Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol104.Issue1.1999:

Journal of Computational and Applied Mathematics 104 (1999) 63–88

GB-splines of arbitrary order1
Boris I. Ksasov a; ∗ , Pairote Sattayatham b
a

Institute of Computational Technologies, Russian Academy of Sciences, Lavrentyev Avenue 6, 630090,
Novosibirsk, Russia
b
School of Mathematics, Suranaree University of Technology, University Avenue 111, 30000,
Nakhon Ratchasima, Thailand
Received 6 April 1998; received in revised form 23 October 1998

Abstract
Explicit formulae and recurrence relations for the calculation of generalized B-splines (GB-splines) of arbitrary order
are given. We derive main properties of GB-splines and their series, i.e. partition of unity, shape-preserving properties,
invariance with respect to ane transformations, etc. It is shown that such splines have the variation diminishing property
c 1999 Elsevier Science B.V. All rights reserved.
and are Chebyshevian splines.
Keywords: Spline; GB-spline; Weak Chebyshevian system; Variation diminishing; Shape-preserving approximation

1. Introduction
Fitting curves and surfaces to functions and data requires the availability of methods which preserve the shape of the data. In practical calculations, we usually deal with data given with prescribed
accuracy. Therefore, we need to develop methods for constructing fair-shape-preserving approximations that satisfy given tolerances and inherit major geometric properties of the data such as positivity,
monotonicity, convexity, presence of linear sections, etc. Such approximations, based on GB-splines
[12] with automatic choice of tension parameters are suggested in [11].
Until recently, local support bases for computations with generalized splines have been available
only for some special types of splines [3, 13, 15]. This limits the choice of methods when using
generalized splines. In [7–9] local support basis functions for exponential splines were introduced
and their application to interpolation problems was considered. A recurrence relation for rational
B-splines with prescribed poles was obtained in [5]. In [10, 12] one of the authors constructed
∗
1

Corresponding author. E-mail: boris@math.sut.ac.th.
Supported by grant BRG=16=2540 of the Thailand Research Fund.

c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 2 6 5 - 9

64

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

GB-splines for tension generalized splines allowing the tension parameters to vary from interval to
interval.
In this paper, we expand the main results in [12] to GB-splines of arbitrary order. These
GB-splines are nonnegative functions with supports of minimal length which form a partition of
unity. We get explicit formulae for such GB-splines and develop recurrence algorithms for their
calculation. In the particular case of polynomial B-splines, we recover the well-known recurrence
relation for such B-splines [1]. The main properties of GB-splines and their series such as shapepreserving properties, invariance with respect to ane transformations, etc., are investigated. It is
shown that the GB-spline series is a variation diminishing function and the systems of GB-splines
are weak Chebyshevian systems.
2. GB-splines of arbitrary degree
Let a partition
: a = x0 ¡x1 ¡ · · · ¡xN = b of the interval [a; b] be given to which we associate
a space of splines SnG whose restriction to a subinterval [xi ; xi+1 ]; i = 0; : : : ; N −1 is spanned by the
system on linearly independent functions {1; x; : : : ; x n−3 ; i; n ; i; n }, n¿2, and where every function
in SnG has n−2 continuous derivatives.
Denition 2.1. A generalized spline of order n is a function S ∈ SnG such that
(i) for any x ∈ [xi ; xi+1 ], i = 0; : : : ; N − 1,

S(x) = Pi; n−2 (x) + S (n−2) (xi )i; n (x) + S (n−2) (xi+1 ) i; n (x);

(1)

where Pi; n−2 is a polynomial of order n−2, and
i;(r)n (xi+1 ) = i;(r)
n (xi ) = 0;
i;(n−2)
(xi )
n

=

i;(n−2)
(xi+1 )
n

r = 0; : : : ; n − 2;
= 1;

(2)

(ii) S ∈ C n−2 [a; b].
The functions i; n and i; n depend on tension parameters. In practice, we choose i; n (x)=i; n (pi ; x);
i; n (x) = i; n (qi ; x); 06pi ; qi ¡∞. In the limiting case when pi ; qi → ∞ we require that limpi →∞
i; n (pi ; x) = 0, x ∈ [xi ; xi+1 ] and limqi →∞ i; n (qi ; x) = 0, x ∈ [xi ; xi+1 ] so that the function S in formula
(1) turns into a polynomial of the order n−2. Additionally, we require that if pi =qi =0 for all i we get
n−1
n−1
i)
i+1 )
a conventional polynomial spline of degree n with i; n (x)=− (x−x
; i; n (x)= (x−x
; hi =xi+1 −xi .
(n−1)!hi
(n−1)!hi
G
Consider now the problem of constructing a basis in the space Sn consisting of functions with
local supports of minimal length. For this, it is convenient to extend the mesh
by adding points
x−n+1 ¡ · · · ¡x−1 ¡a; b¡xN +1 ¡ · · · ¡xN +n−1 . As dim(SnG ) = (n)N − (n − 1)(N − 1) = N + n − 1, it

is sucient to construct a system of linearly independent GB-splines Bj; n ; j = −n + 1; : : : ; N − 1; in
SnG such that Bj; n (x)¿0 if x ∈ (xj ; xj+n ) and Bj; n ≡ 0 outside (xj ; xj+n ).
For n¿2 we require the fulllment of the normalization condition
N
−1
X

j=−n+1

Bj; n (x) ≡ 1

for x ∈ [a; b]:

(3)

65

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

According to (1), on the interval [xj+l ; xj+l+1 ], l = 0; : : : ; n − 1, the GB-spline Bj; n has the form

Bj; n (x) = Pj; l; n−2 (x) + Bj;(n−2)
(xj+l )j+l; n (x) + Bj;(n−2)
(xj+l+1 ) j+l; n (x);
n
n

(4)

where Pj; l; n−2 is a polynomial of order n−2.
Taking into account the continuity conditions for neighboring polynomials Pj; l−1; n−2 and Pj; l; n−2 ,
l = 1; : : : ; n − 1, in (4), we have the relations
Pj; l; n−2 (x) = Pj; l−1; n−2 (x) +

Bj;(n−2)
(xj+l )
n

n−3
X

(r)
r
zj+l;
n (x − xj+l ) =r!;

l = 1; : : : ; n − 1

(5)

r=0

(r)
(r)
(r)
with zj+l;
n = j+l−1; n (xj+l ) − j+l; n (xj+l ), r = 0; : : : ; n−3.
As Bj; n (x) = 0 if x 6∈ (xj ; xj+n+2 ) and by (5), the polynomials Pj; l; n−2 are identical to zero when
l = 0 and l = n. Then by repeated application of formula (5) we have

Pj; l; n−2 (x) =

l
X

Bj;(n−2)
(xj+l′ )
n

(r)
r
zj+l
′ ; n (x − xj+l′ ) =r!

r=0

l′ =1

n
X

=−

n−3
X

Bj;(n−2)
(xj+l′ )
n

n−3
X

(r)
r
zj+l
′ ; n (x − xj+l′ ) =r!;

l = 1; : : : ; n − 2:

(6)

r=0

l′ =l+1

In particular, the following identity is valid:
n−1
X

n−3
X

Bj;(n−2)
(xj+l )
n

(r)
r
zj+l;
n (x − xj+l ) =r! ≡ 0:

(7)

r=0

l=1

Using the expansion of polynomials by powers of x we can rewrite (7) in the form
n−1
X

Bj;(n−2)
(xj+l )
n

n−3
n−3
X
x X
=0

l=1

!

(r)
zj+l;
n

r=

(−xj+l ) r−
≡ 0:
(r−)!

(8)

Now by equating the coecients of the monomials x ,  = 0; 1; : : : ; n−3; in (8) to zero, we arrive
at a system of n−2 linear algebraic equations which denes the unknown quantities Bj;(n−2)
(xj+l );
n
l = 1; : : : ; n − 1,
n−1
X

Bj;(n−2)
(xj+l )
n

l=1

n−3
X

(r)
zj+l;
n

r=

(−xj+l ) r−
= 0;
(r − )!

