General Mathematics Vol. 13, No. 2 (2005), 95–104
Estimations of the Error for Two-point
Formula via Pre-Grüss Inequality
J. Pečarić
and
A. Vukelić
Dedicated to Professor Dumitru Acu on his 60th anniversary
Abstract
Generalization of estimation of two-point formula is given, by
using pre-Grüss inequality.
2000 Mathematical Subject Classification: 26D10, 41A55.
Key words and phrases: Pre-Grüss inequality, generalization, two-point
formula.
1
Introduction
In the recent paper [4] N. Ujević use the generalization of pre-Grüss inequality to derive some better estimations of the error for Simpson’s quadrature
rule. In fact, he proved the next as his main result:
95
96
J. Pečarić and A. Vukelić
Theorem 1. If g, h, Ψ ∈ L2 (0, 1) and
R1
0
Ψ(t)dt = 0 then we have
|SΨ (g, h)| ≤ SΨ (g, g)1/2 SΨ (h, h)1/2 ,
(1)
where
Z
Z
1
SΨ (g, h) =
g(t)h(t)dt−
0
Z
1
g(t)dt
0
Z
1
h(t)dt−
0
Z
1
1
g(t)Ψ0 (t)dt
0
h(t)Ψ0 (t)dt
0
and Ψ0 (t) = Ψ(t)/kΨk2 .
Further, he gave some improvements of the Simpson’s inequality. For
example he get:
Theorem 2. Let I ⊂ R be a closed interval and a, b ∈ IntI, a < b. If
f : I → R is an absolutely continuous function with f 0 ∈ L2 (a, b) then we
have
Z b
a+b
b−a
(b − a)3/2
f (a) + 4f
+ f (b) −
f (t)dt ≤
K1 ,
6
2
6
a
(2)
where
(3)
K12
=
kf 0 k22
and Ψ(t) = t −
1
−
b−a
a+b
, Ψ0 (t)
2
Z
b
2
0
f (t)dt
a
Z
b
2
0
f (t)Ψ0 (t)dt
−
a
= Ψ(t)/kΨk2 .
In this paper using the Theorem 1 we will give the similar result for Euler
two-point formula and for functions whose derivative of order n, n ≥ 1, is
from L2 (0, 1) space. We will use interval [0, 1] because of simplicity and
since it involves no loss in generality.
Estimations of the Error for Two-point Formula...
2
97
Estimations of the error for Euler twopoint formula
In the recent paper [3] the following identity, named Euler two-point for mula, has been proved. For n ≥ 1, x ∈ 0, 12 and every t ∈ [0, 1] we
have
Z
1
(4)
f (t)dt = D(x) − Tn (x) + Rn (x)
0
where
D(x) =
1
[f (x) + f (1 − x)] ,
2
T0 (x) = 0 and
m
1 X B̃k (x) (k−1)
(5)
Tm (x) =
f
(1) − f (k−1) (0) ,
2 k=1 k!
for 1 ≤ m ≤ n and x ∈ 0, 12 , while
B̃k (x) = Bk (x) + Bk (1 − x), k ≥ 1,
Z 1
1
Rn (x) =
Gx (t) f (n) (t)dt
2(n!) 0 n
and
Gxn (t) = Bn∗ (x − t) + Bn∗ (1 − x − t) , t ∈ R.
The identity holds for every function f : [0, 1] → R such that f (n−1) is
a continuous function of bounded variation on [0, 1]. The functions Bk (t)
are the Bernoulli polynomials, Bk = Bk (0) are the Bernoulli numbers, and
Bk∗ (t), k ≥ 0, are periodic functions of period 1, related to the Bernoulli
polynomials as
Bk∗ (t) = Bk (t), 0 ≤ t < 1
and
Bk∗ (t + 1) = Bk∗ (t), t ∈ R.
98
J. Pečarić and A. Vukelić
The Bernoulli polynomials Bk (t), k ≥ 0 are uniquely determined by the
following identities
Bk0 (t) = kBk−1 (t), k ≥ 1; B0 (t) = 1, Bk (t + 1) − Bk (t) = ktk−1 , k ≥ 0.
For some further details on the Bernoulli polynomials and the Bernoulli
numbers see for example [1] or [2]. We have B0∗ (t) = 1 and B1∗ (t) is a
discontinuous function with a jump of −1 at each integer. It follows that
Bk (1) = Bk (0) = Bk for k ≥ 2, so that Bk∗ (t) are continuous functions for
k ≥ 2. We get
∗
Bk∗0 (t) = kBk−1
(t), k ≥ 1
(6)
for every t ∈ R when k ≥ 3, and for every t ∈ R \ Z when k = 1, 2.
Theorem 3. If f : [0, 1] → R is such that f (n−1) is absolutely continuous
function with f (n) ∈ L2 (0, 1) then we have
1/2
Z 1
1 2(−1)n−1
(7)
[B2n + B2n (1 − 2x)] K,
f (t)dt − D(x) + Tn (x) ≤
2
(2n)!
0
where
Z
(8)
2
K =
For n even
kf (n) k22
1
f
−
0
(n)
2 Z
(t)dt −
1
f
(n)
2
(t)Ψ0 (t)dt .
0

