# Revisited Optimal Error Bounds for Interpolatory Integration Rules

Authors: François Dubeau
Journal: Advances in Numerical Analysis
 (2016)

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## Abstract

We present a unified way to obtain optimal error bounds for general interpolatory integration rules. The method is based on the Peano form of the error term when we use Taylor’s expansion. These bounds depend on the regularity of the integrand. The method of integration by parts “backwards” to obtain bounds is also discussed. The analysis includes quadrature rules with nodes outside the interval of integration. Best error bounds for composite integration rules are also obtained. Some consequences of symmetry are discussed.


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## Body

## 1. Introduction

Establishing numerical integration rules and their error bounds is an old subject; for example, see the following classic textbooks [1–4]. However, in recent years, several papers have presented error bounds for the midpoint, trapezoidal, and Simpson’s rules; see [5–10]. In these papers, questions were raised about the generality of the results in terms of the optimality of the bounds and regularity of the function. They also suggest that consequences of the symmetry on integration rules should be more fully investigated. Based on Peano form of the error term when we use Taylor’s expansion, we present a unified way to obtain optimal error bounds for general interpolatory integration rules. These bounds depend on the regularity of the integrand. Other similar approaches based on kernels have been recently presented in [11] for Newton-Cotes quadrature rules and in [12] for Gaussian weighted quadrature rules.Let us start by a transformation of the definite integral as follows:(1)∫abFXdX=∫-hhfxdx,where X=a+b/2+x, h=b-a/2, and f(x)=Fa+b/2+x. Then we consider the method of undetermined coefficients to approximate the following expression:(2)Qf;h=1h∫-hhfxdx.The method of undetermined coefficients consists in finding a (n+1)-dimensional weight vector a→=a0,a1,a2,…,an associated with a given (n+1)-dimensional vector of distinct coordinates (or nodes) x→=x0,x1,x2,…,xn such that Q(f;h) is approximated by its discrete version Qnd(f;h) given by(3)Qndf;h=∑i=0naifhxi.The truncation error of the process is defined by(4)Rnf;h=Qf;h-Qndf;h,and the method is based on the requirement that(5)Rnf;h=ohn.A general analysis of the method of undetermined coefficients was recently published in [13].To study the truncation error, two approaches both based on Taylor’s expansions for absolutely continuous functions and Peano’s kernels are presented. In the first approach, the “direct method,” we use (4) for Rn(f;h) and a Taylor expansion of the integrand. While in the second approach, the method of integration by parts “backwards” [4, 14–16], we use Taylor’s expansion not only for integrand but also for Wn(f;h)=hRn(f;h). It is shown that both methods lead to the same best error bounds.Coming back to the definite integral given in (1), we get(6)∫abFXdX≈b-a2∑i=0naiFXi,where Xi=a+b/2+b-a/2xi, for i=0,1,2,…,n. However, to compute this definite integral we can also use a composite rule, for which we also present optimal error bounds.The plan of the paper is the following. In Section2, we obtain best error bounds using Taylor’s expansions and Peano’s kernels. Section 2.3 is devoted to the “direct method,” and Section 2.4 presents the “method of integration by parts backwards.” Examples are presented in Section 3. Composite integration rules is the object of Section 4. Finally we consider symmetric rules in Section 5.We usef(l)(x) for the lth derivative of f(x) for l=0,1,2,…, where f(0)(x)=f(x). Let 1≤p≤∞; if f(x) is defined on a set E, fp,E is its p-norm on E, and if v→ is a vector in Rn, its p-norm is v→p.


## 2. Truncation Error

### 2.1. Introduction

Let(7)τ=max⁡1,x0,x1,x2,…,xn=max⁡1,x→∞≥1,and set H=τh. Let us observe that H>h, or τ>1, means that at least one xi is strictly greater that 1. Hence it allows the possibility of having numerical integration formulae with nodes hxi outside the interval of integration [-h,h]; see [17], for example. In that situation, the function f(x) has to be defined on an interval [-H,H] which contains [-h,h].For the method of undetermined coefficients it is required thatRn(f;h)=0 at least for polynomial of degree less than or equal to n, but Rn(f;h)=0 might hold for some polynomials f(x) of degree higher than n; see [13]. It happens for Simpson’s rule (n=2) which is also exact for polynomials of degree 3, or also for n+1 points Gaussian rule which is exact for polynomials of degree ≤2n+1. Let us define the degree of accuracy (or precision) n0 of the approximation process (3) as the largest integer n0≥n such that Rn(f;h)=0 holds for any polynomial f(x) of degree l≤n0. These rules are also called interpolatory quadrature formulae.


