Chap
.
11
Fourier Series
,
Integrals
,
and Transforms
Sec
.
11
.
1
Fourier Series
Content
:
u2022
Fourier series (5) and their coefficients (6)
u2022
Calculation of Fourier coefficients by integration (Example 1)
u2022
Reason why (6) gives the Fourier coefficients (Theorem 1)
u2022
Great generality of Fourier series (Theorem 2)
Problem Set 11
.
1
.
Page 485
5
.
Linear combinations of periodic functions
.
Addition of periodic functions with the same period
p
gives a periodic function with that period
p
, and so does the multiplication of such a function by a
constant. Thus all functions of period
p
form an important example of a vector space. This is what you
are supposed to prove. Now, by assumption, given any periodic functions
f
and
g
that have period
p
, that
is,
f
ue0a2
x
u002b
p
ue0a3
u003d
f
ue0a2
x
ue0a3
,
g
ue0a2
x
u002b
p
ue0a3
u003d
g
ue0a2
x
ue0a3
,
(A)
form a linear combination of them, say,
h
u003d
af
u002b
bg
ue0a2
a
,
b
constant
ue0a3
and show that
h
is periodic with period
p
; that is, you must show that
h
ue0a2
x
u002b
p
ue0a3
u003d
h
ue0a2
x
ue0a3
. This follows by
calculating and using the definition of
h
and then (A):
h
ue0a2
x
u002b
p
ue0a3
u003d
af
ue0a2
x
u002b
p
ue0a3
u002b
bg
ue0a2
x
u002b
p
ue0a3
u003d
af
ue0a2
x
ue0a3
u002b
bg
ue0a2
x
ue0a3
u003d
h
ue0a2
x
ue0a3
.
13
.
Fourier series
.
In Probs. 1320 you first have to find a formula for the given function. Then you obtain
the Fourier coefficients from (6) by integration.
In Probs. 13 and 14 the integrands of (6) are constants times cosines or sines. In Probs. 1520 they
are
t
times cosine or sine. Such integrals are evaluated by integration by parts, just as in calculus. Of
course you have noticed that they depend on
n
, which gives factors 1/
n
, 1/
n
2
etc. in the coefficients; these
are essential with respect to convergence.
In Prob. 13 you have to integrate from
u2212
u03c0
/2 to
u03c0
/2 only. (Why?) From (6a) you obtain the constant
term of the series (which is the mean value of the given function over the interval from
u2212
u03c0
to
u03c0
ue0a3
. Indeed,
you obtain
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110
Fourier Analysis. Partial Differential Equations (PDEs)
Part C
a
0
u003d
1
2
u03c0
u222b
u2212
u03c0
/2
u03c0
/2
1
u2219
dx
u003d
1
2
.
Next you use (6b) to get
a
n
u003d
1
u03c0
u222b
u2212
u03c0
/2
u03c0
/2
1
u2219
cos
nx dx
u003d
1
n
u03c0
sin
nx
u2212
u03c0
/2
u03c0
/2
u003d
1
n
u03c0
sin
n
u03c0
2
u2212
sin
u2212
n
u03c0
2
u003d
2
n
u03c0
sin
n
u03c0
2
u003d
2/
n
n
u003d
1,5,9,
u22ef
0
n
u003d
2,4,6,
u22ef
u2212
2/
n
n
u003d
3,7,11,
u22ef
.
Hence the sequence of the cosine coefficients is
2
u03c0
,
0,
u2212
2
3
u03c0
,
0,
2
5
u03c0
,
0,
u22ef
.
Finally, show that the sine coefficients are zero,
b
n
u003d
1
u03c0
u222b
u2212
u03c0
/2
u03c0
/2
1
u2219
sin
nx dx
u003d
u2212
1
n
u03c0
cos
nx
u2212
u03c0
/2
u03c0
/2
u003d
1
n
u03c0
u2212
cos
n
u03c0
2
u002b
cos
u2212
n
u03c0
2
u003d
0.
Conclude that the Fourier series is a “Fourier cosine series”
f
ue0a2
x
ue0a3
u003d
1
2
u002b
2
u03c0
cos
x
u2212
1
3
cos 3
x
u002b
1
5
cos 5
x
u2212
u002b
u22ef
.
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 Spring '12
 M.lee
 Fourier Series, Cos, Discrete cosine transform

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