 1, t ∈ 0, 1 2
,
Ψ(t) =
 −1, t ∈ 1 , 1
2
while for n odd we have

1
Bn+1 ( 12 +x)

 t+
,
t
∈
0, 2 ,
1
2(Bn+1 (x)−Bn+1 ( 2 +x))
Ψ(t) =
1
B
+x −2B
(x)

 t + n+1 ( 2 ) n+1
, t ∈ 21 , 1 .
2(Bn+1 (x)−Bn+1 ( 12 +x))
Estimations of the Error for Two-point Formula...
99
Proof. It is not difficult to verify that
Z 1
(9)
Gn (t)dt = 0,
0
Z
1
(10)
Ψ(t)dt = 0,
0
Z
1
(11)
Gn (t)Ψ(t)dt = 0.
0
From (4), (9) and (11) it follows that
Z 1
(12)
f (t)dt − D(x) + Tn (x) =
Z 1
1
Gx (t)f (n) (t)dt−
2(n!) 0 n
Z 1
Z 1
x
Gn (t)dt
f (n) (t)dt−
0
−
−
1
2(n!)
1
2(n!) 0
0
Z 1
Z 1
Gxn (t)Ψ0 (t)dt
f (n) (t)Ψ0 (t)dt =
0
0
1
=
SΨ (Gxn , f (n) ).
2(n!)
Using (12) and (1) we get
Z 1
(13)
f (t)dt − D(x) + Tn (x) ≤
0
1
SΨ (Gxn , Gxn )1/2 SΨ (f (n) , f (n) )1/2 .
2(n!)
We also have (see [3])
Z
SΨ (Gxn , Gxn )
=
kGxn k22
−
0
= (−1)n−1
(14)
1
2
Gxn (t)dt
Z
1
−
0
2
Gxn (t)Ψ0 (t)dt
=
2(n!)2
[B2n + B2n (1 − 2x)]
(2n)!
and
(15)
SΨ (f
Z
(n)
,f
(n)
)=
kf (n) k22
2
1
−
f
0
From (13)-(15) we easily get (7).
(n)
(t)dt
Z
2
1
−
f
0
(n)
(t)Ψ0 (t)dt
= K 2.
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J. Pečarić and A. Vukelić
Remark 1.Function Ψ(t) can be any function witch satisfies conditions
R1
R1
Ψ(t)dt = 0 and 0 Gxn (t)Ψ(t)dt = 0. Because Gxn (1 − t) = (−1)n Gxn (t)
0
(see [3]), for n even we can take function Ψ(t) such that Ψ(1 − t) = −Ψ(t).
For n odd we have to calculate Ψ(t)
lost in generality in our
 and with not
 t + a, t ∈ 0, 1 ,
2
theorem we take the form Ψ(t) =
 t + b, t ∈ 1 , 1 .
2
Remark 2.For n = 1 in Theorem 3 we have
1/2
Z 1
1 1
2
(16)
f (t)dt − D(x) ≤
− 2x + 4x
K,
2 3
0
while


t ∈ 0, 21 ,
Ψ(t) =
 t + 12x2 −24x+5 , t ∈ 1 , 1 .
24x−6
2
t+
1−12x2
,
24x−6
Also, for n = 2 we have
1/2
Z 1
x2 4x3 4x4
1 1
(17)
−
+
−
K,
f (t)dt − D(x) ≤
2 180
3
3
3
0
while

 1, t ∈ 0, 1 ,
2
Ψ(t) =
 −1, t ∈ 1 , 1 .
2
If in Theorem 3 we choose x = 0, 1/2, 1/3, 1/4 we get inequality related
to the trapezoid, the midpoint, the two-point Newton-Cotes and the twopoint MacLaurin formula:
Corollary 1. If f : [0, 1] → R is such that f (n−1) is absolutely continuous
function with f (n) ∈ L2 (0, 1) then we have
Z
1
(18)
0
1/2
(−1)n−1
1
f (t)dt − [f (0) + f (1)] + Tn (0) ≤
B2n
K,
2
(2n)!
Estimations of the Error for Two-point Formula...
101
where T0 (0) = 0,
bn/2c
Tn (0) =
and
X B2k f (2k−1) (1) − f (2k−1) (0)
(2k)!
k=1
Z
2
K =
For n even
kf (n) k22
2
1
−
f
(n)
(t)dt
Z
−
0
2
1
f
(n)
(t)Ψ0 (t)dt
.
0