 = 0; : : : ; n − 3:

To obtain the unique solution of this system we can use the normalization condition (3). Substituting formula (4) into the identity (3) written for x ∈ [xi ; xi+1 ], we obtain
i
X

Bj; n (x) = i; n (x)

j=i−n+1

i−1+1
X

Bj;(n−2)
(xi )
n

+ i; n (x)

j=i−n+1

Since according to (3)
i−1
X

j=i−n+1

Bj;(n−2)
(xi ) =
n

i
X

j=i−n+2

Bj;(n−2)
(xi+1 ) = 0;
n

i
X

j=i−n+2

Bj;(n−2)
(xi+1 )
n

+

i−1
X

j=i−n+2

Pj; i−j; n−2 (x) ≡ 1:

66

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

it follows from (6) that
i−1
X

Pj; i−j; n−2 (x) =

j=i−n+2

i−j
i−1 X
X

Bj;(n−2)
(xj+l )
n

j=i−n+2 l=1

n−3
X

(r)
r
zj+l;
n (x − xj+l ) =r! ≡ 1:

r=0

This gives us the system of linear equations
i−j
i−1 X
X

Bj;(n−2)
(xj+l )
n

n−3
X

(r)
zj+l;
n

r=

j=i−n+2 l=1

(−xj+l ) r−
= 0;  ;
(r−)!

 = 0; : : : ; n − 3;

where 0;  is the Kronecker symbol.
We can eliminate the unknowns analogously as has been done in [10, 12]. Having computed the
unknowns Bj;(n−2)
(xj+l ); l = 1; : : : ; n − 1, we nd the coecients of the polynomials Pj; l; n−2 ; l =
n
1; : : : ; n − 2, in (4) by using formulae (6). In this way, the computation of the coecients of the
polynomials Pj; l; n−2 can be realized starting from either the left or right endpoint of the support
interval.
3. Recurrence algorithm for the calculation of GB-splines
Let us dene the function
 (n−2)

(x); xj 6x6xj+1 ;


 j; n


(n−2)
xj+1 6x6xj+2 ;
j+1;
n (x);

Bj; 2 (x) =






(9)

x 6∈ (xj ; xj+2 );

0;

(n−2)
and j+1;
where the functions j;(n−2)
n
n are assumed to be positive and monotone on (xj ; xj+1 ) and
(xj+1 ; xj+2 ) respectively.
We will consider the sequence of GB-splines dened by the recurrence formula

Z

Bj; k (x) =

x

Bj; k−1 ()
d −
cj; k−1

xj

Z

x

xj+1

Bj+1; k−1 ()
d;
cj+1; k−1

k = 3; : : : ; n;

(10)

where
cj; k−1 =

Z

xj+k−1

Bj; k−1 () d:

xj

In practical calculations, an alternate representation of formula (10),
Bj; k (x) = −

Z

x

xj+k−1

Bj; k−1 ()
d +
cj; k−1

Z

x

xj+k

Bj+1; k ()
d;
cj+1; k−1

k = 3; : : : ; n

is useful.
By dierentiating formula (10) we obtain
Bj;′ k (x) = Bj; k−1 (x)=cj; k−1 − Bj+1; k−1 (x)=cj+1; k−1 ;

k = 3; : : : ; n:

(11)

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

67

Theorem 3.1. The recurrence formulae (9) and (10) dene the sequence of GB-splines of the form

Bj; k (x) =



Bj;(k−2)
(xj+1 ) j;(n−k)
(x);

n
k





(n−k)

(xj+l )j+l;
Pj; l; k−2 (x) + Bj;(k−2)

k
n (x)




+B(k−2) (x

xj 6x6xj+1 ;

) (n−k) (x)

xj+l 6x6xj+l+1 ; l = 1; : : : ; k − 2;

j+l+1
j; k
j+l; n




(k−2)
(n−k)


B
(xj+k−1 )j+k; n (x);


 j; k


 0;

(12)

xj+k−1 6x6xj+k ;
x 6∈ (xj ; xj+k );

k = 2; : : : ; n, where
Pj; l; k−2 (x) =

l
X

Bj;(k−2)
(xj+l′ )
k

l′ =1

=−

n−3
X

(r)
r−n+k
zj+l
=(r − n + k)!
′ ; n (x − xj+l′ )

r=n−k

k−1
X

Bj;(k−2)
(xj+l′ )
k

l′ =l+1

n−3
X

(r)
r−n+k
zj+l
=(r − n + k)!
′ ; n (x − xj+l′ )

(13)

r=n−k

and
k−1
X

Bj;(k−2)
(xj+l )
k

l=1

n−3
X

(r)
r−n+k
zj+l;
=(r − n + k)! ≡ 0;
n (x − xj+l )

k = 3; : : : ; n:

(14)

r=n−k

Proof. For k = 2 the formula (12) takes the form

Bj; 2 (x) =


Bj; 2 (xj+1 ) j;(n−2)
(x);

n








xj 6x6xj+1 ;

(n−2)
Bj; 2 (xj+1 )j+1;
n (x);

xj+1 6x6xj+2 ;

0;

x 6∈ (xj ; xj+2 ):

We choose Bj; 2 (xj+1 ) = 1, and then by (2), this formula coincides with (9).
Using mathematical induction, we assume that the assertion of the theorem is fullled for some
k ′ = k−1¡n − 1 (k = 3; : : : ; n−1). Let us show its validity for k6n. According to (10) and (12)
we have
Bj; k (x) =

1
cj; k ′

Bj; k (x) = −

′

′

−1)
Bj;(kk ′−2) (xj+1 ) j;(n−k
(x);
n

1
cj+1; k ′

′

′

x ∈ [xj ; xj+1 ];

(k −2)
(n−k −1)
Bj+1;
(x);
k ′ (xj+k ′ )j+k ′ ; n

x ∈ [xj+k ′ ; xj+k ′ +1 ]:

By virtue of (11) and because GB-splines have local supports,
′

Bj;(k−2)
(xj+1 ) = Bj;(kk ′−2) (xj+1 )=cj; k ′ ;
k
′

(k −2)
Bj;(k−2)
(xj+k−1 ) = −Bj+1;
k
k ′ (xj+k ′ )=cj+1; k ′ :

68

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

Therefore,
Bj; k (x) = Bj;(k−2)
(xj+1 ) j;(n−k)
(x);
n
k

x ∈ [xj ; xj+1 ];

(n−k)
(xj+k−1 )j+k−1;
Bj; k (x) = Bj;(k−2)
k
n (x);

(15)

x ∈ [xj+k−1 ; xj+k ]:

Let now formula (12) be fullled in [xj+l ; xj+l+1 ] for some l′ = l − 1¡k − 2 (l = 1; 2; : : : ; k − 2).
We must show its validity for l6k−2. According to (10), and by the induction assumption for
x ∈ [xj+l ; xj+l+1 ], we have
Bj; k (x) = Pj; l−1; k−2 (xj+l )
(n−k)
(k−2)
(n−k)
+ Bj;(k−2)
(xj+l−1 )j+l−1;
(xj+l ) j+l−1;
k
n (xj+l ) + Bj; k
n (xj+l )

+

1

Z

cj; k−1

x

Bj; k−1 () d −

xj+l

1
cj+1; k−1

Z

x

Bj+1; k−1 () d

xj+l

(n−k)
= Pj; l−1; k−2 (xj+l ) + Bj;(k−2)
(xj+l ) j+l−1;
k
n (xj+l ) +


l
X

×

′

′
Bj;(kk ′−2) (xj+l′ )

l =1

n−3
X

x

(n−k −1)
+ Bj;(kk ′−2) (xj+l )j+l;
()

n
′

xj+l

−
where
I=

l−1
X




1

I+
cj+1; k−1 

(k ′ −2)
Bj+1;
k ′ (xj+1+l′ )

l′ =1

=

l
X

(k ′ −2)
Bj+1;
k′

′

l′ =1

n−3
X

′

x


xj+l

(r)
zj+l
′; n

r=n−k ′

x

( − xj+1+l′ )r−n+k +1 

(r − n + k ′ + 1)! x

x
( − xj+l′ )r−n+k 

(r − n + k)! 