### 2.2. Taylor’s Expansions

LetH>0 and IH=-H,H. For H=1, we simply use I=-1,1, IH+=0,H, and IH-=-H,0. Let p and q be two extended real numbers such that 1≤p,q≤∞ and 1/p+1/q=1. Let Cl(IH) be the set of continuously differentiable functions up to order l on IH and let p∈[1,+∞]. Let ACl+1,p(IH) be the set of absolutely continuous function on IH defined by f∈ACl+1,p(IH) if and only if (8)f∈ClIHand (a)
f(l+1)∈Lp(IH),(b)
f(l)(s)=f(l)(r)+∫rsf(l+1)(ξ)dξ,∀r,s∈IH.Taylor’s expansion of f(x)∈ACl+1,p(IH) around x=0 of order l+1 is(9)fx=∑j=0lfj0jxj+∫-HHfl+1yKT,lx,y;Hdy,where KT,l(x,y;H) is its associated kernel(10)KT,lx,y;H=1l!x-y+l1IH+y+-1l+1y-x+l1IH-y,for any x, y in IH; see [18, 19]. This kernel is a piecewise polynomial function of degree l. In this expression, if E is a subset of R, then (11)1Ey=1ify∈E0ify∉E,and for any nonnegative integer l(12)z+l=zlifz>00ifz≤0.If we set H=hτ, x=hξ, and y=hη, then the kernel becomes(13)KT,lx,y;H=KT,lhξ,hη;hτ=hlKT,lξ,η;τ,for any ξ and η in Iτ.


### 2.3. Direct Method

For the truncation error analysis, letf(x)∈ACl+1,p(IH), for 0≤l≤n0. Using Taylor’s expansion (9) of f(x) of order l+1 and the fact that the process is exact for polynomials of degree ≤l, we obtain(14)Rnf;h=∫-HHfl+1yKnly;H,hdy. Here Knl(y;H,h) is the Peano kernel associated with the process Qnd(f;h), given by(15)Knly;H,h=RnKT,l·,y;H;h. Let q∈[1,+∞] be the conjugate of p such that 1/p+1/q=1. From Holder’s inequality, we obtain(16)Rnf;h≤fl+1p,IHKnl·;H,hq,IH,because Knl(y;H,h)∈L∞(IH)⊆Lq(IH), for any 1≤q≤∞.Let us observe that ifH=hτ and y=hη, then Knl(y;H,h)=hlKnl(η;τ,1). It follows that(17)Knl·;H,hq,IH=hl+1-1/pKnl·;τ,1q,Iτ,(18)Rnf;h≤hl+1-1/pCnl,pfl+1p,IH,where the constant(19)Cnl,p=Knl·;τ,1q,Iτdoes not depend on h. So we have established the following results.Theorem 1.
Let the real numberτ≥1 be fixed and H=hτ. If Rn(f;h)=0, for any polynomial of degree ≤n0, then (16), and equivalently (18), holds for any f∈ACl+1,p(IH), where 0≤l≤n0.Theorem 2.
In (17), the kernel Knl(y;H,h) is given by (20)Knly;H,h=1h∫-hhKT,lx,y;Hdx-∑i=0naiKT,lhxi,y;H=1l+1!hh-y+l+11IH+y+-1l+1y+h+l+11IH-y-1l!∑i=0naihxi-y+l1IH+y+-1l+1y-hxi+l1IH-y,and, in (19), the kernel Knl(η;τ,1) is given by (21)Knlη;τ,1=1l+1!1-η+l+11Iτ+η+-1l+1η+1+l+11Iτ-η-1l!∑i=0naixi-η+l1Iτ+η+-1l+1η-xi+l1Iτ-η.Remark 3.
It can be proved that the bound given by (16) and (17), or equivalently by (18) and (19), is the best possible one; see [13].Remark 4.
For the functionF(X) used in (1), we have (22)Fl+1p,a′,b′=fl+1p,IH,where a,b⊆a′,b′=a+b/2+-H,H. Obviously, F(X) is supposed to be defined for any X∈a′,b′.