 1, t ∈ 0, 1 ,
2
Ψ(t) =
 −1, t ∈ 1 , 1 ,
2
while for n odd we have

 t+
Ψ(t) =
 t+
t ∈ 0, 12 ,
1
2−n −3
,
t
∈
,
1
.
1−n
4−2
2
2−n −1
,
4−21−n
Remark 3.For n = 1 in Corollary 1 we have
Z 1
1
K
f (t)dt − [f (0) + f (1)] ≤ √ ,
2
2 3
0
while

 t − 1 , t ∈ 0, 1 ,
6
2
Ψ(t) =
 t − 5 , t ∈ 1 , 1 .
6
2
Corollary 2. If f : [0, 1] → R is such that f (n−1) is absolutely continuous
function with f (n) ∈ L2 (0, 1) then we have
1/2
1
1
(−1)n−1
(19)
f (t)dt − f
+ Tn
≤
B2n
K,
2
2
(2n)!
0
where T0 12 = 0,
Z
1
bn/2c
X (21−2k − 1)B2k 1
Tn
f (2k−1) (1) − f (2k−1) (0)
=
2
(2k)!
k=1
102
J. Pečarić and A. Vukelić
and
Z
2
K =
kf (n) k22
For n even
1
−
f
(n)
2 Z
(t)dt −
0
1
f
(n)
2
(t)Ψ0 (t)dt .
0

 1, t ∈ 0, 1 ,
2
Ψ(t) =
 −1, t ∈ 1 , 1 ,
2
while for n odd we have

 t+
Ψ(t) =
 t+
t ∈ 0, 21 ,
1
3−21−n
,
t
∈
,
1
.
1−n
2
−4
2
1
,
21−n −4
Remark 4.For n = 1 in Corollary 2 we have
Z
1
0
while
1
K
f (t)dt − f
≤ √ ,
2
2 3

 t − 1 , t ∈ 0, 1 ,
3
2
Ψ(t) =
 t − 2 , t ∈ 1 , 1 .
3
2
Corollary 3. If f : [0, 1] → R is such that f (n−1) is absolutely continuous
function with f (n) ∈ L2 (0, 1) then we have
Z
1
(20)
0
where T0
1
3
1
2
1
1
f (t)dt −
f
+f
+ Tn
≤
2
3
3
3
1/2
1 (−1)n−1
1−2n
(1 + 3
)B2n
K,
≤
2
(2n)!
= 0,
bn/2c
1
1 X (31−2k − 1)B2k (2k−1)
Tn
f
(1) − f (2k−1) (0)
=
3
2 k=1
(2k)!
Estimations of the Error for Two-point Formula...
and
Z
2
K =
kf (n) k22
2
1
−
f
(n)
(t)dt
Z
For n even
2
1
−
0
103
f
(n)
(t)Ψ0 (t)dt
.
0

 1, t ∈ 0, 1 ,
2
Ψ(t) =
 −1, t ∈ 1 , 1 ,
2
while for n odd we have

 t+
Ψ(t) =
 t+
t ∈ 0, 12 ,
1−3·2n
1
,
t
∈
,
1
.
2+n
2
−2n
2
1−2n
,
22+n −2
Remark 5. For n = 1 in Corollary 3 we have
Z 1
1
2
K
1
f
+f
≤ ,
f (t)dt −
2
3
3
6
0
while

 t − 1 , t ∈ 0, 1 ,
6
2
Ψ(t) =
 t − 5 , t ∈ 1 , 1 .
6
2
Corollary 4. If f : [0, 1] → R is such that f (2m−1) is absolutely continuous
function with f (2m) ∈ L2 (0, 1) then we have
(21)
Z
−4m
1/2
1
3
−2
1
1
f
+f
+ T2m
≤
B4m
K,
f (t)dt −
2
4
4
4
(4m)!
0
where T0 14 = 0,
1
X
m
1
2−2k (21−2k − 1)B2k (2k−1)
T2m
=
f
(1) − f (2k−1) (0)
4
(2k)!
k=1
and
Z
2
K =
kf (2m) k22
1
−
f
0
(2m)
2 Z
(t)dt −
2
1
f
0
(2m)
(t)Ψ0 (t)dt
,
104
J. Pečarić and A. Vukelić
while

 1, t ∈ 0, 1 ,
2
Ψ(t) =
 −1, t ∈ 1 , 1 .
2
References
[1] M. Abramowitz and I. A. Stegun (Eds), Handbook of mathematical functions with formulae,graphs and mathematical tables, National Bureau of
Standards, Applied Math.Series 55, 4th printing, Washington 1965.
[2] I.S. Berezin and N.P. Zhidkov, Computing methods, Vol. I. Pergamon
Press, Oxford-London-Edinburgh-New York-Paris-Frankfurt, 1965.
[3] J. Pečarić, I. Perić, A. Vukelić, Sharp integral inequalities based on general Euler two-point formulae, ANZIAM J. 46(2005), 555-574.
[4] N. Ujević, New bounds for Simpson’s inequality. Tamkang J. of Math.,
Vol. 33, No. 2, (2002), 129-138.
Faculty of Food Technology and Biotechnology,
Mathematics Department, University of Zagreb,
Pierottijeva 6, 10000 Zagreb, Croatia
E-mail:avukelic@pbf.hr