Directory UMM :Data Elmu:jurnal:J-a:Journal of Computational And Applied Mathematics:Vol104.Issue1.1999:
Journal of Computational and Applied Mathematics 104 (1999) 19–25
A new proof of the Euler–Maclaurin expansion for quadrature
over implicitly de ned curves
Dr. Muhammed I. Syam ∗
Department of Mathematics & Computer Science, Faculty of Science, P.O. Box 17557, United Arab Emirates
University, Al-Ain, UAE
Received 23 March 1998; received in revised form 22 July 1998
Abstract
In this paper we describe and justify a method for integrating over implicitly de ned curves. This method does not
require that the Jacobian be known explicitly. We give a proof of an asymptotic error expansion for this method which
c 1999 Elsevier Science B.V. All rights reserved.
is a modi cation of that of Lyness [4].
AMS classi cation: 65
Keywords: Continuation methods; Implicitly de ned curve; Modi ed trapezoidal rule
1. Introduction
The predictor–corrector methods for numerically tracing implicitly de ned curves have been developed and investigated in a number of papers and books. For recent surveys, see., e.g. [1, 2]. Siyyam
and Syam [5], describe the idea of the predictor–corrector method for tracing an implicitly de ned
closed curves and they presented a device for reliable stopping. They used the Euler–Predictor and
the Gauss–Newton Corrector. They presented the modi ed trapezoidal rule for approximating the
line integrals over implicitly de ned closed curve. This rule was developed by Georg [3] to approximate surface integrals for implicitly de ned surfaces. Siyyam and Syam [5] proved that the
modi ed trapezoidal rule has an asymptotic error expansion in even powers of 1=m over a scalar
and vector eld. But their proof was very long and complicated. In this paper, we present another
proof which is shorter and easier than their proof. Moreover, our new proof depends on the idea of
the Euler–Maclaurin summation.
∗
E-mail: [email protected].
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 9 8 - 8
20
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
We shall mean that a map is smooth if it has continuous derivatives as the discussion requires. For
a smooth map f : Rn → Rn and a smooth curve C with a parametrization,
say = (’1 ; ’2 ; : : : ; ’n ) :
R
S → Rn ; S = [0; 1] and C = {(u) | u ∈ [0; 1]}, the line integral C f dC can be written as an ordinary
integral, i.e.;
Z
f dC =
Z
1
f((u)) : ′ (u) du:
0
C
For the latter integral the trapezoidal rule takes the form
Z
1
f((u)) : ′ (u) du =
0
X
1 m−1
[f((ui )) : ′ (ui ) + f((ui+1 )) : ′ (ui+1 )][ui+1 − ui ];
2 i=0
over the partition points ui = i=m in [0; 1]. Let Xi = (ui ) for i = 0 : m, the modi ed trapezoidal rule
takes the form
Z
f dC =
C
X
1 m−1
[f(Xi ) + f(Xi+1 )] · [Xi+1 − Xi ];
2 i=0
(1.1)
where X0 and Xm are the endpoints of C
Theorem 1.1. The modi ed trapezoidal rule (1:1) admits an asymptotic error expansion in even
powers of 1=m.
In this paper, we present a new proof for Theorem 1.1.
2. Asymptotic error expansion for the modi ed trapezoidal rule
q
In this paper we use the following norm k( 1 ; 2 ; : : : ; n )k = 12 + 22 + · · · + n2 . First, we want
to prove Theorem 1.1 for a scalar eld f : Rn → R. Let uj = j=m and Xj = (uj ) for j = 0 : m. De ne
the quadrature
Q(m) f =
X
1 m−1
[f(Xi ) + f(Xi+1 )]kXi+1 − Xi k:
2 j=0
(2.1)
The main result in this paper is that Q(m) f has an h2 -expansion. That is, setting h = 1=m, when
f ∈ C (2p) (C) and (u) is suciently well behaved, we have
Q(m) f =
Z
f dc + B2 h2 + B4 h4 + · · · + B2p h2p +
(h2p+1 );
C
where the coecients B2q are independent of h for q = 1 : p.
(2.2)
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
However, the line integral
Z
f dc =
Z
R
C
f dc can be written as
f((u))J (u) du =
Z
1
g(u) du;
(2.3)
0
S
C
21
where J 2 (u) = (’′1 (u))2 + (’′2 (u))2 + · · · + (’′n (u))2
and
g(u) = f((u))J (u):
(2.4)
In Theorem 2.1, we want to study the asymptotic expansion of kXi+1 − Xi k for any i = 0 : m − 1.
