Asymptotic formulas are derived for incomplete elliptic integrals of all three kinds when the arguments are real and tend to infinity or to zero. Practical error ...
Asymptotic formulas for elliptic integrals by John L. Gustafson A Dissertation Submitted to the Graduate Faculty in Partial Fulfillment of the Requirements for the Degree of DOCTOR OF PHILOSOPHY Department: Mathematics Major: Applied Mathematics
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Iowa State University Ames, Iowa 1982 [Revised and re-typeset, February 2002]
ii
TABLE OF CONTENTS ABSTRACT............................................................................................................iii INTRODUCTION....................................................................................................1 TABLES OF ASYMPTOTIC FORMULAS............................................................5 PRELIMINARY THEOREMS................................................................................ 9 PROOFS OF THE FORMULAS FOR LARGE ARGUMENTS.......................... 14 PROOFS OF THE FORMULAS FOR SMALL ARGUMENTS.......................... 30 ACKNOWLEDGMENTS......................................................................................39 LITERATURE CITED...........................................................................................40 APPENDIX A: NUMERICAL EXAMPLES........................................................ 41 APPENDIX B: PROGRAMS USED TO CREATE EXAMPLES........................ 51
iii
ABSTRACT1
Asymptotic formulas are derived for incomplete elliptic integrals of all three kinds when the arguments are real and tend to infinity or to zero. Practical error bounds are found for the asymptotic formulas. Several techniques are used, including a method recently discovered by R.!Wong for finding asymptotic expansions with remainder terms for integral transforms. Most of the asymptotic formulas and all of the error bounds appear to be new. We use incomplete elliptic integrals that possess a high degree of permutation symmetry in the function arguments. The asymptotic formulas are applicable to complete elliptic integrals as a special case; some of the error bounds are treated separately in the complete case. Numerical examples are given to demonstrate the typical accuracy that can be expected from the formulas, as well as the closeness of the error bounds.
DOE Report IS-T-1014. This work was performed under Contract No. W-7405-Eng-82 with the U.S. Department of Energy.
1
INTRODUCTION Elliptic integrals are classically defined as integrals of the form
Ú r(x, y)dx
(1.1)
where r is a rational function of x and y, and y 2 is a cubic or quartic polynomial in x. If y 2 is linear or quadratic in x, then the † integral may be evaluated using logarithms and rational 2 functions of x and y, but if y is cubic or quartic then the integral is said to be elliptic and is not † in general expressible in terms of elementary functions. Legendre [9] showed † that only three nonelementary functions are needed to express (1.1) in the elliptic case. † We will choose as our three basis functions RF , RD , and RD , which have the integral definitions •
RF (x, y,z) = † †
1 2
RD (x, y,z) =
3 2
- 12
- 12
- 12
- 32
Ú ( x + t) ( y + t) ( z + t)
dt
(1.2a)
dt
(1.2b)
0
•
†
- 12
0
•
†
- 12
Ú ( x +†t) ( y + t) (z + t)
RJ (x, y,z,) =
3 2
Ú (x + t)
- 12
-1
-3
(y + t) 2 (z + t) 2 ( r + t)-1 dt
(1.2c)
0
We assume x, y, z, and r are real and positive. The RF function is the elliptic integral of the first kind and is symmetric in x, y, and z. The elliptic integral of the second kind, RD , is symmetric in † x and y only; it is related to RF by † ∂ RD (x, y,z) = -6 RF (x, y,z). † (1.3) ∂x † The RJ function is symmetric in x, y, and z only. The constant in front of each integral is chosen so that † †
RF (x, x, x) = x RD (x, x, x) = x RJ (x, x, x, x) = x
†
- 12 - 32 - 32
(1.4)
2
Furthermore, the functions are homogeneous of degrees - 12 , - 32 , and - 32 ; that is, -1
RF (Cx,Cy,Cz) = C 2 RF (x, y,z) -3 † =† RD (Cx,Cy,Cz) C 2 RD (x,†y,z)
(1.5)
- 32
RJ (Cx,Cy,Cz,Cr ) = C RJ (x, y,z, r ) These basis functions are not the classical ones used by Legendre that became established as standard after his work † was published in 1825. Legendre’s basis functions are denoted by F(j,k), E(j,k), and P(j,k,n); however, this notation hides the underlying permutation symmetry in the variables, introduces unnecessary linear transformations, † and makes the quadratic Gauss and Landen transformations as well as other identities more cumbersome than with the present choice of basis functions. Carlson [3], [4] has shown elliptic integrals to be hypergeometric functions of several variables and has shown many classical results concerning properties of elliptic integrals to be special cases of the general properties of R functions. References [2], [3], [4], and [14] provide methods of converting between the classical basis functions and the ones used here. In particular, F(j,k) = (sin j ) RF (cos 2 j,1- k 2 sin 2 j,1),
(1.6)
E(j,k) = F(j,k) - 13 k 2 (sin 3 j )RD (cos 2 j,1- k 2 sin 2 j,1),
(1.7)
† P(j,k,n) = F(j,k) - p3 (sin 3 j ) RJ (cos2 j,1- k 2 sin 2 j,1,1+ n sin 2 j ).
