iterated exponentials

Iterated ExponentialsAuthor(s): Joel AndersonSource: The American Mathematical Monthly, Vol. 111, No. 8 (Oct., 2004), pp. 668-679Published by: Mathematical Association of AmericaStable URL: http://www.jstor.org/stable/4145040 .

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Iterated Exponentials Joel Anderson

1. INTRODUCTION. Fix a > 0 and consider the sequence

a, aa (aa"), a(a(aa))9 9

which we call the sequence of iterated exponentials generated by a. A natural problem is to determine the values of a for which this sequence converges. This question, which was first considered in the eighteenth century, is remarkable both for its unexpected answer and for the different threads that interweave in the development of its solution.

The problem was solved first by Euler [7], who referred to the work of Condorcet [4] as his motivation. Since then solutions have also been published by Eisenstein [6] and by several modern authors (including Barrow [1], Creutz and Sternheimer [5], Knoebel [10], and Linger [11]). It is notable that among these later authors, only Knoebel seemed to be aware of the previous solutions.

This story ostensibly began with a letter from Daniel Bernoulli to Christian Gold- bach dated 29 June 1728 that concerns an apparently unrelated question [2, p. 262]. The end of the letter translates as follows:

I will finish with a problem which I found very interesting and which I have solved. Here it is: Find two unequal numbers x and y such that xy = yx. There is only one case when these numbers are integers, namely x = 2 and y = 4 (since 24 = 42), but one can find an infinity of broken numbers which satisfy the equation. There are also other types of quantities about which I will say nothing.

(Apparently the phrase "broken numbers" refers to rational numbers.) Six months later Goldbach responded [8], including in his reply a derivation of parametric representa- tions for x and y. His argument may be stated as follows. Suppose xy = yx and y > x so that y = sx for some s > 1. Then x'x = (sx)x, which says that x" = sx and thus that

X - s/(-) y - sx =- s'/(-1)

These relations have been rediscovered by a number of later authors (including Euler [7, secs. 13 and 14]), none of whom seemed to have knowledge of Goldbach's original solution. Although it may not be apparent at this point, Bernoulli's problem is related to the determination of the domain of convergence of the sequence of iterated exponentials. It is therefore useful to introduce the following terminology: if x and y are distinct positive numbers such that xx = yY, then we say the set {x, y } is a Bernoulli pair.

In the next three sections we present a solution to the problem of convergence of iterated exponentials, along with some historical notes. The arguments used here are elementary in the sense that they require only a knowledge of freshman calculus. In section 2 we show that, if the sequence converges to b, then a = bl /b. The case when a > 1, which is relatively straightforward, is treated in section 3. When a < 1, the analysis is far more interesting. It is here that the unexpected twists occur. This case

668 @ THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 111



is studied in section 4. Some of the material in this section seems to be new-or at least unpublished. The paper concludes with some comments and additional historical notes.

2. THE FUNCTION f(x) = x)- x. We write al") = a, a{2} _= a" and recursively de- fine a[n") by

a{n} a(aln-1))

when n > 2. We also put

lim a In) = b, ----> OO

provided this limit exists.

Theorem 2.1. If lim,,,, at") = b, then ab = b and a = bl/b.

Proof If limsf00 ant) = b, then we have

ab a(lim a{n}) lim aafn) = lim a n+1) = b. /7---00 --+00o

The second equality is then clear. U

Hence the function given by x -+ lim,,,0 xth) is inverse to the function f(x) =

x 1/x on its domain, which suggests that it would be useful to study the behavior of f. Since f (0) = 0 and

1 f'(x) =

1 (1 - log x)xllx x

it follows that f is increasing on [0, e] and decreasing on [e, ~o), implying that f has an absolute maximum at x = e. Thus we have

0 < f(x) < f(e) =- el/e a 1.4447

for each nonnegative x, and we see that f is one-to-one on each of the intervals (0, e] and [e, oo). Because lim,,,0 x1/

-- 1, f decreases from e to 1 as x increases to infin-

ity. Therefore, for each x in (1, e) there is a unique y in (e, oc) such that f (x) = f(y).

Further, since x' = yx if and only if x /x --

yl/y, the set {x, y} constitutes a Bernoulli pair, as defined in the introduction. These facts are recorded in the following lemma.

