Outre Math: "Splitting" Exponentiation

Factual Questions
1

Here’s a question I’ve always wanted to ask on some forum or other:

I assume that the reader knows what a composite function is. For example the square root of the sine of x.

The next very basic (it seems to me) concept is a single (re- ?)iteration of a function. That is, the composite of a function with itself (to continue use of the terminology above).

For example z=exp y; y=exp x; therefore z=exp(exp x).

Now, let’s reverse the concept of a “double function” or a function (one time) composite with itself. What would a “split” or a “half-function” be? It doesn’t seem to be a problem with squaring or square rooting.

But let’s take “natural” exponentiation, or y=exp x. (As opposed to raising 10, say, to the power of x.)

The inverse function, of course, is the natural logarithm function or y=log (base e) x,
often written as y = ln x.

So, can they be “split”? Let’s call the first “eta” so that
eta(eta x) = exp x…

and the second “lambda” so that
lambda(lambda x) = ln x…

And eta and lambda would be inverse functions, of course.

(At least I think such would always be true. If we want to be exact, and Mathematics is exact, we would have to talk about boundary conditions of domain(s) and range(s). But the whole thing is moot if they do not exist.)

If they do not exist, as I have had reason to believe, can there be a proof that they do not exist?

True Blue Jack

2

If eta exists, it would be referred to as the composite root of exp. I believe that there is such a function, but I don’t know the form of it. Beware that you’re getting into pretty advanced stuff here, and might find yourself in over your head pretty quickly.

3

A quick search makes it look like neither mathworld or wiki have anything on this topic. This is all I could find on short notice.

4

I suspect that there will be an uncountably infinite number of such functions. Pick a value for some arguement of the function: Let’s say, for instance, that eta(1) = 2. We know that eta(eta(1)) = exp(1) = e, so we now know that eta(2) = e. So we take the next step: If eta(2) = e, and eta(eta(2)) = exp(2), then eta(e) = e[sup]2[/sup], and so on. We can also work in reverse: If 1 = eta(x) for some x, then eta(eta(x)) = exp(x) = eta(1) = 2, so x = ln(2), or eta(ln(2)) = 1. Proceeding in this manner, we can get as many known values for eta(x) as we like, and I think thereby define the eta function. But this is just the eta function for eta(1) = 2, an arbitrary choice. I think that for almost all choices of starting point, we could proceed without contradiction, so there would be a different eta function for almost every real number.

I might be mistaken about this working for almost all values of eta(1), and it might be that there is some unique function eta (or at least, a unique well-behaved one). In that case, if we can find any single value of that function that we trust, then my method will at least give us many other trusted values of the function.

5

Thinking further about this: We know that eta has no fixed points (that is, x such that eta(x) = x), because any fixed point of eta would be a fixed point of exp as well, and exp has no fixed points. This also implies that there is no x for which eta(x) = exp(x): If there were, then eta(x) = eta(eta(x)), so eta(x) would be a fixed point of eta, which we know does not exist.

Further, we know that as x approaches infinity, eta(x) must grow faster than any polynomial. If eta is bounded by some polynomial, say, eta(x) < P(x) for all x, then eta(eta(x)) < P(P(x)). But P(P(x)) is also a polynomial, and eta(eta(x)) = exp(x), which eventually exceeds any polynomial.

Now, let us assume that there is at least one eta function which is well-behaved: Call it Eta. At the very minimum, a well-behaved function ought to be continuous. Since eta has no fixed points, this means that either Eta(x) is always greater than x, or Eta(x) is always less than x. But eta is not bounded by any polynomial, so Eta(x) cannot always be less than x. So Eta(x) is always greater than x.

Similarly, Eta(x) can never equal exp(x), so it’s either always less than exp, or always greater than exp. The case Eta > exp leads to some obvious contradictions, so we know that x < Eta(x) < exp(x) is true for all x. In particular, 0 < Eta(0) < 1.

It seems like similar reasoning might lead one to zoom in on the value of Eta(0), at which point my previous method would apply. But the margin of my brain is too small to contain such similar reasoning.

6

If you look at the link I gave earlier, they mention that the known composite root of exp is not single-valued, which implies that we’re looking at complex-valued functions here. I don’t know what that does to your order lemmas, although they look right to me over R.

