Jonathan Sondow a , Diego Marques b
4. Appendix: The infinite power tower functions
We use the Gelfond-Schneider and Hermite-Lindemann Theorems to find algebraic, irrational, and transcendental values of three classical functions, whose analytic properties were studied by Euler [9], Eisenstein [7], and many others.
Definition 4.1. Theinfinite power tower (oriterated exponential)functionh(x)is the limit of the sequence offinite power towers (orhyperpowers)x, xx, xxx, . . ..
For x > 0, the sequence converges if and only if (see [1], Cho and Park [5], De Villiers and Robinson [6], Finch [10, p. 448], and [12])
0.06598. . .=e−e6x6e1/e= 1.44466. . . ,
Algebraic and transcendental solutions of some exponential equations 159 and in that case we write
h(x) =xxx··
·
. By substitution, we see thathsatisfies the identity
xh(x)=h(x). (4.1)
Thusy=h(x)is a solution of the equationsxy=y and, hence,x=y1/y. In other words, g(h(x)) = x, whereg(u) =u1/u foru >0. Replacing xwithg(x), we get g(h(g(x))) =g(x)ifg(x)∈[e−e, e1/e]. Sincegis one-to-one on(0, e], and sincehis bounded above bye(see [12] for a proof) and g([e,∞))⊂(1, e1/e](see Figure 5), it follows that
h(g(x)) =x (e−16x6e), h(g(x))< x (e < x <∞). (4.2) Therefore,his a partial inverse ofg, and is a bijection (see Figure 6)
h: [e−e, e1/e] →[e−1, e].
Figure 6: y=h(x) =xxx··
·
For example, takingx= 1/2 and2in (4.2) gives
(1/4)(1/4)(1/4)··
·
= 1
2, √
2
√2
√2···
= 2, (4.3)
while choosingx= 3yields
√3
3
√3
33
√3···
<3.
Recall that the Hermite-Lindemann Theorem says that if A is any nonzero algebraic number, then eA is transcendental. We claim thatif in addition A lies
160 J. Sondow, D. Marques in the interval (−e, e−1), thenh(eA)is also transcendental. To see this, setx=eA andy =h(x). Then (4.1) yieldseAy=y, and Theorem 1.1 implies y∈T, proving the claim. For instance,
√3
e3
√e√3e··
·
= 1.85718. . .∈T, (4.4)
where the value ofh(√3e)can be obtained by computing a solution tox1/x =√3e, using a starting value ofxbetweene−1ande.
Here is an application of Proposition 2.2.
Corollary 4.2. Given A ∈
e−e, e1/e
, if either An ∈ A\Q for all n ∈ N, or A∈Q\ {1/4,1}, then
AAA··
·
∈T. (4.5)
Proof. From (4.1), we have A1 := 1/A= (1/h(A))1/h(A). The hypotheses imply that A1 satisfies condition (i) or (ii) of Proposition 2.2. Thus 1/h(A)and, hence,
h(A)are transcendental.
For example,h (√
2 + 1)/2
= 1.27005. . .∈Tand (1/2)(1/2)(1/2)··
·
= 0.64118. . .∈T.
It is easy to give an infinite power tower analog to the examples in Section 2 of powersTT ∈Awith T ∈T. Indeed, Theorem 1.2 and relation (4.1) imply that if A∈(A\Q)∩(e−1, e), then
T := 1/AA∈T, TTT··
·
= 1/A∈A. (4.6)
Notice that (4.3), (4.4), (4.5), (4.6) represent the four possible cases(x, h(x))∈ A×A,T×T, A×T,T×A, respectively.
We now define two functions each of which extendshto a larger domain.
Definition 4.3. The odd infinite power tower function ho(x) is the limit of the sequence of finite power towers of odd height:
x, xxx, xxx x
x
, . . . −→ ho(x).
Similarly, theeven infinite power tower functionhe(x)is defined as the limit of the sequence of finite power towers of even height:
xx, xxx x, xxx x
x x
, . . . −→ he(x).
Both sequences converge on the interval0< x6e1/e (for a proof, see [1] or [12]).
Algebraic and transcendental solutions of some exponential equations 161 It follows from Definition 4.3 thathoandhesatisfy the identities
xxho(x) =ho(x), xxhe(x) =he(x) (4.7) and the relations
xhe(x)=ho(x), xho(x)=he(x) (4.8) on(0, e1/e]. From (4.7), we see thaty=ho(x)andy =he(x)are solutions of the equationy=xxy. So isy=h(x), sincey=xy impliesy=xxy.
It is proved in [1] and [12] that on the subinterval [e−e, e1/e] ⊂ (0, e1/e] the three infinite power tower functions h, ho, heare all defined and are equal, but on the subinterval(0, e−e)onlyhoandheare defined, and they satisfy the inequality ho(x)< he(x) (0< x < e−e) (4.9) and are surjections (see Figure 7)
ho: (0, e1/e]→(0, e], he: (0, e1/e]→[e−1, e].