(1.8)
† We will also require the following definitions of complete elliptic integrals: †
•
RK (x, y) =
1 p
Út
- 12
-1
-1
-1
-3
(x + t) 2 (y + t) 2 dt
(1.9a)
0
•
†
RQ (x, y) =
2 p
†
- 12
(x + t) 2 (y + t) 2 dt
(1.9b)
0
•
†
Út
RM (x, y, r) =
2 p
Út 0
- 12
-1
-3
(x + t) 2 (y + t) 2 ( r + t)-1 dt
(1.9c)
3
where the constants in front of the integrals are chosen so that properties analogous to (1.4) are satisfied. These complete cases are related to the incomplete cases by
†
RK (x, y) = p2 RF (x, y,0)
(1.10a)
RQ (x, y) =
4 3p
RD (x,0,z)
(1.10b)
RM (x, y, r) =
4 3p
RJ (x, y,0, r ).
(1.10c)
† We will find it convenient to state some of the formulas in terms of the elementary function RC : †
•
RC (x, y) =
1 2
- 12
-1
Ú (x + t) ( y + t)
dt
†
0
Ï Ê 1 Ô( x - y )- 2 logÁ x + ( x1 - y ) Á Ô y2 Ë =Ì Ê x 12 ˆ Ô - 12 Ô ( y - x ) arccosÁÁ 12 ˜˜ Ëy ¯ Ó 1 2
1 2
ˆ ˜, 0 < y < x, ˜ ¯
(1.11)
0 £ x < y.
This is simply a special case of the RF function, as can be seen from (1.2a): †
RC (x, y) = RF (x, y, y).
(1.12)
† There is a similar relationship between RD and RJ , evident from (1.2b) and (1.2c): †
RD (x, y,z) = RJ (x, y,z,z).
(1.13)
† † Despite the intense activity that surrounded elliptic integrals and their inverses (elliptic functions) during the last century, much was left undiscovered concerning the behavior of † elliptic integrals as the arguments tend to zero or to infinity. Kaplan [8, page 13] gives an asymptotic series for F(j,k) that implies the formula Ê 4z 12 ˆ RF (x, y,z) ~ z logÁÁ 1 1 ˜ ˜ Ë x2 + y2 ¯ - 12
as z Æ •.
(1.14)
Kaplan also gives an asymptotic series for E(j,k) that can be shown to imply the one-term † asymptotic formula for RD derived in the sections that follow. Both formulas also appear in † reference [11, page 228]. No series is given in either work for the elliptic integral of the third kind, and no error bounds are given for truncations of the F(j,k) and E(j,k) series. † † †
4
†
Convergent series expansions for functions F(j,k), E(j,k), and P(j,k,n) , for 0!
5
TABLE 1 Incomplete Elliptic Integrals Behavior for Large Arguments Function
RF (x, y,z)
†
† RJ (x, y,z, r )
†
zƕ
Ê 4z 12 ˆ z logÁÁ 1 1 ˜ ˜+ r Ë x2 + y2 ¯
0 < r < 18 z
- 32
† zÆ•
Ê Ê 4z 12 ˆ † ˆ 3z ÁÁ logÁÁ 1 1 ˜ ˜ -1˜˜ + r 2 2 x + y Ë ¯ ¯ Ë
0 < r < 98 z
- 52
† y Æ•
Ê 3 ˆ † y Á 1 1 ˜+ r Ë x 2z2 + z¯
- 12
† RJ (x, y,z, r )
- 32
† r Æ•
†
- 12
-1
3r RF (x, y,z) -
3p 2
Ê
Ê
Ë
Ë
Ê
Ê
Ë
Ë
( x + y )ÁlogÁ1+
4z ˆ ˆ ˜ + 1˜ x + y¯ ¯
( x + y )ÁlogÁ1+
4z ˆ ˆ ˜ + 1˜ x + y¯ ¯
zƕ
† r +r
1 2
X = ( xy ) + r, 1
† †
1
0 < r < 3 2 r -2 ( x + y + z) 2
Ê Ê ˆ 2z ˜ Á Á - z logÁ1+ + 1 1 Á 4 2 ˜ Ë Ë ( xy ) r ¯ 3 4
Y = ( xr) 2 + ( yr) 2
†
Ê Ê 2y ˆ ˆ -3 - 43 y 2 Á logÁ1+ 1 3 ˜ + 12 ˜ < r < 0 Ë Ë x 4z4 ¯ ¯
- 32
-1 3z 2 RC (X 2 ,Y 2 ) + r,where†
1
†
Error Bound
† RD (x, y,z)
†
Asymptotic Formula
† RD (x, y,z)
†
Argument
†
- 32
1 2
ˆ ˜
bk , z-k -v k= 0
M[h;z] = M[j n ;z] - Â
(2.8)
Analogous results are needed for f(t). Define † M[ f ;1- z] =
1
•
Ú t-z f (t)dt +
Út
0
-z
f (t)dt ,
(2.9)
1
As t Æ +• , we assume that †
f (t) = O(t-a ),
a + v > 1.
