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1 Kangweon-Kyungki Math. Jour. 12 (2004), No. 1, pp THE DIFFERENTIAL PROPERTY OF ODD AND EVEN HYPERPOWER FUNCTIONS Yunhi Cho Abstract. Let h e (y), h o (y) denote the limits of the sequences { 2n }, { 2n+1 }, respectively. From these two functions, we obtain a function y = p() as an inverse function of them. Several differential properties of y = p() are induced. 1. Introduction The related problems for hyperpower function.. h(y) = y yy. have been studied by many mathematicians: Euler, Eisenstein, Siedel, etc, because its shape invokes curiosity. Reader can find many information and references (more than one hundred) from the paper of Knoebel [5]. The function h(y) converges for each real y in [e e, e 1 e ] (see [5]), and diverges elsewhere. Also we can naturally etend the convergence problem from real to comple (see [1]). For each real y in (0, e e ), the hyperpower function h(y) diverges but we can introduce limit type two functions h o (y), h e (y). In order to understand the functions h o (y) and h e (y), let s introduce usual notation: 1 y = y, 2 y = y y, 3 y = y yy,. Then we can easily imply that for 0 < y < 1 (1) 1 y < 3 y < 5 y < 7 y < and < 8 y < 6 y < 4 y < 2 y. Received February 10, Mathematics Subject Classification: 26A18, 26A24. Supported by Research Fund 2001 of University of Seoul. The author deeply thanks Professor Young-One Kim for his help in the subject.

2 56 Yunhi Cho Fig. 1 On the other hand, it is not hard to see that if (0, e e ], then 2n+1 y < e 1 < 2n y (see [5]). Thus we can define two functions h o (y) = lim n n y, h n odd e (y) = lim n n y for y in (0, e e ], in fact n even two functions h o (y), h e (y) converge and coincide with the hyperpower function h(y) for each real y in [e e, e 1 e ]. Let s, however, restrict the domain of definition of h o (y), h e (y) to (0, e e ] for convenience sake. yy... The hyperpower function = y induces = y, hence the function y = 1 has the hyperpower function as an inverse function in the interval [e 1, e] (see Fig. 1). Also we can find other eplicit representation for the inverse functions of the remaining part of y = 1 in [3]. The functions h o (y), h e (y), and y = 1 satisfy a formula y y = trivially. And the implicit function y y = is composed of above three functions by easy calculation (see [4], [5]). Now we will define a function y = p() which is defined on 0 < < 1 and eactly composed of two disjoint (but with one common point (, y) = (e 1, e e )) functions

3 The differential property of odd and even hyperpower functions 57 = h o (y) and = h e (y) (see Fig. 1). The function y = p() is our main consideration. Then the following are established in [4]. i) h o (y) and h e (y) are real analytic for y (0, e e ), and h o > 0 and h e < 0 for y (0, e e ). ii) h o (y) and h e (y) are continuous for y [0, e e ], with additional conditions h o (0) = 0, h e (0) = 1, and h o (e e ) = h e (e e ) = e 1. Hence we know that the function p() is continuous for [0, 1] and is analytic for (0, 1 e ) ( 1 e, 1). This paper shows the analyticity at 1 e of p(). The function p() has three singular points 0, 1, 1 e and has good differential formula (3) at non-singular points. So our problem is to find the values of p (0), p (1), p ( 1 e ), p (0), p (1), p ( 1 e ), and p ( 1 e ). The differential coefficients of p() at singular points is not obtained directly, so we have to use indirect methods. In fact, there is a more deep and theoretical approach to the subject for the function y = p() (see [2]). However in this paper, we will use methods as elementary as possible. Now let s see an eplicit representation of p() from [3], that is ( ) ( )..., (0, 1 p() = e ] ( log ( )log ( )e) 1 1, [ e, 1). 2. Analyticity of p() at 1 e The function y = p() is equal to the y y = 0 with y 1 from Introduction. By substitution w = y, the implicit function y y = is changed to a symmetric implicit function w w = (see Fig. 2) which is divided into two functions w = and w = k(). Let s define q(, w) = log wlogw. We know that log(1 + ) = , so we have

