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Title: Fourier Series Tutorial
Description: Well comprehensive notes on Fourier Series Tutorial
Description: Well comprehensive notes on Fourier Series Tutorial
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Series
FOURIER SERIES
Graham S McDonald
A self-contained Tutorial Module for learning
the technique of Fourier series analysis
● Table of contents
● Begin Tutorial
c 2004 g
...
mcdonald@salford
...
uk
Table of contents
1
...
3
...
5
...
7
...
Theory
● A graph of periodic function f (x) that has period L exhibits the
same pattern every L units along the x-axis, so that f (x + L) = f (x)
for every value of x
...
For example, adding
c2 sin(2kx + α2 ) = a2 cos(2kx) + b2 sin(2kx)
c3 sin(3kx + α3 ) = a3 cos(3kx) + b3 sin(3kx)
(the 2nd harmonic)
(the 3rd harmonic)
Here, symbols with subscripts are constants that determine the amplitude and phase of each harmonic contribution
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Section 1: Theory
5
One can even approximate a square-wave pattern with a suitable sum
that involves a fundamental sine-wave plus a combination of harmonics of this fundamental frequency
...
Their fundamental frequency is then
k = 2π
L = 1, and their Fourier series representations involve terms like
a1 cos x ,
b1 sin x
a2 cos 2x ,
b2 sin 2x
a3 cos 3x ,
b3 sin 3x
We also include a constant term a0 /2 in the Fourier series
...
With a sufficient number of harmonics included, our approximate series can exactly represent a given function f (x)
f (x) = a0 /2
Toc
+ a1 cos x + a2 cos 2x + a3 cos 3x +
...
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Section 1: Theory
7
A more compact way of writing the Fourier series of a function f (x),
with period 2π, uses the variable subscript n = 1, 2, 3,
...
This process is broken down into three steps
STEP ONE
a0
=
1
π
=
1
π
=
1
π
Z
f (x) dx
2π
STEP TWO
an
Z
f (x) cos nx dx
2π
STEP THREE
bn
Z
f (x) sin nx dx
2π
where integrations are over a single interval in x of L = 2π
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Section 1: Theory
8
● Finally, specifying a particular value of x = x1 in a Fourier series,
gives a series of constants that should equal f (x1 )
...
Exercises
Click on Exercise links for full worked solutions (7 exercises in total)
...
Let f (x) be a function of period 2π such that
1, −π < x < 0
f (x) =
0, 0 < x < π
...
2 π
3
5
c) By giving an appropriate value to x, show that
π
1 1 1
= 1 − + − +
...
Let f (x) be a function of period 2π such that
0, −π < x < 0
f (x) =
x, 0 < x < π
...
4
π
3
5
1
1
+ sin x − sin 2x + sin 3x −
...
2
and (ii) π8 = 1 + 312 + 512 + 712 +
...
Let f (x) be a function of period 2π such that
x, 0 < x < π
f (x) =
π, π < x < 2π
...
4
π
3
5
1
1
− sin x + sin 2x + sin 3x +
...
and (ii)
π2
8
= 1+
1
32
+
1
52
+
1
72
+
...
Let f (x) be a function of period 2π such that
f (x) =
x
over the interval 0 < x < 2π
...
2
2
3
c) By giving an appropriate value to x, show that
π
1 1 1 1
= 1 − + − + −
...
Let f (x) be a function of period 2π such that
π − x, 0 < x < π
f (x) =
0,
π < x < 2π
a) Sketch a graph of f (x) in the interval −2π < x < 2π
b) Show that the Fourier series for f (x) in the interval 0 < x < 2π is
π
1
2
1
+
cos x + 2 cos 3x + 2 cos 5x +
...
2
3
4
c) By giving an appropriate value to x, show that
π2
1
1
= 1 + 2 + 2 +
...
Let f (x) be a function of period 2π such that
f (x) = x in the range − π < x < π
...
2
3
c) By giving an appropriate value to x, show that
π
1 1 1
= 1 − + − +
...
Let f (x) be a function of period 2π such that
f (x) = x2 over the interval − π < x < π
...
3
2
3
c) By giving an appropriate value to x, show that
π2
1
1
1
= 1 + 2 + 2 + 2 +
...
Answers
The sketches asked for in part (a) of each exercise are given within
the full worked solutions – click on the Exercise links to see these
solutions
The answers below are suggested values of x to get the series of
constants quoted in part (c) of each exercise
1
...
(i) x =
3
...
x =
π
2,
π
2,
(ii) x = 0,
(ii) x = 0,
π
2,
5
...
x =
π
2,
7
...
