FOURIER SERIES PART II: CONVERGENCE
[Pages:19]FOURIER SERIES PART II: CONVERGENCE
We have seen in the previous note how to associate to a 2-periodic function f
a Fourier series
a0 2
+
an cos(nx)
+ bn sin(nx)
.
n=1
Now we are going to investigate how the Fourier series represents f . Let us first
introduce the following notation. For N = 0, 1, 2, ? ? ? , we denote by SN f (x) the N -th partial sum of the Fourier series of f . That is,
Hence
SN f (x)
=
a0 2
+
N an
cos(nx)
+
bn
sin(nx)
.
n=1
S0f (x) S1f (x) S2f (x)
= a0 ;
= =
a20 a20 2
+ +
a1 a1
cos x cos x
+ +
b1 b1
sin x sin x
; +
a2
cos 2x
+
b2
sin 2x
...
The infinite series is therefore limN SN f . The Fourier series converges at a point x if limN SN f (x) exists.
We consider the functions and their Fourier series of examples 1, 2, and 3 of the
previous note and see how the graphs of partial sums SN f compare to those of f .
1. Examples
Example 1. For f (x) = |x| on [-, ], we found
4 cos(2j + 1)x
|x| - 2
(2j + 1)2
j=0
Thus,
S0f (x) S1f (x) S3f (x) S5f (x)
=;
2 4 cos x = 2 - 4 cos x ; 4 cos 3x = 2 - 4 cos x - 4 c9os 3x ; 4 cos 5x = 2 - - 9 - 25
It appears that as N gets larger, the graph of SN f gets closer to that of f .
Date: February 24, 2016. 1
2
FOURIER SERIES PART II: CONVERGENCE
Sf
0
Sf
1
Sf
Sf
3
9
Figure 1. Graphs of SN f for N = 0, 1, 3, 9.
Examp{le 2. For the 2-periodic function f of example 2 defined by
f (x) =
1 -1
if 0 < x < ; if - < x < 0
, we found the Fourier series
4 sin(2j + 1)x
f (x)
.
(2j + 1)
j=0
Thus,
S1f (x) S3f (x) S5f (x) S7f (x)
4 sin x
=
;
4 sin x 4 sin 3x
=
+
;
3
4 sin x 4 sin 3x 4 sin 5x
=
+
+
4 sin x 4 s3in3x 4 s5in5x 4 sin 7x
=
+
+
+
3
5
7
Again it appears that as N increases SN f gets closer to f at the points where f is continuous.
Examp{le 3. For the 2-periodic function f of example 3 defined by
f (x) =
x 0
if 0 < x < ; if - < x < 0 .
with Fourier series
2 cos(2j + 1)x (-1)j-1 sin jx
4-
(2j + 1)2 +
. j
j=0
j=1
FOURIER SERIES PART II:
Sf
1
CONVERGENCE
3
Sf
3
Sf
Sf
5
7
Sf
9
Sf
27
Figure 2. Graphs of SN f for N = 1, 3, 5, 7, 9, 27.
The first partial sums are
S0f (x) S1f (x) S2f (x) S3f (x)
=
4 2 cos x
= 4 - + sin x ;
2 cos x
sin 2x
=-
+ sin x -
;
4
2
2 cos x
sin 2x 2 cos 3x sin 3x
=-
+ sin x -
-
+
.
4
2
9
3
2. Pointwise Convergence of Fourier series
The above examples suggest that the N -th partial sums SN f converge to f . This is indeed the case at each point where f is continuous. At each discontinuity, the
partial sums approach the average value of f . To be precise, we define the average
of f at a point x0 as
(
)
fav (x0 )
=
f (x+0 ) + f (x-0 ) 2
=
1 2
lim f (x) + lim f (x)
xx+ 0
xx- 0
.
Hence if f is continuous at x0, then fav(x0) = f (x0){. For example for the 2x if 0 < x < ;
periodic function f of example 3 defined by f (x) = 0 if - < x < 0 . we
4
FOURIER SERIES PART II: CONVERGENCE
Sf
0
Sf
1
Sf
2
Sf
3
Sf
10
Sf
15
Figure 3. Graphs of SN f for N = 0, 1, 2, 3, 10, 15.
have fav(x) = f (x) for x = (2k + 1) (with k Z) and
f ((2k + 1)+) + f ((2k + 1)-)
fav((2k + 1)) =
2
= 2
The graph of fav is the following
k = ?1, ?2, ?3, ? ? ?
Figure 4. Graphs of fav.
We have the following theorem.
Theorem (Pointwise convergence) Let f Cp1(R) be 2-periodic. Then the Fourier series of f converges to fav at each point of R. That is,
fav (x)
=
a0 2
+
an cos nx +
bn sin nx
,
n=0
FOURIER SERIES PART II:
CONVERGENCE
5
where
1
a0 =
f (x)dx , and
-
an bn
} =
1
{
f (x)
-
cos nx sin nx
dx
Again this means that at all points x where f is continuous, we have
f (x)
=
a0 2
+
an
cos nx
+
bn
sin nx
,
n=0
and at the points x0 where f is discontinuous we have
f (x+0 ) + f (x-0 ) 2
=
a0 2
+
an cos nx0
+ bn
sin nx0
.
n=0
To prove this theorem, we will need two lemmas
Lemma 1. (Riemann-Lebesgue Lemma) If f is piecewise smooth on an interval [a, b], then
b lim f (x) cos(rx)dx = 0 r ab lim f (x) sin(rx)dx = 0
r a
Proof. Since f is piecewise smooth, then there are finitely many points
c0 = a < c1 < c2 < ? ? ? < cn-1 < b = cn
such that both f and its derivative f are continuous in each interval (cj-1, cj) (j = 1, ? ? ? , n). Furthermore, f (c?k ) and f (c?k ) are finite. Thus the integrals of f and f exist in each subinterval. We have,
b
c1
cn
f (x) cos(rx)dx = f (x) cos(rx)dx + ? ? ? +
f (x) cos(rx)dx
a
c0
n
cj
cn-1
=
f (x) cos(rx)dx
j=1 cj-1
We use integration by parts in each subinterval [cj-1, cj] to obtain
cj
( f (x) sin(rx) )cj
f (x) cos(rx)dx =
cj -
f (x) sin(rx) dx
cj-1
r
cj-1
cj-1
r
( we are assuming that r > 0). Let M > 0 such that
sup |f (x)| < M and sup |f (x)| < M .
a ................
................
In order to avoid copyright disputes, this page is only a partial summary.
To fulfill the demand for quickly locating and searching documents.
It is intelligent file search solution for home and business.
Related download
- euler s formula and trigonometry
- trigonometry laws and identities
- fourier series part ii convergence
- formulas to know
- chapter 4 oscillatory motion
- section 7 2 advanced integration techniques trigonometric
- trigonometric integrals solutions
- section 7 3 some trigonometric integrals
- assignment previewer https v4cgi
- double angle power reducing and half angle formulas
Related searches
- convergence divergence calculator
- fourier transform infrared spectroscopy pdf
- determine convergence calculator
- series convergence calculator with steps
- interval of convergence calculator with steps
- sequence convergence calculator
- fourier transform infrared ftir spectroscopy
- integral test for convergence khan
- integral test convergence calculator
- convergence of integral calculator
- integral convergence divergence calculator
- interval of convergence calculator