Bi Unique Range Sets-A Further Study

abstract: The purpose of the paper is to obtain a new bi-unique range sets, as introduced in [4] with smallest cardinalities ever for derivative of meromorphic functions. Our results will improve all the results in connection to the bi-unique range sets to a large extent. Some examples have been exhibited to justify our certain claims. At last an open question have been posed for future investigations.

Definition 1.2.[13] Let S be a set of distinct elements of C ∪ {∞} and k be a nonnegative integer or ∞.We denote by E f (S, k) the set ∪ a∈S E k (a; f ).
Recently to study the possible answer of Question A the present first author [4] have introduced the notion of bi unique range sets for entire or meromorphic function with weight p, m as follows : Definition 1.3.[4] A pair of finite sets S 1 and S 2 in C is called bi unique range sets for meromorphic (entire) functions with weights p, m if for any two non-constant meromorphic (entire) functions f and g, E f (S 1 , p) = E g (S 1 , p), E f (S 2 , m) = E g (S 2 , m) implies f ≡ g.We write S i 's i = 1, 2 as BURSMp, m (BURSEp, m) in short.As usual if both p = m = ∞, we say S i 's i = 1, 2 as BURSM (BURSE).
In [4] the present first author manipulated the above definition in order to get the possible answer of Question A for two finite sets in C, which significantly improved the results obtained in [20] and [19].Below we are recalling the result in [4].The purpose of the paper is to investigate this fact.
It is to be observed that in [4] we were unable to diminish the cardinalities of the range sets as mentioned in [19].So it is natural to ask the following question.Question 1: Can there exists any pair of range sets in the sense of Definition 1.3 whose cardinalities(s) are less than that given in Theorems A-B ?
Possible answer of the above question is the motivation of the paper.We shall show that if we take the set sharing problem of derivatives of meromorphic functions, in stead of the original functions, a pair of range sets with cardinalities 2 and 3 different from those used in Theorems A-B provide the answer of Question 1. Till date this is the best result obtained in terms of bi-unique range sets.
Throughout the paper for an integer n and a nonzero constant a we shall denote . Below we are giving our main theorem.
, where n(≥ 3) be an integer and a and b be two nonzero constants such that b = β, β 2 .Then S i 's i = 1, 2 are bi-unique range sets with weights 1 and 3 for f (k) and g (k) .
The following example shows that in Theorem 1.4 a = 0 is necessary.
From the following example we see that if in our main result we discard −a n−1 n in S 1 and replace f (k) and g (k) simply by f and g then the conclusion ceases to hold.In other words, the presence of the element −a n−1 n in S 1 is essential in that case.
Example 1.6.Let S 1 = {0}, S 2 = {z : z 3 + az 2 + b = 0} where a = 0, b be so chosen that S 2 has distinct elements.Let f and g be two non-constant meromorphic functions such that f (z) = −a e z +e 2z 1+e z +e 2z , g(z) = −a 1+e z 1+e z +e 2z .Then they share So natural question would be whether the cardinality of the set S 1 in Theorem 1.4 can further be diminished ?
It is seen from the next example that the sets S i , (i = 1, 2) in Theorem 1.4 can not be replaced by two arbitrary sets.
Example 1.7.Let f (z) = e z and g(z) = (−1) k αe −z and for a constant α = 0, 1  2 , 1 we take Though for the standard definitions and notations of the value distribution theory we refer to [10], we now explain some notations which are used in the paper.
Definition 1.8.[12] For a ∈ C ∪ {∞}we denote by N (r, a; f |= 1) the counting function of simple a points of f .For a positive integer m we denote by N (r, a; f |≤ m)(N (r, a; f |≥ m)) the counting function of those a points of f whose multiplicities are not greater(less) than m where each a point is counted according to its multiplicity.
N (r, a; f |≤ m) (N (r, a; f |≥ m)) are defined similarly, where in counting the a-points of f we ignore the multiplicities.
Also N (r, a; f |< m), N (r, a; f |> m), N (r, a; f |< m) and N (r, a; f |> m) are defined analogously.Definition 1.9.[14] We denote by Definition 1.10.[13,14] Let f , g share a value a IM.We denote by N * (r, a; f, g) the reduced counting function of those a-points of f whose multiplicities differ from the multiplicities of the corresponding a-points of g.Clearly N * (r, a; f, g) ≡ N * (r, a; g, f ) and in particular if f and g share (a, p) then counting function of those a-points of f , counted according to multiplicity, which are not the b i -points of g for i = 1, 2, . . ., q.