Toc
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Section 4: Integrals
17
4
...
Useful trig results
When calculating the Fourier coefficients an and bn , for which n =
1, 2, 3,
...
results are useful
...
, can be deduced from
the graph of sin x or that of cos x
1
● sin nπ = 0
s in (x )
x
−3π
−2π
−π
0
π
2π
3π
π
2π
3π
−1
1
● cos nπ = (−1)n
c o s(x )
x
−3π
−2π
−π
0
−1
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Section 5: Useful trig results
1
20
s in (x )
1
c o s(x )
x
−3π
−π
−2π
0
π
3π
2π
x
−3π
−2π
−π
−1
π
● sin n =
2
π
0
3π
2π
−1
0 , n even
0 , n odd
π
1 , n = 1, 5, 9,
...
2
−1 , n = 3, 7, 11,
...
Areas cancel when
when integrating
overR whole periods
●
sin nx dx = 0
2π
R
●
cos nx dx = 0
1
+
−3π
−2π
s in (x )
+
+
−π
0
π
−1
2π
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2π
x
3π
Section 6: Alternative notation
21
6
...
Tips on using solutions
● When looking at the THEORY, ANSWERS, INTEGRALS, TRIG
or NOTATION pages, use the Back button (at the bottom of the
page) to return to the exercises
● Use the solutions intelligently
...
1, −π < x < 0
f (x) =
0, 0 < x < π, and has period 2π
a) Sketch a graph of f (x) in the interval −2π < x < 2π
f(x )
1
−2π
Toc
−π
JJ
0
II
2π
π
J
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Solutions to exercises
26
b) Fourier series representation of f (x)
STEP ONE
1
a0 =
π
Z
1 π
f (x)dx =
f (x)dx +
f (x)dx
π 0
−π
−π
Z
Z
1 0
1 π
=
1 · dx +
0 · dx
π −π
π 0
Z
1 0
=
dx
π −π
1 0
=
[x]
π −π
1
=
(0 − (−π))
π
1
=
· (π)
π
i
...
a0 = 1
...
e
...
nπ
II
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Solutions to exercises
28
STEP THREE
bn
1
π
=
1
π
=
1
π
=
i
...
bn
=
=
=
i
...
bn
=
Z
π
f (x) sin nx dx
−π
Z 0
f (x) sin nx dx +
−π
Z 0
1 · sin nx dx +
−π
1
π
1
π
Z
π
Z
f (x) sin nx dx
0
π
0 · sin nx dx
0
0
1 − cos nx
sin nx dx =
π
n
−π
−π
1
1
− [cos nx]0−π = −
(cos 0 − cos(−nπ))
nπ
nπ
1
1
− (1 − cos nπ) = −
(1 − (−1)n ) , see Trig
nπ
nπ
0
, n even
1 , n even
n
,
since
(−1)
=
2
−1 , n odd
− nπ
, n odd
1
π
Toc
Z
0
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Solutions to exercises
29
We now have that
∞
a0 X
f (x) =
+
[an cos nx + bn sin nx]
2
n=1
with the three steps giving
a0 = 1, an = 0 , and bn =
0
2
− nπ
, n even
, n odd
It may be helpful to construct a table of values of bn
n
bn
1
− π2
2
3
0 − π2 13
4
5
0 − π2 15
Substituting our results now gives the required series
1
1
2
1
f (x) = −
sin x + sin 3x + sin 5x +
...
Comparing this series with
1
2
1
1
f (x) = −
sin x + sin 3x + sin 5x +
...
So we need sin x = 1, sin 3x = −1, sin 5x = 1, sin 7x = −1, etc
The first condition of sin x = 1 suggests trying x =
This choice gives
i
...
sin π2
1
+
−
1
3
sin 3 π2
1
3
+
+
1
5
π
2
...
Toc
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1
7
sin 7 π2
1
7
Solutions to exercises
Picking x =
31
π
2
thus gives
h
0 = 21 − π2 sin π2 +
i
...
0 =
1
2
−
2
π
h
1
−
1
3
sin 3π
2 +
1
5
sin 5π
2
+
1
7
sin 7π
2 +
...
i
i
A little manipulation then gives a series representation of π4
2
1 1 1
1
1 − + − +
...
3 5 7
4
Return to Exercise 1
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Solutions to exercises
Exercise 2
...
e
...
