Quadrature formulae of Gauss type based on Euler identities

Mathematical and Computer Modelling 45 (2007) 355–370www.elsevier.com/locate/mcm

Quadrature formulae of Gauss type based on Euler identities

I. Franjica,∗, I. Perica, J. Pecaricb

a Faculty of Food Technology and Biotechnology, Department of Mathematics, University of Zagreb, Pierottijeva 6, 10000 Zagreb, Croatiab Faculty of Textile Technology, University of Zagreb, Pierottijeva 6, 10000 Zagreb, Croatia

Received 31 May 2005; received in revised form 16 May 2006; accepted 24 May 2006

Abstract

The aim of this paper is to derive quadrature formulae of Gauss type based on Euler identities. First, we derive quadratureformulae where the integral over [0, 1] is approximated by values of the function in three points: x, 1/2 and 1 − x . As specialcases, the Gauss 2-point formula, Simpson’s formula, dual Simpson’s formula and Maclaurin’s formula are obtained. Next,corrected Gauss 2-point formulae are derived and finally, the Gauss 3-point formulae and the corrected Gauss 3-point formulae areconsidered. We call “corrected” such quadrature formulae where the integral is approximated both with the values of the integrandin certain points and the values of its first derivative in the end points of the interval. Corrected formulae have a degree of exactnesshigher than the adjoint original formulae.c© 2006 Elsevier Ltd. All rights reserved.

Keywords: Quadrature formulae; Corrected quadrature formulae; Gauss formulae; Corrected Gauss formulae; Bernoulli polynomials; ExtendedEuler formulae

1. Introduction

The aim of this paper is to derive quadrature formulae of Gauss type based on Euler identities. First, we derivequadrature formulae where the integral over [0, 1] is approximated by values of the function in three points: x, 1/2and 1 − x . As special cases, the Gauss 2-point formula, Simpson’s formula, dual Simpson’s formula and Maclaurin’sformula are obtained.

Next, corrected Gauss 2-point formulae are derived. We call “corrected” such quadrature formulae where theintegral is approximated not only with the values of the integrand in certain points but with the values of its firstderivative in the end points of the interval as well. The term “corrected” was first introduced in [1]. Corrected formulaehave a degree of exactness higher than the adjoint original formulae.

In the final section, the Gauss 3-point formulae are considered. The Gauss formulae are quadrature formulae of thehighest degree of precision 2n − 1, which is why they are so interesting to study, and have for their n nodes the rootsof the Legendre polynomial of degree n (cf. [9]). For some recent results on Gaussian type quadrature formulae seee.g. [2,3].

∗ Corresponding author.E-mail addresses: [email protected] (I. Franjic), [email protected] (I. Peric), [email protected] (J. Pecaric).

0895-7177/$ - see front matter c© 2006 Elsevier Ltd. All rights reserved.doi:10.1016/j.mcm.2006.05.009

356 I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370

The main tool used are the extended Euler formulae, obtained in [4]: for f : [a, b] → R such that f (n−1) iscontinuous of bounded variation on [a, b], for some n ≥ 1, for every x ∈ [a, b] we have

1b − a

∫ b

af (t)dt = f (x) − Tn(x) +

(b − a)n−1

n!

∫ b

aB∗

n

(x − t

b − a

)d f (n−1)(t) (1)

1b − a

∫ b

af (t)dt = f (x) − Tn−1(x) +

(b − a)n−1

n!

∫ b

a

[B∗

n

(x − t

b − a

)− Bn

(x − a

b − a

)]d f (n−1)(t) (2)

where T0(x) = 0 and for 1 ≤ m ≤ n

Tm(x) =

m∑k=1

(b − a)k−1

k!Bk

(x − a

b − a

)[f (k−1)(b) − f (k−1)(a)

]where Bk(t) is the kth Bernoulli polynomial and B∗

k (t) = Bk(t − btc), t ∈ R. Recall that Bernoulli numbers aredefined by Bk = Bk(0), k ≥ 0. For further details on Bernoulli polynomials see [10,11].

We will use standard notation when working with extended Euler formulae, but it should be noted that definitionsof these standard notions vary from section to section.

2. Some general considerations on 3-point quadrature formulae

Let x ∈ [0, 1/2) and f : [0, 1] → R be such that f (2n) is continuous of bounded variation on [0, 1] for somen ≥ 0. Put x ≡ x, 1/2, 1 − x in (1), multiply by A(x), 1 − 2A(x), A(x), respectively, where

A(x) =1

6(2x − 1)2 (3)

and add. The following formula is obtained:∫ 1

0f (t)dt − D(0, 1) + T2n(x) =

1(2n + 1)!

∫ 1

0G2n+1(t)d f (2n)(t), (4)

where

D(0, 1) =1

6(2x − 1)2

[f (x) + 24B2(x) · f

(12

)+ f (1 − x)

](5)

T2n(x) =

2n∑k=1

1k!

Gk(0) [ f (k−1)(1) − f (k−1)(0)] (6)

and

Gk(t) =1

6(2x − 1)2

[B∗

k (x − t) + 24B2(x) · B∗

k

(12

− t

)+ B∗

k (1 − x − t)

](7)

for k ≥ 1 and t ∈ R.It is easy to see that G2k−1(0) = 0 for k ≥ 1, no matter which symmetrical convex combination it was obtained by.

Thus, applying (2) instead of (1), we would get identity (4) again. The reason these specific coefficients were chosenis because they give G2(0) = 0 and that is an interesting case to study since, in general, G2k(0) 6= 0. Moreover, thisprocedure produces formulae with a degree of exactness at least 3 (which in the case when the middle coefficient isequal to 0 is the highest possible, see Section 3.

Analogously, assuming f (2n−1) is continuous of bounded variation on [0, 1] for some n ≥ 1, from (1) we get:∫ 1

0f (t)dt − D(0, 1) + T2n(x) =

1(2n)!