(2.10)
† The first integral in (2.9) is analytic for Re z < u, and the second is analytic for Re z > 1–a. By analytic continuation of the first†integral, M[f;1–z] can be extended to a meromorphic function in the half plane Re z > 1–a. For n = 1,2,º define n-1
f n (t) = f (t) - Â ak t k +u-1
†
(2.11)
k= 0
and †
Ï f (t), 0 < t < 1, y n (t) = Ì Ó f n (t), 1 £ t < •,
(2.12)
Lemma 2. Let f(t) be a locally integrable function on (0,∞) satisfying (2.2) and (2.10). Then for 1 – a < Re z < n + u, †
11 n-1
ak . k -u-z k= 0
M[ f ;1- z] = M[y n ;1- z] - Â
(2.13)
We can now give the asymptotic expansion of I(l) and the remainder term. † Theorem 1 (Wong). Let f(t) satisfy (2.2) and (2.10). Let h(t) satisfy (2.3) and (2.5). Then for any n ≥ 1, n-1
n-1
I( l) = Â ak M[h;k + u]l
-k-u
+ Â bk M[ f ;1- k - v]l- k- v + dn ( l )
k= 0
(2.14)
k= 0
for the case u ≠ v , and †
n-1
n-1
I( l) = Â c k (v) l
-k-v
†
+ log l  akbk l-k-v + dn ( l )
k= 0
(2.15)
k= 0
for the case u = v , where †
n-1
akb j - a j bk . k- j k= 0
c k = ak M[j n ;k + v] + bk M[yn ;1- k - v] - Â
†
(2.16)
In both cases, the remainder term is given by † •
dm ( l ) =
Ú
f n (t)hn ( lt)dt .
(2.17)
0
Proof. We write †
I( l) = I1 (l ) + I 2 ( l),
(2.18)
where I1 and I2 correspond to the intervals (0,1) and (1,∞), respectively. We break up the remainder term similarly: †
dn ( l ) = d n,1 ( l ) + dn,2 ( l ). By applying (2.6) and (2.11) to I1 and I 2 , † †
†
(2.19)
12 n-1
n-1
1
I1 ( l ) = Â ak
Út
k +u-1
0
k= 0
1
h( lt)dt + Â bk l
-k-v
Út
-k- v
f n (t)dt +dn,1 (k)
0
k= 0
and †
n-1
n-1
•
I 2 ( l) = Â ak
Út
hn ( lt)dt + Â bk l
1
k= 0
•
k + u-1
-k-v
Út
k +u-1
f (t)dt +dn,2 (k).
1
k= 0
Assume for the moment that u ≠ v . Since y n (t) = f n (t) for 0 < t < 1 and j n (t) = f (t) for 1 < t < •,†adding the last two identities together gives †
† È 1 k + u-1 I( l) = Â ak Í Ú t h( lt)dt + Î0 k= 0 n-1
†
•
Út
k +u-1
1
˘ † hn ( lt)dt˙ ˚
n-1
(2.20)
+ Â bk M[y n ;1- k - v]l- k- v + dn ( l ). k= 0
The presence of l in the quantity in square brackets prevents us from similarly combining h and hn into a Mellin†transform of j n ; however, we can rewrite that quantity as a Mellin transform of j n plus other terms if we substitute s = lt : l
†
•
† [º] = l-k- u Ú sk + u-1h(s)ds + l-k-u Ú sk +u-1hn (s)ds † 0 l
†
-k- u
=l
- k- u
M[j n ;k + u] + l
Ê l k +u-1 h(t)dt + ÁÚt Ë1
=l
- k- u
M[j n ;k + u] + l
Út
k + u-1
1
n-1
Út 1
ˆ hn (t)dt ˜ ¯
k + u-1
n-1
l -k- u
l
Âb t j
- j-v
(2.21)
dt
j= 0
- j- v
n-1 l bj bj - l-k-u  k + u- j -v k + u- j -v j= 0 j= 0
= l-k- u M[j n ;k + u] + Â
By Lemma 1, the sum of the first term and the third term in (2.21) is equal to l-k-u M[h;k + u]. So (2.20) † now becomes n-1
I( l ) = Â ak M[h;k + u]l k= 0
n-1 † l- j- v akb j + ÂÂ + Â bk M[y n ;1- k - v]l-k-v + dn ( l ), (2.22) k + u - j - v k= 0 k= 0 j= 0 n-1 n-1
-k-u
By applying Lemma 2 to (2.22), the double sum is completely canceled, leaving †
13 n-1
n-1
I( l) = Â ak M[h;k + u]l
-k-u
k= 0
+ Â bk M[ f ;1- k - v]l-k- v + dn ( l),
(2.23)
k= 0
which proves the case u ≠ v . For the case u = v , we apply a limiting process. Since † Ê l-k-u - l- k- v ˆ -k-v limÁ ˜ = l (-log l ), †uÆvË u-v ¯
†
(2.24)
this limit applies to the terms of the sums in (2.21) for which k = j : † • Ê1 ˆ limÁ Ú t k + u-1h( lt)dt + Ú t k + u-1hn ( lt)dt˜ uÆv † ¯ Ë0 1 n-1 -k-v
=l
-k-v
M[j n ;k + v] + bk (log l ) l
+Â
b j ( l- j- v - l-k- v ) k- j
j= 0 j≠ k
(2.25) .
