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Fourier Series
7
...
Then, for any f, g E C[a, b] we have that the product j g is also continuous
on [a, b] and hence integrable on [a, b]
...
The inner product ( , ) is defined on
C[a, b] by
b
(j, g)
=
l j x g x dx
( ) ( )
The three properties of an inner product are satisfied because
b
b
(1) (f, f) J f(x)f(x) d x 2: 0 for all f E C[a, b] and (j, f) J f(x)f(x) dx 0 if
and only if f (x)
0 for all x E [a, b]
...
One interesting consequence is that the norm of a function f with respect to this
(
inner product is
11!11
=
b
l
) 1/2
j2(x) dx
Intuitively, this is quite satisfactory as a measure of how far the function is from the
zero function
...
Fourier Series
Let CP2rr denote the space of continuous real-valued functions of a real variable that
are periodic with period 2n
...
Examples
cos x, sin x, cos 2x, sin 3x, etc
...
However, its "fundamentai (smallest) period" is n
...
5
...
y
7r
2
Figure 7
...
5
A continuous periodic function
...
, cos nx, sin nx,
...
We formulate the questions and ideas as follows
...
)
(i) For any n, the set of functions { l, cos x, sin x, cos 2x, sin 2x,
...
This subspace will be denoted CP2rr,n·
(ii) Given an arbitrary function f in CP2rr, how well can it be approximated by a
function in CP2rr,n? We expect from our experience with distance and subspaces
that the closest approximation to f in CP2rr,n is projcP:z
...
The coefficients
for Fourier's representation off by a linear combination of { 1, cos x, sin x,
...
}, called Fourier coefficients, are found by considering this
projection
...
Since the distance
fromf to the n-th approximation prokP:z
...
n /II, to test if the ap
proximation improves, we must examine whether II perpCP:z
...
Let us consider these statements in more detail
...
, cosnx, sinnx} is orthogonal
...
nf in CP2rr,n to an
...
2 and 7
...
That is, we use the projection formula, given an orthogonal basis {v1,
...
vk
proJs 1 1iJ v 1 +
+
2
···
llvkll
=
There is a standard way to label the coefficients of this linear combination:
prokp2"
...
Thus, we have
rr
(f, 1)
1
ao = li = ;
rr
f(x) dx
(f, cos mx) 1
= am =
7r
11 cos mxll2
(f, sin mx)
1
bm -2
7r
II sin mxll
_
_
(iii) Is projcPin,, fequal to fin the limit
rr
I-rr
rr f( )
...
The
question being asked is a question about the convergence of series-and in fact, about
series of functions
...
(The short answer is "yes, the series converges to f provided
that f is continuous
...
) Questions about convergence are important in physical and
engineering applications
...
Solution: We have
rr lxl dx
I-rr
1 rr
I-rr
rr lxl 2x dx
I-rrrr
1
lxl
3x dx
I-rrrr
1
lxl x dx
I-rr
1 rr
lxl
I
rr 3x dx
I
l
ao = a,= rr
a1= -
cos
7r
7r
=
0
4
cos
bi = -
sin
b1 = -
sin2xdx = 0
7r
= --
9rr
=0
- rr
I
b3 = -
�
rr and
4
a3 = 7r
Hence, prokPin
...
5
...
�
lxlcos xdx = --
1
7r
x
= rr
7r
7r
�
lxl sin
= 0
-rr
- ; cos x -
t,; cos 3x
...
5
...
1 fx
( )
-
Y
=
( )
projCP:u,,3 fx
7r
t
Graphs of projCP,,,, f and prokP,, , f compared to the graph of f(x)
...
J f= ; sinx
-
1
7r
1
7r
1
7r
1
7r
1
7r
1
7r
1
rr
-
x
lf
Jlf
Jlf
Jlf
Jlf
Jlf
Jlf
J
if
-
7r � x �
/2
if -rr/2 < x � rr/2
ifrr/2 < x � rr
fdx = 0
-;r
fcosxdx = 0
-;r
fcos 2xdx = 0
-;r
fcos 3xdx = 0
-;r
-;r
fsinxdx =
4
7r
fsin 2xdx = 0
-;r
-;r
fsin 3xdx = -
4
9rr
-
tr sin 3x
...
5
...
y
1r
2
-
--
-�
Figure 7
...
7
Graphs of prokPi
...
1 f(x)
Y projCPin,J f(x)
=
f and projCPi
...
...
5
Computer Problems
Cl Use a computer to calculate projCPin,,, f for
n =3,7, and 11 for each of the following functions
...
(a)
f(x) =x2,
-;r
�
x
�
f(x) =
1
-
x
n �
if
-
ifO
:5
7r
n :5
<
x
x
�
:5
7r
o
7r
PROBLEMS 7 5
...
1)