1 Fourier Cosine and Sine Series By Ng Tze Beng Consider the series 0 1 1 cos( ) 2 n n a a nx ------------- (C) and the series 1 sin( ) n n a nx . ------------- (S) for the case that the sequence ( a n ) is a non-negative sequence converging to 0. We investigate the convergence of the above series and when they do converge whether the series is the series of a Lebesgue integrable function. When they do converge to a Lebesgue integrable function, we investigate sufficient condition so that the series also converge in the L 1 norm. We recall the following definitions. Suppose f is a function Lebesgue integrable on (π, π). We assume that the function is periodic with period 2π, that is, f (x) = f (x+ 2π) whenever anyone of f (x) or f (x+ 2π) is defined and that f (π) = f (π). Note that f (π) or f (π) need not necessarily be defined and the restriction of f to the interval [π, π] need not necessarily be continuous at the end points. It is convenient to assume that f is defined for all values of x in [π, π] and by periodicity to all of R. We may need to define values of f appropriately where it is not defined in [π, π] and extend to R by periodicity. Then we have the following formula for the definition of the coefficients of a Fourier series of f: 1 ( )cos( ) n a f x nx dx , n = 0, 1, 2, …. --------- (1) 1 ( )sin( ) n b f x nx dx , n = 1, 2, …. --------- (2)
52
Embed
Fourier Cosine and Sine Series By Ng Tze ... - My Calculus Web · Fourier Cosine and Sine Series By Ng Tze Beng Consider the series 0 1 1 cos( ) 2 n n a a nx ----- (C) and the series
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
1
Fourier Cosine and Sine Series
By Ng Tze Beng
Consider the series
0
1
1cos( )
2n
n
a a nx
------------- (C)
and the series
1
sin( )n
n
a nx
. ------------- (S)
for the case that the sequence ( an) is a non-negative sequence converging to 0.
We investigate the convergence of the above series and when they do converge
whether the series is the series of a Lebesgue integrable function. When they
do converge to a Lebesgue integrable function, we investigate sufficient
condition so that the series also converge in the L1 norm.
We recall the following definitions. Suppose f is a function Lebesgue
integrable on (π, π). We assume that the function is periodic with period 2π,
that is, f (x) = f (x+ 2π) whenever anyone of f (x) or f (x+ 2π) is defined
and that f (π) = f (π). Note that f (π) or f (π) need not necessarily be defined
and the restriction of f to the interval [π, π] need not necessarily be
continuous at the end points. It is convenient to assume that f is defined for all
values of x in [π, π] and by periodicity to all of R. We may need to define
values of f appropriately where it is not defined in [π, π] and extend to R by
periodicity.
Then we have the following formula for the definition of the coefficients
of a Fourier series of f:
1( )cos( )na f x nx dx
, n = 0, 1, 2, …. --------- (1)
1( )sin( )nb f x nx dx
, n = 1, 2, …. --------- (2)
2
Consider the series
0
1
1cos( ) sin( )
2n n
n
a a nx b nx
. ------------------ (A)
When an and bn are given by (1) and (2), (A) is called a Fourier series of the
function f . When f is even, then bn = 0 for n ≥ 1 and the series (A) is just (C).
When f is odd, then an = 0 for n ≥ 0 and the series (A) is just (S).
Note that we assume that f is integrable in [π, π] so that (1) and (2) are
meaningfully defined. Thus (A) or (C) or (S) is a Fourier series if it is the
Fourier series of some integrable function f . (However for (2) to be defined it
is sufficient to have the integrability of f (x) sin(x) over [0, π] and we call (S)
the generalized Fourier sine series.)
Note that (A) may or may not converge and may not be the Fourier series of its
limiting function. And when (A) is a Fourier series, it may or may not converge
at all points. Indeed there exists a Lebesgue integrable function f whose
Fourier series diverges at every point.
If we assume nice convergence, we do have some positive result. This is
Theorem S below.
Theorem S. If the series (A) converges uniformly to a function f , then it is the
Fourier series of its sum function f. More is true, if (A) converges almost
everywhere to a function f and the n-th partial sums of (A) are absolutely
dominated by a Lebesgue integrable function, then (A) is the Fourier series of f.
More precisely the n-th partial sum converges to f in the L1 norm.
We note that in all two cases of Theorem S, f is Lebesgue integrable and the
series (A), by using either the consequence of uniform convergence or the
Lebesgue Dominated Convergence Theorem, can be shown to be the Fourier
series of f by the method of Theorem 11. The convergence to f in the L1
norm is a consequence of uniform convergence for the first case and in the other
of being absolutely dominated by a Lebesgue integrable function. We shall not
prove this result but only for the special case that this note is concerned with.