Lemmas
In this section we present some lemmas which will be needed in the sequel.Let F and G be two non-constant meromorphic functions defined in C as follows ) where n(≥ 2) and k are two positive integers and for a meromorphic function h we put P (h) = (h) n + a(h) n−1 .Henceforth we shall denote by H and Φ the following two functions Lemma 2.2.Let S 1 and S 2 be defined as in Theorem 1.4 and F , G be given by (2.1).If for two non-constant meromorphic functions f and where N 0 (r, 0; f (k+1) ) is the reduced counting function of those zeros of f (k+1) which are not the zeros of ) is similarly defined.
Proof: We note that −b and Clearly F and G share (1, 0).Since H has only simple poles, the lemma can easily be proved by simple calculation.✷ Lemma 2.3.[6] Let f and g be two meromorphic functions sharing (1, m), where Lemma 2.4.[18] Let f be a non-constant meromorphic function and let be an irreducible rational function in f with constant coefficients {a k } and {b j }where a n = 0 and b m = 0 Then where d = max{n, m}.
Lemma 2.5.Let S 1 and S 2 be defined as in Theorem 1.4 with n ≥ 3 and F , G be given by (2.1).If for two non-constant meromorphic functions f and g ).
Proof: By the given condition clearly F and G share (1, m).Also we see that

✷
Lemma 2.6.Let S 1 , S 2 be defined as in Theorem 1.4 and F , G be given by (2.1).
If for two non-constant meromorphic functions f and ).
Bi Unique Range Sets -A Further Study
Using Lemmas 2.1, 2.2, 2.3 and 2.4 we note that Using (2.5) in (2.4) and noting that the lemma follows.✷ Lemma 2.7.Let f (k) , g (k) be two non-constant meromorphic functions such that k) , where n (≥ 2) is an integer, k is a positive integer and a is a nonzero finite constant.
Proof: Let z 0 be a zero of f (k) (g (k) ).Then z 0 must be either a 0-point or a c 1 -point of g (k) (f (k) ).But from the given condition if z 0 is not a zero of g (k) , then it must be a zero of g (k) + a, which is impossible.So we conclude that here f (k) and g (k)  share (0, ∞) and f , g share (∞, ∞).We also note that Θ ∞; Now the lemma can be proved in the line of proof of Lemma 3 [16].✷ Lemma 2.8.Let F , G be given by (2.1) and they share (1, m).Also let ω 1 , ω 2 . . .ω n are the members of the set S 2 as defined in Theorem 1.4.Then Proof: First we note that since S 1 has distinct elements, c 1 can not be a member of S 2 .So Lemma 2.10.Let S 1 , S 2 be defined as in Theorem 1.4 with n ≥ 3 an integer.If for two non-constant meromorphic function f and g, E ).
Proof: Using Lemma 2.5 for p = 0 and Lemma 2.8 we get N r, 0; From above the lemma follows.✷ Bi Unique Range Sets -A Further Study  k) ] + S(r, f (k) ) + S(r, g (k) ).
(3.1) gives a contradiction for n ≥ 3. Subcase 1.2 Let H ≡ 0. Then where A = 0, B are constants.Also T (r, In view of Lemma 2.9 it follows that F and G share (1, ∞).We now consider the following cases.Subcase 1.2.1.Let B = 0. From (3.2) we get i.e., Φ ≡ 0, where g (k) − α i 's (i = 1, 2, . . ., n − 2) are the distinct simple factors of G − B−A B .Since f (k) , g (k) share {0, c 1 } and F , G share (1, ∞) it follows that c 1 points of g (k) can not be a pole of f and so it must be an e.v.P. of g (k) .Therefore α i 's are neutralised by the poles of f .Now if z 0 is a zero of g (k) − c 1 of order p, then it would be pole of f (k) of order q such that p = nq ≥ n(k + 1).So in view of the second fundamental theorem and (3.3) we get where g Using the same argument as in Subcase 1.2.2.1.1.we get that 0 is an e.v.P. of g and N (r, −a; g (k) ) ≤ 1 n(k + 1) T (r, f (k) ).
So by the second fundamental theorem and (3.3) we get T (r, g (k) ) ≤ N (r, −a; g (k) ) + N (r, 0; g (k) ) + N (r, ∞; g) + S(r, g (k) ) T (r, g (k) ) + S(r, g (k) ), a contradiction for n ≥ 3. Case 2. Suppose that Φ ≡ 0. On integration we get for some nonzero constant A.Here also in view of Lemma 2.4, (3.3) holds.Since by the given condition of the theorem E f (S 1 , 0) = E g (S 1 , 0), we consider the following cases.Subcase 2.1.Let us first assume f (k) and g (k) share (0, 0) and (c 1 , 0).If one of 0 or c 1 is an e.v.P. of both f (k) and g (k) , then we get A = 1 and we have F ≡ G, which in view of Lemma 2.7 implies f (k) ≡ g (k) .If both 0 and c 1 are e.v.P. of f (k)