Rev. Anal. Num´er. Th´eor. Approx., vol. 37 (2008) no. 1, pp. 53–57 ictp.acad.ro/jnaat
DOUBLE INEQUALITIES FOR QUADRATURE FORMULA OF GAUSS TYPE WITH TWO NODES
MARIUS HELJIU∗
Abstract. In this paper upper and lower error bonds for Gauss’s quadrature rule with two nodes are given.
MSC 2000. 65D32.
Keywords. Double integral inequalities, numerical integration.
1. INTRODUCTION
In a series of papers (see [2], [3], [8]) the authors establish bounds for the quadrature rules such as the trapezoid, Simpson and Newton quadrature rules.
In this work we will consider Gauss’s quadrature rule with two nodes:
(1)
Z b a
f(x)dx= b−a
2 [f(x1) +f(x2)] +R[f]
where f : [a, b] → R, x1 = a+b2 − b−a2 ·ξ, x2 = a+b2 + b−a2 ·ξ, ξ = √1
3 = 0,57735027. . .
Iff ∈C4[a, b], the error R[f] from the formula (1) is given by:
(2) R[f] =
Z b a
ϕ(x)f(4)(x)dx
(see [7], pp. 137–138 and 283–284), where the function ϕhas the form:
(3) ϕ(x) =
(x−a)4
4! if x∈[a, x1],
(x−a)4
4! −b−a2 (x−x3!1)3 if x∈]x1, x2[,
(b−x)4
4! if x∈[x2, b].
It is easy to see that the functionϕ has the following properties:
a) ϕ∈C4[a, b];
b) ϕa+b2 −h=ϕa+b2 +h, for anyh∈h0,b−a2 i;
∗ Department of Mathematics, University of Petro¸sani, Romania, e-mail:
c) Z b
a
ϕ(x)dx= 1351 b−a2 5.
2. MAIN RESULTS
In the following theorem double inequalities for−R[f] are presented where R[f] is the error in the quadrature formula (1) given by the relation (2).
Theorem 1. If f ∈C4[a, b] then
1
17280(b−a)5(41γ4−45S3+ 180ξ3S3−180ξ3γ4)≤ (4)
≤ b−a2 [f(x1) +f(x2)]− Z b
a
f(x)dx
≤ 172801 (b−a)5(41Γ4−45S3+ 180ξ3S3−180ξ3Γ4)
where γ4,Γ4 ∈ R, γ4 ≤f(4)(x) ≤Γ4, for all x∈ [a, b] and S3 = f000(b)−fb−a000(a). Moreover,
(5) γ4 = min
x∈[a,b]f(4)(x), Γ4= max
x∈[a,b]f(4)(x), and the inequalities (4) are sharp.
Proof. From (2), using the properties of the function ϕand integrating by parts, we obtain:
(6)
Z b a
ϕ(x)f(4)(x)dx= Z b
a
f(x)dx−b−a2 [f(x1) +f(x2)].
By using the equality c) in the formula (6) and the assumptions of the theorem, we have:
Z b a
[f(4)(x)−γ4]ϕ(x)dx= Z b
a
f(x)dx−b−a2 [f(x1) +f(x2)]
(7)
−1351 b−a2 5γ4 and
Z b a
[Γ4−f(4)(x)]ϕ(x)dx=− Z b
a
f(x)dx+b−a2 [f(x1) +f(x2)]
(8)
+1351 b−a2 5Γ4 On the other hand, we have:
(9)
Z b a
[f(4)(x)−γ4]ϕ(x)dx≤ max
x∈[a,b]|ϕ(x)|
Z b a
|f(4)(x)−γ4|dx, where
(10) max
x∈[a,b]|ϕ(x)|= 3841 (1−4ξ3)(b−a)4= 9−4
√ 3
3456 (b−a)4,
and
Z b a
|f(4)(x)−γ4|dx= Z b
a
(f(4)(x)−γ4)dx (11)
=f000(b)−f000(a)−γ4(b−a)
= (S3−γ4)(b−a).
From the relations (7), (9), (10) and (11) we obtain:
Z b a
f(x)dx−b−a2 [f(x1) +f(x2)]
(12)
≤ −172801 (b−a)5(41γ4−45S3+ 180ξ3S3−180ξ3γ4).
In the same way we have:
(13)
Z b a
[Γ4−f(4)(x)]ϕ(x)dx≤ max
x∈[a,b]|ϕ(x)|
Z b a
|Γ4−f(4)(x)|dx and
Z b a
|Γ4−f(4)(x)|dx= Z b
a
(Γ4−f(4)(x))dx (14)
= Γ4(b−a)−f000(b) +f000(a)
= (Γ4−S3)(b−a).
Using (8), (10), (13) and (14) we obtain the inequality:
Z b a
f(x)dx−b−a2 [f(x1) +f(x2)]
(15)
≥ −172801 (b−a)5(41Γ4−45S3+ 180ξ3S3−180ξ3γ4).
Inequalities (4) follow from the inequalities (12) and (15).
To prove the second part of the theorem we consider the function f(x) = (x−a)4. It is easy to show that all the three members of the double inequality (4) have in common the value−1801 (b−a)5. This completes the proof.
In the following theorem double inequalities forR[f] are presented.
Theorem 2. If the functionf ∈C4[a, b]then:
1
17280(b−a)5(49γ4−45S3+ 180ξ3S3−180ξ3γ4) (16)
≤ Z b
a
f(x)dx−b−a
2 [f(x1) +f(x2)]
≤ 172801 (b−a)5(49Γ4−45S3+ 180ξ3S3−180ξ3Γ4), where γ4,Γ4, ξ and S3 are given in Theorem 1. Moreover,
γ4 = min
x∈[a,b]f(4)(x), Γ4= max
x∈[a,b]f(4)(x), and the inequalities (16) are sharp.
Proof. By using the relations (7), (9), (10) and (11) it follows:
− Z b
a
f(x)dx+b−a2 [f(x1) +f(x2)]
(17)
≤ −172801 (b−a)5(49γ4−45S3+ 180ξ3S3−180ξ3γ4).
Analogously, using the relations (8), (13), (14) and (15) we obtain:
Z b a
f(x)dx−b−a2 [f(x1) +f(x2)]
(18)
≤ 172801 (b−a)5(49Γ4−45S3+ 180ξ3S3−180ξ3Γ4).
From the relations (17) and (18) result the inequalities (16). To prove that the double inequalities (16) are exact we follow the steps of the proof for
Theorem 1.
Theorem 3 gives us the inequalities which do not depend onS3. Theorem 3. In the assumptions of Theorem1, we have:
1
34560(b−a)5(41γ4−49Γ4−180ξ3γ4+ 180ξ3Γ4) (19)
≤ b−a2 [f(x1) +f(x2)]− Z b
a
f(x)dx
≤ 345601 (b−a)5(41Γ4−49γ4+ 180ξ3γ4−180ξ3Γ4).
If
γ4 = min
x∈[a,b]f(4)(x), Γ4= max
x∈[a,b]f(4)(x), then the inequalities (19) are sharp.
Proof. Multiplying the inequality (16) with (−1) and adding it with the inequality (4) we obtain the double inequalities (19). Considering the function f(x) = (x−a)4 and calculating all the three members of the inequality (19) the value obtained is −1801 (b−a)5. The double inequalities are sharp. This
completes the proof.
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Received by the editors: March 11, 2008.
