The Fourth Operation
Introduction · Groping for a Solution · Uniqueness · A Related Problem · Facts About Functions · Asymptotic Power Series · To Proceed · Uniqueness Again · The Original Problem · Is it Unique? · Other Terminology and Notation · Computer Program · Bases Other Than e · Afterword
Given whole numbers A and B we can add them together to get another whole number, The definition of this operation is simply to count together what A and B count separately.
Then we can define multiplication as repeated addition:
|A · B =|| A + A + A + ...|
| || B times|
And exponentiation as repeated multiplication:
|AB =|| A · A · A · ...|
| || B times|
Those are the first, second, and third operations: addition, multiplication, and exponentiation. Presented in that way a prolonged series of operations comes to mind, each operation iterating the preceding one. The fourth operation would be (using a subscript to denote the second operand):
The series of operations can be defined recursively on the second operand.Addition:
A · 1 = AExponentiation:
A · (B + 1) = A + A · B
A1 = AThe fourth operation:
AB + 1 = A · AB
A1 = AThe nth operation, for n > 1:
AB + 1 = AAB
A [n] 1 = A
A [n] (B + 1) = A [n – 1] ( A [n] B )
Historical note: A [n] B when A and B are whole numbers is the same as the function φ(A, B, n – 1) of Wilhelm Ackermann (1896 – 1962), used in a paper of 1928.
Thus we can define a series of operations between positive whole numbers. It is well known how to extend the definition of the first three operations to include all real numbers (except that in the case of exponentiation the base, A in AB, must be non-negative if the operation is to be well-defined and continuous as the exponent, B, varies).
The problem we investigate here is how to define the fourth operation, AB or A raised to itself B times, for fractional B. What, for example, is A½ or A raised to itself one half times? It might sound like a crazy question but no crazier than asking what is A multiplied by itself one half times. In that case the answer is the square root of A but the situation is not so simple with the fourth operation.
Later it will make some formulas simpler if we let A = e, the base of the natural logarithms.
e0 = 1
e1 = e = 2.71828…
e2 = ee = 15.15426...
e3 = eee = 3814279.10476…
Again, what is
e˝e to itself one half times? And in general what is
esfor any real number s ?
... And what sort of nutball would ask such a question ?
Because it is there. Now Froggy, please stay out of the Algorithms section.
In order to solve the problem we will reformulate the fourth operation in terms of functional iteration, then we can use techniques of functional analysis to try to solve the problem. As well as our own ideas we will use some ideas from the paper by G. Szekeres “Fractional Iteration of Exponentially Growing Functions” (J. of the Australian Mathematical Soc., vol. 2, 1962), which relies on his “Regular iteration of real and complex functions” (Acta Math., vol. 100, 1958). Note that Szekeres’s point of view differs from ours in that his primary interest is (see below for definitions) in exps(x) as a function of x, where s is thought of as a parameter, whereas to us that function is only a stepping stone to exps(1) where s is the variable. Jumping ahead, note that Szekeres claims to have proved that the second coefficient in what we call Φ(x) is 1/2 but unless I’m mistaken he hasn’t done so or even made it plausible.
Sometimes we will write “exp” for the exponentiation of e function:
exp(x) = exand place a subscript on exp to denote the level of its iteration. The functional iterates of exp(x) are closely related to es. First observe that
exp0(x) = xetc. Thus
exp1(x) = exp(x)
exp2(x) = exp(exp(x))
exp3(x) = exp(exp(exp(x)))
exp n (exp m (x) ) = exp n+m (x)holds for all non-negative integers n and m. If it is to hold for all integers then exp-1 must be the inverse of exp:
exp-1(x) = log(x)etc. where “log” is the natural logarithm function (in Calculus textbooks usually written ln ). Setting x = 1 gives us the fourth operation without the “socket”:
exp-2(x) = log(log(x))
expn(1) = ee...e, n times = en
Thus our goal is to define the function
exps(x)for all real numbers s, then
exps(1)will answer our original question, what is es, that is, e raised to itself s times.
If that isn’t clear consider the third operation, exponentiation. It is iterated multiplication and can be obtained by iterating the function “mul”:Thus
etc. In this case we know the continuous iterate of mul, it is
|mul (mul (x)) = e e x|| or ||mul2(x) = e2 x|
|mul (mul (mul (x))) = e e e x|| or ||mul3(x) = e3 x|
muls(x) = es xso that
muls(1) = esIterates of the function of mul(x) where x is a sort of socket for the last multiplier gives a different way to look at exponentiation.
Groping for a Solution
First we will examine how the third operation, exponentiation, is defined when the exponent is a fraction and see if something analogous can be done for the fourth operation.
Unlike the operations of addition and multiplication, each of which has just one inverse (subtraction and division), exponentiation has two inverses because it is not commutative, the order of the operands matters. Given
AB = Cone operation is required to obtain the base A and another the exponent B. Taking a root gives the base
A = Bth root of C = B√ Cand taking a logarithm gives the exponent:
B = logarithm (base A) of C = logACwhere contrary to usage later the subscript A does not denote iteration.
Consider the root inverse. Using facts about exponents with which I’ll assume you are familiar, given that we have defined exponentiation when the exponent is an integer we can use the root operation to define it when the exponent is any rational number. In what follows let m and n be integers, n positive. Then let
C1/n = n√ Cand
Cm/n = n√ C mThis comports with the definition of exponentiation when m/n is an integer and otherwise obeys the expected behavior of exponentiation. (If you investigate you will see it does this because multiplication is commutative and associative, multiplication doesn’t depend on the order or grouping of its operands.) Then by insisting that exponentiation be continuous we can define it for real s by taking any sequence of rational numbers ri that converges to s as i → ∞ (here the subscript on r is an index, not the fourth operation):
lim ri = sand defining
i → ∞
Cs = lim C r i(one would have to prove that this is well defined, that it does not depend on the sequence).
i → ∞
The trouble with generalizing this procedure (using root, the inverse for the first operand) is that, to repeat, it uses the fact that multiplication is commutative and associative. These two properties give a certain homogeneity to multiplication which allows us to turn a fraction 1/n into an integral root, and a fraction m/n into an integral power of that root. This will not work for the fourth operation because exponentiation is so different from multiplication.