∫ 1

0G2n(t)d f (2n−1)(t), (8)

I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370 357

and if f (2n+1) is continuous of bounded variation on [0, 1] for some n ≥ 0, from (2) we get∫ 1

0f (t)dt − D(0, 1) + T2n(x) =

1(2n + 2)!

∫ 1

0F2n+2(t)d f (2n+1)(t), (9)

where Fk(t) = Gk(t) − Gk(0), k ≥ 1. It is easy to check that Gk(1 − t) = (−1)k Gk(t).The key step for obtaining the best possible estimates of the error for this type of quadrature formulae is the

following lemma.

Lemma 1. For x ∈ {0} ∪ [1/6, 1/2) and k ≥ 1, G2k+1(t) has no zeros in the interval (0, 1/2). The sign of thatfunction on (0, 1/2) is determined by

(−1)k G2k+1(t) > 0 for x ∈ [1/6, 1/2),

(−1)k+1G2k+1(t) > 0 for x = 0.

Proof. Observe G3(t). For 0 ≤ t ≤ x < 1/2, it takes the form

G3(t) = −t3,

so its only zero is obviously t = 0. For 0 ≤ x ≤ t ≤ 1/2, it takes the form

G3(t) = −t3+

(x − t)2

2(2x − 1)2 .

Here it has three zeros:

t1 =12, t2 =

x − x2−

√2x3 − 3x4

(2x − 1)2 , t3 =x − x2

+√

2x3 − 3x4

(2x − 1)2 .

It is easy to check that t2 < x which is opposite from our assumption. On the other hand, t3 ≥ x for all choices of x ,but t3 ≥ 1/2 for x ∈ [1/6, 1/2). Thus, our assertion is true for k = 1 (for x = 0 the assertion is trivial). Assuming theopposite, by induction, it follows easily that the assertion is true for all k ≥ 2.

It is elementary to determine the sign of G3(t) since we know its form. In order to determine the sign of G2k+1(t)in general, use their second derivatives and the fact that they don’t have zeros on (0, 1/2). �

Remark 1. From Lemma 1 it follows immediately that, for k ≥ 1 and x ∈ [1/6, 1/2), (−1)k+1 F2k+2(t) is strictlyincreasing on (0, 1/2) and strictly decreasing on (1/2, 1). Since F2k+2(0) = F2k+2(1) = 0, it has constant sign on(0, 1) and obtains its maximum at t = 1/2. The analogous statement, but with the opposite sign, is valid in the casewhen x = 0.

Theorem 1. Assume (p, q) is a pair of conjugate exponents, that is 1 ≤ p, q ≤ ∞, 1/p + 1/q = 1. Letf : [0, 1] → R be such that f (2n)

∈ L p[0, 1] for some n ≥ 1. Then we have∣∣∣∣∣∫ 1

0f (t)dt − D(0, 1) + T2n(x)

∣∣∣∣∣ ≤ K (2n, q) · ‖ f (2n)‖p. (10)

If f (2n+1)∈ L p[0, 1] for some n ≥ 0, then we have∣∣∣∣∣

∫ 1

0f (t)dt − D(0, 1) + T2n(x)

∣∣∣∣∣ ≤ K (2n + 1, q) · ‖ f (2n+1)‖p. (11)

If f (2n+2)∈ L p[0, 1] for some n ≥ 0, then we have∣∣∣∣∣

∫ 1

0f (t)dt − D(0, 1) + T2n(x)

∣∣∣∣∣ ≤ K ∗(2n + 2, q) · ‖ f (2n+2)‖p, (12)

358 I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370

where

K (m, q) =1

m!

[∫ 1

0|Gm(t)|q dt

] 1q

and K ∗(m, q) =1

m!

[∫ 1

0|Fm(t)|q dt

] 1q

.

These inequalities are sharp for 1 < p ≤ ∞ and best possible for p = 1.

Proof. Inequalities (10)–(12) follow immediately after applying Holder’s inequality to the remainders in formulae(8), (4) and (9). To prove inequalities are sharp, put

f (m)(t) = sgn Gm(t) · |Gm(t)|1/(p−1) for 1 < p < ∞ and

f (m)(t) = sgn Gm(t) for p = ∞ in (10) and (11),

f (m)(t) = sgn Fm(t) · |Fm(t)|1/(p−1) for 1 < p < ∞ and

f (m)(t) = sgn Fm(t) for p = ∞ in (12).

The proof that these inequalities are the best possible for p = 1 is the same as the proof of Theorem 2 in [5]. �

Corollary 1. Let f : [0, 1] → R be such that f (2n+2)∈ L1[0, 1] for some n ≥ 1. Then for x ∈ {0} ∪ [1/6, 1/2) we

have ∣∣∣∣∣∫ 1

0f (t)dt − D(0, 1) + T2n(x)

∣∣∣∣∣≤

1(2n + 2)!

∣∣∣∣ 1

3(2x − 1)2

[B2n+2

(x +

12

)− B2n+2(x) − (2 − 2−2n−1)B2n+2

]+ (2 − 2−2n−1)B2n+2

∣∣∣ · ‖ f (2n+2)‖1. (13)

Proof. Inequality (13) follows immediately after taking p = 1 in (12) and using Remark 1. �

Corollary 2. Let f : [0, 1] → R be such that f (2n+1)∈ L∞[0, 1] for some n ≥ 1. Then for x ∈ {0} ∪ [1/6, 1/2) we

have ∣∣∣∣∣∫ 1

0f (t)dt − D(0, 1) + T2n(x)

∣∣∣∣∣≤

2(2n + 2)!

