An Infinitely Differentiable Extension for hyper4

(The presentation that follows is a light introduction. A more austere article is Manuscript 2).

While looking for an analytic continuation of the hyper4 operator, one is tempted to search for series expansions for ⁿx. Having such an expansion at hand may offer additional insight into what an analytic ^yx can look like, for y in R.

Instead of searching for an expansion of ⁿx, it is often convenient to look for an expansion for ⁿ(e^x) first. We summarize the results found in "A Series Expansion for ^m(e^x)":

For m = 1:

^m(e^x) = e^x, of course.

For m > 1:

^m(e^x) =

with:

Having those coefficients at hand, we can now construct a C^oo extension of the hyper4 operator as follows:

We first extend the coefficients above to conform with the hyper4 definition, i.e. ⁰x=1, so we define:
a_m,_n={1, if n=m=0,
            0, if m=0 and n=/=0,
          1/n!, if m=1,
           Sum(j*a_m,n-j*a_m-1,j-1,j=1..n)/n, otherwise}     (6.1)

We will use the well-known C^oo function,

f(x)={exp(-1/x) if x>0,
0, if x<=0}

Elementary calculus shows that f(1-x²) is symmetric about the origin and there it attains its maximum, f(1)=1/e. We first change the support of f to be the interval [-1/2,1/2], so we consider the function

phi(x)=f(1/4-x²)={exp(4/(4x²-1)), if |x|<1/2,
0, otherwise}

phi(x) is shown below.

Set A_m=Int(phi(t-m-1/2),t=m-1..m). Since phi(x-m-1/2) is simply a right translation of phi, it follows that for all m,n in N, A_m=A_n. We first normalize phi with respect to its integral, so we set:

psi_m(x)=(a_m,n-a_m-1,n)phi(x-m-1/2)/A_m and finally,

Definition:
For all m in N, n in N U {0}, x>=0 and with initial values for am,n as in recursion (6.1):
alpha_n(x)={1, if n=0,
, if n<>0}

If x=m>0, then since all the psi_m are non-zero on disjoint subsets, alpha_n(m)=a_m,n, while if x>=n, then alpha_n(x)=a_n,n. The next figure is the graph of alpha₃(x).

Lemma 1:
If x>=0, alpha_n(x) is infinitely differentiable with respect to x.
Proof:
If x>= n>0, alpha_n(x)=a_n,n = constant, so derivatives of all orders exist and are 0. If x<n, then the existence of the p-th derivative of alpha depends on the existence of the p-1-th derivative of psi_m(x), which is simply a right translation of phi, for which derivatives of all orders exist and the Lemma follows.

We are ready for the extension. With alpha_n(x) as above and y>=0,
^y(e^z)= (6.2)

If we fix z and call F(y)=^y(e^z), then for y in N, the above function satisfies the functional equation, F(y+1)=(e^z)^F(y). We have to prove convergence.

Lemma 2:
If y>=0, then S_k(z)=Sum(alpha_n(y)*zⁿ, n=0..k) converges uniformly on compact subsets of C.
Proof:
If y=0 then S_k(z)=1-> 1. Fix y>0 and z in U subset C, U compact. Then y in [m-1,m) for some m in N, and there |alpha_n(y)|<=a_m,n, for all n in N, therefore for all z in U and each y>0, |alpha_n(y)zⁿ|<=a_m,n|z|ⁿ=M_n and Sum(M_n,n=0..+oo)=^m(e^|z|) by the analysis of coefficients above, so by the Weierstrass M-test, the series S_k(z) converges (absolutely and) uniformly on compact subsets and the Lemma follows.

If y=0, then ^y(e^z)=1 as required by the definition of hyper4, while if y=m in N, then ^y(e^z) coincides with the corresponding expansion for the tower ^m(e^z) in the analysis above. Therefore ^y(e^z) interpolates all finite towers of iterates of e^z. The important question now is if this interpolation is not only continuous, but C^oo with respect to y. We are ready for the second result.

Lemma 3:
If y>=0, then ^y(e^z) is infinitely differentiable with respect to y.
Proof:
Since S_k(z) converges uniformly on compact subsets, we can differentiate term by term. But d^p/dy^p{alpha_n(y)} exists in the domain of alpha_n, for all p>=1 by Lemma 1, therefore d^p/dy^p{^y(e^z)} also exists for y>= 0 for all p>= 1 and the Lemma follows.

We can now define a corresponding C^oo function that interpolates between all the finite power iterates of z, as
^yz=

Afterthoughts:

It is interesting to note that if we consider lim_y->+oo^y(e^z), then the limit of the summation becomes a summation of limits, and lim_y->+ooalpha_n(y)=a_n,n, therefore,
lim_y->+oo^y(e^z)=

,
which is the series expansion of W(-z)/(-z), where W is Lambert's W function. The last series converges for |z|<=1/e, as shown in the article for "A Series Expansion for ^m(e^z)".

Because the radius of convergence of the above series is 1/e, the behavior of any numerical computations for the extension ^y(e^z) that depend on the coefficients (6.1) for convergence, will most likely be bad past this radius of convergence. Indeed, we can set up some Maple code to calculate ^y(e^z) for |z|<=1/e, but past these bounds (or disk in the complex case) any code will likely produce garbage. And this is to be expected. If convergence of the series expansions of the iterates ^y(e^z) was everywhere nice, the limit function would be entire, but it's not, although all finite iterates are entire.

In general, calculating ^y(e^z) (resp. ^yz) even when |z|<1/e (resp. |Log(z)|<=1/e), is quite difficult, because it depends on explicitly defining alpha_n(y). The alpha_n functions are all defined piecewise and as n grows larger, they become harder and harder to define explicitly. A numerical approximation for ^y(e^-0.34) using just 6 terms in the extension (n=5), coupled with the extension of the article "A Continuous Extension for the hyper4 Operator", is shown in the next figure, with y in [0,5], where it is known that lim_y->+oo^y(e^-0.34)=W(0.34)/0.34 ~ 0.76973.

The above extension suffers from the fact that the functional equation of tetration is only satisfied at the naturals and 0. For what appears to be a C^∞ solution which doesn't suffer thus, check Andrew Robbins' solution. Another (possibly Cⁿ) solution is given by Robert Munafo.

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