On the Brannan s conjecture

Transcription

1 On the Brannan s conjecture arxiv:7.953v [math.cv 5 Oct 7 Róbert Szász Abstract We will prove the Brannan conjecture for particular values of the parameter. The basic tool of the study is an integral representation published in a recent work [3. Introduction We consider the following Mac-Laurin development (xz α ( z = A β n (α,β,xz n, ( n= where α >, β >, x = e iθ, θ [ π,π, and z U. It is easily seen, that the radius of convergence of the series ( is equal to. In [5 the author conjectured, that if α >, β > and x =, then A n (α,β,x A n (α,β,, where n is a natural number. Partial results regarding this question were already proved in [, [, [5, [8. The case β =, α (, is still open. Regarding this case partial results were obtained in [3, [4, [6, [7. We will prove some partial results regarding the case β =, and α (,. In order to do this we will use an integral representation proved in [3, and the fact that the conjecture was proved for n 5 in [6. Keywords: Integral representation, Brannan s conjecture AMS classification: 3C5

2 Preliminaries We inroduce the notation: It is easily seen that A n (α,x = n k= ( α k ( x k k! α( α( α(3 α 4! We denote A n (α,,x = A n (α,x. = α! x α( α! x α( α( α x 3 ( 3! x 4... α( α( α...(n α x n. (n! B(t,θ = cosθttn cosnθ t n cos(n θ, t tcosθ C(t,θ = sinθtn sinnθt n sin(n θ. t tcosθ We need the following lemmas in our study. Lemma. For α (, the following equality holds: Φ(θ = Γ(αΓ( αa n (α,e iθ = = where F (t = t α ( t α, and F( =. [ n α ( t k e kiθ k= [ α B(t,θiC(t,θ, Proof. We use two equalities in order to prove the assertion of the lemma, the first one is the following: A n (α,x = Γ(αΓ( α ( n ( tx k k= k t α ( t α, (3 whichhasbeendeducedin[3,whilethesecondoneisthewell-knownequality B(p,q = tp ( t q = Γ(pΓ(q. Replacing p = α and q = α, we Γ(pq get t α ( t α = Γ(αΓ( α Γ( = αγ(αγ( α = t α ( t α. This equality and (3 imply that Γ(αΓ( αa n (α,x = α tf (t ( n ( tx k k= k F (t,

3 or equivalently Γ(αΓ( αa n (α,x = ( t n F (t α k= ( Now integrating by parts and using that lim t ( and lim t =, we infer that t n α k= ( tx k k Γ(αΓ( αa n (α,x = ( tx k k. t n α k= ( n α ( t k x. k k= We get the desired equality replacing n by n. ( tx k = k Remark. It is easily seen that the condition α (, implies the existence of the integrals in the previous lemma and its proof. We denote [ Φ(θ = Γ(αΓ( αa n (α,e iθ = α cosθttn cosnθt n cos(n θ t tcosθ i sinθtn sinnθt n sin(n θ t tcosθ. (4 Since Φ(θ R it follows that = [ ( Φ(θ = Φ(θΦ(θ = α cosθ ttn cosnθ t n cos(n θ t tcosθ i sinθtn sinnθt n sin(n θ t tcosθ [ ( F(v α cosθv vn cosnθv n cos(n θ v vcosθ i sinθ vn sinnθv n sin(n θ dv v vcosθ [( F(v α cosθttn cosnθt n cos(n θ t tcosθ ( α cosθ vvn cosnθ v n cos(n θ v vcosθ ( sinθt n sinnθt n sin(n θ t tcosθ ( sinθv n sinnθv n sin(n θ. v vcosθ (5

