Applied Mathematics E-Notes, 13(2013), 77-91 c
Available free at mirror sites of http://www.math.nthu.edu.tw/ amen/
ISSN 1607-2510
Asymptotic Expansions Of Iterates Of Some
Classical Functions
Sukrawan Mavechay, Vichian Laohakosolz
Received 18 February 2013
Abstract
Asymptotic expansions of iterates of …ve functions, namely, the logarithmic
function, the inverse tangent function, the inverse hyperbolic sine function, the
hyperbolic tangent function and the Fresnel integral are derived with explicit
parameters using a re…nement of a 1994 method of Bencherif and Robin.
1
Introduction
As mentioned at the beginning of [2, Chapter 8], many problems in asymptotic analysis
can be stated as follows: let f (x) : R ! R and u0 2 R be given. De…ne the sequence
(un )n 0 by
u1 = f (u0 ); un = f (un
1)
= f2 (un
2)
= f (f (un
2 ))
=
= fn (u0 )
(n
1):
The problem is to …nd an asymptotic expansion of un as n ! 1. In [3, Problem 173],
the case of f (x) = sin x was considered.
p Writing u1 = sin x and un := sinn x (n 1);
the problem is to show that limn!1 n=3 sinn x = 1. In the book [2, Section 8.6], de
Bruijn improved this result by showing that
r
3 log n C(x)
log2 n + log n +
3
log3 n
sinn x =
1
+
+
O
(n ! 1);
n
10 n
2n
n2
n3
where ; ; are explicit parameters depending on C(x), which in turn depends on x,
but is independent of n. In 1994, Bencherif and Robin, [1], generalized this result by
deriving an asymptotic expansion for iterates of a general continuous function f and
applied their result to the function f (x) = sin x (x 2 (0; )) to obtain an even more
precise result.
In the present work, we re…ne Bencherif-Robin’s 1994 work to obtain as many explicit parameters as possible, and apply them to derive asymptotic expansions
of the itR
erates of …ve important classical functions log(1+x); tan 1 x; sinh 1 x; cos t2 =2 dt
and tanh x.
Mathematics Subject Classi…cations: 41A60, 26A18, 40A99.
of Mathematics, King Mongkut’s Institute of Technology Ladkrabang, Bangkok
10520, Thailand. email: ktsukraw@kmitl.ac.th
z Department of Mathematics, Kasetsart University, Bangkok 10900, Thailand.
e-mail: fscivil@ku.ac.th
y Department
77
78
2
Asymptotic Expansions of Iterates of Classical Functions
Preliminaries
Our main tool is the following result of Bencherif and Robin, [1, Theorem 2].
PROPOSITION 1. For k 2, let fa1 ; a2 ; : : : ; ak ; tg be a set of real numbers with
a1 < 0; t > 0, let = 1=ta1 , and let b1 = 1 + t 2a2 =a21 =2t. Then there exists a
sequence of k + 1 polynomials (Pm )0 m k with coe¢ cients in Q(a1 ; a2 ; : : : ; am+1 ; t)[X]
satisfying the di¤erential-di¤erence equation
0
0
Pm+1 = b1 Pm + (tm + 1)Pm (0
m
k
1)
with the following property: if (un ) is a real positive sequence converging to 0 and
satis…es
k
X
un+1 = un +
am umt+1
+ O(u(k+1)t+1
) (n ! 1);
n
n
m=1
then it has an asymptotic expansion of the form
(
k
1=t
X
1
1
Pm
1+
(b1 log n C)
+O
un =
n
t
nm
m=1
where C = limn!1 ( un t
n
1
nk
)
(n ! 1); (1)
b1 log n) ; P0 = 1; P1 = X; and
Pk 2 Q(a1 ; a2 ; : : : ; ak ; t)[X]:
Since our main objective is to derive asymptotic results with explicit parameters,
following Bencherif-Robin, consider vn = =utn . The sequence (vn ) was shown in [1] to
have the following properties.
