Journal of Inequalities in Pure and
Applied Mathematics
SOME CONVERGENCE RESULTS ON SINC INTERPOLATION
1
MOHAMAD S. SABABHEH, 1 ABDUL-MAJID NASAYR AND
1,2
KAMEL AL-KHALED
1 Department
of Mathematics and Statistics
Jordan University of Science and Technology
Irbid 22110, JORDAN
E-Mail: nusayr@just.edu.jo
volume 4, issue 2, article 32,
2003.
Received 26 February, 2003;
accepted 08 May, 2003.
Communicated by: T.M. Mills
2 Department
of Mathematics,
United Arab Emirates University
Al-Ain, P.O. Box 17551, U.A.E.
E-Mail: kamel.alkhaled@uaeu.ac.ae
Abstract
Contents
JJ
J
II
I
Home Page
Go Back
Close
c 2000 Victoria University
ISSN (electronic): 1443-5756
023-03
Quit
Abstract
This paper is devoted to the investigation of sinc interpolation properties corresponding to a sequence of functions having the sinc function as a basis, the
interpolation is taken over the dyadic partition of the interval [0, 2π]. In particular,
a new class of functions for which the interpolation converge is introduced. The
convergence of our interpolation processes is studied and answered in quite a
comprehensive way. In fact, the paper aims to provide a guideline towards a
large number of problems of interest in applied sciences.
2000 Mathematics Subject Classification: 39A10, 30B10, 15A45, 15A24, 11B37.
Key words: Recurrence, Restricted Coefficients, Power Series, Triangular Matrices.
Contents
1
2
3
4
5
Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
The Interpolation Processes . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
The Functional Properties Of Sn . . . . . . . . . . . . . . . . . . . . . . . . 11
Special Classes Of Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
Some Convergence Results . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 2 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
1.
Introduction
The sinc approximation method is a very promising method for function approximation, for approximation of derivatives, for approximate definite and indefinite integration, for solving initial value problems, for approximation and
inversion of Fourier and Laplace transforms. The sinc method is an attractive alternative for numerical solutions to those problems which have no closed form.
The theory of sinc series on the whole real line is developed in [8]. There are
several reasons to approximate by sinc functions. Firstly, they are easily implemented and give good accuracy for problems with singularities; approximations
by sinc function are typified by errors of the form O(exp(−c/h)) where c > 0
is a constant and h is a step size. Secondly, approximation by sinc functions
handles singularities in the problem. The effect of any such singularities will
appear in some form in any scheme of numerical solution, and it is well known
that polynomial methods do not perform well near singularities. Finally, these
kinds of approximation yield both an effective and rapidly convergent scheme
for solving the problem, and so circumvents the instability problems that one
typically encounters in some difference methods. Numerical processes of interpolation on the real line, with the help of adroitly selected conformal maps
is adapted to handle these same processes on finite intervals, or in general on
other subsets of the real line. For more details see, [3, 4, 5]. Also, it is worthy to
mention the work by Stenger [9], where he presents practically useful constructive linear methods of approximation of analytic functions by polynomials, sinc
functions and rational functions. In [6], the author proves some convergence
results on finite intervals,using the
linear
combination of the basis functions
1
−1
−1
Bn,k = S(k, h) ◦ sin h
cos h
where k = −n, . . . , n, h = log n/n,
|x|
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 3 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
and S(k, h) is the sinc translated function, to be defined later.
Although there is no unique choice for the conformal map, and so one will
not guarantee an exponential decay of the convergence rate using the sinc method.
It should be pointed out that it might be possible that the selection of the conformal mapping does not lead to a symmetric discrete system. While a symmetric
approximation system is not necessary for a good approximation, it is computationally efficient and analytically advantageous for solving the discrete system.
As a final note on selection availability of the conformal mapping. In problems
where two (or more) maps are applicable, the use of either of the maps leads
to a smaller size of the discrete system, for example, in the case of the domain
(0, ∞) there are available the selections ln(x) and ln(sinh(x)). The map ln(x)
often leads to a smaller discrete system that does the map ln(sinh(x)) for equivalent accuracy. To avoid these difficulties and as an alternative for the extension
(using conformal maps) made by Stenger [8], this paper is devoted to the investigation of sinc interpolation on the interval [0, 2π] (see, [7]). The paper is organized as follows. In Section 2 we define our interpolation processes Sn (f ; x),
where the nodes are taken to be the diadic partition of the interval [0, 2π]. We
then study some basic properties of the interpolating function Sn (f ; x). In Section 3 we take up the functional properties of Sn (f ; x). Section 4 deals with
new classes of functions for which the interpolation processes converges. In the
last section of this chapter, we give the most important convergence results in
this paper.
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 4 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
2.
