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Information Technology papers
Bond Business School
May 2004
View factor for inclined plane with Gaussian source
Stephen J. Sugden
Bond University, [email protected]
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Stephen J. Sugden. (2004) "View factor for inclined plane with Gaussian source" ,, .
http://epublications.bond.edu.au/infotech_pubs/18
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View Factor for Inclined Plane
with Gaussian Source
Stephen J Sugden
School of Information Technology, Bond University
February 24, 2004
Abstract
The view factor (angle factor) for a di¤erential inclined plane in the
case of a radiating source of radially Gaussian intensity is considered. This
information is useful for modelling of solar radiation in certain applications. The view factor is expressed in terms of two integrals, one of which
is obtained in closed form in terms of special functions, and the other is
approximated. A compact estimate for the view factor is presented which
is suitable for machine computation. While the relative error associated
with the …nal estimate is typically less than 0.01%, and in all cases, less
than 0.2%, the method is easily extended to yield even greater accuracy.
Keywords: solar radiation, view factor, Gaussian source.
1
Introduction
View factors for inclined planes for the case of a uniformly radiating source
are well-known and can be found in standard texts on radiative transfer, for
example [1] or [2]. The purpose of the present work is to derive an accurate
estimate of the view factor for an in…nite plane source whose intensity falls o¤
radially in accordance with a Gaussian law and a small di¤erential receiving
plane surface inclined at some acute angle to the source plane. The …nal result
is intended to be of use for work on modelling the intensity distribution of the
apparent solar disc by the Gaussian function; the receiving plane being the
Earth’s surface. The Gaussian source assumption follows a suggestion of Peck
[4] in his work on parabolic solar collector design. The paper is organized as
follows. We …rst consider the geometry of the problem and obtain an exact
integral expression for the view factor, as a function of the angle of inclination,
, of the di¤erential receiving plane. Following this, some standard estimation
techniques are employed to derive a useful approximation to the view factor.
Finally, this estimate is used as the basis of a computational model in Microsoft
Excel, which computes, tabulates, and graphs the view factor as a function of
.
1
2
Methods of computing view factors
According to Modest1 [1], view factor evaluation techniques may be classi…ed
broadly as follows.
1. Direct integration.
2. Statistical: sampling with the well-known Monte-Carlo class of methods.
3. Special methods.
(a) Algebraic: application of symmetry, reciprocity, and other properties.
(b) Crossed-string method: applied to long enclosures with constant
cross-section.
(c) Unit sphere method: used between one in…nitesimal and one …nite
area, such as we have here2 .
(d) Inside sphere method3 .
Direct integration may be accomplished by any of a number of e¢ cient methods, analytical or numerical, with the common Simpson’s Rule being very accurate for most shapes, even with modest mesh size. Methods in the Monte-Carlo
class are p
typically poorer cousins to the other, preferred methods, and errors are
of order h; where h is the mesh size. Such techniques should be chosen only
when others are, for whatever reason, very awkward to apply. Industry Standard scienti…c software such as FACET [3] employs a combination of several of
these methods to achieve satisfactory results, in general.
The present method falls into category 1, with the integrals here being evaluated partially analytically, and partially estimated. There is no particular
numerical technique used; rather the whole exercise here is primarily one of
estimation of semi-tractable integrals in terms of known special functions, and
their …nal estimation using asymptotic techniques. Thus, it makes little sense to
compare the present work with some of the more advanced numerical techniques
for view factor estimation.
1 This work appears to be both the most modern and the most complete text dealing
with radiative heat transfer, devoting an entire chapter to a discussion of view factors and a
comparison of computational methods.
2 This method is of little value here, given the unusual intensity distribution of the source.
3 Similar comments as in footnote 2 apply.
2
3
Net incident intensity for arbitrary source distribution function
With reference to Figure 1, the vector ' is the position vector to an arbitrary
point P in the plane A1 : With respect to the right-handed orthonormal system
fi; j; kg emanating from O2 , we have
' =xi+yj+zk
(1)
The vector '0 is normal to the plane A1 :
'0 = '0 j
(2)
Figure 1: Geometry
Since P lies in A1 , we have y = '0 , so eq (1) becomes:
' =xi + '0 j+zk
(3)
The angles 1 and 2 are respectively the angles of emission from P and
incidence at O2 of radiation from the source plane A1 . The angle
is the
(acute) angle between the normals, respectively n1 and n2 , to the planes A1
and dA2 .