Now consider the other inverse of exponentiation, the logarithm. Though its basic definition is that it is the exponent there is an interesting theorem which allows us to compute it more directly. Let
f (x) = ∫1x 1/t dtThen for all real x and y(On the right side, in the second integral make the change of variables u = xt .) Thus for any whole number p
f (xp) = p f (x)Also
q f (x1/q) = f (x1/qq) = f (xq1/q) = f (x)so that
f (x1/q) = 1/q f (x)Thus for any rational number r
f (xr ) = r f (x)and since f is continuous (indeed differentiable) this holds for any real number r. Finally, let e be the solution to f (x) = 1 (there is one and only one solution because f (1) = 0 and f (x) increases to infinity). Then
f (er ) = rThus f (x) is the logarithm for exponentiation using the base e and we have a very interesting formula for it:
log x = ∫1x 1/t dtAnd we have ex as well because it is the inverse of log x.
Note that the derivative of log x is:a simple function independent of log x and of ex. Thus we can use 1/x to determine log x. This contrasts with the derivative of ex , which is ex again.
Though the second procedure (using logarithm, the inverse for the second operand) doesn’t generalize to higher operations, for the same reason the first doesn’t, it does furnish an important clue: if one seeks to continuously iterate a function – any function, not necessarily exp – then instead of trying to find it directly, look for the derivative of its inverse – the inverse that gives the “higher” operand. In other words, try to define the derivative of the generalized “logarithm” that counts the number of times the function was applied in its argument.
The hard part of the problem of defining the fourth operation es isn’t finding a definition it is finding a natural, best definition. There are any number of “low quality” solutions: in the interval 0 ≤ s < 1 set
es = any function of s that is one when s is zerothen use the fact that
es+1 = eesto translate the values at 0 ≤ s < 1 to any higher unit interval. In effect, for s ≥ 1 find its integral part n and fractional part t and set
es = expn(et)What is wanted is a condition on our definition of es that makes it the unique best definition.
The problem is analogous to defining the factorial function
n ! = n (n – 1) (n – 2) ... 1for fractional n. One could define 0 ! = 1 and s ! as anything when s is between 0 and 1, then use
(s + 1) ! = s ! (s + 1)to define it for other values of s. However one particular definition of s! (we won’t give the details) has two recommendations: (1) it arises by a simple and natural argument, (2) it is the only function defined for s ≥ 0 such that 0 ! = 1, s ! = s (s – 1) !, and that is log convex. It is also the only such function that is asymptotically equal to a logarithmico-exponential function. As Szekeres points out in the first reference above, this analogy suggests that in the search for a best solution of fractional functional iterates it is “better to concentrate on the real variable and asymptotic properties than on the complex-analytic character of the solution.” We seek a real variable condition that will make the fourth operation the fourth operation.
A Related Problem
Finding a continuous iterate of ex that is natural and arguably the best is a difficult problem to attack directly, so instead at first we will try to find such an iterate for a slightly different function
e(x) = ex – 1This function has a fixed point, and only one fixed point, at zero:
e(0) = 0It is much easier to continuously iterate a function with a unique fixed point.
... I knew a man who dropped his latchkey at the door of his house and began looking for it underneath the street light. I asked him why he was looking for it there when he knew very well it was over by the door. He replied “Because there isn’t any light over there.” LOL.
In our case, Froggy, we might find a key to the key where the light is. Perhaps if we solve the continuous iteration problem for e(x) we will be able to use it to solve the problem for exp(x).
Our choice of notation can be confusing in that sometimes e stands for the number 2.718 ... and sometimes the function e(x). es(x) is a function and the function it iterates is exp(x) – 1, yet if we were to write the function without its argument, es, it would look like the fourth operation — e^(e^(...^e)...) s times — rather than the sth iterate of the function e(x) = exp(x) – 1. To avoid confusion we will never do that, that is, we will always include at least a placeholder dummy variable when referring to the function e(x) and its iterates.
Finding the “logarithm” for es(1) is simpler than finding es(1). We mean logarithm in a general sense, as counting the number of times something has been done. The ordinary logarithm of Calculus counts the number of factors of e in a product, for example
log(e e e e) = 4and in general
log(es) = swhere s can be fractional as well as whole. The crux of the definition, see the discussion of mul(x) = e x above, is that
log(mul (x) ) = log(x) + 1
Facts About Functions
When speaking of a general function we shall always assume it is strictly increasing and smooth (infinitely differentiable). In general the “logarithm” for iterates of a function f (x) is a function A(x) defined for x > 0 such that
A( f (x) ) = A(x) + 1and
A is strictly increasing and smoothThe functional equation is called Abel’s equation, after Niels Henrik Abel (1802 – 1829). A(x) counts the number of times f (x) is used in x, it is a sort of “functional logarithm.” (Some authors don’t include A(1) = 0 in the definition, and when it is posited call the resulting A(x) a “normalized” Abel function.)
A(1) = 0
You can see from Abel’s equation that:
A(f2(x)) = A( f ( f (x) ) ) = A( f (x) ) + 1 = A(x) + 2and in general
A( fn(x) ) = A(x) + nThus if we had such a function A we could define fs for real s as the solution to
A( fs(x) ) = A(x) + sthat is,
fs (x) = A -1( A(x) + s )where A -1 is the inverse of A. Such an fs (x) would have the properties
f0 (x) = xNote that
fs (fr (x)) = fs + r (x)
fs (1) = A -1(s)
That works in reverse. If fs(x) is a continuum of iterates for f (x), that is, f0(x) = x and fs (fr (x)) = fs + r (x) , then fs can be obtained from the inverse of the function
B(s) = fs(1)as above. First replace s with s + r :
B(s + r) = fs + r (1) = fs(fr(1))Given x let r = B -1(x). Then
B(B -1(x) + s) = fs(x)Let A = B -1 and we have our original formula.
If we knew nothing more about A than Abel’s equation then A would not be unique. If (given the function f ) A is one solution of Abel’s equation then so would be B(x) = A(x) + g(A(x)) for any periodic function g with period 1 and g(0) = 0. This is easily seen by substituting B into Abel’s equation. (Abel showed that this describes all solutions of Abel’s equation.)
As might be expected from the formula for the logarithm for exponentiation the derivative of A(x) might be useful. Call
a(x) = A’(x)a “derived Abel function” of f. Let
Φ(x) = 1 / a(x)Then because of the condition A(1) = 0 we have
which generalizes the famous formula for the natural logarithm (third operation). At this point the equation looks contrived but if we can determine Φ(x) we solve the iteration problem for f(x). It turns out we will be able to do it when f (x) = ex – 1 .
If we take the derivative of each side of Abel’s equation, after some manipulation we getCall this Schröder’s equation for f (x) , after Ernst Schröder (1841 – 1902).