The Interpolation Processes
Let E1 = {0, π, 2π}, and E2 = {0, π/2, π, 3π/2, 2π}. In general let
2kπ
n
(2.1)
En =
, 0≤k≤2 .
2n
In the following Lemma we state, without proof, some properties of the partition
En
Lemma 2.1. For the sets En the following holds true
1. The sequence {En } is an increasing sequence, i.e, E1 ⊂ E2 ⊂ . . . .
2. E = ∪∞
n=1 En is dense subset of [0, 2π].
Definition 2.1. Let f : [0, 2π] → R be any function. For each natural number
n we define,
X
(2.2)
Sn (f, x) =
f (xk )Ln,k (x),
xk ∈En
where
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back

 sin[2(n−1) (x − xk )]
, x 6= xk
Ln,k (x) =
2(n−1) (x − xk )

1,
x = xk
(2.3)
and xk =
Some Convergence Results on
Sinc Interpolation
2kπ
2n
Close
Quit
Page 5 of 34
for 0 ≤ k ≤ 2n .
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
In the following sequence of lemmas we will give the basic properties of
Sn (f, x) as an interpolating function.
Lemma 2.2. For any natural numbers n, k and j where 0 ≤ k, j ≤ 2n we have
Ln,k (xj ) = δj,k .
Proof. If j = k then Ln,k (xj ) = 1 by the definition of Ln,k (x). Now if j 6= k,
we have
sin[2n−1 (xj − xk )]
Ln,k (xj ) =
2n−1 (xj − xk )
sin 2n−1 2jπ
− 2kπ
2n
2n
=
2n−1 (xj − xk )
sin[(j − k)π]
= n−1
= 0.
2 (xj − xk )
This completes the proof.
Proof.
Title Page
JJ
J
II
I
Go Back
n
Sn (f, xk ) =
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Contents
Lemma 2.3. Sn (f, x) interpolates f on En for any function f defined on [0, 2π].
i.e., Sn (f, xk ) = f (xk ) for all xk ∈ En .
2
X
Some Convergence Results on
Sinc Interpolation
f (xj )Ln,j (xk )
Close
j=0
= f (xk )Ln,k (xk ) +
X
f (xj )Ln,j (xk ) = f (xk ).
xj 6=xk
Since xk is an arbitrary element of En , the result follows.
Quit
Page 6 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
We have shown that Sn (f, x) interpolates f on the set En , the following
lemma is a generalization:
Lemma 2.4. For any real valued function f on [0, 2π], if limn→∞ Sn (f, x) =
g(x) then g(x) = f (x) for all x ∈ E.
Proof. Let x be arbitrary element of E. Since E is the union of the sets En0 s
there must be n0 such that x ∈ En for all n ≥ n0 . Now for n ≥ n0 we have
Sn (f, x) = f (x) therefore limn→∞ Sn (f, x) = f (x), thus g(x) = f (x). Since
x is an arbitrary element of E the result follows.
Lemma 2.5. Sn (f, x) = Sm (f, x) for all x in Ek , where k = min(n, m).
Proof. Let n and m be any two natural numbers, and k = min(m, n). Let
x` ∈ Ek for some `, then x` ∈ En because k ≤ n. But Sn interpolates f
on En , so that Sn (f, x` ) = f (x` ). By the same argument, we can show that
Sm (f, x` ) = f (x` ).
The following sequence of lemmas give some result on the derivative of the
basis of the interpolation, from which one can approximate the solution for
some differential equations.
Lemma 2.6. For any natural numbers n and k with 0 ≤ k ≤ 2n we have
L0n,k (xk ) = 0.
(2.4)
Proof. For x = xk , we have
Ln,k (x) − Ln,k (xk )
L0n,k (xk ) = lim
x→xk
x − xk
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 7 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
sin[2n−1 (x − xk )] − 2n−1 (x − xk )
x→xk
2n−1 (x − xk )2
2n−1 cos[2n−1 (x − xk )] − 2n−1
= lim
x→xk
22n−1 (x − xk )
−2n−1 2n−1 sin[2n−1 (x − xk )]
=0
= lim
x→xk
22n−1 (1)
= lim
and the lemma is proved.
Lemma 2.7. For any natural numbers n and k with 0 ≤ k ≤ 2n we have
(2.5)
L0n,k (xj )
2n × (−1)j−k
=
.
2π(j − k)
Proof.
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Ln,k (x) − Ln,k (xj )
x→xj
x − xk
sin[2n−1 (x − xk )]
= lim n−1
x→xj 2
(x − xk )(x − xj )
n−1
2
cos[2n−1 (x − xj )]
1
= lim
lim
n−1
x→xj
x→xj x − xk
2
j−k
n
j−k
(−1)
2 × (−1)
=
=
.
xj − xk
2π(j − k)
L0n,k (xj ) = lim
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 8 of 34
Now the proof is complete.