Lemma 2.2. The range of the function f(x) = x lx is the interval [0, el/e]. It is increasing on [0, e], is decreasing on [e, oc), and has an absolute maximum at x = e. If 1 < x < e, then there is a unique y with y > e such that f(x) = f(y). Further, the set {x, y} is a Bernoulli pair if and only if x A y and f (x) = f (y).

The graph of f is shown in Figure 1.1 Note that, since 2 is the only integer that lies between 1 and e, this analysis shows that the set {2, 4} is the only Bernoulli pair with integer values, just as Bernoulli asserted in his letter to Goldbach.

1Although the graph appears to coincide with the x-axis for x < .2, this is not the case. This apparent anomaly is the result of the fact that if x < .2, then x1/x < .00035, which is smaller than the resolution available on most printers and computers.

October 2004] ITERATED EXPONENTIALS 669



el/e

f (x) = x'I/

0 1 2 e 3 4

Figure 1.

Recall that Goldbach showed that if xy = yX, then we may write x = s'/("-1) and y = ss/(s-I) for some s > 1. Euler derived Goldbach's parameterizations in sections 13 and 14 of [7], and in section 17 he noted that if we put

I

s-= +-,

n

then we get

(n + 1)n (n + 1)n+' x nn nn+!

Accordingly, x and y are rational. Hence, as Bernoulli asserted, there are an infinite number of Bernoulli pairs with rational values. Hurwitz showed in [9] that every Bernoulli pair with rational values has this form.

3. THE CASE a > 1. Let us now return to the problem of determining the domain of convergence for the sequence of iterated exponentials. Since this sequence can only converge to a value in the range of x /X, its domain of convergence must lie in the interval [0, e /e] (Lemma 2.2). Moreover, I In

-- 1 for all n, so the sequence of iterated

exponentials converges at a = 1. Because aIn) > 1 if a > 1, a natural first guess might be that the sequence does not converge for points larger than one. This is incorrect, as we now see.

Theorem 3.1. If a > 1, then the sequence of iterated exponentials converges if and only if a < el •e. If I < a < el/" and

limn,,,• = a"I = b (so that a = bl/b), then 1 <

b < e.

Proof. Fix a with a > 1. As noted, if the sequence converges, then a < el/e (i.e.,

the sequences diverges if a > el/e). Suppose that 1 < a < e1/e. If 0 < p < q, then




aP < aq. Thus, if we use the fact that a > 1 and repeatedly exponentiate this inequality by a, we get 1 < a, a < a, aa121 < a{3}, ..., and it follows by induction that

an) < n+l1

for each positive integer n. Hence the sequence of iterated exponentials is increasing in this case.

Next observe that

ax < (el/e)x < (el/e)e = e

when 1 < x < e. Since a < e /e < e, we may apply this observation with x = a to get aa < e. A similar argument shows that a{3} < e, and a simple induction argument now establishes that a{I) < e for each n. We conclude that, if 1 < a < el/e, the sequence of iterated exponentials is increasing and bounded, so it converges to some limit b. By Theorem 2.1 we have a = bl/b.

Finally, since each al"I is less than e, we get

b = lim a{n) < e.

It is clear that b > 1. M

The material in this section is not new. In fact, the argument is essentially the same as Euler's original analysis in [7, secs. 13-15]. It also appears in several other papers (including [5, Theorem 1], [10, p. 241], and [11, p. 76]).

4. THE CASE a < 1. Throughout this section we assume that 0 < a < 1. In view of the preceding section it is natural to guess that the sequence of iterated exponentials converges for each such a. Remarkably, this is also incorrect, as we now show.