Although we won’t need it here, it’s interesting that if we let
fs(x) = Φ -1( f ’(s x) Φ(x) )then if f ’(0) = 1 the above is a system of continuous iterates of f (x).
Recall that if a function f (x) is smooth there is a power series that might not converge but at least will be asymptotic to the function at 0 :
We will explain what asymptotic means in the next section. The general term iswhere f (i)(x) denotes the ith derivative of f (x).
(0) + f ’
(0) x + 1/2 f ’’
+ 1/6 f ’’’
as x → 0
If the function is not only smooth but real analytic the series converges and we needn’t specify “as x goes to zero.” Some authors write ≈ instead of = when the series diverges or isn’t known to converge. We shall let “as x goes to zero” serve to indicate that the series is asymptotic but might not converge.
Asymptotic Power Series
Suppose we are given, out of the blue, an infinite series
a0 + a1(x – x0) + a2(x – x0)2 + a3(x – x0)3 + ...Regard this as a “formal” series, effectively just a sequence of coefficients provenance unknown.
Given a particular x we could try to sum the series. The series might converge, that is, if we compute the partial sums
Sn(x) = a0 + a1(x – x0) + a2(x – x0)2 + ... an(x – x0)nthen
lim Sn(x) = [some number] as n → ∞for the particular x. In such cases we can think of the series as a function defined where it converges. The series would stand alone and could be used to define the function it converges to.
The series always converges at x = x0, but except for that value it might not converge. If it doesn’t converge anywhere besides at x0 it is useless, by itself.
Scrap the above series and suppose we are given a function F(x) instead of a series. We might not know how to compute F(x) but suppose we have an abstract definition, know what its properties are. And suppose we know it is smooth at x0, any number of its derivatives exist there. Then we can form the Taylor expansion of F at x0
F(x0) + F’(x0) (x – x0) + 1/2 F’’(x0) (x – x0)2 + ... 1 / i ! F(i)(x0)(x – x0)i + ...As before we regard this as a formal series that is not necessarily summable, in effect just a sequence of coefficients. For a finite number of terms we can compute the partial sum
Sn(x) = F(x0) + F’(x0) (x – x0) + 1/2 F’’(x0) (x – x0)2 + ... 1 / i ! F(n)(x0)(x – x0)nThe sequence might diverge, that is, if we fix x ≠ x0 and let n → ∞ the limit of Sn(x) might not exist. But even so the series tells something about F near x0. Instead of fixing x and letting n get arbitrarily large, if we fix n and let x get arbitrarily close to x0, then by definition of the derivative Sn(x) will approximate F(x).
But what good is that? We already know the series gives the correct value when x = x0, what use is an approximation for x near x0 if what we want to know is F(x) for x well away from x0? No use at all, unless – and this is the point – the function possesses a property that allows us to put its value at x in terms of its value at points near x0.
If the property allows us to approach x0 arbitrarily close then the asymptotic series, despite the fact that it doesn’t converge, can be used to compute F(x) to any desired accuracy. We will see an example of this in the next section.
To repeat, with a convergent series, given any x (in the domain of convergence) the series converges as n → ∞. With an asymptotic series, given any n the truncated series converges as x → x0. A convergent series defines a function, an asymptotic series by itself does not – however it can be used to compute a function whose value near x0 determine its value far from x0.
Again, convergence is concerned with Sn(x) for fixed x as n → ∞ whereas being asymptotic is concerned with Sn(x) for fixed n as x → x0 when one knows the series is the Taylor expansion of a function that we know other properties of besides the Taylor coefficients. Convergence depends on only the coefficients, asymptotic depends on the function as well as the coefficients.
We have been using “asymptotic” as if the Taylor series diverges but since we might not know that in advance it is useful to subsume convergent series under asymptotic series, considering that the definition of convergence generally includes the condition for asymptotic.
As before, let
e(x) = ex – 1It has a fixed point at 0. Its power series expansion about 0 is
Let A(x) be an Abel function for e(x),and a(x) the corresponding derived Abel function for e(x),Let
(x) = x
Φ(x) = 1 / a(x)
Schröder’s equation for e(x) isand using the fact that e’(x) = ex
We will try to find the Taylor series for Φ(x) at 0. We can expect only that it converges asymptotically, nonetheless, as we shall see later, knowing Φ(x) when x is near 0 will allow us to compute it for any x .
Φ(x) = c0 + c1 x + c2 x2 + c3 x3 + ... as x → 0+It is fairly easy to show that both c0 and c1 must be 0. Take the derivative of both sides of Schröder’s equation to eventually obtain
Φ’(e(x)) = Φ’(x) + Φ(x)Then by considering the particular case x = 0 we can conclude that
Φ(0) = 0Then, returning to the last equation in x , take the derivative of each side again, again evaluate at x = 0 and we can conclude that
Φ’(0) = 0Continuing in the same vein yields nothing about Φ’’(0) but at least now we know that the series begins
Φ(x) = c2 x2 + c3 x3 + c4 x4 + ... as x → 0+where we have yet to determine the remaining coefficients c2, c3, c4, ...
To determine them we will plug the known power series for e(x) and the unknown one for Φ(x) into Schröder’s equation, then equate coefficients of the same power of x and solve for c2, c3, etc. one after the other. The trouble is that when we do it there is one degree of freedom. That is, we need to know at least one of the ci in order to solve for the others. If we knew c2 we could solve for c3, then for c4, etc. We will explain in detail when we have figured out what c2 should be and actually perform the procedure.
Finding c2 in the Taylor series for Φ(x) will take a bit of work. What we will do is find the beginning of the Taylor series for es(x) in two different ways, first thinking of x as the variable then s. In the first case the coefficients will be functions of s, in the second functions of x. Looking ahead, in the first case we will be able to determine those functions, in the second they will involve Φ(x). By equating the series and comparing the same powers of s, we can make a plausible choice for c2 in Φ(x). However, in the last analysis it looks like it will have to be a condition, an imposed value, rather than something we can deduce. But given this one value everything else is determined. Φ(x) will be unique given c2 .
To begin, let’s see how far we can go finding a power series for es(x). We can try to find one in powers of x and in powers of s.
First consider es(x) as a function of x with s as a parameter. Consider its Taylor expansion at x = 0
es(x) = c0 + c1 x + c2 x2 + c3 x3 + ... as x → 0The coefficients ci are functions of s , namely the ith derivative of es(x) with respect to x divided by i ! evaluated at x = 0 .
