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In this paper, we prove a uniqueness theorem for entire functions sharing values on a finite set. The result extends and improves some theorems obtained earlier by Fang, Zhang-Lin and Zhnag-Xiong. Mathematics Subject Classification 2010: 30D35. Key words: uniqueness, meromorphic function, entire function, share value. 1. Introduction and main results In this paper, we will use the standard notations of Nevanlinna, s value distribution theory as in [3]. Let f be a nonconstant meromorphic function in the whole complex plane C, we set E(a, f ) = {z|f (z) - a = 0, counting multiplicties}, and E(S, f ) = aS E(a, f ), where S denotes a set of complex numbers. Let p be a positive integer. Set Ep (S, f ) = aS {z|f (z) - a = 0, i, 0 < i p, s.t. f (i) (z) = 0}, where each zero of f (z) - a with multiplicity m is counted m times when m p in E(S, f ). The research of authors was supported by NSF of Guangxi China (0728041) and NSF of China (Grant No. 10671109), and the Scientific Research Foundation for the Returned Overseas Chinese Scholars, State Education Ministry. Let f and g be two nonconstant entire functions, n, m, l, t and k be positive integers, we set (1.1) (1.2) F = [f n (f l - 1)t ](k) , G = [g n (gl - 1)t ](k) , Hm = (F m ) (Gm ) (F m ) (Gm ) - , -2 m +2 m (F m ) F -1 (Gm ) G -1 and Sm = {1, , 2 , · · · , m-1 }, where = e m i . Fang [1] proved the following result. Theorem A ([1]). Let f and g be two nonconstant entire functions, and let n, k be tow positive integer with n > 2k + 8. If [f n (z)(f (z) - 1)](k) and [gn (z)(g(z) - 1)](k) share 1 CM, then f (z) g(z). Zhang and Lin [7] improved Theorem A and obtained the following results. Theorem B ([7]). Let f and g be two nonconstant entire functions, and let n, m and k be three positive integers with n > 2k + m1 + 4, and a, b be constants such that |a| + |b| = 0. If [f n (z)(af m (z) + b)](k) and [gn (z)(ag m (z) + b)](k) share 1 CM, then: (i) when ab = 0, f (z) g(z); (ii) when ab = 0, either f (z) tg(z), where t is a constant satisfying tn+m1 = 1, or f (z) = c1 ecz , g(z) = c2 e-cz , where c1 , c2 and c are three constants satisfying (-1)k a2 (c1 c2 )n+m1 {(n+m1 )c}2t = 1, or(-1)k b2 (c1 c2 )n+m1 {(n+m1 )c}2t = 1, when a = 0, m1 = 0, when a = 0, m1 = m. Theorem C ([7]). Let f and g be two nonconstant entire functions, and let n, m and k be three positive integers with n > 2k + m + 4. If [f n (z)(f (z)-1)m ](k) and [gn (z)(g(z)-1)m ](k) share 1 CM, then f (z) g(z), or f and g satisfy the algebraic equation R(f, g) 0, where R(1 , 2 ) = n n 1 (1 - 1)m - 2 (2 - 1)m . Zhang and Xiong [8] improved Theorem B and Theorem B and obtained the following results. Theorem D ([8]). Let f and g be two transcendental entire functions, n, m, t, l, p be positive integers. If