### 2.4. Integration by Parts “Backwards”

The method of integration by parts “backwards,” which is reported to have initially appeared in [4, 14–16], was used in [6–10] to find truncation error estimates for the midpoint, trapezoidal, and Simpson’s rules. These rules possess a property of symmetry, which help in finding optimal bounds in these cases. However, the method of integration by parts “backwards” can be applied to any rule that can be obtained by the method of undetermined coefficients. We present an analysis of the truncation error based on this method.The process is based on Taylor’s expansions of(23)Wnf;h=hRnf;h,and we suppose that f∈ACl+1,p(IH) for 0≤l≤n0. We have(24)Wnf;h=hRnh;f=∫-hhfxdx-h∑i=0naifhxi,so Wn(f;0)=0. For 1≤j<l, we have(25)Wnjf;h=fj-1h+-1j-1fj-1-h-j∑i=0naixij-1fj-1hxi-h∑i=0naixijfjhxi,(26)limh→0⁡Wnjf;h=fj-101+-1j-1-j∑i=0naixij-1=0. Also, for j=l,(27)Wnlf;h=fl-1h+-1l-1fl-1-h-l∑i=0naixil-1fl-1hxi-h∑i=0naixilflhxi.Using Taylor’s expansions of order 2 for f(l-1)(x) which is in AC2,p(IH) and of order 1 for f(l)(x) which is in AC1,p(IH), we obtain(28)Wnlf;h=∫-HHfl+1yKW,ly;H,hdy,where(29)KW,ly;H,h=KT,1h,y;H+-1l-1KT,1-h,y;H-l∑i=0naixil-1KT,1hxi,y;H-h∑i=0naixilKT,0hxi,y;H.We remark that (30)KW,ly;H,h=KW,lhη;τh,h=hKW,lη;τ,1.Taylor’s expansion of orderl for Wn(f;h) or integration by parts “backwards” leads to (31)Wnf;h=∫0hWnlf;zh-zl-1l-1!dz=∫0h∫-τzτzfl+1yKW,ly;τz,zdyh-zl-1l-1!dz=∫-τhτhfl+1y∫y/τhKW,ly;τz,zh-zl-1l-1!dzdy=∫-HHfl+1yK^nly;H,hdy.As mentioned in Remark3, the construction used in [13] to show the optimality of the bounds leads also to the following result.Theorem 5.
Leth>0 and H=τh be given; the kernels Knl(·;H,h) and K^nl(·;H,h) are such that (32)hKnly;H,h=K^nly;H,h almost everywhere.As a consequence both methods lead to the same best error bounds.


## 2.1. Introduction

Let(7)τ=max⁡1,x0,x1,x2,…,xn=max⁡1,x→∞≥1,and set H=τh. Let us observe that H>h, or τ>1, means that at least one xi is strictly greater that 1. Hence it allows the possibility of having numerical integration formulae with nodes hxi outside the interval of integration [-h,h]; see [17], for example. In that situation, the function f(x) has to be defined on an interval [-H,H] which contains [-h,h].For the method of undetermined coefficients it is required thatRn(f;h)=0 at least for polynomial of degree less than or equal to n, but Rn(f;h)=0 might hold for some polynomials f(x) of degree higher than n; see [13]. It happens for Simpson’s rule (n=2) which is also exact for polynomials of degree 3, or also for n+1 points Gaussian rule which is exact for polynomials of degree ≤2n+1. Let us define the degree of accuracy (or precision) n0 of the approximation process (3) as the largest integer n0≥n such that Rn(f;h)=0 holds for any polynomial f(x) of degree l≤n0. These rules are also called interpolatory quadrature formulae.


## 2.2. Taylor’s Expansions

LetH>0 and IH=-H,H. For H=1, we simply use I=-1,1, IH+=0,H, and IH-=-H,0. Let p and q be two extended real numbers such that 1≤p,q≤∞ and 1/p+1/q=1. Let Cl(IH) be the set of continuously differentiable functions up to order l on IH and let p∈[1,+∞]. Let ACl+1,p(IH) be the set of absolutely continuous function on IH defined by f∈ACl+1,p(IH) if and only if (8)f∈ClIHand (a)
f(l+1)∈Lp(IH),(b)
f(l)(s)=f(l)(r)+∫rsf(l+1)(ξ)dξ,∀r,s∈IH.Taylor’s expansion of f(x)∈ACl+1,p(IH) around x=0 of order l+1 is(9)fx=∑j=0lfj0jxj+∫-HHfl+1yKT,lx,y;Hdy,where KT,l(x,y;H) is its associated kernel(10)KT,lx,y;H=1l!x-y+l1IH+y+-1l+1y-x+l1IH-y,for any x, y in IH; see [18, 19]. This kernel is a piecewise polynomial function of degree l. In this expression, if E is a subset of R, then (11)1Ey=1ify∈E0ify∉E,and for any nonnegative integer l(12)z+l=zlifz>00ifz≤0.If we set H=hτ, x=hξ, and y=hη, then the kernel becomes(13)KT,lx,y;H=KT,lhξ,hη;hτ=hlKT,lξ,η;τ,for any ξ and η in Iτ.


