−nt cos nt + sinnt).

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,

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Examples of Fourier series

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]π0

=

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 + sinnt]π0

=

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

tdt =

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

sinnπ

}

=

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

,

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Examples of Fourier series

61

and hence

∞∑

p=0

1

(2p + 1)2

=

3

4

· π

2

6

=

π2

8

.

We get by insertion the sum

π

4

+

2

π

∞∑

p=0

1

(2p + 1)2

=

π

4

+

2

π

· π

2

8

=

π

4

+

π

4

=

π

2

.

Sum function of Fourier series

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Examples of Fourier series

62

2 Fourier series and uniform convergence

Example 2.1 The function f ∈ K2π is given by

f(t) = π2 − t2,

−π < t ≤ π.

1) Find the Fourier series for f .

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.

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Examples of Fourier series

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 sinnt dt

= 0 +

4

πn2

[−t cosnt]π0 +

4

πn2

∫ π

0

cosnt dt = − 4

πn2

π cosnπ + 0 =

4(−1)n+1

n2

.

In the exceptional case n = 0 we get instead

a0 =

2

π

∫ π

0

(π2−t2)dt = 2

π

[

π2x − x

3

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 cosnt =

2π2

3

+

∞∑

n=1

(−1)n−1 · 4

n2

cosnt.

2) The estimate

∣∣∣∣(−1)n−1 4n2 cosnt

∣∣∣∣ ≤ 4n2 shows that

2π2

3

+ 4

∞∑

n=1

1

n2

=

2π2

3

+ 4 · π

2

6

=

4π2

3

is a convergent majoring series. Hence the Fourier series is uniformly convergent.

Fourier series and uniform convergence

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Examples of Fourier series

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

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Examples of Fourier series

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.

2) 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 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 sinnt 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 sinnt 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 sinnt dt =

1

πn

[−t2 cos nt]2π

0

+

2

πn

∫ 2π

0

t cos nt dt

= −4π

2

πn

+

2

πn2

[t sin nt]2π

0 −

2

πn2

∫ 2π

0

sinnt dt = −4π

n

− 2

πn3

[cos nt]2π

0 = −

4π

n

.

The Fourier series is (NB. Remember the term

1

2

a0)

f ∼ 4π

2

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).

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Examples of Fourier series

66

Example 2.3 Let the function f ∈ K2π be given by

f(t) = et sin t

for − π < t ≤ π.

1) Prove that the Fourier series for f is given by

sinhπ

π

{

1

2

+

∞∑

n=1

(

(−1)n(4−2n2)

n4 + 4

cos nt+

(−1)n−14n

n4 + 4

sin nt

)}

.

We may use the following formulæ without proof:

∫ π

−π

et cos mtdt =

2(−1)m sinhπ

1 + m2

,

m ∈ N0,

∫ π

π

et sin mtdt =

2m(−1)m+1 sinhπ

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 without vertical half tangents, so f ∈ K∗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) ,

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Examples of Fourier series

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)}

]

=

sinhπ

π

· i(−1)n

{

1

1 + i(n + 1)

−

1

1 + i(n − 1)

}

=

sinhπ

π

· (−1)n i · 1 + i(n − 1) − {1 + i(n + 1)}

1 − (n2 − 1) + i 2n

=

sinhπ

π

(−1)n · i ·

−2i

−(n2 − 2) + 2in =

sinhπ

π

· (−1)n · 2 · −(n

2 − 2) − 2in

(n2 − 2)2 + 4n2

=

sinhπ

π

· (−1)n · 4 − 2n

2 − 4in

n4 + 4

.

Fourier series and uniform convergence

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Examples of Fourier series

68

When we split into the real and the imaginary part we find

an =

sinhπ

π

· (−1)n · 4 − 2n

2

n4 + 4

,

n ≥ 0,

and

bn =

sinhπ

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 sinhπ

1 + (n + 1)2

−2(n − 1)(−1)

n sinhπ

1 + (n − 1)2

}

=

sinhπ

π

(−1)n (n+1)(n

2−2n+2)−(n−1)(n2+2n+2)

(n2+2−2n)(n2+2+2n)

=

sinhπ

π

· (−1)n · 4 − 2n

2

n4 + 4

,

and

bn =

1

π

∫ π

−π

et sin t sinnt dt =

1

2π

∫ π

−π

et{cos(n − 1)t − cos(n + 1)t} dt

=

1

2π

{

2(−1)n−1 sinhπ

1 + (n − 1)2 −

2(−1)n−1 sinhπ

1 + (n + 1)2

}

=

sinhπ

π

· (−1)n−1 · n

2+2n+2−(n2−2n+2)

(n2−2n+2)(n2+2n+2)

=

sinhπ

π

· (−1)n−1 ·

4n

n4 + 4

.

In both cases we see that a0 =

sinhπ

π

, hence we get the Fourier series (with equality by the remarks

above)

f(t) =

sinhπ

π

{

1

2

+

∞∑

n=1

(

(−1)n(4 − 2n2)

n4 + 4

cos nt +

(−1)n−14n

n4 + 4

sinnt

)}

.