Recall that if 0 < p < q and a > 1, then a" < a<. On the other hand, if 0 < a < 1, then this inequality is reversed and aq < aP. Since the sequence of iterated exponentials behaves quite differently as a result of this reversed inequality, we call this relation the fundamental inequality. Next observe that if we use the fact that a < 1, repeatedly exponentiate this inequality by a, and use the fundamental inequality, then we get

0 < a < 1, O < a < aa < 1,

0 < a <a131 <a1{2} < 1,

and a simple induction argument shows that

0 < a{2n-11 < a{2n+ll < . . < a{2n+21 < a{2n} < 1

for each positive integer n. Thus, the sequence {(a2n+ll) is increasing and bounded and

{a 2n} } is decreasing and bounded. Hence each of these sequences converges. Let

bo = lim a2n+l}, be - lim a{2n"} n"/--- > oo n --- *0oo

(Here the subscripts are meant to remind the reader that these numbers are the limits of the odd and even sequences, respectively.) Clearly bo < be and the sequence {aIn) } converges if and only if bo = be. Since aa{2n+l} = a{f2n+2} and aa{2 n} - a{[2n+1, we get abo = be and abe = bo. Thus bl/be - a = bl/bo, so bbo = bbe. This suggests that we




consider the function f (x) = xx. Now

f'(x) = (log x + 1)xx,

so f'(1/e) = 0, f'(x) < 0 if x < 1/e, and f'(x) > 0 if x > 1/e. Hence f has an absolute minimum at x = 1/e and is two-to-one on the interval [0, 1]. Its graph is shown in Figure 2. If {an")} fails to converge (i.e., if bo < be), then the preceding analysis confirms that

1 0 < bo < - <be < 1.

e

In order to determine the set for which the sequence of iterated exponentials con-

verges, it is useful to consider the question of whether a pair {bo, be} with the property that bbo =

bebe

is unique. That is, we may ask: If 0 < c < d and clId - dI/C, then must it be the case that c = bo and d = be?

IIl

f(x) = x 2

0 /e 1

Figure 2.

Suppose that there are numbers 0 < c < d with c' = dd so that 0 < c < 1/e < d < 1. In this case, if we set a = c~ld - d/c', then we have a' = d and ad1 - c. If c < a, then by the fundamental inequality

a = cda <aIld <a

because I/d > 1. Since this is impossible, we infer that a < c. It now follows from the fundamental inequality that d = a' < a" and therefore that

I a <c < - <d <a

e

and

d a3 < ad =C < -< d = a' < a{2}

e

Repeatedly applying this argument, we see that

0 < a2n-1 < c < - <d < a2n < 1 e




and therefore that

1 0 < a{2n+l} < b < c < - < d < be < a2n < 1

e

for n = 1, 2, 3 .... Let us now pause to summarize the data that we have collected so far.

Theorem 4.1. If 0 < a < 1, then the following statements hold:

(1) The sequence {aInj } satisfies

a < a{3} {2n-1} {2n} {4} {2} a< a < .. < a < ... <a < .. < a <

(2) If bo = limn--,

a 2n+1) and be

= limn-,,, a{2n, then

bo < be and {aIn)} con-

verges if and only if bo = be.

(3) If {aInI} fails to converge (so that bo < be), then

abo = be, abe = bo, bl/be = a = b/bo,

and

1 0 < bo < - <be < 1.

e

(4) If O < c < d and cc = dd, then

1 0 < a{2n+l} <

bo < c < - < d < be < a2n < ,

e

for each n, where a = c/ld = d1/c.

If 0 < c < d < 1 and cc = dd, then we say that the set {c, d} is an Eulerpair. Recall that if 0 < x < y and xy = yX, then the set {x, y} is called a Bernoulli pair. It turns out that there is a one-to-one correspondence between Euler pairs and Bernoulli pairs. For example if we start from Bernoulli's observation that 24 = 16 = 42 and write c = 1/4 and d = 1/2, then {c, d} is an Euler pair. In this case we have

a = (1/4)2 = (1/2)4 = 1/16 = .0625.

Proposition 4.2. The set {c, d} is an Euler pair if and only if {1/d, 1/c} is a Bernoulli pair

Proof If 0 < c < d and we write x = 1/d and y = 1/c, then

cC = dd (iY 1 )/X

1 1 X~

I +x' = yx, yx xY

so the assertion holds. U

The relation between Euler pairs and Bernoulli pairs was noticed by Euler in [7, secs. 30-35]. It is surprising that the literature seems to contain no other references to




the connection between Euler pairs and Bernoulli pairs-even from those who were aware of Euler's work.