From …gure 1, we have
n2 = (sin ) i+ (cos ) j
3
(4)
so that
cos
2
= nT2 ' =
x sin
+ '0 cos
'
(5)
Also,
'0
'
Now the intensity of illumination, I2 , on dA2 , is given by
Z Z
I(P ) cos 1 cos 2
dA1
I2 =
0
'2
A1 A 1
cos
1
=
(6)
(7)
where I(P ) is the intensity of emission at P on A1 and A01 = fP 2 A1 : cos 2 > 0g,
i.e., A01 is the set of all points on A1 giving rise to an acute angle of incidence
at O2 .
Using eqs (5), (6) in (7), we …nd that:
Z Z
I(P ) (x sin + '0 cos )
I2 =
dA1
(8)
0
'4
A1
Fubini’s theorem permits the conversion of this double integral to the following iterated integral:
Z 1Z 1
I(P ) (x sin + '0 cos )
I2 = '0
dzdx
(9)
'4
x0
1
0
where A1 = [x0 ; 1) <. It is a simple matter to determine x0 from problem
constants. From eq (5) we deduce that
x0 =
'0 cot
(10)
Hence, using this result and (3), eq (9) becomes:
Z 1
Z 1
I(P ) (x sin + '0 cos )
I2 = '0
dzdx
2
(x2 + '20 + z 2 )
'0 cot
1
4
(11)
Net incident intensity for Gaussian source
In this section we take
I(P ) =
2
2
I
p0 e (x +z )=2
2
2
Substituting for I(P ) in equation (11), using the transformations
and = z=', and writing = '20 =2 2 , we obtain:
p
Z 1
Z 1
2
2
2 I2
'0 cos
e =2 d d
=2
=
e
2
2 + 2 )2
'0 I0
('0 cot )=
1 (2 +
Z 1
Z 1
2
2
sin
e =2 d d
+
e =2
2 + 2 )2
('0 cot )=
1 (2 +
4
(12)
= x=
(13)
Eq (13) may be rewritten as follows:
r
Z 1
p
p
2
I2
= 2 cos
e =2 J
2 + 2 d
p
2
I0
2 cot
Z 1
p
2
+ sin
e =2 J
2 + 2 d
p
(14)
2 cot
where,
J(t) =
Z
1
2
e
(t2
0
=2
d
(15)
2 )2
+
and the evenness of the integrand has been exploited. The function J, de…ned by the integral (15) may be expressed in terms of the function K, where
Z
K(x) =
1
xz 2
dz
(16)
2
(1 + z 2 )
0
The substitution
e
= tz in (15) yields
J(t) =
Z
1
t4
0
tz 2 =2
e
tdz
2
(1 + z 2 )
so that
1
t2
K( )
3
t
2
It is possible to express K in terms of common functions [5]:
p
p
x
x
K(x) = (1 2x)e 1 erf x +
4
2
J(t) =
(17)
(18)
Combining equations (17) and (18) we …nd that
1
J(t) = 3
t
4
Writing
S( ; ) =
(1
Z
and
R( ; ) =
2
Z
t2 =2
t )e
1
p
e
t
erf p
2
1
2
J
p
J
p
2 +
=2
2 cot
1
p
e
2 cot
2
=2
+
2 +
r
t
22
(19)
2
d
(20)
2
d
(21)
we may express eq (14) more concisely as:
r
p
I2
= 2 cos R( ; ) + sin S( ; )
2
Io
5
(22)
Turning our attention now to the integral S( ; ), as de…ned by (20), it will
be found that substitution for J from eq (19) in the integrand of eq (20) yields
the following expression for S( ; ):
e
S( ; ) =
4
+
p
p
2
p
Z1
2 cot
Z1
2 cot
1
2
2
1
(2 +
e
2
erf
p
+
2 )3=2
2 =2
d
(23)
=2
p
2 +
2
d
Appropriate p
changes of variable reduce eq (23) to a somewhat
p more tractable
= sin and in the …rst integral, take u =
+ 2 =2, while
form. Put =
2
in the second, u = + =2, yielding:
p
2
1
e p
+
(1 erf ) e
(24)
S( ; ) = p
2
2 2
p
For the remaining integral in eq (22), R( ; ), the substitution =
2 cot
reduces eq (21) to the following angular integral:
Z
p
2
R( ; ) =
e cot J( 2 = sin )d
and using eq (17), we obtain:
Z
1
R( ; ) =
sin e
2
cot2
K( = sin2 )d
(25)
Eq (18) may now be used to write the integrand of (25) in terms of elementary
functions:
Z
p
e
4 R( ; ) =
(sin
2 = sin ) erf c
= sin d
(26)
2
Z
p
2
+
e cot d
In order to establish an asymptotic representation for R( ; );valid as !