2
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Solutions to exercises
34
STEP TWO
an =
1
π
Z
π
f (x) cos nx dx =
−π
1
π
1
π
Z
0
f (x) cos nx dx +
−π
Z 0
1
π
Z
Z
π
f (x) cos nx dx
0
π
1
0 · cos nx dx +
x cos nx dx
π
−π
0
π Z π
Z π
1
sin nx
sin nx
1
x cos nx dx =
x
−
dx
i
...
an =
π 0
π
n
n
0
0
=
(using integration by parts)
1
sin nπ
1 h cos nx iπ
π
−0 −
−
π
n
n
n
0
1
1
(
0 − 0) + 2 [cos nx]π0
π
n
1
1
{cos nπ − cos 0} =
{(−1)n − 1}
2
2
πn
πn
0
, n even
, see Trig
...
e
...
e
...
e
...
e
...
e
...
5
1
1
+ sin x − sin 2x + sin 3x −
...
π
3
5
1
1
+ sin x − sin 2x + sin 3x −
...
Comparing this series with
π
2
1
1
f (x) =
−
cos x + 2 cos 3x + 2 cos 5x +
...
,
2
3
the required series of constants does not involve terms like 312 , 512 , 712 ,
...
The Trig section shows that cos n π2 = 0 when n is odd, and note also
that cos nx terms in the Fourier series all have odd n
i
...
cos x = cos 3x = cos 5x =
...
e
...
= 0
Toc
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when x =
I
π
2,
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Solutions to exercises
39
Setting x = π2 in the series for f (x) gives
π
π
2
π
1
1
3π
5π
f
=
−
cos + 2 cos
+ 2 cos
+
...
+ sin − sin
2
2
2
3
2
4
2
5
2
π
2
=
− [0 + 0 + 0 +
...
+ 1 − sin
2 | {z } 3
4 | {z } 5
=0
=0
The graph of f (x) shows that f
π
2
π
i
...
4
Toc
JJ
=
=
π
2
=
π
2,
so that
π
1 1 1
+ 1 − + − +
...
3 5 7
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Solutions to exercises
40
Pick an appropriate value of x, to show that
(ii)
π2
8
=1+
1
32
+
1
52
+
1
72
+
...
π
3
5
1
1
+ sin x − sin 2x + sin 3x −
...
e
...
4
π
3
5
7
sin 0 sin 0
+ sin 0 −
+
−
...
+ 0 − 0 + 0 −
...
=
π
3
5
7
1
1
1
and
1 + 2 + 2 + 2 +
...
8
Return to Exercise 2
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Solutions to exercises
42
Exercise 3
...
e
...
2
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Solutions to exercises
44
STEP TWO
an
=
=
=
Z
1
π
Z
1
π
"
f (x) cos nx dx
0
π
0
1
x cos nx dx +
π
sin nx
x
n
|
=
2π
1
π
π
Z
−
0
0
{z
π
Z
2π
π · cos nx dx
π
#
2π
sin nx
π sin nx
dx +
n
π
n
π
}
using integration by parts
"
π #
− cos nx
1 1
π sin nπ − 0 · sin n0 −
π n
n2
0
+
Toc
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1
(sin n2π − sin nπ)
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Solutions to exercises
i
...
an
45
=
"
#
1 1
cos nπ cos 0
1
0−0 +
0−0
− 2
+
π n
n2
n
n
=
1
(cos nπ − 1), see Trig
n2 π
=
1
(−1)n − 1 ,
2
n π
(
− n22 π
, n odd
0
, n even
...
e
...
e
...
e
...
n
We now have
∞
f (x) =
where a0 =
3π
2 ,
a0 X
+
[an cos nx + bn sin nx]
2
n=1
(
0
, n even
an =
,
bn = − n1
− n22 π , n odd
Constructing a table of values gives
n
an
bn
Toc
1
− π2
−1
JJ
2
0
− 21
3
− π2 312
− 13
II
J
4
0
− 14
5
− π2 512
− 15
I
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Solutions to exercises
48
This table of coefficients gives
f (x) =
1
2
i
...
f (x) =
3π
2
3π
4
+
−
+
h
−1
sin x +
−
−
2
π
2
π
h
cos x + 0 · cos 2x +
1
32
cos 3x +
...
i
h
cos x +
1
32
cos 3x +
1
52
cos 5x +
...
i
and we have found the required series
...
Compare this series with
3π
2
1
1
f (x) =
−
cos x + 2 cos 3x + 2 cos 5x +
...
2
3
Here, we want to set the cos nx terms to zero (since their coefficients
are 1, 312 , 512 ,
...
Note also that f ( π2 ) = π2
This gives
π
2
=
3π
4
−
−
2
π
cos π2 +
1
32
sin π2 +
1
2
Toc
JJ
cos 3 π2 +
sin 2 π2 +
II
1
3
1
52
cos 5 π2 +
...