It follows from parts (3) and (4) of Theorem 4.1 that {a"11} fails to converge if and only if there is an Euler pair {c, d} such that c1/d = a = dl/c. Also, note that if we combine Proposition 4.2 with Goldbach's parameterization of Bernoulli pairs, we conclude that each Euler pair {c, d} has the form

c = c(s) = ss/(1-s), d = d(s) = s-

for some s > 1. Thus we have

a = a(s) = c(s)1/d(s) = d(s)l/c(s),

which implies that {a(n)} fails to converge precisely on the range of a(s). It is clear from the graph of xx (Figure 2) that

1 lim c(s) = lim d(s) = -, lim c(s) = 0, lim d(s) = 1.

s--+ 1+ s

--1+ e s--o s--+

(This may also be verified directly from the parameterizations of c and d.) Since a(s) - c(s)1/d(s), it follows immediately that

1 lim a(s) = - .0660, lim a(s) = 0.

s- 1+ ee s*0

As a(s) is continuous on its domain, we may apply the intermediate value theorem to conclude that the range of a(s) contains the interval (0, e-e), which naturally leads one to guess that the range of a(s) is precisely this interval. For once, this conjecture turns out to be correct, but establishing it is not entirely straightforward. The problem is that none of the obvious estimates work. For example, if 0 < c < 1/e < d < 1 and cC = dd, then

a=cI <( - d> -

e ee

which provides no information. (Here, the first inequality follows from the fact that c < Il/e, while the second follows from the fundamental inequality.)

On the other hand, the graph of a(s) strongly indicates that the function is decreasing. For example, if we plot a(s) on (1, 100], we get the picture in Figure 3.

There are several ways to identify formally the range of a(s). We will show that a(s) does indeed decrease from e-e to 0 as s increases from I to infinity. It follows that

1 0 < a(s) <-- ee

when s > 1. Another method of identifying the values of a for which the sequence of iterated exponentials diverges will be sketched in the next section.

Lemma 4.3. The function a(s) = c(s)'/Id(s) is decreasing on its domain.

Proof We show that a'(s) < 0. Since

a(s) -= c(s)l/d(s) - (sS/(-s)) l/(sI/(-)




e-e = .066

.0367 a(s)

0-

1 10 50 100

Figure 3.

this does not appear to be an appealing task. Fortunately some simplifications are pos- sible. First note that the relation d = s1/(l-s) leads to

d' logs d = d

(slogs - (s - 1)). 1 - s s(s - 1)2

Next, observe that, since c(s) = ss/(1-s) = (s'/('-s))s = ds, we can write a = ds/d

With this, we are able to compute a' as follows:

a' (ds/d)' (eslogd/d) (s logd )'

(log d sd' sd' log d ) a

= d d (d log d + sd' - sd' log d). d d2 d2 d2

If we substitute the formula for d' into the latter equation and perform a bit of algebra, we arrive at the relation

a' = a

(s log2 - ( - 1)2 (s - 1)3d

Since a, d, and s - 1 are positive, a is decreasing if and only if

f(s) = slog2s - (s - 1)2

is negative. Differentiating, we find that

f'(s) = log2 s + 2log s - 2(s - 1),

2 logs 2 f " (s) = + - -2,

s s

log s f"'(s) = -2 2 < 0.

s2

Thus f" is decreasing, and since f"(1) = 0, f" is negative on (1, 00oo). This shows that f' is decreasing. Because f'(1) = 0, f' is negative on (1, 00). Hence a(s) is

decreasing, as desired. 0




The delicacy of the relationship between of a, bo, and be is indicated by the fact that we are forced to examine the third derivative of f in order to establish that a is decreasing.

With these preparations, we can now present the main results in this section. From the fact that a(s) is decreasing we learn that

1 0 < a(s) < -

ee

when s > 1. Therefore, the range of a is precisely (0, e-e). This tells us that the sequence of iterated exponentials diverges on (0, e-e) and converges on [e-e, eI/e]. Thus, if 0 <a < e-, we have

bo = lim a{2n+l} < lim a2n) = be, n--+oo -n--* oo

where {bo, be} is an Euler pair. Conversely, suppose that {c, d} is an Euler pair. In this case, a - clId - d/Ic is

uniquely determined. The number a in turn determines the Euler pair {bo, be} with the property that

bl-ae = = bl/bo e

In other words there are Euler pairs {c, d} and {bo, be} with

blhe = ci/d = a = d'/c = bI/hbo 0 e

Since a is one-to-one by Lemma 4.3, these pairs cannot be distinct: if they were, they would determine distinct values of a. Hence, bo = c and be = d.