+1 through real values, and with as a parameter, the following well-known
asymptotic expansion for the complementary error function will be employed;
see, for example, Abramowitz & Stegun, [6, sec 7.1.23, 7.1.24].
2
1
e x
erf x = p
1
x
(x ! +1)
6
1
+ r(x)
2x2
(27)
where
jr(x)j <
3
4x4
Using eq (27) in eq (26), we obtain:
r
Z
2
1
4
R( ; ) =
e cot sin2
2
Z
2
1
+
e cot
sin2
2
(28)
2
1
sin2
d
2
p
2 r( = sin )d
(29)
P( ; ) + r ( ; )
In eq (29), the …rst integral, P ( ; ), is our estimate and the second, r ( ; ),
is the remainder. P ( ; ) may be split into two integrals, each of which may be
handled by the method of Laplace [7, p81] to yield asymptotic representations.
We have:
Z
Z
1
1
P( ; ) =
q1 ( )e p( ) d
q2 ( )e p( ) d
4 2
in which q1 ( ) = sin2 ; q2 ( ) = sin4 ; p( ) = cot2 :
The minimum of p( ) occurs at = =2, an interior point of the interval of
integration. Accordingly, expand the functions p( ); q1 ( ); q2 ( ) about = =2
and retain only dominant terms, to yield:
r
r
1
1
3=2
P( ; ) =
+ o(
)
+ o( 5=2 )
4 2
r
1
=
+ o( 3=2 )
( ! +1)
(30)
As for the remainder, r ( ; ), we have:
Z
p
2
sin2
jr ( ; )j =
e cot
1 r( = sin )d
2
Z
p
2
1
<
e cot r( = sin ) d
for >
2
Z
2
3
<
sin4 e cot d
using eq (28)
4 2
Z
2
3
<
sin4 e cot d
2
4
0
The substitution u =
cot leads to:
Z 1
3
jr ( ; )j <
1 + u2
2 2 0
7
3
e
u2
du
(31)
A crude estimate for this last integral is adequate for the present purpose.
r
Z 1
3
3
u2
jr ( ; )j <
e
du
=
2 2 0
2 2
We conclude that
jr ( ; )j = o(
2
)
(32)
Combining (29), (30) and (32), we have:
R( ; ) =
5
2
4
2
+ o(
)
(33)
An estimate for R( ; ) with error bounds
The asymptotic estimate of eq (33) for R( ; ) may be useful for some applications, but does not provide error bounds suitable for the construction of a
reasonably accurate tabulation of R( ; ) versus , with …xed. Since the primary aim of the present work is to derive computationally useful approximations
to the net view factor, we now consider alternative methods. One possibility is
integration by parts, which certainly gives an explicit expression for the error
term; another is the use of Watson’s Lemma [7, p71], which yields a useful approximation plus an acceptable bound for the error. Here, we choose integration
by parts, because of its simplicity and ability to obtain approximations for a
large range of the parameter of interest, with quite a small relative error.