4 58 Yunhi Cho Fig. 2 log(a + ) = loga + log(1 + a ) = loga + a 2 2a a 3 4 4a 4 + (a + )log(a + ) = aloga + (1 + loga) + 2 2a 3 6a 2 + Hence we have (2) q(a +, a + v) = ( v)[1 + loga + + v 2a 2 + v + v 2 6a 2 + ] =: ( v)g a (, v). and 4 12a 3. From the equality between w w = and q(, w) = 0, we know that the functions w = k() and v = k( + 1 e ) 1 e are the same to the implicit

5 The differential property of odd and even hyperpower functions 59 functions g 0 (, w) = 0 and g 1 (, v) = 0, respectively. It is trivial that e v (n) (0) = k (n) ( 1 e ). Analytic version of implicit function theorem induces the analyticity of two functions w = k() and y = p() = (k()) 1 at = 1 e. The analytic version of implicit function theorem needs only analytic condition instead of differential condition without change of first derivative condition. In this case, the conditions are satisfied by the following facts that g 1 (, v) is analytic for two variables, v, and e v g 1 (0, 0) = e e 2 0. Also we can find another proof of the analyticity of k() and p() at 1 e. In the paper [2], it was proved that two functions G() and H() are analytic in (0, 1). Those functions G() and H() are, in fact, different representations of k() and p(), respectively. Anyway we conclude that the function y = p() is continuous on [0, 1] and is analytic on (0, 1). 3. The values of p (0), p (1), and p ( 1 e ) The function y = p() satisfies y y = and y 1, then we can induce p () by using z = y y and z = y (logy) 2 1, yy z y = y 1 (1 + logy), so yy (3) p () = z / z y = y y+1 (logy) 2 y (1 + logy ). Hence p () at a given point on the graph y = p() is obtained by above formula, but p (0), p (1), and p ( 1 e ) are not obtained from the formula. In order to get the differential coefficients at 0, 1, 1 e, we will mainly use the following squeezing lemma. The proof is trivial. Lemma 1. Let f, f 1, f 2 be continuous and f 1 f f 2 on (a, b) and f 1, f 2 be differentiable at c, and suppose that f 1 (c) = f 2 (c) and f 1(c) = f 2(c) for some c (a, b). Then f is differentiable at c and f (c) = f 1(c) = f 2(c). We already know that the function y y = can be transformed into w w = by letting w = y. The ± signs in Fig. 2 means positiveness or negativeness of w w at that place, and will be used at the proof of Proposition 7.

6 60 Yunhi Cho Proposition 2. h e(0) =, so p (1) = 0. Proof. From the relation (1), we know even y < 2 y for y (0, 1). Since lim y 0 y y = 1, we set y y (0) := 1, and by the Mean Value Theorem, lim y 0 (y y ) = induces (y y ) (0) =. Hence the function = h e (y) is squeezed by y = 0 and = y y at (, y) = (1, 0), and consequently h e(0) = by the squeezing lemma. Then p (1) = 0 is trivial. Proposition 3. p ( 1 e ) = 0. Proof. In section 2, we know that k ( 1 e ) is the same to the differential coefficient v (0) induced by the implicit function g 1 (, v) = 0. So we e have g + v g dv dv d = 0 and d =0 = g (0, 0)/g v (0, 0) = 1. Hence k ( 1 e ) = 1. Finally p ( 1 e ) = 0 is deduced from p() = k() 1. The remaining thing is to get p (0), this calculation is more difficult than p (1), and p ( 1 e ). We need the following lemma. Lemma 4. Let f be continuous and differentiable on (0, ). If lim 0 f() = 0, lim 0 f() = 1, and lim 0 f () eists with finite value. Then lim 0 f() = 1. Proof. Since f = e (f 1)log, we need lim f 1 0 (log) 1 L Hôpital s law, we have lim 0 f 1 f (f log + f 1 = lim ) (log) 1 0 (log) 2 1 = lim f (f (log) 3 + f (log) 2 ) 0 = 0. By = lim 0 f ( f (log) 3 + f (log) 2 ). It is easy that lim 0 (log) 3 = lim 0 (log) 2 = 0. Also f L Hôpital s law deduces lim 0 = lim 0 2 f, so the two limits take the same finite value. Hence we can conclude lim 0 ( f 1)log = 0. Since The function f() = satisfies the three conditions of Lemma 4, we have the following property;