′


x 

−1)
()

xj+l

j+l

:

xj+l

Using the formula of dierentiation (11) we obtain
′

′

(k −2)
Bj;(k−2)
(xj+l ) = Bj;(kk ′−2) (xj+l )=cj; k ′ − Bj+1;
k
k ′ (xj+l )=cj+1; k ′ :

This permits us to transform expression (16) into the form
Bj; k (x) =

l−1
X
l′ =1

Bj;(k−2)
(xj+l′ )
k

n−3
X

r=n−k

(r)
zj+l
′; n

x


(k ′ −2)
(n−k ′ −1)
+Bj+1;
()
k ′ (xj+l+1 ) j+l; n

′

(r)
zj+1+l
′; n

j+l

(n−k
+ Bj;(kk ′−2) (xj+l+1 ) j+l;
n


(n−k ′ −1)
(xj+l )j+l;
()
n

r=n−k ′

(k −2)
Bj+1;
k ′ (xj+l′ )

(−xj+l′ )r−n+k +1 

(r − n + k ′ + 1)! x



n−3
X

x

′

(r)
zj+l
′; n

r=n−k ′
′

1
cj; k−1

(x j+1 − xj+l′ )r−n+k
(r − n + k)!

(n−k)
(k−2)
(n−k)
(k−2)
(n−k)
(x j+1 )zj+l;
(x j+1 )j+l;
(xj+l+1 ) j+l;
+ Bj;(k−2)
k
n + Bj; k
n (x) + Bj; k
n (x)

xj+l





;

(16)

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

+

l
X
l′ =1

−

l
X

l
X

(r)
zj+l
′; n

(x − xj+l′ )r−n+k
(r − n + k)!

(r)
zj+l
′; n

(x j+1 − xj+l′ )r−n+k
(r − n + k)!

r=n−k ′

Bj;(k−2)
(xj+l′ )
k

l′ =1

=

n−3
X

Bj;(k−2)
(xj+l′ )
k

n−3
X

r=n−k ′

69

(n−k)
(k−2)
(n−k)
Bj;(k−2)
(xj+l′ )zj+l
(x j+1 )j+l;
′ ; n + Bj; k
k
n (x)

l′ =1

(n−k)
(xj+l+1 ) j+l;
+ Bj;(k−2)
k
n (x)

+

l
X

Bj;(k−2)
(xj+l′ )
k

l′ =1

= Pj; l; k−2 (x) +

(n−k)
Bj;(k−2)
(x j+1 )j+l;
n (x)
k

n−3
X

(r)
zj+l
′; n

r=n−k ′

+

(x − xj+l′ )r−n+k
(r − n + k)!

(n−k)
Bj;(k−2)
(xj+l+1 ) j+l;
k
n (x):

We have now proved formula (12) with
Pj; l; k−2 (x) =

l
X

Bj;(k−2)
(xj+l′ )
k

l′ =1

n−3
X

(r)
zj+l
′; n

r=n−k

(x − xj+l′ )r−n+k
;
(r − n + k)!

l = 1; : : : ; k − 2:

(17)

Taking into account the conditions for continuity we obtain the validity of the formula (17) for
l=k −1. However, according to (15), Pj; k−1; k−2 ≡ 0. So from (17) for l=k −1 we obtain the identity
(14). By subtracting this identity from (17) we arrive at the second formula in (13). This proves
the theorem.
To use formulae (12) and (13) for calculations we rst need to nd the quantities Bj;(k−2)
(x j+1 );
k
l = 1; : : : ; k − 1; k = 3; : : : ; n. According to (11),
(k−3)
Bj;(k−2)
(x j+1 ) = Bj;(k−3)
k
k−1 (x j+1 )=cj; k−1 − Bj+1; k−1 (x j+1 )=cj+1; k−1 ;

l = 1; : : : ; k − 1; k = 3; : : : ; n:

(18)

In particular, it follows from here with Bj; 2 (xj+1 ) = 1 that
Bj;′ 3 (xj+1 ) =
Bj;′ 3 (xj+2 )

1
;
cj; 2

=−

Bj;′′4 (xj+1 ) =

1
cj+1; 2

;

1
;
cj; 2 cj; 3

Bj;′′4 (xj+2 )

=−

Bj;′′4 (xj+3 ) =

1
cj+1; 2

1
1
+
cj; 3 cj+1; 3

!

;

1
;
cj+2; 2 cj+1; 3

etc. Therefore, to nd the required values of the derivatives of the GB-splines in the interior nodes of
their support intervals, it is necessary to know the quantities cj; k , i.e., the integrals of the GB-splines
Bj; k ; k = 2; : : : ; n − 1.
Theorem 3.2. The integrals cj; k =
cj; k =

k−1
X
l=1

Bj;(k−2)
(x j+1 )
k

n−3
X

r=n−k−1

R xj+k
xj

(r)
zj+l;
n

Bj; k () d of GB-splines are given by the formula

(xj+ − x j+1 )r−n+k+1
;
(r − n + k + 1)!

 = 1; : : : ; k − 1; k = 2; : : : ; n − 1: (19)

70

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

Proof. For k = 2 according to (9) we obtain
cj; 2 =

xj+2

Z

xj

(n−3)
Bj; 2 () d = Bj; 2 (xj+1 )zj+1;
n;

Bj; 2 (xj+1 ) = 1;

which corresponds to the formula (19). Let us suppose by induction that the formula (19) holds for
all k ′ =k −1¡n−1 (k =3; : : : ; n−1). We must prove its validity for k ′ +1=k 6 n−1. By formula (12),
cj; k =

Z

xj+k

xj

+

Bj; k () d = Bj;(k−2)
(xj+1 ) j;(n−k−1)
(xj+1 )
n
k

k−2 Z
X
l=1

xj+l+1

x j+1

h

(n−k)
Pj; l; k−2 () + Bj;(k−2)
(x j+1 )j+l;
k
n ()

i

(n−k)
(k−2)
(n−k−1)
(xj+l+1 ) j+l;
(xj+k−1 )j+k−1;
+ Bj;(k−2)
k
n () d − Bj; k
n (xj+k−1 ):

(20)

Choosing the knot xj+ ∈ supp Bj; k ; 1 6  6 k − 1 and using formulae (13) we can transform expression (20) into the form
cj; k =

k−1
X

(n−k−1)
Bj;(k−2)
(x j+1 )zj+1;
n
k

+

l=1

+

" l
−1 X
X
l=1

"
k−2
X
l=

Bj;(k−2)
(xj+l′ )
k

l′ =1

n−3
X

(r)
zj+l
′; n

r=n−k

( − xj+l′ )r−n+k+1
(r − n + k + 1)!

#xj+l+1
k−1
n−3
X
X
( − xj+l′ )r−n+k+1 
(k−2)
(r)
:
−
Bj; k (xj+l′ )
zj+l′ ; n

(r − n + k + 1)! 
′
l =l+1

r=n−k

#xj+l+1




x j+1

x j+1

Collecting here the terms with Bj;(k−2)
(x j+1 ); l = 1; : : : ; k − 1, we have
k
cj; k =

k−1
X

(n−k−1)
Bj;(k−2)
(x j+1 )zj+1;n
+
k

l=1

+

Bj;(k−2)
(x j+1 )
k

n−3
X

(r)
zj+l;
n

r=n−k

l=+1

=

Bj;(k−2)
(x j+1 )
k

l=1

k−1
X

k−1
X

−1
X

Bj;(k−2)
(x j+1 )
k

l=1

n−3
X

(r)
zj+l;
n

r=n−k−1

n−3
X

(r)
zj+l;
n

r=n−k

(xj+ − x j+1 )r−n+k+1
(r − n + k + 1)!

(xj+ − x j+1 )r−n+k+1
(r − n + k + 1)!

(xj+ − x j+1 )r−n+k+1
(r − n + k + 1)!