Proceeding with the analysis, we have from eq (29):
r
R( ; ) = Q( ; ) + r+ ( ; )
4
where
Q( ; ) =
1
Z
cot2
e
(34)
sin2 d
(35)
and
+
r ( ; )=
+
Z
1
2
4
Z
Now from (36), we obtain:
Z
r+ ( ; ) <
e
+
1
4
2
sin2
2
cot2
e
Z
cot2
e
cot2
e
p
1 r( = sin )d
sin4 d
(36)
p
r( = sin ) d
cot2
8
sin4 d ;
if
> 1=2
Using (28), we have, for > 1=2:
Z
Z
3 sin4
1
cot2
r+ ( ; ) <
e
d
+
4 2
4 2
Z
2
1
= 2
e cot sin4 d
<
1
2
Z
e
cot2
cot2
e
sin4 d
sin2 d
Comparing this last inequality with eq (35), we have:
r+ ( ; ) <
1
Q( ; )
(37)
We have, therefore,
r
R( ; ) = Q( ; ) (1 + ( ; ))
4
where
( ; )=
(38)
jr+ ( ; )j
Q( ; )
and
0< ( ; )<
(39)
1
(40)
It is convenient to make the substitution u = cot in eq (35), so that
Z 1
2
Q( ; ) =
e u (1 + u2 ) 2 du
cot
or,
Q( ; ) =
Z
cot
e
u2
(1 + u2 )
2
du +
0
Z
1
e
u2
(1 + u2 )
2
du
(41)
0
Using notation de…ned by eqs (16) and (74), we may write eq (41) as:
Q( ; ) = Y ( ; cot ) + K( )
(42)
Appendices A and B provides estimates with error bounds for Y ( ; x) for
x
1 and 0
x
1 respectively, while Appendix C has an asymptotic representation with error bound for K( ). In order to make use of these results,
in eq(42), it is necessary to dissect the interval in which cot lies, viz., [0; 1)
into the two subintervals [0; 1] and (1; 1). Thus, we consider the following two
cases.
1. 0
2. cot
cot
1, i.e., =4
> 1, i.e., 0 <
=2, and
< =4.
Note that if
= 0, eq (42) implies that Q( ; 0) = 2K( ), which may
be conveniently included in case 2 above by allowing ! 0. The details will
emerge later.
9
5.1
Case 1: 0
cot
1
From eqs (42), (109) and (114), we have:
p
p
Q ( ; ) = p 1 + erf
cot +
2
( )
erf
p
cot
(43)
We de…ne the error term
( ; )
( )
erf
p
cot
(44)
and bounds on ( ) and are given by (116) and (110) respectively. These
may be used to obtain a useful bound for ( ; ).
p
j ( ; )j < j ( )j + erf
cot
< j ( )j +
1
1
< +
1
2
<
1
In fact, 0 <
in the form:
( ; ) < 2=(
1) since
< 0. We may now write eq (43)
p
p
cot +
(45)
Q ( ; ) = p 1 + erf
2
where
0<
( ; )<
2
(
(46)
1)
Returning to eq (102) and using (109), we have:
p
1 + erf
cot +
(1 + ( ; ))
R( ; ) =
8 2
De…ne our error term
=
(47)
:
p
1 + erf
cot
+
1 + erf
p
(1 + ( ; ))
We may now write
R( ; ) =
0
2
8
cot
cot
+
(48)
1
Using eqs 46 and 40, and the fact that erf < 1, we may obtain a bound for
the error term as follows.
j
j<2 +
<2 +
<
2
+
(1 + )
2
(1 + )
1
2
1
1+
1
10
Thus, we have:
j
5.2
Case 2: cot
4
j<
(49)
1
>1
From eq (41), we have
Z
Q( ; ) = 2K ( )
1
u2
e
(1 + u2 )
2
du
(50)
cot
De…ne
e( ; ) =
Z
1
u2
e
(1 + u2 )
2
du
(51)
cot
We now consider 2K ( ) as the approximation to Q( p
; ) and e ( ; ) as
= , a bound of
the error, and seek a bound for e ( ; ). Since 2K ( )
lesser order is sought. Fortunately, such a bound is easy to obtain. Eq (51)
yields
Z 1
2
e ( ; ) < e cot
(1 + u2 ) 2 du
<e
Z
cot
1
(1 + u2 )
2
du
0
Thus, we have
e( ; ) <
e
(52)
4
Since is large, it is clear that Q( ; ) is approximated very well by 2K ( ),
whenever cot > 1. In the physical situation, i.e., Sun-Earth radiation path,
> 20; 000, and so there seems little point in working through a complete
error analysis. It is more expedient to simply ignore the exponentially small
term e
in comparison to other terms like 1= . So, for cot > 1, we commit
an exceedingly small error by taking Q( ; )
2K ( ). For cot > 1 and
1= < ( ) < 0, this leads to:
R( ; ) =
(1 + ( ))
(53)
4 2
Indeed, eq (48) is valid (within the given
p bound for ) for unrestricted
positive cot , since if cot > 1;then erf (
cot ) di¤ers from unity by less
than the following quantity [6, sec 7.1.23, 7.1.24].
p
1
e
cot
cot2
<e
From
eq (48), it is clear that the uniform 2 [0; =2] approximation 1 +
p
erf (
cot ) to the expression 8 2 R( ; )= incurs a relative error of less than
j j, since 0 erf 1. Finally then, we have the result:
p
R( ; ) '
1 + erf
cot
(54)
8 2
This last estimate has a relative error of less than 4= (
1).