∣∣∣∣ 1

3(2x − 1)2

[B2n+2

(x +

12

)− B2n+2(x) − (2 − 2−2n−1)B2n+2

]+ (2 − 2−2n−1)B2n+2

∣∣∣ · ‖ f (2n+1)‖∞. (14)

If f (2n+2)∈ L∞[0, 1] for some n ≥ 1, then we have∣∣∣∣∣

∫ 1

0f (t)dt − D(0, 1) + T2n(x)

∣∣∣∣∣≤

‖ f (2n+2)‖∞

(2n + 2)!

∣∣∣∣ 1

3(2x − 1)2

[B2n+2(x) + (1 − 2−2n−1)B2n+2

]− (1 − 2−2n−1)B2n+2

∣∣∣∣ . (15)

Proof. Inequality (14) is produced after taking p = ∞ in (11) and using Lemma 1. Similarly, to obtain (15), takep = ∞ in (12) and use Remark 1. �

Estimations for the case when f (n)∈ L p[0, 1] where n = 1, 2 and p = ∞ or p = 1 can also be calculated (as

special cases of Theorem 1), but the expressions are somewhat cumbersome so we won’t state them.

I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370 359

Next, we’ll consider which x gives the best estimation in inequalities (14) and (15). Assume x ∈ [1/6, 1/2). Denotethe function on the right-hand side of (15) with

H(x) = (−1)n[

1

3(2x − 1)2 [B2n+2(x) + (1 − 2−2n−1)B2n+2] − (1 − 2−2n−1)B2n+2

].

Notice that H(x) = |G2n+2(0)|. Then

H ′(x) =(−1)n

3(2x − 1)3

[(2x − 1)(2n + 2)B2n+1(x) + 4

(B2n+2

(12

)− B2n+2 (x)

)].

We claim H ′(x) > 0. To prove that, it is enough to check that h(x) < 0 where

h(x) = (−1)n[(2x − 1)(2n + 2)B2n+1(x) + 4

(B2n+2

(12

)− B2n+2 (x)

)].

Now,

h′′(x) = (2n + 2)(2n + 1)(2n) · (2x − 1) · (−1)n B2n−1(x) < 0.

From there we conclude h′ is a decreasing function and since h′(1/2) = 0, we have h′(x) > 0. Therefore, his increasing and since h(1/2) = 0, we conclude h(x) < 0. Thus, our statement is true, which means H(x) isan increasing function so it obtains its minimal value at x = 1/6. Further, it is easy to see (by induction) that|H(1/6)| < |H(0)|. This shows that Maclaurin’s formula, i.e. its generalization — the Euler–Maclaurin’s formulae— gives the best estimation out of all quadrature formulae of the form∫ 1

0f (t)dt ≈ D(0, 1)

where D(0, 1) is defined by (5).It can be proved analogously that the function on the right-hand side of (14) is also increasing and obtains its

minimum at x = 1/6, so the same conclusion is derived again. Furthermore, the same conclusion follows from (13)since the functions on the right-hand sides of (13) and (14) are equal.

Denote:

R2n+2( f ) =1

(2n + 2)!

∫ 1

0F2n+2(t) f (2n+2)(t)dt. (16)

Theorem 2. If f : [0, 1] → R is such that f (2n+2) is continuous on [0, 1] for some n ≥ 1 and x ∈ {0} ∪ [1/6, 1/2),then there exists a point ξ ∈ [0, 1] such that

R2n+2( f ) = −G2n+2(0)

(2n + 2)!· f (2n+2)(ξ). (17)

Proof. From Remark 1 we know that function F2n+2(t) has constant sign on (0, 1), so the claim follows from themean value theorem for integrals. �

When we apply (17) to the remainder in formula (9) for n = 1, we obtain:∫ 1

0f (t)dt −

1

6(2x − 1)2

[f (x) + 24B2(x) · f

(12

)+ f (1 − x)

]=

12880

(−10x2+ 10x − 1) · f (4)(ξ). (18)

Theorem 3. Let f : [0, 1] → R be such that f (2n+2) is a continuous function on [0, 1] for some n ≥ 1 and does notchange sign on [0, 1] and let x ∈ {0} ∪ [1/6, 1/2). Then there exists a point θ ∈ [0, 1] such that

R2n+2( f ) =θ

(2n + 2)!· F2n+2

(12

)·

[f (2n+1)(1) − f (2n+1)(0)

]. (19)

360 I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370

Proof. Let x ∈ [1/6, 1/2) and suppose f (2n+2)(t) ≥ 0, 0 ≤ t ≤ 1. Then we have

0 ≤

∫ 1

0(−1)n+1 F2n+2(t) f (2n+2)(t)dt ≤ (−1)n+1 F2n+2(1/2) ·

∫ 1

0f (2n+2)(t)dt,

which means there exists θ ∈ [0, 1] such that

(2n + 2)! · R2n+2( f ) = θ · F2n+2(1/2)[ f (2n+1)(1) − f (2n+1)(0)].

When x = 0 or f (2n+2)(t) ≤ 0, 0 ≤ t ≤ 1, the statement follows similarly. �

Remark 2. For x = 0, results of this section produce results obtained in [6], where Euler–Simpson’s formulae werederived. For x = 1/4, results from [7] are produced, i.e. dual Euler–Simpson’s formulae and all related results, andfinally, as we’ve mentioned already, for x = 1/6 Euler–Maclaurin’s formulae are obtained together with all the resultsfrom [8]. Of course, from those formulae, as special cases, follow classical Simpson’s, dual Simpson’s and Maclaurin’sformula, respectively (cf. [12]).

We’ll finish this section by considering the limit process when x tends to 1/2. Formula (18) then becomes:∫ 1

0f (t)dt − f

(12

)−

124

f ′′

(12

)=

11920

f (4)(ξ). (20)

Of course, all other related results can be obtained as well. We have

Gk(t) = B∗

k

(12

− t

)+

k(k − 1)

24B∗

k−2

(12

− t

), k ≥ 2.

