(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 ^{n}x. Having
such
an expansion at hand may offer additional insight into what an analytic
^{y}x
can look like, for y in R.

Instead of searching for an expansion of ^{n}x, 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. ^{0}x=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-x^{2}) 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-x^{2})={exp(4/(4x^{2}-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 alpha_{3}(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)*z^{n},
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)z^{n}|<=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

^{y}z=

Afterthoughts:

It is interesting to note that if we consider lim

lim

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

Because the radius of convergence of the above series is 1/e, the behavior of any numerical computations for the extension

In general, calculating

The above extension is obsolete as it suffers from the fact that the functional equation of tetration is only satisfied at the naturals and 0 (As noted by Andrew Robbins).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