11
6
An estimate for S( ; ) with error bounds
Eq (93) gives a closed form expression for S( ; ) in terms of commonly tabulated functions. Since this expression makes reference to the error function
erf(x), in order to obtain an estimate
p for S( ; ) in terms of exponential and
= sin , we now need some estimate of this
rational functions of , where =
function. It is found that, if the expression of eq (24) is naïvely evaluated; for
example, by coding into some programming language with a library call to erf,
signi…cant loss of accuracy results because of cancellation of high order terms. It
is preferred therefore to use an appropriate asymptotic expansion for erf so that
we may observe this cancellation algebraically, and hence determine how many
terms we need for the …nal estimate. The asymptotic expansion of Abramowitz
& Stegun, [6, sec 7.1.23] is used, taking three terms plus remainder. The two
terms of highest order cancel with other terms in the expression (93) for S( ; ).
The relevant approximation to erf ( ) is given by:
p
(1
erf ( )) =
e
2
1
1
2
where
S( ; ) =
where
( )=
3
16
p
p
e
p
4 2
2
4
R3 ( )
(55)
(56)
(1
erf ( )) gives:
2
[1 + ( )]
4
1+2
2
3
4
15
8 6
0 < R3 ( ) <
Substitution of eq (55) in eq (93) for
+
2
2
R3 ( )
(57)
(58)
Using the triangle inequality, and (56), we have:
j ( )j <
Finally, since
=
p
15
63
+ 4
2
16
8
(59)
= sin , eq (57) becomes:
S( ; ) =
with
j j<
p
cot
e
p
4 2
63 sin2
16
12
+
sin4
2
15 sin4
8 2
[1 + ]
(60)
(61)
7
Final approximation for the view factor I2 =I0
We are now in a position to use the approximate representation for S ( ; )
and R ( ; ) (eqs (60) and (54) respectively) to construct a fairly accurate approximation to the net view factor I2 =I0 . The expression for I2 =I0 in terms of
S ( ; ) and R ( ; ) is given by eq (22). It is
r
p
I2
= 2 cos R ( ; ) + sin S ( ; )
(62)
2
I0
Using the expressions for S ( ; ) and R ( ; ) from eqs (60) and (54), eq
(62) becomes:
r
p
p
I2
= 2 cos
1 + erf
cot + ( ; )
2
2
I0
8
p
2
sin4
e cot (1 + ( ; ))
(63)
+ sin p
2
4 2
in which
j
( ; )j <
j ( ; )j <
=
p
4
(64)
1
63
15
+ 4
16 2
8
(65)
= sin
(66)
Eq (63) may be simpli…ed somewhat:
p
4 I2
p
cos
=
2I0
1 + erf
p
1
+ p sin5 e
cot
cot2
+&( ; )
(67)
In this last equation, we have made use of the total error term, & ( ; ),
where:
p
2
1
( ; ) + p sin5 e cot
( ; )
(68)
&( ; )=
From eqs (64) and (65) the following upper bound for j& ( ; )j may be
derived:
p
2
4
1
63
15
j& ( ; )j <
+ p sin5 e cot
+ 4
1
16 2
8
p
and since =
= sin , we have:
j& ( ; )j <
<
=
4
p
4
p
4
p
cot2
1
1
+ p sin5 e
63 sin2
16
+
15 sin4
8 2
cot2
1
1
+ p sin5 e
63 sin2
16
+
15 sin2
8
1
+
93
16
3=2
sin7 e
13
cot2
(69)
Figure 2: View Factor versus angle of incidence, :
p
The approximation (67) for 4 I2 = 2I0 is associated with an absolute error
j& ( ; )j, a bound for which is given by (69). From the next section, it is seen
that, for almost all values of , the relative error of this approximation is less
than 10 4 .