This note is concerned mainly with the special case of the two series (C) and (S)
when the coefficients ( an) is a sequence of non-negative terms and an 0.
3
1. The Main Results.
For the sine series (S) we have the following result giving a necessary and
sufficient condition for (S) to be a Fourier series.
Theorem 1. Suppose ( an) is a sequence of nonnegative terms, an = an – an+1
≥0 and an 0. Then the limit function or sum function of (S), g, is Lebesgue
integrable if and only if 1
n
n
a
n
. If 1
n
n
a
n
, then (S) is the Fourier series of
g and ( ) ( ) 0ns x g x dx
, where sn(x) is the n-th partial sum of the series (S),
that is, sn(x) converges to g in the L1 norm.
The situation with the cosine series is somewhat different. We state the result
as follows.
Theorem 2. Suppose the sequence (a0 ,a1, ) is convex and an 0. Then
(1) The cosine series (C) converges except possibly at x = 0 to a non-
negative Lebesgue integrable function f .
(2) The series (C) is the Fourier series of f.
(3) ( ) ( ) 0n x f x dx
, where n(x) is the Cesaro 1 or (C, 1) means of the
series (C) .
(4) ( ) ( ) 0nt x f x dx
if and only if 1
ln( )( )n n
a o or equivalently anln(n) 0.
Here tn(x) is the n-th partial sum of the series (C) .
We now elaborate on the terms in italic in Theorem 2..
Suppose ( an ) is a sequence and an = an – an+1 . Then (an ) is also a sequence.
The sequence ( an ) = (a0 ,a1, ) is said to be convex if 2 an = an – an+1 ≥
0 for all n ≥ 0. The Cesaro 1 or (C,1) means of the sequence is defined to be
1 0 1
1
1n ns s s
n
,
where 0
n
n k
k
s a
for n ≥ 0. The (C,1) means of the cosine series is then given
by
4
1 0 1
1( ) ( ) ( ) ( )
1n nx t x t x t x
n
,
where 0
1
1( ) cos( )
2
n
n k
k
t x a a kx
for n ≥ 1 and 0 0
1( )
2t x a .
If the series ( an ) = (a0 ,a1, ) is only decreasing and an 0, then we may not
have Lebesgue integrability of the sum function for the series (C) but improper
Riemann integrability.
Theorem 3. Suppose the sequence (a0 ,a1, ) is decreasing and an 0. Then
the cosine series (C) converges except possibly at x = 0 to a function f on [π,
π], which is continuous at x for x ≠ 0 in [π, π] . The sum function f is, in
general, improperly Riemann integrable. Thus if we use improper integral in
the formula for the Fourier coefficients an , then (C) is the Fourier Riemann
series of f .
In the next section we collect together the useful technical results such as
summation techniques and properties of special sums for the proofs of these
three theorems.
2. Technical and Useful Results.
We recall first the Riemann Lebesgue Theorem:
Theorem R L. Suppose f is Lebesgue integrable on [-π, π]. Then
( )cos( ) 0limn
f x nx dx
and ( )sin( ) 0limn
f x nx dx
.
In view of Theorem RL, the condition that the sequence ( an ) be a null
sequence, that is, an 0, is a necessary condition in Theorem 1 and 2.
2.1 Summation formula
Summation technique features prominently in the proof. We use
predominantly Abel’s summation formula, which we describe below.
5
Abel’s Summation Formula.
Suppose (an ) and (bn) are two sequences. Let 1
n
n k
k
s b
. Then we have the
following summation formula:
1
1
1 1
( )n n
k k k k k n n
k k
a b a a s a s
1 1
1
( )n
k k k n n
k
a a s a s
----------------------------- (3)
For the truncated sum we have:
1
1( ) ' 'q q
k k k k k q q
k p k p
a b a a s a s
, ------------- (4)
where 'k
k j
j p
s b
, k ≥ p.
We have similar formula when the summation starts from 0 instead of 1. We
interpret formula (3) and (4) accordingly. Formula (3) is sometimes called
summation by parts.
Formula (3) or (4) is used to give an alternative useful way to sum the series (C)
or (S). And we have the following estimate of the sum particularly useful in
showing convergence of Fourier series.
Lemma 4. Suppose ( an ) is a decreasing sequence and an ≥ 0 for all n. Then
11
1
max | |n
k k kk n
k
a b a s
------------------- (5)
and max | ' |q
k k p kp k q
k p
a b a s
. ------------------ (6)
Proof.
By the summation formula (3), we have
1
1
1 1
n n
k k k k k n n
k k
a b a a s a s
6
1
11 1
1
max maxn
k k j n jj n j n
k
a a s a s
since ( an ) is a decreasing sequence and an ≥ 0 for all n.