7 The differential property of odd and even hyperpower functions 61 Corollary 5. lim 0 = 1. Corollary 6. lim ( 1) 0 = 1. Proof. Let f() = 1, then lim 0 f() = 0 is trivially checked, and Corollary 5 induces lim 0 f = 1, and easily lim 0 f = lim 0 (1 + log) = 0. Therefore all conditions of the lemma are satisfied. Let s prove the p (0) = 1. Proposition 7. p (0) = 1. Proof. The positiveness of is obtained by the formula (1) for (0, 1) and trivially for (1, ). So we have 0 (equals when = 1) for all positive. Therefore so y = is an upper squeezing function of y = p() at = 0. A function w = = 4 satisfy w w < 0 for (0, δ) (sufficiently small δ), because we get w w = 3 ( + 2 ) and < 1 for (0, δ), which is induced from lim 0 ( ) = 1 and lim 0 ( ) =. So w = 4 lies on the lower of the decreasing part of w w = around the zero. The fact that 0 < f 1 < f 2 implies f 1 1 < f 1 2 for positive induces that y = w 1 = ( 4 ) 1 is located in the lower part of y = p() when is near 0. We have to show that if y =, then y (0) = 1. By supposing y(0) = lim 0 y() we have y(0) = 0 by corollary 6 or 0 y() p() with p(0) = 0. Then it 1) 0 can be determined that y (0) = lim ( ɛ 0 = 1 by corollary 6. Therefore we can conclude p (0) = 1 by the two squeezing functions y = and y = and the squeezing lemma. Also we can find other methods from [2] for finding of values p (0), p (1), and p ( 1 e ). The formula (2.5) and Proposition 4.4 in [2] show the methods. We know that lim y 0+ h o(y) = 1) from Proposition 2.5 in [2]. This implies lim 0+ p () = 1. And the formulas h e (y) = y ho(y) and lim y 0+ h o(y) = 1 induces lim y 0+ h e(y) = and lim 1 p () = 0. Hence the function p() is C 1 -function on [0, 1].

8 62 Yunhi Cho 4. The values of p (0), p ( 1 e ), p (1), and p ( 1 e ) Now we can induce the second or third derivative of p() at singular points. In particular, the evaluation of p (0) is more difficult than others, so we need a result of paper [2]. In this section, the variable w denotes k() and we simplify g 1 (, v) as g(, v) for convenience sake. e Proposition 8. k ( 1 e ) = 2 3 e. Proof. The value k ( 1 e ) is obtained from g(, v()) = 0 and we can etract the value of v (0) from it. We induce that g + g v v = 0 and g + (2g v + g vv v )v + g v v = 0 from d dg(, v()) = 0 and d 2 d g(, v()) = 0, so 2 v = g + (2g v + g vv v )v g v = 2g v g g vv g v. The values g v (0, 0) = 1 2a, g v(0, 0) = 1 6a, and g 2 (0, 0) = g vv (0, 0) = 1 3a are evaluated by the formula (2). Hence v (0) = 2 2 3a = 2e 3. Proposition 9. p ( 1 e ) = 1 3 e3 e. Proof. We know y = p() = k() 1 = w 1. So y = w 1 and more w wlogw w 2 y = (y w log ) w = y w log w = w 1 w log w 2 = w 1 w log, w + y (w 1 )w (w log)(w + w ) w 2 2. Already we get y ( 1 e ) = 0 by Proposition 3, w ( 1 e ) = 2e 3 by Proposition 8, and w( 1 e ) = 1 e, w ( 1 e ) = 1, y( 1 e ) = e e. Therefore y ( 1 e ) = 1 3 e3 e. Proposition 10. k ( 1 e ) = 2 3 e2.