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
For the second derivative, we have
Lemma 2.8. For any natural numbers n and k with 0 ≤ k ≤ 2n we have
(2.6)
L00n,k (xk ) =
−22n−2
.
3
Proof.
L00n,k (xk )
L0n,k (x) − L0n,k (xk )
= lim
x→xk
x − xk
2n−1 (x − xk ) cos[2n−1 (x − xk )] − sin[2n−1 (x − xk )]
= lim
x→xk
2n−1 (x − xk )3
−2n−1 2n−1 (x − xk ) sin[2n−1 (x − xk )]
= lim
x→xj
3.2n−1 (x − xk )2
−2n−1 .2n−1
=
3
−22n−2
=
.
3
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
The following is a connection between our interpolation and the Lagrange
interpolation.
Close
Lemma 2.9. Let f be a real-valued function on [0, 2π] and let Hn denote the
Lagrange interpolation function on En , then Sn (f, x) = Sn (Hn , x).
Page 9 of 34
Quit
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Proof. We recall that Hn (xk ) = f (xk ) for all xk ∈ En . Now,
X
Sn (f, x) =
f (xk )Ln,k (x)
xk ∈En
=
X
Hn (xk )Ln,k (x)
xk ∈En
= Sn (Hn , x).
Some Convergence Results on
Sinc Interpolation
In fact the previous lemma is a particular case of the following lemma.
Lemma 2.10. Let f and g be two real-valued functions defined on [0, 2π] such
that g(xk ) = f (xk ) for all xk ∈ En for some natural number n, then Sn (f, x) =
Sn (g, x) for all x ∈ [0, 2π].
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 10 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
3.
The Functional Properties Of Sn
Notice that Sn can be considered as an operator on the space of all real-valued
functions on [0, 2π]. So that, it is convenient to study the functional properties
of Sn as an operator on the dual space of [0, 2π]. Note that Sn (f, x) is a linear
operator on the space of all functions defined on [0, 2π].
Lemma 3.1. For each natural number n, the operator Sn is a bounded linear
operator.
Proof. In order to show the boundedness of Sn , we find a real number cn such
that kSn (f )k ≤ cn kf k. This is an immediate result.
Corollary 3.2. For each natural number n, Sn is a continuous linear operator.
We have shown that Sn is a continuous function on [0, 2π] and is a continuous linear operator on the dual space of [0, 2π]. Therefore, Sn is a continuous
maping in both its components, i.e., if we consider
Sn : X × [0, 2π] −→ R,
then Sn is continuous on X × [0, 2π], where X is the dual space of [0, 2π].
The following few results give us some fixed points for Sn .
Lemma 3.3. For any natural numbers n and k where 0 ≤ k ≤ 2n we have
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
(3.1)
Sn (Ln,k , x) = Ln,k (x).
Page 11 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Proof.
Sn (Ln,k , x) =
X
Ln,k (xj )Ln,j (x)
xj ∈En
= Ln,k (xk )Ln,k (x) +
X
Ln,k (xj )Ln,j (x)
xj 6=xk
= Ln,k (x).
Some Convergence Results on
Sinc Interpolation
Lemma 3.4. For any natural number n and for any function f on [0, 2π] we
have Sn (Sn (f ), x) = Sn (f, x).
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Proof.
Sn (Sn (f ), x) =
X
Sn (f, xk )Ln,k (x)
xk ∈En
=
X X
f (xj )Ln,j (xk )Ln,k (x)
xk ∈En xj ∈En
=
X X
f (xk )δk,j Ln,k (x)
xk ∈En xj ∈En
=
X
xk ∈En
f (xk )Ln,k (x) = Sn (f, x).
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 12 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Thus Sn as a linear operator on the class of all real valued functions defined
on [0, 2π] has at least 2n + 2 fixed points.
Talking about fixed points of a linear operator leads to talking about the
contraction which is considered in the following corollary.
Theorem 3.5. For any function f on [0, 2π], the operator Sn (f, x) is not a
contraction.
Proof. We firstly recall that the space of all real valued functions on [0, 2π] is a
complete metric space with respect to the metric
(3.2)
d(f, g) = sup{f (x) − g(x) : x ∈ [0, 2π]}.
Now, if we assume on the contrary that Sn (f, x) is a contraction, then Sn will
satisfy the requirments of the “Banach fixed point theorem”, and, hence, Sn has
a unique fixed point. The last statement is a contradiction because Sn has at
least 2n + 2 fixed points. Thus, we conclude that Sn is not a contraction.
Since Sn is not a contraction, one may ask about the relation between d(f, g)
and d(Sn (f ), Sn (g)). The following theorem answers this question.
Theorem 3.6. Let f and g be any two functions on [0, 2π]. For each natural
number n, we have
(3.3)
d(Sn (f ), Sn (g)) ≤ (2n + 1)d(f, g).
