Journal of Shanghai University (English Edition), 2003, 7(3) :201- 205

Article liD: 1007-6417(20~)03-0201-05

z'"

W h e n Does x Exist ?

Lt CHENGuan-Rong($ ) 1. Department of M ~ , College of,Science, ~ m g h a i g r a t i S , Shanghai 200436, C~na 2. Department of Elevtronic Engineering, City Uniwrsa~/ of Hong Kong , Otina

Abstract This paper ftwther explores some interesting properties of the infinite exponem~ series, or called tower of powers, with a

rigorous proof of its convergence in a positive interval.

Key words Infinite exponent/al, implicit function, convergence.

MSC 2000 26A03, 26A18, 26B10

1 Introduction l~rst o f all, let us define a sequence of functions,

y~ = x,y.÷~ = xY. ,n E Z* (1)

For every n > 1, y, is an ' exponen t -power ' composi te

function. The domains of these exponent functions and

power functions of x will be restricted to be positive in

this paper . For negative values of x , power functions

are mostly undefined, except for a few trivial cases

s u c h a s x = - I .

It is clear that lim y~ exists if x = 1. For other values n ee

of x , does this limit exist? In other words , how to de-

termine a set or an interval of values of x for which the

l imit lim y~ exis ts? n

This question seems very elementary but actually is

non-Wivial and quite intereslivg. In Section 2, we do

find an interval, on which we can guarantee that the

limit lim y. exists, and is strictly incre~,dng in this in- n w

terval.

2 When Does lim y . Exist ?

Obviously, l imy. = 1 i f x = 1. In the following, we a ~ e l

first consider the case x > 1, for which we want to find

the ~ u m x , if it exists, such that lim y~ exists.

2 . 1 C a s e 1 : x > 1

It is easy to see that

Received Dec.2,2002; Revised Feb.14,2003

M Chang-Pin,Associate Professor,E-man:leecp@ onltne.sh.cn

y2 = x" > yl = x > 1

Suppose that

Y, = x y'-' > Y,-I > 1

Then

Y.+I = xY~ > x 7--t = y, > l

Hence, { y, } is strictly increasing for any fixed x > 1 by

mathematical induction. It then follows that !im~ y, ex-

ists ff and only if y, ~< M, for a positive number M, ,

possibly depending on x . Now, suppose that there

indeed exists a finite number , M, ( > 1) , such that for

every n ,

y . < M .

It is already known that y. is a strictly increasing func-

tion in n for each fixed x ( > I ) . Therefore, since

Y.÷t = x~" ~< xS" ~< M=

we have

x ~< M._~ ~

Set lim y. = A ( x ) . Then, it follows that m

y. <~ A ( x ) < ~ M , , ¥ n E Z*

and

A(x) = x ~( ' ) , x > 1 ( 2 )

202 Journal of Shanghai University

Assume that for x~ > x2 > 1, the two limits lim y. a eD

exist. Then, we must have

A ( x , ) > A ( x 2 )

due to (2) and the fact that

y. I .o., > y. I .~ . , V n E Z *

From this result , one can asser t that A ( x ) is a strictly

increasing function of x ( > 1) , so x is the inverse func-

tion of A, i . e . ,

x ( A ) = A " s , A > 1 (3)

where x is also strictly increasing as A increases.

In the following, we consider an auxiliary function,

l / J y = x , x > 1 (4)

Let y ' = O, i . e . ,

1 - lnx Y ' = Y 2 - 0

Then

reached in El ] .

( i i ) This theorem exptains that there exists no x

such that lira y . = 8, which was suggested in [ 2 ] , be- I t oo

cause A ( x ) <. e .

(iii) I n [ 3 ] , a potent paradox (4 = 2) was con-

sU'ucted by showing that lira+ y. = 2 and also seemingly

lim y. = 4. Theorem 1 above discloses the "secret " of n

the paradox: lira y. <~ e ff this limit exists, so lim y. n ~ m n ~ e e

4. ( iv) e has another expression associated with itself

and its powers l /e , that is ,

I / e ." I1® ( e )"

e = ( e ' ) ( ° ) ( * )

This expression is new, and was not found in the lit-

erature before.

(v) For all x E ( 1, e t~') , lira y, can be theoretical- n

ly and uniquely determined by Eq. (2 ) .

The next question is: for x E ( 0 , 1 ) , does lira y. exo a

ist? This will be discussed in the following subsection.