A Microsoft Excel model was constructed to evaluate and graph I2 =I0 for
speci…c = 22961 and 0
900 with increments of 10 using the estimate of
eq (67). The graph appears in Figure 2.
8
The special case of parallel source and absorber planes ( = 0)
The view factor for the case = 0 has been determined by Peck [4, App C, eq
C.13]. Denoting the exponential integral function by E1 (), Peck’s estimate is
given by:
r
1
I2
=
(1 E1 ( ))
(70)
I0 =0
2
He approximates this expression by:
I2
I0
=0
'
p
2
'20
14
=
r
1
2
(71)
From eq (22), we have, using (54), and …nally, ignoring :
r
p
p
p
I2
= 2 R ( ; 0) = 2
(1 + 1 + ) ' 2
2
2
I0 =0
8
4
From this we arrive at:
I2
I0
=0
'
2
p
p
p
2
4
2
=
r
2
1
2
This is in agreement with Peck’s estimate, given by the present eq (71).
9
Discussion and conclusion
We have presented a technique for the estimation of an unusual view factor in
which the source is of Gaussian intensity, as suggested in [4]. The receiving body
is a di¤erential plane surface element. The necessary integrals have been partially evaluated exactly and partially estimated, with the …nal result computed
very rapidly in terms of known functions and accurate in most cases to 0.01%.
The …nal estimates are simple enough to quickly enter into a spreadsheet for
tabulation and graphing of the view factor as a function of angle of incidence.
The whole exercise is done for a …xed geometry and orientation, and moreover, is not intended to be immediately applied to the case of moving bodies.
Recomputation after designated time intervals is therefore not required and a
one-o¤ estimation and calculation are all that is needed. This computation is
rapid and may be easily implemented in a common programming language such
as C, Java, Pascal, or as we have done, in Microsoft’s Excel spreadsheet. Further,
the complexity inherent in the present analysis stems largely from the assumed
intensity distribution of the source, rather than any geometric features. Thus,
we have a somewhat unusual approach to a very speci…c and highly regular
problem.
10
Acknowledgment
The author gratefully acknowledges the useful advice of John Gough, Faculty
of Information Technology, Queensland University of Technology, and that of
Douglas McGregor, Ken Smith, and Vincent Hart, all from the Department of
Mathematics, University of Queensland, Australia.
References
[1] Modest, M.F. (2003). Radiative Heat Transfer, 2nd edition. Academic Press,
California. ISBN 0-12-503163-7.
[2] Sparrow, E.M. and Cess, R.D. (1966). Radiation Heat Transfer, Brooks-Cole.
ISBN 0-89116-923-7.
15
[3] — — –, Features of FACET view factor software are described at URL:
www.osti.gov/estsc/PDFs/facet.pdf.
[4] Peck, M.K. (1976). A Ray-Trace Technique for Solar Concentrators, MEngSc
Thesis, Department of Mechanical Engineering, University of Queensland.
[5] Sugden, S.J. (2002). Estimating a Planar View Factor for Source Radiation of Radially Gaussian Intensity. Bond University School of Information
Technology Technical Report #TR02-03.
[6] Abramowitz , M. and Stegun, I.A. (1972). Handbook of Mathematical Functions, Dover, 10th printing, with correction, ISBN 0-486-61272-4.
[7] Olver, F.W.J. (1974). Asymptotics & Special Functions. Academic Press,
ISBN 0-12-525850-X.
A
( ; x)
Estimation of
We estimate the integral
Z
( ; x) =
1
(1 +
x
for x
(u2 x2 )
e
(2
5)
du
(72)
2
u2 )
1=4
=
(73)
The method of integration by parts is used to obtain an approximation for
this integral, where is regarded as a large positive real constant. Two successive integrations by parts yield the required approximation, with a relative error
of less than 4 (3 + 14 2 + 35 4 ) when x
, and less than 13= 2 x2 1 + x2
when x 1. Proceeding with the method, we …rst de…ne
Z x
2
2
Y ( ; x) =
e u 1 + u2
du
(74)
0
Yc ( ; x) =
Z
1
u2
e
1 + u2
2
du
(75)
x
We have:
Y ( ; x) + Yc ( ; x) = K( )
where the function K is de…ned by eq (16).