Since e0(x) = x for all x, including when x = 0, it must be that c0 = 0 and we can write
es(x) = c1 x + c2 x2 + c3 x3 + ...Now c1 is the first derivative of es(x) evaluated at x = 0. To try finding the first derivative we use a trick: before taking the derivative break es(x) into e(es–1(x) ) then use the chain rule to find a recursive relation for the derivative. In what follows the apostrophe indicates differentiation with respect to x.
es(x) = e(es–1(x))Then since es–1(0) = 0 and e’ (0) = 1
e’s (x) = e’(es–1(x)) e’s–1 (x)
e’s (0) = e’s–1 (0)Repeating that formula within itself on the right we get down to
e’s (0) = e’t (0)where 0 ≤ t < s is the fractional part of s, that is, the number left over after subtracting the largest whole number less than s from s.
We have failed to determine e’s (0) in general. However in the particular cases when s is a whole number, that is when t = 0, we have determined it:
e’n (0) = 1for every whole number n. We shall impose the natural, simplest, condition on es(x) that generalizes that formula:
With that condition c1 = 1 independent of s. Thus the power series for es(x) begins:
es(x) = x + c2 x2 + c3 x3 + ...Now try the same trick with the second derivative. If we can find it c2 will be it divided by 2. For typographical reasons in what follows let D as well as apostrophe denote d/dx, differentiation by x, and let D2 as well as double apostrophe denote differentiation twice.
D2[ es(x) ] = D[ e’(es–1 (x) ) e’s–1 (x) ]Setting x = 0 and using the fact that es – 1(0) = 0 and e’s – 1 (0) = 1 and e’’ (0) = 1, then simplifying, we get
= e’’(es–1 (x) ) e’s–1 (x) e’s–1 (x) + e’s–1 (x) e’’s–1 (x)
e’’s (0) = 1 + e’’s–1 (0)Repeating that formula within itself on the right
e’’s (0) = 1 + 1 + e’’s–2 (0)etc. so that
e’’s (0) = n + e’’t (0)where n + t = s, n being the largest whole number ≤ s and 0 ≤ t < s .
We have failed to determine e’’s (0) in general. However in the particular cases when s is an integer, that is, t = 0, n = s, we have determined it:
e’’n (0) = nWe impose the natural, simplest, condition on es(x) that generalizes that formula:
Thus we have the power series for es(x) beginning:
es(x) = x + 1/2 s x2 + c3 x3 + ... as x → 0+We will not need any more conditions. The rest of the coefficients can be determined by breaking es+1(x) two ways
e(es(x)) = es(e(x))and plugging in the known power series for e(x) and what we know of the power series for es(x) and equating coefficients of like powers of x. It helps to have a symbolic algebra calculator such as PARI/GP by the PARI Group of the Institute of Mathematics at the University of Bordeaux. The next three coefficients are
c3 = 1/4 s2 – 1/12 sso that
c4 = 1/8 s3 – 5/48 s2 + 1/48 s
c5 = 1/16 s4 – 13/144 s3 + 1/24 s2 – 1/180 s
For any finite number of terms we can rearrange in powers of s.
|es(x)2||2 = ||x + 1/2 s x2 + (1/4 s2 – 1/12 s) x3 + (1/8 s3 – 5/48 s2 + 1/48 s) x4 + (1/16 s4 – 13/144 s3 + 1/24 s2 – 1/180 s) x5 + ...|
as x → 0
|es(x)2||2 = ||x + (1/2 x2 – 1/12 x3 + 1/48 x4 – 1/180 x5 + ...) s + (1/4 x3 – 5/48 x4 + 1/24 x5 + ...) s2 + (1/8 x4 – 13/144 x5 + ...) s3 + (1/16 x5 + ...) s4 + ...|
Now consider es(x) as a function of s with x as a parameter. Let its Taylor expansion at s = 0 be
es(x) = c0 + c1 s + c2 s2 + c3 s3 + ... as s → 0The coefficients ci are functions of x , namely the ith derivative of es(x) with respect to s evaluated at s = 0 divided by i ! .
Since e0(x) = x right away we know c0 = x .
Now c1 is the derivative of es(x) with respect to s evaluated at s = 0. To find that consider an Abel function for e(x), and we say “an” rather than “the” because at this point we do not know if it is uniquewhere A-1 is the inverse of A. We will differentiate each side with respect to s, using the fact that the derivative of the inverse of a function is the reciprocal of its derivative evaluated at the inverse. We will also use the fact that the derivative of A(x) + s with respect to s is 1 . In the next few equations an apostrophe indicates differentiation with respect to s.
Thus since e0(x) = x
(x) = 1 / A’
(A(x) + s)
= 1 / A’
= 1 / a(es
e0’ (x) = 1 / a(x) = Φ(x)So the power series in s begins
es(x) = x + Φ(x) s + c2 s2 + c3 s3 + ... as s → 0It’s easy to get all the higher derivatives with respect to s from the formula for the first derivative, es’ (x) . You find that they form a sequence of odd powers of Φ(x) times a constant (depending on i ) that alternates in sign. By way of a footnote (we won’t need it) the exact formula for the ith derivative of es(x) with respect to s when i > 1 is
– (–1)i (2i–3)!! Φ2i–1(es(x))where !! indicates the double factorial: N!! = N (N – 2) (N – 4) (N – 6) ... and the last factor is 1 or 2 depending on whether N is odd or even; in the case here it’s always odd. That ith derivative divided by i! and evaluated at s = 0 gives ci for i > 1
ci = – (–1)i (2i–3)!! / i! Φ2i–1(x)
We are trying to find Φ(x) . We proved that its Taylor expansion begins
Φ(x) = c2 x2 + c3 x3 + c4 x4 + ... as x → 0+and if we can determine just c2 we can determine the others. Now we found formulas for es(x) in two ways and the second involved Φ(x).
es(x) = x + 1/2 s x2 + [higher powers of s and x] as x → 0It would solve our problem if we could equate coefficients of the first power of s and conclude that
es(x) = x + Φ(x) s + ... as s → 0
Φ(x) = 1/2 x2 – 1/12 x3 + 1/48 x4 + ... as x → 0+but it isn’t obvious that they must be equal. For one thing the first series is as x goes to 0 and the second as s goes to 0. Still it makes plausible that c2 in the power expansion of Φ(x) is 1/2. Furthermore, equating coefficients of the first power of x rather than of s results in the very same value. Therefore we impose the following condition, which supersedes our previous conditions #1 and #2 because alone it will allow us to determine Φ(x).