Back
Solutions to exercises
50
and
π
2
=
3π
4
2
π
−
[0 + 0 + 0 +
...
then
π
2
=
3π
4
− 1−
1
3
+
1
5
−
1
7
1−
1
3
+
1
5
−
1
7
+
...
=
π
4,
(ii)
π2
8
+
...
To show that
=1+
1
32
+
1
52
+
1
72
+
...
= 1 + 2 + 2 + 2 +
...
e
...
The graph of f (x) shows a discontinuity (a “vertical jump”) at x = 0
The Fourier series converges to a value that is half-way between the
two values of f (x) around this discontinuity
...
e
...
4
π
3
5
7
1
1
− sin 0 + sin 0 + sin 0 +
...
− [0 + 0 + 0 +
...
...
32
52
72
1
1
1
1 + 2 + 2 + 2 +
...
f (x) = x2 , over the interval 0 < x < 2π and has period 2π
a) Sketch a graph of f (x) in the interval 0 < x < 4π
π
f(x )
0
Toc
JJ
3π
2π
π
II
J
I
4π
Back
x
Solutions to exercises
54
b) Fourier series representation of f (x)
STEP ONE
a0
i
...
a0
Toc
JJ
1
π
Z
Z
=
1
π
=
1
π
=
1
π
=
2π
f (x) dx
0
2π
x
dx
2
0
2 2π
x
4 0
(2π)2
−0
4
= π
...
e
...
JJ
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Solutions to exercises
56
STEP THREE
bn
=
=
=
1
π
Z
2π
f (x) sin nx dx =
0
1
π
Z
0
2π
x
2
sin nx dx
2π
1
2π
Z
1
2π
(
2π Z 2π
)
− cos nx
− cos nx
x
−
dx
n
n
0
0
|
{z
}
1
2π
(
x sin nx dx
0
using integration by parts
=
−2π
cos(n2π)
2πn
1
= − cos(2nπ)
n
1
= −
, since 2n is even (see Trig)
n
Toc
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=
i
...
bn
)
1
1
(−2π cos n2π + 0) + · 0 , see Trig
n
n
Solutions to exercises
57
We now have
∞
a0 X
f (x) =
+
[an cos nx + bn sin nx]
2
n=1
where a0 = π, an = 0, bn = − n1
These Fourier coefficients give
f (x)
=
i
...
f (x)
=
Toc
∞
π X
1
+
0 − sin nx
2 n=1
n
π
1
1
− sin x + sin 2x + sin 3x +
...
and
π
=
4
π
=
4
1 1 1 1
1 − + − + −
...
e
...
=
3 5 7 9
Toc
−
II
π
1
1
− 1 + 0 − + 0 + + 0 −
...
2
3 5 7 9
π
4
π
...
π−x , 0
0
, π < x < 2π, and has period 2π
a) Sketch a graph of f (x) in the interval −2π < x < 2π
f(x )
π
−2π
Toc
−π
JJ
0
II
π
J
I
2π
x
Back
Solutions to exercises
60
b) Fourier series representation of f (x)
STEP ONE
a0
=
=
i
...
a0
Toc
1
π
Z
1
π
Z
f (x) dx
0
π
1
(π − x) dx +
π
0
π
1
πx − x2 + 0
2
0
2
π
π2 −
−0
2
=
1
π
=
1
π
=
π
...
e
...
e
...
e
...
n
Toc
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=
i
...
bn
1
π
1
π
h
Solutions to exercises
In summary, a0 =
π
2
63
and a table of other Fourier cofficients is
n
an =
2
πn2
bn =
1
n
∴ f (x) =
(when n is odd)
1
2
3
4
5
2
π
0
2 1
π 32
0
2 1
π 52
1
1
2
1
3
1
4
1
5
∞
X
a0
+
[an cos nx + bn sin nx]
2
n=1
π
2
2 1
2 1
+ cos x +
cos 3x +
cos 5x +
...
2
3
4
π
2
1
1
i
...
f (x) =
+
cos x + 2 cos 3x + 2 cos 5x +
...
2
3
4
Toc
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=
Solutions to exercises
64
π2
8
c) To show that
=1+
1
32
+
1
52
+
...
...
32
52
1
1
sin 0 + sin 0 + sin 0 +
...
+ 0
4
π
3
5
2
1
1
1 + 2 + 2 +
...
3
5
Return to Exercise 5
2
π
II
cos 0 +
J
I
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Solutions to exercises
65
Exercise 6
...
e
...