2 . 2 C a s e 2: x E ( 0 , 1 )

F o r x E ( 0 , 1 ) , w e h a v e

x - - e

Observe that y ' > 0 i f x E ( 1 , e ) , so y ( x ) is strictly in-

creasing in ( 1 , e ) , while y ' < 0 i f x E ( e , ~ ) , s o t h a t

y ( x ) is strictly decreasing in ( e , ~ ) . Therefore,

y.~ = y (e ) = e TM

Consider the function (4) in ( 1, e ) and compare it

with (3 ) . We have

x m = x ( A ) IA=. = e u+

Hence, x ~ e " and (2) - (4) together imply that

A ( x ) <~ e

Therefore, the aforementioned M, can be chosen as

e. It then also folows that A ( e ' " ) = e .

Theorem I Assume x > I . I f x ~< e TM , then lim y, tt

exists, and this limit functon A ( x ) ( > 1 ) is a strictly

incre~qing function of x , i . e . , Am = A (e t~') = e.

Remark 1

( i ) From Theorem I , one can see that lim y, does

not exist ff x > e"" , which agrees with the conclusion

Yl < Y2, Y3 < Y2

We next determine the relation between y, and Y3. For

E ( o , t ) ,

y'2 = y2(Inx + I )

For x E ( I / e , l ) , y2 is strictly incre~r, dng ,and

while for x E (0 , l / e ) , Y2 is s t r ic t~ decreasing and, at

the end,

1 / 11.

Y2 ~ ( e l

On the other hand, by convention in calculus,

~my2_ = limx',_.0 = 1

Therefore, for all x E ( 0 , 1 ) , we have

0 < ~<Y2 < 1

Vol.7 No.3 Sep.2003 LIC.P. , e ta / . : WhenDoes~ r b"xi~? 203

Since 0 < x < 1, and 0 < 3.2 < 1, it follows that

3.t : X < X 72 : X :~ : 3.3

Hence,

O < Yt < Y3 < Y2 < 1

Similarly,

3.4 = xy3 > ~ 2 : 3.3 a n d 3.4 = xy3 < xyl : 3.2

Therefore,

3.1 < Y3 < 3"4 < 3.2

Next) suppose that for n = k )

3.2t-1 < Y2k+i < Y2/,+2 < 3.2k

Then, for n = k + 1, we have

3.21.+1 : xy2t <~ XYZ*÷2 : 3 .2t+3 '

3.21'+4 "~ XY2t+3 < XY2t÷I ---- 3.21'+2

(5)

(6)

To proceed) we need the following Lemmas. Here

and throughout, x -~ 0 means x - - 0 ÷ .

First, consider a general equation: z = x"

Lemma 1 The function equation z = x" has a unique

solution z = z ( x ) for x E ( 0 , 1 ) ) and l imz(x ) = ' 0 . s~0

Proof Taldng the derivative gives

z ' ( x ) - > O) x E (0)1) (1 - x ' l n x ) x

It is easy to show, by the implicit function theorem)

that there exists only one solution z ( x ) that solves z =

x" . Here, we can also see that z ( x ) is strictly increas-

~ g i n ( 0 , 1 ) , a n d z ( x ) E (0 )1 ) .

It can be verified that

Us las(s)/s(z) X = X = e ( 9 )

Let x approach 0 f3~m both sides of (9 ) . and note that

z( x ) is strictly increasivg in ( 0 , 1 ) . Then)

Inz (x ) lira ~ - • --o z ( x )

3.2k+3 : XY2t÷2 ~ xY2b*I ~" 3 .2t+2 '

3.2t,+3 = XY2t÷2 ~ XY2t÷3 ---- 3.2k+4

Consequently,

( 0 < 3.1 < )3.2t+1 < 3.2t+3 < 3.2t'÷4 < 3.2t+2( < 3.2 < l )

(7)

By mathematical induction and (5) - (7 ) , we find that

13.~,_. ~ is strictly incre~-~ng, whilst { 3.., I is mict iy de-

creasing, So, there exist O <~ a ( x ) <~ b ( x ) ~ I such

that

lim3.2t_ t = a ( x ) , lim3.2t = b ( x ) t-*:* t-*w

Moreover, by n~ng

3.21.÷ 1 : xzy2t - I XK'2t ) 3.2k+2 : ' 3.2.1 : x'2, and

3.2t+2 : XY2t÷I )

we have

a ( x ) = x *'(') b ( x ) = x ~"') 9

a ( x ) = x ' ( ' ) , b ( x ) = x "(') (S)

Up to now. we still do not know whether or not a ( x )

= b ( x ) f o r x E ( 0 , 1 ) .

l n z ( x ) where ~ is also sUict~ increasiv~ in x E ( 0 , 1 ) ,

because

( l n z ( x ) ~ , z ' ( x ) ( 1 - l n z ( x ) ) z ( x ) ] = z ~ > 0

Therefore,

l imz(x) : 0 z-*O

Remark 2 The function equation z = x" in z E (0,

1) determines a unique solufionx = x ( z ) , which is a

strictly increasing function of z, and x = x (z ) E (0,

1). Lemma 2 The solution to z = x' and the solution to

x" = x : are identical for x E ( 0 , 1 ) .