I. As n ! 1, we have
vn+1
vn+1
vn
log
1
i
vn+1
II. For k
III. For k
vn = 1 +
=
k
X1
k
X1
bj
j
j=1 vn
a0j
j
j=1 vn
1
vnk
+O
1
vnk
+O
k
X1 aij
1
+O
=
j
vni
j=i+1 vn
2, the limit limn!1 (vn
1
vnk
n
(k
2)
(2)
(k
2)
(3)
(1
i
k
b1 log n) = C exists.
3, there is a unique family of real numbers
(cm )1
m k 2;
cm 2 Q(b1 ; b2 ; : : : ; bm+1 )
2; k
3):
(4)
S. Mavecha and V. Laohakosol
79
for which the function
(y) = y
(b1 log y + C) +
c1
+
y
+
ck
yk
2
2
(5)
is well-de…ned in the neighborhood of in…nity, and the inverse function
and satis…es
1
vn =
(n) + O n k+1 (k 3):
1
exists
(6)
To apply Proposition 1, consider a function f continuous in a neighborhood of the
origin and is of the form
f (x) = x +
k
X
am xmt+1 + O(x(k+1)t+1 )
(x ! 0)
m=1
with a1 < 0 and t > 0. For a given su¢ ciently small u0 = x0 2 R, de…ne un+1 =
f (un ) (n 1). Thus,
un+1 = un +
k
X
am umt+1
+ O(u(k+1)t+1
):
n
n
(7)
m=1
With vn = =utn , to derive asymptotic estimates for vn , Bencherif and Robin, [1,
Proposition 8], showed that if f is increasing over (0; ], then the limiting function (in
the right hand expression of (1))
C(x0 ) = lim
n!1
fn (x0 )
t
n
b1 log n
exists, is continuous over (0; ] and satis…es the asymptotic expansion
C(x) = x
t
where di = ci
1])
b1 log
i
x
t
+ d1 xt +
(i = 1; : : : ; k
j 1
X
+ dk
2x
(k 2)t
+ O(x(k
1)t
) (x ! 0; k
3)
2). The numbers ci de…ned in (5), satisfy ([1, Lemma
aij ci =
bj
(1
j
k
1)
(8)
i=0
where the parameters bj and aij are de…ned via (2), (3) and (4). The next three lemmas
give explicit shapes of bj and aij .
LEMMA A. For 1
j
k
1 (k
2), we have:
j
X
( t)j+1 s X
j+1 s
an1 1 an2 2
ank k ;
(9)
(k;j+1)
(j
+
1
s)!
n
;
n
;
:
:
:
;
n
1
2
k
s=0
P
where ( t)r := ( t)( t 1)
( t r + 1), the sum (k;j+1) is over nonnegative
integers n1 ; n2 ; : : : ; nk such that n1 + 2n2 +
+ knk = j + 1 and
(
(j+1 s)!
if n1 + n2 +
+ nk = j + 1 s
j+1 s
= n1 !n2 ! nk !
n1 ; n2 ; : : : ; nk
0
otherwise.
bj =
j+1
80
Asymptotic Expansions of Iterates of Classical Functions
PROOF. Using vn+1 = =utn+1 and (7), we get
vn+1 =
=
utn
k
X
1+
am umt
n
+O
m=1
8
<
1+
utn :
(
k
X
u(k+1)t
n
am umt
n
m=1
!
!
t
t
+ O u(k+1)t
n
9
=
;
!p
)
k
k
X
( t)p X
mt
(k+1)t
am un
+ O un
= t 1+
un
p!
p=1
m=1
(
k m
X
X1 ( t)m s X
m s
(a1 utn )n1
= t 1+
(k;m) n1 ; n2 ; : : : ; nk
un
(m
s)!
m=1 s=0
)
nk
(ak ukt
n )
+ O(u(k+1)t
)
n
= vn
(
k
X
1+
m
X1
m
m=1
+ O vn k
1
s=0
)
( t)m s X
m s
(k;m) n1 ; n2 ; : : : ; nk
(m s)!
ank k
an1 1 an2 2
!
