(The presentation that follows is a light introduction. For more austere results, consult Paper 1).

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 are then faced with the following problem:

Let a series s with coefficients a_n be given:

Determine the coefficients b_n of the series expansion (initially disregarding issues of convergence) of s1 = e^s*x:

I posted the above problem to sci.math a little while ago, and here are two answers:

The first one, by Robert Israel:

where the sum is over all sequences s = [s₁, s₂, s₃, ... ] of nonnegative integers with:

and the product is over those k for which s_k > 0 (a finite set).

For example, the partitions of 4 correspond to s = [4,0...], [2,1,0...], [0,2,0...], [1,0,1,0...] and [0,0,0,1,0...], so:
b₄ = a₀⁴/4! + (a₀²*a₁)/(2!*1!) + a₁²/2! + a₀*a₂ + a₃.

Robert has also provided Maple code to explicitly calculate the b_n.
> b:= (n::nonnegint) ->
  add( mul(a[q[1]-1]^q[2]/q[2]!,
  q = map(j -> [j, numboccur(j,P)], convert(P,set))),
  P = combinat[partition](n));

The second one by Leroy Quet, from a simple differentiation property of exp(z):

b_n=1, if n=0 else:

Leroy derived the above recursive expression by noticing that if B(x) = exp(A(x)), then:
B'(x) = B(x) *A'(x). Ignoring temporarily issues of convergence inside a specified range, the expression follows by equating coefficients and changing indexes.

The recursion then becomes:

It is immediately clear that a_m,₀ = 1, for all m in N.

Since we are also in possession of the coefficients of e^x, the recursion is complete, and we can immediately write Maple code to generate any a_m,n.

> a:=proc(m,n)
> option remember;
> local j;
> if n=0 then 1
> elif m=1 then 1/n!
> else add(j*a(m-1,j-1)*a(m,n-j),j=1..n)/n;
> fi;
> end:

Here's a table with the coefficients of xⁿ in the series expansions for the first few ^m(e^x):
 

The keen eye will notice the pattern that emerges. But before we actually investigate it, let us prove that _there is_ a pattern, whatever it is:

Proof:
By induction on n: It is clear that a₁,₁ = 1 and a_m,₀ = 1 for all m in N.
If a_k,₁ = 1 then:
a_k+1,₁ = Sum(j*a_k+1,_1-j*a_k,_j-1,j=1..1)/1 = a_k+1,₀*a_k,₀ = 1 = a₁,₁.

Therefore, a_m,₁ = a₁,₁ for all m in N.

Now assume a_m,_k = a_k,_k is true for all m >= k.

We have to show that a_m,_k+1 = a_k+1,_k+1 for all m >= k+1.

However:

a_m,_k+1 = Sum(j*a_m,_k+1-j*a_m-1,_j-1,j=1..k+1)/(k+1)
a_k+1,_k+1 = Sum(j*a_k+1,_k+1-j*a_k,_j-1,j=1..k+1)/(k+1)

m >= k+1 => m >= k+1-j, for all j in {1..k+1}, so a_m,_k+1-j = a_k+1-j,_k+1-j, by the induction hypothesis. (1)

k+1 >= k+1-j, for all j in {1..k+1}, so a_k+1,_k+1-j = a_k+1-j,_k+1-j, again by the induction hypothesis. (2)

(1) and (2) establish: a_m,_k+1-j = a_k+1,_k+1-j.

j <= k+1 => j-1 <= k and m >= k+1 => m-1 >= k, so j-1 <= m-1, so a_m-1,_j-1 = a_j-1,_j-1, by the induction hypothesis as well. (3)

And finally: j <= k+1 => j-1 <= k, so a_k,_j-1 = a_j-1,_j-1, by the induction hypothesis. (4)

(3) and (4) establish: a_m-1,_j-1 = a_k,_j-1 and we are done.

W(-x)/(-x).

Now, knowing that the series expansion for the Lambert's W function is:

it follows immediately that whenever we have convergence, W(-x)/(-x) is:

Now we know exactly what those coefficients a_m,_n in the expansion of ^m(e^x) will be. To summarize:

For m = 1:

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

For m > 1:

^m(e^x) = 

with:

.

> a:=proc(m,n)   #coefficients calculation
> option remember;
> local j;
> if n=0 then 1
> elif m=1 then 1/n!
> else add(j*a(m-1,j-1)*a(m,n-j),j=1..n)/n; #recursion
> fi;
> end:

> f:=proc(m,x)  #expansion as per above
> local s1,s2;
> s1:=1+sum('(n+1)^n/(n+1)!*x^n','n'=1..m);
> s2:=sum('a(m,n)*x^n','n'=m+1..100);
> s1+s2;
> end:

> g:=proc(m,x) #for comparison
> if m=1 then exp(x)
> else exp(x*g(m-1,x));
> fi;
> end:

> evalf(g(2,0.12345));
                             1.149894868
> evalf(f(2,0.12345));
                             1.149894869

It is easy to see via the Ratio Test that the radius of convergence of the limit function: f_∞(x) is: R = 1/e. (Compare with the radius of convergence of the LambertW, here).

Because of the above fact, it is natural to expect "bad" convergence behavior around 1/e in the corresponding series expansions of the functions f_m(x).

What this means, is that while the Maple code above works perfectly for |x| <= 1/e, all the corresponding series expansions will most likely misbehave for values of x of the form: x = 1/e + dx(m), where dx(m) > 0, and dx(m) is a function of m.

If we were to describe the situation in very rough terms, we could say that the series expansion for f_m(x) will "misbehave" around 1/e + dx(m), with dx(m) -> 0 as m -> +∞.

And this is to be expected. If convergence of the series expansions of the iterates: ^m(e^x) was everywhere nice, the limit function would be entire. But it's not.

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