So, when f (m)∈ L p[−1, 1] for p = ∞ or p = 1 and m = 2, 3, 4 we get the following estimations:∣∣∣∣∣

∫ 1

0f (t)dt − f

(12

)−

124

f ′′

(12

)∣∣∣∣∣ ≤ C(m, q) · ‖ f (m)‖p

where

C(2, 1) =1

24, C(3, 1) =

1192

, C(4, 1) =1

1920,

C(2, ∞) =18, C(3, ∞) =

148

, C(4, ∞) =1

480.

Comparing these estimations with the ones obtained for the midpoint and the trapezoid formula (cf. [5]) shows thatthese are better in the case when m = 4.

3. Gauss 2-point formulae of Euler type

In this section we consider a special case of the results obtained in the previous section. Namely, take A(x) = 1/2,where A(x) was defined by (3). This gives us

x =12

−1

2√

3.

Since 16 < 1

2 −1

2√

3< 1

2 , we can apply all previously derived results in this special case.

For this choice of the point x , formula (18) becomes the classical Gauss 2-point formula but stated on the interval[0, 1]. Since it is customary to study Gauss formulae on the interval [−1, 1], in order to make use of the symmetry ofthe nodes and coefficients, by a simple linear transformation we transform the interval [0, 1] we have worked with sofar, into [−1, 1]. Denote

QG2 =

∫ 1

−1f (t)dt − f

(−

√3

3

)− f

(√3

3

)+ T2n .

I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370 361

Formulae (8), (4) and (9) now become:

QG2 =

22n−1

(2n)!

∫ 1

−1G2n(t)d f (2n−1)(t), (21)

QG2 =

22n

(2n + 1)!

∫ 1

−1G2n+1(t)d f (2n)(t), (22)

QG2 =

22n+1

(2n + 2)!

∫ 1

−1F2n+2(t)d f (2n+1)(t), (23)

where

T2n =

n∑k=2

22k

(2k)!· B2k

(3 −

√3

6

)[ f (2k−1)(1) − f (2k−1)(−1)],

Gk(t) = B∗

k

(−

√3

6−

t

2

)+ B∗

k

(√3

6−

t

2

),

Fk(t) = Gk(t) − Gk(−1), k ≥ 1, t ∈ R.

We have Gk(−t) = (−1)k Gk(t). From Lemma 1, it follows that

(−1)k G2k+1(t) > 0 for − 1 < t < 0,

so (−1)k+1 F2k+2(t) is increasing on (−1, 0), decreasing on (0, 1) and thus reaches its maximum at t = 0, sinceF2k+2(−1) = F2k+2(1) = 0.

Theorem 1, Corollaries 1 and 2, and Theorems 2 and 3 now become:

Corollary 3. Assume (p, q) is a pair of conjugate exponents, that is 1 ≤ p, q ≤ ∞, 1/p + 1/q = 1. Letf : [−1, 1] → R be such that f (2n)

∈ L p[−1, 1] for some n ≥ 1. Then we have∣∣∣QG2

∣∣∣ ≤ K (2n, q) · ‖ f (2n)‖p. (24)

If f (2n+1)∈ L p[−1, 1] for some n ≥ 0, then we have∣∣∣QG

2

∣∣∣ ≤ K (2n + 1, q) · ‖ f (2n+1)‖p. (25)

If f (2n+2)∈ L p[−1, 1] for some n ≥ 0, then we have∣∣∣QG

2

∣∣∣ ≤ K ∗(2n + 2, q) · ‖ f (2n+2)‖p, (26)

where

K (m, q) =2m−1

m!

[∫ 1

−1|Gm(t)|q dt

] 1q

and K ∗(m, q) =2m−1

m!

[∫ 1

−1|Fm(t)|q dt

] 1q

.

These inequalities are sharp for 1 < p ≤ ∞ and best possible for p = 1.

Corollary 4. Let f : [−1, 1] → R be such that f (2n+1)∈ L∞[−1, 1] for some n ≥ 1. Then we have∣∣∣QG

2

∣∣∣ ≤22n+3

(2n + 2)!

∣∣∣∣∣B2n+2

(√3

6

)− B2n+2

(3 −

√3

6

)∣∣∣∣∣ · ‖ f (2n+1)‖∞. (27)

If f (2n+2)∈ L∞[−1, 1] for some n ≥ 1, then we have∣∣∣QG

2

∣∣∣ ≤22n+3

(2n + 2)!

∣∣∣∣∣B2n+2

(3 −

√3

6

)∣∣∣∣∣ · ‖ f (2n+2)‖∞. (28)

362 I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370

Finally, if f (2n+2)∈ L1[−1, 1] for some n ≥ 1, then we have∣∣∣QG

2

∣∣∣ ≤22n+2

(2n + 2)!

∣∣∣∣∣B2n+2

(√3

6

)− B2n+2

(3 −

√3

6

)∣∣∣∣∣ · ‖ f (2n+2)‖1. (29)

Corollary 5. If f : [−1, 1] → R is such that f (2n+2) is continuous on [−1, 1] for some n ≥ 1, then there exists apoint ξ ∈ [−1, 1] such that

R2n+2( f ) = −22n+3

(2n + 2)!· B2n+2

(3 −

√3

6

)· f (2n+2)(ξ). (30)

Corollary 6. If f : [−1, 1] → R is such that f (2n+2) is a continuous function on [−1, 1], for some n ≥ 1, and doesnot change sign on [−1, 1], then there exists a point θ ∈ [0, 1] such that

R2n+2( f ) = θ ·22n+2

(2n + 2)!

[B2n+2

(√3

6

)− B2n+2

(3 −

√3

6

)]·

[f (2n+1)(1) − f (2n+1)(−1)

]. (31)

As we’ve mentioned earlier, applying (30) to the remainder in (23) for n = 1, produces the classical Gauss 2-pointformula:∫ 1

−1f (t)dt = f

(−

√3

3

)+ f

(√3

3

)+

1135

· f (4)(ξ), ξ ∈ [−1, 1].