2) The Fourier series has the convergent majoring series

sinhπ

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 =

sinhπ

π

{

1

2

− 2

∞∑

n=1

n2 − 2

n4 + 4

}

,

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Examples of Fourier series

69

hence by a rearrangement

∞∑

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)

The function f is continuous and piecewise C1 without vertical half tangents, so f ∈ K∗2π. The Fourier

series is by the main theorem pointwise convergent with the sum function f ∗(t) = f(t).

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Examples of Fourier series

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.

In the exceptional case n = 1 we get instead

a1 =

2

π

∫ π/2

0

cos2 tdt =

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.

The Fourier series has the convergent majoring series

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.

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Examples of Fourier series

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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Examples of Fourier series

72

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

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Examples of Fourier series

73

We obtain by some partial integrations,

bn =

2

π

∫ π

0

(π2 − t3)sinnt dt = − 2

πn

[

t(π2 − t2)cos nt]π

0

+

2

πn

∫ π

0

(π2 − 3t2)cos nt dt

=

2

πn2

[(

π2 − 3t2) sinnt]π

0

+

12

πn2

∫ π

0

t sin nt dt =

12

πn3

[−t cos nt]π0 +

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

sinnt.

The Fourier series has the convergent majoring series

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 − π

2

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 .

Then by a rearrangement,

∞∑

n=1

(−1)n+1

(2n − 1)3 =

π3

32

.

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Examples of Fourier series

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) Prove that f has the Fourier series

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.

2) Find the sum of the series

∞∑

n=0

(−1)n

(2n − 1)(2n + 1)(2n + 3) .

The function f is continuous and piecewise C1 without vertical half tangents, so f ∈ K∗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.

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Examples of Fourier series

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.

The Fourier series has the convergent majoring series

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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2009

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Examples of Fourier series

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

.

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Examples of Fourier series

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.

The function f is continuous and piecewise C1 without vertical half tangents, so f ∈ K∗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 4tdt =

[

− 1

4π

cos 4t

]π/2

0

= 0.

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Examples of Fourier series

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

{

2cos(4n−1)t

(4n−3)(4n−1) +

cos 4nt

(2n−1)(2n+1) +

2cos(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

{

8sin(4n − 1)t

(4n−3)(4n−1)(4n+1)

+

sin 4nt

(2n−1)n(2n+1) +

8sin(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

]

.

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Examples of Fourier series

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

−π

2

6

+

∞∑

n=1

cos(nx)

n2

.

4) Prove that g is that solution of the differential equation

d2y

dx2

− y = −f(x) − π

2

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.

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Examples of Fourier series

80

2) The series

∑∞

n=1

cos(nx)

n2 + 1

has the convergent majoring series

∞∑

n=1

1

n2 + 1

,

so it is also uniformly convergent.

3) Finally, the termwise differentiated series,

∑∞

n=1

− sin(nx)

n(n2 + 1)

has the convergent majoring series

∞∑

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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Examples of Fourier series

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) = −π

2

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

d2y

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) − π

2

6

,

and we have shown that they fulfil the differential equation.

Now,

g′(x) = −

∞∑

n=1

sinnx

n(n2 + 1)

,

so g′(0) = g′(π) = 0. It follows that g is a solution of the boundary value problem

(7)

d2y

dx2

− y = −f(x) − π

2

6

= −1

4

x2 +

π

2

x − π

2

6

,

with the boundary conditions y′(0) = y′(π) = 0 [notice, over half of the interval].

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Examples of Fourier series

82

The corresponding homogeneous equation without the boundary conditions has the complete so-

lution,

y = c1 coshx + c2 sinhx.

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

d2y

dx2

− y = 2a−ax2−bx−c = −ax2−bx+(2a−c).

This equal to

−f(x) − π

2

6

= −1

4

x2 +

π

2

x − π

2

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 coshx + c2 sinhx.

Since

y′ =

1

2

x − π

2

+ c1 sinhx + 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 sinhπ + c2 coshπ = c1 sinhπ +

π

2

coshπ = 0,

hence c1 = −π2 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 · coshx + π

2

sinhx =

∞∑

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

.

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Examples of Fourier series

83

Example 2.9 Let f ∈ K2π be given by

f(t) = |t|3

for − π < t ≤ π.

1) Sketch the graph of f in the interval [−π, 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

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Examples of Fourier series

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]π0 −

∫ π

0

sin nt dt

}

=

6π2

n2

(−1)n − 12

πn4

[cos nt]π0 = 6π

{

(−1)n

n2

+

2(1 − (−1)n)

π2n4

}

.

Then by insertion (remember 12 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.

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Examples of Fourier series

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 · π

2

6

+

12

π

∞∑

p=1

2

(2p − 1)4 (−1)

2p−1

=

π3

4

+ π3 − 24

π

∞∑

p=1

1

(2p − 1)4 ,

hence by a rearrangement,

∞∑

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 without vertical half tangents, so f ∈ K∗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,

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Examples of Fourier series

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

[sinnt]ππ/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,

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Examples of Fourier series

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.

Example 2.11 Let f ∈ K2π be given by

f(t) = e|t|

for − π < t ≤ π.

1) Sketch the graph of f and explain why f ∈ K∗2π.