1
vnm
The formula for bj follows at once by comparing with (2).
LEMMA B. For 1
a0j =
where the sums
j
k
j 1
X
( 1)j 1
j s
s=0
P
and
(k;j)
1 (k
2), we have
sX
(k;j)
j s
n1 ;n2 ;:::;nk
j s
n1 ; n2 ; : : : ; nk
1n1 bn1 2 bn2 3
bnk k 1
are similarly de…ned as in Lemma A.
PROOF. From (2), we obtain
0
1
k
1
X
bj
1
vn+1
+
+ O vn k 1 A
= log @1 +
log
vn
vn j=1 vnj+1
0
1p
k
k
1
X1 ( 1)p+1
X
bj A
@1 +
=
+ O vn k
j+1
p
v
v
n
p=1
j=1 n
=
k
X1
m=1
m
X1
s=0
+ O vn k
( 1)m 1
m s
(10)
sX
(k;m)
m s
n1 ; n2 ; : : : ; nk
1n1 bn1 2
bnk k 1
!
1
vnm
S. Mavecha and V. Laohakosol
81
The formula for a0j follows at once by comparing with (3).
Explicitly, the …rst four a0j -terms are
a01 = 1; a02 = b1
1=2; a03 = b2
LEMMA C. Let k
aij =
jX
i 1
s=0
where the sum
b1 + 1=3 and a04 =
3. For i + 1
( i)j
(j i
P
(k;j i)
i s
s)!
and
X
j
(k;j i)
j i s
n1 ;n2 ;:::;nk
k
b21 =2 + b3
b2 + b1
1n1 bn1 2 bn2 3
bnk k 1
1=4:
1, we have
j i s
n1 ; n2 ; : : : ; nk
(11)
are similarly de…ned as in Lemma A.
PROOF. Using (2), we get
1
i
vn+1
1
1
= i
vni
vn
=
1
vni
=
1
vni
=
1
vni
=
vni
1
i
vn+1
8
0
1 i9
>
>
k
1
=
<
X bj
1
1
@
A
+
O
1+ 1+
+
>
>
vn j=1 vnj+1
vnk+1
;
:
8
9
0
1 i
>
>
k
1
<
=
X
1
bj A
1
1 + @1 +
+
+
O
j+1
k+1
>
>
vn j=1 vn
vn
:
;
8
9
1p
0
k
<kX1 ( i)
=
X1 bj
p @ 1
A +O 1
+
:
p!
vn j=1 vnj+1
vnk ;
p=1
kX
1 i
m=1
+O
m
X1
s=0
1
vnk
( i)m s X
m s
(k;m) n1 ; n2 ; : : : ; nk
(m s)!
1n1 bn1 2 bn2 3
bnk k 1
!
1
vnm+i
The formula for aij follows by comparing with (4).
The …rst four aij -terms are
ai;i+1 =
ai;i+3 =
i; ai;i+2 =
i3
+ b1
6
1
2
i2
+
2
1
2
i2 + b1
b1 i;
b2
1
3
i
and
ai;i+4 =
i4
+
24
b1
1
+
2
4
i3 +
b21
+ b2
2
3b1
11
+
2
24
i2 +
b3 +
b21
+ b2
2
b1 +
1
4
i:
82
Asymptotic Expansions of Iterates of Classical Functions
Using (9), (10) and (11), we solve for ci from the system (8) to get
!
i 1
X
1
c0 = b1 and ci =
bi+1 +
at;i+1 ct
(1 i k 2; k
i
t=0
LEMMA D. We have, for k
1
3):
(12)
3,
k
X2
Tm+1
+O
ym
m=1
(y) = y + T1 +
1
yk
(y ! 1)
1
(13)
and
1=t
un =
n
(
1+
X
m 1
1 X ( 1=t)m s
nm s=0 (m s)!