E1 (Sm , [f n (f l - 1)t ](p) ) = E1 (Sm , [gn (g l - 6 1)t ](p) ) and n > m + 3tl + 4p, then f (z) bg(z), where bl = 1. In this article, we prove Theorem 1. Let f and g be two transcendental entire functions, n, m, t, l, k and p( 2) be positive integers. If Ep (Sm , [f n (f l - 1)t ](k) ) = Ep (Sm , p+1 4 2tl [gn (gl - 1)t ](k) ) and n > max{ p-1 [ m + 2k + tl + p+1 ], 3t }, then f (z) bg(z), 2 where bl = 1. Remark 1. Under the condition of Theorem 1, let p and m = 1, one can check that the result of theorem 1 is still valid if E(1, [f n (f l - 1)t ](k) ) = E(1, [g n (gl - 1)t ](k) ) and n > max{4 + 2k + tl, 3t }. Note that as p 2 goes to our Theorem 1 includes Theorem A , Theorem B and Theorem C as special cases. We also note that our Theorem 1 together with Theorem D gives the complete solution to the uniqueness problem of entire functions sharing a set of values. 2. Lemmas To prove the theorem, we need the following lemmas. Lemma 1 ([4]). Let f (z) be a nonconstant meromorphic functions and let R(f ) = n ak f k / m bj f j be an irreducible rational function in f k=0 with constant coefficient {ak } and {bj }, where an = 0 and bm = 0. Then T (r, R(f )) = dT (r, f ) + S(r, f ), where d =max{n, m}. Lemma 2 ([6]). Let f (z) be a nonconstant meromorphic function,k be 1 1 positive integer, if f (k) 0, then N (r, f (k) ) N (r, f ) + kN (r, f ) + S(r, f ). Lemma 3 ([6, Second Fundamental Theorem]). Let f (z) be a nonconstant meromorphic function, a1 , · · · , an (n 3) be complex numbers such that when k = j, ak = aj , then (n - 2)T (r, f ) N 1 +N f - a1 - N1 (r) + S(r, f ), r, r, 1 f - a2 + ··· + N r, 1 f - an 1 where N1 (r) = 2N (r, f ) - N (r, f ) + N (r, f ). By second fundamental theorem, we have 1 1 1 ) + N (r, ) + · · · + N (r, ) (n - 2)T (r, f ) N(r, f - a1 f - a2 f - an 1 - N0 (r, ) + S(r, f ), f 1 where N0 (r, f ) is the counting function which only counts those points such = 0 but f = a , k = 1, · · · , n. that f k Lemma 4. Let F , G and Hm be defined as in (1.1) and (1.2), p( 2) be a positive integer. If Ep (Sm , F ) = Ep (Sm , G), and n > k + 2, Hm 0, then m[m(r, 1 1 1 1 ) + m(r, )] N (r, m ) + N (r, m ) F G F G 1 1 - 2(m(n - k) - 2)[N (r, ) + N (r, )] f g 2m + (T (r, F ) + T (r, G)) + S(r), p+1 where S(r) = max{S(r, f ), S(r, g)}. Proof. Since Ep (Sm , F ) = Ep (Sm , G),we have Ep (1, F m ) = Ep (1, Gm ). Suppose that z0 is a common simple zero-point of F m - 1 and Gm - 1. It follows from (1.2) that z0 is a zero-point of Hm . Moreover, we know that the zero-points of F m - 1 and Gm - 1 with multiplicity q( p) are not poles of Hm , the simple pole and simple zero-points of F m or Gm also are not poles of Hm . Thus, we have N1) (r, 1 1 1 ) = N1) (r, m ) N (r, ) Fm - 1 G -1 Hm T (r, Hm ) + O(1) N (r, Hm ) + S(r). Furthermore, by the definition of Hm , we