Therefore, instead of (2.22) and (2.23), we obtain the second case of the theorem: †
n-1
n-1
I( l) = Â c k (v) l
-k-v
k= 0
+ log l  akbk l-k-v + dn ( l ), k= 0
where †
n-1
akb j - a j bk .n k- j k= 0
c k = ak M[j n ;k + v] + bk M[yn ;1- k - v] - Â
Both cases will be of considerable value to us in deriving asymptotic formulas and error bounds for elliptic†integrals.
14
PROOFS OF THE FORMULAS FOR LARGE ARGUMENTS We now prove the asymptotic formulas and error bounds summarized in Tables 1 and 3, and give more detailed results for certain cases. In attempting to bound the error term, we seek the following features in a bound: 1. The bound should be simple, i.e. an elementary function of the argument in question. It should not involve limits or integrals. 2. There should be little “waste” in the bound. The bound should not exceed the actual error by more than, say, a factor of three. Ideally, the bound should asymptotically agree with the actual error. These features tend to compete directly with one another, since the actual error always involves an integral that is not expressible using elementary functions, and all simplifications of that integral introduce a difference between the actual error and the bound. Hence, the bounds presented here are the result of compromise. All satisfy the first condition in that they involve only logarithms and rational powers of the argument; the second condition is generally satisfied also, based on numerical tests (see Appendices). Theorem 2. If x, y ≥ 0 and x+y, z > 0, then Ê 4z 12 ˆ RF (x, y,z) = z logÁÁ 1 1 ˜ ˜ + r, 2 2 x + y Ë ¯ - 12
(3.1)
where the error term, r, is bounded by † 0 < r < 18 z
- 32
Ê
Ê
Ë
Ë
( x + y )ÁlogÁ1+
4z ˆ ˆ ˜ + 1˜. x + y¯ ¯
Proof. If we lets s = 1 t in the integral representation (1.2a) of RF (x, y,z), † †
†
(3.2)
15 •
RF (x, y,z) =
1 2
- 12
Ú (( x + t )( y + t )(z + t ))
dt
0
1
ÊÊ 1 ˆÊ 1 ˆÊ 1 ˆˆ 2 Ê -1ˆ Ú ÁËÁË x + s ˜¯ÁË y + s ˜¯ÁË z + s ˜¯˜¯ ÁË s2 ˜¯ds 0
•
=
1 2
=
1 2
•
- 12
Ú (s(1+ xs)(1+ ys)(1+ zs))
ds
0
Hence, RF has the alternative representation † •
RF (x, y,z) =
†
1 2
- 12
Ú ( t(1+ xt)(1+ yt)) (1+ zt)
1 2
dt,
(3.3)
0
which is in the form (2.1) for which Theorem 1 applies, with z = l . Let † f (t) = t
- 12
- 12
((1+ xt)(1+ yt)) †
and †
1
- 12
- 12
h(zt) = (1+ zt) = ( zt)
Ê 1 ˆ- 2 Á1+ ˜ . Ë zt ¯
The functions f and h have asymptotic formulas † Ê x + yˆ ˆ -1 Ê + f (t) ~ t 2 Á1- Á ˜ t ˜ as t Æ 0 , Ë ¯ 2 Ë ¯ - 12 1 -1 h(t) ~ t (1- 2 t ) as t Æ +•,
(3.4)
which means that u = v = - 12 , and the second case of Theorem 1 applies. Here, a0 = 1, b0 = 1, -3 a1 = - 12 ( x + y ) , and † b1 = - 12 . Observe that f (t) = O(t 2 ) as t Æ •, and h(t) = O(1) as t Æ 0, so conditions (2.5) and (2.10) are satisfied. Using (2.15) with n = 1, we obtain the first term of the † † † expansion: asymptotic † † † † † 1 1 1 R (x, y,z) = 1 c ( 1 )z 2 † + log z 2 z 2 + 1 d (z), (3.5)
†
F
2
0 2
( )
2 1
where c 0 and d1 (z) will be derived as follows: By (2.7) and (2.12), † †
†
16
Ï h(t), 0 < t (1+ w ) + 1 = 2 + w.