## 2.3. Direct Method

For the truncation error analysis, letf(x)∈ACl+1,p(IH), for 0≤l≤n0. Using Taylor’s expansion (9) of f(x) of order l+1 and the fact that the process is exact for polynomials of degree ≤l, we obtain(14)Rnf;h=∫-HHfl+1yKnly;H,hdy. Here Knl(y;H,h) is the Peano kernel associated with the process Qnd(f;h), given by(15)Knly;H,h=RnKT,l·,y;H;h. Let q∈[1,+∞] be the conjugate of p such that 1/p+1/q=1. From Holder’s inequality, we obtain(16)Rnf;h≤fl+1p,IHKnl·;H,hq,IH,because Knl(y;H,h)∈L∞(IH)⊆Lq(IH), for any 1≤q≤∞.Let us observe that ifH=hτ and y=hη, then Knl(y;H,h)=hlKnl(η;τ,1). It follows that(17)Knl·;H,hq,IH=hl+1-1/pKnl·;τ,1q,Iτ,(18)Rnf;h≤hl+1-1/pCnl,pfl+1p,IH,where the constant(19)Cnl,p=Knl·;τ,1q,Iτdoes not depend on h. So we have established the following results.Theorem 1.
Let the real numberτ≥1 be fixed and H=hτ. If Rn(f;h)=0, for any polynomial of degree ≤n0, then (16), and equivalently (18), holds for any f∈ACl+1,p(IH), where 0≤l≤n0.Theorem 2.
In (17), the kernel Knl(y;H,h) is given by (20)Knly;H,h=1h∫-hhKT,lx,y;Hdx-∑i=0naiKT,lhxi,y;H=1l+1!hh-y+l+11IH+y+-1l+1y+h+l+11IH-y-1l!∑i=0naihxi-y+l1IH+y+-1l+1y-hxi+l1IH-y,and, in (19), the kernel Knl(η;τ,1) is given by (21)Knlη;τ,1=1l+1!1-η+l+11Iτ+η+-1l+1η+1+l+11Iτ-η-1l!∑i=0naixi-η+l1Iτ+η+-1l+1η-xi+l1Iτ-η.Remark 3.
It can be proved that the bound given by (16) and (17), or equivalently by (18) and (19), is the best possible one; see [13].Remark 4.
For the functionF(X) used in (1), we have (22)Fl+1p,a′,b′=fl+1p,IH,where a,b⊆a′,b′=a+b/2+-H,H. Obviously, F(X) is supposed to be defined for any X∈a′,b′.