This proves the theorem.
Theorem 3.3. If cj; k ; k = 2; : : : ; n − 1; are integrals of GB-splines Bj; k then the following equalities
are valid:
k−1
X
l=1

Bj;(k−2)
(x j+1 )
k

n−3
X

r=n−k−1+

(r)
zj+l;
n

(−x j+1 )r−n+k+1−
= cj; k · 0; 
(r − n + k + 1 − )!

 = 0; : : : ; k − 2; k = 2; : : : ; n − 2;

where 0;  is the Kronecker symbol.

(21)

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

71

Proof. We can write these identities
cj; k ≡ Fj; k (x) =

k−1
X

Bj;(k−2)
(x j+1 )
k

l=1

n−3
X

(r)
zj+l;
n

r=n−k−1

(x − x j+1 )r−n+k+1
;
(r − n + k + 1)!

k = 2; : : : ; n − 1:

(22)

For k = 2, formula (22) does not depend on x and coincides with (19). According to Theorem 3.2,
the polynomial Fj; k − cj; k ; 2 6 k 6 n − 1, of order k − 1 takes zero values at the points xj+ ;  =
1; : : : ; k − 1. Therefore, by the Fundamental Theorem of Algebra it must be identically equal to zero.
Using the expansion of polynomials in (22) by powers of x we obtain
cj; k =

k−1
X

Bj;(k−2)
(x j+1 )
k

k−2 
X
x
=0

l=1

n−3
X

! r=n−k−1+

(r)
zj+l;
n

(−x j+1 )r−n+k+1−
;
(r − n + k + 1 − )!

k = 2; : : : ; n − 1:

The right-hand side is a polynomial of order k − 1 for xed j; k while the left-hand side is constant.
It follows that the coecient of x equals cj; k when  = 0 and equals zero otherwise. From this we
obtain the equalities (21). This proves the theorem.
To construct the GB-spline Bj; k ; k = 3; : : : ; n, we can formulate an algorithm by applying formulae
(18) and (19), and requiring the calculation of the following quantities for GB-splines:
Bj;(k−2)
(xj+ )
k
..
.

···
..
.

Bj; 2 (xj+1 )
..
.

···
Bj;(n−2)
(xj+ )
n
··
·
( = 1; : : : ; n − 1)

(k−2)
Bj+n−k; 2 (xj+n−k+1 ) · · · Bj+n−k;
k (xj+n−k+ )
·
..
( = 1; : : : ; k − 1)
.
··
Bj+n−2; 2 (xj+n )

and the integrals of GB-splines
cj; 2
cj+1; 2
..
.

···
···
..
.

cj+n−k; 2
..
.

· · · cj+n−k; k
·
··

cj; k
cj+1; k
..
.

· · · cj; n−1
· · · cj+1; n−1
·
··

cj+n−2; 2 :
Algorithm 1. (a) Form the diagonal matrix A = {aij }; i; j = 1; : : : ; n − 1, with diagonal elements
al+1; l+1 = Bj+l; 1 (xj+l+2 ) = 1; l = 0; : : : ; n − 2. Attach to the matrix A at the left an additional column
with elements al+1; 0 = cj+l; 2 ; l = 0; : : : ; n − 2 calculated by formula (19).
(k−2)
(b) For k = 3; : : : ; n, using formula (18) we nd the elements a; l+1 = Bj+l−(k−2);
k (xj+ );  = l +
3 − k; : : : ; l + 1; l = n − 2; : : : ; k − 2, and place them on the main diagonal and on the rst k − 2
upper o-diagonals of the matrix A. At every step k (for k¡n − 1) we also calculate the elements
al+1; k−2 = cj+l−(k−2); k ; l = n − 2; : : : ; k − 2, using formula (19) and place them into the (k − 2)nd
column of the lower triangular part of the matrix A.

72

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

As a result the matrix A is transformed to the form


cj; 2
..
.





A =  cj+k−2; 2

..


.

cj+n−2; 2

Bj; 2 (xj+1 )
..
.

Bj;(k−2)
(xj+1 )
k
..
.

:::

cj+k−3; 3
..
.

· · · Bj;(k−2)
(xj+k−1 )
k
..
.

cj+n−3; 3

···

···





· · · Bj;(n−2)
(xj+k−1 )  :
n

..


.

···

cj+n−1−k; k+1

Bj;(n−2)
(xj+1 )
n

..

.

Bj;(n−2)
(xj+n−1 )
n

The (k−1)st column of the upper triangular part of the matrix A contains the quantities Bj;(k−2)
(xj+ );
k
 = 1; : : : ; k − 1; k = 2; : : : ; n. This permits us to construct the GB-splines Bj; k ; k = 2; : : : ; n, using
formulae (12) and (13). The integrals for these GB-splines are located along the main diagonal of
the matrix A.
The supports of the GB-splines Bj; k ; k = 2; : : : ; n, begin at the point xj . We can also consider an
alternative version of the above algorithm in which the GB-splines Bj+k; n−k ; k = n − 2; : : : ; 0, whose
supports end at the point xj+n are calculated.
Algorithm 2. (a) Form the diagonal matrix A of dimension (n − 1) × (n − 1) with diagonal elements
al+1; l+1 = Bj+l; 2 (xj+l+1 ) = 1; l = 0; : : : ; n − 2. Attach to A the additional (n)th row with elements
an; l+1 = cj+l; 2 ; l = 0; : : : ; n − 2, calculated by formula (19).
(k−2)
(b) For k = 3; : : : ; n; l = 0; : : : ; n − k, we nd al+1; l+ = Bj+l;
k (xj+l+ );  = k − 1; : : : ; 1 and (for
k¡n − 1) an−k+2; l+1 = cj+l; k by formulae (18) and (19).
As a result the matrix A takes the form
Bj;(n−2)
(xj+1 ) · · ·
n

..

.





A=





cj; k+1
..
.
cj; 3
cj; 2

Bj;(n−2)
(xj+1+n−k )
n
..
.

···

(k−2)
· · · Bj+n−k;
···
k (xj+1+n−k )
..
.

···
···

Bj;(n−2)
(xj+n−1 )
n

..

.




(k−2)
Bj+n−k;
k (xj+n−1 ) 
:
..


.



· · · Bj+n−2; 2 (xj+n−1 ) 
···
cj+n−2; 2

cj+n−k; 3
cj+n−k; 2

The elements of the (n − k + 1)th row in the upper triangular part of the matrix A permit us to
construct the GB-splines Bj+n−k; k ; k = 2; : : : ; n, using formulae (12) and (13). The integrals of these
GB-splines are located along the main diagonal of the matrix A.
4. Another representation for GB-splines
According to (12) and (13) the expressions for Bj; k ; k = 3; : : : ; n, in the subintervals [xj+l−1 ; x j+1 ]
and [x j+1 ; xj+l+1 ] dier by the quantity
(n−k)
(k−2)
−j+l−1;
(xj+l−1 )
n (x)Bj; k

+

(n−k)
(k−2)
+ j+l;
(xj+l+1 ):
n (x)Bj; k

"

(n−k)
j+l;
n (x)

−

(n−k)
j+l−1;n
(x)

+

n−3
X

r=n−k

#

− x j+1 )r−n−k
Bj;(k−2)
(x j+1 )
k
(r − n + k)!

(r) (x
zj+l;
n

73

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

By summing over the jumps we arrive at the representation
Bj; k (x) =

k−1
X


j+l; k (x)Bj;(k−2)
(x j+1 );
k

k = 3; : : : ; n

(23)

l=1

with
(n−k)

j+l; k (x) = j+l−1;
n (x)(x

− xj+l−1 ) +

"

(n−k)
j+l;
n (x)

−

(n−k)
j+l−1;
n (x)

+

n−3
X

− x j+1 )r−n+k
(r − n + k)!

