INEQUALITIES INVOLVING INVERSE CIRCULAR AND INVERSE HYPERBOLIC FUNCTIONS

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1 Univ. Beograd. Publ. Elektrotehn. Fak. Ser. Mat , Available electronically at http: //pefmath.etf.bg.ac.yu INEQUALITIES INVOLVING INVERSE CIRCULAR AND INVERSE HYPERBOLIC FUNCTIONS Edward Neuman Inequalities connecting inverse circular inverse hyperbolic functions are established. These results are otained with the aid of an elementary transcendental function which belongs to the family of R-hypergeometric functions discussed in detail in Carlson s monograph []. 1. INTRODUCTION AND NOTATION In this paper we offer several inequalities involving inverse circular inverse hyperbolic functions. The main results are derived from the inequalities satisfied by the R-hypergeometric function R C,. Let x 0 y > 0. Following [] 1.1 R C x, y = 1 0 t + x 1/ t + y 1 dt. It is well-known that R C λx, λy = λ 1/ R C x, y λ > 0, i.e., R C is a homogeneous function of degree 1/ in its variables also that R C x, x = x 1/ 1. R C 0, y = π y y > 0. For later use let us record the following formula { y x 1/ arccosx/y 1/, x < y 1.3 R C x, y = x y 1/ arccoshx/y 1/, x > y 000 Mathematics Subject Classification: Primary 6D07, 33B10 Keywords Phrases: Inequalities, inverse circular inverse hyperbolic functions, R- hypergeometric functions, total positivity, logarithmic convexity. 3

2 Inverse circular inverse hyperbolic functions 33 see [, ]. Other inverse circular inverse hyperbolic functions also admit representations in terms of the R C function [, Ex ] 1.4 arcsinx = xr C 1 x, 1, x arctanx = xr C 1, 1 + x, x R 1.6 arcsinhx = xr C 1 + x, 1, x R 1.7 arctanhx = xr C 1, 1 x, x < 1. Bounds for the inverse circular inverse hyperbolic functions can be obtained using the following inequalities x n + y n R C x, y x n y n 1/3, n 0 see [5, 3.10.] where the sequences {x n } 0 {y n } 0 are generated using the Schwab-Borchardt algorithm x 0 = x, y 0 = y, x n+1 = x n + y n /, y n+1 = x n+1 y n, n = 0, 1,... see [1], []. It has been shown in [5, 3.3] that the sequences {3/x n + y n } 0 {x n yn 1/3 } 0 converge monotonically to the common limit R Cx, y. It is worth mentioning that Carlson s inequalities 61 x 3 1/ 41 x 1/ x < arccosx <, 0 < x < 1 1/ 1 + x 1/6 see, e.g., [4, ] follow from 1.8 with n = used with x := x y = 1. Lower bounds for the function arcsinx see [4, ] can be derived using the first inequality in 1.8 with n = 0, n = 1 x 0 = 1 x 1/ followed by application of 1.4. We omit further details. For later use, let us record three inequalities 1.9 y R C y, x 1 RC x, y 1 RC y, x + y, 1.10 RC x, y R C y, A A y, 1.11 RC x, A RC y, x A = x + y/ which have been established in [6, Theorem 3.1]. The main results of this note are contained in the next section.

3 34 Edward Neuman Our first result reads as follows. Theorem. The following inequalities. MAIN RESULTS.1 arcsinx arctanhx x x 1/ arcsinx x, x < 1 1 x. arcsinhx arctanx 1/ arcsinhx x x x, x R 1 + x hold true. Inequalities.1. become equalities if x = 0. Proof. For the proof of inequalities.1 we shall employ the following one.3 R Cx, y R Cy, x y 1 RC x, y 1/. y x The first inequality in.3 follows from the first inequality in 1.9 while the second one is obtained from the first inequality by interchanging x with y, i.e., by letting x := y y := x. Substituting x := 1 x y = 1 in.3 we obtain the desired result using In order to prove the inequalities. it suffices to use.3 with x := 1 + x y = 1 followed by application of Companion inequalities to.1. are contained in the following. Theorem. Let x < 1. Then.4 arctanhu arcsinx u x 1/ arctanhu u1 u, where u = x. If x R, then.5 arctanv arcsinhx v x 1/ arctanv v1 + v, where v = x 1. Equalities hold in.4.5 if x = 0. Proof. There is nothing to prove when x = 0. Since all members of.4.5 are even functions in x, we will always assume that x > 0. Inequalities.4.5 follow easily from the following one.6 RC x, A 1/ RC y, x RC x, A A, x