Here we use the fact that w = 1 zt is positive for t in the range of integration, (0, ∞). For the -1 same reason, the quantity in (3.10) is positive. Hence, 1- (1+ w ) 2 < w (2 + w ) , or † 1
Ê 1 ˆ- 2 1 0 < 1- Á1+ ˜ < . Ë † zt ¯ 1+ 2zt
†
(3.11)
It will turn out to be convenient to specialize further this inequality according to whether t is small or large. Since 1 (1+†2zt ) < 1 (2zt ) ,
†
Ê 1- Á1+ Ë Ê 1- Á1+ Ë
-1
1ˆ 2 1 ˜ < zt ¯ 1+ 2zt - 12 1ˆ 1 ˜ < zt ¯ 2zt
"t > 0, close for small t, (3.12) "t > 0, close for large t.
The term “close” here means that the ratio of the terms being compared approaches unity as t becomes small†or large, appropriately. The first term in square brackets in (3.9) is bounded differently; we use -1
1- (1+ ( x + y ) t + xyt 2 ) 2 < 1 - 12
1- (1+ ( x + y ) t + xyt 2 )
0, close for small t.
(3.13)
18
The first inequality is obvious; the second follows from the inequality of arithmetic and geometric means:
( xy )
1 2
£
x+y for x, y ≥ 0. 2
(3.14)
As a result, †
1
1- (1+ ( x + y ) t + xyz
1 2 -2
)
-2 2 Ê x + y) t 2 ˆ ( ˜˜ £ 1- ÁÁ1+ ( x + y ) t + 4 Ë ¯
Ê ( x + y ) t ˆ-1 = 1- Á1+ ˜ 2 ¯ Ë Ê (x + y)t ˆ ( x + y ) t < 1- Á1. ˜= Ë 2 ¯ 2 By partitioning the interval of integration into (0,s) and (s,∞) for any s > 0, we may bound the remainder † term r defined by (3.9): r < 12 z
s
• 1 Ê x + y ˆÊ 1 ˆ Ê 1 ˆ 1 2 dt + z Ú ÁË 2 ˜¯ÁË1+ 2zt ˜¯ 2 Ú ÁË 2zt 2 ˜¯dt 0 s Ê 3 Ê ˆ x+y 1ˆ = 14 z 2 ÁÁ ˜ log(1+ 2zs) + ˜. s¯ ËË 2 ¯
- 12
(3.15)
As s tends to infinity, the log term dominates and the bound in (3.15) tends to infinity; as s tends to zero, the 1 s term † dominates and the bound tends to infinity. In either case, the bound is not useful. The minimum occurs for a particular value of s that is a function of x, y, and z. Minimizing the quantity in square brackets in (3.15) by differentiation, † ( x + y )z - 1 = 0, 1+ 2zs s2 which implies † x+y=
1 2 + . zs2 s
(3.16)
Since we are mainly interested in large z, drop the 1 zs2 term to obtain smin ~ 2 ( x + y ) . By † †
†
19
using this value of s in (3.15), the stated bound (3.2) is obtained. n
(
†
-1
)
(
-3
)
Since RF is O z 2 log z and the error bound is O ( x + y )z 2 log z , the relative error is O(( x + y ) z) . Therefore, one expects to be able to approximate RF to about three significant figures if z ( x + y ) is on the order of 1000. Numerical tests bear this out; see Appendix A. † Since†there is complete symmetry in the arguments x, y, and z, one can always permute the † † arguments so that z is the largest argument. † As a corollary to Theorem 2, we can find an asymptotic formula and error bound for the complete case RK (y,z) as z Æ •. Using (1.10a) gives Ê16z ˆ -1 RK (y,z) = p1 z 2 logÁ ˜ + r, Ë y ¯
†
†
(3.17)
where † 0 0, then Ê Ê 4z 12 ˆ ˆ RD (x, y,z) = 3z ÁÁ logÁÁ 1 1 ˜ ˜ -1˜˜ + r , 2 2 x + y Ë ¯ ¯ Ë - 32
(3.20)
where the error term, r, is bounded by † 0 < r < 98 z
- 52
Ê
Ê
Ë
Ë
( x + y )ÁlogÁ1+
4z ˆ ˆ ˜ + 1˜ . x + y¯ ¯
(3.21)