4 Lemma. (a Let f,g : [, R be two continuous function. If there is a point t (, such that f is decreasing on (t,, and the equation g(t = has a unique root t [t,, such that g(t, t [t,, g(t, t [,t, and f(v f(t for v [,t, then we have f(tg(t f(t g(t. (b Let f,g : [, R two continuous functions. If f is a decreasing function, and if there is a point t (, such that g(t, t (,t and g(t, t (t,, then f(tg(t f(t g(t. The statement (b is a particular case of (a. (c A well-known result is the following statement. (Chebyshev s inequality If f and g are monotonic functions with different monotony, then f(tg(t and in case of the same monotony we have f(t g(t Proof. We have f(tg(t f(t g(t. f(tg(t = t f(tg(t t f(tg(t t f(tg(t = t f(tg(t f(tg(t. Lemma 3. If θ [, π, and n 7, then (a ( B(t, B(t,θ ( cosθtn ( cosnθt n ( cos(n θ, (6 (t(t tcosθ (b ( B(t,θ (t(cosθtn (cosnθt n (cos(n θ. (7 t tcosθ

5 Proof. According to Lemma (c, we have t n t t n We use assertion (b of Lemma putting f(t = ln = t n. (8 t tcosθ and g(t = t tn t. If θ [, π, then the mapping f : [, [, is strictly decreasing and we get t tn (t(t tcosθ F(t n ( t t n t ncosθ t tn F(t n ( 3 ln ln >, (9 t t n t n cosθ n where t n denotes the unique root of the equation t tn =, in the interval t (,. The following equality holds: ( B(t, B(t,θ = (tn t n ( cosnθ (t(t tcosθ The equality ( and the inequality (9 imply that ( B(t, B(t,θ t n ( cosnθ (t(t tcosθ ( cosθ (t(t tcosθ (tn t n ( cos(n θ (t(t tcosθ ( t tn ( cosθ. ( (t(t tcosθ ( cosθ (t(t tcosθ tn ( cos(n θ. ( (t(t tcosθ We use assertion (b of Lemma putting f(t = and g(t = t tcosθ t t n t n. If θ [, π, then f is strictly decreasing and we get t t n t n t tcosθ F(t n = t n t ncosθ F(t n t n t n cosθ (t t n t n ( 3 n n >. ( In order to finish the proof of the second inequality, we take notice of the fact that in case θ [, π each member of the following sum is positive: ( B(t,θ = (t(cosθ t tcosθ t n (cosnθ (t(t tcosθ t t n t n t tcosθ tn (cos(n θ t tcosθ tcosθ t tcosθ.

6 Thus we get ( B(t,θ t n (cosnθ (t(t tcosθ and the proof is done. Lemma 4. If θ [ π, π 3 5 tcosθ, then t tcosθ 5 3 Proof. The inequality (3 is equivalent to f(t = 5 ( 65 t t 3 (t(cosθ (t(t tcosθ tn (cos(n θ (t(t tcosθ, (t(cosθ, ( t [,. (3 t tcosθ 7 cosθ cosθ, ( t [,. (4 3 The [ function f has a minimum point at t = 65 π θ, π. Thus we get 3 5 cosθ 7 5 (,, for every f(t f(t = cosθ ( 7 65cosθ, [ π ( t [,andθ 3, π 3 Consequently in order to prove (3 we have to show that cosθ ( 7 65cosθ [ π, ( θ 3, π 3.. Using the notation x = 3 35cosθ 35, we get 5 3 cosθ 5 The equality implies 5 3 cosθ 5 ( 3 35cosθ 6 min [ x ( 3 35cosθ 6 [ π θ, π x 3 3,3 5 min = x x [ 37 3,9 46 [ 37 3, { x } x = min = g x. { x } x { ( 3 g,g ( 3 } 5 ( 3 = 3 [ π 5 5, θ, π 3 where g(x = x x. and consequently (3 holds.,