## 2.4. Integration by Parts “Backwards”

The method of integration by parts “backwards,” which is reported to have initially appeared in [4, 14–16], was used in [6–10] to find truncation error estimates for the midpoint, trapezoidal, and Simpson’s rules. These rules possess a property of symmetry, which help in finding optimal bounds in these cases. However, the method of integration by parts “backwards” can be applied to any rule that can be obtained by the method of undetermined coefficients. We present an analysis of the truncation error based on this method.The process is based on Taylor’s expansions of(23)Wnf;h=hRnf;h,and we suppose that f∈ACl+1,p(IH) for 0≤l≤n0. We have(24)Wnf;h=hRnh;f=∫-hhfxdx-h∑i=0naifhxi,so Wn(f;0)=0. For 1≤j<l, we have(25)Wnjf;h=fj-1h+-1j-1fj-1-h-j∑i=0naixij-1fj-1hxi-h∑i=0naixijfjhxi,(26)limh→0⁡Wnjf;h=fj-101+-1j-1-j∑i=0naixij-1=0. Also, for j=l,(27)Wnlf;h=fl-1h+-1l-1fl-1-h-l∑i=0naixil-1fl-1hxi-h∑i=0naixilflhxi.Using Taylor’s expansions of order 2 for f(l-1)(x) which is in AC2,p(IH) and of order 1 for f(l)(x) which is in AC1,p(IH), we obtain(28)Wnlf;h=∫-HHfl+1yKW,ly;H,hdy,where(29)KW,ly;H,h=KT,1h,y;H+-1l-1KT,1-h,y;H-l∑i=0naixil-1KT,1hxi,y;H-h∑i=0naixilKT,0hxi,y;H.We remark that (30)KW,ly;H,h=KW,lhη;τh,h=hKW,lη;τ,1.Taylor’s expansion of orderl for Wn(f;h) or integration by parts “backwards” leads to (31)Wnf;h=∫0hWnlf;zh-zl-1l-1!dz=∫0h∫-τzτzfl+1yKW,ly;τz,zdyh-zl-1l-1!dz=∫-τhτhfl+1y∫y/τhKW,ly;τz,zh-zl-1l-1!dzdy=∫-HHfl+1yK^nly;H,hdy.As mentioned in Remark3, the construction used in [13] to show the optimality of the bounds leads also to the following result.Theorem 5.
Leth>0 and H=τh be given; the kernels Knl(·;H,h) and K^nl(·;H,h) are such that (32)hKnly;H,h=K^nly;H,h almost everywhere.As a consequence both methods lead to the same best error bounds.


## 3. Examples

In this section we present several examples. In some cases, constantsCnl,p=Knl(·;τ,1)q,Iτ are computed for p=∞ (and q=1) and are compared to constants already existing in the references.Example 1.
Midpoint rule (or one point Gauss rule):n=0, x→=(0), and hence τ=1. Also, a→=(2), and n0=1. The quadrature formula is (33)∫-11fxdx≈2f0. For l=0,1, we have (34)K0lη;1,1=1l+1!1-ηl+11I+η+-1l+1η+1l+11I-η,so we obtain (35)K0l·;1,1q,I=1l+1!21+l+1q1/qfor1≤q<∞1l+1!forq=∞.For example, C00,∞=1 and C01,∞=1/3, which correspond to values obtained in [7, 9].Example 2.
Trapezoidal rule:n=1, x→=(-1,1), and τ=1. Also, a→=(1,1) and n0=1. The quadrature formula is (36)∫-11fxdx≈f-1+f1. For l=0,1, we have (37)K1lη;1,1=-1-ηll+1!η+l1I+η+-1l+1η+1ll+1!η-l1I-η.We obtain (38)
K
1
l
·
;
1,1
q
,
I
=
2
l
+
1
l
+
2
!
f
o
r
q
=
1
1
l
+
1
!
2
∫
0
1
1
-
η
l
q
η
+
l
q
d
η
1
/
q
f
o
r
1
<
q
<
∞
1
l
+
1
!
f
o
r
q
=
∞
.For example, C0,10,∞=1 and C0,11,∞=2/3, which correspond to values obtained in [5, 6, 9].Example 3.
First Simpson’s rule:n=2, x→=(-1,0,1), and τ=1. Also, a→=1/3,4/3,1/3 and n0=3. The quadrature formula is (39)∫-11fxdx≈13f-1+43f0+13f1.For l=0,1,2,3, we have (40)K2lη;1,1=-1-ηll+1!η+l-231I+η+-1l+1η+1ll+1!η-l-231I-η.We obtain (41)
K
0,2
l
·
;
1,1
q
,
I
=
1
l
+
1
!
2
∫
0
1
1
-
η
l
q
η
+
l
-
2
3
q
d
η
1
/
q
f
o
r
1
≤
q
<
∞
1
l
+
1
!
m
a
x
η
∈
I
+
1
-
η
l
η
+
l
-
2
3
f
o
r
q
=
∞
.For p=∞ and q=1, we get C0,20,∞=5/9, C0,21,∞=8/81, C0,22,∞=1/36, and C0,23,∞=1/90, which correspond to constants obtained in [5, 7–10].Example 4.
Second Simpson’s rule:n=3, x→=-1,-1/3,1/3,1, and τ=1. Also, a→=1/4,3/4,3/4,1/4 and n0=3. The quadrature formula is (42)∫-11fxdx≈14f-1+34f-13+34f13+14f1. For l=0,1,2,3, we have (43)K3lη;1,1=-1-ηll+1!η+l-34+34l!13-η+l1I+η+-1l+1η+1ll+1!η-l-34-34l!η+13+l1I-η.Example 5.
A2-point Gauss rule: n=1, x→=-1/3,1/3, and τ=1. Also, a→=(1,1) and n0=3. The quadrature formula is (44)∫-11fxdx≈f-13+f13.For l=0,1,2,3, we have (45)K1lη;1,1=1l+1!1-ηl+1-l+113-η+l1I+η+-1l+1l+1!η+1l+1-l+1η+13+l1I-η.Example 6.
A3-point Gauss rule: n=2, x→=-3/5,0,3/5, and τ=1. Also, a→=5/9,8/9,5/9 and n0=5. The quadrature formula is (46)∫-11fxdx≈59f-35+89f0+59f35. For l=0,1,2,3,4,5, we have (47)K0,2lη;1,1=1l+1!1-ηl+1-59l+135-η+l1I+η+-1l+1l+1!η+1l+1-59l+1η+35+l1I-η.Example 7.
First nonstandard rule:n=1, x→=-2,2, and τ=2. Also, a→=(1,1) and n0=1. The quadrature formula is (48)∫-11fxdx≈f-2+f2.For l=0,1, we have (49)K1lη;2,1=1l+1!1-η+l+1-l+12-η+l1I2+η+-1l+1l+1!η+1+l+1-l+1η+2+l1I2-η.Example 8.
Second nonstandard rule:n=2, x→=-2,0,2, and τ=2. Also, a→=1/6,10/6,1/6 and n0=3. The quadrature formula is (50)∫-11fxdx≈16f-2+106f0+16f2. For l=0,1,2,3, we have (51)K2lη;2,1=1l+1!1-η+l+1-l+162-η+l1I2+η+-1l+1l+1!η+1+l+1-l+16η+2+l1I2-η.Example 9.
Third nonstandard rule:n=2, x→=(2,0,2), and τ=2. Also, a→=1/12,11/6,1/12 and n0=3. The quadrature formula is (52)∫-11fxdx≈112f-2+116f0+112f2.For l=0,1,2,3, we have (53)K2lη;2,1=1l+1!1-η+l+1-l+1122-η+l1I2+η+-1l+1l+1!η+1+l+1-l+112η+2+l1I2-η.Example 10.
Gauss-Radau rule:n=1, x→=-1/3,1, and τ=1. Also, a→=(3/2,1/2) and n0=2. The quadrature formula is (54)∫-11fxdx≈32f-13+12f1.For l=0,1,2, we have (55)K1lη;1,1=-1-ηll+1!η+l-121I+η+-1l+1l+1!η+1l+1-32l+1η+13+l1I-η.