7

…Right. I was assuming that, since exp(x) is real for x real, that the same would be true of eta. But there’s no real justification for that assumption. And if eta can be complex-valued, then my results about the ordering go out the window, since eta could go continuously from less than exp to greater than exp without ever actually equalling exp, if it’s ever even reasonable to say that eta < exp in the first place (it’s conceivable that exp(x) - eta(x) always has nonvanishing imaginary part for real x, for instance, in which case there wouldn’t be any natural order comparison between eta and exp at all). It is, at least, the case that eta(x) never equals x nor exp(x), though.

We might at least be able to extend my reasoning to prove that Eta must be complex-valued, which might match the OP’s criteria for suspecting that it does not exist.

8

I can already tell you that this may well be asking for far too much. See what happens if you try to take the square root of a differentiation operator, which is nice and local. Any possible result is hideously nonlocal. There isn’t really any good reason why the composition-square root of a continuous function need be continuous.

That said, has anyone tried to write down a power series and see what it has to look like?

9

Thank you. At least now I have a name for it for online research.

True Blue Jack

10

Yes, but this will require quite a bit of writing. I don’t have much time, now, but perhaps tomorrow at this time.

Meanwhile, I appreciate all the feedback. Thanks to you all!

TBJ

11

I never said that a well-behaved function must exist, only that continuity is necessary for a function to be well-behaved. Of course, I’m using a physicist’s definition of “well-behaved” here, which amounts to no definition at all, just “I know it when I see it”. And as a physicist, I do perhaps have an unreasonable expectation that most functions I’ll encounter will be well-behaved (for whatever that means in context).

But I can accept the possibility that there might not be any continuous form of eta.

Oh, and incidentally, my result that eta have no fixed points also fails, since exp does indeed have fixed points in the complex numbers. One such is at (0.3181315052 ± 1.337235701 i), which might therefore conceivably be a fixed point of eta, as well.

12

Doing this for f(f(x))=exp(x) is ugly; the nonzero constant term f[sub]0[/sub] (where f(x)=sum[sub]n[/sub]f[sub]n[/sub]x[sup]n[/sup]) gives a nonlocal dependency, 1=exp(0)=f(f(0))=f(f[sub]0[/sub]).

But doing the expansion about a zero of the function — e.g., for f(f(x))=log(1+x) or f(f(x))=exp(x)-1 — is easy because the set of equations for the f[sub]n[/sub] is lower-triangular: If f[sub]0[/sub]=0, then the coefficient of x[sup]n[/sup] in f(f(x)), for n>=1, is a sum over all 2[sup]n-1[/sup] ordered natural partitions n=p[sub]1[/sub]+…+p[sub]m[/sub] of n (i.e., those with p[sub]i[/sub]>0 and order considered important) of f[sub]m[/sub]f[sub]p[sub][/sub]1[/sub]…f[sub]p[sub]m[/sub][/sub]. This sum can be written as f[sub]n/sub+(terms involving only f[sub]1[/sub],…,f[sub]n-1[/sub]), so knowing earlier values f[sub]k[/sub] lets us calculate f[sub]n[/sub] (as long as f[sub]1[/sub]!=0,-1). (Of course, there are exponentially many terms in this sum, so this is not an efficient calculational method for large n.)

For f(f(x))=log(1+x), the first few Maclaurin coefficients of f computed in this way (starting from f[sub]0[/sub]=0) are 0, 1, -1/4, 5/48, 5/96, 109/3840, -497/30720, 127/13440, -11569/2064384, 312757/92897280, -1219255/594542592, 165677473/130799370240, -885730939/1121137459200, 20163875141/40809403514880, -252312616027/816188070297600.

For f(f(x))=exp(x)-1, they are 0, 1, 1/4, 1/48, 0, 1/3840, -7/92160, 1/645120, 53/3440640, -281/30965760, -1231/14863564800, 87379/24222105600, -13303471/7847962214400, -54313201/40809403514880, 10142361989/5713316492083200.

I’m not sure if either of these has a nonzero radius of convergence; I haven’t thought about the asymptotics.