Theorem 2.1. Let (u) be C (p) (S); and let = kXi+1 − Xi k. Then;
= h(0 + h1 + · · · + hp−1 p−1 ) +
(hp+1 );
(2.5)
where 0 = J (uj ) and s = s; 1 (uj ) where s; 1 (u) is a function of u having continuous derivatives of
order p.
Proof. It is easy to see that 2 = kXi+1 −Xi k2 =
about uj to get
2
=
p−1
n
X
X
i=1
h
’i(s) (uj )
s
s!
s=1
+
h
’(p)
i (j )
p
p!
!2
Pn
i=1 (’i (uj+1 )−’i (uj ))
2
. Expand ’i (uj+1 ) = ’i (uj +h)
;
(2.6)
where j lies between uj and uj+1 . Simplify Eq. (2.6) to obtain
p−1
2 = h2 J 2 (uj ) +
X
!
cs (uj )hs +
(hp+2 ):
s=1
(2.7)
For suciently small h, we may take the square root of Eq. (2.7) to get
= h(0 + h1 + · · · + hp−1 p−1 ) +
(hp+1 );
where 0 = J (uj ). Since the derivatives of ’1 ; ’2 ; : : : ; ’n of order p are continuous functions of u, it
follows that so are the derivatives of cs of order p − 1 − s. Finally, so long as J (u) is bounded a
way from zeros in S, we nd that s also has this order of continuity.
We should note that the proof of Theorem 2.1 is for kXi+1 −Xi k. The proof is similar for kXi−1 −Xi k,
but
s; 1 (u) = (−1)s s; −1 (u):
(2.8)
Moreover, we should note that the functions s; (u) do not depend on m for = − 1 and 1. As for
0 (u) = J (u), it depends only on ’1 ; ’2 ; : : : ; ’n .
To clarify our discussion, we need the following de nition.
22
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
De nition 2.2. Let ∈ {−1; 1} and B be any interval subset of R. Then de ne (m)
j; (B) = 1 or 0
depending on whether or not the interval with endpoints uj and uj+ is a subset of B.
(m)
For ∈ {−1; 1}, de ne the quadratures R(m)
g and Q f as
R(m) g =
m
1X
(m)
(S)g(uj );
m j=0 j;
m
X
Q(m) f =
(2.9)
(m)
j; (S)f(Xj )j; ;
(2.10)
j=0
where j; = kXj+ − Xj k. Let h = 1=m. From Eq. (2.4), we can rewrite the quadrature (2.9) as
R(m)
g=
m
X
h(m)
j; (S)f(Xj )J (uj ):
(2.11)
j=0
The expression of Theorem 2.1 allows us to establish the following.
(m)
Theorem 2.3. Let ∈ C (p) (S) and let R(m)
g and Q f be given by (2:9) and (2:10); respectively.
Then;
p−1
Q(m) f =
X
p
hs R(m)
gs; +
(h );
(2.12)
s=0
where
gs; (u) = s; (u)f((u))
(2.13)
and s; (u)is as in Theorem 2:1.
Proof. Using formulas (2.5) and (2.10), we obtain
Q(m) f
=
m
X
(m)
j; (S)
j=0
p−1
=
X
s=0
p−1
=
X
hs
(
h
m
1 X
m
p−1
X
s
s; h +
(h
s=0
(m)
j; (S)s; f(Xj )
j=0
p
hs R(m)
gs; +
(h );
s=0
where gs; (u) = s; (u)f((u)).
p+1
)
) f(Xj );
+
(hp )
23
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
(m)
We should note that g0; (u) coincides with g(u). Also, R(m)
gs; depend on h via m, so Q f is
not an h-expansion. If g ∈ C (p) (S), the following Euler–Maclaurin summation formula is valid,
p−1
R(m)
g=
X
h j Bj (R ; g) +
(hp );
(2.14)
j=0
where
B0 (R; g) =
Z
1
g(u) du
(2.15)
0
and Bj (R ; g) is independent of m. Substitute formula (2.14) in (2.12) to establish the following:
p−1
Q(m) f =
X
hs
s=0
=
Z
1
s
X
h j Bj (R ; gs; ) +
(hs+1 )
j=0
g(u) du +
0
p−1
es (; f)
XB
s=0
ms
+
(hp )
+
(hp );
(2.16)
where
Bes ( ; f) =
s
X
h j Bj (R ; gs−j; )
(2.17)
j=0
if (u) and f((u)) are C (p) (S). In Theorem 2.4, we want to state a very useful property of the
coecients in the Euler–Maclaurin expansion.