m=1
k
X1
m s
n1 ; n2 ; : : : ; nk
(k 1;m)
T1n1 T2n2
1
for n ! 1; where T1 = X; T2 = b1 X
n
Tk k 1 1
+O
1
(14)
nk+1=t
c1 ;
b1 X 2 =2 + (b21 + c1 )X
T3 =
)
b1 c1
c2 ;
X = b1 log y + C;
C := x
t
b1 log
and, in general, for m
Tm+1 = b1
m
X1
s=0
m
X1
e=1
ce
x
d=1
+ c1
1 t
x +
+ ck
(k 2) (k 2)t
+ O(x(k
T1n1 T2n2
Tk k 1 1
x
2
1)t
)
2
( 1)m 1
m s
m
Xe
t
sX
(k 1;m)
( e)d X
(k
d!
1;m e)
m s
n1 ; n2 ; : : : ; nk
1
d
m1 ; m2 ; : : : ; mk
1
T1m1 T2m2
n
mk
1
Tk
cm
1
;
with the two sums and the special multinomial symbols being similarly de…ned as in
Lemma A.
PROOF. Using (5) and
1
n
1
( (n)) = n, we get
(b1 log n + C) +
c1
+
n
+
ck
nk
2
2
= n;
which yields
1
(y) = y + T1 +
k
X2
Tm+1
+O y
ym
m=1
k+1
(y ! 1):
S. Mavecha and V. Laohakosol
To determine Tm (1
m
k
83
1
2), we use
k
X2
k
X2
Tm+1
b1 log 1 +
+O
y m+1
m=0
Tm+1
y = y + T1 b1 log y C +
ym
m=1
! 1
k
X2 Tm+1
c1
1
+
+O
1+
y
y m+1
yk
m=0
ck
+ k
y
+
2
2
k
X2
1
yk
Tm+1
1+
+O
y m+1
m=0
(y) = y and (5) to get
!
1
yk
!
(k 2)
:
(15)
Substituting
k
X2
Tm+1
log 1 +
+O
y m+1
m=0
=
k
X1
m=1
+O
m
X1
s=0
1
yk
( 1)m 1
m s
1
yk
sX
!
(k 1;m)
m s
n1 ; n2 ; : : : ; nk
T1n1 T2n2
1
n
Tk k 1 1
!
1
ym
and
! p
Tm+1
1
1+
+O
y m+1
yk
p=1
m=0
! p
k
k
X2 Tm+1
X2 cp
1
=
1+
+O
p
m+1
k 1
y
y
y
m=0
p=1
(
!s )
k
k
2
X cp
X2 ( p)s kX2 Tm+1
=
1+
+O
yp
s!
y m+1
s=1
m=0
p=1
(
k
m
X2
X1 m
Xe ( e)d
c1
=
+
cm +
ce
y
d!
m=2
e=1
k
X2
cp
yp
k
X2
1
yk
d=1
X
(k 1;m e)
d
m1 ; : : : ; mk
1
T1m1
m
Tk k1
1
)
1
1
+O
ym
1
yk
1
;
84
Asymptotic Expansions of Iterates of Classical Functions
into (15) we get
1
y = y + (T1 b1 log y C) + (T2 b1 T1 + c1 )
y
(
k
2
m
1
m
1
s
X
X
X ( 1)
m s
+
Tm+1 b1
(k
1;m)
m s
n1 ; : : : ; n k
m=2
s=0
+cm +
m
X1
m
Xe
ce
e=1
1
yk
+O
1
d=1
( e)d X
(k
d!
1;m e)
d
m1 ; : : : ; mk
m
Tk k1
T1m1
1
n
T1n1
1
Tk k 1 1
1
)
1
ym
:
The shape of Tm follows from comparing the coe¢ cient of 1=y m on both sides. Next,
using (6) and (13), we get
1=t
un =
=
=
vn
8
1=t <
n
1=t
=
1+
1+
n
1=t
=
:
(
1=t
(
1+
n
+O n
k
)
1
(n) + O
k
X2
Tm+1
m+1
n
m=0
k
X1
p=1
!