7 Lemma 5. If θ 5 [ π, π, and n 7, then 3 ( B(t, B(t,θ 7 5 ( cosθtn ( cosnθt n ( cos(n θ (5 (t(t tcosθ ( B(t,B(t,θ 7 (t(cosθtn (cosnθt n (cos(n θ t tcosθ (6 and (t(t tcosθ Proof. We use assertion (b of Lemma, putting f(t = g(t = 7 5 t tn. If θ [ π, π 3 the mapping f : [, [, is strictly decreasing and we get 7 5 t tn (t(t tcosθ F(t n ( 7 (t n (t n t n cosθ 5 t t n F(t n ( 7 > (t n (t n t ncosθ 5 > (7 n where t n denotes the unique root of the equation 7 5 t tn =, in the interval (,. ( B(t, B(t,θ = (tn t n ( cosnθ (t(t tcosθ 3 5 ( cosθ (t(t tcosθ (tn t n ( cos(n θ (t(t tcosθ ( 7 The equality (8 and the inequality (7 imply (5. In order to prove the second inequality, we remark that = ( B(t,B(t,θ = t tn ( cosθ 5 v (8 (t(t tcosθ ( 5 tn t B(t,θ ( 5 = tn t tcosθtn cos(n θt n cos(nθ (t t tcosθ ( 5 tcosθ t tcosθ tn t tn cos(n θt n cos(nθ (t t tcosθ ( 5 = tcosθ t tcosθ tn (cos(n θt n (cos(nθ t tcosθ ( tn t t (t tn (cos(n θt n (cos(nθ t tcosθ (9

8 We put g (t = t t t 3 4t n g (t = t 3 8t n, and f(t = in the assertion (b of Lemma, and we get (t 3 (t tn (cos(n θt n (cos(nθ t tcosθ ( t (t 3tn 3t n ( t t tcosθ (t ( ( tt t 3 t n (t (t 3 ( tt t 3 4t n (t 3 F(t ( tt t 3 4t n ((t 3 F(t 5 = ((t ( 3 [ π >, θ n, π. 3 ( t 6tn t t Finally equality (9, Lemma 4 and inequality ( imply (6, and the proof is done. 3 The Main Result Theorem. If n is a natural number, n 5 and α (, then the following inequality holds A n (α,e iθ A n (α,, for all θ [ π, π. ( Proof. According to (4 and (5 the inequality ( is equivalent to Φ( Φ(θ, θ [ π,π Φ( Φ(θ, θ [ π, π. ( We denote B(t,θ = cosθttn cosnθ t n cos(n θ, t tcosθ C(t,θ = sinθtn sinnθt n sin(n θ. t tcosθ The equality (5 implies that Φ ( Φ(θ = = [( ( F(v α B(t, α B(v, ( ( α B(t,θ α B(v,θ C(t,θC(v,θ dv B(v, B(v,θ B(t,B(v, B(t,θB(v,θ C(t,θC(v,θ dv. [ F(v B(t, B(t,θ α( ( α (3

9 It is easily seen that Φ ( Φ(θ is an even function with respect to θ, consequently we have to prove ( only for θ [, π. Lemma 3, (a implies that in case θ [, π, the following inequality holds (B(t, B(t,θ. Thus we infer that Φ ( Φ(θ = ( ( B(v, B(v,θ B(t, dv [( ( F(v B(t, B(t,θ B(v,θ ( F(v B(t, B(t,θ = ( ( B(v, B(v,θ B(t, dv ( B(t, B(t,θ ( ( B(v, B(v,θ B(t, dv This inequality is equivalent to Φ ( Φ(θ F(vC(t, θc(v, θdv ( B(v, B(v,θ dv [( ( F(v B(t, B(t,θ B(v,θ [( F(v B(t, B(t,θ ( B(t, B(t,θ ( B(t,θ F(vC(t, θc(v, θdv ( F(v B(v, B(v,θ dv ( B(v,θ F(vC(t, θc(v, θdv ( F(v B(v,θ dv ( F(v B(v, B(v,θ dv (4 F(vC(t, θc(v, θdv

10 The inequality between the arithmetic and geometric means leads to ( B(t, B(t,θ ( B(t,θ [ = ( B(t, B(t,θ ( B(t,θ ( F(v B(v,θ dv ( F(v B(v, B(v,θ dv (5 F(vC(t, θc(v, θdv ( B(t,θ ( F(v B(v,θ dv ( F(v B(v, B(v,θ dv F(vC(t, θc(v, θdv ( B(t, B(t,θ ( The Cauchy-Schwarz inequality for integrals implies ( { C(t, θ Lemma 3 implies { ( B(t,θ [( B(t,θ C(t, θ ( B(t, B(t,θ ( B(t, B(t,θ ( C(t, θ [( ( B(t,θ B(t, B(t,θ } (7 t tcosθ ( C(t,θ ( ( cosθtn ( cosnθt n ( cos(n θ (t(t tcosθ (t(cosθt n (cosnθt n (cos(n θ Putting a = ( cosθ, (v(v vcosθ b = (t(cosθ v vcosθ and so an, in the inequality (. C(t,θ a a a 3 b b b 3 a b a b a 3 b 3, } (6