ConditionThus we have
The second order coefficient in the power expansion of
Φ(x) = 1/2
... as x → 0+
Now to find the remaining coefficients. Consider Schröder’s equation
Φ(e(x)) = ex Φ(x)We know the power series for ex
ex = 1 + x + 1/2 x2 + 1/6 x3 + ...and for e(x)
e (x) = x + 1/2 x2 + 1/6 x3 + ...Begin by thinking of Φ(x) as
Then plug these power series into Schröder’s equation. (You might want to use a symbolic algebra computer program such as PARI/GP mentioned earlier.) The left side is
Φ(x) = 1/2
+ [higher powers of x]
1/2 x2 + (c + 1/2) x3 + (3/2 c + 7/24) x4 + [higher powers of x]and the right
1/2 x2 + (c + 1/2) x3 + (c + 1/4) x4 + [higher powers of x]Thus, focusing on the fourth order term,
3/2 c + 7/24 = c + 1/4we have
c = – 1/12Then think of Φ(x) as
and again plug into Schröder’s equation. As before the first few coefficients are tautological but equating the ones for x5
Φ(x) = 1/2
+ [higher powers of x]
2 c + 1/48 = c + 1/24and solving for c
c = 1/48Continuing in this manner we can determine the coefficients one after the other.
Remarkably, this appears to be the first degree coefficient of the power series of es(x) in s. Positing that the second order term in Φ(x) be the same as the corresponding term in es(x) seems to entail that all of them are the same. I say “seems” because we can only expand to a finite number of terms and so far the conjecture holds. If we could prove the conjecture then we would not need our condition on “c2.” In any case, expanding Φ(x) directly is much easier than expanding es(x); it has the advantage of being a function of only one variable.
|Φ(x)2|| = ||1/2 x2 – 1/12 x3 + 1/48 x4 – 1/180 x5 + 11/8640 x6 – 1/6720 x7 – 11/241920 x8 ...|
as x → 0+
The coefficients continue (order 9 to 13): 29/14551520, 493/43545600, –2711/239500800, –6203/3592512000, 2636317/373621248000.
Now that we have an asymptotic power series for Φ(x) as x → 0+ we can use it, together with known properties of the function Φ(x), to calculate Φ(x) for any x .
First note that the inverse e–1(x) of e(x) = exp(x) – 1 is
e–1(x) = log(1 + x)Repeated application of this function takes a number arbitrarily close to zero:
lim e– n(x) = 0Now suppose we are given x = x0 > 0. Here a subscript on x is a sequential index, not the fourth operation. Then the sequence
n → ∞
x0can be continued far enough (n can be made large enough) so that xn is as small as desired.
x1 = e–1(x0)
x2 = e–2(x0)
xn = e– n(x0)
Now apply Schröder’s equation
Φ(e(x)) = ex Φ(x)to x1
Φ(x0) = Φ(e(x1)) = ex1 Φ(x1)then to x2
Φ(x1) = Φ(e(x2)) = ex2 Φ(x2)Thus
Φ(x0) = ex1 ex2 Φ(x2)And so on through xn
Φ(x0) = ex1 ex2 ... exn Φ(xn)Since xi+1 = e–1(xi) = log(1 + xi) we have
We can use this equation to, in a manner of speaking, drop any x = x0 > 0 down to near zero. Thus with our asymptotic expansion of Φ(x) at 0 we can compute Φ(x) for any x.
) = (1 + x0
) (1 + x1
) ... (1 + xn–1
Having Φ(x) we can compute the “functional logarithm” for e(x) from
A (x) = ∫1x 1 / Φ(t) dtThen the functional iterates of e(x) areNote that
es(1) = A – 1(s)And because es(1) = en(es – n(1) ) we have
A – 1(s) = en(A – 1(s – n) )so if we know A – 1(s) for 0 ≤ s < 1 we know it for all s.
Φ(x) is unique (given the “c2 condition”) because we have calculated what it must be. Thus also A(x) and its inverse A – 1(x) are unique.
Since the uniqueness of Φ(x) is so important we shall prove it again without resorting to an asymptotic expansion, simplifying an argument of Szekeres.
Let Φ(x) be the reciprocal of a derived Abel function for e(x). We shall prove that if Φ(0) = Φ’(0) = 0, and Φ’’(0) exists and ≠ 0 then Φ(x) is unique.
The condition says
lim Φ(x) / x2 exists and ≠ 0Call this constant c .
x → 0
Suppose there were another derived Abel function for e(x) whose reciprocal, call it Φ*(x), satisfied our condition. We want to show it must be that Φ* = Φ.
Choose one x = x0 and consider the sequence
yn = e– n(x0)It goes to 0 (see earlier text). We will use Schröder’s equation for each Φ to express yn in terms of x0. It’s easy to see from the equation, by substituting e– n(x) for x, that
Φ(e– 1(x)) = e–x Φ(x)Thus
Φ(e– n(x)) = e–n Φ(x)So for each Φ
Φ(yn) = e–n Φ(x0)Thus their ratio doesn’t depend on n. Call the ratio r.
Φ*(yn) = e–n Φ*(x0)
so thatAt this point it looks like r depends on x0.
r = Φ*(yn
) / Φ(yn
) = Φ*(x0
) / Φ(x0
By the condition we knowwhere c ≠ 0. Hence
The left side is a constant so r is the product of two constants. Thus r is constant and independent of x0, and
= r / cn → ∞
Φ*(x) / Φ(x) = ror
Φ*(x) = r Φ(x)for every x.
Now recall that for each Φ and all x
∫1x 1 / Φ(t) dt = A (x)Letting x = e(1)
dt = A
(e(1)) = 1
dt = 1
1 = 1 / rThus r = 1 and Φ*(x) = Φ(x) for all x.
The Original Problem
We began by asking what is the number e to itself a fractional number of times. We reformulated the problem as finding the fractional iterates of the function exp(x). Then instead of solving that problem we found the fractional iterates of a somewhat different function e(x) = exp(x) – 1.
Knowing how to compute the fractional iterates of e(x) will enable us to compute the fractional iterates of exp(x). The idea is to use integral iterations of exp(x), which we know already, then introduce a fractional iteration of e(x) which will differ a little from the desired fractional iteration of exp, then take an iterated log so the error is reduced and the original integral iteration of exp is undone. The error goes to zero as the number of integral iterations increases. We now give the details.
We need define exps(x) only for 0 ≤ s < 1 because exps(x) for larger values of s can then be found by repeated application of
and for smaller values by repeated application of
(x) = exp(exps
(x) = exp–1
) = log(exps
So in groping for a definition we will at first suppose 0 ≤ s < 1. At this point all we impose on exps(x) is that it be an increasing function of s and of x, and
expr(exps(x) ) = expr+s(x)for r and s fractional just as when they are integral.