Now make the substitution t = u2 x2 in equation (72), yielding:
Z 1
( ; x) =
e t q(t)dt
(76)
(77)
x
where
q(t) =
p
t + x2
2
2 (1 + t + x2 )
16
(78)
and x is to be regarded as a …xed, but otherwise arbitrary positive parameter.
The expression in eq (77) for ( ; x) is a Laplace integral for each …xed
x. Standard techniques such as integration by parts or application of Watson’s
Lemma [7, p71] may be used to obtain an asymptotic expansion for the integral,
as ! 1, provided that the function q(t) satis…es certain conditions (see, for
example, [7, pp67-72]. The method used here is that of integration by parts
(twice). We have:
Z 1
1
q(0) q 0 (0)
( ; x) =
+ 2 + 2
e t q 00 (t)dt
(79)
0
Now write
= t + x2 . Since x is constant, we have d=dt = d=d . We have:
2
2q(t) = (1 + )
and hence
1
5+
4
q 0 (t) =
1
1=2
1=2
(80)
3
(1 + )
(81)
Also,
1 5=2
4
(1 + )
35 2 + 14 + 3
(82)
8
Since q 00 (t) is a positive, decreasing function of t, and for x > 0, its maximum
occurs at t = 0, i.e., at = x2 . That is, we have, for x > 0:
q 00 (t) =
1
x
8
0 < q 00 (t)
5
1 + x2
4
35x4 + 14x2 + 3
(83)
From eqs (80) and (81), we derive:
q(0) = x
q 0 (0) =
1
4
1
2
1 + x2
1
x2
5+
(84)
1
1 + x2
x
3
(85)
We may therefore write eq (79) as:
( ; x) = A( ; x) + r( ; x)
(86)
where the approximation is:
A( ; x) =
1
1
2
x (1 + x2 )
!
5 + 1=x2
1
2
3
4x (1 + x2 )
!
(87)
and the error (remainder) is:
r( ; x) =
1
2
Z
1
0
17
e
t 00
q (t)dt
(88)
Using eq 83, a useful upper bound may be obtained for r( ; x) as follows.
Z 1
1
r( ; x) < 2 max q 00 (t)
e t dt
[0;1)
0
4
=
2
35x + 14x + 3
8
3 x5
(89)
4
(1 + x2 )
It is clear from eqs (88) and (82) that r( ; x) > 0, but perhaps not immediately obvious that A( ; x) is also positive (for su¢ ciently large ). Indeed, the
estimate A( ; x) for the integral ( ; x) would be entirely useless if this were
not the case. To show that A( ; x) > 0, it su¢ ces to show that:
5 + 1=x2
<4
1 + x2
(90)
Put y = x2 , so that (90) reduces to
y2 + (
5=4)y
1=4 > 0
(91)
This last equation tells us that A( ; x) > 0 for all x, provided
A.1
Relative error of the approximation to
> 9=4.
( ; x)
We have ( ; x) = A( ; x) + r( ; x) and A( ; x) > 0; r( ; x) > 0. Now relative
error is given by:
r( ; x)
A( ; x) + r( ; x)
r( ; x)
<
A( ; x)
Relative Error =
Thus, using eq (89), we have:
Relative Error <
35x4 + 14x2 + 3
8
3 x5
(92)
4
(1 + x2 ) A( ; x)
Using the expression for A( ; x) from (87), eq (92) becomes:
35x4 + 14x2 + 3
2
(1 + x2 ) (2 x4 + (2
5) x2
35x4 + 14x2 + 3
<
2 x4 (1 + x2 ) (2
5)
Relative Error =
x2
1)
(93)
This last inequality holds if
2 x4 > 1
18
(94)
Assuming (94) to be satis…ed, proceed to:
35x4 + 14x2 + 3
Relative Error <
(95)
2
5) x4
(2
1=4
Now restrict x to x > (2
5)
, which is compatible with (94). Then,
observing that 35x4 + 14x2 + 3 =x4 is a decreasing function of x, eq (95) then
yields:
1
1=2
35 (2
5) + 14 (2
5)
+3
(96)
Relative Error <
2
1
(2
5) (2
5)
Writing
= (2
5)
1=4
, we have, for x > :
4
Relative Error <
3 + 14
2
+ 35
4
(97)
From the line immediately preceding inequality (93), we see that, for x 1, a
much better bound may be obtained. Since the decreasing function 35x4 + 14x2 + 3 =x4
for x 1 is greatest at x = 1. That is, for x 1, we have:
Relative Error <
35x4 + 14x2 + 3
1
2 x2 (1 + x2 )
2 x4
1
52
2 x2 (1 + x2 ) 2
Thus,
Relative Error <
B
2 x2
13
;
(1 + x2 )
for x
1
(98)
Evaluation of Y ( ; x)
Here, we consider the function Y ( ; x), de…ned by the following integral.