## 4. Composite Rules

For an integral∫abf(X)dX, where f∈ACl+1,p([a,b]), a composite rule uses a partition of [a,b] in M subintervals and applies a formula on each subinterval. To simplify, we consider subintervals of equal length h=b-a/2M. To allow the possibility that τ>1, which cause an overlap between subintervals, we suppose also that f∈ACl+1,pa′,b′, where a′<a<b<b′ and a-τ-1h,b+τ-1h∈a′,b′, for h small enough or equivalently M large enough. Let us set ξm=a+2mh for m=0,1,2,…,M, and cm=ξm-1+ξm/2, for m=1,2,3,…,M. Then (56)∫abfXdX=∑m=1M∫ξm-1ξmfξdξ=∑m=1M∫-hhfcm+xdx=h∑m=1MQfcm+·;h. The composite rule CQn(f) is then defined by(57)CQnf=h∑m=1MQndfcm+·;h.The truncation error is (58)∫abfXdX-h∑m=1MQndfcm+·;h=h∑m=1MRnfcm+·;h. But (59)∑m=1MRnfcm+·;h≤∑m=1MRnfcm+·;h≤∑m=1Mhl+1-1/pCnl,pfl+1p,cm+IH≤hl+1-1/pCnl,p∑m=1Mfl+1p,cm+IH.To measure the overlap for 1≤p<∞, let us define Np(τ) by (60)Npτ=τfor1≤p<∞1forp=∞; then (61)
∑
m
=
1
M
f
l
+
1
p
,
c
m
+
I
H
≤
M
1
-
1
/
p
∑
m
=
1
M
f
l
+
1
p
,
c
m
+
I
H
p
1
/
p
1
≤
p
<
∞
M
f
l
+
1
∞
a
′
,
b
′
,
p
=
∞
≤
M
1
-
1
/
p
N
p
τ
f
l
+
1
p
,
a
′
,
b
′
, and we obtain(62)∑m=1MRnfcm+·;h≤hlb-a21-1/pCnl,pNpτfl+1p,a′,b′.