(r) (x
zj+l;
n

r=n−k

(n−k)
× (x − x j+1 ) − j+l;
n (x)(x − xj+l+1 );

(x − y) =

(

1;

#

(24)

x ¿ y;

0; x¡y:

As (x − y) = 1 − (y − x) we obtain
Bj; k (x) = −

k−1
X


ˆ j+l; k (x)Bj;(k−2)
(x j+1 ) + Rj; k (x);
k

(25)

l=1

where
ˆ j+l; k is derived from
j+l; k by replacing (x − xj+m ) with (xj+m − x); m = l − 1; l; l + 1.
Now according to (14)
Rj; k (x) =

k−1
X

Bj;(k−2)
(x j+1 )
k

l=1

n−3
X

(r)
zj+l;
n

r=n−k

(x − x j+1 )r−n+k
≡ 0;
(r − n + k)!

k = 3; : : : ; n

and it follows from formulae (23) and (25) that Bj; k (x) ≡ 0 if x ∈= (xj ; xj+k ). Any of these formulae
can be used to dene the GB-spline of order k; 3 6 k 6 n.
We will transform the expression for the function
j+l; k in (24). Using the Taylor expansion of
(n−k)
(n−k)
the functions j+l;
n ; j+l; n with the remainder in integral form and the properties (2) we have

j+l; k (x) =

"Z

x

xj+l−1

+

"Z

x

x j+1

#

(x − )k−2 (n−1)
()d (x − xj+l−1 )

(k − 2)! j+l−1; n
(x − )k−2 (n−1)

() d
(k − 2)! j+l; n

×(x − x j+1 ) −

"Z

x

xj+l+1

Z

x

x j+1

(x − )k−2 (n−1)

() d
(k − 2)! j+l−1; n
#

(x − )k−2 (n−1)
() d (x − xj+l+1 ):

(k − 2)! j+l; n

From here we obtain for polynomial splines with
j; n (x) = −

(x − xj+1 )n−1
;
(n − 1)! hj

#

j; n (x) =

(x − xj )n−1
(n − 1)! hj

74

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

that

m; k (x) = (xm+1 − xm−1 )gk [x; xm−1 ; xm ; xm+1 ];
gk (x; y) = (−1)k (x − y)+k−1 =(k − 1)!;

m = j + l;

z+ = max(0; z)

and thus (23) is transformed to
Bj; k (x) = (xj+k − xj )gk [x; xj ; : : : ; xj+k ];

k = 3; : : : ; n:

The recurrence formula (10) takes the form [14]
Bj; k (x) =

xj+k − x
x − xj
Bj; k−1 (x) +
Bj+1; k−1 (x)
xj+k−1 − xj
xj+k − xj+1

with
cj; k =

Z

xj+k

Bj; k (x) d x =

xj

xj+k − xj
:
k

5. Properties of GB-splines
Let us formulate some properties of GB-splines which are similar to those of polynomial B-splines
[14].
Theorem 5.1. The functions Bj; k ; k = 2; : : : ; n; have the following properties:
(i) Bj; k (x)¿0 if x ∈ (xj ; xj+k ) and Bj; k (x) ≡ 0 if x 6∈ (xj ; xj+k );
(ii) the splines Bj; k have k − 2 continuous derivatives;
PN −1
(iii) for k ¿ 3 and x ∈ [a; b];
j=−k+1 Bj; k (x) = 1;
(iv) for x ∈ [xj ; xj+1 ];
j;(r)n (x)

=

n−r−1
Y
k=2

!

cj; k Bj; n−r (x); j;(r)n (x) =

n−r−1
Y

j = 0; : : : ; N − 1; r = 0; : : : ; n − 2;

(−cj−k+1; k )Bj−n+1+r; n−r (x)

k=2

where cj; k =

Z

xj+k

Bj; k () d:

xj

Proof. The functions Bj+; 2 given by formula (9) are positive if x ∈ (xj+ ; xj++2 ), while Bj+; 2 (x) = 0
if x 6∈ (xj+ ; xj++2 ), and are monotone on the intervals [xj++l ; xj++l+1 ]; l = 0; 1;  = 0; : : : ; k − 2.
Let us suppose by induction that Bj+; k−1 (x)¿0 if x ∈ (xj+ ; xj++k−1 ); Bj+; k−1 ≡ 0 if x 6∈ (xj+ ;
(k−3)
xj++k−1 ) and Bj+;
k−1 (x) is monotone on the intervals [xj++l ; xj++l+1 ]; l = 0; : : : ; k − 2 with
(k−3)
l+1
(−1) Bj+; k−1 (xj++l )¿0, l = 1; : : : ; k − 2;  = 0; 1. Then by formula (11) the function Bj;(k−2)
k
l+1 (k−2)
is monotone in [xj+l ; xj+l+1 ]; l = 0; : : : ; k − 1, and in addition (−1) Bj; k (xj+l )¿0; l = 1; : : : ; k − 1.
has exactly k − 2 zeros in (xj ; xj+k ) and by Rolle’s theorem, Bj; k does not vanish
Therefore, Bj;(k−2)
k
on (xj ; xj+k ). Taking formula (10) into account, we have Bj; k (x)¿0 if x ∈ (xj ; xj+k ) and Bj; k (x) = 0 if
x 6∈(xj ; xj+k ).

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

75

Property (ii) is obvious by virtue of the recurrence formula (10) and by continuity of the function Bj; 2 .
According to (12) for x ∈ [xj ; xj+1 ],
(n−2)
Bj; n (x) = Bj;(n−2)
(xj+1 ) j; n (x); Bj−n+1; n (x) = Bj−n+1;
n
n (xj )j; n (x):

(26)

Applying the dierentiation formula (11) we obtain
Bj;(r)n (x)

=

r
Y

cj;−1n−k Bj; n−r (x);

k=1

(r)
Bj−n+1;
n (x)

=

(27)

r
Y

−1
(−cj−n+1+k;
n−k )Bj−n+1+r; n−r (x);

x ∈ [xj ; xj+1 ]; r = 1; 2; : : : ; n − 2

k=1

and in particular,
Bj;(n−2)
(xj+1 )
n

=

n−2
Y

(n−2)
Bj−n+1;
n (xj )

cj;−1
n−k ;

=

k=1

n−2
Y

−1
(−cj−k;
k+1 ):

(28)

k=1

Therefore, if x ∈ [xj ; xj+1 ] then according to (26)–(28) we have
j;(r)n (x) =

n−r−1
Y

cj; k Bj; n−r (x); j;(r)n (x) =

k=2

n−r−1
Y

(−cj−k+1; k )Bj−n+1+r; n−r (x);

r = 0; : : : ; n − 2:

k=2

Applying the recurrence formula (10) for k¿2, we have for x ∈ [a; b]:
N
−1
X

N
−1
X

Bj; k (x) =

j=−k+1

j=−k+1

=

Z

x

x−k+1

"Z

x

xj

Bj; k−1 ()
d −
cj; k−1

B−k+1; k−1 ()
d −
c−k+1; k−1

Z

Z

x

x

BN; k−1 ()
d =
cN; k−1

xj+1

xN

Bj+1; k−1 ()
d
cj+1; k−1
Z

#
x0

x−k+1

B−k+1; k−1 ()
d = 1;
c−k+1; k−1

i.e.,
N
−1
X

Bj; k+1 (x) ≡ 1;

k = 3; : : : ; n if x ∈ [a; b]:

(29)

j=−k+1

This proves the theorem.
Corollary 5.2. The following identities are valid:
i−1
X

i−j
X

j=i−k+2 l=1

Bj;(k−2)
(xj+l )
k

n−3
X

r=n−k

(r)
zj+l;
n

(x − xj+l )r−n+k
≡ 1;
(r − n + k)!

k = 3; : : : ; n:

76

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

Proof. Substituting formula (12) into the identity (29) written for x ∈ [xi ; xi+1 ], we obtain
i
X

i−1
X

Bj; k (x) = i;(n−k)
(x)
n

j=i−k+1

j=i−k+1

+

i−1
X

i
X

Bj;(k−2)
(xi ) + i;(n−k)
(x)
n
k

Bj;(k−2)
(xi+1 )
k

j=i−k+2

Pj; i−j; k−2 (x) ≡ 1:

j=i−k+2

According to (29)
i−1
X

Bj;(k−2)
(xi ) =
k

j=i−k+1

i
X

Bj;(k−2)
(xi+1 ) = 0:
k

j=i−k+2

Then using (13) one obtains
i−1
X

Pj; i−j; k−2 (x) =

j=i−k+2

i−j
i−1
X
X

Bj;(k−2)
(xj+l )
k

j=i−k+2 l=1

n−3
X

r=n−k

(r)
zj+l;
n

(x − xj+l ) r−n+k
≡ 1:
(r − n + k)!