As immediate consequences of Corollaries 3 and 4, we obtain the following estimations for p = ∞ and p = 1:∣∣∣∣∣∫ 1

−1f (t)dt − f

(−

√3

3

)− f

(√3

3

)∣∣∣∣∣ ≤ C(m, q) · ‖ f (m)‖p, m = 1, 2, 3, 4,

where

C(1, 1) =5 − 2

√3

3≈ 0.511966, C(1, ∞) =

√3

3≈ 0.57735,

C(2, 1) =49

√26

√3 − 45 ≈ 0.0811291, C(2, ∞) =

2 −√

33

≈ 0.0893164,

C(3, 1) =9 − 4

√3

108≈ 0.0191833, C(3, ∞) =

2 −√

39

√2√

3 − 3 ≈ 0.0202823,

C(4, 1) =1

135≈ 0.00740741, C(4, ∞) =

9 − 4√

3216

≈ 0.00959165.

Remark 3. The constant C(1, ∞) was obtained in [2]. See Remark 5.

Remark 4. Gauss 2-point formulae of Euler type (21)–(23) were also obtained in [5], as a special case of the two-pointformulae that were studied there, but this wasn’t explicitly mentioned.

4. Corrected Gauss 2-point formulae of Euler type

The topic of this section are corrected Gauss 2-point quadrature formulae which, as we’ve mentioned in theIntroduction, approximate the integral not only with the values of the integrand in certain points but with the valuesof its first derivative in the end points of the interval as well. These formulae have a higher degree of exactness thanthe Gauss 2-point formulae of Euler type.

The idea is similar as earlier: instead of taking nods that give G2(−1) = 0, i.e. the nods of the Gauss 2-pointformula, we’ll take nods that give G4(−1) = 0.

I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370 363

Thus, assuming f : [−1, 1] → R is such that f (2n−1) is continuous of bounded variation on [−1, 1] for some

n ≥ 1, put −

√1 −

2√

3015 and

√1 −

2√

3015 in (1), then add those formulae. Denote

QCG2 =

∫ 1

−1f (t)dt − f

−

√1 −

2√

3015

− f

√1 −2√

3015

+ T2n .

The following identity is produced:

QCG2 =

22n−1

(2n)!

∫ 1

−1G2n(t)d f (2n−1)(t), (32)

where, for k ≥ 1 and t ∈ R,

Gk(t) = B∗

k

−12

√1 −

2√

3015

−t

2

+ B∗

k

12

√1 −

2√

3015

−t

2

(33)

T2n =

n∑k=1

22k−1

(2k)!G2k(−1)[ f (2k−1)(1) − f (2k−1)(−1)]

=5 −

√30

15[ f ′(1) − f ′(−1)] +

n∑k=3

22k−1

(2k)!G2k(−1)[ f (2k−1)(1) − f (2k−1)(−1)]. (34)

Assuming f (2n) is continuous of bounded variation on [−1, 1] for some n ≥ 0 we obtain

QCG2 =

22n

(2n + 1)!

∫ 1

−1G2n+1(t)d f (2n)(t). (35)

Finally, if f (2n+1) is continuous of bounded variation on [−1, 1] for some n ≥ 0, from (2) we get

QCG2 =

22n+1

(2n + 2)!

∫ 1

−1F2n+2(t)d f (2n+1)(t), (36)

where

Fk(t) = Gk(t) − Gk(−1), k ≥ 1. (37)

Lemma 2. For k ≥ 2, G2k+1(t) has no zeros in the interval (0, 1). The sign of that function is determined by

(−1)k+1G2k+1(t) > 0, 0 < t < 1.

Proof. Analogous to the proof of Lemma 3. To determine the sign of this function, it is enough to calculate its value

in any point of the interval (0, 1), i.e. t =

√1 −

2√

3015 . �

Theorem 4. Assume (p, q) is a pair of conjugate exponents, that is 1 ≤ p, q ≤ ∞, 1/p + 1/q = 1. Letf : [−1, 1] → R be such that f (2n)

∈ L p[−1, 1] for some n ≥ 1. Then we have∣∣∣QCG2

∣∣∣ ≤ K (2n, q) · ‖ f (2n)‖p. (38)

If f (2n+1)∈ L p[−1, 1] for some n ≥ 0, then we have∣∣∣QCG

2

∣∣∣ ≤ K (2n + 1, q) · ‖ f (2n+1)‖p. (39)

If f (2n+2)∈ L p[−1, 1] for some n ≥ 0, then we have∣∣∣QCG

2

∣∣∣ ≤ K ∗(2n + 2, q) · ‖ f (2n+2)‖p, (40)

364 I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370

where

K (m, q) =2m−1

m!

[∫ 1

−1|Gm(t)|q dt

] 1q

and K ∗(m, q) =2m−1

m!

[∫ 1

−1|Fm(t)|q dt

] 1q

.

These inequalities are sharp for 1 < p ≤ ∞ and best possible for p = 1.

Proof. Analogous to the proof of Theorem 1. �

Corollary 7. Let f : [−1, 1] → R be such that f (2n+1)∈ L∞[−1, 1] for some n ≥ 2. Then we have

∣∣∣QCG2

∣∣∣ ≤22n+3

(2n + 2)!

∣∣∣∣∣∣B2n+2

12

√1 −

2√

3015

− B2n+2

12

−12

√1 −

2√

3015

∣∣∣∣∣∣ · ‖ f (2n+1)‖∞. (41)

If f (2n+2)∈ L∞[−1, 1] for some n ≥ 2, then we have

∣∣∣QCG2

∣∣∣ ≤22n+3

(2n + 2)!

