New York Journal of Mathematics
New York J. Math. 10 (2004) 69–81.
Error inequalities for a corrected interpolating
polynomial
Nenad Ujević
Abstract. A corrected interpolating polynomial is derived. Error inequalities
of Ostrowski type for the corrected interpolating polynomial are established.
Some similar inequalities are also obtained.
Contents
1. Introduction
2. Corrected interpolating polynomial
3. Error inequalities
References
69
70
76
80
1. Introduction
Many error inequalities in polynomial interpolation can be found in [2] and
[12]. These error bounds for interpolating polynomials are usually expressed by
means of the norms ·p , 1 ≤ p ≤ ∞. Some new error inequalities in polynomial
interpolation can be found in [16]. In this paper we derive error inequalities for a
corrected interpolating polynomial. Similar inequalities are obtained in numerical
integration. For example see [3]–[11], [14] and [15]. In some of the mentioned papers
we can find estimations for errors of quadrature formulas which are expressed by
means of the differences Γk − γk , S − γk , Γk − S, where Γk , γk are real numbers
such that γk ≤ f (k) (t) ≤ Γk , t ∈ [a, b] (k is a positive
integer while [a, b] is an
interval of integration) and S = f (k−1) (b) − f (k−1) (a) /(b−a). It is shown that the
estimations expressed in such a way can be much better than estimations expressed
by means of the norms f (k) p , 1 ≤ p ≤ ∞. Furthermore, it is well-known that
corrected quadrature formulas have better estimations of errors than corresponding
original formulas.
As we know there is a close relationship between interpolation polynomials and
quadrature rules. Thus, it is a natural try to establish similar error inequalities in
Received November 5, 2003.
Mathematics Subject Classification. 26D10, 41A80, 65D05.
Key words and phrases. corrected interpolating polynomial, representation of remainder, Ostrowski type inequalities.
ISSN 1076-9803/04
69
70
Nenad Ujević
polynomial interpolation. The usual procedure is to find an interpolating formula
and then we write a corresponding quadrature formula. Here we reverse the procedure. We have some results for quadrature formulas and we derive corresponding
results for interpolating polynomials.
We first establish general error inequalities, expressed by means of f (k) − Pm ,
where Pm is any polynomial of degree m and then we obtain inequalities of the
above mentioned types. For that purpose we derive a representation of remainder
in corrected interpolating polynomial. This is done in Section 2. In the same
section we give a relationship between the corrected interpolating polynomial and
a corresponding quadrature rule. In Section 3 we obtain the error inequalities.
Finally, we emphasize that the usual error inequalities in polynomial interpolation (for the Lagrange interpolating polynomial Ln (x)) are given by means of the
(n+1)th derivative while in this paper we can find these error inequalities expressed
by means of the kth derivative for k = 1, 2, . . . , n.
2. Corrected interpolating polynomial
Let π = {a = x0 < x1 < · · · < xn = b} be a given subdivision of the interval [a, b]
and let f : [a, b] → R be a given function. The Lagrange interpolation polynomial
is given by
Ln (x) =
(2.1)
n
pni (x)f (xi ),
i=0
where
(2.2)
pni (x) =
(x − x0 ) · · · (x − xi−1 )(x − xi+1 ) · · · (x − xn )
,
(xi − x0 ) · · · (xi − xi−1 )(xi − xi+1 ) · · · (xi − xn )
for i = 0, 1, . . . , n. We have the Cauchy relations ([12, pp. 160-161]),
n
(2.3)
pni (x) = 1
i=0
and
(2.4)
n
pni (x)(x − xi )j = 0, j = 1, 2, . . . , n.
i=0
Let π = {x0 = a < x1 < · · · < xn = b} be a given uniform subdivision of the
interval [a, b], i.e., xi = x0 + ih, h = (b − a)/n, i = 0, 1, 2, . . . , n. Then the Lagrange
interpolating polynomial is given by
Ln (x) = Ln (x0 + th)
n t(t
= (−1)
n
− 1) · · · (t − n) i n f (xi )
,
(−1)
n!
i t−i
i=0
where t ∈
/ {0, 1, 2, . . . , n}, 0 < t < n.
