# Tetration extended to real exponents

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### Ingolf Dahl

Dec 13, 2006, 3:36:08 AM12/13/06
to
Tetration (also hyperpower, power tower, super-exponentiation,
and hyper4) can be seen as the fourth operation in the chain addition,
multiplication, power...
I have proposed a way to extend tetration to the range of real numbers,
see
http://forums.wolfram.com/mathgroup/archive/2006/Dec/msg00207.html and
http://forums.wolfram.com/mathgroup/archive/2006/Dec/msg00133.html. I
have also made a Mathematica notebook, with an extended PowerTower
http://web.telia.com/~u31815170/Mathematica/.
I got into this business by solving the equations f(f(x)) = x^2 +1 and
f(f(x)) = exp (x). (Solving for f(x)). The zero point value for the
first equation is
f(0) = 0.6420945043908285
and for the second equation
f(0) = 0.498743364531670879182837375041686052396842401697695
with derivatives
f'(0) = 0.87682850111901739234161348908444926737095344685410,
f''(0) = 0.489578037438247862041903165991088781934758253721,
f'''(0) = 0.1222616742183662064640316453590024071635142693
I am grateful for any comments on this, and hope that someone can check
if my ideas are sound and relevant.

Thus Pi to the superpower e (or "Pi tetrated to e") is
1885451.906681809677772360465630708698

Math is fun

Ingolf Dahl
ingol...@telia.com

### bo198214

Dec 13, 2006, 7:26:09 AM12/13/06
to
Because this topic seems often to be requested, let me mention some key
points and references.

There are two problems, that look equal, but are quite different
1. Extension of the hyperpower to fractional/real hyperexponents.
2. Fractional/real iteration, i.e. extension of natural numbered
iterates to fractional/real numbered iterates.

The problem of 1. is not in finding a continuous/differentiable
extension but in finding uniqueness criterions. There are some proposed
Ioannis Galidakis
http://ioannis.virtualcomposer2000.com/math/papers/Extensions.pdf
Robert Munafo

The problem is that the usual compatibility with the multiplication,
which is given for multiplication and exponentiation

1x = x; (mn)x = m(nx)
x^1 = x; x^(mn) = (x^m)^n

is no more given for tetration

x^^1 = x; x^^(mn) != (x^^m)^^n

But this compatibility allowed us to uniquely define for example
x^(1/n)

(x^(1/n))^n = x^(n/n) = x

thatswhy is x^(1/n) the inverse function of x^n, and further
x^(m/n)=(x^m)^(1/n) = (x^(1/n))^m. This breaks all down using
tetration.

We see that the problem is the non-associativity of the power. Which
also would allow other bracketings for the definition of tetration. I
thatswhy developed a domain which can deal with different bracketings
and non-associativity and in which indeed
(ab) ** x = a ** ( b ** x ) for all higher operations **.

As a byproduct of this research there emerged a fairly recursive
structure "fractional trees", which is a (non-associative)
generalization of fractional numbers, for which I wrote a little
visualization and experimentation web application
http://math.eretrandre.org/cgi-bin/ftc/ftc.pl

Now to 2. fractional iteration.

Here again is the main problem the uniqueness.

For the fractional iteration of e^x one should at least have read the
papers of Szekeres on this topic
G. Szekeres, Fractional iteration of exponentially growing functions,
1961
G. Szekeres, Abel's equation and regular growth: Variations ..., 1998

For me the results look rather unpromising to find "the" fractional
iteration of e^x.

For fractional iteration of real-analytic functions:
G. Szekeres, Regular iteration of real and complex functions, 1958

The topic of analytic iteration (i.e. fractional/real/complex iteration
in the set of powerseries developable functions) seems to me the most
interesting field regarding uniqueness of the solutions.