9 The differential property of odd and even hyperpower functions 63 Proof. The function w = k() is symmetric to the line w =, so its inverse function is = k(w). Inverse function theorem says (f 1 ) (y) = 1 f (), and this formula makes (f 1 ) (y) = 3(f ()) 2 f () f () (f ()) 5. So we get k ( 1 e ) = 3(k ( 1 e ))2 k ( 1 e ) k ( 1 e ) (k ( 1 e ))5. Hence k ( 1 e ) = 2 3 e2. Proposition 11. p ( 1 e ) = 1 3 e4 e. Proof. The function y = w 1 induces y = y w log w =: y α(). By α( 1 e ) = 0 and y ( 1 e ) = 0, we get y ( 1 e ) = y( 1 e )α ( 1 e ). Let s define l() := w log and m() := w, then α() = l() m(). By l( 1 e ) = 0 and m ( 1 e ) = 0, we have α ( 1 e ) = l ( 1 e ) m( 1 e ). So y ( 1 e ) = y(1 e ) l ( 1 e ) m( 1 e ) = y( 1 e ) m( 1 e )(w ( 1 e ) + e2 ) = 1 3 e4 e. For n 4, the values of k (n) ( 1 e ) and p(n) ( 1 e ) also can be deduced by essentially the formula (2) too. But the author cannot find the eact or inductive representations of them. Now we evaluate p (0) and p (1). Proposition 12. p (0) =. Proof. Proposition 2.5 in [2] say h o(0) =. We know (f 1 ) (y) = f () (f ()) 3, so we get p (0) =. Lemma 13. lim 1 ( 1)logw = 0.

10 64 Yunhi Cho Proof. We use L Hôpital s law and w = 1+log 1+logw. Then we have lim 1 1 = lim (logw) 1 1 w(logw)2 w = lim 1 w(logw)2 + w(logw) log = 0. Lemma 14. lim 1 w 1 w = 1. Proof. Since lim 1 w 1 = 1 (by Lemma 13) and lim ( ɛ ) = 1 1 ɛ > 1 for fied ɛ with 0 < ɛ < 1, we get that w 1 < ( 1 1 ɛ ) is satisfied on (δ, 1) for some δ (depending on ɛ) with 0 < δ < 1. We know that (1 ɛ)w < w 1 and < 1 implies w 1 < w. Hence we conclude that 1 ɛ < w 1 w < 1 for (δ, 1). By successive taking 1 and ɛ 0 at three sides of the inequality, we obtain the result. Proposition 15. p (1) =. Proof. From the formula (3), we know p () = p() 1 loglogw w(1+logw). So p p () () = lim 1 1 = lim p()(1 loglogw) 1 ( 1)w(1 + logw). We know that lim 1 loglogw = 0 and lim 1 ( 1)logw = 0 and p() lim 1 w = 1 by Lemma 4.2 in [2] and Lemma 13 and Lemma 14, respectively. Therefore we have p (1) =. Proposition 2.5 and Lemma 4.2 in [2] are used in the proofs of Proposition 12, 15, respectively. Lemma 4.2 is elementary, so Proposition 15 is eventually elementary, but Proposition 12 is not. The author left several unsolved questions for readers. Other unsolved questions related the subject can be found in [2]. Q1. Find the eact or inductive representations for the values of k (n) ( 1 e ) and p(n) ( 1 e ). Q2. Prove that the function y = p() has a unique inflection point. Q3. Prove that the derivative function y = p () is strictly conve.

11 The differential property of odd and even hyperpower functions 65 References 1. I.N. Baker and P.J. Rippon, A note on comple iteration, Amer. Math. Monthly 92 (1985), Yunhi Cho and Young-One Kim, Analytic properties of the limits of the even and odd Hyperpower sequences, (will be appear in Bull. of the Korean Math. Society). 3. Yunhi Cho and Kyeongwhan Park, Inverse functions of y = 1/, Amer. Math. Monthly 108 (2001), J.M. De Villiers and P.N. Robinson, The interval of convergence and limiting functions of a hyperpower sequence, Amer. Math. Monthly 93 (1986), R. Arthur Knoebel, Eponentials reiterated, Amer. Math. Monthly 88 (1981), Department of Mathematics University of Seoul Seoul , Korea yhcho@uoscc.uos.ac.kr