This proves the corollary.
Corollary 5.3. Let Fj; k ; k = 2; : : : ; n − 1; be polynomials as dened in (22). Then the following
equalities are valid:
N
−1
X

j=−k+1

Z

b

Bj; k (x) dx =

a

N
−1
X

Fj; k (x) = b − a;

k = 2; : : : ; n − 1:

j=−k+1

Proof. Integrating the identity (29) on the interval [a; b] and using (22) we obtain
N
−1
X

j=−k+1

Z

b

Bj; k (x) dx =

a

N
−1
X

Z

N
−1
X

cj; k =

j=−k+1

=

j=−k+1

xj+k+1

Bj; k (x) dx

xj
N
−1
X

Fj; k (x) = b − a:

j=−k+1

This proves the corollary.
Theorem 5.4. The GB-splines Bj; k ; k = 2; : : : ; n; have supports of minimum length.
Proof. It follows from the explicit formula (9) that the support of the GB-spline Bj; 2 cannot be
reduced. Let us suppose that the assertion of the theorem is fullled for some k ′ = k − 1¡n (k =
3; : : : ; n). Using mathematical induction we will prove its validity for k ′ + 1 = k6n.
By the properties of the functions j+l; n , j+l; n ; 0 6 l 6 k − 1, and by formula (12), a GB-spline
Bj; k cannot be dierent from zero on only a part of the subinterval [xj+l ; xj+1+l ]; 06l6k − 1. If we
suppose that Bj; k is zero on interval [xj+l ; xj+1+l ]; 06l6k − 1, then due to the continuity of Bj;(k−2)
,
k
(k−2)
(k−2)
we have Bj; k (xj+l ) = Bj; k (xj+l+1 ) = 0.

77

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Using formula (11) one can show, however, that (−1)l+1 Bj;(k−2)
(xj+l )¿0; l = 1; : : : ; k − 1. For
k
′
l+1 (k −2)
k = 2 we have Bj; 2 (xj+1 ) = 1. Suppose by induction that (−1) Bj; k ′ (xj+l )¿0; l = 1; : : : ; k ′ − 1,
for 16k ′ 6k − 1. By formula (11) we get
(−1)l+1 Bj;(k−2)
(xj+l ) = (−1)l+1
k



′



′

(k −2)
B (k ′−2) (xj+l ) Bj+1;
k ′ (xj+l ) 
 j; k
−
cj; k ′
cj+1; k ′
′

′

= (−1)

l+1

B(k −2) (xj+1+(l−1) )
Bj;(kk ′−2) (xj+l )
l j+1; k ′
+ (−1)
¿0;
cj; k ′
cj+1; k ′

l = 1; : : : ; k − 1:

We have obtained a contradiction. This proves the theorem.
Theorem 5.5. The GB-splines Bj; k ; j = −k + 1; : : : ; N − 1; k = 2; : : : ; n; are linearly independent and
form a basis for the space SkG of generalized splines.
Proof. Let us assume to the contrary that there exist constants bj; k ; j =−k +1; : : : ; N −1; k =2; : : : ; n,
which are not all equal to zero and such that
b−k+1; k B−k+1; k (x) + · · · + bN −1; k BN −1; k+1 (x) = 0;

x ∈ [a; b]:

(30)

According to formula (9), and taking into account the properties of the functions i; n and i; n in
(2), we obtain from (30) for k = 2
N
−1
X

bj; 2 Bj; 2 (xi ) = bi−1; 2 = 0;

i = 0; : : : ; N:

j=−1

Thus, bj; 2 = 0; j = −1; : : : ; N − 1, and the functions Bj; 2 ; j = −1; : : : ; N − 1, are linearly independent.
Suppose by induction that the functions Bj; k ′ ; j = −k ′ ; : : : ; N − 1 are linearly independent for some
k ′ = k − 1¡n (k = 3; : : : ; n). We will prove the assertion of theorem for k ′ + 1 = k 6 n. Dierentiating
the equality (30) and using the recurrence formula (10) we have
N
−1
X

bj; k Bj;′ k (x) =

j=−k+1

N
−1
X

bj; k

j=−k+1

= b−k+1; k

"

Bj; k−1 (x) Bj+1; k−1 (x)
−
cj; k−1
cj+1; k−1

#

N
−1
X
BN; k−1 (x)
B−k+1; k−1 (x)
Bj; k−1 (x)
− bN −1; k
= 0:
+
(bj; k − bj−1; k )
c−k+1; k−1
cj; k−1
cN; k−1
j=−k+2

(31)

The supports of the GB-splines Bj; k−1 ; j = −k + 1; N , however, are outside the interval [a; b]. By
the induction assumption, the GB-splines Bj; k−1 ; j = −k + 2; : : : ; N − 1 are linearly independent and
thus from (31) we get bj; k − bj−1; k = 0; j = −k + 2; : : : ; N − 1 or bj; k = c = const; j = −k + 1; : : : ; N − 1.
By Theorem 5.1, the GB-splines Bj; k ; j = −k + 1; : : : ; N − 1 give a partition of unity on the interval
[a; b]. Using this property, and Eq. (29), we arrive at the equality
N
−1
X

j=−k+1

bj; k Bj; k (x) = c

N
−1
X

j=−k+1

Bj; k (x) = c · 1 = 0:

78

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

Therefore, bj; k = 0; j = −k + 1; : : : ; N − 1 and the GB-splines Bj; k ; j = −k + 1; : : : ; N − 1, k = 2; : : : ; n,
are linearly independent.
Since by Denition 2.1, dim(SkG ) = (k)N − (k − 1)(N − 1) = N + k − 1, we see that the GB-splines
Bj; k ∈ SkG ; j = −k + 1; : : : ; N − 1 form a basis of this space. This proves the theorem.
By virtue of this theorem, any spline S ∈ SkG ; k = 2; : : : ; n, can be uniquely written in the form
S(x) =

N
−1
X

bj; k Bj; k (x) for x ∈ [a; b]

(32)

j=−k+1

for some constant coecients bj; k .
Corollary 5.6. Any spline S 6≡ 0 in SkG ; k = 2; : : : ; n; with nite support of minimal length coincides
with a GB-spline up to a constant multiplier.
Proof. By Theorem 5.4 the minimal support of a spline S ∈ SkG , k = 2; : : : ; n dierent from identical
zero, is an interval (xi ; xi+k ); i = 0; : : : ; N − k. Using representation (32) we get
S(x) = bi−k+1; k Bi−k+1; k + · · · + bi+k−1; k Bi+k−1; k (x):
As S ≡ 0 for x 6∈ (xi ; xi+k ), when choosing sequentially x ∈ (xp ; xp+1 ); p = i − k + 1; : : : ; i − 1, we
obtain bp; k = 0. In the same manner, bp; k = 0 for p = i + k − 1; : : : ; i + 1. Therefore, S(x) = bi; k Bi; k (x).
This proves the corollary.

6. Series of GB-splines
In practical applications such as approximation of functions, discrete data etc., one considers linear
combinations of GB-splines. According to Theorem 5.5, any generalized spline S ∈ SkG ; k = 2; : : : ; n,
can be uniquely represented in the form
S(x) =

N
−1
X

bj; k Bj; k (x) for x ∈ [a; b]

(33)

j=−k+1

with some constant coecients bj; k .
Let us study how the behaviour of a spline S depends on the coecients bj; k . Since GB-splines
are local, from (33) we obtain the inequalities
min

i−k+16j6i

bj; k 6 S(x) =

i
X

j=i−k+1

bj; k Bj; k (x) 6

max

i−k+16j6i

bj; k ;

xi 6 x 6 xi+1 ; k = 2; : : : ; n:

(34)

Hence it follows that the behaviour of the spline S on the interval [xi ; xi+1 ] is determined by the
coecients bi−k+1; k ; : : : ; bi; k . In particular, in order for a spline S to be zero at a point of the interval
[xi ; xi+1 ], it is necessary that bj; k bj+1; k ¡0 for some i − k + 1 6 j 6 i.