1=t
1
nk
1
1=t
+O n
k
X2
Tm+1
nm+1
m=0
( 1=t)p
p!
k
X1
k
!p
9
=
;
+O n
m 1
1 X ( 1=t)m s X
(k
nm s=0 (m s) !
m=1
1;m)
k
)
m s
n1 ; : : : ; n k
1
T1n1
n
Tk k 1 1
:
As a useful by-product of the explicit forms so derived above, we use them to derive
combinatorial identities which seems di¢ cult to prove by other means.
PROPOSITION 2. Let s; k (
2) 2 N. Then, for j = 1; 2; : : : ; k
j
X
X
( 1)s
(k;j+1)
s=0
and
j 1
X
( 1)j 1
j s
s=0
jX
i 1
s=0
( i)j
(j i
sX
(k;j)
i s
s)!
X
j+1 s
n1 ; n2 ; : : : ; nk
j s
n1 ; n2 ; : : : ; nk
(k;j i)
j i s
n1 ; n2 ; : : : ; nk
1n1 0n2
1n1 0n2
1, we have
= 0;
0nk =
( 1)j+1
j
0nk =
( i)j i
:
(j i)!
S. Mavecha and V. Laohakosol
85
PROOF. Consider a rational function of the form
x
f (x) =
= x Ax2 + A2 x3 + + ( A)k xk+1 + O xk+1
1 + Ax
(A > 0; x 2 (0; 1)):
By direct computation, its iterates are u0 = x0 2 (0; 1); u1 = f (u0 ) =
un = f (un
1)
=
x0
1
=
1 + nAx0
nA
1
+
(nA)2 x0
+
( 1)k
+O n
(nA)k+1 xk0
x0
1+Ax0 ,
k 2
(n; k
2):
Referring to the notation of (7) and Lemma A, from
un+1 = f (un ) = un +
k
X
( A)m um+1
+ O uk+2
n
n
(n ! 1);
m=1
we have t = 1; aj = ( A)j (1
vn+1 vn =
1
Aun+1
1
1
=
Aun
A
j
k);
1
un+1
1
un
comparing with (2), we get bj = 0 (1
the …rst identity follows. Next, using
log
vn+1
vn
= log 1 +
= 1=A: Since
=
j
k
1
vn
=
1
A
1 + (n + 1)Ax0
x0
1 + nAx0
x0
= 1;
1). Putting these values of bj into (9),
k
X1
j=1
( 1)j+1
j vnj
+ O vn k ;
and comparing with (3), we get a0j = ( 1)j+1 =j (1 j k 1). Substituting these
values of a0j into (10), we get the second identity. Since
!
k
i
X1 ( 1)j i 1
1
1
1
1
+ O vn k ;
=
1
+
1
=
i
j
vni
vni
vn
(j
i)!
vn+1
v
n
j=i+1
comparing with (4), we get
aij = ( i)j i =(j
i)!
(1
i
k
2; i + 1
j
k
1):
Putting these values of aij into (11), we get the third identity.
3
Asymptotic Formulas
Our asymptotic expansions of the …ve classical functions are:
THEOREM 1. Let k 2 N; k 3.