obtain N1) (r, Fm 1 1 1 1 ) = N1) (r, m ) N (2 (r, m ) + N (2 (r, m ) -1 G -1 F G 1 1 + N 0 (r, m ) + N 0 (r, m ) (F ) (G ) 1 1 ) + N (p+1 (r, m ) + S(r) + N (p+1 (r, m F -1 G -1 1 1 1 1 N (2 (r, m )+N (2 (r, m )+N 0 (r, m )+N 0 (r, m ) F G (F ) (G ) m [T (r, F ) + T (r, G)] + S(r). + p+1 (2.1) By the second fundamental theorem, we have m(T (r, F ) + T (r, G)) = T (r, F m ) + T (r, Gm ) N (r, F m ) 1 1 1 ) + N (r, Gm ) + N(r, m ) + N (r, m ) + N (r, m F F -1 G 1 1 1 ) - [N0 (r, m ) + N0 (r, m )] + S(r) + N (r, m G -1 (F ) (G ) 1 1 1 1 ) + N (r, m ) + N(r, m ) = N (r, m ) + N (r, m F F -1 G G -1 1 1 - [N0 (r, m ) + N0 (r, m )] + S(r). (F ) (G ) (2.2) By Lemma 2, we get N (r, (G1 ) ) N (r, G1 ) + S(r). Thus m m N 0 (r, 1 1 1 1 ) + N(2 (r, m ) - N (2 (r, m ) ) + N (2 (r, m m ) (G G -1 G G 1 1 N (r, m ) N (r, m ) + S(r). (G ) G 1 1 1 ) N (r, m ) + S(r). ) + N (2 (r, m m ) (G G -1 G It follows that (2.3) N 0 (r, Similarly, we have (2.4) N 0 (r, 1 1 1 ) + N (2 (r, m ) N (r, m ) + S(r, f ). m ) (F F -1 F From (2.1)-(2.4), we have m(T (r, F ) + T (r, G)) 2[N (2 (r, (2.5) Since N (r, and N(3 (r, 1 1 1 1 1 ) + N (2 (r, m ) N (r, m ) - [N(3 (r, m ) - 2N (3 (r, m )] m F F F F F 1 1 1 ) - 2N (3 (r, m ) [m(n - k) - 2]N (r, ), Fm F f 1 1 1 ) + N (2 (r, m )] + 2[N (r, m ) m F G F 2m 1 + N (r, m )] + [T (r, F ) + T (r, G)] + S(r). G p+1 we have (2.6) N (r, 1 1 1 1 ) + N (2 (r, m ) N (r, m ) - [m(n - k) - 2]N (r, ). Fm F F f 1 1 1 1 ) + N (2 (r, m ) N (r, m ) - [m(n - k) - 2]N (r, ). Gm G G g 1 1 ) - (m(n - k) - 2)N (r, )] Fm f 1 1 + 2[N (r, m ) - (m(n - k) - 2)N (r, )] G g 2m [T (r, F ) + T (r, G)] + S(r), + p+1 Similarly, (2.7) N (r, Combine (2.5)-(2.7), we have m[T (r, F ) + T (r, G)] 2[N (r, thus m[m(r, 1 1 1 1 1 )+m(r, )]N (r, m )+N (r, m )-2(m(n - k) - 2)[N (r, ) F G F G f 2m 1 (T (r, F ) + T (r, G)) + S(r), + N (r, )] + g p+1 which completes the proof of Lemma 4. Lemma 5. Let F , G and Hm be defined as in (1.1) and (1.2), k( 2) be p+1 4 2tl positive integer. If Ep (Sm , F ) = Ep (Sm , G), and n > p-1 ( m +2k+tl+ p+1 ), then Hm 0. Proof. Let F1 = f n (f l - 1)t , G1 = gn (gl - 1)t . Since Ep (Sm , F ) = Ep (Sm , G), we get Ep (1, F m ) = Ep (1, Gm ). If Hm 0, by Lemmas 1 and 4, we have mT (r, F ) + mT (r, G) = T (r, F m ) + T (r, Gm ), m[m(r, 1 ) F 1 1 1 + m(r, )] N (r, m ) + N (r, m ) G F G 1 1 - 2(m(n - k) - 2)[N (r, ) + N (r, )] f g 2m + [T (r, F ) + T (r, G)] + S(r). p+1 (2.8) 7 Since F1 (2.9) (k) = F , G1 = G, thus 1 1 1 1 ) m(r, ) + S(r, f ), m(r, ) m(r, ) + S(r, g). F1 F G1 G (k) m(r, By Lemma 2, we have (2.10) N (r, 1 1 1 1 ) N (r, ) + S(r, f ), N (r, ) N (r, ) + S(r, g). F F1 G G1 Combining (2.8)-(2.10) we have m(1 - 2 1 1 2 1 1 )[m(r, ) + m(r, )] m(1 + )[N (r, ) + N (r, )] p+1 F1 G1 p+1 F1 G1 1 1 - 2(m(n - k) - 2)[N (r, ) + N (r, )] + S(r). f g Thus m(1 - 2 1 