J
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Solutions to exercises
67
STEP TWO
an
=
=
=
Z
1 π
f (x) cos nx dx
π −π
Z π
1
x cos nx dx
π −π
(
π
)
Z π
1
sin nx
sin nx
x
−
dx
π
n
n
−π
−π
|
{z
}
using integration by parts
Z
1 1
1 π
(π sin nπ − (−π) sin(−nπ)) −
sin nx dx
π n
n −π
1 1
1
(0 − 0) − · 0 ,
π n
n
Z
since sin nπ = 0 and
sin nx dx = 0,
i
...
an
=
=
2π
i
...
an
=
0
...
e
...
n
JJ
II
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68
Solutions to exercises
69
We thus have
f (x)
=
∞
i
a0 X h
an cos nx + bn sin nx
+
2
n=1
with a0 = 0, an = 0, bn = − n2 (−1)n
and
n
1
2
bn
2 −1
3
2
3
Therefore
f (x)
i
...
f (x)
= b1 sin x + b2 sin 2x + b3 sin 3x +
...
2
3
and we have found the required Fourier series
...
Setting x = π2 gives f (x) = π2 and
π
π 1
2π 1
3π 1
4π 1
5π
= 2 sin − sin
+ sin
− sin
+ sin
−
...
e
...
3
5
7
1 1 1
2 1 − + − +
...
3 5 7
Return to Exercise 6
Toc
JJ
II
J
I
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Solutions to exercises
71
Exercise 7
...
e
...
3
J
I
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Solutions to exercises
73
STEP TWO
an
=
=
=
Z
1 π
f (x) cos nx dx
π −π
Z π
1
x2 cos nx dx
π −π
(
π
)
Z π
1
sin
nx
sin
nx
2x
x2
−
dx
π
n
n
−π
−π
|
{z
}
using integration by parts
=
=
=
)
Z
2 π
1 2
2
x sin nx dx
π sin nπ − π sin(−nπ) −
n
n −π
)
(
Z
1 1
2 π
x sin nx dx , see Trig
(0 − 0) −
π n
n −π
Z π
−2
x sin nx dx
nπ −π
1
π
(
Toc
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I
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Solutions to exercises
i
...
an
74
−2
nπ
(
π
)
Z π
− cos nx
− cos nx
x
−
dx
n
n
−π
−π
|
{z
}
=
−2
nπ
(
=
−2
nπ
=
−2
nπ
=
−2
nπ
=
using integration by parts again
Toc
)
Z
1
1 π
π
− [x cos nx]−π +
cos nx dx
n
n −π
)
(
1
1
−
π cos nπ − (−π) cos(−nπ) + · 0
n
n
(
)
1
−
π(−1)n + π(−1)n
n
(
)
−2π
(−1)n
n
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Solutions to exercises
i
...
an
75
(
=
−2
nπ
=
+4π
(−1)n
πn2
=
4
(−1)n
n2
(
i
...
an =
2π
−
(−1)n
n
4
n2
, n even
−4
n2
, n odd
...
e
...
e
...
Toc
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II
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Solutions to exercises
78
∞
a0 X
+
[an cos nx + bn sin nx]
2
n=1
(
4
, n even
2π 2
n2
a0 = 3 ,
an =
,
−4
, n odd
n2
∴ f (x) =
where
n
an
1
i
...
f (x) =
2
2π 2
3
1
2
−4(1) 4
1
22
3
−4
bn = 0
4
1
32
4
1
42
1
1
1
− 4 cos x − 2 cos 2x + 2 cos 3x − 2 cos 4x
...
]
π2
1
1
1
i
...
f (x) =
− 4 cos x − 2 cos 2x + 2 cos 3x − 2 cos 4x +
...
,
(
1 , n even
use the fact that cos nπ =
−1 , n odd
c) To show that
cos x −
1
22
cos 2x +
1
32
cos 3x −
1
42
cos 4x +
...
1
32
· (−1) −
i
...
i
...
(−1) −
i
...
−1 −
1
22
· (1) +
1
22
1
32
−
−
1
42
· (1) +
...
1
1
1
= −1 · 1 + 2 + 2 + 2 +
...
Setting x = π in the Fourier series thus gives
π2
1
1
1
2
π =
− 4 cos π − 2 cos 2π + 2 cos 3π − 2 cos 4π +
...
3
2
3
4
π2
1
1
1
2
+ 4 1 + 2 + 2 + 2 +
...
3
2
3
4
π2
1
1
1
i
...
= 1 + 2 + 2 + 2 +
Title: Fourier Series Tutorial
Description: Well comprehensive notes on Fourier Series Tutorial
Description: Well comprehensive notes on Fourier Series Tutorial