2) Prove that the Fourier series for f is given by

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∞ without vertical half tangents, we have f ∈ K∗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)neπ − 1} + 2n

π

[

et sin nt

]π

0

− 2n

2

π

∫ π

0

et cos nt dt

=

2

π

{(−1)neπ − 1} + 0 − n2an,

so by a rearrangement,

an =

2

π

· (−1)

neπ − 1

n2 + 1

,

n ∈ N0,

specielt a0 =

2

π

{eπ − 1}.

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Examples of Fourier series

88

5

10

15

20

y

–4

–2

0

2

4

x

Alternatively it follows directly by complex calculations with cosnt = Re eint that

an =

2

π

∫ π

0

et cos nt dt =

2

π

Re

∫ π

0

e(1+in)tdt =

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)

neπ − 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.

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Examples of Fourier series

89

Example 2.12 Let f ∈ K2π be given by

f(t) = (t − π)2

for − π < t ≤ π.

1) Sketch the graph of f in the interval [−π, π].

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.

4) Prove that the Fourier series for f is given by

4π2

3

+ 4

∞∑

n=1

{

(−1)n

n2

cosnt +

(−1)nπ

n

sinnt

}

,

t ∈ R.

(1) and (2) 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 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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Examples of Fourier series

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

sinnt · (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)2dt = 1

π

[

(t − π)3

3

]π

−π

=

1

π

· 8π

3

3

=

8π2

3

,

and for n ∈ N,

bn =

1

π

∫ π

−π

(t − π)2 sinnt 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

[sinnt · (t − π)]π−π −

2

πn

∫ π

−π

sin nt dt =

4π

n

· (−1)n.

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Examples of Fourier series

91

Summing up we get with pointwise equality,

f∗(t) =

4π2

3

+ 4

∞∑

n=1

{

(−1)n

n2

cos nt +

(−1)nπ

n

sin nt

}

,

t ∈ R.

Example 2.13 The periodic function f : R → R of period 2π is defined by

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 f̃ 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 2tdt +

1

2π

∫ π

0

{cos 2t + 1}dt = 0 + 1

2π

{0 + π} = 1

2

.

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Examples of Fourier series

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

f̃(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

sin t,

t ∈ ]− π, 0[,

1/2,

t = 0,

cos t,

t ∈ ]0,π[,

−1/2,

t = π,

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Examples of Fourier series

93

which is continued periodically.

Since f (and f̃) 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

f̃+(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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Examples of Fourier series

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.

Example 2.14 Let f ∈ K2π be given by

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.

4) Prove that the Fourier series for f is given by

2

3

− 1

π2

∞∑

n=1

{

1

n2

cos nt +

π

n

sinnt

}

,

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.

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Examples of Fourier series

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

sinnt 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

sinnt

}

,

t ∈ R.

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Examples of Fourier series

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 = rneint = rn{cos nt + i sin nt}.

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Examples of Fourier series

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{cosnt + i sinnt}.

Here we take two times the imaginary part,

2

∞∑

n=1

rn sinnt = 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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Examples of Fourier series

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 sinnt},

where

an =

1

π

∫ 2π

0

fr(t)cos nt dt,

bn =

1

π

∫ 2π

0

fr(t)sinnt 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 = −2πr

5

5

,

∫ π

−π

fr(t) sin(5t)dt = πb5 = 0.

Fourier series and uniform convergence

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Examples of Fourier series

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 · r

5

5

∫ 2π

0

cos2(5t)dt = −2πr

5

5

,

∫ π

−π

fr(t) sin(5t)dt = −2

∞∑

n=1

rn

n

∫ π

−π

cos nt · sin 5tdt = 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) = −12 ln(1 + r

2 − 2r cos t),

we get by choosing r =

1

2

and t = 0 that

∞∑

n=1

1

n2n

= −1

2

f1/2(0) = −12 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) = −12 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+1rn

n

= ln(1 + r),

r ∈ ]− 1, 1[,

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Examples of Fourier series

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 sinnt.

If 0 < r < 1, then 2

∑∞

n=1 r

n =

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 sinnt =

d

dt

ln(1+r2−2r cos t) =

2t sin t

1+r2−2r cos t .

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Examples of Fourier series

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)2dt =

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 without vertical half tangents, we have f ∈ K∗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

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Examples of Fourier series

Calculus 4c-1

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Examples of Fourier series – Calculus 4c-1

© 2008 Leif Mejlbro & Ventus Publishing ApS

ISBN 978-87-7681-380-2

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Examples of Fourier series

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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Examples of Fourier series

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

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Examples of Fourier series

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

sinn

π

2

=

{

(−1)(n−1)/2

for n ulige,

0

for n lige.

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Examples 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).

Sum function of Fourier series

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Examples of Fourier series

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

[sinnt]π0 = 0, n ≥ 1,

bn =

1

π

∫ π

−π

f(t)sin nt dt =

1

π

∫ π

0

sinnt dt = − 1

nπ

[cos nt]π0 =

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} = 12 +

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 tdt =

1

π

[− cos t]π0 =

2

π

,

a1 =

1

π

∫ π

0

sin t · cos tdt = 1

2π

[

sin2 t

]π

0

= 0,

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Examples of Fourier series

9

Sum function of Fourier series

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 tdt =

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

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Examples of Fourier series

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

The function f is piecewise C1 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

)}

.