1 1 1 )[T (r, ) + T (r, )] 2m[N (r, ) + N (r, )] p+1 F1 G1 F1 G1 1 1 - 2(m(n - k) - 2)[N (r, ) + N (r, )] + S(r), f g we get m(1 - 2 1 1 )(n + lt)[T (r, f ) + T (r, g)] (2mk + 4)[N (r, ) + N (r, )] p+1 f g 1 1 ) + N (r, l )] + S(r) + 2mt[N (r, l f -1 g -1 (2mk + 4 + 2mtl)[T (r, f ) + T (r, g)] + S(r), p+1 4 2tl which contradicts the assumption that n > p-1 ( m +2k+tl+ p+1 ) . Therefore Hm 0, which completes the proof of Lemma 5. Lemma 6 ([6]). Let f be a transcendental entire function, k be a positive integer, and c be a nonzero finite complex number. Then 1 1 1 ) - N (r, (k+1) ) + S(r, f ) T (r, f ) N (r, ) + N (r, (k) f f -c f 1 1 1 Nk+1 (r, ) + N (r, (k) ) - N0 (r, (k+1) ) + S(r, f ), f f -c f where N0 (r, 1/f (k+1) ) is the counting function which only counts those points such that f (k+1) = 0 but f (f k - c) = 0. Lemma 7 ([6]). Let f be a transcendental meromorphic function, a1 and a2 be two meromorphic functions such that T (r, aj ) = S(r, f )(j = 1, 2) and a1 a2 , then T (r, f ) N (r, f ) + N (r, 1 1 ) + N (r, ) + S(r, f ). f - a1 f - a2 Lemma 8 ([2]). Let f and g be two entire functions. If there exists two nonconstant polynomials p and q such that p f (z) = q g(z), then there exists entire function h and rational functions U (z) and V (z) such that f (z) = U h(z), g(z) = V h(z). Lemma 9. Let U and V be two rational functions, n and t be two positive integers such that n > 3t , and set U n (U - 1)t aV n (V - 1)t . If 2 there exists z0 such that U (z0 ) = 0, and z0 is a zero of V -1 with multiplicity q < 4, then U j (U - 1) akV j (V - 1), where j = 2 or j = 3, kt = 1. Proof. Suppose that z0 is a zero of U (z) with multiplicity p, by U n (U - aV n (V - 1)t , we have np = qt. If q = 1, then np = t, which contradicts with n > 3t . 2 If q = 2, then np = 2t. Since n > 3t , we get p = 1, so n = 2t and 2 U 2 (U - 1) akV 2 (V - 1), where kt = 1. If q = 3, then np = 3t. Since n > 3t , we get p = 1, so n = 3t and 2 3 (U - 1) akV 3 (V - 1), where k t = 1. Which completes the proof of U Lemma 9. 1)t 3. Proof of Theorem 1 Let F , G and Hm be defined as in (1.1) and (1.2). By Lemma 5, we have m m (F m ) (Gm ) Hm 0, that is (F m)) - 2 F m -1 (G m)) - 2 Gm -1 . Thus (F (G (3.1) A 1 m + B, Gm - 1 F -1 where A = 0 and B be two constants. Hence E(1, F m ) = E(1, Gm ), T (r, F ) = T (r, G) + S(r, F ). We will prove the theorem by the following four steps. Step I. We claim that (3.2) f n (f l - 1)t ag n (gl - 1)t . To see this, we consider the following two cases. Case 1. When B = 0, by (3.1), we have (3.3) ag n (g l (3.4) F m = AGm + (1 - A). Case 1.1. If A = 1, by (3.3), we have F m = Gm , and hence f n (f l -1)t - 1)t . Case 1.2. If A = 1, by (3.3) we have F m-1 F = AGm-1 G . From (3.3) and (3.4), we get: when F = 0, we have Gm = 0, 1 and G = 0; when G = 0, we have m = 