I. For f (x) = log(1 + x) (x 2 (0; 1)), we have
fn (x) =
1
2
1+
n
n
+C 2 (x) +
1
1
log n + C(x) + 2
3
n
2C(x) 1
+
+O
3
9
log3 n
n3
1
log2 n +
9
(n ! 1)
2
C(x)
3
2
9
log n
(16)
86
Asymptotic Expansions of Iterates of Classical Functions
with
C(x) =
2 1
+ log
x 3
2
x
+
x
191x2
+
+
36
2160
II. For f (x) = arctan x (x 2 (0; 1)), we have
r
3 log n C(x) 1
3
1+
+
fn (x) =
2n
40
2
n
+
3C 2 (x) 3C(x)
47
+
+
8
40
5600
1
+O
n2
(log)
2
+ ck
k 2
x
2
+ O(xk
27 log n
3200
9
C(x)
80
log3 n
n3
(n ! 1)
9
800
1
):
log n
(17)
with
C(x)
=
3
3
+
log
2x2
20
+O(x2(k
III. For f (x) = sinh
r
3
fn (x) =
1+
n
+
1)
1
3
2x2
+
47x2
x4
+
+
4200 12000
2x2
3
(arctan)
2
+ ck
k 2
):
x (x 2 (0; 1)), we have
3 log n
10
C(x)
2
3C 2 (x) 3C(x)
79
+
+
8
10
700
1
+
n
1
+O
n2
27 log n
200
9C(x)
20
9
50
log3 n
n3
(n ! 1)
log n
with
C(x)
=
3
3
+ log
x2
5
3
x2
+
79x2
1050
11567x4
+
459
(arc sinh)
2
+ ck
+O(x2(k 1) ):
Rx
IV. For f (x) = 0 cos t2 =2 dt (x 2 (0; 1)), we have
r
10
X 1
5X 2
55X
127
1
4
fn (x) =
1
+
+
+O
2n
4 n
32
432
2992 n2
where X =
55 log n
108
C(x)
log3 n
n3
k 2
(n ! 1);
+ C(x),
=
10
2 x4
+
55
log
108
(Fresnel)
2
+ ck
10
2 x4
2 4
x
10
V. For f (x) = tanh x (x 2 (0; =2)), we have
r
3
b1 log n + C(x)
3 log2 n
1
+
fn (x) =
2n
2n
8
+
x2
3
3C(x) log n
4
C(x) 1
+
2
4
b1 +
+
127 2 x4
416 4 x8
+
7480
228500
k 2
+ O(x4(k
log n
2
b2
3C 2 (x)
+
2
8
1
2
1)
):
b21
1
+O
n2
log3 n
n3
(n ! 1);
S. Mavecha and V. Laohakosol
87
where
C(x) =
3
2x2
3
2x2
b1 log
+2b2 + 2b3 ) x4 +
bj =
3
2
j+1
and
aj =
+
1
2
b21 +
(tanh)
2
+ ck
b1
1
+ b2 x 2 +
2
4
2b31
b21 +
b1
3
4b1 b2
k 2
2x2
3
+ O(x2(k
1)
);
j
X
( 2)j+1 s X
j+1 s
(k;j+1) n1 ; n2 ; : : : ; nk
(j
+
1
s)!
s=0
22(j+1) 22(j+1) 1 B2(j+1)
(1
(2(j + 1))!
j
k
ank k
an1 1
1)
with Bk being Bernoulli numbers.
PROOF. I. Here,
f (x) = log(1 + x) =
X ( 1)m xm+1
m+1
m 0
(x 2 (0; 1)) ;
choose x0 su¢ ciently small in (0; ) with 0 <
< 1, and de…ne u0 = x0 ; un =
f (un 1 ) (n 1). Since 0 < log(1 + x0 ) < x0 , by induction, we see that 0 < un < 1
and u1 > u2 >
. Thus, limn!1 un = 0, and
un+1 = log(1 + un ) = un +
k
X
( 1)m um+1
n
+ O uk+2
:
n
m
+
1
m=1
Comparing with (7), we have here
aj = ( 1)j =(j + 1) (1
and so
j
k); t = 1
= 2. By Lemma A, the …rst four explicit bj -terms are
b1 =
1=3; b2 = 1=3; b3 = 3=10; b4 = 3=5:
By Lemma B, the …rst four a0j -terms are
a01 = 1; a02 =
5=6; a03 = 1; a04 =
121=180:
By Lemma C, we have
ai;i+1 =
and
ai;i+3 =
i
3
i; ai;i+2 =
2 ( i)( i
3
2!