Proof. By (1.3), RD can be obtained through differentiation of RF with respect to z. Hence, we obtain the first-order† term and remainder by applying -6∂ ∂z to the expressions for RF and r, (3.1) and (3.9): † † 1 Ê ˆ • Ê 4z 2 ˆ † -3 † ∂ - 12 - 32 - 12 2 Á ˜ RD (x, y,z) = 3z Á logÁÁ 1 -1 3z 3 t f (t) zt - h(zt) dt . (3.22) ( ) 1 ˜ Ú ˜ ˜ 2 2 ∂z 0 Ë Ëx + y ¯ ¯
) (
(
)
To justify the differentiation of (3.9) under the integral sign, we need to majorize the integrand of † (3.22) for z in any closed interval [K,L], 0 < K < L, by a function that is integrable and does not depend on z; see [4, Appendix B.3]. This will be done below in the process of bounding the integral. To bound the error term, we need to bound Ï Ê 1 ˆ- 32 ¸Ô 1 3Ô ∂ - 12 - 12 3 -2 -2 -3 (zt ) - (1+ zt ) = 2 t z Ì1- Á1+ ˜ ˝ . ∂z ÔÓ Ë zt ¯ Ô˛
(
)
(3.23)
We rewrite the quantity in curly brackets above by substituting w = 1 (1+ zt ) : †
3
3 Ê 1+ zt ˆ- 2 Ê 1 ˆ- 2 1- Á1+ ˜ = 1- Á ˜ † zt ) -1¯ Ë zt ¯ Ë (1+
3
= 1- (1- w ) 2 , where 0 < w < 1. † This function has the same value and derivative as † †
†
3 2
w at w = 0, but is concave down since its
21 3
second derivative is negative for 0 < w < 1. Hence, = 1- (1- w ) 2 < 32 w , or 3
3 Ê 1 ˆ- 2 = 1- Á1+ ˜ < 2 . 1+ zt †Ë zt ¯ -1
-3
The above inequality can be used to bound (3.23) for 0 < K ≤ z ≤ L by 32 t 2 K 2 ( 32 (1+ Kt )) . This provides an integrable majorizing function for the integrand of (3.22) that is independent of z, as † required to justify differentiation under the integral sign. The inequality is specialized for large and small values of zt ; † 3
3 Ê 1 ˆ- 2 1- Á1+ ˜ < 2 Ë zt ¯ 1+ zt - 32 3 Ê 1ˆ 1- Á1+ ˜ < 2 Ë zt ¯ zt
†
"t > 0, close for small t, (3.24) "t > 0, close for large t.
By combining inequalities (3.13) from the previous proof and (3.24), † 3 Ê x + y ˆÊ z- 2 ˆ Ú ÁË 2 ˜¯ÁÁ1+ zt ˜˜dt + 49 0 Ë ¯
s
0 0, close for large t.
(3.35)
Combining these inequalities in (3.33) gives † 3 2
0 < -r < y
- 12
s
3 1 3 1 Ú 1+ 2yt dt + 43 y - 2 x - 2 z- 2 0
= 43 y
- 32
(log(1+ 2ys) +
1 2
•
Út
-3
dt (3.36)
s
- 12 - 32 -2
x z s
).
The quantity in square brackets in (3.36) can be minimized by differentiation with respect to s: †
2y -1 - 3 - x 2 z 2 s-3 = 0. 1+ 2ys
(3.37)
-1 - 3
As y Æ •, smin ~ x 4 z 4 , which when applied to (3.36) yields the bound (3.30). n
† †
† As x Æ 0, the bound in (3.30) approaches infinity, and hence a slightly different approach is needed † for the complete case.
24
Theorem 5. If y and z are positive, then -1
RQ (y,z) = p4 y 2 z-1 + r,
(3.38)
Ê 2y ˆ ˆ - 3Ê 0 < -r < p1 y 2 Á logÁ1+ ˜ + 23 ˜. z ¯ ¯ Ë Ë
(3.39)
where †
Proof. Equation (3.38) is easily obtained from (1.10b) and (3.29). To obtain a bound, apply (1.10b) and (3.33) with†x = 0:
2 p
r=- y
†
- 12
•
- 32
Ú (1+ zt ) 0
1 Ê Ê ˆ- 2 ˆ 1 Á1- Á1+ ˜ ˜ dt . Á Ë yt ¯ ˜ Ë ¯
(3.40)
Here we use a slightly different bound on the first factor: †
-3
- 3 - 32
(1+ zt) 2 < z 2 t -3 (1+ zt) 2 < 1
"t > 0, close for large t, "t > 0, close for small t.