0, 1 and F = 0. Hence F (3.5) N (r, 1 1 1 1 )-N0 (r, m ) = S(r, F ), N (r, )-N0 (r, m ) = S(r, F ). F (G ) G (F ) By the second fundamental theorem, we have T (r, F m ) N (r, F m ) + N (r, + N (r, 1 ) Fm 1 1 ) - N0 (r, m ) + S(r, F ) F m - (1 - A) (F ) 1 1 1 N (r, m ) + N (r, m ) - N0 (r, m ) + S(r, F ) F G (F ) 1 1 1 = N (r, ) + N (r, ) - N0 (r, m ) + S(r, F ). F G (F ) Similarly, we have T (r, Gm ) N (r, Combining (3.5), we get 2mT (r, F ) [N (r, 1 1 ) + N (r, )] + S(r, F ) 2T (r, F ) + S(r, F ). G F 1 1 1 ) + N (r, ) - N0 (r, m ) + S(r, G). G F (G ) Hence m = 1. By (3.3) we get (3.6) f n (f l - 1)t ag n (g l - 1)t + P (z), where P (z) is a polynomial. If P (z) 0, then by (3.6), we get f n (f l - 1)t ag n (gl - 1)t . If P (z) 0, then by (3.6) and Lemma 6, we have T (r, f n (f l - 1)t ) N (r, f n (f l - 1)t ) + N (r, + N (r, 1 ) f n (f l - 1)t 1 ) + S(r, f ) f n (f l - 1)t - P 1 1 = N (r, n l ) + N (r, n l ) + S(r, f ) t f (f - 1) g (g - 1)t 1 1 1 ) + N (r, ) N (r, ) + N (r, l f f -1 g 1 ) + S(r, f ) 2(1 + l)T (r, f ) + S(r, f ), + N (r, l g -1 thus n + tl 2(1 + l), which contradicts the assumption that n > 2tl 2k + tl + p+1 ) . Case 2. When B = 0, by (3.1), we have (3.7) and (3.8) p+1 4 p-1 ( m A 1 F m + ( B - 1) Gm - ( B + 1) 1 A =B , m = -B Gm - 1 Fm - 1 F -1 Gm - 1 Gm-1 G (Gm -1)2 F F = A (F m -1)2 . Thus m-1 Fm + ( A - 1) = 0, B Gm - ( 1 + 1) = 0. B Case 2.1. If A = B, by (3.7), we have F = 0. Since F = (f n (f l -1)t )(k) and n > k, thus f = 0. Let f = e , where is a nonconstant entire function. j j Thus f n (f l - 1)t = en t (-1)t-j Ct elj = t (-1)t-j Ct e(n+lj) . Let j ((-1)t-j Ct e(n+lj) )(k) = Pj ( , , · · · , (k) )e(n+lj) , where Pj ( , , · · · , (k) )(j = 0, 1, 2, · · · , t) are differential polynomials. Thus t t F= Pj ( , , · · · , (k) )e(n+lj) =en Pj ( , , · · · , (k) )elj =en F0 , where F0 = t Pj ( , , · · · , (k) )elj . Obviously, there exists j(0 j t), such that Pj ( , , · · · , (k) ) 0. Suppose P0 ( , , · · · , (k) ) 0. Since F = 0, thus F0 = 0. Since f is a nonconstant entire function, we use Lemma 7 and obtain ltT (r, e ) = T (r, F0 ) N (r, 1 ) F0 1 ) + N (r, F0 ) + S(r, e ) + N (r, F0 - P0 ( , , · · · , (k) ) 1 = N (r, t ) + S(r, e ) , , · · · , (k) )elj Pj ( j=1 1 = N (r, t ) , , · · · , (k) )el(j-1) j=1 Pj ( + S(r, e ) l(t - 1)T (r, e ) + S(r, e ), which is a contradiction. Case 2.2. If A = B and B = -1, by (3.7), we have G = 0. Since G = (gn (gl - 1)t )(k) and n > k, we have g = 0. Set g = e , where is a nonconstant entire function. Similar to Case 2.1, we also have ltT (r, e ) l(t - 1)T (r, e ) + S(r, e ), which is a contradiction. Case 2.3. If A = B and B = -1, we consider the following two subcases. Case 2.3.1. When m > 1, by (3.8) and the second fundamental theorem, we have T (r, Gm ) N (r, 1 1 1 ) + N (r, Gm ) - N0 (r, m ) ) + N (r, m 1 m G (G ) G - ( B + 1) 1 1 + S(r, G) N (r, ) - N0 (r, ) + S(r, G) G G and T (r, F m ) N (r, 1 1 1 ) + N (r, m ) + N (r, F m ) - N0 (r, m ) A Fm (F ) F - (1 - B ) 1 1 + S(r, F ) N (r, ) - N0 (r, ) + S(r, F ), F F thus m[T (r, G) + T (r, F )] T (r, F ) + T (r, G) + S(r, G), which is a contradiction. A Case 2.3.2. When m = 1, by (3.8), we have F + ( B - 1) = 0, thus t )(k) + ( A - 1) = 0. Since f is a nonconstant entire function, we - 1) B use Lemma 6 to obtain 1 ) (n + lt)T (r, f ) = T (r, f n (f l - 1)t ) Nk+1 (r, n l f (f - 1)t 1 ) + N (r, n l A (f (f - 1)t )(k) + ( B - 1) 1 ) + S(r, f ) - N0 (r, n l (f (f - 1)t )(k+1) 1 1 Nk+1 (r, n l ) + S(r, f ) (k + 1)N (r, ) t f (f - 1) f 1 ) + S(r, f ) (k + 1 + lt)T (r, f ) + S(r, f ), + Nk+1 (r, l (f - 1)t (f n (f l 4 thus n k + 1, which contradicts the assumption that n > p+1 ( m + 2k + p-1 2tl tl + p+1 ) . Combining case 1 and case 2, we get (3.2). Step II. By the first step, we claim that if f l gl , then l = 1. By (3.2), we have n n (3.9) f n-1 (f l - 1)t-1 (f l - )f = ag n-1 (gl - 1)t-1 (g l - )g . n + tl n + tl From (3.2) and (3.9), we obtain the following three cases: (i) when f = 0, we get g = 0 or g l = 1, (ii) when f l = 1, we get g l = 1 or g = 0, n n n (iii) when f l = n+tl , we get g l = n+tl or g = 0 (such that g l = n+tl , l = 1). g = 0, g Combining (3.2), (i) and (ii) we have (3.10) (3.11) 1 1 1 t 1 ) - N (r, l , l ) N (r, l ) -1 f -1 g -1 n f -1 t 1 1 1 1 ). N (r, ) - N (r, , ) N (r, l f f g n g -1 N (r, fl Using the second fundamental theorem we have 1 1 1 2lT (r, f ) N (r, ) + N (r, l ) + N (r, l n ) f f -1 f - n+tl 1 + N (r, f ) - N0 (r, ) + S(r, f ) (3.12) f 13 and (3.13) 1 1 1 ) + N (r, l 2lT (r, g) N (r, ) + N (r, l n ) g g -1 g - n+tl 1 + N (r, g) - N0 (r, ) + S(r, g). g If f l gl , then by (3.10)-(3.13),(i)-(iii), we have 1 1 4lT (r, f ) = 2l[T (r, f ) + T (r, g)] + S(r, f ) 2N (r, , ) f g 1 1 1 1 , ) + 2N (r, l + 2N (r, l n , n ) f - 1 gl - 1 f - n+tl g l - n+tl 1 2t 1 2t ) + N (r, l ) + 2N (r, f ) + S(r, f ) + N (r, l (3.14) n f -1 n g -1 2t 1 2t 1 1 ) + N (r, l ) + N (r, l ) + S(r, g) 2N (r, l l f -g n f -1 n g -1 4tl )T (r, f ) + S(r, f ). (2l + n When l 2, by (3.14), we obtain that 2l 4tl , which contradicts the n p+1 4 2tl assumption that n > p-1 ( m + 2k + tl + p+1 ). Therefore, we get l = 1. Step III. We claim that if l = 1, then f g. In fact, we consider the following two cases. Case 1. We shall prove that f g, or there exists positive integer j such that f j (f - 1) agj (g - 1), where j = 2 or j = 3. Since l = 1, by (3.2), we have (3.15) f n (f - 1)t = ag n (g - 1)t . By Lemma 8 and (3.15), then there exists entire function h and rational functions U (z) and V (z) such that f = U (h), g = V (h) and (3.16) and (3.17) U n-1 (U - 1)t-1 (U - n n )U aV n-1 (V - 1)t-1 (V - )V . n+t n+t