As we know the divided difference of the first order of the function f is given by
f [x0 ; x1 ] =
f (x1 ) − f (x0 )
.
x1 − x0
Error inequalities
71
The divided difference of order n is defined via the divided differences of order n − 1
by the recurrence formula
f [x1 ; x2 ; . . . ; xn ] − f [x0 ; x1 ; . . . ; xn−1 ]
f [x0 ; x1 ; . . . ; xn ] =
.
xn − x1
The following lemma is valid ([2, p. 86]):
Lemma 2.1. The nth-order divided difference satisfies the relation
n
f (xi )
.
f [x0 ; x1 ; . . . ; xn ] =
(x
−
x
)
·
·
·
(x
−
x
i
0
i
i−1 )(xi − xi+1 ) · · · (xi − xn )
i=0
The interpolating polynomial can be written in the Newton form as
Ln (x) = f (x0 ) +
n−1
(x − x0 ) · · · (x − xi )f [x0 ; . . . ; xi+1 ]
i=0
= f (x0 ) +
n−1
ωi (x)f [x0 ; . . . ; xi+1 ] ,
i=0
where
ωi (x) = (x − x0 )(x − x1 ) · · · (x − xi ),
(2.5)
for i = 0, 1, 2, . . . , n.
We also recall some properties of Euler polynomials. The Euler polynomials are
defined by the relation
∞
tk
2ext
=
, |t| < π,
E
(x)
k
et + 1
k!
k=0
such that
1
E0 (x) = 1, E1 (x) = x − , E2 (x) = x2 − x, . . .
2
(2.6)
We have
Ek (x) = kEk−1 (x) or
(2.7)
1
(2.8)
0
and
Ek (x)dx =
Ek+1 (x)
, k = 1, 2, . . . ,
k+1
En (t)Em (t)dt = (−1)n 4(2m+n+2 − 1)
(2.9)
0
1
En (t)dt =
m!n!
Bm+n+2
(m + n + 2)!
En+1 (1) − En+1 (0)
, n = 1, 2, . . . .
n+1
We also have
(2.10)
Ek (0) = −Ek (1) = −2
2k+1 − 1
Bk+1 ,
k+1
where Bk are Bernoulli numbers such that
1
1
E1 (0) = − , E2 (0) = 0, E3 (0) = , E4 (0) = 0, . . .
2
4
or generally E2n (0) = 0. Further properties of Euler polynomials can be found in
[1].
72
Nenad Ujević
Lemma 2.2. Let Pm (t) be any polynomial of degree ≤ m and let π be a given
subdivision of the interval [a, b]. Then
x
n
t − xi
k
dt = 0,
pni (x)(x − xi )
Pm (t)Ek
x − xi
xi
i=0
for 0 ≤ k + m ≤ n − 1 and x ∈ [a, b], where Ek (t) are Euler polynomials.
Proof. Let x be a real number. Then we have
m
Pm (t) =
cj (x − t)j ,
j=0
for some coefficients cj = cj (x), j = 0, 1, . . . , m. (This is a consequence of the
Taylor formula.) Thus,
x
x
m
t − xi
t − xi
j
dt =
dt.
Pm (t)Ek
cj
(x − t) Ek
x − xi
x − xi
xi
xi
j=0
We now calculate
x
x−xi
t − xi
u
j
j
dt =
du
(x − t) Ek
(x − xi − u) Ek
x − xi
x − xi
xi
0
j
x−xi u
u
= (x − xi )j
1−
du
Ek
x − xi
x − xi
0
1
j+1
= (x − xi )
(1 − v)j Ek (v)dv.
0
Integrating by parts and using the property (2.7), we obtain
1
1
1
j
Ek+1 (0) +
(1 − v)j Ek (v)dv = −
(1 − v)j−1 Ek+1 (v)dv.
k+1
k+1 0
0
In a similar way we get
1
1
j−1 1
(1 − v)j−1 Ek+1 (v)dv = −
(1 − v)j−2 Ek+2 (v)dv.
Ek+2 (0) +
k+2
k+2 0
0
Hence, we have
1
1
j
Ek+1 (0) −
Ek+2 (0)
(1 − v)j Ek (v)dv = −
k+1
(k + 1)(k + 2)
0
1
j(j − 1)
(1 − v)j−2 Ek+2 (v)dv.
+
(k + 1)(k + 2) 0
Continuing in this way we get
1
j−1
k!j!
k!j!Ek+j+1 (0)
,
(1 − v)j Ek (v)dv = −
Ek+l+1 (0) − 2
(k
+
l
+
1)!(j
−
l)!