Recall that e(x) = exp(x) – 1. Consider the identities
(x) = e0
) + 0
exp1(x) = e1(exp0(x) ) + 1The first just says x = x and the second exp(x) = e(x) + 1. Since exps(x) and es(x) are both increasing functions of s the following condition on exps(x) is reasonable.
If 0 ≤ s < 1 then
(x) = es
(x) + ε
where 0 ≤ ε < 1.
Now let n be a whole number and consider
exps(x) = exp–n(expn+s(x) ) = exp–n(exps(expn(x) ) )Since exp–n is log iterated n times,
exps(x) = logn(exps(expn(x) ) )The idea is to replace the exps() on the right with es() and let the iterated exp / log wash out the difference as n grows large.
By our assumption above, replacing x with expn(x)
exps(x) = logn(es(expn(x) ) + ε)where 0 ≤ ε < 1 depends on n and x as well as s. We want to prove that the difference between the right side andgoes to 0 as n → ∞.
This is a consequence of the increasingly slow growth of logn(x) as either n or x increases. First note that
log’x = 1 / xfor x > 0, and in general
logn’x = 1 / x log x log2x log3x ... logn–1xfor x large enough that the arguments of all the logarithms exceed zero. Define the sequence of numbers
an = expn –1(x)Then
logn’an = 1 / expn–1x expn–2x expn–3x ... exp0x → 1as n → ∞. And since logn’x is a decreasing function of x the above limit is true for any sequence whose terms are greater than an, such as es(expn(x)) > es(expn-1(x)) ≥ expn-1(x).
Letting y = es(expn(x)) we have
0 < logn(y + ε) – logn(y) < ε logn’(y)and the right side goes to zero as n goes to infinity.
exps(x) = lim logn(es(expn(x) ) )In practice the limit converges very rapidly. So rapidly in fact that n = 2 or 3 is enough for eight significant figures.
n → ∞
Here is how much exps(x) differs from es(x) when s is between 0 and 1. To three decimal places:
| x = ||1||2||3||4||5||6||7||8||9||10|
| 0.00 ||0.000||0.000||0.000||0.000||0.000||0.000||0.000||0.000||0.000||0.000|
Naturally if s = 1 the difference is always one but if s < 1 not only is the difference less than one it looks like it goes to zero, and rapidly, with increasing x .
exps(x) thus defined is a continuum of iterates for exp(x). Even for finite n we have
logn⚬ e0 ⚬ expn = logn ⚬ I ⚬ expn = logn ⚬ expn = Iand in
logm⚬ er ⚬ expm ⚬ logn⚬ es ⚬ expnsince the limit on m and on n exists and is unique we can choose n = m so in the limit on m the above is
logm⚬ er ⚬ es ⚬ expm = logm⚬ er+s ⚬ expm
Now to answer the question we started with, what is e to itself one half times, we compute exp˝(1) using the above limit formula.
| 0 || 1.2710274 |
| 1 || 1.6100969 |
| 2 || 1.6451508 |
| 3 || 1.6451508 |
the last two accurate to seven decimal places. As a check that this great web of arithmetic hangs together we compute exp˝(exp˝(1) ). By n = 2 it has stabilized at seven decimal places to 2.7182818, which is as it should be exp(1) = e, the base we are using.
The following table shows how fast the fourth operation grows as the second operand increases. The numbers in the right column are to three decimal places. The zeros at the end of the higher numbers are just placeholders due to the limited precision of the computer.
Here is a graph of y = ex with the scale of the y axis shrunk in order to show more of the graph.
Features to note:
is undefined for
x ≤ –2. (It would involve the logarithm of a non-positive number.)
x → –2+
x = –1, ex
x = 0, ex
x = 1, ex
= 2.71828 ...
There is a point of inflexion between
0. Numerical analysis (specifically, examining the second differences) shows it is at about
x = –.5838, y = .1710 .
By definition a graph is fairly straight near a point of inflexion. This graph is fairly straight throughout the entire unit interval from
x = –1 to 0. Throughout that range
x + 1
by less than
Is it Unique?
The question of uniqueness bedevils attempts to define the fourth operation. Earlier we showed that, given some justified assumptions, es(x) is the unique continuum of iterates of e(x). Then we defined exps(x) in terms of es(x). It isn’t clear though that this exps(x) is unique. Without proof, there might be another way to define it, using e(x) or not.
Other Terminology and Notation
The fourth operation is known by various names: tetration, super-exponentiation, hyperpowers.
Instead of our subscript notation es some authors use the pre-superscript notation se (Hans Maurer [1868 – 1945] used it in a paper of 1901). Other notation is Donald Knuth’s up arrow notation e ↑↑ s and e^^s.
The following computer program is written for PowerBasic PBCC and can easily be modified for PBWin, Visual Basic, FORTRAN and any other BASIC-like compiler.
This glossary translates the computer program functions into the mathematical functions of the above analysis when e(x) = exp(x) – 1.
| DeLog(x) || A’(x) = a(x) = 1 / Φ(x)|
| AsympDeLog(x) || 1 / asymptotic power series for Φ(x) |
| eLog(x)|| A(x)|
| eExpInteger(n, x)|| en(x)|
| eExp(x)|| A – 1(x)|
| ExpInteger(n, x)|| expn(x)|
| ExpReal(s, x)|| exps(x)|
' The Fourth Operation
' © 2018
'Double precision: plus or minus 4.19E-307 to 1.79*E+308 with 15 or 16 digits of precision
Macro Precision = Double
Macro expRealTol = .00000001
Macro eExpTolNewton = .00000001
Macro eExpTolBisection1 = .00000001
Macro eExpTolBisection2 = .00000001
Macro CheckQuit = If WaitKey$ = $Esc Then End
Global e, em1 As Precision
%ht = 8 'no more than 8, see Dim coef and Assign coef
Function AsympDeLog(x As Precision) As Precision
Local i As Long
Local sum As Precision
Static coef() As Precision :Dim coef(8) 'static so need only assign once
If coef(2) = 0 Then 'first call?