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

79

The estimate (34) can be substantially improved on. Applying the dierentiation formula (11), we
obtain for r 6 k − 2
N
−1
X

S (r) (x) =

bj;(r)k Bj; k−r (x);

(35)

j=−k+r+1

where
bj;(l)k

=



b ;

 j; k

l = 0;

(l−1)

bj; k




(l−1)
− bj−1;
k

cj;k−l

(36)
;

l = 1; 2; : : : ; r:

Lemma 6.1. If bj; k ¿ 0 (60); j = −k + 1; : : : ; N − 1; k = 2; : : : ; n; then S(x) ¿ 0(6 0) for all x.
The conclusion is obvious, because the GB-splines Bj; k are nonnegative.
Lemma 6.2. If bj; k ¿bj−1; k (bj; k ¡bj−1; k ); j = −k + 2; : : : ; N − 1; k = 3; : : : ; n; then the function S is
monotonically increasing (decreasing).
Proof. According to formulae (35) and (36), we have
S ′ (x) =

N
−1
X

bj;(1)k Bj; k−1 (x);

bj;(1)k =

j=−k+2

bj; k − bj−1; k
:
cj; k−1

Because the GB-splines Bj; k−1 ; k = 3; : : : ; n, are nonnegative, the formula above and Lemma 1 imply
that S is monotonic. This proves the lemma.
(1)
(1)
(1)
Lemma 6.3. If bj;(1)k ¿bj−1;
k (bj; k ¡bj−1; k ); j = −k + 3; : : : ; N − 1; k = 4; : : : ; n; then the function S is
convex downwards (upwards).

Proof. By virtue of (35) and (36), we have
′′

S (x) =

N
−1
X

bj;(2)k Bj; k−2 (x);

j=−k+3

bj;(2)k

(1)
bj;(1)k − bj−1;
k
:
=
cj; k−2

(37)

Because the GB-splines Bj; k−2 ; k = 4; : : : ; n, are nonnegative, taking into account Lemma 1 we obtain
that S is convex. This proves the lemma.
Let Z[a; b] (f) denote the number of isolated zeros of a function f on the interval [a; b].
Lemma 6.4. If the spline S(x) =
terval of [a; b]; then
Z[a; b] (S) 6 N + k − 2:

PN −1

j=−k bj; k Bj; k (x);

k = 2; : : : ; n; is not identically zero on any subin-

80

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

Proof. According to (35) and (9), for x ∈ [xi ; xi+1 ], we have
S (k−2) (x) =

N
−1
X

(k−2) (n−2)
bj;(k−2)
Bj; 2 (x) = bi−1;
(x) + bi;(k−2)
i;(n−2)
(x):
n
k
k i; n
k

j=−1

This function has at most one zero on [xi ; xi+1 ], because the functions i;(n−2)
and i;(n−2)
are monoton
n
(k−2)
nous and nonnegative on this subinterval. Hence Z[a; b] (S
) 6 N . Then, according to Rolle’s
theorem [14], we nd Z[a; b] (S) 6 N + k − 2. This proves the lemma.
Denote by supp Bj; k = {x|Bj; k (x) 6= 0}; k = 2; : : : ; n, the support of GB-spline Bj; k , i.e. the interval
(xj ; xj+k ).
Theorem 6.5. Assume that −k+1 ¡−k+2 ¡ · · · ¡N −1 ; k = 2; : : : ; n. Then
D = det(Bj; k (i )) 6= 0;

i; j = −k + 1; : : : ; N − 1

if and only if
j ∈ supp Bj; k ;

j = −k + 1; : : : ; N + 1:

(38)

If condition (38) is satised; then D¿0.
Proof. Let us prove the theorem by induction. It is clear that the theorem holds for a single basis
function. Assume that it also holds for l−1 basis functions. Let us show that if (38) is satised,
then D 6= 0 for l basis functions.
Let l ∈= supp Bl; k . If l lies to the left (right) with respect to the support of Bl; k then the last
column (line) of the determinant D consists of zeros, i.e., D = 0. If l ∈ supp Bl; k and D = 0, then
there exists a nonzero vector c = (c−k+1; k ; : : : ; cl−k; k ) such that
S(p ) =

l−k
X

cj; k Bj; k (p ) = 0;

p = −k + 1; : : : ; l − k;

j= −k+1

i.e., the spline S has l isolated zeros. But this contradicts Lemma 6.4, which states that S can have
no more than l−1 isolated zeros. Hence c = 0 and D 6= 0.
Now it only remains to prove that D¿0 if (38) is satised. Let us choose xp ¡p ¡xp+1 for all p.
Then the diagonal elements of D are positive and all the elements above the main diagonal are
zero, i.e., D¿0. It is clear that D depends continuously on p , p = −k + 1; : : : ; l − k, and D 6= 0
for p ∈ supp Bp; k . Hence the determinant D is positive if condition (38) is satised. This proves the
theorem.
The following three statements follow immediately from Theorem 6.5.
Corollary 6.6. The system of GB-splines {Bj; k }; j = −k + 1; : : : ; N − 1; k = 2; : : : ; n; is a weak
Chebyshevian system in the sense of [6]; i.e., for any −k+1 ¡−k+2 ¡ · · · ¡N −1 we have D ¿ 0;
and D¿0 if and
only if condition (38) is satised. If the latter is satised; then the generalized
PN −1
spline S(x) = j=
−k+1 bj; k Bj; k (x); k = 2; : : : ; n; has no more than N + k − 2 isolated zeros.

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

81

Corollary 6.7. If the conditions of Theorem 6.5 are satised; the solution of the interpolation
problem
S(i ) = fi ;

i = −k + 1; : : : ; N − 1;

fi ∈ R

(39)

exists and is unique.
Let A = {aij }; i = 1; : : : ; m; j = 1; : : : ; n; be a rectangular (m × n) matrix with m 6 n. The matrix
A is said to be totally nonnegative (totally positive) [4] if the minors of all orders of the matrix are
nonnegative (positive), i.e., for all 16l6m we have
det(aip jq ) ¿ 0 (¿0)

for all 1 6 i1 ¡ · · · ¡il 6 m; 1 6 j1 ¡ · · · ¡jl 6 n:

Corollary 6.8. For arbitrary integers −k + 16−k+1 ¡ · · · ¡l−k 6N − 1 and −k+1 ¡−k+2 ¡ · · ·
¡l−k ; k = 2; : : : ; n; we have
Dl = det{Bj ; k (i )} ¿ 0;

i; j = −k + 1; : : : ; l − k;

and Dl ¿0 if and only if
j ∈ supp Bj ; k ;

j = −k + 1; : : : ; l − k;

i.e., the matrix {Bj; k (i )}; i; j = −k + 1; : : : ; N − 1 is totally nonnegative.
The last statement is proven by induction on the basis of Theorem 6.5 and the recurrence relations
for the minors of the matrices {Bj; k (i )}; k = 2; : : : ; n. The proof does not dier from that described
by Schumaker [14].
Since the supports of GB-splines are compact, the matrix of the system (39) is a banded matrix
and has 2k − 1 nonzero diagonals in general. If the knots of the spline xi ; i = −k + 1; : : : ; N − 1; are
placed in a suitable manner, then the number of nonzero diagonals of this matrix can be reduced
to k − 1.
De Boor and Pinkus [2] proved that linear systems with totally nonnegative matrices can be
solved by Gaussian elimination without choosing a pivot element. Thus, the system (39) can
be solved eciently by the conventional Gauss method.
Denote by S − (C) the number of sign changes (variations) in the sequence of components of the
vector C = (v1 ; : : : ; vn ), with zero being neglected. Karlin [6] showed that if a matrix A is totally
nonnegative, then it decreases the variation, i.e.
S − (AC) 6 S(C):
By virtue of Corollary 6.8, the totally nonnegative matrix {Bj; k (i )}; i; j = −k + 1; : : : ; N − 1; k =
2; : : : ; n, formed by the GB-splines decreases the variation.
For a bounded real function f, let S − (f) be the number of sign changes of the function f on
the real axis R without taking into account the zeros
S − (f) = sup S − [f(1 ); : : : ; f(p )];
p