From Lemmas 1 and 2, we have the following:

Lemma 3 The unique solution x ( z ) for z E (0, 1 )

( o r z ( x ) f o r x E ( 0 , 1 ) ) t o : = x ~ solves z ffi x ~ .

Lemma 3 does not give an answer to the question of

whether or not z = x" determines a unique solution for

z E ( 0 , 1 ) . Nevertheless, we have the following result.

Theorem 2 The function equation z = x : has a

unique solution x (z ) for z E ( 1/e, 1 ) , and this solution

204 Journal of Shanghai University

salisfles x E ( e - ' , l ) .

P r o o f F o r 0 < x ~ 1 , i f z > 1 t h e n x" ~ 1 . F u r t h e r -

z t

more, x: <~ I( < z ) , which conlradicts x = z .

Hence, 0 < z ~< 1.

Now let

•s

F ( z , x ) = z - x = 0 ,

where/ ' / = { ( z , x ) 10 < z <~ 1,0 < x ~< 1}.

(i) F( z, x) is continuous in the given domain D =

I ( z , x ) I 1 - - < z ~ < 1 , e -" < x ~< 1 } . e

(ii) The partial derivatives

F ' , ( z , x ) = 1 - xCx'ln2x,

F ' , ( z , x ) = - x'-tx" (zlnx + 1)

are continuous in D.

(iii) F ( 1 , 1 ) = O.

(iv) F ' ~ ( z , x ) ~ Oin D (we will show it below).

Thus, by the implicit function theorem, there exists

one and only one x (z ) such that

(a) x(1) = 1 , F ( z , x ( z ) ) ~_0.

(b) x (z ) is continuous in z E ( 1 , 1 ] . e

( c ) x' ( z ) is continuous in z E ( 1 , 1 ] , and e

x ' ( z ) = 1 - xS'xSln 2x

x'-t x¢ ( zlnx + 1)

Conditions (i) - ( iii ) are easy to be verified. Hence,

we only prove ( i v ) . Hrst , we show that F ' , ( z , x) is

by no memm equal to 0 in D. Suppose that there e x i t s

one ( Zo, Xo) E 0 such that zo in Xo + 1 = O. Then, x~o

= l i e . From this equation and z = x ; , it follows that

• 0'o ue "o Xo~" t h a t i s , xo~" = l / e , Zo = xo = xo andx0 = ,

where Xo E (0 ,1 ) .

Now, consider the following function:

I/e

g ( x ) = x" , x E (0 ,1)

Differentiating it with respect to x leads to

l i e - 1

g ' ( x ) = g ( x ) x (lnx + e) e

Obviously, g ' ( x ) < O w h e n x E (O,e-e) , i . e . , g ( x )

is strictly decreasing, so g ( x ) > g ( e - ' ) = __l if x E e

( 0 , e - ' ) , while g ' ( x ) > 0 w h e n x E ( e - ' , 1 ) , i . e . ,

g( x ) is strictly increasing. Therefore, g ( x ) > g(e-" )

_ 1 f ix E ( e - ' , l ) . Moreover, g (1) = 1, l i m g ( x ) = e x-~0

1. Hence, g .~ = g ( e-e ) = 1 Consequently, there e

1 exists one and only one solution to g ( x ) = e ' i , e . ,

F ' , ( z, x) = 0 has one and only one zero point, which

1 is (z0, x0) = ( e ' e-" ) . Hnaliy, since the solution

x (z) is extended at the initial value ( 1,1 ) , D is deter-

mined as defined above.

Remark 3

(i) From the above proof, we do not know whether

or not F( z , x ) = z - x : = 0 determines a unique solu-

tion x ( z ) in D I D .

(ii) From Remark 2 and Lenuna 3, the tmique solu-

tion x ( z ) in D is also increasing for z E ( l /e ,1 ] .

