SOME RESULTS ON INFINITE POWER TOWERS

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1 NNTDM ) 3, SOME RESULTS ON INFINITE POWER TOWERS Mladen Vailev - Miana 5, V. Hugo Str., Sofia 1124, Bulgaria miana@abv.bg Abtract To my friend Kratyu Gumnerov In the paper the infinite power tower which are generated by an algebraic number belonging to the cloed interval [1, e 1 e ] are invetigated and an anwer i given to the quetion when they are trancendental or rational number. Alo a neceary condition for an infinite power tower to be an irrational algebraic number i propoed. Keyword: Infinite power tower, Algebraic number, Trancendental number Below the following variant of Gelfond-Schneider theorem ee[1]) hall be ued: Theorem 1. If a a 0, 1) i an algebraic number and b i an irrational algebraic number, then a b i a trancendental number. Further we hall ue the denotation x x for x 1 x where x 1 def x = e ln x x ) and a uual e for John Napier number e = Let u note that e e = Firt we need the following: Lemma 1. Let for every real x 1 fx) = x x. Then the function f ha the following propertie: a) f : 1, + ) 1, e e) b) f1) = 1 ; fe) = e e; lim x + fx) = 1 18

2 c) f i a continuou function trictly increaing on the interval 1, e) and trictly decreaing on the interval e, + ) and f ha an abolute maximum at x = e, i.e. for x 1, e) we have: 1 = f1) < fx) < fe) = e e 1) and for x e, + ) we have We omit the elementary proof of thi lemma. e e = fe) > fx) > 1 = lim x fx). A an obviou corollary of the above lemma we obtain: Lemma 2. Let a 1, + ) be a real number. Then: a) For a 1, e e) the equation x x = a 2) ha exactly two different olution x 1 1, e) and x 2 e, + ). b) For a = e e the equation 2) ha the unique olution x = e c) For a > e e the equation 2) ha no olution. Our firt important reult in the paper i ee alo [2, Theorem 2]): Theorem 2. Let a 1, e e) be an algebraic number that cannot be repreented in the form a = b b 3) for any rational number b > 1. Then: a) The equation a x = x 4) ha exactly two different olution: x 1 1, e) and x 2 e, + ); b) x i i = 1, 2) are trancendental number; c) The algebraic number a admit the following two different repreentation, uing the trancendental number x 1 and x 2 : a = x 1 x1 and a = x 2 x2. 5) 19

3 Proof. We note that 4) i equivalent to 2). Hence a) immediately follow from Lemma 2 a). Let x = b be any olution of 4). Then x = b i a olution of 2) too. Therefore, b atifie 3). Hence b i an irrational number becaue of the condition of the theorem. Let u aume that b i an algebraic number. Then Theorem 1 yield that a b i a trancendental number. But a b = b, ince x = b i a olution of4). Hence b i a trancendental number too. The lat contradict to the aumption that b i an algebraic number. Therefore, our aumption that b i an algebraic number i wrong. Hence b i a trancendental number and b) i proved. Now, c) in particular5)) hold from a) and b). The Theorem i proved. Remark 1. If a > 1 i an algebraic number given by 3), then either b i a rational number or b i a trancendental number. Indeed, if we aume that b i an irrational algebraic number, then according to Theorem 1 b b i a trancendental number which mean that a i a trancendental number becaue of 3)) in contradiction to the fact that a i an algebraic number. Let a 1 be a real number. Then we conider an infinite equence {K n a)} n=1 given by K 1 a) = a, K n+1 a) = a Kna), for n 1 6) Definition. If there exit lim n K n a) we denote it by Ka), i.e Ka) def.. = a aa. and we call Ka) infinite power tower generated by a. Let u uppoe that for a given a 1, Ka) exit. Then putting Ka) = x, from 6) after paage to the limit, we obtain: i.e. 4)) Hence, i.e. 2)) and from 1) we obtain Therefore, uing Lemma 2 c), we get: x = a x. a = x x a = x x e e 20