1
1max | |j
j na s
.
Inequality (6) is derived from (4) in exactly the same way.
2.2 Properties of Convex Sequence
Recall that a sequence ( an ) = (a0 ,a1, ) is said to be convex if 2 an = an –
an+1 ≥ 0 for all n ≥ 0. For most of the time the sequence that we deal with is
usually convergent or a null sequence. Hence it is always bounded. For convex
sequence that is bounded we have :
Lemma 5. If ( an ) = (a0 ,a1, ) is convex and bounded, then it is
decreasing, i.e., an = an –an+1 ≥ 0 for all n ≥ 0.
Proof. By hypothesis, the sequence ( an ) is decreasing. Then we claim that
an ≥ 0 for all n ≥ 0. We show this by contradiction.
Suppose there is an integer N ≥ 0, such that aN = aN – aN+1 < 0. Then since
( an ) is decreasing, for all n ≥ N, an = an – an+1 ≤ aN < 0. Thus
aN+1 = aN aN , aN+2 = aN+1 aN+1 ≥ aN+1 aN ≥ aN 2 aN , ,
aN+p ≥ aN p aN . Since aN > 0, ( aN p aN ) is unbounded and so ( an )
is unbounded. This contradicts that ( an ) is bounded. Hence an ≥ 0 for all n
≥ 0.
More is true:
Lemma 6. If ( an ) = (a0 ,a1, ) is convex and bounded, then n an 0
and the series 2
0
( 1) n
n
n a
Converges to 0 lim n
na a
.
Proof. By Lemma 5, ( an ) is decreasing and bounded, and so by the Monotone
Convergence Theorem, ( an ) is convergent.
7
Observe that 0 1
0
n
k n
k
a a a
so that 0
k
k
a
is convergent and 0
0
limk nn
k
a a a
.
That is, 0
k
k
a
is a Cauchy series. Therefore, given any > 0, there exists an
integer N such that for all n ≥ N,
2
1
n
k
k n
a
.
Since ak ≥ 0 and 2 ak ≥ 0 for all k ≥ 0, for all n ≥ N,
2
2 2 2 1 2
1
n
n n n n n k
k n
n a a a a a a
.
Hence n a2n 0 . It follows that 2n a2n 0 .
Now 2 1 2 2(2 1) (2 1) 3n n nn a n a n a for n > 0 and since n a2n 0, by the
Comparison Test, (2n+1) a2n+1 0. Therefore, n an 0.
Let 0
n
n k
k
s a
for n ≥ 0. Then by Abel’s summation formula (3)
2
1
0
( 1) ( 1)n
n k n
k
s a k a n
.
Since (n+1) an+1 0 and (sn) is convergent, 2
0
( 1) n
n
n a
is convergent and
2
0
0
( 1) lim limn n nn n
n
n a s a a
.
2.3 Summing the sine and cosine series.
Consider the (n+1)-th partial sum of the cosine series (C),
0
1
1( ) cos( )
2
n
n k
k
t x a a kx
. ---------------------- (7)
Applying the Abel summation formula (3), we have
8
1
0
( ) ( ) ( )n
n k k n n
k
t x D x a a D x
, --------------------- (8)
where,
1
1( ) cos( )
2
n
n
k
D x kx
------------------------------- (9)
for n > 0 and D0(x) = ½.
Dn(x) is called the Dirichlet kernel. Note that Dn(x) is defined and continuous
for all x in [π, π]. We shall use this form of the (n+1)-th partial sum of (C) to
investigate convergence of (C).
Now consider the n-th partial sum of the sine series (S):
1
( ) sin( )n
n k
k
s x a kx
. -------------------------------- (10)
Applying the Abel summation formula (3) to (10) gives
1
1
( ) ( ) ( )n
k nn k n
k
s x D x a a D x
, ------------------------- (11)
where 1
( ) sin( )n
n
k
D x kx
-------------------------------- (12)
for n ≥ 1.
( )nD x is called the conjugate Dirichlet kernel. Observe that ( )nD x is defined and
continuous in [π, π]. The name conjugate Dirichlet kernel has its origin in
considering complex Fourier series as a power series on the unit circle so that
(A) is the real part of 0
1
1
2
inx
n n
n
a a ib e
and if (C) is the series, then the sine
series is the conjugate series appearing as the imaginary part of the power
series.
Now we proceed to investigate the properties of the Dirichlet kernels. Before
we do that we introduce a second summation formula involving the Dirichlet
kernels.
If anDn(x) 0, this will then bring us by taking limits of (8) to the series
9
0
( )k k
k
D x a
and the problem of the convergence of this series.