4 Inverse circular inverse hyperbolic functions 35 where A = x + y/ is the arithmetic mean of two positive numbers x y. The first inequality in.6 is 1.11 while the second one follows from 1,10 after interchanging x with y. Letting y = 1 x x = 1 in.6 we obtain RC 1, A RC 1 x, 1 RC 1, A A 1/, where A = x. Writing A = 1 u we obtain RC 1, 1 u RC 1 x RC 1, 1 u 1/, 1 1 u. Application of completes the proof of.4. Inequalities.5 can be established in an analogous manner. We use.6 with y = 1 + x, x = 1 to obtain RC1, 1 + v R C 1 + x RC 1, 1 + v 1/, v. Making use of we obtain the desired result. Our next result reads as follows. Theorem 3. The following inequalities arcsinx arcsinx.7 +, x < 1 arctanh x x.8 arcsinhx arcsinhx +, x R arctan x x.9.10 arccosx arccosh1/x arccosx +, x < 1, x 0 1 x arccoshx arccoshx +, x 1 arccos1/x x 1 are valid. Inequalities.7.8 become equalities if x = 0. Equalities hold in.9.10 if x = 1. Proof. Inequalities.7.19 can be regarded as special cases of the inequality.11 RC x, y R C y, x + y x > 0, y > 0 which follows from the second inequality in 1.9. Equality holds in.11 if x = y. In order to prove.7 we put x := 1 x y = 1 in.11 next we use Similarly, letting x := 1 + x y = 1 in.11 applying we obtain the inequalities.8. For the proof of the inequalities.9 we use.11 with y = 1 together with two formulas R C x, 1 = arccosx 1 x

5 36 Edward Neuman.1 R C 1, x = arccosh1/x 1 x x 1 which follow easily from 1.3. If x 1, then R C x, 1 = arccoshx x 1.13 R C 1, x = arccos1/x x 1 Letting y = 1 in.11 next using the last two formulas we obtain the inequalities.10. We shall prove now the following. Theorem 4. If 0 < y 1 x, then..14 arccosh x x 1 arccosy 1 y with equality if x = y = 1. Also, if 0 x 1, then.15 1 x arctanhx 1 + x arctanx with the inequality reversed if 1 < x 0. Inequality.15 becomes an equality if x = 0. Proof. B. C. Carlson J. L. Gustafson [3] have proven a result which in a particular case states that the function R C is strictly totally positive. Thus if 0 x 1 < x 0 < y 1 < y, then R C x 1, y R C x, y 1 < R C x 1, y 1 R C x, y. Letting above x 1 = 0, x = x > 0 next using 1. we obtain.16 y1 R C x, y 1 < y R C x, y. Assume that 0 < y < 1 < x. Putting in.16 y 1 = 1/x, y = 1/y, x = 1 we obtain 1.17 x R C 1, 1x < 1 y R C 1, 1y. Application of.1, with x := 1/x, to the first member of.17 use of.13, with x := 1/y, on the second member of.17 completes the proof of.14. In

6 Inverse circular inverse hyperbolic functions 37 order to establish the inequality.15 we use.16 with y 1 = 1 x, y = 1 + x 0 < x < 1, x = 1, to obtain 1 x R C 1, 1 x < 1 + x R C 1, 1 + x. Making use of we obtain the assertion. The proof is complete. We close this section with the following. Theorem 5. Let ft denote one of the following functions arcsint, arctant, arcsinht, arctanht let x y belong to the domain of ft. If z = x +y /, then the following inequality.18 is valid. fz z fx x fy y Proof. It follows from Proposition.1 in [6] that the function R C, is logarithmically convex in its variables x1 + x.19 R C, y 1 + y R C x 1, y 1 R C x, y x 1 0, x 0, y 1 > 0, y > 0. Letting in.19 x 1 = 1 x, x = 1 y, y 1 = y = 1 next using 1.4 we obtain the desired result when ft = arcsint. The remaining cases can be established in the same way. Using one can establish inequalities similar to.18 when ft = arccos t ft = arccosh t. We omit further details. REFERENCES 1. B. C. Carlson: Algorithms involving arithmetic geometric means. Amer. Math. Monthly, , B. C. Carlson: Special Functions of Applied Mathematics. Academic Press, New York, B. C. Carlson, J. L. Gustafson: Total positivity of mean values hypergeometric functions. SIAM J. Math. Anal., , D. S. Mitrinović: Analytic Inequalities. Springer-Verlag, Berlin, E. Neuman, J. Sándor: On the Schwab-Borchardt mean. Math. Pannonica, , E. Neuman, J. Sándor: On the Schwab-Borchardt mean II. Math. Pannonica, , Department of Mathematics, Received April 11, 006 Mailcode 4408, Southern Illinois University, 145 Lincoln Drive, Carbondale, IL 6901, USA edneuman@math.siu.edu Url address: