International Journal of Analysis and Applications
ISSN 2291-8639
Volume 4, Number 1 (2014), 21-25
http://www.etamaths.com

ON THE GROWTH OF ITERATED ENTIRE FUNCTIONS

DIBYENDU BANERJEE∗ AND NILKANTA MONDAL

Abstract. We consider iteration of two entire functions of (p, q)-order and

study some growth properties of iterated entire functions to generalise some

earlier results.

1. Introduction
For any two transcendental entire functions f(z) and g(z), lim

r→∞
M(r,f◦g)
M(r,f)

= ∞

and Clunie [2] proved that the same is true for the ratio
T(r,f◦g)
T(r,f)

. In [7] Singh

proved some results dealing with the ratios of log T(r,f◦g) and T(r,f) under some
restrictions on the orders of f and g. In this paper, we generalise the results of
Singh [7] for iterated entire functions of (p,q)-orders. Following Sato [6], we write

log[0] x = x, exp[0] x = x and for positive integer m, log[m] x = log(log[m−1] x),
exp[m] x = exp(exp[m−1] x).

Let f(z) =
∞∑
n=0

anz
n be an entire function. Then the (p,q)-order and lower

(p,q)-order of f(z) are denoted by ρ(p,q)(f) and λ(p,q)(f) respectively and defined
by [1]

ρ(p,q)(f) = lim
r→∞

sup
log[p] T(r,f)

log[q] r

and λ(p,q)(f) = lim
r→∞

inf
log[p] T(r,f)

log[q] r
, p ≥ q ≥ 1.

According to Lahiri and Banerjee [4] iff(z) and g(z) be entire functions then the

iteration of f with respect to g is defined as follows:
f1(z) = f(z)
f2(z) = f(g(z)) = f(g1(z))
f3(z) = f(g(f(z))) = f(g2(z))
. . . . . . . . . . . . . . . . . . .
fn(z) = f(g(f(g(....(f(z) or g(z) according as n is odd or even))))
and so are gn(z).
Clearly all fn(z) and gn(z) are entire functions.

The main purpose of this paper is to study growth properties of iterated entire
functions to that of the generating functions under some restriction on (p,q)-orders
and lower (p,q)-orders of f and g.

Throughout we assume f, g etc., are non-constant entire functions having finite
(p,q)-orders.

2010 Mathematics Subject Classification. 30D35.

Key words and phrases. entire functions, iteration, growth.

c©2014 Authors retain the copyrights of their papers, and all
open access articles are distributed under the terms of the Creative Commons Attribution License.

21



22 DIBYENDU BANERJEE AND NILKANTA MONDAL

2. Lemmas
Following two lemmas will be needed throughout the proof of our theorems.
Lemma 1[5]. Let f(z) and g(z) be entire functions. If M(r,g) > 2+�

�
|g(0)| for

any � > 0, then T(r,f ◦g) ≤ (1 + �)T(M(r,g),f).
In particular, if g(0) = 0 then T(r,f ◦g) ≤ T(M(r,g),f)for all r > 0.

Lemma 2[3]. If f(z) be regular in |z| ≤ R, then for 0 ≤ r < R
T(r,f) ≤ log+ M(r,f) ≤ R+r

R−rT(R,f).
In particular if f be entire, then
T(r,f) ≤ log+ M(r,f) ≤ 3T(2r,f).

3. Main Results
First we shall show that if we put some restriction on (p,q)-orders of f and g

then the limit superior of the ratio is bounded above by a finite quantity. The
following two theorems admit the results.

Theorem 1. Let f(z) and g(z) be two entire functions with f(0) = g(0) = 0
and ρ(p,q)(g) < λ(p,q)(f). Then for even n

lim sup
r→∞

log[p+(n−2)(p+1−q)] T(r,fn)

log[q−1] T(2n−2r,f)
≤ ρ(p,q)(f).

Proof. We have by Lemma 1 and Lemma 2

log[p] T(r,fn) ≤ log[p] T(M(r,gn−1),f)
< (ρ(p,q)(f) + �) log

[q] M(r,gn−1), for all large values of r

and � > 0
≤ (ρ(p,q)(f) + �) log

[q−1]{3T(2r,gn−1)}

= (ρ(p,q)(f) + �) log
[q−1] T(2r,gn−1) + O(1).

So, log[p+(p+1−q)] T(r,fn) < log
[p] T(2r,gn−1) + O(1)

< (ρ(p,q)(g) + �) log
[q−1] T(22r,fn−2) + O(1).

Proceeding similarly after (n− 2) steps we get
log[p+(n−2)(p+1−q)] T(r,fn) < log

[p] T(2n−2r,f(g)) + O(1)

≤ log[p] T(M(2n−2r,g),f) + O(1)
< (ρ(p,q)(f) + �) log

[q] M(2n−2r,g) + O(1)

< (ρ(p,q)(f)+�){exp[p−q](log
[q−1](2n−2r))ρ(p,q)(g)+�}+O(1)

(3.1)
for all large values of r

< (ρ(p,q)(f)+�){exp[p−q](log
[q−1](2n−2r))λ(p,q)(f)−�}+O(1)

by choosing � > 0 so small that ρ(p,q)(g) + � < λ(p,q)(f) − �.
On the other hand,

T(r,f) > exp[p−1](log[q−1] r)λ(p,q)(f)−� , for all r ≥ r0
or, log[q−1] T(r,f) > exp[p−q](log[q−1] r)λ(p,q)(f)−� , for all r ≥ r0.
Therefore, from above

log[p+(n−2)(p+1−q)] T(r,fn)

[log[q−1] T(2n−2r,f)
<

(ρ(p,q)(f)+�){exp
[p−q](log[q−1](2n−2r))

λ(p,q)(f)−�}+O(1)
exp[p−q](log[q−1](2n−2r))

λ(p,q)(f)−�
,

for all r ≥ r0.
Hence,



ON THE GROWTH OF ITERATED ENTIRE FUNCTIONS 23

lim sup
r→∞

log[p+(n−2)(p+1−q)] T(r,fn)

log[q−1] T(2n−2r,f)
≤ ρ(p,q)(f) + �.

The theorem now follows since � (> 0) is arbitrary.
Note 1. From the hypothesis it is clear that f must be transcendental.
Theorem 2. Let f and g be two entire functions with f(0) = g(0) = 0 and
ρ(p,q)(f) < λ(p,q)(g). Then for odd n

lim sup
r→∞

log[p+(n−2)(p+1−q)] T(r,fn)

log[q−1] T(2n−2r,g)
≤ ρ(p,q)(g).

The proof of the theorem is on the same line as that of Theorem 1.
If ρ(p,q)(g) > ρ(p,q)(f) holds in Theorem 1 we shall show that the limit superior

will tend to infinity. Now we prove the following two theorems.
Theorem 3. Let f(z) and g(z) be two entire functions of positive lower (p,q)-
orders with ρ(p,q)(g) > ρ(p,q)(f). Then for even n

lim sup
r→∞

log[p+(n−2)(p+1−q)] T(r,fn)

log[q−1] T( r
4n−1

,f)
= ∞.

Proof. We have,
T(r,fn) = T(r,f(gn−1))

≥ 1
3

log M( 1
8
M( r

4
,gn−1) + o(1),f) { see [7], page 100 }

≥ 1
3

log M( 1
9
M( r

4
,gn−1),f)

≥ 1
3
T( 1

9
M( r

4
,gn−1),f)

> 1
3

exp[p−1]{log[q−1] 1
9
M( r

4
,gn−1)}λ(p,q)(f)−�, for all r ≥ r0

= 1
3

exp[p−1]{log[q−1] M( r
4
,gn−1)}λ(p,q)(f)−� + O(1), for all r ≥ r0.

Therefore,

log[p] T(r,fn) > log{log[q−1] M( r4,gn−1)}
λ(p,q)(f)−� + O(1),

= (λ(p,q)(f) − �) log
[q] M( r

4
,gn−1) + O(1).

(3.2)
So, we have for all r ≥ r0
log[p+(p+1−q)] T(r,fn) > log

[p][log M( r
4
,gn−1)] + O(1)

≥ log[p] T( r
4
,gn−1) + O(1)

> (λ(p,q)(g) − �) log
[q] M( r

42
,fn−2) + O(1), using (3.2)

or, log[p+2(p+1−q)] T(r,fn) > log
[p] T( r

42
,fn−2) + O(1)

> (λ(p,q)(f) − �) log
[q] M( r

43
,gn−3) + O(1), using (3.2).

Proceeding similarly after some steps we get

log[p+(n−2)(p+1−q)] T(r,fn) > (λ(p,q)(f) − �) log
[q] M( r

4n−1
,g) + O(1)

> (λ(p,q)(f) − �) exp[p−q](log
[q−1]( r

4n−1
))ρ(p,q)(g)−� + O(1)

(3.3)
for a sequence of values of r →∞.
On the other hand for all r ≥ r0 we have,

T(r,f) < exp[p−1](log[q−1] r)ρ(p,q)(f)+�

or, log[q−1] T(r,f) < exp[p−q](log[q−1] r)ρ(p,q)(f)+�.
(3.4)

So, from (3.3) and (3.4) we have for a sequence of values of r →∞,
log[p+(n−2)(p+1−q)] T(r,fn)

log[q−1] T( r
4n−1

,f)
>

(λ(p,q)(f)−�) exp
[p−q](log[q−1]( r

4n−1
))
ρ(p,q)(g)−�

exp[p−q](log[q−1] r
4n−1

)
ρ(p,q)(f)+�

+ o(1)

and so,



24 DIBYENDU BANERJEE AND NILKANTA MONDAL

lim sup
r→∞

log[p+(n−2)(p+1−q)] T(r,fn)

log[q−1] T( r
4n−1

,f)
= ∞

since we can choose � (> 0) such that ρ(p,q)(g) − � > ρ(p,q)(f) + �.
This proves the theorem.

An immediate consequence of Theorem 3 for odd n is the following theorem.
Theorem 4. Let f(z) and g(z) be two entire functions of positive lower (p,q)-
orders with ρ(p,q)(g) < ρ(p,q)(f). Then for odd n

lim sup
r→∞

log[p+(n−2)(p+1−q)] T(r,fn)

log[q−1] T( r
4n−1

,g)
= ∞.

Next if we consider the ratios
log[p+(n−1)(p+1−q)] T(r,fn)

log[p] T(2n−2r,g)
or

log[p+(n−2)(p+1−q)] T(r,fn)

log[p] T( r
4n−1

,g)

we have obtained the following four theorems.
Theorem 5. Let f(z) and g(z) be two transcendental entire functions with f(0) =
g(0) = 0 and let λ(p,q)(g) > 0. Then for even n

lim sup
r→∞

log[p+(n−1)(p+1−q)] T(r,fn)

log[p] T(2n−2r,g)
≤ ρ(p,q)(g)

λ(p,q)(g)
.

Proof. We get from (3.1), for all large values of r

log[p+(n−2)(p+1−q)] T(r,fn) < (ρ(p,q)(f)+�){exp[p−q](log
[q−1](2n−2r))ρ(p,q)(g)+�}+

O(1)

or, log[p+(n−1)(p+1−q)] T(r,fn) < (ρ(p,q)(g) + �) log
[q](2n−2r) + O(1).

On the other hand,

log[p] T(r,g) > (λ
(p,q)

(g) − �) log[q] r, for all r ≥ r0.
Thus for all r ≥ r0
log[p+(n−1)(p+1−q)] T(r,fn)

log[p] T(2n−2r,g)
<

(ρ(p,q)(g)+�) log
[q](2n−2r)+O(1)

(λ
(p,q)

(g)−�) log[q](2n−2r) .

Therefore,

lim sup
r→∞

log[p+(n−1)(p+1−q)] T(r,fn)

log[p] T(2n−2r,g)
≤ ρ(p,q)(g)

λ(p,q)(g)
.

Hence the theorem is proved.
Theorem 6. Let f(z) and g(z) be two transcendental entire functions with f(0) =
g(0) = 0 and let λ(p,q)(f) > 0. Then for odd n

lim sup
r→∞

log[p+(n−1)(p+1−q)] T(r,fn)

log[p] T(2n−2r,f)
≤ ρ(p,q)(f)

λ(p,q)(f)
.

The proof is omitted.
Theorem 7. Let f(z) and g(z) be two transcendental entire functions of positive
lower (p,q)-orders with ρ(p,q)(g) > 0. Then for even n

lim sup
r→∞

log[p+(n−2)(p+1−q)] T(r,fn)

log[p] T( r
4n−1

,g)
= ∞.

Proof. From (3.3), we have for a sequence of values of r →∞
log[p+(n−2)(p+1−q)] T(r,fn) > (λ(p,q)(f)−�) exp[p−q](log

[q−1]( r
4n−1

))ρ(p,q)(g)−�+
O(1).

Also, log[p] T(r,g) < (ρ(p,q)(g) + �) log
[q] r, for all r ≥ r0.

Thus
log[p+(n−2)(p+1−q)] T(r,fn)

log[p] T( r
4n−1

,g)
≥ (λ(p,q)(f)−�)

(ρ(p,q)(g)+�)

exp[p−q](log[q−1]( r
4n−1

))
ρ(p,q)(g)−�

log[q] r
4n−1

which tends to infinity as r →∞, through this sequence since ρ(p,q)(g) > 0.
Theorem 8. Let f(z) and g(z) be two transcendental entire functions of positive
lower (p,q)-orders with ρ(p,q)(f) > 0. Then for odd n



ON THE GROWTH OF ITERATED ENTIRE FUNCTIONS 25

lim sup
r→∞

log[p+(n−2)(p+1−q)] T(r,fn)

log[p] T( r
4n−1

,f)
= ∞.

The proof is omitted, since it follows easily as in Theorem 7.
Note 2. If we put n = 2, p = q = 1 in the Theorem 1 and Theorem 5 we get the
results of A.P. Singh [7].

References

[1] Walter Bergweiler, Gerhard Jank and Lutz Volkmann, Results in Mathematics 7 (1984),
35-53.

[2] J. Clunie, The composition of entire and meromorphic functions, Mathematical Essays dedi-

cated to A. J. Macintyre, Ohio University Press, 1970, 75-92.
[3] W. K. Hayman, Meromorphic functions, Oxford University Press, 1964.

[4] B. K. Lahiri and D. Banerjee, Relative fix points of entire functions, J. Indian Acad. Math.,

19 (1), (1997), 87-97.
[5] K. Ninno and N. Suita, Growth of a composite function of entire functions, Kodai Math. J.,

3 (1980), 374-379.
[6] D. Sato, On the rate of growth of entire functions of fast growth, Bull. Amer. Math. Soc., 69

(1963), 411-414.

[7] A. P. Singh, Growth of composite entire functions, Kodai Math. J., 8 (1985), 99-102.

Department of Mathematics, Visva-Bharati, Santiniketan-731235, West Bengal, India

∗Corresponding author