Applying the Abel summation formula (3) to the (n+1)-th partial sum Fn(x) of
this series, we get
1
2
0 0
( ) ( ) ( ) ( )n n
n k k k k n n
k k
F x D x a E x a E x a
, ----------------------- (13)
where 0
( ) ( ) ( 1) ( )k
k k k
j
E x D x k K x
and 0
1( ) ( )
1
k
k j
j
K x D xk
----------------------- (14)
is called the Fejér kernel and is actually the mean of the Dirichlet kernel. It is
also the (C,1) mean of the sequence 1
,cos( ),cos(2 ),cos(3 ),2
x x x
.
In view of (13), we then have
2
2
1 1
0
( ) ( 1) ( ) ( ) ( )n
n k k n n n n
k
t x k K x a nK x a a D x
, ---------- (15)
the result of applying the summation formula twice to the (n+1)-th partial sum
of the series (C).
We shall use formula (8), (11) and (15) to study the convergence of (C) and (S).
Thus we derive below some properties of the Dirichlet and Fejér kernels.
2.4 Dirichlet, Fejér and Conjugate Kernels
1 1 12 2 2
1
2sin( ) ( ) sin( ) 2sin( )cos( )n
n
k
x D x x x kx
1 1 12 2 2
1
sin( ) sin( ) sin(( 1) )n
k
x kx x k x x
1 1 1 12 2 2 2
sin( ) sin(( ) ) sin( ) sin(( ) )x n x x n x .
10
Thus for x ≠ 0 and x in [π, π], or 0 < x < 2π,
12
12
sin(( ) )( )
2sin( )n
n xD x
x
. --------------------------- (16)
Observe that 1 1 12 2 2 1
21 10 02 2
sin(( ) ) ( )cos(( ) )lim lim (0)
2sin( ) cos( )n
x x
n x n n xn D
x x
and so the Dirichlet kernel in its functional form (16) is continuous at 0.
For the estimate of the Dirichlet kernel it is useful to consider the modified
Dirichlet kernel defined by
* 12
( ) ( ) cos( )n nD x D x nx
1 1 12 2 21
21 12 2
sin(( ) ) sin(( ) ) cos( )sin( )cos( )
2sin( ) 2sin( )
n x n x nx xnx
x x
12
12
sin( )cos( )
2sin( )
nx x
x
12
sin( )
2 tan( )
nx
x . ------------------------------- (17)
Note that the modified Dirichlet kernel is continuous on [π, π] and
*(0)nD n and *( ) 0nD ------------------------ (18)
The Fejér kernel has too a useful functional form. Using (16),
12
10 0 2
1 1 sin(( ) )( ) ( )
1 1 2sin( )
n n
n k
k k
k xK x D x
n n x
1 12 22 1
02
1 1sin(( ) )sin( )
1 2sin ( )
n
k
k x xn x
2 1
02
1 1 cos( ) cos(( 1) )
1 2sin ( ) 2
n
k
kx k x
n x
2 1
2
1 1 cos(( 1) )
1 4sin ( )
n x
n x
11
2 1
2
2 12
1 2sin ( ( 1) )
1 4sin ( )
n x
n x
212
12
2 sin( ( 1) )
1 2sin( )
n x
n x
. -------------------------------- (19)
Observe that the Fejér kernel in its functional form (19) is continuous in [π, π].
Since 12
(0)kD k , 0 0
1 1 1 1(0) (0) ( )
1 1 2 2 2
n n
n k
k k
nK D k
n n
. -------- (20)
Note that from (9)
1
1( ) cos( )
2
n
n
k
D x dx dx kx dx
------------------- (21)
and so 0 0
1 1( ) ( )
1 1
n n
n k
k k
K x dx D xn n
and 1
( ) 1nK x dx
. ---------------- (22)
We have similar derivations for the conjugate kernels.
Now 1 12 2
1
2sin( ) ( ) 2sin( )sin( )n
n
k
x D x x kx
1 12 2
1
cos(( ) ) cos(( ) )n
k
k x k x
1 12 2
cos( ) cos(( ) )x n x ,
so that for x ≠ 0 and in [π, π],
1 12 2
12
cos( ) cos(( ) )( )
2sin( )n
x n xD x
x
. --------------------- (23)
Observe that 1 12 2
102
cos( ) cos(( ) )lim 0 (0)
2sin( )n
x
x n xD
x
so that in its functional form
(23), the conjugate Dirichlet kernel is continuous in [π, π]. We shall also use
the modified conjugate Dirichlet kernel particularly because it is nonnegative.
The modified conjugate Dirichlet kernel is defined by
12
* 1( ) ( ) sin( )
2n nD x D x nx . ----------------------- (24)
Using (23) we have for 0 < x < 2π or x in [π, π] – {0},