Theorem 2.4. The coecients in the Euler–Maclaurin expansion (2:14) satisfy
Bj (R ; g) = (−1) j Bj (R− ; g):
(2.18)
Proof. One can establish this result by direct evaluation of the coecients in terms of Bernoulli
function. But our approach is based on the fact that a symmetric rule has an m2 -expansion. It is
(m)
easy to see that Rg = 21 (R(m)
g + R− g) is symmetric. Thus,
Bj (R− ; g) =
−Bj (R ; g) if j is odd
:
Bj (R ; g)
if j is even
Hence, we establish the relation (2.18).
From Eqs. (2.8) and (2.13), we see that
gs; (u) = (−1)s gs; − (u):
(2.19)
De ne the quadrature
(m)
Q(m) f = 21 Q1(m) f + 21 Q−1
f:
(2.20)
24
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
Thus,
Q(m) f =
m
X
1 m−1
1X
f(Xj ) kXj+1 − Xj k +
f(Xj ) kXj − Xj−1 k
2 j=0
2 j=1
X
1 m−1
=
f(Xj ) + f(Xj+1) kXj+1 − Xj k :
2 j=0
(2.21)
Hence, quadrature (2.21) is the modi ed trapezoidal rule. In Theorem 2.5, we establish the asymptotic error expansion of quadrature (2.21).
Theorem 2.5. Let and f() be C (p) (S). The quadrature (2:21) has an h2 -error expansion.
Proof. From Eqs. (2.17), (2.18) and (2.19) we obtain
Bes (1; f) =
=
s
X
h j Bj (R1 ; gs−j;1 )
j=0
s
X
j=0
(−1) j (−1)s−j h j Bj (R−1 ; gs−j;−1 ) = (−1)s Bes (−1; f):
(2.22)
Thus, the coecients of h s in the expansion is (Bes (1; f) + Bes (−1; f))=2. From Eq. (2.18), this
coecient vanishes when s is odd.
Theorem 2.5 gives us the asymptotic error expansion for the modi ed trapezoidal rule for a scalar
eld. To prove Theorem 1.1, let f = (f1 ; f2 ; : : : ; fn ) : Rn → Rn be a given function and let C1 be a
subcurve of C with endpoint X = (x1 ; x2 ; : : : ; xn ) and Y = (y1 ; y2 ; : : : ; yn ). Then the line integral of f
over C1 is de ned by
Z
f dC1 =
C1
Z
0
1
′
f((u)) · (u) du =
n Z
X
i=1
0
1
fi ((u))i′ (u) du;
(2.23)
while the modi ed trapezoidal rule is given by
n
1X
1
[f(X ) + f(Y )] · [X − Y ] =
[fi (X ) + fi (Y )][xi − yi ]
2
2 i=1
(2.24)
which follows from the approximation that
Z
0
1
1
fi ((u))i′ (u) du ≈ [fi (X ) + fi (Y )][xi − yi ]
2
(2.25)
for all i = 1 : n. The principle idea of the proof is to apply the techniques we used to prove that
Eq. (2.25) has an asymptotic error expansion of even power of 1=m for all j = 1 : n. Since the nite
addition preserves the asymptotic expansion, we conclude that our modi ed trapezoidal rule for
approximating the line integral of a vector eld still has an asymptotic expansion of even power
of 1=m.
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
25
References
[1] E.L. Allgower, K. Georg, Numerical Continuation Methods, Springer, New York, 1990.
[2] E.L. Allgower, K. Georg, Continuation and path following, Acta Numerica, Cambridge Univ. press, Cambridge, 1993,
pp. 1– 64.
[3] K. Georg, Approximation of integrals for boundary element methods, SIAM J. Sci. Statist. Comput. 12 (1991) 443 –
453.
[4] J.N. Lyness, Quadrature over curved surfaces by Extrapolation, SIAM J. Numer. Anal. 15 (1993).
[5] H.I. Siyyam, M.I. Syam, The modi ed trapezoidal rule for line integrals, J. Comput. Appl. Math. 84 (1997) 1–14.