(3.41)
So the bound on r here takes the form † -r
0 and x, y, z ≥ 0 with at most one of†x, y, z equal † to 0, then † †
25 -3
RJ (x, y,z, r ) = 3r -1RF (x, y,z) - 32p r 2 + r,
(3.43)
where † 1
0 < r < 3 2 r -2 ( x + y + z) 2 .
(3.44)
Proof. By using the substitution s = 1 t in (1.2c), † •
RJ (x, y,z, r ) = †
3 2
- 12
- 12
- 12
= r
- 12
•
- 12
Ú ((1+ xs)(1+ ys)(1+ zs)) 0
3 2
r
- 12
Ú
• 0
dt
0
3 2
This is of the form
-1
Ú ( x + t ) ( y + t ) (z + t ) ( r + t )
( rs)
(3.45)
1 2
1+ rs
ds.
f (t)h( pt)dt , where
† - 12
f (t) = ((1+ xt )(1+ yt )(1+ zt ))
~ 1 as t Æ 0 +
(3.46)
† and † 1
t2 -1 h(t) = ~t 2 1+ t
as t Æ +•.
(3.47) -3
So the first case of Theorem 1 applies, with u = 1 and v = 12 . Since f (t) = O(t 2 ) as t Æ +• and 1 h(t) = O(t 2 ) as t Æ 0 + , conditions (2.5) and (2.10) are satisfied. By (2.14), †
†
-1 -1 RJ (x, y,z, r†) = 32 r 2 † M [ h;1] r -1 + † M [ f ; 12 ] r 2 + r, †
{
†
}
where by (3.3), † •
M [ f ; 12 ] =
- 12
- 12
Ú t ((1+ xt )(1+ yt)(1+ zt ))
dt = 2RF (x, y,z).
0
Using the beta function, † •
M [ h;1] =
0
†
-1
Ú (1+ t )
1
t 2 dt = B( 32 ,- 12 ) = -p ,
(3.48)
26
we arrive at the first asymptotic term, equation (3.43). The error term, r, is by (2.17) equal to • 3 2
- 12
Ú {1- ((1+ xt)(1+ yt)(1+ zt )) 0
Ï
¸
}t ÌÓ r1t - 1+1rt ˝˛dt, 1 2
(3.49)
where we bound each factor in curly brackets as follows: †
1 1 1 1 = < 2 2 rt 1+ rt rt (1+ rt ) r t
"t > 0,
(3.50)
and † -1 1- ((1+ xt )(1+ yt )(1+ zt )) 2
0, close for large t. (3.51b)
It might at first appear that we should specialize the inequality in (3.50) by bounding 1 rt (1+ † rt ) by 1 rt for small t; however, this turns out to yield an error bound that is only - 47 O( r ). Inequality (3.51a) follows from the following general inequality for b, x i > 0: † †
n
†
n
-b
’ (1+ x ) i
-b
=e
 log(1+xi ) i =1
i=1
n
n
> 1- b log(1+ x i ) > 1-†b x i . i=1
(3.52)
i=1
Applying these bounds to (3.49) gives †
s
0 0, close for small t, 1 2
r -1 "t > 0, close for large t,
(3.58)
and † - 12
(zt )
- 12
(zt )
- 12
- 12
( zt)
- 12
1+ 2zt -3 < 12 ( zt ) 2
- (1+ zt ) < - (1+ zt )
"t > 0, close for small t,
(3.59)
"t > 0, close for large t. -1
The latter inequality follows from multiplying (3.12) by ( zt ) 2 . Applying these inequalities to the error term, † for any s > 0, gives s
• 3 1 - 12 -1 † 3 -2 0 < -r < z Ú dt + 4 z ( xy ) r Ú t -3 dt, 0 1+ 2zt s 3 2
- 12
= 43 z
- 32
(log(1+ 2zs) +
1 2
- 12
( xy ) r-1s-2
(3.60)
).