4 Lemma 3. Let a 1 be a real number. Then the neceary condition for the exitence of Ka) i a [1, e e]. Lemma 3 yield Corollary 1. For a > e e the infinite power tower Ka) doe not exit. In [2], with the help of Theorem 1, the following reult i etablihed: Theorem 3. Let a 1, e e) be a real number. Then the infinite power tower Ka) exit, Ka) belong to 1, e) and x = Ka) atifie the equation 2). If a atifie the condition of Theorem 2, then Ka) i a trancendental number. Remark 2. We note that K1) = 1 and K e e) = e. Thu, the quetion that remain to be anwered i what happen when a 1, e e) i an algebraic number which admit the repreentation a = b b, where b > 1 i a rational number. In thi cae, from 4) we obtain: Further we will conider the following two cae: Cae 1 b 1, e). Cae 2 b e, + ). x x = b b 7) Let Cae 1 hold. The following conideration are valid not only for the cae when b i a rational number but alo when b i an arbitrary real number. In thi cae, from 1), 3), Lemma 2 a) and Theorem 3, it follow that. b.. b b Ka) = b b b = b 8) From 8) it i een that in Cae 1 Ka) coincide with the rational number b. Let Cae 2 hold. In thi cae we have to conider two poibilitie for x = Ka) : i) x i a rational number belonging to 1, e). ii) x i an irrational number belonging to 1, e). Let i) hold. Then the equation 7) i atified with rational number x 1, e) and rational number b e, + ). According to [3, problem 124, p.28] all uch rational olution of 7) are given by: x = ; b = 1 + ) 1 ) +1 21

5 = 1, 2, 3,... Therefore, each one of them i obtained for an appropriate integer 1. In thi cae a = 1+ 1 ) ) and Ka) = ) i a rational number. { If b e, + ) i a rational number that doe not belong to the infinite equence 1 ) } , then x, that atifie 7), i not a rational number. Therefore, x atifie ii). =1 Let ii) be fulfilled. Then it follow that x i a trancendental number. Indeed, if we aume that x i an irrational algebraic number, then according to Theorem 1 a x i a trancendental number and ince a x = x, x i a trancendental number too, which contradict to the aumption that x i an algebraic number. So when ii) hold, Ka) i a trancendental number. Thu we proved the following Theorem 4. Let b > 1 be a rational number and a = b b. Then: a) If b 1, e), then the infinite power tower Ka) i a rational number, and moreover Ka) = b b) If b = 1 + ) 1 +1 for ome integer 1, then we have b e, + ), a = 1+ 1 ) ) and the infinite power tower Ka) i a rational number given by: Ka) = ) c) If b e, + ) and b i not a term of the equence power tower Ka) i a trancendental number. { 1 ) } , then the infinite =1 Now by combining the reult from Theorem 3 and Theorem 4 we are ready to formulate the main reult of the paper which give u the anwer what i the nature of the infinite power tower Ka) when a 1, e e) i an algebraic number. Theorem 5. Let a 1, e e) be an algebraic number. Then: a) If a b b for every rational b > 1, then the infinite power tower Ka) i a trancendental number. b) If a = b b for ome rational number b > 1, then: 22

6 b 1 ) if b 1, e), then the infinite power tower Ka) i the rational number b; b 2 ) if b e, + ), then b 21 ) if b = 1 + ) 1 +1 for ome integer 1, then the infinite power tower Ka) i the rational number 1 + ) 1 ; { 1 ) } b 22 ) if b i not a term of the equence , then the infinite power =1 tower Ka) i a trancendental number. Remark 3. Since K1) = 1, the infinite power tower K1) generated by 1 i the rational number 1. Since, K e e) = e, ee Remark 2) and e i a trancendental number, the infinite power tower K e e) = e i a trancendental number. Let a 1, e e) be a trancendental number the cae which i not invetigated in Theorem 5). Then we put Ka) = x and the equality 2) yield that x i not a rational number. Therefore, in thi cae the infinite power tower Ka) i an irrational algebraic number or a trancendental number. Thu we obtain Corollary 2. A neceary condition for the infinite power tower Ka) to be irrational algebraic number i a 1, e e] to be a trancendental number. Remark 4. Thu we ee that if a i an algebraic number then the infinite power tower Ka) can not be irrational algebraic number and if a i a trancendental number then the infinite power tower Ka) cannot be a rational number. A a corollary from the reult in the paper we obtain that = = 2 and for every integer n different from 1, 2 and 4 the infinite power tower i a trancendental number. In particular n n n n n... n i a trancendental number. Finally, in the preent paper an anwer ha been given to the Open Problem from [2]: To decribe all rational number a 1, e) and b e, + ) which are olution of the equation: a a = b b. Namely, all rational olution of the above type ) of the above equation are given by: a = 1 + 1, b = 1 + ) 1 +1, = 1, 2, 3,... ) 23

7 Reference [1] Baker, A. Trancendental Number Theory. London, Cambridge Univerity Pre, [2] Vailev-Miana, M. A hort remark on trancendental number. Note on Number Theory and Dicrete Mathematic, Vol. 14, No. 4, 2008, 1-3. [3] Shkliarki D., N. Chentzov, I. Yaglom. The USSR Olympiad Problem Book: Selected Problem and Theorem of Elementary Mathematic, New York, Dover Publication,