11 we get ( ( cosθtn ( cosnθt n ( cos(n θ (t(t tcosθ (t(cosθt n (cosnθt n (cos(n θ ( On the other hand we have t tcosθ ( C(t,θ (8 (t sinθ tn sinnθ t n sin(n θ (t tcosθ t (. C(t,θ ( (t sinθ tn sinnθ t n sin(n θ (9 (t tcosθ t ( C(t,θ ( (t sinθ t n sinnθ t n sin(n θ (t tcosθ t ( C(t,θ ( sinθ tn sinnθ t n sin(n θ t tcosθ ( sinθ tn sinnθ t n sin(n θ π, θ [, t tcosθ, Finally the inequalities (4, (5, (6, (7, (8 and (9 imply that Φ ( Φ(θ θ [, π, and consequently inequalities ( and ( hold in case θ [ π, π. Theorem. If n is a natural number, n 5 and α (,, then the following inequality holds A n (α,e iθ A n (α,, for all θ [ π 3, π [π, π 3 (3

12 Proof. Equality (3 can be rewritten as follows = ( B(t, B(t,θ ( F(v B(v, B(v,θ dv (Φ ( Φ(θ ( F(v α B(v,B(v,θ dv (3 ( α B(t,B(t,θ F(vC(t, θc(v, θdv. Since (Φ ( Φ(θ defines an even function in order to prove (3 it is enough to show that [ π Φ ( Φ(θ, θ, π. (3 3 The inequality between the arithmetic and geometric means and the condition α (, imply [ (Φ ( ( ( Φ(θ B(t, B(t,θ F(v B(v, (33 B(v, θ dv = ( F(v B(v, B(v,θ dv ( B(t, B(t,θ Lemma 5 and inequality (33 lead to 5 ( B(t,B(t,θ F(vC(t, θc(v, θdv ( B(t,B(t,θ ( C(t, θ Φ ( Φ(θ 5 ( cosθtn ( cosnθt n ( cos(n θ (34 (t(t tcosθ (t(cosθtn (cosnθt n (cos(n θ t tcosθ ( C(t, θ We apply twice the Cauchy-Schwarz inequality jest like in the proof of the previous theorem and we get that in order to prove (34 it is enough to show that Φ ( Φ(θ ( t sinθ t n sinnθ t n sin(n θ (t tcosθ t ( sinθtn sinnθt n sin(n θ t tcosθ

13 This inequality is equivalent to Φ ( Φ(θ ( sinθ t n t tcosθ sinnθ (t tcosθ (35 t t n sin(n θ (t tcosθ t ( sinθ t tcosθ t n sinnθ (t tcosθ and the proof is done. t n sin(n θ [ π (t tcosθ t, θ, π 3 Theorem 3. If n is a natural number, n 7, and α [,, then the 3 following inequality holds. A n (α, A n (α,e iθ, for all θ [ π, π 3 [π 3,π (36 Proof. We will use the Taylor formula with an integral remainder. Let f : ( a,a R be a n times derivabile function, such that f (n is continous. If x ( a,a, then f(x = f( f (! x f ( x... f(n (! xn (n! (n! xn ( t n f (n (xt. Letf bethefunctiondefinedbyf : (, R, f(x = (x α, α (, and we get (x α = α! x α(α! x nα(α...(α n (n! x... α(α...(α n (n! x n ( t n (xt α n. Since the mapping f : U R, f(z = (z α is well defined(we take the principal branch of the multi valued function it follows that the equality (z α = α! z α(α! z nα(α...(α n (n! z... α(α...(α n (n! z n ( t n (zt α n. holds for every z U. The mapping f : U R, f(z = (z α is radially continuous and so we infer that (e iθ α = α! eiθ α(α e iθ... α(α...(α n! (n! e niθα(α...(α n (n! e (n iθ ( t n (e iθ t α n, θ ( π,π.