The first interesting thing is that in the ring of formal powerseries
with a_0=0 and a_1=1 there is only one solution to the equation
fof...of=F. There is even a widely unknown explicit formula describing
the fractional/real/complex iterates:
(f^os)_n = sum_{k=0}^{n-1} (-1)^{n-1-k}(s over k)(s-k-1 over n-k-1)
(f^ok)_n

where f^os denotes the s-iterated f (s complex), and f^ok the k times
iterated f. _n denotes the nth coefficient of the powerseries.

And there exists an "iterational logarithm" L. I.e. L maps a
powerseries to a powerseries such that L(f^or) = r(Lf).

For all this see:
Jabotinsky & Erdoes, On analytic iteration, 1961
Jabotinsky, Analytic iteration, 1963

Now it can though happen, that the corresponding iterates do not
converge. For example the iterates of e^x-1 only converge for integer
(composition) exponents. However Szekeres showed in the above mentioned
paper that for real-coefficient powerseries, a formal iterate in 0 can
always uniquely approximated by an analytic function iterate defined on
x>0.

### Ingolf Dahl

Dec 14, 2006, 7:46:16 AM12/14/06
to
Thank you a lot for your engagement and elaborated answer, I really
love it. Nothing is worse than indifference.
In Mathematica notation, which I find convenient here:

The PowerTower (tetration) function for integers is defined this way in
Mathematica

PowerTower[(y_)?NumericQ, (n_Integer)?Positive] := Nest[y^#1 & , y, n -
1]

We then easily find that

PowerTower[y, n] == y^PowerTower[y, n - 1]

Thus

Subscript[g, y][x_] := y^x

is a "step-up" function:

PowerTower[y, n] = Subscript[g, y][PowerTower[y, n - 1]]

I have used these step-up functions for my algorithm. If we start with
the PowerTower function for some given y, we can note that just some
point values of the step-up function is used. We could thus define an
infinite number of step-up functions that do the same job, as long as
the function values in these isolated points are the same. And for each
such function, we might find a way of extending the hyper exponent to
real values. So we remove a lot of ambiguity if we associate PowerTower
for real hyper exponents both with the PowerTower for integer hyper
exponents and a specified step-up function. I find it quite natural to
demand that we should choose y^x as "the" step-up function of interest.

If we find a function f[x] for which f[f[x]] = g[x] when g[x] =
Subscript[g, y][x], we can see that f[x] can act as a "halfway" step-up
function.

Then, one thing that I have used a lot is the following observation:

Suppose that we have obtained a solution f[x] to the equation f[f[x]]
== g[x]. Let h[x] be a well-behaved function with the inverse hinv[x].
Then f1[x] = h[f[hinv[x]]] should be a solution to the equation
f1[f1[x]] == h[g[hinv[x]]].

Another thing I have used is the assumption that if f1 is the solution
for g(x) = 2 Sinh[x] and f2 is the solution for g(x) = Exp[x], we could
with very good precision approximate f2 with f1 if x is large enough
(say x > 20 or so). It seems reasonable that the relative error we do
in that approximation should have an upper bound Exp[-2x]. The function
f1 can be found by using the series expansion of 2 Sinh[x] for small x.

With these assumptions there is not very much ambiguity left in the
definition of the halfway function for exp[x]. If there are some other
candidates, obtained by some other procedure, one might check the
asymptotic behaviour.

Considering your first point, you imply that (in Mathematica notation)

PowerTower[PowerTower[2.17, 1./3], 3]

should be different from 2.17. Let us try with my algorithm: Evaluating

PowerTower[PowerTower[2.17, 1./3], 3]

returns 1.51434, so there it is an agreement between your opinion and
the algorithm I have made. When I posted a strategy for solving the
equation f[f[x]] == exp[x] December 1, I did not even know the word
tetration, and have never been tempted to fall into that trap.
But we can instead do the following to get back to 2.17, using the
two-third step-up function when Exp[x Log[2.17]] is the step-up
function for a unit step:

fexpk[PowerTower[N[217/100, 100],
N[1/3, 100]], N[Log[217/100], 100], N[2/3, 100]]