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Examples of Fourier series

11

It follows immediately that if n > 1 is odd, n = 2p + 1, p ≥ 1, then b2p+1 = 0 (note that b1 = 12 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.

Sum function of Fourier series

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Examples of Fourier series

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 2tdt =

1

2π

[− cos 2t]π0 = 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),

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Examples of Fourier series

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 cosnπ =

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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Examples of Fourier series

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.

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Examples of Fourier series

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

)

,

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Examples of Fourier series

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

tdt =

[

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

)

,

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Examples of Fourier series

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

)

.

Sum function of Fourier series

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Examples of Fourier series

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{2sin 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) .

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Examples of Fourier series

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].

1) Prove that f has the Fourier series

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 without vertical half tangents, so f ∈ K∗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]π0 +

2

πn

∫ π

0

(π − 2t)cos nt dt

= 0 +

2

πn2

[(π − 2t)sin nt]π0 +

4

πn2

∫ π

0

sin nt dt = 0 − 4

πn3

[cos nt]π0 =

4

π

· 1

n3

{1 − (−1)n}.

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Examples of Fourier series

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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Examples of Fourier series

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 .

Then by a rearrangement,

∞∑

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].

2) Prove that f has the Fourier series

−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 · sinnt 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 2tdt = − 1

2π

[t cos 2t]π0 +

1

2π

∫ π

0

cos 2tdt = −1

2

.

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Examples of Fourier series

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 sinnt.

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

1) Since f is piecewise C1 without vertical half tangents, we get f ∈ K∗2π. Then by the main

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]π0 −

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.

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Examples of Fourier series

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 without vertical half tangents, we have f ∈ K∗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]π0 −

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 = π

2

3

− π +

∞∑

n=1

4

πn2

{1 + (−1)n(π − 1)} cos nt

for t ∈ [0,π]

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Examples of Fourier series

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) cosnt dt

= − 2

πn

π(π−2) · (−1)n + 4

πn2

[(t−1) sinnt]π0 −

4

πn2

∫ π

0

sinnt dt

=

2

n

(π − 2) · (−1)n−1 + 0 + 4

πn3

[cosnt]π0 =

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]

}

sinnt

for t ∈ [0,π[.

Sum function of Fourier series

2009

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Examples of Fourier series

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]π0 +

1

2π

∫ π

0

sin 2tdt =

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.

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Examples of Fourier series

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

}

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Examples of Fourier series

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

· π

2

6

=

π2

64

.

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Examples of Fourier series

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[.

1) Find the Fourier series for f .

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 without vertical half tangents, we have f ∈ K∗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 4tdt =

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}.

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Examples of Fourier series

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

= sinnπ · 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.

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Examples of Fourier series

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 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) = 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 2tdt =

1

2π

[−t cos 2t]π0 +

1

2π

∫ π

0

cos 2tdt =

−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 ,

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Examples of Fourier series

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) sinnt,

for t ∈ [−π, π].

Hence by a rearrangement for t ∈ [−π, π],

∞∑

n=2

(−1)n sinnt

(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 ]t0 =

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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Examples of Fourier series

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 sinnt

]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 without vertical half tangents, hence f ∈ K∗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

· π

3

3

− 1

π

· (−1)2 · 2π

4

=

π2

3

− 1

2

.

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Examples of Fourier series

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

}

,

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Examples of Fourier series

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].

2) Find the sum of the series

∞∑

n=0

(−1)n(2n + 1)

(4n + 1)(4n + 3)

.

Since f is piecewise C1 without vertical half tangents, we have f ∈ K∗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

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Examples of Fourier series

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 sinnt dt − 2

π

∫ π

π/2

sin t sinnt 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

⎫⎬

⎭ .

Sum function of Fourier series

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Examples of Fourier series

36

Hence bn+1 = 0 for n ≥ 1, and

b2n =

2

π

⎧⎨

⎩

sin

(

nπ − π

2

)

2n − 1

−

sin

(

nπ +

π

2

)

2n + 1

⎫⎬

⎭ = 2π (−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 tdt−

∫ π

π/2

sin2 tdt

}

=

2

π

{∫ π/2

0

sin2 tdt−

∫ π/2

0

sin2 tdt

}

= 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

.

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Examples of Fourier series

37

Example 1.18 1) Given the infinite series

a)

∞∑

n=1

(−1)n+1n

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

2sinhπ

π

∞∑

n=1

(−1)n+1n

n2 + 1

sinnt.

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.

3) 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 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]π0 +

2

πn

∫ π

0

cosh t · cos nt dt

=

2

πn

sinhπ · (−1)n+1 + 2

πn2

[cosh t · sinnt]π0 −

2

πn2

∫ π

0

sinh t · sin nt dt

= (−1)n+1 · 2

n

· sinhπ

π

− 1

n2

bn,

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Examples of Fourier series

38

–10

–5

0

5

10

–4

–2

2

4

x

hence by a rearrangement,

bn =

(

1 +

1

n2

)−1

· (−1)n+1 · 2

n

· sinhπ

π

=

2sinhπ

π

· (−1)

n+1n

n2 + 1

.

The Fourier series is (with equality sign, cf. the above)

f∗(t) = sinh t =

2sinhπ

π

∞∑

n=1

(−1)n+1n

n2 + 1

sinnt for t ∈ ]− π, π[.