(k + j + 1)!
0
l=0
for j ≥ 1, since
k+1 (0)
− 2Ek+1
.
Thus,
1
0
Ek+j (v)dv = −2
x
xi
(x − t)j Ek
Ek+j+1 (0)
k+j+1 .
t − xi
x − xi
For j = 0,
1
(1
0
dt = (x − xi )j+1 Dkj ,
− v)0 Ek (v)dv =
73
Error inequalities
where
Dkj
⎧
2Ek+1 (0)
⎪
,
j=0
⎨− k+1
j−1
=
k!j!Ek+j+1 (0)
k!j!
⎪
⎩−
(k+l+1)!(j−l)! Ek+l+1 (0) − 2 (k+j+1)! , 1 ≤ j ≤ m.
l=0
It follows that
x
xi
Pm (t)Ek
t − xi
x − xi
dt =
m
cj Dkj (x − xi )j+1 .
j=0
Finally, we get
n
k
pni (x)(x − xi )
i=0
=
m
j=0
cj Dkj
n
x
xi
Pm (t)Ek
t − xi
x − xi
dt
pni (x)(x − xi )k+j+1 = 0,
i=0
for 0 ≤ k + m ≤ n − 1, since (2.4) holds.
Theorem 2.1. Under the assumptions of Lemma 2.2 suppose that f ∈ C k+1 (a, b).
Then
(2.11)
f (x) = Ln (x) + ωn (x)
k
(−1)j Ej (0)
j=1
j!
gj [x0 ; x1 ; . . . ; xn ] + Rk,m (x),
where
(2.12)
Rk,m (x) =
x
n
t − xi
(−1)k dt
f (k+1) (t) − Pm (t) Ek
pni (x)(x − xi )k
k! i=0
x − xi
xi
and
gj (t) = (x − t)j−1 f (j) (t),
Proof. We have
Rk,m (x) =
j = 1, 2, . . . , k.
x
n
t − xi
(−1)k dt
pni (x)(x − xi )k
f (k+1) (t)Ek
k! i=0
x − xi
xi
x
n
t − xi
(−1)k k
dt.
pni (x)(x − xi )
Pm (t)Ek
−
k! i=0
x − xi
xi
From the above relation and Lemma 2.2 it follows that
x
n
t − xi
(−1)k k
(k+1)
Rk,m (x) = Rk (x) =
dt.
pni (x)(x − xi )
f
(t)Ek
k! i=0
x − xi
xi
74
Nenad Ujević
Integrating by parts, we obtain
x
t − xi
(−1)k
k
(k+1)
(x − xi )
dt
f
(t)Ek
k!
x − xi
xi
(−1)k
=
(x − xi )k Ek (1)f (k) (x) − Ek (0)f (k) (xi )
k!
x
t − xi
(−1)k−1
k−1
(k)
dt,
f (t)Ek−1
+
(x − xi )
(k − 1)!
x − xi
xi
since (2.7) holds. In a similar way we get
x
(−1)k−1
t − xi
k−1
(k)
(x − xi )
dt
f (t)Ek−1
(k − 1)!
x − xi
xi
(−1)k−1
=
(x − xi )k−1 Ek−1 (1)f (k−1) (x) − Ek−1 (0)f (k−1) (xi )
(k − 1)!
x
t − xi
(−1)k−2
(x − xi )k−2
dt.
f (k−1) (t)Ek−2
+
(k − 2)!
x − xi
xi
Continuing in this way we get
x
t − xi
(−1)k
k
(k+1)
(x − xi )
dt
f
(t)Ek
k!
x − xi
xi
k
x
(−1)j
=
f (t)dt
(x − xi )j Ej (1)f (j) (x) − Ej (0)f (j) (xi ) +
j!
x
i
j=1
=
k
(−1)j
j=1
j!
(x − xi )j Ej (1)f (j) (x) − Ej (0)f (j) (xi ) + f (x) − f (xi ).