(returns 2.1700000000000000000000000000000000000000000)

and we might do it in two equal steps

fexpk[fexpk[
PowerTower[
N[217/100, 100], N[1/3, 100]], N[Log[217/100], 100], N[1/
3, 100]], N[Log[217/100], 100], N[1/3, 100]]

(returns 2.1700000000000000000000000000000000000000000)

or two unequal steps (here with some loss in precision, have to check
why)

fexpk[fexpk[
PowerTower[
N[217/100, 100], N[1/3, 100]], N[Log[217/100], 100], N[1/
2, 100]], N[Log[217/100], 100], N[1/6, 100]]

( returns 2.170000000000000000000)

(Note that my PowerTower routines presently use the real value also of
rational numbers.)
Such experiments have convinced me that there is some sense in what I
am doing. If you have Mathematica available, try to play with my
routines yourself. I think that my egg still is standing.

The formula describing the fractional/real/complex iterates seems very
handy. I will try to implement it. Thanks a lot.

Best regards

Ingolf Dahl

### bo198214

Dec 14, 2006, 2:23:09 PM12/14/06
to
First of all it would make my life much easier if you were *not* using
Mathematica notation, but normal mathematical/semi-TeX notation. I can
program a bit Maple but am not especially inclined to learn
Mathematica.
So let me rephrase what I understood despite the notation. We have the
power-tower, which I will denote as ^^ in ascii notation. (Btw. we can
discuss this topic also on my forum http://math.eretrandre.org/mybb/
which supports LaTeX, I would be glad to welcome you there.)

The hyperpower/tetration is usually defined as
x^^1 = x ; x^^(n+1) = x^(x^^n)

Now you say if we take the step function g_y(x) = y^x then we can
represent
y^....y^x = g_y o g_y o ... o g_y (x)
x^^{n+1} = (g_x)^{n}(x) (where g^n means the n-times iteration of g)

And hence if we can appropriately define (g_y)^r then we have a
definition of ^^, i.e. x^^r = (g_x)^{r-1}(x).

I guess your Mathematica function fexpk[x,y,k] is just (g_y)^k(x);
the function x -> y^x iterated k times. Now you demonstrate

(2.17 ^^ 1/3) ^^ 1/3 != 2.17
but
(g_2.17)^{2/3} (2.17 ^^ 1/3) = 2.17
(g_2.17)^{1/2}o(g_2.17)^{1/6}(2.17 ^^ 1/3) = 2.17

But this is no wonder, this is how fractional iteration is defined,
i.e. the fractional iterates f^r of f must satisfy the so called
translation equation:

f^0(x)=x and f^{r+s}(x) = f^r(f^s(x))

form a set of iterates.

(g_2.17)^{2/3} (2.17 ^^ 1/3) = (g_2.17)^{2/3}((g_2.17)^{1/3-1}(2.17)) =
g_2.17^{2/3 + 1/3 - 1}(2.17) = g_2.17^0 (2.17) = 2.17

Unfortunately I am still not sure how you compute fexpk (due to
Mathematica notation and my laziness). So it would be nice to present
it in "normal" notation and also to explain why there is not much
ambiguity.

For example the "normal" method to construct exp^{1/2} is to imagine
any function f on the interval (0,1] which lies between x->x and
x->exp(x) and is continuous and strictly increasing and exp(f(0))=f(1).
>From this start you can then define f on (1,e] by f(x):=exp(f(ln(x)))
and so on you can define f on (exp^{n}(0),exp^{n+1}(0)] and in a
similar way in the other direction.

So you can start with nearly any function f on (0,1] (or even on any
interval (x,exp(x)]) and extend it to a continuous strictly increasing
half-iterate of exp on (-oo,+oo). Though of course some functions
oscillate a bit stronger than others (draw it on paper!).

A bit more difficult it becomes if you demand that the fractional
iterates to be analytic (as exp is). But even then fractional iterates
are not unique (as you can read in Szekeres latest paper; the formula
for unique analytic iteration I gave in my previous post is only for
functions with fixed point at 0 (a_0=0) and a_1=1) and there is no
well-established criterion which set of iterates should be considered
the best (for example what is should mean to minimize that oscillation).