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Proof.
Page 13 of 34
d(Sn (f ), Sn (g)) = sup |Sn (f, x) − Sn (g, x)|
[0,2π]
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
= sup |Sn (f − g)|
[0,2π]
n
= sup |
[0,2π]
≤ sup
2
X
[f (xk − g(xk )] Ln,k (x)|
k=0
n
2
X
[0,2π] k=0
2n
X
≤ sup
[0,2π] k=0
2n
X
≤ sup
|f (xk ) − g(xk )||Ln,k (x)|
Some Convergence Results on
Sinc Interpolation
|f (xk ) − g(xk )|
n
|f (x) − g(x)| = (2 + 1)d(f, g).
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
[0,2π] k=0
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 14 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
4.
Special Classes Of Functions
Since the nodes of the interpolation are of the form 2kπ
, one can think about
2n
P2n
2kπ
the limit; limn→∞ k=0 f 2n . In fact this idea introduces the following
definition.
Definition 4.1. Let U [0, 2π] be the class of all real valued-functions f on [0, 2π]
and for which
2n
X
2kπ
lim
f
< ∞.
n
n→∞
2
k=0
Lemma 4.1. The condition in the last lemma is equivalent to the condition
X
|f (xk )| < ∞.
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
xk ∈E
Proof. We show that
Title Page
2n
lim
n→∞
X
k=0
f
2kπ
2n
=
X
Contents
|f (xk )|
xk ∈E
in order to prove the lemma. For, numerate the countable set E as following:
The elements of E1 are x0 , x1 and x2 .
The elements of E2 − E1 are x3 , x4 .
The elements of E3 − E2 are x5 , ..., x9 .
In general, the elements of En+1 − En are x2n +1 , ..., x2n+1 . Now,
2n
2n
X
X
X
2kπ
|f (xk )| = lim
|f (xk )| = lim
f
n→∞
n→∞
2n
x ∈E
k=0
k=0
k
JJ
J
II
I
Go Back
Close
Quit
Page 15 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
the last equation is valid because of our choice of the numeration.
The reader must realize that any rearrangement of the above sums is not
important because we are dealing with absolute sums.
Example 4.1. Here we give an example of a function that belongs to the class
U [0, 2π], i.e, we show that U [0, 2π] 6= ∅. For, let f : [0, 2π] −→ R be defined
by
( 1
, x = xk ,
k2
f (x) =
1, x 6= xk .
Here we consider some numeration for the countable set E. It is clear that
n
lim
n→∞
2
X
f
k=0
2kπ
2n
=
∞
X
1
π2
|f (xk )| =
=
.
k2
6
∈E
k=1
xk
an =
k=0
f
Title Page
Contents
Example 4.2. In this example we show that the class U [0, 2π] doesn’t contain
any polynomial of the form f (x) = axm . For, consider
n
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
X
Therefore, f ∈ U [0, 2π]
2
X
Some Convergence Results on
Sinc Interpolation
2kπ
2n
n
=
2
X
k=0
JJ
J
II
I
n
2
(2kπ)m
2m π m X m
a nm = |a| nm
k .
2
2
k=0
Although the exact formula for the last sum needs more complicated computations, we know
that this sum will be a polynomial in 2n of degree m + 1. Thus,
(2π)m
an = |a| 2nm × g(m + 1), where g(m + 1) is the indicated polynomial. Now it
is clear that, limn→∞ an = ∞. Therefore f 6= U [0, 2π].
Go Back
Close
Quit
Page 16 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Lemma 4.2. If f is any real valued function on [0,
R 2π2π] such that |f | is integrable
in the sense of Reimann and f ∈ U [0, 2π] then 0 |f (x)|dx = 0.
Proof. For each natural number n, En = 2kπ
, 0 ≤ k ≤ 2n is a partition for
2n
[0, 2π]. The subintervals of this partition are
n
2π
2π 4π
2(2 − 1)π
, 2π .
0, n , n , n , . . . ,
2
2 2
2n
Now consider the Riemann sum of f over this partition,
n
2
2π X
Rn (f ) = n
|f (x∗k )|,
2 k=1
where x∗k is any point of the k−th interval of the partition. Since |f | is integrable
(in the sense of Riemann) we can take x∗k to be xk = 2kπ and, hence,
2n
2n
2π X
2π X
2kπ
Rn (|f |) = n
|f (xk )| = n
f
.
