(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 *very quickly* 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. For the final word on the matter, see this math.SE post.