### bo198214

Dec 18, 2006, 7:27:04 AM12/18/06
to

As it was just pointed out to me, I made quite a blunder in the
arbitrary construction of exp^{1/2}(x). You dont start defining f on
(0,1] but you start defining f on (0,f(0)] with f(f(0))=exp(0)=1. This
determines f on (f(0),1], (1,f(1)], (f(1),e], (e,f(e)], (f(e),e^e], ...
etc.

### Ingolf Dahl

Dec 18, 2006, 8:15:00 AM12/18/06
to

A letter very similar to this had been sent to MathGroup.

I think I have to back-off a half step from my claims. But I do not
regret what I have written: I think it is a really good math lesson for
me, and maybe for other. And I think the conclusions are quite
interesting: It is possible to extend PowerTower to real exponents in a
consistent way, BUT the solution will not be unique. And that is not by
our ignorance, that is by mathematical principle. The case is similar
to 0^0, but without a clear "best choice". So we are free to let UN
make a law and decide which solution to choose. And such constants as
Pi^^E will become somewhat fuzzy, and the value I have published is
possible but not necessary. It is not a very common situation in
mathematics that constants have intrinsic uncertainness, even if we are
used to the fact that the primitive function to an arbitrary function
always contains a constant of integration.

And in the end of this letter I think I anyway see the opening of the
tunnel...

What is the problem? One thing that I had assumed was that if f1 is the
solution to the equation f[f[x]] == g[x] for g[x] = 2*Sinh[x] and f2 is
the solution for g[x] = E^x, we could with very good precision

approximate f2 with f1 if x is large enough (say x > 20 or so). It
seems reasonable that the relative error we do in that approximation

should have an upper bound E^(-2*x). The function f1 can be found by
using the series expansion of 2*Sinh[x] for small x, since x=0 is a
fixed point of 2*Sinh[x].

But the mathematics is not so kind and well-behaved. If we compare f1
with the corresponding solution f3 for g[x] = Exp[x]-1, we see that
they also should have the same asymptotic behaviour. Both are derived
from the analytical expression at small x, but anyway the difference
between f1 and f3 will show oscillations when x gets large. When we
pick out the analytic expression for the solution f1 and f3, these
solutions are special and unique just at the fixed point, but a bit
away from the fixed point the mathematics do not discern between that
special solution and other solutions (which might oscillate near to the
fixed point, maybe similar to x*Sin[1/x]). I have obtained a solution
f2 for g[x] = E^x by identifying the asymptotic behaviour with that for
f1, but I could as well have used f3, and would then obtain another
good solution. f2 is thus not an unique solution, even if it seems to
be a nice solution.

This is also illustrated if we solve the equation f[f[x]] == g[x] for
g[x] = Exp[k*x] for small k (0 < k < 1/E). There g[x] has two fixed
point, and the smaller one can be used to find the solution for x < E,
and the larger fixed point can be used for x > E, by identification of
Taylor expansion coefficients. It is thus not needed to compare
asymptotes in this case. If we then calculate the function f, we will
find that the curves do not connect at x = E. There is a jump of size
approx. 0.003 for k=0.01. I plan to include a curve describing this in
my next version of the PowerTower notebook on my web site. The
PowerTower function is not affected by this jump, since this function
will just sample values on the left side of the smaller fixed point.
Thus we can anyway use this fixed point to define PowerTower for real
exponents for 1 < x <= Exp[1/E].

For larger values of k, there are thus several solutions f(x)
available, where it is difficult to say that one is "unique", since
there are no fixed points for Exp[k * x]. (Maybe one could find a
solution that is the "best", according to some criterion.) Then this
spills over to PowerTower too. If x > Exp[1/E], the range of
PowerTower[x, n] for real values of n >= -2 is from -Infinity up to
(approaching) +Infinity. Then if we would have PowerTower defined in
some unique way, the relation

PowerTower[y, n] = f[PowerTower[y, n - 1/2]]

should define an unique solution f(x) to f[f[x]] == exp[k*x] with k =
Log[y], and thus we get a contradiction if we state that there is no
unique solution to this equation.