2 k=1
2 k=1
2n
Now write the integral of f as the limit of a Riemann sum to get
Z 2π
2n
2π X
2kπ
|f (x)|dx = lim Rn (|f |) = lim n
f
.
n
n→∞
n→∞ 2
2
0
k=1
P2n
2kπ
But since f ∈ U [0, 2π] we have limn→∞ k=1 f 2n is finite. Thus,
Z 2π
2π
|f (x)|dx = lim n × (finite value) = 0.
n→∞
2
0
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 17 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Lemma 4.3. If f ∈ U [0, 2π] and |f | is integrable in the sense of Riemann, then
[0, 2π] does not contain any interval I such that f (x) 6= 0 for all x ∈ I.
Proof. Assume that there is an interval I ⊂ [0, 2π] such that f (x) 6= 0 for all
x ∈ I, then
Z 2π
Z
Z
|f (x)|dx = |f (x)dx +
|f (x)|dx
0
I
[0,2π]−I
Z
≥ |f (x)|dx
I
>0
but this contradicts the last lemma, and the lemma is proved.
Lemma 4.4. The only continuous function in U [0, 2π] is the zero function.
Proof. Let f be a non-zero continuous function on [0, 2π], then there is at least
one x ∈ [0, 2π] such that f (x) 6= 0. Since f is continuous at x, there must be
an interval I containing x for which f (x) 6= 0 for all x ∈ I. But m(I) > 0 and
this contradicts the last lemma.
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 18 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
5.
Some Convergence Results
Lemma 5.1. For any natural numbers n and k with 0 ≤ k ≤ 2n , we have
Z 2n−1
1
2πkit −ixt
(5.1)
Ln,k (x) = n
exp
e dt.
2 −2n−1
2n
Proof.
Z 2n−1
2πkit
1
e 2n −ixt dt
n
2 −2n−1
Z 2n−1
2kπ
1
= n
ei( 2n −x)t dt
2 −2n−1
i2n−1
h 2kπ
1
1
i( 2n −x)t
e
= n 2πk
2 i 2n − x
−2n−1
h 2kπ
i
2kπ
1
1
n−1
n−1
ei( 2n −x)2 − e−i( 2n −x)2
= n 2πk
2 i 2n − x
1
1
2kπ
n−1
cos[2
= n 2πk
−x ]
2 i 2n − x
2n
1
1
2kπ
2kπ
n−1
n−1
i sin 2
+ n 2πk
− x − cos 2
−x
2 i 2n − x
2n
2n
1
1
2kπ
n−1
i sin 2
+ n 2πk
−x
2 i 2n − x
2n
sin 2n−1 x − 2kπ
n
2
=
= Ln,k (x).
2n−1 x − 2kπ
2n
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 19 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
The last proof is valid whenever x 6= xk ; the case where x = xk is easy to
seen.
Corollary 5.2. The Fourier transform of Ln,k (x) is
(
(5.2)
F (t) =
2π
2n
exp( 2kiπt
), |t| < 2n−1
2n
0,
|t| > 2n−1
Proof. Let F be the Fourier transform of f , then F must satisfy the equation
Z
1
Ln,k (t) =
e−ixt F (x)dx.
2π R
By the last lemma we find that the function defined in equation (5.2) satisfies
this condition, and since the Fourier transform is unique, the result follows.
n
Corollary 5.3. For any natural numbers n, k and j where 0 ≤ k, j ≤ 2 we
have
Z
2π
(5.3)
Ln,k (x)Ln,j (x)dx = n δk,j .
2
R
Proof. Let F and G denote the Fourier transforms of Ln,k and Ln,j respectively,
then by Parseval’s theorem,
Z
Z
1
Ln,k (x)Ln,j (x)dx =
F (t)G(t)dt.
2π R
R
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 20 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Firstly, if k = j, then
Z
Z 2n−1
1
4π 2
2π
2π
Ln,k (x)Ln,j (x)dx =
dt = n = n δk,k .
2n
2π −2n−1 2
2
2
R
Secondly, if k =
6 j then
Z 2n−1
Z
4π 2 2πit( k−j
1
2n ) dt
Ln,k (x)Ln,j (x)dx =
e
2π −2n−1 22n
R
h
i2n−1
k−j
2n
2π
= 2n
e2πit( 2n )
2 2iπ(k − j)
−2n−1
1
= n
2i sin[(k − j)π]
2 i(k − j)
= 0.
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Corollary 5.4. Let f : [0, 2π] −→ R. For each natural nuber n we define

2n

2kπ
2kπix
 2π X
f
exp
, |x| < 2n−1 ,
(5.4)
Fn (x) =
2n k=0
2n
2n


0,
|x| > 2n−1 ,
then for any natural number n, we have
Z 2n−1
1
(5.5)
Sn (f, t) =
Fn (x)e−ixt dx.
2π −2n−1
In fact, Fn is the Fourier transform of Sn .
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 21 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Proof.