Finally, recall that for each a satisfying 0 < a < e/ve there is a unique b with 0 < b < e such that a - hi/b. Further, we have

I 1 1 1 a< b<-, a> b > -

ee e ee -e

As a result, the sequence of iterated exponentials converges if and only if b > 1/e. These results are formally recorded in Theorem 4.4.

Theorem 4.4. If 0 < a < 1 and a = bl/h, then the sequence a") of iterated exponentials converges if and only if e-'e a < 1, and this condition holds precisely when I l/e < b < 1. If a < e-", then b < 1/e and there is a unique Euler pair {bo, be} such that

0 < bo

<b < - < < I < b<1, abe = bo, abo = be. e

In this case

a < a(3) <---<a{2n- I <

< < bo <b < 1/e< be <"" < a{2n} .. {4} {2} < a <a

Also limnoo a{2n-'1 - bo and limn., a12n - =

be.




To the best of the author's knowledge the proof of Theorem 4.4 has not appeared in print before. In particular, the proof that a is a decreasing function of s seems to be new.

5. CONCLUDING REMARKS. We now briefly sketch another method for determining the domain of convergence of iterated exponentials. If we argue as in the first part of section 4, then we may assume that the results in Theorem 4.1 hold. Now suppose that the sequence of iterated exponentials diverges. Then there are numbers bo and be with bo < be such that abe = bo and abo - be (i.e., {bo, be} is an Euler pair). In this case we have aabo - abe = bo and aabe - abo = be. Also, if a - bl/b, then

ab _ ab a=a=b

Thus, if we write f(x) = ax and g(x) = aax, then the graphs of f and g meet the line y = x at (b, b), and the graph of g touches the graph of this line in at least two other places, namely, (bo, bo) and (be, be).

On the other hand, if the sequence of iterated exponentials converges, then the graphs of f and g still meet the line y = x at (b, b), but g has no other fixed points. To see this, suppose that the sequence of iterated exponentials converges for some a = bl/b (< 1) and that aac = c for some c different from b. Since f is decreasing, this function is one-to-one. Hence b is the unique number such that f(b) = ab = b and therefore ac : c. If we write d = ac, then we get d : c and aad = ac = d. In other words, {c, d} is an Euler pair. But in this case, it would follow from Theorem 4.1(4) that the sequence of iterated exponentials diverges, which is not the case.

Next let us examine the first and second derivatives of g. We have

g'(x) = (log a)2axaax, g"(x) = (log a)3(1 + ax log a)axax.

Thus g'(x) > 0, so g is increasing. If a > 1/e, then log a > -1 and 1 + ax log a > 0 whenever 0 < x < 1. But if a < 1/e, then it can be shown that 1 + ax log a changes sign as x varies from 1 to 0. Thus, if a < 1/e, then the graph of g changes from being concave up to being concave down as x increases from 0 to 1.

A closer analysis shows that when a < e-e the slope of the graph of g at x = b is greater than the slope of the diagonal y = x. Since the graph of g is concave up for small values of x and g (0) = 0, it follows that there is a point c with c < b such that the graph of g crosses the diagonal y = x at (c, c), dips below the diagonal, and then rises to cross the diagonal once more at (b, b). Similarly, since the graph of g is concave down for large x and g(1) < 1, it must be the case that the graph of g continues to rise above the diagonal and then turns down to cross it again at some point (d, d) with d > b.

Thus, when a < e-e there are precisely three values of x for which g(x) = x, implying that the sequence of iterated exponentials must diverge for these values of a. On the other hand, if a > e-e, then the slope of the graph of g at x = b is less than the slope of the diagonal and it follows in this case that x = b is the unique point for which g(x) = b. In this instance the sequence of iterated exponentials converges for these values of a. This change in behavior is illustrated in Figures 4 and 5 for the cases a = .03 and a = 0.1, respectively. This method for determining the domain of convergence for iterated exponentials was used in [5].