4) When we put t =

π

2

into the Fourier series, we get

sinh

π

2

= 2

sinhπ

π

∞∑

n=1

(−1)n+1n

n2 + 1

sin

(

n

π

2

)

= 2

sinhπ

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

,

hence by a rearrangement

∞∑

n=1

(−1)n+1(2n − 1)

(2n − 1)2 + 1 =

π

4cosh

π

2

.

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Examples of Fourier series

39

Example 1.19 The even and periodic function f of period 2π is given in the interval [0,π] by

f(x) =

⎧⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎩

k − k2x, 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 − k2x)cos nxdx = 2

πn

[

(k − k2x)sinnx]1/k

0

+

2k2

πn

∫ 1/k

0

sinnxdx

=

2k2

πn2

{

1 − cos

(n

k

)}

.

Since

a0 =

2

π

∫ 1/k

0

(k − k2x)dx = 2

π

[

kx − 1

2

k2x2

]1/k

0

=

1

π

,

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Examples of Fourier series

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 → ∞.

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Examples of Fourier series

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) Prove that f has the Fourier series

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.

3) Find the sum of the series

∞∑

n=1

(−1)n+1

n2 − 1

4

.

1) Clearly, f is piecewise C1 without vertical half tangents, so f ∈ K∗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

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Examples of Fourier series

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

cosnt dt =

1

π

∫ π

0

{

cos

(

n+

1

2

)

t+cos

(

n− 1

2

)

t

}

dt

=

1

π

[

1

n+ 12

sin

(

n+

1

2

)

t+

1

n− 12

sin

(

n− 1

2

)

t

]π

0

=

1

π

{

(−1)n

n+ 12

− (−1)

n

n− 12

}

=

(−1)n

π

· (n −

1

2 ) − (n + 12 )

n2 − 14

=

(−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 cosnt =

2

π

+

1

π

∞∑

n=1

(−1)n+1

n2 − 1

4

cosnt.

Sum function of Fourier series

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Slangerupgade 69

3400 Hillerød

Tel. +45 70103370

www.foss.dk

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Examples of Fourier series

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

)

.

1) Since f is piecewise C1 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.

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Examples of Fourier series

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

)

sinnt 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

sinn

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

sinn

π

2

)

cos nt,

t ∈ R.

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Examples of Fourier series

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

sinn

π

n

)

cosn

π

2

=

π2

24

+ 2

∞∑

n=1

(

1

n2

cosn

π

2

− 2 sinn

π

2 cosn

π

2

πn3

)

=

π2

24

+ 2

∞∑

n=1

(

1

n2

cosn

π

2

− 1

πn3

sinnπ

)

=

π2

24

+ 2

∞∑

n=1

1

n2

cosn

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

,

Sum function of Fourier series

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Examples of Fourier series

46

because cos

(π

2

+ pπ

)

= 0 for p ∈ Z.

Then by a rearrangement,

∞∑

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):

sinu + sin v = 2 sin

(

u + v

2

)

cos

(

u − v

2

)

.

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Examples of Fourier series

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

· 2sin(pπ) · cos p · π

2

= 0.

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Examples of Fourier series

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

− −···

}

,

hence by a rearrangement,

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

.

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Examples of Fourier series

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.

3) Find the sum of the series

∞∑

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

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Examples of Fourier series

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 − 2π2t2 + π4) dt

=

2

π5

[

1

5

t5 − 2

3

π2t3 + π4t

]π

0

= 2

{

1

5

− 2

3

+ 1

}

=

16

15

,

hence

1

2

a0 =

8

15

.

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Examples of Fourier series

51

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

sinnt

]π

0

− 2

π5

· 4

n

∫ π

0

t(t2−π2)sin nt dt

= 0 +

8

π5

· 1

n2

[t(t2−π2)cos nt]π0 −

8

π5

· 1

n2

∫ π

0

(3t2−π2)cos nt dt

= 0 − 8

π5

· 1

n3

[(3t2−π2)sin nt]π0 +

48

π5

∫ π

0

t sinnt dt

= −48

π5

· 1

n4

[t cos nt]π0 +

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.

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Examples of Fourier series

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 t2

∣∣∣∣, 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

sinnt +

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 t2

∣∣∣∣ cos nt dt = 2π

∫ 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.

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Examples of Fourier series

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

}

,

hence by a rearrangement,

∞∑

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

.

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Examples of Fourier series

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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Examples of Fourier series

55

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 − 3cos 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 − 3cos 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 = π2 .

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.

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Examples of Fourier series

56

Since

∣∣∣∣eix3

∣∣∣∣ = 13 < 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 − e

ix

3

=

eix

3 − eix ·

3 − e−ix

3 − e−ix

=

3eix − 1

9 − 6cos x + 1 =

1

2

· 3cos x − 1 + 3i sin x

5 − 3cos 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

· 3cos x − 1

5 − 3cos x

=

1

4

· (5 − 3cos x) + (3 cos x − 1)

5 − 3cos x

=

1

5 − 3cos x = f(x),

and

2

3

∞∑

n=1

1

3n

sin nx =

2

3

Im

{ ∞∑

n=1

1

3n

einx

}

=

2

3

· 1

2

·

3sin x

5 − 3cos x =

sinx

5 − 3cos x = g(x).