All this does not seem to be common knowledge today (e.g. it is not
clear for me from Wikipedia), but much of the behaviour follows
directly from the description by G. Szekeres in "Abel's equation and
regular growth: Variations of a theme by Abel", see
www.emis.de/journals/EM/restricted/7/7.2/szekeres.ps. I have just found
it myself (by a tip by the previous poster) and have not yet understood
all details there.

Anyway, even if the solution f is not unique, there could be useful to
have available some nice solutions from the solution set. Then we might
at least be able to choose "the best" or "the best known so far". I
have solved the equation by letting f[x] ride on the asymptotic
behaviour of some "horse function", with fixed points. (I think that
this is my contribution to this field of mathematics.) In the paper by
Szekeres it is indicated that Exp[k*x] - 1 should display "regular
growth" for k > 1. If that is true, it could be the ideal horse
function. Another similar alternative, which also might have regular
growth behaviour, is Exp[k*x] - (1 + Log[k])/k, which becomes identical
to Exp[k*x] at the k value 1/E. So just now it appears to me as the
case is almost solved. Are there more small devils hiding in the bush?

I have got some nice letters and posts commenting my proposals, but
but I am working at it.

Best regards

Ingolf Dahl

### Ingolf Dahl

Dec 18, 2006, 9:45:01 AM12/18/06
to

If g(x) = exp(x) -1

and

h(x) = x/k - Log[k]/k

with the inverse function hinv, then

h(g(hinv(x))) = exp[k*x] - (1 + Log(k))/k

This implies that if f(x) is a solution to f(f(x)) = exp(x) -1, the
function h(f(hinv(x))) is a solution to f(f(x)) = exp[k*x] - (1 +
Log(k))/k.
If the function f has regular growth, we could thus expect that
h(f(hinv(x))) also has regular growth. Thus exp[k*x] - (1 + Log(k))/k
can be used as a "horse function" (according to my previous post)
for exp[k*x] with regular growth.
Wow!

Ingolf Dahl

### bo198214

Dec 18, 2006, 11:04:51 AM12/18/06
to

Ingolf Dahl wrote:
> interesting: It is possible to extend PowerTower to real exponents in a
> consistent way, BUT the solution will not be unique.

The question is even what "consistent" should mean. For fractional
iteration one has at least the condition of the translation equation
F^0=id, F^{s+t}= F^s o F^t
while for extension of the power tower I dont see any algebraic rules.
What one could demand is that x^^t should be at least continuous in x
and t, or stronger infinitly many times differentiable (what Galidakis
showed) or even stronger, analytic.

If we use analytic iterates of g_y(x):=y^x for the extension of the
power tower, i.e. x^^t = (g_x)^{t-1}(x), then it looks as if x^^t is
indeed analytic in x and in t.

But extending the power tower by analytic iterates (which arent already
unique) is just one way. If I remember correctly, Galidakis followed an
other path.

> But the mathematics is not so kind and well-behaved. If we compare f1
> with the corresponding solution f3 for g[x] = Exp[x]-1, we see that
> they also should have the same asymptotic behaviour. Both are derived
> from the analytical expression at small x, but anyway the difference
> between f1 and f3 will show oscillations when x gets large. When we
> pick out the analytic expression for the solution f1 and f3,

Though all formal fractional iterates of exp-1 (i.e. formal solution to
the powerseries, which are even unique) do not converge at the fixed
point 0, except the trivial iterates (exp-1)^k, k integer.

> All this does not seem to be common knowledge today (e.g. it is not
> clear for me from Wikipedia),

Maybe I will start to write an article on wikipedia about analytic
iteration.

### Ingolf Dahl

Dec 21, 2006, 7:44:50 AM12/21/06
to

Now the bumblebee is flying...

I have uploaded a new version of my notebook PowerTower.nb.