Then we have
Rk (x)
=
n
pni (x) [f (x) − f (xi )]
i=0
+
n
i=0
pni (x)
k
(−1)j
j!
j=1
= f (x) − Ln (x) +
k
n
(−1)j j!
j=1
= f (x) − Ln (x) −
(x − xi )j Ej (1)f (j) (x) − Ej (0)f (j) (xi )
k
j=1
pni (x)(x − xi )j Ej (1)f (j) (x) − Ej (0)f (j) (xi )
i=0
n
(−1)j
Ej (0)
pni (x)(x − xi )j f (j) (xi ),
j!
i=0
since (2.3) holds and
n
i=0
pni (x)(x − xi )j Ej (1)f (j) (x) = 0,
Error inequalities
75
for 1 ≤ j ≤ n. We have
n
pni (x)(x − xi )j f (j) (xi )
i=0
n
= ωn (x)
i=0
(x − xi )j−1 f (j) (xi )
(xi − x0 ) · · · (xi − xi−1 )(xi − xi+1 ) · · · (xi − xn )
and
gj [x0 ; x1 ; . . . ; xn ] =
n
i=0
(x − xi )j−1 f (j) (xi )
,
(xi − x0 ) · · · (xi − xi−1 )(xi − xi+1 ) · · · (xi − xn )
by Lemma 2.1, so that
Rk (x) = f (x) − Ln (x) − ωn (x)
k
(−1)j
j=1
j!
Ej (0)gj [x0 ; x1 ; . . . ; xn ] .
This completes the proof.
Remark 2.1. Since E2j (0) = 0, j = 1, 2, . . . , we can write
f (x) = Ln (x) − ωn (x)
k−1
[
2 ]
j=0
E2j+1 (0)
g2j+1 [x0 ; x1 ; . . . ; xn ] + Rk,m (x).
(2j + 1)!
Example 2.1. If we choose n = 1 in (2.11) then we get
x − x0
x − x1
f (x0 ) +
f (x1 )
f (x) =
x0 − x1
x1 − x0
1
f (x1 ) − f (x0 )
+ (x − x0 )(x − x1 )
+ R.
2
x1 − x0
Thus,
x1
f (x0 ) + f (x1 )
f (x)dx =
(x1 − x0 )
2
x0
x1
(x1 − x0 )2 [f (x1 ) − f (x0 )] +
−
Rdx,
12
x0
since
x1
x − x1
dx =
x0 − x1
x0
1
2
x1
x0
x1
x0
x − x0
dx = x1 − x0 ,
x1 − x0
(x − x0 )(x − x1 )
(x1 − x0 )2
dx = −
.
x1 − x0
12
We see that the above quadrature formula is the well-known corrected trapezoidal
rule.
Remark 2.2. If we generalize the above example then we find that the following
conclusions are valid: If we integrate (2.11) from x0 to xn then we get a corrected
Newton-Cotes formula. (Of course, we suppose that the partition π is uniform.)
It is well-known that the corrected formulas have better estimations of errors than
corresponding original quadrature formulas.
76
Nenad Ujević
3. Error inequalities
We now introduce the notations
(3.1)
Ck (x) =
n
i=0
(3.2)
Fk (x) =
n
i=0
(3.3)
Dk (x) =
n
i=0
k
|x − xi |
,
|xi − x0 | · · · |xi − xi−1 | |xi − xi+1 | · · · |xi − xn |
k
(Ski − γk+1 ) |x − xi |
,
|xi − x0 | · · · |xi − xi−1 | |xi − xi+1 | · · · |xi − xn |
k
(Γk+1 − Ski ) |x − xi |
,
|xi − x0 | · · · |xi − xi−1 | |xi − xi+1 | · · · |xi − xn |
where Ski = f (k) (x) − f (k) (xi ) /(x − xi ), i = 0, 1, . . . , n and γk+1 , Γk+1 are real
numbers such that γk+1 ≤ f (k+1 (t) ≤ Γk+1 , t ∈ [a, b], k = 0, 1, . . . , n − 1. We also
define the constant
|B2k+2 | 2k+2
δk = (3.4)
2
− 1.
(2k + 2)!
Let g ∈ C(a, b). As we know among all algebraic polynomials of degree ≤ m there
∗
exists the only polynomial Pm
(t) having the property that
∗
g − Pm
∞ ≤ g − Pm ∞ ,
where Pm ∈ Πm is an arbitrary polynomial of degree ≤ m. We define
(3.5)
∗
Gm (g) = g − Pm
=
inf
Pm ∈Πm
g − Pm ∞ .