∣∣∣∣∣∣B2n+2

12

−12

√1 −

2√

3015

∣∣∣∣∣∣ · ‖ f (2n+2)‖∞. (42)

Finally, if f (2n+2)∈ L1[−1, 1] for some n ≥ 2, then we have

∣∣∣QCG2

∣∣∣ ≤22n+2

(2n + 2)!

∣∣∣∣∣∣B2n+2

12

√1 −

2√

3015

− B2n+2

12

−12

√1 −

2√

3015

∣∣∣∣∣∣ · ‖ f (2n+2)‖1. (43)

Proof. Analogous to the proof of Corollaries 1 and 2. �

Let R2n+2( f ) be as in (16) with the function F2n+2 defined by (37).

Theorem 5. If f : [−1, 1] → R is such that f (2n+2) is continuous on [−1, 1] for some n ≥ 2, then there exists apoint ξ ∈ [−1, 1] such that

R2n+2( f ) = −22n+3

(2n + 2)!· B2n+2

12

−12

√1 −

2√

3015

· f (2n+2)(ξ). (44)

Proof. Analogous to the proof of Theorem 2. �

Theorem 6. If f : [−1, 1] → R is such that f (2n+2) is a continuous function on [−1, 1], for some n ≥ 2, and doesnot change sign on [−1, 1], then there exists a point θ ∈ [0, 1] such that

R2n+2( f ) = θ ·22n+2

(2n + 2)!

[f (2n+1)(1) − f (2n+1)(−1)

]

·

B2n+2

12

√1 −

2√

3015

− B2n+2

12

−12

√1 −

2√

3015

. (45)

Proof. Analogous to the proof of Theorem 3. �

Applying (44) to the remainder in (36) for n = 2, produces the following formula:∫ 1

−1f (t)dt = f

−

√1 −

2√

3015

+ f

√1 −2√

3015

−

5 −√

3015

[ f ′(1) − f ′(−1)] +14

√30 − 90

70 875· f (6)(ξ), ξ ∈ [−1, 1].

I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370 365

We call this formula the corrected Gauss 2-point formula. Notice that

14√

30 − 9070 875

≈ −0.00018792.

Now, let us consider the case when p = ∞ and p = 1. Then, for m = 1, 2 from Theorem 4 and Corollary 7 weobtain the following estimations:∣∣∣∣∣∣

∫ 1

−1f (t)dt − f

−

√1 −

2√

3015

− f

√1 −2√

3015

∣∣∣∣∣∣ ≤ C(m, q) · ‖ f (m)‖p,

where

C(1, 1) = 3 −4√

3015

− 2

√1 −

2√

3015

≈ 0.500747,

C(1, ∞) =

√1 −

2√

3015

≈ 0.51933,

C(2, 1) =245

3√

30 − 15 + 4

√15(

2√

30 − 15 + 2√

585 − 106√

30)

− 2

√30√

225 − 30√

30 − 225

≈ 0.073765,

C(2, ∞) = 1 −

√30

15−

√1 −

2√

3015

≈ 0.115522,

and for 2 ≤ m ≤ 6∣∣∣∣∣∣∫ 1

−1f (t)dt − f

−

√1 −

2√

3015

− f

√1 −2√

3015

+5 −

√30

15[ f ′(1) − f ′(−1)]

∣∣∣∣∣∣≤ C(m, q) · ‖ f (m)

‖p,

where

C(2, 1) ≈ 0.0650208, C(2, ∞) =1

15

(20 − 2

√30 −

√225 − 30

√30)

≈ 0.083707,

C(3, 1) =1

180

(339 − 60

√30 − 4

√585 − 106

√30)

≈ 0.0109032,

C(3, ∞) =1

45

√2√

25 653 825 − 4649 430√

30 + 1226√

30 −22 745

3≈ 0.010905,

C(4, 1) ≈ 0.00209577, C(4, ∞) =524

−1

√30

−1

90

√585 − 106

√30 ≈ 0.002415,

C(5, 1) =2√

392 625 − 71 670√

30 + 150√

30 − 82527 000

≈ 0.000503075,

C(5, ∞) ≈ 0.000523942, C(6, 1) =90 − 14

√30

70875≈ 0.00018792,

C(6, ∞) =1

3600

2

√1745 −

4778√

3015

+ 10√

30 − 55

≈ 0.000251537.

366 I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370

The exact values of C(2, 1) (for the case when the values of the first derivative are included in the quadrature formula),C(4, 1) and C(5, ∞) can be calculated (with the help of Wolfram’s Mathematica) but obtained expressions are rathercumbersome so we state only their approximate values.

Also, notice that when p = ∞ and m = 1, 2, and when p = 1 and m = 1, the corrected Gauss 2-point formulaegive better estimations than the Gauss 2-point formulae.

5. Gauss 3-point formulae of Euler type

In this final section, following the same idea as previously, the Gauss 3-point formula is generalized. We start froma three point formula:∫ 1

−1f (t)dt ≈ A f (−x) + (2 − 2A) f (0) + A f (x). (46)

Let us consider what happens if conditions G2(−1) = 0 and G4(−1) = 0 are imposed simultaneously. The solutionof this system produces the following quadrature formula:∫ 1

−1f (t)dt ≈

19

[5 f

(−

√155

)+ 8 f (0) + 5 f

(√155

)],

which is exactly the classical Gauss 3-point formula.Therefore, assume f : [−1, 1] → R is such that f (2n−1) is continuous of bounded variation on [−1, 1] for some

n ≥ 1. Put −√

15/5, 0,√

15/5 in (1), multiply by 5/9, 8/9, 5/9, respectively, and add. Denote

QG3 =

∫ 1

−1f (t)dt −

19

[5 f

(−

√155

)+ 8 f (0) + 5 f

(√155

)]+ T2n .

The following formula is produced:

QG3 =

22n−1

(2n)!