Since the last version, I have changed "horse function" to (Exp[x] -
1), to avoid oscillating functions when x->+Infinity This means
slightly changed values of the constants I have calculated. I have also
included some theory and some graphs in this file, and have structured
it better.

Thus the new value of Pi to the hyper power e (or "Pi tetrated to e")
is
1921616.48318907465

Now I want to comment on the posts by bo198214:

The reference to Szekeres has been of great help, thanks again. But the
books by Szekeres and by Jabotinsky are hard to find.

"For example the iterates of e^x-1 only converge for integer
(composition) exponents."

Reference to this? My numerical tests of the convergence indicate a
convergence circle of radius around pi/2, which seems reasonable due to
the relationship with sin(x), where the derivative gets negative at
pi/2. I need good approximations for x < 0.07 or so.

"First of all it would make my life much easier if you were *not* using
Mathematica notation, but normal mathematical/semi-TeX notation. I can
program a bit Maple but am not especially inclined to learn
Mathematica. "

And I am not inclined to learn TeX. OK, I can do it, but maybe not as
first task use it in such contexts as this, since the probability of
errors will approach one. And TeX notation was designed for making nice
the mathematical content), while the Mathematica notation was designed
for giving unambiguous specification of mathematical relations. And a
simple guess is that you should gain more by learning Mathematica than
I should gain by learning to master TeX. Think about it, if Mathematica
can help a lazy bum like me produce so much mathematical trash in three
weeks, think of what you could produce...

"I guess your Mathematica function fexpk[x,y,k] is just (g_y)^k(x);
the function x -> y^x iterated k times."

It is the function x -> exp[x y] = exp(y)^x iterated k times (0<= k <
1) (y is a constant parameter).

"Unfortunately I am still not sure how you compute fexpk (due to
Mathematica notation and my laziness). So it would be nice to present
it in "normal" notation and also to explain why there is not much
ambiguity.

For example the "normal" method to construct exp^{1/2} is to imagine
any function f on the interval (0,1] which lies between x->x and
x->exp(x) and is continuous and strictly increasing and exp(f(0))=f(1).

>From this start you can then define f on (1,e] by f(x):=exp(f(ln(x)))
and so on you can define f on (exp^{n}(0),exp^{n+1}(0)] and in a
similar way in the other direction.

So you can start with nearly any function f on (0,1] (or even on any
interval (x,exp(x)]) and extend it to a continuous strictly increasing
half-iterate of exp on (-oo,+oo). Though of course some functions
oscillate a bit stronger than others (draw it on paper!). "

I am not doing it exactly so, and I believe that this is my
contribution to this field. It is well-known how to get the principal
solution near fixed points, and in those few cases I have tested, the
series expansion works nicely near the fixed point. In some cases an
expansion in x^(Sqrt(2)) might be better. But when there is no fixed
point available, my idea is to look at very large x. If I can define
the function there, I could use this to extend the function to any x,
exactly as you propose, with the exception that I might have to iterate
with the inverse function. At large x, I might approximate f(x) with
the corresponding function for exp(x) - 1, or some other function with
the same asymptotic behaviour, with an error that I can estimate. I
have denoted such a function a "horse function", since I let my
solution ride on it. Then of course the rider will have to follow all
oscillations of the horse.

To calculate the iterate for any k, I write k in base two. Then I
iterate binary decomposition a sufficient number of times (maybe 188),
and apply the iterate if there is an "1" in the corresponding place in
the binary expansion. Nice and fast. But I cannot do that in TeX.

"there is no well-established criterion which set of iterates should be
considered the best (for example what is should mean to minimize that
oscillation). "

How about the criterion by Szekeres? He seems quite sure that exp(x) -
1 displays regular growth. I have not tested yet if my (new) solutions
have inherited this, but I find it reasonable. Anyway, my first
solution involved Sinh(2 x), which seems to give violent oscillations,
but anyway gave function values at low x very similar to those I obtain
now. (It is hard to see any differences in graphs, I have to look at
the decimals). If there are several solutions available with regular
growth (I doubt that), the difference for small x are probably even
less. And then it might be useful to have one solution available
anyway.