Theorem 3.1. Under the assumptions of Theorem 2.1 we have
k
j
(−1) Ej (0)
f (x) − Ln (x) − ωn (x)
g
[x
;
x
;
.
.
.
;
x
]
j
0
1
n j!
j=1
≤ 2δk Gm (f (k+1) )Ck (x) |ωn (x)| ,
where Ck (·), δk and Gm (·) are defined by (3.1), (3.4) and (3.5), respectively.
Proof. We have
x
2
2 x
t − xi
t − xi
dt ≤ |x − xi | Ek
Ek
dt
x
−
x
x
−
x
i
i
xi
xi
1
2
= (x − xi )2
(Ek (u)) du
0
(k!)2
=4
(22k+2 − 1)(x − xi )2 |B2k+2 | ,
(2k + 2)!
since (2.8) holds.
Error inequalities
77
∗
∗
Let Pm (t) = Pm
(t), where Pm
(t) is defined by (3.5) for the function g(t) =
(t). Then we have
f
x
n
(−1)k t − xi
k
(k+1)
∗
dt
f
pni (x)(x − xi )
(t) − Pm (t) Ek
|Rk,m (x)| = k!
x
−
x
i
xi
i=0
n
x (k+1)
1 t − xi
∗
≤
dt
f
pni (x)(x − xi )k (t) − Pm
(t) Ek
k! i=0
x
−
x
i
xi
(k+1)
∗
f
− Pm ∞ 2k! |B2k+2 | ≤
22k+2 − 1Ck (x) |ωn (x)|
k!
(2k + 2)!
Gm (f (k+1) ) |B2k+2 | 2k+2
=2
2
− 1Ck (x) |ωn (x)|
(2k + 2)!
(k+1)
and
Rk,m (x) = f (x) − Ln (x) − ωn (x)
k
(−1)j
j=1
j!
Ej (0)gj [x0 ; x1 ; . . . ; xn ] .
This completes the proof.
Remark 3.1. The above estimate has only theoretical importance, since it is difficult to find the polynomial P ∗ . In fact, we can find P ∗ only for some special cases
of functions. However, we can use the estimate to obtain some practical estimations
— see Theorem 3.2.
Theorem 3.2. Let the assumptions of Theorem 2.1 hold. If γk+1 , Γk+1 are real
numbers such that γk+1 ≤ f (k+1) (t) ≤ Γk+1 , t ∈ [a, b], k = 0, 1, . . . , n − 1, then
k
j
(−1)
E
(0)
j
f (x) − Ln (x) − ωn (x)
gj [x0 ; x1 ; . . . ; xn ]
j!
j=1
≤ δk [Γk+1 − γk+1 ] Ck (x) |ωn (x)| ,
where ωn , δk and Ck (·) are defined by (2.5), (3.4) and (3.1), respectively. We also
have
k
j
(−1) Bj
f (x) − Ln (x) − ωn (x)
g
[x
;
x
;
.
.
.
;
x
]
j
0
1
n ≤ 2δk |ωn (x)| Fk (x)
j!
j=1
and
k
j
(−1)
B
j
f (x) − Ln (x) − ωn (x)
gj [x0 ; x1 ; . . . ; xn ] ≤ 2δk |ωn (x)| Dk (x),
j!
j=1
where Fk (·) and Dk (·) are defined by (3.2) and (3.3), respectively.
78
Nenad Ujević
Proof. We set Pm (t) = (Γk+1 + γk+1 )/2 in (2.12). Then we have
k
j
(−1)
f (x) − Ln (x) − ωn (x)
Ej (0)gj [x0 ; x1 ; . . . ; xn ]
j!
j=1
= |Rk,m (x)|
x
n
t − xi
1 Γk+1 + γk+1 k (k+1)
dt .
pni (x)(x − xi ) f
≤
−
Ek
k! i=0
2
x − xi
xi
∞
We also have
(k+1) Γk+1 + γk+1 ≤ Γk+1 − γk+1
f
−
2
2
∞
and
x
xi
Ek
t − xi
x − xi
2k!
dt ≤ 22k+2 − 1 |x − xi | |B2k+2 |.
(2k + 2)!
From the above three relations we get
k
j
(−1)
f (x) − Ln (x) − ωn (x)
Ej (0)gj [x0 ; x1 ; . . . ; xn ]
j!
j=1
n
Γk+1 − γk+1 2k+2
k+1
≤ 2
− 1 |B2k+2 |
|pni (x)| |x − xi |
(2k + 2)!
i=0
Γk+1 − γk+1 2k+2
= 2
− 1 |B2k+2 |Ck (x) |ωn (x)| .