1
2π
Z
2n−1
2n
2kπix
2kπ
2π X
f
e 2n e−itx dx
n
n
2
−2n−1 2 k=0
Z 2n−1
2n
X
2kπix
2kπ
=
f
e 2n e−itx dx
n
2
−2n−1
k=0
2n
X
2kπ
=
f
Ln,k (t)
n
2
k=0
1
Fn (x)e−itx dx =
2π
−2n−1
Z
2n−1
= Sn (f, t).
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Corollary 5.5. Let f be a real-valued function such that both f and |f | are
integrable in the sense of Riemann, and such that f = 0 outside [0, 2π], also let
F be the Fourier transform of f , then
(5.6)
lim Fn (x) = F (x),
n→∞
where Fn is defined in Corollary 5.4. Moreover Fn −→ F uniformly.
Proof. For fix x ∈ [−2n−1 , 2n−1 ], let
(5.7)
g(t) = f (t) exp(itx),
Some Convergence Results on
Sinc Interpolation
Title Page
Contents
JJ
J
II
I
Go Back
t ∈ [0, 2π]
and consider the n-th Reimann sum of g over [0, 2π];
2n
2π X
2kπ
2kπix
f
exp
(5.8)
Rn (g) = n
2 k=1
2n
2n
Close
Quit
Page 22 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
therefore,
n
2
2kπ
2kπix
2π X
f
exp
lim Rn (g) = lim n
n
n→∞
n→∞ 2
2
2n
k=1
= lim Fn (x),
n→∞
which implies
Z
2π
f (t) exp(itx)dt = lim Fn (x)
Some Convergence Results on
Sinc Interpolation
n→∞
0
thus,
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
F (x) = lim Fn (x)
n→∞
The uniformly convergence fact follows because
n
(5.9)
n
2
2
X
X
2π
2π
2kπ
2kπix
2kπ
lim
≤ lim
f
exp
f
n→∞
n→∞
2n
2n
2n
2n
2n
k=0
k=0
and the last series converges to a real number because |f | is integrable. By the
M −test we have the result. And the corollary is proved.
Corollary 5.6. For any natural numbers n and k with 0 ≤ k ≤ 2n , we have
(5.10)
f
2kπ
2n
=
1
2π
Z
2n−1
Fn (x) exp
−2n−1
−2kπix
2n
dx,
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 23 of 34
where {Fn } as defined in corollary 5.4.
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Proof. Consider the Fourier series representation for Fn in (−2n−1 , 2n−1 );
∞
X
Fn (x) =
ck exp
k=−∞
2kπix
2n
,
|x| < 2n−1 .
Compare this withthe definition of Fn (x) to get ck = 0 for k < 0 and k > 2n
and ck = 2π
f 2kπ
for 0 ≤ k ≤ 2n . Also we know that
2n
2n
1
ck = n
2
Z
2n−1
F (x) exp
−2n−1
−2kπix
2n
Some Convergence Results on
Sinc Interpolation
dx,
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
and hence
f
2kπ
2n
1
=
2π
Z
2n−1
F (x) exp
−2n−1
−2kπix
2n
dx.
Title Page
Contents
The last corollary gives rise to the following new class of functions.
Definition 5.1. Let J[0, 2π] be the class of all real-valued functions, f on [0, 2π]
and 0 outside this interval) for which there is a function Fn satisfying the following conditions:
• Fn (x) = 0 for all x outside (−2n−1 , 2n−1 ) for some natural number n
R 2n−1
1
• f (t) = 2π
F (x)e−ixt for all t ∈ [0, 2π].
−2n−1 n
JJ
J
II
I
Go Back
Close
Quit
Page 24 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Although the class J[0, 2π] seems to be very complicated, it has many nice
properties. In the following sequence of theorems we give the most important
properties for this class.
Theorem 5.7. For any function f ∈ J[0, 2π] we have the series representation
f (x) = Sn (f, x),
where n is as in the definition of J[0, 2π] and
Sn (f, x) =
(5.11)
2n
X
f
k=0
2kπ
2n
Some Convergence Results on
Sinc Interpolation
Ln,k (x).
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Proof. Since f ∈ J[0, 2π], there exists n0 such that
1
f (t) =
2π
Z
Title Page
2n−1
−ixt
e
Fn (x)dx
Contents
−2n−1
for all n ≥ n0 . The function F in the last equation can be represented on the
interval (−2n−1 , 2n−1 ) by its Fourier series representation, i.e.
Fn (x) =
∞
X
ck exp
k=−∞
2kπix
2n
JJ
J
II
I
Go Back
, −2n−1 < x < 2n−1
Close
Quit
with
1
ck = n
2
Z
2n−1
2kπix
Fn (x) exp − n
2
−2n−1
Page 25 of 34
dx.
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
By our choice of Fn we have, ck = 2π
f 2kπ
.