U n (U - 1)t aV n (V - 1)t Hence T (r, U ) = T (r, V ) + S(r, U ). Since f and g are entire functions, thus U and V are polynomials or be rational functions and only have one common pole. By the second fundamental theorem, we have 2T (r, U ) N (r, (3.18) and 2T (r, V ) N(r, (3.19) 1 1 ) + N (r, ) + N (r, V ) V V -1 1 1 + N (r, n ) - N0 (r, ) + S(r, U ). V - n+t V 1 1 ) + N (r, ) + N (r, U ) U U -1 1 1 + N (r, n ) - N0 (r, ) + S(r, U ) U - n+t U By Lemma 9 and f = U (h), g = V (h), we get f j (f - 1) akgj (g - 1), where j = 2 or j = 3, kt = 1, and - when there exists z0 such that U (z0 ) = 0, we have V (z0 ) = 0 or V (z0 ) = 1 with multiplicity q 3; - when there exists z0 such that U (z0 ) = 1, we have V (z0 ) = 1 or V (z0 ) = 0 and U (z0 ) = 1 with multiplicity q1 3; n n - when there exists z0 such that U (z0 ) = n+t , we have V (z0 ) = n+t or n U (z0 ) = 0 such that V (z0 ) = n+t and U (z0 ) = 0, 1. If U V , by (3.18) and (3.19), we have 2T (r, U ) N (r, 1 1 1 1 , ) + N (r, , ) U V U -1 V -1 1 1 1 1 1 1 + N (r, , ) + N (r, , ) n , n ) + N (r, U - n+t V - n+t U V -1 V U -1 1 1 1 ) + (N (r, ) + N (r, U ) + S(r, U ) N (r, U -V 3 V -1 1 )) + N (r, U ) + S(r, U ) + N (r, U -1 2 (1 + )T (r, U ) + N (r, U ) + S(r, U ). 3 Thus (3.20) T (r, U ) 3N (r, U ) + S(r, U ). Since U is a polynomial or rational function which has only one pole, then by (3.20), we have dU 3, where m1 m2 n1 n2 U (z) = k=0 ak z / bj z , V (z) = k=0 ck z / dj z j , dU =max{m1 , m2 }, dV =max{n1 , n2 }. Combining (3.16), we get dV 3. If there exists z0 , such that U (z0 ) = 0 and V (z0 ) = 1, since dV 3, by Lemma 8 and f = U (h), g = V (h), we get f j (f - 1) akgj (g - 1), where j = 2 or j = 3, kt = 1. If U and V IM 0, by (3.16), we obtain that U and V CM 0. Since U and V CM , there exists constant A such that U AV , hence, we get f Ag. By (3.15), we have A = 1, thus f g. Summarizing the above discussion we obtain f g or there exists positive integer j, such that f j (f - 1) ag j (g - 1), where j = 2 or j = 3, which completes the proof Case 1. Case 2. We shall prove that if U j (U - 1) akV j (V - 1), where j = 2 or j = 3, kt = 1, then f g. By f = U (h), g = V (h) and U j (U - 1) aV j (V - 1), we have (3.21) f j (f - 1) akgj (g - 1). 1 1 Let h1 = f , then by (3.21) we have hj (h1 - g ) = c(1 - g ), thus (h3 - 1 1 g ak)g = hj - ak. 1 If h1 is constant, we have h1 = 1, thus f g. If h1 is nonconstant, we have g = tion that g be transcendental entire function. Step IV. If f l gl , then there exists constant b, such that f bg, where bl = 1. Summarizing the above discussion we obtain the proof of Theorem 1. hj -ak 1 , hj -ak 1 which contradicts the assump-
Annals of the Alexandru Ioan Cuza University - Mathematics – de Gruyter
Published: Nov 24, 2014
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