(2k + 2)!
The first inequality is proved.
We now set Pm (t) = γk+1 in (2.12). Then we have
k
j
(−1)
f (x) − Ln (x) − ωn (x)
(0)g
[x
;
x
;
.
.
.
;
x
]
E
j
j
0
1
n j!
j=1
= |Rk,m (x)|
≤
n
x (k+1)
1 t − xi
dt .
f
pni (x)(x − xi )k (t) − γk+1 Ek
k! i=0
x − xi
xi
We also have
x (k+1)
f
(t) − γk+1 dt = f (k) (x) − f (k) (xi ) − γk+1 (x − xi )
xi
= (Ski − γk+1 ) |x − xi | .
Error inequalities
Thus,
79
k
j
(−1)
f (x) − Ln (x) − ωn (x)
(0)g
[x
;
x
;
.
.
.
;
x
]
E
j
j
0
1
n j!
j=1
n
2k! |B2k+2 | 2k+2
1 k+1
≤
|pni (x)| |x − xi |
(Ski − γk+1 ) 2
−1
k! i=0
(2k + 2)!
|ωn (x)| 2k+2
2
− 1 |B2k+2 |Fk (x).
= 2
(2k + 2)!
The second inequality is proved. The third inequality holds by a similar proof. Lemma 3.1. Let π = {x0 = a < x1 < · · · < xn = b} be a given uniform subdivision
of the interval [a, b], i.e., xi = x0 + ih, h = (b − a)/n, i = 0, 1, 2, . . . , n. If
x ∈ (xj−1 , xj ), for some j ∈ {1, 2, . . . , n}, then
|ωn (x)| ≤ j!(n − j + 1)!hn+1 ,
(3.6)
Ck (x) ≤
(3.7)
2n
n!
k
1
[n + 1 + |n − 2j + 1|] hk−n ,
2
and
Ck (x) |ωn (x)| ≤ αjnk
(3.8)
where
αjnk =
(3.9)
n − j + 1 2n (b − a)k+1
n ,
n
j
k
1
(n + 1 + |2j − n − 1|) .
2n
Proof. For i < j we have
|x − xi | ≤ (j − i)h
and for i ≥ j we have
|x − xi | ≤ (i − j + 1)h
such that
|(x − x0 ) · · · (x − xj−1 )(x − xj ) · · · (x − xn )| ≤ j!hj (n − j + 1)!hn−j+1
= j!(n − j + 1)!hn+1 .
We have
Ck (x) ≤
j−1
n
(j − i)k hk
(i − j + 1)k hk
+
i!(n − i)!hn i=j i!(n − i)!hn
i=0
j−1
n
(n − j + 1)k hk
j k hk
+
i!(n − i)!hn i=j i!(n − i)!hn
i=0
n 1 k
n
k−n
k
≤h
max j , (n − j + 1)
n! i=0 i
k
2n 1
[n + 1 + |n − 2j + 1|] hk−n ,
=
n! 2
≤
80
Nenad Ujević
since
1
(n + 1 + |2j − n − 1|)
2
such that it is not difficult to show that (3.7) holds.
Finally, from the above relations we find that (3.8) holds, too.
max [j, n − j + 1] =
Remark 3.2. Note that
αjnk ≤ 1
and αjnk = 1 if and only if j = 1 or j = n. If we choose x ∈ [xj , xj+1 ], j =
0, 1, . . . , n − 1, then we get the factor (j + 1)/n instead of the factor (n − j + 1)/n
in (3.8).
Theorem 3.3. Under the assumptions of Lemma 3.1 and Theorem 3.2 we have
k
j
(−1) Ej (0)
f (x) − Ln (x) − ωn (x)
g
[x
;
x
;
.
.
.
;
x
]
j
0
1
n j!
j=1
≤ [Γk+1 − γk+1 ] δk αjnk
n − j + 1 2n (b − a)k+1
n .
n
j
Proof. The proof follows immediately from Theorem 3.2 and Lemma 3.1.
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ujevic@pmfst.hr
This paper is available via http://nyjm.albany.edu:8000/j/2004/10-4.html.