Best regards

Ingolf Dahl

http://web.telia.com/~u31815170/Mathematica/

### bo198214

Dec 22, 2006, 8:51:59 AM12/22/06
to

Ingolf Dahl wrote:
> The reference to Szekeres has been of great help, thanks again. But the
> books by Szekeres and by Jabotinsky are hard to find.

Oh, these arent books, just papers.

Paul Erd�1\$}os and Eri Jabotinsky, On analytic iteration, J. Analyse
Math. 8 (1960/1961),361�1rs376. MR0125943 (23 #A3240)

Eri Jabotinsky, Analytic iteration, Trans. Amer. Math. Soc. 108 (1963),
457�1rs477.MR0155971 (27 #5904)

G. Szekeres, Regular iteration of real and complex functions, Acta
Math. 100 (1958),203�1rs258. MR0107016 (21 #5744)

G. Szekeres, Fractional iteration of exponentially growing functions,

J. Austral. Math. Soc. 2 (1961/1962), 301�1rs320. MR0141905 (25 #5302)

> "For example the iterates of e^x-1 only converge for integer
> (composition) exponents."

Irvine Noel Baker, Zusammensetzungen ganzer Funktionen, Math. Z. 69
(1958), 121�1rs163 (German). MR0097532 (20 #4000)

> "First of all it would make my life much easier if you were *not* using
> Mathematica notation, but normal mathematical/semi-TeX notation. I can
> program a bit Maple but am not especially inclined to learn
> Mathematica. "
>
> And I am not inclined to learn TeX. OK, I can do it, but maybe not as
> first task use it in such contexts as this, since the probability of
> errors will approach one. And TeX notation was designed for making nice
> the mathematical content), while the Mathematica notation was designed
> for giving unambiguous specification of mathematical relations. And a
> simple guess is that you should gain more by learning Mathematica than
> I should gain by learning to master TeX. Think about it, if Mathematica
> can help a lazy bum like me produce so much mathematical trash in three
> weeks, think of what you could produce...
>

That all wasnt my point at all. My points:
* Dont use a proprietary language to describe normal mathematical
content. People understanding standard math notation > p.u. TeX
notation > p.u. Mathematica notation
* Standard ASCII math notation is: functions are written f(x),
subscripts a_i, and superscripts a^i
* I speak already Maple (and dont you think Maple can produce nice
results too?) so why dont you explain it to me in Maple? ;)

> But when there is no fixed
> point available, my idea is to look at very large x.

Ok, thatswhy you used exp(x) - exp(-x) as approximation of exp(x). But
with the benefit that it has a fixed point at 0 with a series
expansion. And one can uniquely take the fractional (or just half)
iterate of a series expansion.
And if one can compute the half iterate then one can easily compute the
value of an binary fraction.

But then I wonder a bit about exp(x)-1 does not approximate exp(x) as
good as exp(x)-exp(-x) does?

> "there is no well-established criterion which set of iterates should be
> considered the best (for example what is should mean to minimize that
> oscillation). "
>
> How about the criterion by Szekeres? He seems quite sure that exp(x) -
> 1 displays regular growth.

Sorry, but that paper of Szekeres is full of conjectures and numerical
"evidence". If some of them are proved/disproved we can speak again
I personally doubt that there is *the* fractional iteration of exp, I
guess it is more a question of personal asthetics (and of human psyche
to have an ideal ;) )

Did you btw already try to move the fixed point at infinity to 0 by
h(x)=1/x, i.e. to consider h^-1 o exp o h? I mean we generally have
the rule (g^-1 o f o g)^(m/n) = g^-1 o f^(m/n) o g, which you can
easily verify by taking the compositional power to n. In this
particular case (h^-1 o exp o h)(x) = exp(-1/x) which has a fixed point
at 0 (though no series expansion there). So if you had an halfiterate f
of exp(-1/x)=f(f(x)) then 1/f(1/x)) would be a half iterate of exp.