∫ 1

−1G2n(t)d f (2n−1)(t), (47)

where, for k ≥ 1 and t ∈ R,

Gk(t) =19

[5B∗

k

(−

√15

10−

t

2

)+ 8B∗

k

(−

t

2

)+ 5B∗

k

(√15

10−

t

2

)](48)

T2n =

n∑k=3

22k−1

(2k)!G2k(−1)[ f (2k−1)(1) − f (2k−1)(−1)]. (49)

Assuming f (2n) is continuous of bounded variation on [−1, 1] for some n ≥ 0 we obtain

QG3 =

22n

(2n + 1)!

∫ 1

−1G2n+1(t)d f (2n)(t). (50)

Finally, if f (2n+1) is continuous of bounded variation on [−1, 1] for some n ≥ 0, from (2) we get

QG3 =

22n+1

(2n + 2)!

∫ 1

−1F2n+2(t)d f (2n+1)(t), (51)

where

Fk(t) = Gk(t) − Gk(−1), k ≥ 1. (52)

Lemma 3. For k ≥ 2, G2k+1(t) has no zeros in the interval (0, 1). The sign of that function is determined by

(−1)k G2k+1(t) > 0, 0 < t < 1.

I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370 367

Proof. We start from G5(t). It is easy to see that G ′′

5(t) = 5G3(t) and that

G3(t) =

1

12t (−9 + 2

√15 + 4t − 3t2), 0 ≤ t ≤

√15/5

14(1 − t)3,

√15/5 ≤ t ≤ 1.

Thus, we conclude G5(t) is concave on I1 =

(0,

2−

√6√

15−233

)and convex on I2 =

(2−

√6√

15−233 , 1

). Also, we

have G ′

4(t) = −2G3(t), so G4(t) is decreasing on I2 and since G4(1) = 0, it follows that G4(t) is positive on thatinterval which implies G5(t) is decreasing there, as G ′

5(t) = −5/2 · G4(t). Finally, since G5(0) = G5(1) = 0, weconclude G5(t) cannot have any zeros on (0, 1). Therefore, the assertion is true for k = 2. Similarly as in the proof ofLemma 1, the assertion for k ≥ 3 follows by induction.

In order to determine the sign of G5(t), by direct calculation for√

155 ≤ t ≤ 1 we get G5(t) =

116 (1 − t)5, so

G5(t) > 0, 0 < t < 1. This tells us that G7(t) is convex on (0, 1) and as it has no zeros, we conclude G7(t) < 0. Bythis procedure we can determine the sign of G2k+1(t) for k ≥ 4 which completes our proof. �

Theorem 7. Assume (p, q) is a pair of conjugate exponents, that is 1 ≤ p, q ≤ ∞, 1/p + 1/q = 1. Letf : [−1, 1] → R be such that f (2n)

∈ L p[−1, 1] for some n ≥ 1. Then we have∣∣∣QG3

∣∣∣ ≤ K (2n, q) · ‖ f (2n)‖p. (53)

If f (2n+1)∈ L p[−1, 1] for some n ≥ 0, then we have∣∣∣QG

3

∣∣∣ ≤ K (2n + 1, q) · ‖ f (2n+1)‖p. (54)

If f (2n+2)∈ L p[−1, 1] for some n ≥ 0, then we have∣∣∣QG

3

∣∣∣ ≤ K ∗(2n + 2, q) · ‖ f (2n+2)‖p, (55)

where

K (m, q) =2m−1

m!

[∫ 1

−1|Gm(t)|q dt

] 1q

and K ∗(m, q) =2m−1

m!

[∫ 1

−1|Fm(t)|q dt

] 1q

.

These inequalities are sharp for 1 < p ≤ ∞ and best possible for p = 1.

Proof. Analogous to the proof of Theorem 1. �

Corollary 8. Let f : [−1, 1] → R be such that f (2n+1)∈ L∞[−1, 1] for some n ≥ 2. Then we have∣∣∣QG

3

∣∣∣ ≤22n+3

9(2n + 2)!

∣∣∣∣∣5[

B2n+2

(√15

10

)− B2n+2

(5 −

√15

10

)]+ (8 − 21−2n)B2n+2

∣∣∣∣∣ · ‖ f (2n+1)‖∞. (56)

If f (2n+2)∈ L∞[−1, 1] for some n ≥ 2, then we have∣∣∣QG

3

∣∣∣ ≤22n+3

9(2n + 2)!

∣∣∣∣∣5B2n+2

(5 −

√15

10

)− (4 − 21−2n)B2n+2

∣∣∣∣∣ · ‖ f (2n+2)‖∞. (57)

Finally, if f (2n+2)∈ L1[−1, 1] for some n ≥ 2, then we have∣∣∣QG

3

∣∣∣ ≤22n+2

9(2n + 2)!

∣∣∣∣∣5[

B2n+2

(√15

10

)− B2n+2

(5 −

√15

10

)]+ (8 − 21−2n)B2n+2

∣∣∣∣∣ · ‖ f (2n+2)‖1. (58)

Proof. Analogous to the proof of Corollaries 1 and 2. �

368 I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370

Let R2n+2( f ) be defined as in (16) where the function F2n+2 is given by (52).

Theorem 8. If f : [−1, 1] → R is such that f (2n+2) is continuous on [−1, 1] for some n ≥ 2, then there exists apoint ξ ∈ [−1, 1] such that

R2n+2( f ) = −22n+3

9(2n + 2)!·

[5B2n+2

(5 −

√15

10

)− (4 − 21−2n)B2n+2

]· f (2n+2)(ξ). (59)

Proof. Analogous to the proof of Theorem 2. �

Theorem 9. If f : [−1, 1] → R is such that f (2n+2) is a continuous function on [−1, 1], for some n ≥ 2, and doesnot change sign on [−1, 1], then there exists a point θ ∈ [0, 1] such that

R2n+2( f ) = θ ·22n+2

9(2n + 2)![ f (2n+1)(1) − f (2n+1)(−1)]

·

[5

[B2n+2

(√15

10

)− B2n+2

(5 −

√15

10

)]+ (8 − 21−2n)B2n+2

]. (60)

Proof. Analogous to the proof of Theorem 3. �

Applying (59) to the remainder in (51) produces the classical Gauss 3-point formula:∫ 1

−1f (t)dt =

19

[5 f

(−

√155

)+ 8 f (0) + 5 f

(√155

)]+

115 750

· f (6)(ξ), ξ ∈ [−1, 1].