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Examples of Fourier series

57

1) From

∣∣∣∣∣14 + 12

∞∑

n=1

1

3n

cos nx

∣∣∣∣∣ ≤ 14 + 12

∞∑

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 − 3cos x = f(x).

The termwise differentiated series,

−1

2

∞∑

n=1

n

3n

sinnx,

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

sinnx = f ′(x) =

d

dx

(

1

5 − 3cos x

)

= −

3sin x

(5 − 3cos 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 23

∑

3−n sinnx 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.

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Examples of Fourier series

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)]x0 =

1

3

ln(5 − 3 cos x) − 1

3

ln 2

=

2

3

∞∑

n=1

1

3n

∫ x

0

sinnx dx = −2

3

∞∑

n=1

1

n · 3n (cosnx − 1),

hence by a rearrangement,

∞∑

n=1

1

n · 3n cosnx =

∞∑

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).

Sum function of Fourier series

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Examples of Fourier series

59

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].

2) Prove that the Fourier series for f is given by

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 sinnt dt =

1

n2

(−nt cos nt + sinnt).

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,

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Examples of Fourier series

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]π0

=

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 + sinnt]π0

=

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

tdt =

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

sinnπ

}

=

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

,

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Examples of Fourier series

61

and hence

∞∑

p=0

1

(2p + 1)2

=

3

4

· π

2

6

=

π2

8

.

We get by insertion the sum

π

4

+

2

π

∞∑

p=0

1

(2p + 1)2

=

π

4

+

2

π

· π

2

8

=

π

4

+

π

4

=

π

2

.

Sum function of Fourier series

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Examples of Fourier series

62

2 Fourier series and uniform convergence

Example 2.1 The function f ∈ K2π is given by

f(t) = π2 − t2,

−π < t ≤ π.

1) Find the Fourier series for f .

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.

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Examples of Fourier series

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 sinnt dt

= 0 +

4

πn2

[−t cosnt]π0 +

4

πn2

∫ π

0

cosnt dt = − 4

πn2

π cosnπ + 0 =

4(−1)n+1

n2

.

In the exceptional case n = 0 we get instead

a0 =

2

π

∫ π

0

(π2−t2)dt = 2

π

[

π2x − x

3

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 cosnt =

2π2

3

+

∞∑

n=1

(−1)n−1 · 4

n2

cosnt.

2) The estimate

∣∣∣∣(−1)n−1 4n2 cosnt

∣∣∣∣ ≤ 4n2 shows that

2π2

3

+ 4

∞∑

n=1

1

n2

=

2π2

3

+ 4 · π

2

6

=

4π2

3

is a convergent majoring series. Hence the Fourier series is uniformly convergent.

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Examples of Fourier series

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

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Examples of Fourier series

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.

2) 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 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 sinnt 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 sinnt 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 sinnt dt =

1

πn

[−t2 cos nt]2π

0

+

2

πn

∫ 2π

0

t cos nt dt

= −4π

2

πn

+

2

πn2

[t sin nt]2π

0 −

2

πn2

∫ 2π

0

sinnt dt = −4π

n

− 2

πn3

[cos nt]2π

0 = −

4π

n

.

The Fourier series is (NB. Remember the term

1

2

a0)

f ∼ 4π

2

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).

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Examples of Fourier series

66

Example 2.3 Let the function f ∈ K2π be given by

f(t) = et sin t

for − π < t ≤ π.

1) Prove that the Fourier series for f is given by

sinhπ

π

{

1

2

+

∞∑

n=1

(

(−1)n(4−2n2)

n4 + 4

cos nt+

(−1)n−14n

n4 + 4

sin nt

)}

.

We may use the following formulæ without proof:

∫ π

−π

et cos mtdt =

2(−1)m sinhπ

1 + m2

,

m ∈ N0,

∫ π

π

et sin mtdt =

2m(−1)m+1 sinhπ

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 without vertical half tangents, so f ∈ K∗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) ,

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Examples of Fourier series

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)}

]

=

sinhπ

π

· i(−1)n

{

1

1 + i(n + 1)

−

1

1 + i(n − 1)

}

=

sinhπ

π

· (−1)n i · 1 + i(n − 1) − {1 + i(n + 1)}

1 − (n2 − 1) + i 2n

=

sinhπ

π

(−1)n · i ·

−2i

−(n2 − 2) + 2in =

sinhπ

π

· (−1)n · 2 · −(n

2 − 2) − 2in

(n2 − 2)2 + 4n2

=

sinhπ

π

· (−1)n · 4 − 2n

2 − 4in

n4 + 4

.

Fourier series and uniform convergence

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Examples of Fourier series

68

When we split into the real and the imaginary part we find

an =

sinhπ

π

· (−1)n · 4 − 2n

2

n4 + 4

,

n ≥ 0,

and

bn =

sinhπ

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 sinhπ

1 + (n + 1)2

−2(n − 1)(−1)

n sinhπ

1 + (n − 1)2

}

=

sinhπ

π

(−1)n (n+1)(n

2−2n+2)−(n−1)(n2+2n+2)

(n2+2−2n)(n2+2+2n)

=

sinhπ

π

· (−1)n · 4 − 2n

2

n4 + 4

,

and

bn =

1

π

∫ π

−π

et sin t sinnt dt =

1

2π

∫ π

−π

et{cos(n − 1)t − cos(n + 1)t} dt

=

1

2π

{

2(−1)n−1 sinhπ

1 + (n − 1)2 −

2(−1)n−1 sinhπ

1 + (n + 1)2

}

=

sinhπ

π

· (−1)n−1 · n

2+2n+2−(n2−2n+2)

(n2−2n+2)(n2+2n+2)

=

sinhπ

π

· (−1)n−1 ·

4n

n4 + 4

.