(iii) Ire-" < x ~< 1, we can see that a ( x ) •_ b ( x )

from (8 ) .

Lemma 4 Let a, = e - ' , a,.i = ( e - ' ) ' . , n E Z +.

1 Then, lim a, -

e n e e

Proof By the definition of the series, one has

-® (o ) = - _ ( e _ , ) ( , ) 2, a 2 t , t ( e _ e ) ( e ") 'zt t , a 2 t . 2 =

a . . , = ( e - ' ) ' 2 , and a~,÷~ = ( e - ' ) ' ~ " ' ,

Define A = limau_~ and B = l ima2,, one has

l - A e I - B e - B e - A e A = e-" , B = e-" , A = e , B = e (10)

Obviously, 1 satisfies the first two equations of e

(10). Substituting A = 1 into the fourth equation of e

1 (10) y~dds B -

e

equation of (10) .

However, we

- - - , which also satisfies the third

cannot yet assert that the first

1 equation of (10) has a unique solution A - e

Assume that this equation has two solutions, and an-

other one is x , . Without the loss of generality, suppose

1 1 t ha t0 < x~ < - - ( the c a s e - - < x~ < 1 c a n b e similarly

e e

Vol.7 No.3 Sep.2003 L I C . P . , e ta l . : WhenDoes~" E m t ? 205

~Ylscussed). Let

I - z e

h ( x ) = x - e-"

By assumpt ion, h ( xL ) = h ( 1 ) = 0 . Applying the e

Rolle Theorem in calculus to h ( x ) over [ x~, 1 ] , we e

can see that h ' ( x ) v a 1 ~ h e s for at least one value ~ in

t h e i n t e r v a l ( x , , 1 ) C ( 0 , 1 ) , i . e . ,

• I - ~ : e l ~-e

h ' ( ~ ) = l - e - e - e = 0

that is,

~ e + e t- ~ = 2

Let

d ( x ) = xe + e I-ze

It is clear tha t d ( $ ) = d ( 1 ) = 2. Applying the Rolle e

Theorem again, to d ( x ) in x E [ ~, 1 ] , s h o w s that e

there exists at least one value x0 in ( ~, 1 ) such that e

d' ( x ) vanishes , i . e . ,

d ' ( x o ) = e(1 - e ' - '0") = 0

It thus follows that

1 X 0 - -

e

which contradicts x0 E ( ~,--1 ) . e

Therefore , h ( x ) has only one zero poin t , x -

Consequent ly ,

1 1 limak - , i . e . , l imy , l,ffi,-. -

Remark 4 F rom Lemma 4 , I has the following in- e

terest ing express ion :

e = ( e - ' ) ~-°)~'-%'" ( * * )

This formula is also new ( s e e ( * ) a b o v e ) , and was

no t found f rom the li terature before .

Up to n o w , w e k n o w tha t lira y~ exists if x E E e - ' ,

l ie ] . We also know that the solution to z = x" and so- z z

lution to z = x are the same for x E [ e -e , 1 ] . On the

o ther hand , w e have a l ready k n o w n that w h e n 0 < x <

e - ' , lira y, does no t exist ( fo r more details, see[ 1] and n m

[4] ) . Nevertheless ,

lira inf( l i r a y. ) = 0 ( 11 ) z- -*O n ee

due to (8) and Lemmas 1 and 3.

Now, based on Theorems 1 and 2, and Lemmas 4

a n d 1, the main results o f this paper are summarized as

fol lows.

Theorem 3

( i ) The limit funct ion lim y. exists ff and only ff x E n

[ e - ' , et~']. And this funct ion is a strictly m o n o t o n e in-

cre~Lging funct ion o f x , wh ich lies in l /e ~< l imy, ~< e. n

(ii) For all x E [ e - ' , 1 ] , lira y, can be uniquely de- m

te rmined by z = x" ; while for all x E [ 1, e 11" ] , lira y, n m

can be uniquely determined by ( 2 ) .

R e f e r e n c e s [1] Barrow D F. Infinite exponentials[J]. Amer. Math.

Mort., 1936, 43(3) : 150- 160. [2] Rippon P J. Infinite exponentials [ J ] . Math. Gazette,

1983, 67: 189- 196. [3] CourantT. Towers of power : apotentpatatk~[J] . Math.

Journa/, 1993,3(3) : 60 -64 . [4] KnoebeJ R. E ~ . n t ~ reiteratedEJ]. A~er. Math.

Mort. 1981, 88(4) : 235 - 252. ( ~ t i v e editor SHEN Me, Fang )