1)
i
( i)( i
+
3
2!
+
( i)( i
1)
1)( i
3!
2)
:
88
Asymptotic Expansions of Iterates of Classical Functions
Using (12), we get
c0 = 1=3; c1 = 1=18; c2 = 191=540:
From Lemma D, with k
we have
1
(n)
=
X=3 1=18; T3 = X 2 =6 + X=6 181=540,
3; T1 = X; T2 =
X=3 1=18 X 2 =6 + X=6
X
+
+
n
n2
n3
Tk 2
+ k 2 + O n k+1
n
n 1+
+
181=540
with
X
=
C(x0 )
=
log n
+ C(x0 );
3
2
1
2
x0
191x20
+ log
+
+
+
x0
3
x0
36
2160
+ ck
2
x0
2
k 2
+ O(xk0
1
):
The shape of un = fn (x) as stated in (16) follows from (14) in Lemma D.
Since the remaining four asymptotic expansions are derived via similar arguments
as in Case I, we simply list their explicit expressions for records.
II. For
X
f (x) = arctan x =
( 1)m x2m+1 =(2m + 1);
m 0
we have
t = 2;
un+1 = arctan un = un +
= 3=2;
k
X
( 1)m u2m+1
n
+ O u2k+3
;
n
2m
+
1
m=1
aj = ( 1)j =(2j + 1) (1
b1 =
3=20; b2 = 4=35; b3 =
a01 = 1; a02 =
j
19=175; b4 = 222=1925;
13=20; a03 = 251=420; a04 =
2
ai;i+1 =
=
3551=5600;
3
10i + 13i
i
13i2
; ai;i+3 =
20
6
20
c0 = 3=20; c1 = 47=2800; c2 = 3=16000;
251i
;
420
i; ai;i+2 =
T1 = X; T2 =
C(x0 )
k);
3
3
+
log
2
2x0
20
2(k 1)
+O(x0
47
3X 2
11X
261
; T3 =
+
+
;
2800
40
280
112000
3X
20
3
2x20
+
47x20
x40
+
+
4200 12000
);
X=
3 log n
+ C(x0 )
20
+ ck
2
2x20
3
k 2
S. Mavecha and V. Laohakosol
89
and
1
(n) = n 1 +
+
Tk
nk
X
+
n
2
2
3X
47
20
2800
1
:
nk 1
+O
III. For
1
f (x) = sinh
1
+
n2
t = 2;
1
un = un +
b1 =
= 3;
k
X
( 1)m (2m)! u2m+1
n
+ O u2k+3
;
n
2m (m!)2
2
2m
+
1
m=1
( 1)j (2j)!
(1
22j (j!)2 (2j + 1)
aj =
3=5; b2 = 31=35; b3 =
a01 = 1; a02 =
j
i; ai;i+2 =
10i + 22i
; ai;i+3 =
20
T1 = X; T2 =
3X
5
3
3
+ log
2
x0
5
3
x20
=
2(k 1)
+O(x0
34101=205;
11=10; a03 = 191=105; a04 =
c0 = 3=5; c1 = 79=350; c2 =
C(x0 )
k);
6317=55; b4 =
2
ai;i+1 =
1
+
n3
1
X
( 1)m (2m)! x2m+1
;
22m (m!)2
2m + 1
m=0
x=
we have
un+1 = sinh
3X 2
11X
261
+
+
40
280
112000
3
i
6
8641=74;
11i2
10
191i
;
105
11567=51;
79
3X 2
41X
7489
; T3 =
+
+
;
350
10
70
33
+
79x20
1050
11567x40
+
459
+ ck
2
x20
3
k 2
);
X=
3 log n
+ C(x0 )
5
and
1
(n)
=
n 1+
1
+
n2
+
:
IV. For
f (x) =
X
3X
79
+
n
5
350
Tk 2
1
+ k 2 +O
n
nk 1
Z