A new proof of the Euler–Maclaurin expansion for quadrature
over implicitly de ned curves
Dr. Muhammed I. Syam ∗
Department of Mathematics & Computer Science, Faculty of Science, P.O. Box 17557, United Arab Emirates
University, Al-Ain, UAE
Received 23 March 1998; received in revised form 22 July 1998
Abstract
In this paper we describe and justify a method for integrating over implicitly de ned curves. This method does not
require that the Jacobian be known explicitly. We give a proof of an asymptotic error expansion for this method which
c 1999 Elsevier Science B.V. All rights reserved.
is a modi cation of that of Lyness [4].
AMS classi cation: 65
Keywords: Continuation methods; Implicitly de ned curve; Modi ed trapezoidal rule
1. Introduction
The predictor–corrector methods for numerically tracing implicitly de ned curves have been developed and investigated in a number of papers and books. For recent surveys, see., e.g. [1, 2]. Siyyam
and Syam [5], describe the idea of the predictor–corrector method for tracing an implicitly de ned
closed curves and they presented a device for reliable stopping. They used the Euler–Predictor and
the Gauss–Newton Corrector. They presented the modi ed trapezoidal rule for approximating the
line integrals over implicitly de ned closed curve. This rule was developed by Georg [3] to approximate surface integrals for implicitly de ned surfaces. Siyyam and Syam [5] proved that the
modi ed trapezoidal rule has an asymptotic error expansion in even powers of 1=m over a scalar
and vector eld. But their proof was very long and complicated. In this paper, we present another
proof which is shorter and easier than their proof. Moreover, our new proof depends on the idea of
the Euler–Maclaurin summation.
∗
E-mail: [email protected].
c 1999 Elsevier Science B.V. All rights reserved.
0377-0427/99/$ – see front matter
PII: S 0 3 7 7 - 0 4 2 7 ( 9 8 ) 0 0 1 9 8 - 8
20
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
We shall mean that a map is smooth if it has continuous derivatives as the discussion requires. For
a smooth map f : Rn → Rn and a smooth curve C with a parametrization,
say = (’1 ; ’2 ; : : : ; ’n ) :
R
S → Rn ; S = [0; 1] and C = {(u) | u ∈ [0; 1]}, the line integral C f dC can be written as an ordinary
integral, i.e.;
Z
f dC =
Z
1
f((u)) : ′ (u) du:
0
C
For the latter integral the trapezoidal rule takes the form
Z
1
f((u)) : ′ (u) du =
0
X
1 m−1
[f((ui )) : ′ (ui ) + f((ui+1 )) : ′ (ui+1 )][ui+1 − ui ];
2 i=0
over the partition points ui = i=m in [0; 1]. Let Xi = (ui ) for i = 0 : m, the modi ed trapezoidal rule
takes the form
Z
f dC =
C
X
1 m−1
[f(Xi ) + f(Xi+1 )] · [Xi+1 − Xi ];
2 i=0
(1.1)
where X0 and Xm are the endpoints of C
Theorem 1.1. The modi ed trapezoidal rule (1:1) admits an asymptotic error expansion in even
powers of 1=m.
In this paper, we present a new proof for Theorem 1.1.
2. Asymptotic error expansion for the modi ed trapezoidal rule
q
In this paper we use the following norm k( 1 ; 2 ; : : : ; n )k = 12 + 22 + · · · + n2 . First, we want
to prove Theorem 1.1 for a scalar eld f : Rn → R. Let uj = j=m and Xj = (uj ) for j = 0 : m. De ne
the quadrature
Q(m) f =
X
1 m−1
[f(Xi ) + f(Xi+1 )]kXi+1 − Xi k:
2 j=0
(2.1)
The main result in this paper is that Q(m) f has an h2 -expansion. That is, setting h = 1=m, when
f ∈ C (2p) (C) and (u) is suciently well behaved, we have
Q(m) f =
Z
f dc + B2 h2 + B4 h4 + · · · + B2p h2p +
(h2p+1 );
C
where the coecients B2q are independent of h for q = 1 : p.
(2.2)
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
However, the line integral
Z
f dc =
Z
R
C
f dc can be written as
f((u))J (u) du =
Z
1
g(u) du;
(2.3)
0
S
C
21
where J 2 (u) = (’′1 (u))2 + (’′2 (u))2 + · · · + (’′n (u))2
and
g(u) = f((u))J (u):
(2.4)
In Theorem 2.1, we want to study the asymptotic expansion of kXi+1 − Xi k for any i = 0 : m − 1.