14 Taking the absolute value, this equality implies that A n (α,e iθ = α! eiθ α(α e iθ... α(α...(α n e (n iθ! (n! α( α( α...(n α ( t n e iθ t α n (n! e iθ α e iθ α α( α( α ( α 3...( α n On the other hand we have α( α( α(3 α 4! ( t n e iθ t α n, θ ( π,π. A n (α, = α! α( α!... α( α...(n α (n! α( α( α 3! α(α Thus in order to prove (36 we have to show that the following inequality holds α(α e iθ α α( α( α ( α 3...( α n ( t n e iθ t α n, θ [ π 3,π. We denote x = cosθ, and the inequality (37 will be equivalent to α(α ( x α α( α( α ( α 3...( α ( n t n ( t tx α, x [,, (37 t tx and this inequality can be rewritten in the following form: ( α x α ( α( α ( α 3...( α n It is easily seen that α ( t n ( t tx t tx α, x [,. (38 ( α( α ( α 3...( α n is decreasing with respect to n and x, and ( t n ( t tx t tx α ( xα α <, for all α (,, and x [,. Thus in order to prove (38 it is enough to prove the inequality 3 ( α( α ( α 3...( α 53 ( α t (39

15 in case α =, that is 4 3 6( 3 ( 3 ( ( 3 7. This inequality holds and the proof is done. 4 Concluding Remarcs The following two corollaries are the proof of the Brannan conjecture in two different particular cases. Theorem, Theorem and the result of [6 imply the following corollary. Corollary. If x C with argx π, and x =, then the inequality 3 holds for every α (,. A n (α,x A n (α, Theorem, Theorem, Theorem 3 and the result of [6 imply the following corollary. This corollary is the solution of the Brannan conjecture in case α [,. Corollary. The inequality A n (α,x A n (α, holds for every α [,, and x C, with x =. 3 Conjecture. Numerical resuls suggest that the inequality α(α ( x α α( α( α ( α 3...( α ( n t n ( t tx t tx α, x [,, (4 holds for every α (, 3. Remark. If the previous conjecture holds, then the conjecture of Brannan holds in case β = and every α (,. References [ D. Aharonov and S. Friedland, On an inequality connected with the coefficient conjecture for functions of bounded boundary rotation, Ann. Acad. Sci. Fenn. Ser. A I 54 (97, 4. MR355 (48, 59 [ R. W. Barnard, Brannans coefficient conjecture for certain power series, Open problems and conjectures in complex analysis, Computational Methods and Function Theory (Valparaso, 989, -6. Lecture notes in Math. 435, Springer, Berlin, 99. MR7758 (9j:

16 [3 R. W. Barnard, Udaya C. Jayatilake, and Alexander Yu. Solynin, Brannan s conjecture and trigonometric sums, Proc. Amer. Math. Soc. Volume 43, Number 5, May 5, Pages 7-8 S (5398- [4 Roger W. Barnard, Kent Pearce, and William Wheeler, On a coefficient conjecture of Brannan, Complex Variables Theory Appl. 33 (997, no. -4, 56. MR64894 (98m:3 [5 D. A. Brannan, On coefficient problems for certain power series, Proceedings of the Symposium on Complex Analysis (Univ. Kent, Canterbury, 973, Cambridge Univ. Press, London, 974, pp. 77. London Math. Soc. Lecture Note Ser., No.. MR44 (54,537 [6 Udaya C. Jayatilake, Brannans conjecture for initial coefficients. Complex Var. Elliptic Equ. 58 (3, no. 5, [7 John G. Milcetich, On a coefficient conjecture of Brannan, J. Math. Anal. Appl. 39 (989, no., 55-5, DOI.6/- 47X(8995-X. MR (9d:36 [8 Stephan Ruscheweyh and Luis Salinas, On Brannans coefficient conjecture and applications, Glasg. Math. J. 49 (7, no., 45-5, DOI.7/S MR (8f:348 address: rszasz@ms.sapientia.ro