We can minimize the quantity in square brackets by differentiation: †
2z -1 - ( xy ) 2 r -1s-3 = 0 , 1+ 2zs -1
(3.61)
-1
which implies that as z Æ •, smin Æ ( xy ) 4 r 2 . Using this value in (3.60) gives the inequality (3.61). n † † formula The asymptotic † and bound given for RD (x, y,z) as y Æ • are special cases of the preceding theorem, in accordance with (1.13). Hence, this gives an independent proof of Theorem 4. † † In the complete case, y = 0, the error bound of Theorem 7 is not useful. An error bound for RM (x,z, r ) may be found by a slight modification of inequality (3.58). Theorem 8. If † x, z ≥ 0 and x+z, r > 0, then † -1
-1
RM (x, y, r) = p4 z 2 r 2 RC ( r, x) + r , where †
(3.62)
29
0 < -r < p1 z
- 32
(log(1+ 2zx
- 13
r
- 23
) + ). 2 3
(3.63)
Proof. The first term on the right side of (3.62 is obtained by setting y = 0 in (3.54) and then -1 using (1.10c) and homogeneity to bring the r 2 outside the RC function. To bound the error † term, we use † 1 1 1 -1 2 † t 2 (1+ xt ) (1+ r†t ) < t 2 "t > 0, close for small t, (3.64) 1 1 1 -2 -1 -1 - 2 -1 2 t (1+ xt ) (1+ rt ) < t x r "t > 0, close for large t. Inequality (3.59) is still useful in this case; by combining it with (3.64), we obtain a variation on (3.60): † s
• 3 1 1 -5 1 - 2 - 2 -1 0 < -r < z Ú dt + p z x r Ú t 2 dt, 0 1+ 2zt s 2 p
- 12
= p1 z
- 32
(log(1+ 2zs) +
2 3
- 12
-1 - 32
z r s
(3.65)
).
We can minimize the quantity in square brackets by differentiation: †
2z -1 -5 - x 2 r -1s 2 = 0 , 1+ 2zs -1
-2
(3.66)
which implies that as z Æ •, smin Æ x 3 r 3 . Using this value in (3.65) gives the bound on the error, (3.63). n † † the proofs This completes † of formulas for large arguments.
30
PROOFS OF THE FORMULAS FOR SMALL ARGUMENTS The formulas for arguments tending to zero are of a different character from those for arguments tending to infinity. In four of the five incomplete cases, setting an argument equal to zero gives a finite value that can be used as an approximation. In three of the cases, that first term is simply the complete version of the elliptic integral. In addition to Theorem 1, a variety of approaches is used to obtain approximations and error bounds. Theorem 9. If x, y >0 and z ≥ 0, then RF (x, y,z) = p2 RK (x, y) + r ,
(4.1)
where † 0 < -r
0. Using inequality (3.11), we bound -1 1- (1+ zt ) 2 by † zt (2 + zt ) for all t > 0. Combining these yields an elementary integral as a bound: † † • -1 -1 -1 † 0 < -r < 1 z xy 2 t 2 2 + zt dt 2
( ) Ú
(
)
0
-1
= ( xy ) 2 RC (0,2 z) † =
p 2 2
1
-1
z 2 ( xy ) 2 . n
†
† †
The asymptotic formula for the complete case RK (x, y) as y Æ 0 is the same as the formulas for RK (x, y) as x Æ •, (3.17) † and (3.18). This follows from homogeneity, since - 12 RK (x, y) = y RK (x y ,1). Hence, by letting z = x in (3.17) and (3.18), we obtain a formula and † † † †
31
an error bound that are O(log y) and O(y log y), respectively:
†
†
Ê16x ˆ -1 RK (x, y) = p1 x 2 logÁ ˜ + r, Ë y ¯
(4.4)
where † 0 0, close for small t, y ¯¯
(4.10)
where C = z x + z y . So for any positive s, † 1 2
0 < -z r
1.000000E-15 then r1 = rc : goto 500 else return 590 ' 600 '''''' Compute RJ(x1,y1,z1,p1): 610 s = 0 620 fa = 3 630 rj = 0 640 r0 = 1 650 la = sqr(x1*y1)+sqr(x1*z1)+sqr(y1*z1) 660 mu = (x1+y1+z1+2*p1)/5 670 x2 = (x1+la)/4
52 680 690 700 710 720 730 740 750 760 770 780 790 800 810 820 830 840 850 860 870
y2 = (y1+la)/4 z2 = (z1+la)/4 p2 = (p1+la)/4 x3 = 1-x1/mu y3 = 1-y1/mu z3 = 1-z1/mu p3 = 1-p1/mu s2 = (x3^2+y3^2+z3^2+2*p3^2)/4 s3 = (x3^3+y3^3+z3^3+2*p3^3)/6 al = ((sqr(x1)+sqr(y1)+sqr(z1))*p1+sqr(x1*y1*z1))^2 be = p1*(p1+la)^2 : gosub 470 s = s+fa*rc fa = fa/4 rj = 1+3*s2/7+s3/3+3*s2^2/22+s2*s3*3/13+3*s4/11+3*s5/13-s2^3/10 rj = s+fa/4*(rj+3*s3^2/10+3*s2*s4/5)*mu^-1.5 x1 = x2 y1 = y2 z1 = z2 p1 = p2 if abs((rj-r0)/rj) > 1.000000E-15 then r0 = rj : goto 650 else return