Theorem 2.1. Let (u) be C (p) (S); and let = kXi+1 − Xi k. Then;
= h(0 + h1 + · · · + hp−1 p−1 ) +
(hp+1 );
(2.5)
where 0 = J (uj ) and s = s; 1 (uj ) where s; 1 (u) is a function of u having continuous derivatives of
order p.
Proof. It is easy to see that 2 = kXi+1 −Xi k2 =
about uj to get
2
=
p−1
n
X
X
i=1
h
’i(s) (uj )
s
s!
s=1
+
h
’(p)
i (j )
p
p!
!2
Pn
i=1 (’i (uj+1 )−’i (uj ))
2
. Expand ’i (uj+1 ) = ’i (uj +h)
;
(2.6)
where j lies between uj and uj+1 . Simplify Eq. (2.6) to obtain
p−1
2 = h2 J 2 (uj ) +
X
!
cs (uj )hs +
(hp+2 ):
s=1
(2.7)
For suciently small h, we may take the square root of Eq. (2.7) to get
= h(0 + h1 + · · · + hp−1 p−1 ) +
(hp+1 );
where 0 = J (uj ). Since the derivatives of ’1 ; ’2 ; : : : ; ’n of order p are continuous functions of u, it
follows that so are the derivatives of cs of order p − 1 − s. Finally, so long as J (u) is bounded a
way from zeros in S, we nd that s also has this order of continuity.
We should note that the proof of Theorem 2.1 is for kXi+1 −Xi k. The proof is similar for kXi−1 −Xi k,
but
s; 1 (u) = (−1)s s; −1 (u):
(2.8)
Moreover, we should note that the functions s; (u) do not depend on m for = − 1 and 1. As for
0 (u) = J (u), it depends only on ’1 ; ’2 ; : : : ; ’n .
To clarify our discussion, we need the following de nition.
22
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
De nition 2.2. Let ∈ {−1; 1} and B be any interval subset of R. Then de ne (m)
j; (B) = 1 or 0
depending on whether or not the interval with endpoints uj and uj+ is a subset of B.
(m)
For ∈ {−1; 1}, de ne the quadratures R(m)
g and Q f as
R(m) g =
m
1X
(m)
(S)g(uj );
m j=0 j;
m
X
Q(m) f =
(2.9)
(m)
j; (S)f(Xj )j; ;
(2.10)
j=0
where j; = kXj+ − Xj k. Let h = 1=m. From Eq. (2.4), we can rewrite the quadrature (2.9) as
R(m)
g=
m
X
h(m)
j; (S)f(Xj )J (uj ):
(2.11)
j=0
The expression of Theorem 2.1 allows us to establish the following.
(m)
Theorem 2.3. Let ∈ C (p) (S) and let R(m)
g and Q f be given by (2:9) and (2:10); respectively.
Then;
p−1
Q(m) f =
X
p
hs R(m)
gs; +
(h );
(2.12)
s=0
where
gs; (u) = s; (u)f((u))
(2.13)
and s; (u)is as in Theorem 2:1.
Proof. Using formulas (2.5) and (2.10), we obtain
Q(m) f
=
m
X
(m)
j; (S)
j=0
p−1
=
X
s=0
p−1
=
X
hs
(
h
m
1 X
m
p−1
X
s
s; h +
(h
s=0
(m)
j; (S)s; f(Xj )
j=0
p
hs R(m)
gs; +
(h );
s=0
where gs; (u) = s; (u)f((u)).
p+1
)
) f(Xj );
+
(hp )
23
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
(m)
We should note that g0; (u) coincides with g(u). Also, R(m)
gs; depend on h via m, so Q f is
not an h-expansion. If g ∈ C (p) (S), the following Euler–Maclaurin summation formula is valid,
p−1
R(m)
g=
X
h j Bj (R ; g) +
(hp );
(2.14)
j=0
where
B0 (R; g) =
Z
1
g(u) du
(2.15)
0
and Bj (R ; g) is independent of m. Substitute formula (2.14) in (2.12) to establish the following:
p−1
Q(m) f =
X
hs
s=0
=
Z
1
s
X
h j Bj (R ; gs; ) +
(hs+1 )
j=0
g(u) du +
0
p−1
es (; f)
XB
s=0
ms
+
(hp )
+
(hp );
(2.16)
where
Bes ( ; f) =
s
X
h j Bj (R ; gs−j; )
(2.17)
j=0
if (u) and f((u)) are C (p) (S). In Theorem 2.4, we want to state a very useful property of the
coecients in the Euler–Maclaurin expansion.