2n
2n
Substitute this value in the Fourier series of F to get

∞
X

2kπ
2kiπx
 2π
f
exp
, |x| < 2n−1 ,
2n k=−∞
2n
2n
Fn (x) =


0,
|x| > 2n−1 .
Now,
1
f (t) =
2π
Z
Some Convergence Results on
Sinc Interpolation
2n−1
e−ixt F (x)dx
−2n−1
Z
2n−1
∞
X
2π
2kπ
2kiπx −ixt
f
exp
e dx
n
n
2
2n
−2n−1 2 k=−∞
Z 2n−1
∞
2kπ
2kiπx
1 X
= n
f
exp
− ixt dx
2 k=−∞
2n
2n
−2n−1
∞
2n
X
X
2kπ
2kπ
Ln,k (t) =
f
Ln,k (t) = Sn (f, t).
=
f
n
n
2
2
k=0
k=−∞
=
1
2π
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Theorem 5.8. If f ∈ J[0, 2π] then for n as in Definition 5.1,
Close
Quit
Z
0
2n
2π
|f (x)|2 dx =
2π X
f
2n k=0
2kπ
2n
2
.
Page 26 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Proof. Since f ∈ J[0, 2π] we have
2n
X
2kπ
f (x) =
f
Ln,k (x)
n
2
k=0
for n as in Definition 5.1 . Now,
Z
|f (x)|2 dx
R
#2
Z "X
2n
2kπ
=
f
Ln,k (x) dx
2n
R k=0
#
2
Z "X
2n
X 2kπ 2jπ 2kπ
L2n,k (x) +
f
=
f
f
Ln,k (x)Ln,j (x) dx
n
n
n
2
2
2
R k=0
k6=j
n
Z
2
2
X
X 2kπ 2jπ Z
2kπ
2
Ln,k (x)dx +
f
Ln,k (x)Ln,j (x)dx
=
f
2n
2n
2n
R
R
k=0
k6=j
2
2n
2π X
2kπ
= n
f
.
2 k=0
2n
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
The following theorem tells us some type of convergence of our interpolation.
Quit
Page 27 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Theorem 5.9. Let f : [0, 2π] −→ R such that f 2 is integrable in the sense of
Riemann, then
Z 2π
Z ∞
2
(5.12)
|f (x)| dx = lim
Sn2 (f, x)dx
n→∞
0
−∞
Proof. Firstly, we notice that
!2
2n
X
2kπ
Sn2 (f, x) =
f
Ln,k (x)
2n
k=0
2
2n
X
X 2kπ 2jπ 2kπ
2
=
f
f
Ln,k (x) +
f
Ln,k (x)Ln,j (x).
n
n
n
2
2
2
k6=j
k=0
Now integrate both sides of the last equation on R, to get
Z
Sn2 (f, x)dx
R
2n
X
!
X
2kπ
2kπ
2jπ
=
f
L2n,k (x) +
f
f
Ln,k (x)Ln,j (x) dx
n
n
2
2
2n
R
k=0
k6=j
2Z
2n
X
X 2kπ 2jπ Z
2kπ
2
=
f
Ln,k dx +
f
f
Ln,k (x)Ln,j (x)dx
n
n
n
2
2
2
R
R
k=0
k6=j
n
2
2
2π X
2kπ
= n
f
.
2 k=0
2n
Z
2
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 28 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
The validity of the last equation arises from corollary 5.3. Now take the limit of
the last equation as n −→ ∞ to get
2
Z
2n
2kπ
2π X
2
lim
Sn (f, x)dx = lim n
f
n→∞ R
n→∞ 2
2n
k=0
but
n
2
2kπ
2π X
lim
f
n→∞ 2n
2n
k=0
2
Z
2π
=
|f (x)|2 dx
0
because it is the limit of the Riemann sum for f 2 on [0, 2π], this completes the
proof of the theorem.
Lemma 5.10. Let f be any real-valued function on [0, 2π], then for any natural
numbers n and k with 0 ≤ k ≤ 2n , we have
Z
2kπ
2π
(5.13)
Sn (f, x)Ln,k (x)dx = n f
.
2
2n
R
Proof. We recall that
2n
X
2jπ
Sn (f, x) =
f
Ln,j (x),
2n
j=0
multiply both sides by Ln,k (x) and integrate on R to get:
Z
Sn (f, x)Ln,k (x)dx
R
Z X
2n
2jπ
=
f
Ln,j (x)Ln,k (x)dx
2n
R j=0
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 29 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
=f
2kπ
2n
Z
L2n,k (x)dx
R
X 2jπ Z
f
+
Ln,k (x)Ln,j (x)dx
2n
R
j6=k
2kπ 2π
+0
2n
2n
2π
2kπ
= nf
.
2
2n
=f
Theorem 5.11. Let f be a real-valued function such that f and |f | are integrable in the sense of Riemann on [0, 2π], then
(5.14)
lim Sn (f, x) = f (x) a.e.