In both cases we see that a0 =

sinhπ

π

, hence we get the Fourier series (with equality by the remarks

above)

f(t) =

sinhπ

π

{

1

2

+

∞∑

n=1

(

(−1)n(4 − 2n2)

n4 + 4

cos nt +

(−1)n−14n

n4 + 4

sinnt

)}

.

2) The Fourier series has the convergent majoring series

sinhπ

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 =

sinhπ

π

{

1

2

− 2

∞∑

n=1

n2 − 2

n4 + 4

}

,

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Examples of Fourier series

69

hence by a rearrangement

∞∑

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)

The function f is continuous and piecewise C1 without vertical half tangents, so f ∈ K∗2π. The Fourier

series is by the main theorem pointwise convergent with the sum function f ∗(t) = f(t).

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Examples of Fourier series

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.

In the exceptional case n = 1 we get instead

a1 =

2

π

∫ π/2

0

cos2 tdt =

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.

The Fourier series has the convergent majoring series

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.

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Examples of Fourier series

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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Examples of Fourier series

72

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

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Examples of Fourier series

73

We obtain by some partial integrations,

bn =

2

π

∫ π

0

(π2 − t3)sinnt dt = − 2

πn

[

t(π2 − t2)cos nt]π

0

+

2

πn

∫ π

0

(π2 − 3t2)cos nt dt

=

2

πn2

[(

π2 − 3t2) sinnt]π

0

+

12

πn2

∫ π

0

t sin nt dt =

12

πn3

[−t cos nt]π0 +

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

sinnt.

The Fourier series has the convergent majoring series

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 − π

2

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 .

Then by a rearrangement,

∞∑

n=1

(−1)n+1

(2n − 1)3 =

π3

32

.

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Examples of Fourier series

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) Prove that f has the Fourier series

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.

2) Find the sum of the series

∞∑

n=0

(−1)n

(2n − 1)(2n + 1)(2n + 3) .

The function f is continuous and piecewise C1 without vertical half tangents, so f ∈ K∗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.

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Examples of Fourier series

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.

The Fourier series has the convergent majoring series

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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2009

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Examples of Fourier series

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

.

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Examples of Fourier series

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.

The function f is continuous and piecewise C1 without vertical half tangents, so f ∈ K∗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 4tdt =

[

− 1

4π

cos 4t

]π/2

0

= 0.

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Examples of Fourier series

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

{

2cos(4n−1)t

(4n−3)(4n−1) +

cos 4nt

(2n−1)(2n+1) +

2cos(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

{

8sin(4n − 1)t

(4n−3)(4n−1)(4n+1)

+

sin 4nt

(2n−1)n(2n+1) +

8sin(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

]

.

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Examples of Fourier series

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

−π

2

6

+

∞∑

n=1

cos(nx)

n2

.

4) Prove that g is that solution of the differential equation

d2y

dx2

− y = −f(x) − π

2

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.

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Examples of Fourier series

80

2) The series

∑∞

n=1

cos(nx)

n2 + 1

has the convergent majoring series

∞∑

n=1

1

n2 + 1

,

so it is also uniformly convergent.

3) Finally, the termwise differentiated series,

∑∞

n=1

− sin(nx)

n(n2 + 1)

has the convergent majoring series

∞∑

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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Examples of Fourier series

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) = −π

2

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

d2y

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) − π

2

6

,

and we have shown that they fulfil the differential equation.

Now,

g′(x) = −

∞∑

n=1

sinnx

n(n2 + 1)

,

so g′(0) = g′(π) = 0. It follows that g is a solution of the boundary value problem

(7)

d2y

dx2

− y = −f(x) − π

2

6

= −1

4

x2 +

π

2

x − π

2

6

,

with the boundary conditions y′(0) = y′(π) = 0 [notice, over half of the interval].

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Examples of Fourier series

82

The corresponding homogeneous equation without the boundary conditions has the complete so-

lution,

y = c1 coshx + c2 sinhx.

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

d2y

dx2

− y = 2a−ax2−bx−c = −ax2−bx+(2a−c).

This equal to

−f(x) − π

2

6

= −1

4

x2 +

π

2

x − π

2

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 coshx + c2 sinhx.

Since

y′ =

1

2

x − π

2

+ c1 sinhx + 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 sinhπ + c2 coshπ = c1 sinhπ +

π

2

coshπ = 0,

hence c1 = −π2 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 · coshx + π

2

sinhx =

∞∑

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

.

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Examples of Fourier series

83

Example 2.9 Let f ∈ K2π be given by

f(t) = |t|3

for − π < t ≤ π.