0
x
cos
2
t2 dt =
3X 2
41X
7489
+
+
10
70
33
2m
1
X
( 1)m 2
x4m+1 ;
(2m)!(4m
+
1)
m=0
1
n3
90
Asymptotic Expansions of Iterates of Classical Functions
we have
2
= 10=
un+1
; t = 4;
2m
k
X
( 1)m 2
;
= un +
u4m+1
+ O u4k+5
n
n
(2m)!(4m
+
1)
m=1
2j
aj =
( 1)j 2
(1
(2j)!(4j + 1)
j
k);
b1 = 55=108; b2 = 245=1404; b3 = 89=12747;
a01 = 1; a02 = 1=108; a03 =
1=702; a04 =
2
ai;i+1 =
i
i
i
i2
i
; ai;i+3 =
+
+
;
2
108
6
108 702
55=108; c1 = 127=748; c2 = 416=2285;
i; ai;i+2 =
c0 =
T1 = X; T2 =
C(x0 )
=
10
2 x4
0
411=10835;
3
55X
108
55
log
108
127
; T3 =
748
55X 2
657X
+
216
1531
127 2 x40
416 4 x80
+
+
7480
228500
10
2 x4
0
+
X=
55 log n
+ C(x0 )
108
714
;
2659
+ ck
2
2 4
x0
k 2
10
4(k 1)
+O(x0
);
and
1
(n)
X
55X
127
+
n
108
748
Tk 2
1
+ k 2 +O
k
n
n 1
= n 1+
+
V. For
f (x) = tanh x =
we have
aj =
b1 =
55X 2
657X
+
216
1531
714
2659
1
n3
:
1
X
22(m+1) 22(m+1) 1 B2(m+1) x2m+1
;
(2(m + 1))!
m=0
t = 2;
un+1 = tanh un = un +
1
+
n2
= 3=2;
k
X
22(m+1) 22(m+1) 1 B2(m+1) u2m+1
n
+ O u2k+3
;
n
(2(m
+
1))!
m=1
22(j+1) 22(j+1) 1 B2(j+1)
(1
(2(j + 1))!
126B6
+ 675B42 ; b2 =
5
j
153B8
+ 1134B4 B6
14
k);
13500B43 ;
S. Mavecha and V. Laohakosol
b3 =
91
1023B10
11907B62
6885B4 B8
+
+
34020B42 B6 + 253125B44 ;
350
25
14
a01 = 1; a02 = b1 1=2; a03 = b2 b1 + 1=3;
ai;i+1 =
ai;i+3 =
c0 =
C(x0 )
b1 ; c1 =
=
3
2x20
b21 +
b1 log
i; ai;i+2 =
i3
+ b1
6
1
2
1
2
b1 i;
i 2 + b1
1
3
b2
b1
b1
+ b2 ; c2 = 2b31 b21 +
2
3
T1 = X; T2 = b1 X c1 ;
3
2x20
+
1
2
4b1 b2 + 2b2 + 2b3 ) x40 +
X=
i2
+
2
b21 +
+ ck
i;
4b1 b2 + 2b2 + 2b3 ;
b1
1
+ b2 x20 +
2
4
2
2x20
3
2b31
b21 +
b1
3
k 2
2(k 1)
+ O(x0
);
126B6
+ 675B42 log n + C(x0 )
5
and
1
(n) = n 1 +
X
+ b1 X
n
b2 + b21
b1
2
1
+
n2
+
Tk
nk
2
2
+O
1
nk
1
:
Acknowledgment. The …rst author was supported by King Mongkut’s Institute
of Technology Ladkrabang Research Fund.
References
[1] F. Bencherif and G. Robin, Sur l’itéré de sin x, Publ. Inst. Math., 56(70)(1994),
41–53.
[2] N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland, Amsterdam,
1981.
[3] G. Pólya and G. Szegö, Problems and Theorems in Analysis I, Springer, New York,
1972.