It follows from the analysis in the previous sections that the function defined by a s-+ limF-,> a") is inverse to the function x + X1/X on the interval [e-e, e]. The question then arises as to the form of the inverse on the intervals [0, e-e) and (e, oc).




y=x

abo = be

-----y -

a a x

_

a = .03- bo b 0.5 be

Figure 4. The case a = 0.3, b 0.3226, bo 1 0.0561, be a 0.8214.

This problem was addressed in [3], where it was found that the "iterated logarithm" functions defined by

y ..ogloge (1 < y < elle), y -+

...loglog,(1/e) (0 < y e-e)

y=x

aa ,

.794 - - y = ax

y = aax

a =

.1-

b 0.5 1

b 0.5

Figure 5. The case a = 0.1, b 0.399013.




are the desired inverses. The values of the first function lie in (e, 00), while those of the second are in (0, e-e).

One may also consider the problem of the convergence of iterated exponentials in the case where a is complex. A good deal of work has also been done in this case. For example, contributions have been made by Macintyre [12], Shell [13], and Thron [14], just to name a few.

As remarked earlier, among the authors of [5], [6], [10], and [11], only Knoebel refers to Euler's original paper. In addition, Knoebel provides well over one hundred references to previous work on this topic, many of which are quite useful in tracing the history of the problem.

The present work began while the author was teaching an honors section of second semester freshman calculus. It seems to be ideally suited as an exercise for bright calculus students. In fact when the author teaches this course again, he plans to pose the problem on the first day of the semester and challenge the students to seek a solution.

ACKNOWLEDGMENTS. Thanks are due to John Roe and Samuel Senti for valuable assistance during the preparation of this paper. Particular thanks are also due to William C. Waterhouse for providing a translation of a portion of Euler's original paper.

REFERENCES

1. D. F. Barrow, Infinite exponentials, this MONTHLY 43 (1936) 150-160. 2. D. Bernoulli, Letter to Goldbach, in Correspondance mathdmatique et physique de quelles ce'dlbres

gdomhtres du XVIII sicle, vol. 2, P.-H. von Fuss, ed., Imperial Academy of Sciences, St. Petersburg, 1843, p. 262.

3. Y. Cho and K. Park, Inverse functions of y = x1/x, this MONTHLY 108 (2001) 963-967. 4. M. J. A. N. C Marquis de Condorcet, Sur quelques series infinies dont la somme puet &tre exprimde par

des fonctions analytique d'une forme particuliere, Acta Scientiarum Petropolitanae 1 (1778) 34-37. 5. M. Creutz and R. M. Sternheimer, On the convergence of iterated exponentials, Fibonacci Quart. 18

(1980) 341-47.

6. G. Eisenstein, Entwicklung von aca , J. Reine Angew. Math. 28 (1844) 49-52. 7. L. Euler, De formulis exponentialibus replicatis, Acta Scientiarum Petropolitanae 1 (1778) 38-60; also

in Opera Omnia, vol. 15, G. Faber, ed., Teubner, Leipzig, 1927, pp. 268-297. 8. C. Goldbach, Letter to Daniel Bernoulli, in Correspondance mathdmatique et physique de quelles

cdl'bres g omhtres du XVIII sihcle, vol. 2, P.-H. von Fuss, ed., Imperial Academy of Sciences, St. Peters- burg, 1843, pp. 280-281.

9. S. Hurwitz, On the rational solutions of mn = nm with m 0 m, this MONTHLY 74 (1967) 298-300. 10. R. A. Knoebel, Exponentials reiterated, this MONTHLY 88 (1981) 235-252. 11. H. Linger, An elementary proof of the convergence of iterated exponentials, Elem. Math. 51 (1996)

75-77. 12. A. J. Macintyre, Convergence of iii, Proc. Amer Math. Soc. 17 (1966) 67. 13. D. L. Shell, On the convergence of infinite exponentials, Proc. Amer Math. Soc. 13 (1962) 678-681. 14. W. J. Thron, Convergence of infinite exponentials with complex elements, Proc. Amer Math. Soc. 8

(1957) 1040-1043.

JOEL ANDERSON received his Ph.D. at Indiana University under the direction of James P. Williams. After

spending five years at Caltech as a Harry Bateman Instructor and assistant professor, he moved to Penn State, where he is currently professor. His interests include operator algebras and operator theory. Department ofMathematics, Penn State University, University Park, PA, 16802 anderson @math.psu.edu




iterated exponentials

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