1) Sketch the graph of f in the interval [−π, 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

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Examples of Fourier series

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]π0 −

∫ π

0

sin nt dt

}

=

6π2

n2

(−1)n − 12

πn4

[cos nt]π0 = 6π

{

(−1)n

n2

+

2(1 − (−1)n)

π2n4

}

.

Then by insertion (remember 12 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.

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Examples of Fourier series

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 · π

2

6

+

12

π

∞∑

p=1

2

(2p − 1)4 (−1)

2p−1

=

π3

4

+ π3 − 24

π

∞∑

p=1

1

(2p − 1)4 ,

hence by a rearrangement,

∞∑

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 without vertical half tangents, so f ∈ K∗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,

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Examples of Fourier series

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

[sinnt]ππ/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,

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Examples of Fourier series

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.

Example 2.11 Let f ∈ K2π be given by

f(t) = e|t|

for − π < t ≤ π.

1) Sketch the graph of f and explain why f ∈ K∗2π.

2) Prove that the Fourier series for f is given by

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∞ without vertical half tangents, we have f ∈ K∗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)neπ − 1} + 2n

π

[

et sin nt

]π

0

− 2n

2

π

∫ π

0

et cos nt dt

=

2

π

{(−1)neπ − 1} + 0 − n2an,

so by a rearrangement,

an =

2

π

· (−1)

neπ − 1

n2 + 1

,

n ∈ N0,

specielt a0 =

2

π

{eπ − 1}.

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Examples of Fourier series

88

5

10

15

20

y

–4

–2

0

2

4

x

Alternatively it follows directly by complex calculations with cosnt = Re eint that

an =

2

π

∫ π

0

et cos nt dt =

2

π

Re

∫ π

0

e(1+in)tdt =

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)

neπ − 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.

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Examples of Fourier series

89

Example 2.12 Let f ∈ K2π be given by

f(t) = (t − π)2

for − π < t ≤ π.

1) Sketch the graph of f in the interval [−π, π].

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.

4) Prove that the Fourier series for f is given by

4π2

3

+ 4

∞∑

n=1

{

(−1)n

n2

cosnt +

(−1)nπ

n

sinnt

}

,

t ∈ R.

(1) and (2) 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 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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Examples of Fourier series

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

sinnt · (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)2dt = 1

π

[

(t − π)3

3

]π

−π

=

1

π

· 8π

3

3

=

8π2

3

,

and for n ∈ N,

bn =

1

π

∫ π

−π

(t − π)2 sinnt 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

[sinnt · (t − π)]π−π −

2

πn

∫ π

−π

sin nt dt =

4π

n

· (−1)n.

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Examples of Fourier series

91

Summing up we get with pointwise equality,

f∗(t) =

4π2

3

+ 4

∞∑

n=1

{

(−1)n

n2

cos nt +

(−1)nπ

n

sin nt

}

,

t ∈ R.

Example 2.13 The periodic function f : R → R of period 2π is defined by

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 f̃ 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 2tdt +

1

2π

∫ π

0

{cos 2t + 1}dt = 0 + 1

2π

{0 + π} = 1

2

.

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Examples of Fourier series

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

f̃(t) =

⎧⎪⎪⎪⎪⎪⎪⎨

⎪⎪⎪⎪⎪⎪⎩

sin t,

t ∈ ]− π, 0[,

1/2,

t = 0,

cos t,

t ∈ ]0,π[,

−1/2,

t = π,

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Examples of Fourier series

93

which is continued periodically.

Since f (and f̃) 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

f̃+(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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Examples of Fourier series

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.

Example 2.14 Let f ∈ K2π be given by

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.

4) Prove that the Fourier series for f is given by

2

3

− 1

π2

∞∑

n=1

{

1

n2

cos nt +

π

n

sinnt

}

,

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.

Fourier series and uniform convergence

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Examples of Fourier series

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

sinnt 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

sinnt

}

,

t ∈ R.

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Examples of Fourier series

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 = rneint = rn{cos nt + i sin nt}.

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Examples of Fourier series

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{cosnt + i sinnt}.

Here we take two times the imaginary part,

2

∞∑

n=1

rn sinnt = 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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Examples of Fourier series

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 sinnt},

where

an =

1

π

∫ 2π

0

fr(t)cos nt dt,

bn =

1

π

∫ 2π

0

fr(t)sinnt 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 = −2πr

5

5

,

∫ π

−π

fr(t) sin(5t)dt = πb5 = 0.

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Examples of Fourier series

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 · r

5

5

∫ 2π

0

cos2(5t)dt = −2πr

5

5

,

∫ π

−π

fr(t) sin(5t)dt = −2

∞∑

n=1

rn

n

∫ π

−π

cos nt · sin 5tdt = 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) = −12 ln(1 + r

2 − 2r cos t),

we get by choosing r =

1

2

and t = 0 that

∞∑

n=1

1

n2n

= −1

2

f1/2(0) = −12 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) = −12 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+1rn

n

= ln(1 + r),

r ∈ ]− 1, 1[,

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Examples of Fourier series

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 sinnt.

If 0 < r < 1, then 2

∑∞

n=1 r

n =

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 sinnt =

d

dt

ln(1+r2−2r cos t) =

2t sin t

1+r2−2r cos t .

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Examples of Fourier series

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)2dt =

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 without vertical half tangents, we have f ∈ K∗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.

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