' 0 1 2 3 4 5 6 7 8
Array Assign coef() = 0, 0, 1/2, -1/12, 1/48, -1/180, 11/8640, -1/6720, -11/241920
If x < 0 Then
Print "AsympDeLog: out of bounds" :CheckQuit
ElseIf x = 0 Then
Function = 0
sum = 0
For i = 2 To %ht 'terms to x^5 give at least five decimal places if x < .05
sum = sum + coef(i)*x^i
Function = 1/sum
%maxLogRep = 20
Function DeLog(x As Precision) As Precision
Dim xx(%maxLogRep) As Precision
Local d As Precision
Local i, n As Long
If x < 0 Then
Print "DeLog: out of bounds" :CheckQuit
ElseIf x = 0 Then
Function = 0
i = 0
xx(0) = x
Incr i :If i > %maxLogRep Then Print "DeLog: failed" :Exit Function
xx(i) = Log(1 + xx(i - 1))
Loop Until xx(i) < .1 'anything smaller doesn't affect output before 6 decimal places
n = i
d = 1
For i = 0 To n - 1
d = d * (1 + xx(i))
Function = AsympDeLog(xx(n)) / d
Function eLog(ByVal x As Precision) As Precision 'integrate DeLog(t) from 1 to x
Local cnt As Long 'use Romburg's Method of integration
Local i, k, nmax As Long
Local sum, h, dx As Precision
If x < 0 Then
Print "eLog: out of bounds" :CheckQuit
If x = 1 Then
Function = 0
cnt = 0 'shorten interval, at best x < e - 1, 0 <= eLog(x)
If x > 0 Then
Do Until x < em1
x = Log(1 + x)
dx = .001 'want nmax so abs(x - 1)/2^nmax <= dx
nmax = Log2(Abs(x - 1)/dx)
If nmax = 0 Then nmax = 1
Dim R(nmax,nmax) As Precision 'typically nmax is no more than 9
'done with dx, all we needed was nmax
h = (x - 1) / 2
R(0,0) = h * (DeLog(1) + DeLog(x))
For i = 1 To nmax
h = (x - 1) / 2^i
sum = 0
For k = 1 To 2^(i-1)
sum = sum + DeLog(1 + (2*k-1)*h)
R(i,0) = R(i-1,0)/2 + h*sum
For k = 1 To i
R(i,k) = R(i,k-1) + (R(i,k-1) - R(i-1,k-1))/(4^k - 1)
Function = R(nmax,nmax) + cnt
Function eExpInteger(ByVal n As Long, ByVal x As Double) As Precision 'e sub n (1)
Local i As Long, a As Precision
For i = 1 To n
x = Exp(x) - 1
If x = 1/0 Then Exit Function 'Function = 0, flag for overflow
Function = x
Function eExp(ByVal y As Precision) As Precision 'inverse of eLog, y >= 0
Local x, xlast, x1, x2 As Precision
Local n, cnt As Long
cnt = 0 'get fractional part of y
Do Until y < 1
Do Until y >= 0
'here 0 <= y < 1, cnt = integral part of original y
'Try Newton's method first because it usually converges and is fast.
'It uses the derivative of eLog.
x = 1
n = 0
xlast = x
x = xlast + (y - eLog(xlast)) / DeLog(xlast)
Incr n :If n > 30 Then GoTo BisectionMethod 'hung, use bisection instead
Loop Until Abs(x - xlast) < eExpTolNewton
Function = eExpInteger(cnt, (x + xlast) / 2)
'bisection method (slow and must specify range of solution)
x1 = 0 'bracket solution
x2 = eExpInteger(CLng(y)+1,1)
x = (x1 + x2) / 2
If Abs(eLog(x) - y) < eExpTolBisection1 Or Abs(x1 - x2) < eExpTolBisection2 Then Exit Do
If Sgn(eLog(x) - y) = Sgn(eLog(x2) - y) Then
x2 = x
x1 = x
Function = eExpInteger(cnt, x)
' n must not exceed 3 or 4 (depends on x) to avoid overflow.
' If n = -1, x must > 0, if n = -2, x must > 1, if = -3, x must > e, ...
' in general if n < 0, x must > ExpInteger(-2-n,1)
Function ExpInteger(n As Long, ByVal x As Precision) As Precision
Local i As Long
If n >= 0 Then
For i = 1 To n
x = Exp(x)
For i = 1 To -n
If x <= 0 Then
Exit Function 'Function = 0, flag for overflow
x = Log(x)
Function = x
Function ExpReal(ByVal s As Precision, ByVal x As Precision) As Precision
Local p, y, z, zlast As Precision
Local cnt, n As Long
cnt = 0 'get fractional part of s
Do Until s < 1
Do Until s >= 0
'here 0 <= s < 1, cnt = integral part of original s
If s = 0 Then
Function = ExpInteger(cnt, x)
If x < 0 Then 'get 0 <= x < e
x = Exp(x)
Do Until x < e
x = Log(x)
n = 0
zlast = z
p = eLog(ExpInteger(n, x))
y = eExp(p + s)
z = ExpInteger(-n, y)
Loop Until Abs(z - zlast) < expRealTol Or z = 0
If z <> 0 Then
Function = ExpInteger(cnt, z)
Function = ExpInteger(cnt, zlast)
Local w, h, a, b As Long
Local s As Precision
Desktop Get Size To w, h
Console Get Size To a, b
Console Set Loc (w - a)\2, (h - b)\2
Console Set Screen 120, 100
e = Exp(1) 'global 2.71828...
em1 = e - 1
Print " "; " s", " e sub s"
Print " "; "----------------------"
For s = -1 To 3.5 Step .25
Print " "; Format$(s,"* .000;-0.000"), Format$(ExpReal(s, 1)," #.00000")
Bases Other Than e
What if we wanted to know 2˝ instead of e˝? And in general bs for other bases b > 1. We will try to proceed as before when b equaled e and find functional iterates of bx.
We will use the same labels for the various functions as before. Note that letting
e(x) = bx – 1will not work because in general it has two fixed points, one at zero and another either greater or less than zero depending on whether b is less or greater than e. (When b = e the graph of y = exp(x) – 1 is tangent to the graph of y = x, otherwise they intersect in two points one of which is zero.) The trouble with having more than one fixed point is that then there could be no Abel’s function for the function. Recall that the Abel function A(x) must be defined for x > 0 and
A(e(x)) = A(x) + 1If for example, e(x) = 2x – 1, which has a fixed point at zero and at one, then A(x) could not be defined at x = 1. The Abel function will always be singular at a fixed point.
Instead of translating the graph of y = bx vertically by minus one we need to translate it by an amount making it tangent to y = x. In general, unlike the case when b = e, the tangent point won’t be at the origin. To simplify expressions in what follows let
r = log bwhere as usual log is the natural logarithm. Note that
bx = e rxWhen b = e, r = 1.