1 ¡2 ¡ · · · ¡p :

82

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

N −1
Theorem 6.9. The generalized spline S(x) = j=
−k+1 bj; k Bj; k (x); k = 2; : : : ; n; is a variation diminishing function; i.e., the number of sign changes of S does not exceed the one in the sequence of
its coecients

P



SR− 

N
−1
X

j= −k+1



bj; k Bj; k 6 S − (b);

b = (b−k+1; k ; : : : ; bN −1; k ):

Proof. We use the approach proposed by Schumaker [14]. Let S − (b) = d − 1. Let us divide the
coecients bj; k into d groups:
b−k+1; k ; : : : bk2 ; k ; bk2 +1; k ; : : : ; bk3 ; k ; : : : ; bkd +1; k ; : : : ; bN −1; k :
In each group at least one coecient is not zero, and all the nonzero coecients have the same
sign.
Putting k1 = −k and kd+1 = N − 1, we dene the function
B̃j; k (x) =

kj+1
X

|bi; k |Bi; k (x);

j = 1; : : : ; d:

i=kj +1

Then for arbitrary 1 ¡2 ¡ · · · ¡d we have
det(B̃j; k (i ))di; j=1

=

k2
X

···

kd+1
X

|b1 ; k | · · · |bd ; k |det(Bj; k (i )) ¿ 0;

d =kd +1

1 =k1 +1

i = 1; : : : ; d; j = 1 ; : : : ; d ; k = 2; : : : ; n
by virtue of Corollary 6.8 and because at least one coecient bi; k is not zero in each group. It is
clear that we can choose 1 ¡2 ¡ · · · ¡d such that det(B̃j; k (i ))¿0. Hence the functions B̃j; k are
linearly independent.
Assume that =±1 is the sign of the rst group of the coecients bi; k . Let us take b̃i; k =(−1)i−1 ,
i = 1; 2; : : : ; d. Then
S̃(x) =

d
X

b̃i; k B̃i; k (x) ≡ S(x) =

i=1

N
−1
X

bj; k Bj; k (x):

j=−k+1

Applying Lemma 6.4, we obtain


Z

N
−1
X

j=−k+1



bj; k Bj; k  = Z

d
X
i=1

!

b̃i; k Bi; k 6 d − 1 = S − (b−k+1; k ; : : : ; bN −1; k );

k = 2; : : : ; n:

This proves the theorem.

The statement of Theorem 6.9 can be rened, namely we can point out a relation between the
point at which the spline changes its sign and the corresponding spline coecient. The coecient
corresponds to the GB-spline whose support includes the point of the sign change [see (34)].

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

83

Theorem 6.10. Assume
that the inequalities (−1)i S(i )¿0; i = 1; 2; : : : ; d; are valid for the generPN −1
alized spline S(x) = j=−k+1 bj; k Bj; k (x) at some 1 ¡2 ¡ · · · ¡d . Then there exist −k + 1 6 j1 ¡j2
¡ · · · ¡jd 6 N − 1 such that
(−1)i bji ; k Bji ; k (i )¿0;

i = 1; : : : ; d:

The proof of this statement does not dier from the proof of the corresponding theorem for polynomial B-splines [14].

7. Invariance of generalized splines with respect to ane transformations
In some applications of spline approximation we encounter ane transformations of the independent variable: x̂ = px + q, where p 6= 0 and q are constant. It is well-known that the usual
Lagrange–Newton, Chebyshev, etc., polynomials are invariant with respect to such transformations.
Let us show that generalized splines also have this property.
Let SÌ‚ nG be a set of generalized splines on the mesh
ˆ = {x̂i | x̂i = pxi + q; i = 0; : : : ; N } which is
obtained from the linear space SnG by ane transformation of the variable x.
Theorem 7.1. An approximating generalized spline S ∈ SnG is invariant with respect to ane transformations of the real axis R = (−∞; ∞).
Proof. The function Bj; 2 in (9) can be written in the form

Bj; 2 (x) =











x − xj
hj

(n−2)
j; n

!


(n−2) x − xj+1

’j+1;

n


hj+1




0

where
j; n

x − xj
hj

!

hjn−2

= j; n (x);

;

xj 6 x 6 xj+1 ;

!

;

xj+1 6 x 6 xj+2 ;
otherwise;

x − xj+1
hj+1

’j+1; n

!

n−2
hj+1
= j+1; n (x):

Using the change of the variable x̂ = px + q we get
(n−2)
j; n

(n−2)
’j+1;
n

!

=

x − xj+1
hj+1

!

x − xj
hj

(n−2)
j; n

=

(n−2)
’j+1;
n

x̂ − x̂j
x̂j+1 − x̂j

!

;

x̂ − x̂j+1
x̂j+2 − x̂j+1

!

:

84

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

Therefore, Bj; 2 (x)= B̂j; 2 (x̂). Let the equality Bj; l (x)= B̂j; l (x̂) be fullled for l=k −1¡n (k =3; : : : ; n).
By virtue of the recurrence relation (10) we have
Z

Bj; k (x) =

x

xj

Bj; k−1 ()
d −
cj; k−1

x

Z

xj+1

Bj+1; k−1 ()
d;
cj+1; k−1

k = 3; : : : ; n;

with
cj; k−1 =

xj+k−1

Z

Bj; k−1 () d:

xj

Using the substitution ˆ = p + q we obtain by induction
cj; k−1 =

x̂j+k−1

Z

B̂j; k−1 (ˆ)

x̂j

Z

Bj; k (x) =

x̂

x̂j

1
1
d ˆ = ĉj; k−1 ;
p
p

B̂j; k−1 (ˆ)
d ˆ −
ĉj; k−1

Z

x̂

x̂j+1

B̂j+1; k−1 (ˆ)
d ˆ = B̂j; k (x̂);
ĉj+1; k−1

k = 3; : : : ; n:

ˆ respectively, conIf now S ∈ SnG and Ŝ ∈ Ŝ nG are approximating splines on the meshes
and
,
nected by an ane transformation x̂ = px + q, then by the uniqueness of the spline representation
as a linear combination of GB-splines we obtain
S(x) =

X

bj Bj; n (x) =

X

bj B̂j; n (x̂) = Ŝ(x̂):

(40)

j

j

Therefore, the approximating generalized spline S is invariant with respect to ane transformations
of its variable. This proves the theorem.
It was shown above that Bj; k (x) = B̂j; k (x̂); k = 2; : : : ; n. By dierentiation of this equality we obtain
d
d
d
d x̂
[Bj; k (x)] = [B̂j; k (x̂)] = [B̂j; k (x̂)] = pB̂′j; k (x̂)]; k = 3; : : : ; n:
dx
dx
d x̂
dx
Dierentiating now the equality (40), we can write down
S ′ (x) =

X

bj Bj;′ n (x) =

X

bj pB̂j;′ n (x̂) = pŜ ′ (x̂):

j

j

By repeated dierentiation of the last and next to last equalities we arrive at
S (r) (x) =

X

bj Bj;(r)n (x) =

j

X

bj pr B̂j;(r)n (x̂) = pr Ŝ (r) (x̂);

r = 1; : : : ; n − 2:

j

8. Local approximation by GB-splines
Using the locality of GB-splines one can reduce the representation of a spline S as a linear
combination of GB-splines (33) for k = n to the form
S(x) =

i
X

j=i−n+1

bj; n Bj; n (x);

x ∈ [xi ; xi+1 ];

i = 0; 1; : : : ; N − 1:

(41)

B.I. Ksasov, P. Sattayatham / Journal of Computational and Applied Mathematics 104 (1999) 63–88

85

Theorem 8.1. The restriction (41) of the spline S to the interval [xi ; xi+1 ] can be written in the
form
(n−2)
(n−2)
S(x) = Pi; n−2 (x) + bi−1;
i; n (x);
n i; n (x) + bi; n

(42)

where
iâˆ