In order to give an analog forho and he to Corollary 4.2 on h, we require a lemma.
Lemma 4.4. AssumeQ, Q1∈Q+. Then
QQQ1 =Q1 (4.10)
if and only if (Q, Q1) is equal to either (1/16,1/2)or (1/16,1/4)or (1/nn,1/n), for somen∈N.
Proof. The “if” part is easily verified. To prove the “only if” part, note first that (4.10) and Theorem 1.2 imply QQ1 ∈Q. Then, writing Q=a/bandQ1 =m/n, where a, b, m, n ∈ N and gcd(a, b) = gcd(m, n) = 1, the FTA implies a = a1n and b = b1n, for some a1, b1 ∈ N. From (4.10) we infer that mb1m = a1na1m and nb1m =bna
m 1
1 .
We show that m = 1. If m 6= 1, then some prime p | m, and hence p | a1. Writem =m′pr and a1 =a2ps, wherer, s ∈Nandgcd(m′, p) = gcd(a2, p) = 1.
Substituting intomb1m =a1na1m, we deduce thatrb1m=sna1m. Sincegcd(a1, b1) = 1, we havea1m|r. Buta1m=a1m′pr > r, a contradiction. Therefore,m= 1.
It follows thata1 = 1, and hence nb1 =b1n. Proposition 3.4 then implies that (n, b1) = (2,4) or(n, b1) = (4,2)or n=b1. The lemma follows.
Proposition 4.5. We have ho(1/16) = 1/4 and he(1/16) = 1/2. On the other hand, if Q∈Q∩(0, e−e] butQ6= 1/16, thenho(Q)and he(Q)are both irrational, and at least one of them is transcendental.
162 J. Sondow, D. Marques
Figure 7: (from [12])x=g(y),y=h(x),y=he(x),y=ho(x)
Proof. Since1/16< e−e, the equation
(1/16)(1/16)y =y
has exactly three solutions (see [12] and Figure 7), namely, y = 1/4, 1/2, and y0, say, where 1/4 < y0 < 1/2. By (4.7) and (4.9), two of the solutions are y=ho(1/16)andhe(1/16). In view of (4.9), eitherho(1/16) = 1/4orho(1/16) = y0. But the latter would imply that he(1/16) = 1/2, which leads by (4.8) to y0 = (1/16)1/2 = 1/4, a contradiction. Therefore ho(1/16) = 1/4. Then (4.8) implieshe(1/16) = (1/16)1/4= 1/2, proving the first statement.
To prove the second, suppose Q1 := ho(Q) is rational. Then (4.7) and Lemma 4.4 imply (Q, Q1) = (1/nn,1/n), for some n ∈ N. Hence QQ1 = Q1. But from (4.8) and (4.9) we see thatQho(Q)=he(Q)> ho(Q), so thatQQ1 > Q1, a contradiction. Therefore, ho(Q) is irrational. The proof that he(Q) 6∈ Q is similar. Now (4.8) and Theorem 1.2 imply that {ho(Q), he(Q)} ∩T6=∅.
Algebraic and transcendental solutions of some exponential equations 163 For example, the numbersho(1/17) = 0.20427. . . and he(1/17) = 0.56059. . . are both irrational, and at least one is transcendental. The values were computed directly from Definition 4.3.
Conjecture 4.6. In the second part of Proposition 4.5 a stronger conclusion holds, namely, that both ho(Q)andhe(Q)are transcendental.
As with Conjecture 3.6, we can give a conditional proof of Conjecture 4.6.
Namely, in view of Proposition 4.5 and the identities (4.7), Conjecture 4.6 is a special case of the following conditional result [15, Theorem 4].
Theorem 4.7. Assume Schanuel’s conjecture and let α 6= 0 and z be complex numbers, with αalgebraic andz irrational. Ifααz =z,thenz is transcendental.
Some of our results on the arithmetic nature of values ofh, ho, andhecan be extended to other positive solutions to the equationsy=xy andy=xxy. As with the rest of the paper, an extension to negative and complex solutions is an open problem (compare [12, Section 4] and [16]).
Acknowledgments. We are grateful to Florian Luca, Wadim Zudilin, and the anonymous referee for valuable comments.
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Jonathan Sondow
209 West 97th Street, New York, NY 10025 USA e-mail: jsondow@alumni.princeton.edu
Diego Marques
Departamento de Matemática, Universidade de Brasília, DF, Brazil e-mail: diego@mat.unb.br
Annales Mathematicae et Informaticae 37(2010) pp. 165–175
http://ami.ektf.hu