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
n→∞
Title Page
on [0, 2π].
Proof. We saw in Corollary 5.4 that Fn is the Fourier transform of Sn , and that
in Corollary 5.5,
lim Fn = F
n→∞
uniformly, where F is the Fourier transform of f .
Therefore,
lim F(Sn ) = F(f )
n→∞
but by our conditions on f and, F is a continuous linear operator, we have,
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 30 of 34
F( lim Sn ) = F(f )
n→∞
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
So, we conclude that
lim Sn = f a.e.
n→∞
As a notation, Let Pn (x) denote the Lagrange interpolating function of Sn (f, x)
with the nodes of En , and let Hn (x) denote the Lagrange interpolating function
of f (x) with the nodes of En .
Lemma 5.12. Let f be any real-valued function on [0, 2π], then Pn (x) =
Hn (x).
Proof. By the definition of the Lagrange interpolation we have,
X
Pn (x) =
Sn (f, xk )Jn,k (x),
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
xk ∈En
x−xj
j6=k xk −xj
where Jn,k (x) =
and j ranges over the integers between 0 and 2n ,
included. But since Sn (f, xk ) = f (xk ) for all xk ∈ En , we would have
X
Pn (x) =
f (xk )Jn,k (x)
Q
j6=k
= Hn (x).
Title Page
Contents
JJ
J
II
I
Go Back
Theorem 5.13. Let f ∈ C 2
x ∈ [0, 2π] we have
(f (x)−Sn (f, x)) =
n +1
Close
[0, 2π] for any natural number n then for each
Y
n
1
n
(x−xk ) Sn(2 +1) (ζ(x)) + f (2 +1) (ξ(x)) ,
n
(2 + 1)! x ∈E
k
n
Quit
Page 31 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
where ζ(x) and ξ(x) are two numbers in the interval (0, 2π) and depend on x
only.
Proof. Let Hn and Pn as in the last lemma, then
(f (x) − Sn (f, x)) = (f (x) − Hn (x) + Hn (x) − Pn (x) + Pn (x) − Sn (f, x))
= (f (x) − Hn (x)) + (Hn (x) − Pn (x)) + (Pn (x) − Sn (f, x))
Y
1
n
= n
(x − xk ) f (2 +1) (ξ(x))
(2 + 1)! x ∈E
n
k
Y
1
n
+ n
(x − xk ) Sn(2 +1) (ζ(x))
(2 + 1)! x ∈E
n
k
Y
n
1
n
= n
(x − xk ) Sn(2 +1) (ζ(x)) + f (2 +1) (ξ(x))
(2 + 1)! x ∈E
k
which completes the proof.
n
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 32 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
References
[1] U.G. ABDULLAEV, Stability of symmetric traveling waves in the Cauchy
problem for the Kolmogorov-Petrovskii-Piskunor equation, Diff. Eqn.,
30(3) (1994), 377–386.
[2] M.F.K. ABUR-ROBB, Explicit solutions of Fisher’s equation with three
zeros, Int. J. Math. Math. Sci., 13(3) (1990), 617–620.
[3] K. AL-KHALED, Sinc numerical solution for solitons and solitary waves,
J. Comput. Appl. Math., 130(1-2) (2001), 283–292.
Some Convergence Results on
Sinc Interpolation
[4] K. AL-KHALED, Numerical study of Fisher’s reaction-diffusion equation
by the Sinc collocation method, J. Comput. Appl. Math., 137(2) (2001),
245–255.
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
[5] D. BORWEIN AND J.M. BORWEIN, Some remarkable properties of sinc
and related integrals, Ramanujan J., 5(1) (2001), 73–89.
Title Page
[6] F. KEINERT, Uniform approximation to X β by Sinc Functions, J. Approximation Theory, 66(1) (1991), 44-52.
[7] M.S. SABABHEH, Sinc Interpolation on [0, 2π], M.Sc. Thesis, Jordan University of Science and Technology, August, 2002.
[8] F. STENGER, Numerical Methods Based on Sinc and Analytic Functions,
Springer-Verlag, New York, (1993).
[9] F. STENGER, Polynomails, sinc and rational function methods for approximation analytic functions, Rational approximation and interpolation, Proc.
Conf. Tama/Fla., 1983, Lect. Notes Math., 1105 (1984), 49–72.
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 33 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au
Some Convergence Results on
Sinc Interpolation
Mohamad S. Sababheh,
Abdul-Majid Nasayr and
Kamel Al-Khaled
Title Page
Contents
JJ
J
II
I
Go Back
Close
Quit
Page 34 of 34
J. Ineq. Pure and Appl. Math. 4(2) Art. 32, 2003
http://jipam.vu.edu.au