Theorem 2.4. The coecients in the Euler–Maclaurin expansion (2:14) satisfy
Bj (R ; g) = (−1) j Bj (R− ; g):
(2.18)
Proof. One can establish this result by direct evaluation of the coecients in terms of Bernoulli
function. But our approach is based on the fact that a symmetric rule has an m2 -expansion. It is
(m)
easy to see that Rg = 21 (R(m)
g + R− g) is symmetric. Thus,
Bj (R− ; g) =
−Bj (R ; g) if j is odd
:
Bj (R ; g)
if j is even
Hence, we establish the relation (2.18).
From Eqs. (2.8) and (2.13), we see that
gs; (u) = (−1)s gs; − (u):
(2.19)
De ne the quadrature
(m)
Q(m) f = 21 Q1(m) f + 21 Q−1
f:
(2.20)
24
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
Thus,
Q(m) f =
m
X
1 m−1
1X
f(Xj ) kXj+1 − Xj k +
f(Xj ) kXj − Xj−1 k
2 j=0
2 j=1
X
1 m−1
=
f(Xj ) + f(Xj+1) kXj+1 − Xj k :
2 j=0
(2.21)
Hence, quadrature (2.21) is the modi ed trapezoidal rule. In Theorem 2.5, we establish the asymptotic error expansion of quadrature (2.21).
Theorem 2.5. Let and f() be C (p) (S). The quadrature (2:21) has an h2 -error expansion.
Proof. From Eqs. (2.17), (2.18) and (2.19) we obtain
Bes (1; f) =
=
s
X
h j Bj (R1 ; gs−j;1 )
j=0
s
X
j=0
(−1) j (−1)s−j h j Bj (R−1 ; gs−j;−1 ) = (−1)s Bes (−1; f):
(2.22)
Thus, the coecients of h s in the expansion is (Bes (1; f) + Bes (−1; f))=2. From Eq. (2.18), this
coecient vanishes when s is odd.
Theorem 2.5 gives us the asymptotic error expansion for the modi ed trapezoidal rule for a scalar
eld. To prove Theorem 1.1, let f = (f1 ; f2 ; : : : ; fn ) : Rn → Rn be a given function and let C1 be a
subcurve of C with endpoint X = (x1 ; x2 ; : : : ; xn ) and Y = (y1 ; y2 ; : : : ; yn ). Then the line integral of f
over C1 is de ned by
Z
f dC1 =
C1
Z
0
1
′
f((u)) · (u) du =
n Z
X
i=1
0
1
fi ((u))i′ (u) du;
(2.23)
while the modi ed trapezoidal rule is given by
n
1X
1
[f(X ) + f(Y )] · [X − Y ] =
[fi (X ) + fi (Y )][xi − yi ]
2
2 i=1
(2.24)
which follows from the approximation that
Z
0
1
1
fi ((u))i′ (u) du ≈ [fi (X ) + fi (Y )][xi − yi ]
2
(2.25)
for all i = 1 : n. The principle idea of the proof is to apply the techniques we used to prove that
Eq. (2.25) has an asymptotic error expansion of even power of 1=m for all j = 1 : n. Since the nite
addition preserves the asymptotic expansion, we conclude that our modi ed trapezoidal rule for
approximating the line integral of a vector eld still has an asymptotic expansion of even power
of 1=m.
M.I. Syam / Journal of Computational and Applied Mathematics 104 (1999) 19–25
25
References
[1] E.L. Allgower, K. Georg, Numerical Continuation Methods, Springer, New York, 1990.
[2] E.L. Allgower, K. Georg, Continuation and path following, Acta Numerica, Cambridge Univ. press, Cambridge, 1993,
pp. 1– 64.
[3] K. Georg, Approximation of integrals for boundary element methods, SIAM J. Sci. Statist. Comput. 12 (1991) 443 –
453.
[4] J.N. Lyness, Quadrature over curved surfaces by Extrapolation, SIAM J. Numer. Anal. 15 (1993).
[5] H.I. Siyyam, M.I. Syam, The modi ed trapezoidal rule for line integrals, J. Comput. Appl. Math. 84 (1997) 1–14.