Setting the derivative of y = bx to 1 (the slope of y = x) and solving for x, gives x = – (log r) / r. Again to simplify future expressions let
u = – (log r) / rNote that
bu = 1/rSo the point (u, 1/r) on the graph of y = bx is where the slope is 1. Thus if we let
e(x) = bx – (1/r – u)then e(u) = u and that is the only fixed point. The derivative is
e’(x) = r bx
Let A(x) be an Abel function for the new e(x)
A(e(x)) = A(x) + 1and let
a(x) = A’(x)be the corresponding derived Abel function. Schröder’s equation for the new e(x) is
Φ(e(x)) = r bx Φ(x)
As before we want to determine the Taylor series of Φ(x) but this time at x = u.
Φ(x) = c0 + c1 (x – u) + c2 (x – u)2 + c3 (x – u)3 + ... as x → u+Taking the first and second derivatives of each side of Schröder’s equation and evaluating at u we can, as before, eventually conclude that Φ’(u) = 0 and Φ’’(u) = 0, so that both c0 and c1 in the Taylor expansion are 0.
Φ(x) = c2 (x – u)2 + c3 (x – u)3 + ... as x → u+Again we need another coefficient and turn to finding the Taylor expansion of es(x), first in powers of (x – u) and then in powers of s.
More or less as before we take the first derivative with respect to x of each side of es(x) = e(es–1(x)) and evaluate at u to eventually obtain
e’s (u) = e’s–1 (u)and so
e’s (u) = e’t (u)where t is the fractional part of s. Thus as before e’s(u) = 1 when s is a whole number and we impose the condition that this holds for all s.
Taking the derivative with respect to x again and evaluating at x = u eventually leads to
e’’s (u) = r + e’’s–1 (u)so that
e’’s (u) = n r + e’’t (u)where n and t are the integral and fractional parts of s respectively. Thus when s is a whole number e’’s(u) = s r and we impose the condition that this holds for all s.
Thus, with our conditions the Taylor expansion begins:
es(x) = (x – u) + 1/2 s r (x – u)2 + ... as x → u
Determining the Taylor expansion of es(x) viewed as a function of s goes exactly as before and we have
es(x) = x + Φ(x) s + c2 s2 + c3 s3 + ... as s → 0
We are trying to find Φ(x) . We proved that its Taylor expansion begins
Φ(x) = c2 (x – u) 2 + c3 (x – u) 3 + c4 (x – u) 4 + ... as x → u+and if, as before, we can determine just c2 we can determine the others. Comparing the two series for es(x), as when b = e (and r = 1) we impose the following condition, which supersedes our previous conditions because alone it will allow us to determine Φ(x).
ConditionThus we have
The second order coefficient in the power expansion of
Φ(x) = 1/2
(x – u) 2
(x – u) 3
(x – u) 4
... as x → u+
Now we use Schröder’s equation to determine the other coefficients. Schröder’s equation is
Φ(e(x)) = r bx Φ(x)or
Φ(erx – 1/r + u) = r erx Φ(x)Substituting in the power series as we did before when b = e we can deduce that
c3 = (1/48 u2 r4 – 1/6 u r3 + 7/24 r2 – 1/3 r3)and
/ (27/8 r2 – 2 r – 7/8 +1/24 u3 r3 + 1/2 u2 r2 + 3/2 u r)
c4 = (25u4r9 – 16u3r9 – 162u2r9 – 256u3r8 – 576u2r8 + 1134ur8 + 1296r8 + 630u2r7 – 480ur7 – 1620r7 –294ur6 + 48r6 + 336r5)and succeeding coefficients can be computed likewise. If b = e then r = 1 and u = 0; and c3 reduces to – 1/12 and c4 to 1/48, the values we found before. When b = 2, c3 = – .053891092257005 and c4 = .00280571012920368.
/ ( –2u7r9 + 16u6r8 – 162u4r8 + 2048u3r8 + 168u5r7 + 96u4r7 + 24576u2r7 + 165888r7 –918u4r6 – 1536u3r6 – 42768u2r6 + 73728ur6 – 255768r6 –3968u3r5 + 16128u2r5 – 31104ur5 + 81408r5 + 19344u2r4 – 9216ur4 + 2304r4 – 19008ur3 + 3840r3 + 5208r2 + 1296u3r7)
Repeated application of
e-1(x) = logarithm to base b of (x– u + 1/r) = (1/r) log (x – u + 1/r)will reduce any x > u to as close to u as desired. Given x0 > u, let xi = e-i(x0), then using Schröder’s equation as before we can eventually conclude
Φ(x0) = rn (x0 + v) (x1 + v) ... (xn-1 + v) Φ(xn)where v = 1/r – u. Since u might exceed zero we also need to consider the case x0 < u. Repeated application of e(x) will increase x0 to as close to u as desired. Let xi = ei(x0), then
Φ(x0) = Φ(xn) / (rn(x1 + v)(x2 + v) ... (xn + v))
Thus we have Φ(x) and consequently A(x) and the continuous iterates of e(x). We can get from continuous iterates of e(x) to those of bx by the same method as before except that exp and log are replaced with exp and log to the base b.
A case can be made that there is no series of operations as presented in the introduction, not when the second operand is fractional. In this view there are only two binary operations, addition and multiplication. Exponentiation is not a binary operation but a unary one (the exponent) with a parameter (the base).
Exponentiation cannot be a true binary operation because it cannot be defined when the base is negative and the exponent varies continuously. As a unary operation its purpose is to turn multiplication into addition; it is a go-between for two binary operations but not a binary operation itself.
If you try to think of exponentiation as a binary operation, its non-associatively is a major drawback. (Its noncommutativity is minor in comparison.)
Another reason exponentiation stands out from addition and multiplication is that it has no geometric construction that is continuous in the exponent. Addition can be thought of as laying two line segments end to end in a straight line, multiplication as the area indicated when line segments are joined perpendicularly. Though x, x squared, x cubed, etc. make sense geometrically – a line segment, square, cube, and so forth in higher dimensions – there is no continuum of dimensions to make sense of x raised to a fraction. (The fractional and continuous geometries of Menger, Birkoff, and von Newman aren’t applicable here because their dimension ranges between zero and one, and also they are restricted to projective geometries.)
The problem with negative bases in exponentiation affects what we have been calling the fourth operation, for there, even when the second operand is integral, each number in the “exponential tower” except the highest is an exponential base.
Some simple fractional (not irrational) exponents are useful in physics, where the denominator of the fraction indicates a root. And in mathematics, when trying to describe the growth of a function as a variable goes to infinity a real (even transcendental) exponent might be useful. Szekeres was interested in continuously iterating e(x) because it would interpolate G. H. Hardy’s logarithmico-exponential scale.