Further, for p = ∞ and p = 1, the following estimations are obtained:∣∣∣∣∣∫ 1

−1f (t)dt −

19

[5 f

(−

√155

)+ 8 f (0) + 5 f

(√155

)]∣∣∣∣∣ ≤ C(m, q) · ‖ f (m)‖p, 1 ≤ m ≤ 6

where

C(1, 1) =1051 − 234

√15

405≈ 0.357338, C(1, ∞) =

49

≈ 0.444444,

C(2, 1) =8

2187(18

√15 − 65)3/2

≈ 0.0374355, C(2, ∞) =9 − 2

√15

18≈ 0.0696685,

C(3, 1) =160(6

√15 − 23)3/2

+ 6516√

15 − 25 17514 580

≈ 0.00548184,

C(3, ∞) =(18

√15 − 65)3/2

− 108√

15 + 4222187

≈ 0.0063794,

C(4, 1) ≈ 0.000908828, C(4, ∞) =4√

15 − 15360

≈ 0.00136648,

C(5, 1) =25 − 6

√15

9000≈ 0.000195789, C(5, ∞) ≈ 0.000227207,

C(6, 1) =1

15 750≈ 0.0000634921, C(6, ∞) =

25 − 6√

1518 000

≈ 0.0000978944.

Integral∫ 1−1 |G4(t)|dt is calculated only approximately, with the help of Wolfram’s Mathematica, because it is difficult

to obtain the exact value of the zero of G4(t) (which is t ≈ 0.280949). Further, |G5(t)| achieves its maximum in thezero of G4(t), so we have the same problem again. Therefore, C(5, ∞) is also calculated with the help of Wolfram’sMathematica.

I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370 369

Remark 5. The constants C(1, ∞) obtained for the Gauss 2- and 3-point formula coincide with the constants %V (RG2 )

and %V (RG3 ), respectively, obtained in Theorem 1.1. in [2].

Remark 6. In Section 2, we have proved that in the class of all 3-point quadrature formulae where the integral isapproximated with (5), the best estimation is obtained for x = 1/6 (when considering an integral over [−1, 1], forx = 2/3), i.e. Maclaurin’s formula. Similar type inequalities as we’ve just derived for the Gauss 3-point formulacan be obtained for Maclaurin’s formula as well, as special cases of results from Section 2 (see [8]). Comparingestimations obtained for Maclaurin’s formula and the Gauss 3-point formula shows that, when p = ∞, the Gauss3-point formula gives a better approximation for m = 2, 3, 4, while when p = 1, it gives a better approximation form = 3 and m = 4.

Analogously as was done in Section 4 for the Gauss 2-point formula, the corrected Gauss 3-point formulae can bederived. Namely, if we consider quadrature formulae (46) and instead of G2(−1) = G4(−1) = 0, impose conditionsG4(−1) = G6(−1) = 0, the following formula is obtained:∫ 1

−1f (t)dt ≈

1977 + 16√

1023465

[f

(−

17

√45 − 2

√102

)+ f

(17

√45 − 2

√102

)]+

2976 − 32√

1023465

f (0).

As the formula is rather cumbersome, we’ll state only the estimations obtained from Holder’s inequality for functionssuch that f (m)

∈ L p[−1, 1] for p = ∞ or p = 1 and 1 ≤ m ≤ 8. To shorten notation, put x0 =17

√45 − 2

√102.

Then, for m = 1, 2∣∣∣∣∣∫ 1

−1f (t)dt −

1977 + 16√

1023465

[ f (−x0) + f (x0)] −2976 − 32

√102

3465f (0)

∣∣∣∣∣ ≤ C(m, q) · ‖ f (m)‖p

where

C(1, 1) ≈ 0.337807, C(1, ∞) ≈ 0.382802,

C(2, 1) ≈ 0.0313153, C(2, ∞) ≈ 0.0609023

and for 2 ≤ m ≤ 8∣∣∣∣∣∫ 1

−1f (t)dt −

1977 + 16√

1023465

[ f (−x0) + f (x0)] −2976 − 32

√102

3465f (0) +

9 −√

102105

[ f ′(1) − f ′(−1)]

∣∣∣∣∣≤ C(m, q) · ‖ f (m)

‖p

where

C(2, 1) ≈ 0.0300722, C(2, ∞) ≈ 0.0504308,

C(3, 1) ≈ 0.00351347, C(3, ∞) ≈ 0.00386067

C(4, 1) ≈ 0.000438656, C(4, ∞) ≈ 0.000610086,

C(5, 1) ≈ 0.0000618111, C(5, ∞) ≈ 0.000077903,

C(6, 1) ≈ 1.00553 × 10−5, C(6, ∞) ≈ 0.0000151755,

C(7, 1) ≈ 2.077547 × 10−6, C(7, ∞) ≈ 2.513835 × 10−6,

C(8, 1) ≈ 6.530965 × 10−7, C(8, ∞) ≈ 1.038773 × 10−6.

Notice that the corrected Gauss 3-point formula (when the quadrature formula doesn’t include derivatives) gives betterestimates than the classical Gauss 3-point formula in cases when m = 1, 2.

370 I. Franjic et al. / Mathematical and Computer Modelling 45 (2007) 355–370

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