Z x
2
2
Y ( ; x) =
e u 1 + u2
du
(99)
0
2
2
Write z = u and = x so that 0 z
x2
variable, eq (99) becomes
Z
1
z 2
Y ( ; x) = p
e z 1+
z
2
0
. With this change of
1=2
dz
(100)
A mean value theorem of the di¤erential calculus enables us to write
1+
where 0 <
z
2
2z
=1
3
1+
. Using eq (101) in eq (100), we have:
Z
Z
p
2
1=2
z
2 Y ( ; x) =
z
e dz
z 1=2 1 +
(101)
<z
0
0
19
3
e
z
dz
so that, writing
A( ) =
Z
1=2
z
e
z
dz
(102)
0
and
R( ) =
2
Z
3
z 1=2 1 +
z
e
dz
(103)
0
we have:
2
p
Y ( ; x) = A ( ) R ( )
(104)
p
where A ( ) is the estimate for 2 Y ( ; x) with remainder term (or error)
R ( ). Note that A ( ) 0 and R ( ) 0.
Now, since < , we have:
Z
2
0 < R( ) <
z 1=2 e z
(105)
0
Integrating by parts once gives:
2
R( ) <
p
1
+ A( )
2
e
This implies the existence of a function
R( ) =
2
p
1
A( )
2
( ) such that
e
( ) > 0 and
( )
(106)
Therefore, the relative error of the approximation satis…es the following
p
2
1
e
2A ( )
R( )
<
p
A( ) R( )
A ( ) 2 12 A ( )
e
( )
p
A( ) 2 e
p
=
(
1) A ( ) + 2 e + 2 ( )
Finally:
Relative Error =
R( )
<
A( ) R( )
1
(107)
1
As for the estimate, A ( ), we have:
A( ) =
Z
z
1=2
e
z
dz =
0
Therefore,
1
Y ( ; x) = p
2
or
Z
p
2e
t2
dt =
p
0
p
p
erf x
p
1 p
Y ( ; x) = p
erf x
2
20
R( )
1
R( )
A( )
erf
p
(108)
But, since R ( ) > 0, eq (107) implies that:
0<
then, …nally:
1
p
p
Y ( ; x) = p erf x
2
where
0<
C
1
R( )
<
A( )
(1
e
(1
4
K( ) =
(109)
1
<
(110)
1
Estimation of K( ) for
We wish to estimate K( ) for
)
1
1. From eq (18) we have:
p
p
2 ) 1 erf
+
2
(111)
and from [6, sec 7.1.23, 7.1.24], we have
1
erf
where 0 < ( ) < 3=4
it is seen that
p
e
=p
1
1
+ ( )
2
2
. Using the expression (112) for 1
p
1
1
K( ) = p
1
( )
4
2
2
(112)
erf
p
in (111),
(113)
We may therefore write
p
K( ) = p (1 +
2
where
1
4
( )=
and
( ) < 0 if
1
2
> 21 . Also, for
j
( ))
( )
(114)
(115)
> 1=2, we have
( )j <
1
3
8
2
(116)
In particular, we see that j
( )j < 1= . From eq (114):
p
K( ) = p (1 + ( ))
(117)
2
p
p
Therefore, if we considerp
as the error term, then the relative
p( ) =2
error of the approximation
=2
to K( ) is given by:
Relative Error =
1+
21
( )
( )
=
( )
1+ ( )
(118)
since 1 < 1= < ( ) < 0. Now, on [ 1= ; 0), the decreasing function
f , where f (x) = x=(1 + x), attains its maximum when x = 1= ; hence
Relative Error <
1=
1 1=
22
=
1
1
(119)