## 5. Symmetric Interpolatory Rules

In case of symmetry with respect tox=0, more precisely when (63)xi=-xn-i,fori=0,1,2,…,n, we have (64)ai=an-i,fori=0,1,2,…,n. As a consequence, we have Knl(-η;τ,1)=(-1)l+1Knl(η;τ,1), for l=0,…,n0. Moreover, Rn(f;h)=0 for any monomials of odd degree; then n0 is odd and K0,nn0(η;τ,1) is an even function. Moreover, for f(x)=xj, since f(l+1)(x)=(j)l+1xj-(l+1), for two nonnegative integers j and l, it follows that (65)Rnxj;h=0forj≤n0,0forj>n0andjodd,2hjjn0+1∫01ηj-n0+1K0,nn0η;τ,1dηforj>n0andjeven.For some polynomials, we can evaluate exactly the truncation error, as mentioned in [10] for Simpson’s rule. Indeed, let (66)fx=pn0+2x=pn0x+αn0+1xn0+1+αn0+2xn0+2, where pn0(x) is a polynomial of degree ≤n0. Then we have (67)pn0+2n0+1x=n0+1!αn0+1+n0+2!αn0+2x. It follows, for the composite rule, that (68)Rncm+·n0+1;h=n0+1!∫-HHK0,nn0y;H,hdy=n0+1!hn0+1∫-ττK0,nn0η;τ,1dη=2n0+1!hn0+1∫0τK0,nn0η;τ,1dη,Rncm+·n0+2;h=n0+2!∫-ττcm+yKnn0y;H,hdy=n0+2!cm∫-ττKnn0y;H,hdy+∫-ττyKnn0y;H,hdy=2n0+2!cmhn0+1∫0τKnn0η;τ,1dη. So we have (69)Rnpn0+2cm+·;h=αn0+1Rncm+·n0+1;h+αn0+2Rncm+·n0+2;h=2n0+1!hn0+1αn0+1+n0+2αn0+2cm∫0τKnn0η;τ,1dη. Moreover we obtain (70)h∑m=1MR0,npn0+2cm+·;h=2n0+1!hn0+2αn0+1M+n0+2αn0+2∑m=1Mcm∫0τKnn0η;τ,1dη=2n0+1!hn0+1b-a2αn0+1+n0+2αn0+2a+b2∫0τKnn0η;τ,1dη,because(71)∑m=1Mcm=a+b2M.Then we have a general explanation of the result mentioned in [10] for Simpson’s rule applied to quartic and quintic polynomials.Example 11.
Midpoint rule:n=0, τ=1, and n0=1; (72)∫01K01η;1,1dη=16, and for p3(x) we have (73)∫abp3ξdξ-CQ0p3=112α2+3α3a+b2b-a3M2.Example 12.
Trapezoidal rule:n=1, τ=1, and n0=1; (74)∫01K11η;1,1dη=-13,and for p3(x) we have (75)∫abp3ξdξ-CQ1p3=-16α2+3α3a+b2b-a3M2.Example 13 (see [10]).
First Simpson(1/3,4/3,1/3) rule: n=2, τ=1, and n0=3; (76)∫01K23η;1,1dη=-1180, and for p5(x) we have (77)∫abp5ξdξ-CQ2p5=-1120α4+5α5a+b2b-a5M4.Example 14.
Second Simpson(1/4,3/4,3/4,1/4) rule: n=3, τ=1, and n0=3; (78)∫01K0,33η;1,1dη=-1405, and for p5(x) we have (79)∫abp5ξdξ-CQ3p5=-1270α4+5α5a+b2b-a5M4.Example 15.
First nonstandard rule:n=1, τ=2, and n0=1; (80)∫02K11η;2,1dη=-176, and for p3(x) we have (81)∫abp3ξdξ-CQ1p3=-172α2+3α3a+b2b-a3M2.Similar results for monomial of degreek≥n0+2 require the evaluation of ∑m=1Mcml for l≥2, which is not simple.



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*Source: 3170595-2016-11-16.xml*